THE ^CALCULATIONS OF GENERAL CHEMISTRY WITH DEFINITIONS, EXPLANATIONS, AND PROBLEMS BY WILLIAM J. HALE, PH.D. ASSISTANT PBOFB8SOB OF CHEMISTBY IN THK UNIVBBSITY OV MICHIGAN THIRD EDITION, REVISED NEW YORK D. VAN NOSTRAND COMPANY 23 MURRAY AND 27 WARREN STS. 1911 COPYBIGHT, 1910, BY D. VAN NOSTRAND COMPANY NEW YORK Stanhope press P. H. GILS ON COM PANT BOSTON, U.S.A. To THE MEMORY OF Jjtll LATE PROFESSOR OP ORGANIC CHEMISTRY IN HARVARD UNIVERSITY IN APPRECIATION OP HIS MOST INSPIRING INFLUENCE THIS BOOK 18 GRATEFULLY DEDICATED c 257707 PREFACE TO THE FIRST EDITION THE justification for a book of this character is to be found in the realization that many of the mathematical applications of our fundamental conceptions in chemistry, even upon the most elementary points, remain uncompre- hended by students several years advanced in the study of the science. The great attention given to the study of Physical Chemistry, and the far-reaching importance attached to the interpretation of chemical phenomena in the light of modern theories, make it absolutely essential that more time be given in the college courses on General Chemistry to these mathematical demonstrations. Unless a clear and concise exposition of the methods of calculation is presented in the very first course of college work, the progress of the student is greatly hindered. On the other hand, too great a mass of mathematical data may impede rather than promote this progress. The aim, therefore, has been to limit these calculations to those subjects generally regarded as fundamental, and in which the student should receive the most complete and thorough drill. Such may rightfully constitute our "Arithmetic of Chemistry." Of the many important subjects omitted, the greater number possess a theoretical bearing which brings them more properly into those courses following General and Analytical Chemistry or into the study of Physical Chemistry itself. The absence of mathematical training as a basis for the study of chemistry constitutes a widely prevailing defect in the education of chemists at the present day. The vii Vlll PKEFACB appearance of works on General Chemistry from the stand- point of Physical Chemistry, with their extensive adoption in our American colleges and universities, is strongly indic- ative of the mathematical trend in modern chemical instruction. Under this influence the presentation of the methods of chemical calculations in their simplest possible form should facilitate and extend the use of these modern texts and at the same time operate favorably in the eradi- cation of former defects. Primarily, however, this ele- mentary presentation is intended to accompany the laboratory work in General Chemistry and to help the student to a better understanding -of the many possible conditions that may arise in the course of experimental work. The ideal plan, therefore, should be to associate these problems and their solutions with the actual labora- tory practice. In the study of the last two, and possibly three, chapters only a very few correlated experiments may be found to fit properly into the laboratory courses on General Chemistry; but the important bearing of the topics here outlined upon the thorough grounding of the student's ideas of chemical reactions will make it impera- tive that a number of the more simple illustrations be given by way of lecture experiments. Especially is this true of those reactions relating to combinations between gases by volume. For the greater part of material presented no originality is claimed, but gratitude is freely expressed to the authors of the many valuable and admirable treatises on Stoi- chiometry. In the manner of presentation a somewhat different plan has been followed from that usually found in books of this nature. This consists in a gradual intro- duction of each new condition properly falling under the consideration of some one subject, and the final develop- ment of the subject in its entirety from all the conditions PREFACE IX thus considered. By this method it is thought to obviate the dogmatic manner of presenting mathematical facts and to make the student realize immediately the connec- tion between all of our fundamental laws of energy and matter. With this train of thought connecting each and every point to that preceding, it is believed that the student will gain a better and firmer hold upon the entire subject-matter rather than upon the isolated points. This book in itself comprises no more than is presented by the author to students in their first year of chemistry at the university. The problems at the end of each chap- ter serve as a guide and basis of selection for examinations throughout the course. Great emphasis, in fact, is laid upon this side of the student's development as the highest and most beneficial for his future advancement. In general the field of use for a book of this type will be among colleges and universities. It is, however, sincerely hoped that at least the first portion of the book may have a helpful and profitable bearing upon the instruction of chemistry in the high schools. For the many valuable suggestions and the interest with which criticism has been freely extended in the revision of the manuscript, the author expresses his sincere apprecia- tion and lasting indebtedness to Professor Charles Loring Jackson of Harvard University, Professor Alexander Smith of the University of Chicago, and Professor S. Lawrence Bigelow of the University of Michigan. WILLIAM J. HALE. ANN ARBOR, May, 1909. PREFACE TO THE SECOND AND THIRD EDITIONS WITH a new and larger edition it has seemed best to make all possible minor changes in both text and problems. The work of revision has been greatly facilitated through the kind suggestions of those who have found the book help- ful, and of others who have shown especial interest in this exposition of the subject. For all of which grateful appreciation is heartily expressed. THE AUTHOR ANN ARBOR, January, 1910. October, 1911. CONTENTS. CHAPTER PAGE I. UNITS OP MEASUREMENT 1 II. DENSITY AND SPECIFIC GRAVITY 5 III. THE EFFECT OF PRESSURE UPON GASES 9 IV. THE EFFECT OF TEMPERATURE UPON GASES 15 V. THE COMBINED EFFECT OF PRESSURE AND TEM- PERATURE UPON GASES. PARTIAL PRESSURES . . 21 VI. AVOGADRO'S HYPOTHESIS AND SOME OF ITS AP- PLICATIONS 33 VII. THE LAW OF DEFINITE PROPORTIONS 46 VIII. THE DERIVATION OF CHEMICAL FORMULA 59 IX. CALCULATIONS DEPENDING UPON CHEMICAL EQUA- TIONS 82 X. NORMAL SOLUTIONS 106 XI. COMBINATIONS BETWEEN GASES BY VOLUME 122 XII. COMPLEX EQUATIONS 148 APPENDIX 167 INDEX . 173 CHEMICAL CALCULATIONS. CHAPTER I. UNITS OF MEASUREMENT. THE fundamental units employed in chemical calcula- tions will be defined at the outset to insure their clear and consistent use. Time. As the unit of time we have adopted the second, g^-i^ part of the mean solar day, or that period elapsing between the successive daily transits of the sun across a meridian. Distance or Length. The meter as the standard of length is fixed as that distance between two marks on a platinum-iridium bar preserved in Paris when this bar is at the temperature of C. It is approximately one forty-millionth of a meridian, or one ten-millionth of the earth's quadrant'' from pole to equator. Larger values in multiples of the meter by ten, one hundred and one thousand are named by use of the Greek prefixes, deca- meter, hectometer and kilometer (km.), while sub mul- tiples by the same values receive names from the Latin prefixes, decimeter (dcm.), centimeter (cm.) and milli- meter (mm.). The one one-hundredth part of the meter, the centimeter, is usually defined as the unit of length. As still smaller values we employ the micron (//), the one one-thousandth of a millimeter, and the millimicron the one one-thousandth of a micron. 1 2 CHEMICAL. CALCULATIONS ; l Volume ; aiid Mass/ The cubic decimeter or liter is the standard of volume. The standard of mass is the kilogram, or a mass of platinum-iridium in block form, preserved in Paris, and originally intended to have the same mass as a cubic decimeter of water at its greatest density, 4 C. It is, however, slightly less than a cubic decimeter of water which at 4 C. weighs 1.000013 kilo- grams. Both the liter and kilogram are somewhat large for general scientific purposes. It is customary, therefore, to use the one one-thousandth part of each, the cubic centimeter (cm. 3 or c.c.) as the unit of volume and the gram (g.) as the unit of mass. The mass of 1 c.c. of water at 4 C. is considered as 1 gram. The slight discrepancy between this value and the true one need be considered only in the most exact calculations. Through a combination of the fundamental units just mentioned we arrive at a standard system in terms of which so many of our important units of measurement may be defined. This system of units, known as the centimeter-gram-second (C.G.S.) system, has met with universal adoption throughout the scientific world. Its applications may be noted in the following paragraph. Force and Energy. When a body moves at a uniform rate through a unit of distance, the centimeter, in a unit of time, the second, it is said to have a unit of velocity (abbreviated cm. /sec.). When the change in velocity during one second is one centimeter per second we have a unit of acceleration. That force which will give to unit mass unit acceleration is called the unit of force or dyne. Now the same force operating upon different bodies does not produce in each the same acceleration. When, how- ever, the same acceleration is attained in the several cases the bodies must have equal masses. The work done by the force of one dyne in producing a displace- UNITS OF MEASUREMENT 3 ment in its own direction of one centimeter is called the unit of work or erg. When a body gains or loses energy the amount of energy may be measured in units of work, and when the work done upon a body imparts to it a certain velocity, the body, by virtue of this, is said to possess kinetic energy. The amount of energy so pos- MV 2 sessed may be expressed by the formula, E = - , in which E represents the energy, M the mass and V the velocity. Mass, then, is in itself but a " measure of the kinetic energy which a body possesses when it has a definite velocity." Ostwald. Standards of Temperature and Pressure. The accuracy with which measurements are conducted requires certain definite and constant conditions of tem- perature and pressure, the effect of changes in which will be studied in a later chapter. The standards adopted with these factors will receive only brief mention at this point. Temperature. The freezing-point of pure water is taken as the normal temperature and made upon the Celsius or centigrade thermometer. The boiling-point of pure water at 760 mm. pressure is registered at 100 on this scale, and the intervening range of temperature between the freezing- and boiling-points of water is di- vided into one hundred equal parts or degrees centigrade. As pressure exerts a considerable influence upon the boiling-points of liquids, all temperatures are deferred to the normal condition of pressure. Pressure. The 'standard condition of pressure, the normal pressure, is taken as. that pressure which the atmosphere exerts at sea level in a latitude of 45. This 4 CHEMICAL CALCULATIONS pressure is sufficient to sustain a column of mercury, at C., 76 cm. (760 mm.) in height. In a column of this height and 1 sq. cm. cross section there are exactly 76 c.c. of mercury, and since, volume for volume, mercury is 13.596 times as heavy as water and 1 c.c. of it therefore weighs this number of grams, we have at once the value 76 X 13.596 = 1033.2 grams as the weight of the atmos- phere per square centimeter. By reference to a barometer the actual weight of the atmosphere in' millimeters of mercury is observed under the various conditions. These barometric readings are usually made at other tempera- tures than C.; the correct readings, therefore, for milli- meters of mercury at C. must be calculated by use of a proper table showing the expansion of mercury with tem- perature (Appendix I). Gases measured in vessels in- verted over a liquid will of course have the same pressure as the atmosphere (recorded by the barometer) when the levels of the liquid in the vessel and outside of it are equal. The temperature of the gases and liquid should of course be alike. CHAPTER II. DENSITY AND SPECIFIC GRAVITY. THE term density is used to designate the mass in unit of volume. As this is a constant for any substance under given conditions of temperature and pressure, we may call it the absolute density. It is expressed simply as D = M/V, where D, M and V represent respectively density, mass and volume. When the absolute density of one substance is referred to that of another, i.e., a relation between these two densities determined upon the basis of one of them as a standard, we have a value known as the specific or rela- tive density, or what has Jbeen more commonly called the specific gravity. The relative density of a substance is not affected by the force of gravity, since any change in this force will operate equally upon the absolute densities of all substances as well as upon the standard. The relative density or specific gravity may be defined simply as the ratio between the weight of any substance and that of an equal volume of the standard substance when this latter is considered as unity. As a standard for the specific gravity of liquids and solids, the most natural choice is distilled water, which is considered at the point of its greatest density, namely 4 C. Determinations made at other than this temperature (e.g. 15) are usually referred to the density of water at 4 C., and expressed thus: sp. gr. at 15/4. In the case of gases, the values obtained for the absolute densities, the number of units of mass per unit of volume, are exceedingly small. By reason of this it is more cus- 5 6 CHEMICAL CALCULATIONS ternary to multiply these values by 1000 or, what is the same thing, to determine the number of units of mass in 1 liter instead of 1 c.c. Though the densities of gases referred to some standard such as air are known as relative densities, the term vapor density meets with constant use when the relative density is determined for a liquid or solid in the state of vapor. Since the composition of the atmosphere is somewhat variable, the relative densities upon this standard cannot possess much accuracy. The employment of a pure gaseous substance, such as hydrogen, gives much better results. As the lightest gas, all other densities referred to it assume values greater than unity. The absolute density of hydrogen, the weight in grams of 1 liter at and 760 mm., is 0.08987. The weight of an equal volume, 1 liter, of oxygen under the same conditions is 1.429 grams; hence with reference to hydrogen as unity, oxygen will be found to have a relative density of 15.90, i.e., 1.429 is 15.90 times 0.08987, or, by simple proportion: 0.08987: 1.429 = 1: x, or x = 15.90. The Oxygen Standard. The intimate relation (to be noted later) between the molecular weight of a substance and the volume which this weight occupies in the state of vapor has led us to refer all densities of gaseous substances to the density of that substance which is to serve as the basis for molecular weights. Formerly hydrogen, as the lightest substance, served this purpose, and consequently the close relationship between densities and molecular weights was apparent. In recent times oxygen, with the value 32, has been adopted as the basis of molecular weights by reason of the great importance of this element in its numerous combinations with other elements and for reasons that will be made clear after further considerations. The DENSITY AND SPECIFIC GRAVITY 7 absolute density of oxygen, therefore, is now taken as the standard of gas densities and made equal to unity. Upon this basis the close relationship between the densities of gases and their corresponding molecular weights will be as well attained as originally, when hydrogen was the standard. The relative densities of a number of sub- stances, however, will fall below unity. Thus the relative density of hydrogen becomes on this scale 0.0629, a value easily derived from the simple proportion: 1.429 : 0.08987 - I: x. In order to avoid these small values some chemists prefer to assign to oxygen the absolute density of 16 instead of unity. From the relations already noted between oxygen on the one hand and hydrogen, as the lightest substance, on the other, we readily see that the relative densities of all other substances are thus brought to values greater than unity. By the adoption of 32 instead of 16, as this basis of densities, the relative densities may be made to coincide at once with the molecular weights.* Under all conditions the measure of the volumes of gases is conducted at definite temperatures and pressures. The influence* of change in these factors will be discussed in the succeeding chapters. Whatever the volume of a gas may be, its relative density will be determined from the ratio of the observed weight per volume to the weight y* * Strictly speaking, an exact agreement on this basis between relative densities and molecular weights is found true only in the case of perfect gases (cf. footnote, p. 34). In the study of actual weights of equal volumes of gases it seems preferable, therefore, to base the comparison upon the weight of the standard as unity. The significance of the theoretical or calculated values, upon the basis of 32 as the chemical unit, will be noted in succeeding chapters. 8 CHEMICAL CALCULATIONS of an equal volume of oxygen under like conditions, when this ratio is brought over to the basis of oxygen as unity. Example 1. The actual weight of 1 liter of carbon dioxide, at standard conditions, is 1.977 grams. One liter of oxygen weighs 1.429 grams. What is the relative density of carbon dioxide? The ratio between these two densities is now to be referred to the density of oxygen as unity. That is, the 1 429 1 ratio ' is to be made equal to the ratio - , where x is i.y it x the relative density of carbon dioxide. We thus have 1.429 1 , = - , or, as more commonly expressed, i. y 1 i x 1.429 : 1.977 = l:z. From which x is found to be 1.384. PROBLEMS. 1. A piece of metal weighing 30 grams displaced 20 c.c. of water (i.e., its own volume). What is the relative density of this metal referred to water? Ans. 1.5. 2. A vessel weighing 6.448 grams weighed 7.963 grams when filled with water and 8.266 grams when filled with a salt solu- tion. What is the relative density of this solution referred to water? (The temperature constant throughout.)^ Ans. 1.2. 3. The weight of 1 liter of aqueous vapor at and 760 mm. is 0.8045 gram. What is its relative density? Ans. 0.5630. 4. Calculate the relative density of mercury vapor, 1 liter of which at standard conditions (0 and 760 mm.) weighs 8.87 grams. Ans. 6.207. 5. Calculate the relative density of hydrogen chloride, 5 liters of which under standard conditions weigh 8.205 grams. Ans. 1.148. 6. Calculate the relative density of chlorine, 100 c.c. of which at standard conditions weigh 0.322 gram. Ans. 2.253. CHAPTER III. THE EFFECT OF PRESSURE UPON GASES. THE LAW OF BOYLE. The Relation of Volume to Pressure. When a gas is subjected to increase of pressure the volume decreases. On release of the pressure the volume increases, regain- ing its former volume only at the original pressure. The Law of Boyle states this as follows: At constant tem- perature the product of the volume of a gas (V) by its pressure (P) is a constant, or P X V = K. Thus a volume of gas (1) at a pressure of one atmosphere (1) will be reduced when under a pressure of two atmos- pheres to one-half of its original volume, or, in the equa- tion P X V = K, the values 1X1 = 1 will become 2 X -J- = 1. This constancy, however, applies to all gases, and hence the product PV in one case will be equal to the product P'V' for any other case when either P or V or both change, the temperature considered constant throughout; hence the equation PV = P'V'. In dividing this equation, PV = P'V', through by the quantity PV and canceling like terms in numerator and PV P'V' V P' denominator, we obtain ^7- = ^^-or^-, = p- . The two fractional quantities may be expressed also by the propor- tion V : V = P' : P. This is exactly what is indicated above, where a volume of gas under a definite pressure will have its volume halved under double this pressure, etc., or at constant temperature the volume of a gas is inversely proportional to its pressure (a general form for 9 10 CHEMICAL CALCULATIONS the statement of Boyle's Law). Strictly speaking, the Law of Boyle holds true only for a perfect gas. For prac- tical purposes the law gives a sufficiently accurate means of studying volume and pressure relations. In order to ascertain what a given volume of gas may become at some new pressure, the temperature constant throughout, we have only to apply the formula PV = P'V, in which we may designate the new values by the prime marks. Example 2. 200 c.c. of hydrogen measured at 750 mm. pressure will occupy what volume at the standard pressure (760 mm.), temperature a constant? PV = P'V, or V' = V (P/F). Substituting here the respective values P' = 760, P = 750 and V = 200, we obtain the expression V = 200 X 750/760, which is easily solved to a value of 197.3. In general terms, we shall say that a change in the volume of any gas by change in pressure is readily calculated from the values given, by multiplying the volume by the fraction formed from the original pressure as numerator and new pressure as denominator. If this new pressure is greater than the original, then the fraction will have a value less than unity and the new volume (the product of fraction by original volume) will be proportionately less; con- versely, if the new pressure is smaller than the original pressure, the fraction (formed from the pressures) be- comes larger than unity in value, and the new volume must increase over that of the original. It comes, there- fore, to the same end if one ask himself the question: " Is the new pressure greater or less than the original? " If greater, then the fraction formed from the pressures must be so arranged that the new volume of the gas will be smaller, i.e., the fraction must have as its numerator THE EFFECT OF PRESSURE UPON GASES 11 the smaller of the two pressure values, and as its denomi- nator the larger. This fraction less than unity in value will, as was just shown, reduce the volume which it mul- tiplies in the correct proportion. If the new pressure is smaller than the original, the diminution in pressure must be accompanied by an expansion in volume, and, accord- ingly, the fraction formed from the pressures, which we may conveniently call the pressure-fraction, takes the larger value in the numerator and smaller value in the denominator. From the product of this fraction, greater than unity, by the original volume we obtain the new and correspondingly increased volume. The Relation of Pressure to Volume. Just as change in pressure produces a change in volume, so also a change in volume will produce a change in the pressure of a gas. The effect of this change in volume upon the pressure is readily determined from the equation PV = P'V, and takes the form of expression: P' = p (V/V). As the new volume V' increases or decreases with reference to the original, the volume-fraction becomes respectively less or greater than unity in value. Examples. A 'volume of gas measuring 100 c.c. at 750 mm. was expanded to 500 c.c. at constant temperature. What was the pressure of the gas at this final volume? In order to arrange the fraction (formed from the volumes) in such a way as to give a value less than unity and one that will reduce the given pressure in the ratio indicated by Boyle's Law, it is necessary here to place the smaller number in the numerator. The product of this volume-fraction by the original pressure P, or 750 mm., gives the new pressure sought: 750 X 100/500 = 150. 12 CHEMICAL CALCULATIONS The Relation of Density to Pressure. The absolute density of a gas, the mass per liter, may be seen now to vary with the pressure to which the gas is subjected. With reference to the formula D = M/V, we are aware that the weight or mass is unaffected by any changes in temperature or pressure. The volume (V), however, is altered by either or both of these influences. Thus with an increase of pressure it decreases and hence the absolute density must increase. In other words, both the absolute density and the pressure of a definite quantity of gas vary inversely with the volume when the temperature is con- stant, and hence are directly proportional to each other. Therefore, in the equation PV = P'V we may substitute D for P throughout and obtain DV = D'V, or D 7 = D (V/V). Example 4- 100 c.c. of hydrogen, the absolute density of which (the weight of 1 liter) is 0.08987, will have what density when expanded to 200 c.c. in volume at a constant tem- perature? Here the increase in volume means a decrease in the pressure and hence in the density. The fraction formed from the volumes, the volume-fraction, must be arranged so as to produce this decrease corresponding to Boyle's Law and, therefore, the numerator must contain the smaller of the two values. When this quantity 100/200 is multiplied by the density 0.08987, we obtain the new density 0.0449. This is identical with saying that at this new and larger volume the actual weight of the gas has not changed, or the original weight of the 100 c.c., 0.008987 gram, has remained a constant. The mass per volume, however, has been changed, in that this weight of gas is now spread THE EFFECT OF PRESSURE UPON GASES 13 out over twice the original volume. The actual weight of 100 c.c. of the rarefied gas is just one-half of the weight of the total volume, 200 c.c., or 0.00449 gram. As a basis for calculations of this nature, that general property of homogeneous substances is of course understood: The weight of any fractional part of a volume bears the same ratio to the total weight as this fractional volume itself does to the total volume. It should be remembered in this connection that the relative density of a solid or liquid may vary with the temperature or pressure (p. 5), since the effect of change in these factors upon the volume of a substance and upon that of the standard chosen may not be the same. The relative density of a gas, however, is constant, since changes in these factors affect all gases alike. PROBLEMS. 7. 512 c.c. of hydrogen at a pressure of 744 mm. will occupy what volume at a pressure of 790.5 mm., the temperature constant? Ans. 481 c.c. 8. 240 c.c. of gas at a pressure of 740 mm. were admitted into an empty vessel of 800 c.c. capacity. What was the pressure of the gas at this new volume, temperature constant? Ans. 222 mm. 9. 500 c.c. of oxygen, absolute density 1.429, were com- pressed to a volume of 125 c.c. at constant temperature. What was the density of the gas at this final volume? What would be the weight of 50 c.c. of the compressed gas? Ans. 5. 71 6, 0.2858 gram. 10. A volume of gas measuring 600 c.c. at 760 mm. pres- sure was expanded to a volume of 1000 c.c. at constant tem- perature. What was the final pressure of the gas, and what fractional change in its absolute density must have followed this expansion? Ans. 456 mm. Density I of former value. 14 CHEMICAL CALCULATIONS 11. 500 c.c. of a gas, the absolute density of which is 6, must be reduced to a density of 0.75 at constant temperature. What will be the volume of this rarefied gas? Calculate also the weight of 400 c.c. of the rarefied gas. Ans. 4000 c.c. ; 0.3 gram. 12. A volume of gas weighing 5 grams was expanded, at a constant temperature, till the pressure was reduced to one-half of its former value. 500 c.c. of the rarefied gas weighed 1.25 grams. What was the original volume of the gas? Calculate also the original density assuming the original observations made at standard conditions. Ans. 1000 c.c. Density 5. CHAPTER IV. THE EFFECT OF TEMPERATURE UPON GASES. THE LAW OF CHARLES. WITH pressure constant the volume of a gas is found to vary with the temperature. The Law of Charles (known also as Gay-Lussac's Law), states this as follows: Under constant pressure the volume of a gas varies directly with the temperature upon the absolute scale. The Determination of the Coefficient of Expansion. The standard condition of temperature, C., becomes on the absolute scale 273. This is calculated from the fact that, given any volume of gas at C., the effect of cooling upon this gas will bring about a diminution in its volume by 1/273 for each degree centigrade below the centigrade zero, consequently at 273 C. the volume will have been diminished by 273/273 of itself, or by its entirety. This point, therefore, is regarded as the zero point or lowest point at which a gas may possibly come under considera- tion. At or below this point no gas could exist as such. Though this temperature has not as yet been reached, it follows that all gases will be liquefied above it. A volume of gas measuring 273 c.c. at C. would lose per degree of cooling 1/273 of itself, or a single unit of volume. At one degree of temperature above the absolute zero (designated as 1 A.) the gas, having cooled through 272 and lost 272/273 of its volume at C., would consequently occupy but 1 c.c. At 2 A. its volume would be 2 c.c., having lost here 271/273 of its volume at C. At 273 A. its volume would be again 273 c.c.; in fact through- 15 16 CHEMICAL CALCULATIONS out the entire range, A. to 273 A., a unit of volume corresponds to a unit of temperature, and the one is there- fore directly proportional to the other, i.e., the volume varies directly with the temperature. Upon passing the point 273 A. the same principle holds. For each degree of temperature there is an expansion in volume corre- sponding to 1/273 of the volume at 273 A. or C. At 283 A. the volume of a gas measuring 273 c.c. at 273 A. will have increased by 10/273 of this volume and measure 283 c.c. The rate of expansion above C. is identical with that of cooling below this temperature. This ratio between the increase in volume per degree and the total volume of a gas at C. called the coefficient of expan- sion and often expressed decimally as 0.00367 in place of the fraction 1/273 was determined by measuring the expansion of gases from C. upward and this upon the centigrade scale; consequently the determination of the zero point on the absolute scale is made in centigrade units, and each degree of temperature centigrade must be equivalent to one degree on the absolute scale. If 273 A. is the zero of centigrade, then 283 A. is equal to 10 C. and so on. In order to preserve this direct pro- portionality between volume and temperature we have only to convert the centigrade temperatures into their corresponding values upon the absolute scale. By reason of the equivalence in the units we add 273 to the read- ings above C. and subtract readings below C. from 273; the results represent absolute temperatures (desig- nated by T in contradistinction to t for the centigrade temperatures). With this one volume, 273 c.c. as above chosen, the absolute temperature coincides numerically throughout with the volume in cubic centimeters, and simplifies the calculations to adding or subtracting single units. Had THE EFFECT OF TEMPERATURE UPON GASES 17 any other volume of gas at C. been considered, the same ratio between volume and temperature would be found a direct proportionality but the variations would not be in single units. The equation which represents this change in volume V under a definite change in tem- perature t (centigrade) may be simply expressed as V = V (1 + at), where a is the coefficient of expansion per degree (1/273) and t the number of degrees centigrade through which it operates. Thus for a definite rise in temperature, t, the total expansion will be equal to that fractional part of the original volume as is indicated by the quantity at, i.e., (1/273 X t) V; and any new volume V must be equal to the original volume V.plus the total expansion : V = V + (1/273 X Vor V' = V (1 4- Z/273) or V = V(l + at). It is regarded as more helpful in this work to arrange the original and new values as members of a simple propor- tion, in which, of course, only the absolute temperatures are considered: V : V = T' : T. The application of the Law of Charles to various cases may be illustrated now by examples. The Relation of Volume to Temperature. Example 5. A volume of gas measuring 200 c.c. at 20 C. will have what volume at the standard condition, C. From the direct proportionality between volume and absolute temperature we have V : V = T : T', or V = V (T'/T), where T and T' represent, respectively, the original and new temperatures expressed in absolute units. In the 18 CHEMICAL CALCULATIONS example above, T = 20 + 273 = 293; the new tem- perature is 273. The change in volume of this gas must follow according to the proportion: 200 : x = 293 : 273, or x = 200 (273/293), or x = 186.35 Here we see the original volume is to be multiplied by a fraction formed from the two temperatures, the tempera- ture-fraction, and again we may apply the question as in the consideration of Boyle's Law: "Is the new volume to be smaller or greater than the original volume?" If greater, then the larger number (the degrees in absolute units) must be set over the smaller in order that the frac- tional quantity, when multiplied by the original volume, may increase that volume in the proportion indicated by these temperature values. If the new volume is to be smaller, then the numerator must contain the smaller of the two numbers and the product accordingly will be proportionately less. The Relation of Temperature to Volume. Example 6. A volume of gas measuring 100 c.c. at 17 C. was expanded by warming to a volume of 250 c.c. at constant pressure. To what temperature must the gas have been heated to bring about this expansion? From the Law of Charles we draw up the equation T' = T (V'/V). In conformity with the direct propor- tionality between volume and absolute temperature, the new temperature (T') must be greater than the original temperature. The volume-fraction thus arranged to bring about this proportional increase when multiplied by the original temperature (T = 273 + 17 = 290) becomes 250/100, and the final temperature accordingly is T' = 290 (250/100) or 725. 725 A. is equal to 452 C. THE EFFECT OF TEMPERATURE UPON GASES 19 The Relation of Density to Temperature. In con- sideration of the change in absolute density under change of temperature, with pressure a constant, we see from the formula D = M/V that, as the volume increases, the fraction M/V (the expression for the density) must become less and less. With the volume of a gas the denominator of this fractional quantity becoming less, we obtain a larger quotient or a larger value for the den- sity. Now as this change in volume is directly propor- tional to a change in absolute temperature, the density of a gas will decrease with an increase in its absolute tem- perature and increase with a decrease in this factor, that is, under constant pressure the absolute density of a definite quantity of gas is inversely proportional to its absolute temperature or D : D' = T' : T. Example 7. A volume of hydrogen at C., with the abso- lute density 0.08987, will have what density when expanded by heating to 10 C. at constant pressure? From the equation above we have D' = D (T/T X ). This signifies that the density given must be multiplied by a fraction formed from the two temperatures (the temperature-fraction). We should now ask the question whether, from the inverse proportionality just mentioned, the new density (D') will be greater or less than the original density (D). Here the new temperature is higher, consequently the new density will be lower, there- fore the smaller number must be placed in the numerator and the value for D' can be readily calculated: . D' = D (273/283), or 0.08987 (273/283), or 0.0867. 20 CHEMICAL CALCULATIONS PROBLEMS. 13. What will be the volume of 250 c.c. of hydrogen meas- ured at 30 when cooled to 10 at constant pressure? Ans. 217 c.c. 14. A rubber balloon containing 400 c.c. of oxygen measured at - 20 is subjected to a temperature of 120. What is the increase in volume of the balloon, the atmospheric pressure constant throughout? Ans. 221.3 c.c. 15. A volume of gas measuring 500 c.c. at was expanded by heating to 600 c.c., at constant atmospheric pressure. What was the temperature which the gas attained? Ans. 54.6. 16. The absolute density of oxygen is 1.429. What would be the weight of one liter of oxygen collected at 40 and 760 mm. pressure? Ans. 1.246. 17. A volume of gas with the absolute density 4 and measur- ing 250 c.c. at was expanded by warming, under constant pressure, to a volume of 600 c.c. What increase in temperature was required, and what would be the weight of 300 c.c. of the rarefied gas? Ans. 382.2; 0.5 gram. CHAPTER V. THE COMBINED EFFECT OF TEMPERATURE AND PRESSURE UPON GASES. PARTIAL PRESSURES. WHEN a volume of gas is subjected to change in both temperature and pressure the relation of the original volume to the new volume must be in accordance with the two principles already enunciated: V : V = P : P' and V 7 : V = T 7 : T or V 7 /V = P/P' and V'/V = T'/T or V 7 = V (P/P 7 ) and V 7 = V (T 7 /T). Considered separately, we know that the original volume under change in pressure must adjust itself in accordance with the ratio denoted by Boyle's Law and attain that value which is expressed by the product of the pressure- fraction by the original volume, V (P/P 7 ). The operation of a change in temperature will modify the original volume in accordance with the Law of Charles, and the resulting value will be expressed, as the formula above indicates, by the product of the original volume by the temperature-fraction, V (T 7 /T). If the original volume, already modified to accord with a change in pressure, is to come in turn under the influence of a change in temperature, which is separate and inde- pendent of the pressure, we shall be able to express the result by substituting this already modified value of the original volume, V (P/P 7 ), for that volume denoted by V in the equation, V 7 = V (T 7 /T), representing the effect of an alteration in temperature. In other words, the vol- 21 22 CHEMICAL CALCULATIONS ume Vin the equation V 7 = V (T 7 /T) may be regarded as already affected by some change in pressure, a change which can have altered it only in accordance with the Law of Boyle and brought it to its new value through the expression V (P/P'). It follows, therefore, that by sub- stituting V (P/P') for V in the equation V 7 = V (T'/T) we shall obtain the value which results from the action of both these agencies, pressure and temperature, upon the gas. Whether we consider the action of pressure before or after that of temperature, the final product will be the same. Thus, when the expression V 7 = V (T 7 /T) is made to include the effect of a change in pressure upon the original volume V, we have the form V = V(P/F) (T'/T),orV = v(J, . J); whereas the expression denoting the influence of pressure alone, V 7 = V (P/P X ), when made to include the effect of a change in temperature upon the volume V, gives the same form V = V (T'/T)

273, and 750 mm. -> 760 mm. With respect to these alterations in temperature and pres- sure the questions may now be asked in regard to the magnitude of the new volume brought about by each of these influences. From the change in temperature the new volume must be smaller, as the gas in cooling undergoes a contraction (direct proportionality), hence 273/283 will represent the temperature-fraction. From the change in pressure the new volume must be smaller, since the new pressure is larger and the volume therefore will be dimin- ished (inverse proportionality), hence 750/760 will repre- sent the pressure-fraction. Bringing these two influences together, the product of the two by the original volume will give the combined effect of the two changes upon this volume: V' = 100 (273/283) X (750/760), or V 7 = 95.2 c.c. * If in this connection the form of equation V' = V (1 -f at) is used, we shall arrive at a similar expression for change in pressure at con- stant volume: P r = P (1 + at). The combined effect of a change in volume and change in pressure is then represented by P'V' = PV (1 + at). By substituting here for t its value on the absolute scale, T 273^ we obtain P'V - PV I V \273 or P'V- T ' ~' J 273 ' ~273 The expression PV/273 is found to be a constant for a given quantity of gas under all values of temperature, pressure and volume, and is usu- ally expressed by R ; hence the general gas equation, PV = RT, which signifies "that the pressure-volume product of a definite quantity of any definite gas is proportional to its absolute temperature." A. A. FOTES. TEMPERATURE AND PRESSURE UPON GASES 25 With these two fractions each less than unity in value, the original volume is materially diminished under the new conditions. The Relation of Temperature to Volume and Pressure. Example 9. At what temperature will a gas 200 c.c. in volume, at 20 and 750 mm., occupy a volume of 300 c.c. when the pressure is 740 mm.? Here the volumes and pressures are known, and the undetermined quantity is one of temperature. The general formula must be arranged so that the undeter- mined quantity is on one side of the equation: V'/V = P/P' X T'/T when divided through by P/P 7 gives T'/T = V'/VX P'/P, hence the new temperature T' is expressed by the equation /V P'\ v = T{ W P/* The entire matter may be simplified and made more apparent if we recall the reasoning advanced in the pre- ceding cases; namely, the unknown quantity is derivable from the known quantity by multiplying this latter by the fractions formed from the related terms, when these terms are arranged so as to give a quotient that will express the magnitude of the term sought (the unknown quantity) over that of the one already existent. At 20 C., or 293 A., the volume of gas in the problem above was 200 -c.c. At the new temperature it is to be 300 c.c., hence this temperature must be larger, else no expansion could occur (direct proportionality); therefore T' = 293 (300/200). Again, the pressure-fraction must be arranged so that its quotient will acquire a value less than unity, i.e., 740/750, since the effect of 26 CHEMICAL CALCULATIONS pressure must necessarily decrease the temperature here required (direct proportionality) (cf. p. 23). The com- bined influence will be expressed by the product of the two fractions by the original temperature: T' = 293 (300/200) X (740/750) or T' = 433.6 A. = 160.6 C. The Relation of Pressure to Volume and Temperature. Example 10. 100 c.c. of a gas at 20 and 740 mm. will exert what pressure when occupying 120 c.c. at 40? Arrange the related terms in fractions in accordance with the questions as to whether the new pressure is to be greater or less than the original, 740 mm., and multiply both by this value. The temperature-fraction becomes 313/293, since at higher temperature the pressure must be greater (direct proportionality); the volume-fraction becomes 100/120, since at increased volume the pressure must be less (inverse proportionality), therefore by arranging all these factors we have: P' = 740 (100/120) X (313/293), or P' = 658.7 mm. This is in accord with the general equation which may be drawn up for changes in both volume and temperature upon the pressure of a gas: D l^'T When the volume is a constant the equation takes the form indicated on p, 23: P' = P/^ The Relation of Density to Temperature and Pres- sure. The variation in the absolute density of a gas through the simultaneous variations in temperature and PAETIAL PRESSURES 27 pressure follows readily from what has been stated in regard to variation in volume under these same influences. In the study of absolute density, D = M/V, the inverse proportionality between the absolute density and the volume of a gas permits, in the general equation denoting the influence of temperature and pressure changes in this connection, of the substitution of the term D' (the new density) and D (the original density) for V and V re- spectively. The general equation V'/V = P/P' X T'/T then becomes D/D' = P/P' X T'/T or, as transposed to indicate the alteration in temperature and pressure upon the original density: P T This is in keeping with what has been already observed, i.e., the absolute density of a definite quantity of gas varies directly with the pressure and indirectly with the absolute temperature. Example 11. A volume of oxygen at 20 and 750 mm. pres- sure has what absolute density under these conditions? The absolute density of oxygen at and 760 mm. pressure is 1.429. The new density will be less under the conditions named, since both the pressure-fraction 750/760, and temperature-fraction 273/293, will reduce the original value according to the equation: D' = 1.429 (273/293) (750/760), or D' = 1.314. PARTIAL PRESSURES. Whether we have to deal with a simple gas or a mixture of several gaseous components, the Law of Boyle is equally applicable. Each component of a gaseous mixture exerts a definite individual pressure, the same that it would exert were it alone present in this volume. According to the Law of Dalton, the sum of these individual pressures, 28 CHEMICAL CALCULATIONS partial pressures, of the several components is equal to the total pressure of the mixture. When two gases of equal volume at the same conditions are brought together and communication made between the two vessels, we find that the gas from each vessel diffuses into the other vessel and becomes uniformly distributed throughout this double volume (the two vessels). In other words, each gas be- haves as if it alone were present in the total space included in the two vessels. Under double the original volume each gas can exert but one-half of its original pressure. The sum of the pressures of the two gases, however, must be again equal to unity or that pressure which each originally exerted. Consequently the total pressure of a gaseous mixture may be regarded as the sum of the individual or partial pressures of the several components. When aqueous vapor is present in a volume of gas we have again the consideration of gaseous mixtures. Water gives off a varying amount of its vapor, independent of any other gaseous substance that may be present in the space about it, but always dependent upon the tempera- ture. At any one temperature we assume an equilibrium between the two tendencies, that of the molecules of the vapor to condense as liquid, and that of the molecules of the latter to fly off as vapor; in other words, we have a condition of saturation with aqueous vapor. The density or concentration of aqueous vapor (the aque- ous tension) as attained at each condition of equilibrium is definite for that temperature at which the equilibrium exists. This density of the vapor is determined most easily by measuring the pressure which it exerts, the vapor pressure. If a little water is admitted into the vacuum at the top of a barometric column of mercury, the vapor evolved will exert a pressure which increases with a rise in the temperature of the surrounding medium. At 100 the PARTIAL PRESSURES 29 pressure of this vapor will have just sufficed to displace all of the mercury (sustained by the atmosphere at sea level), and consequently will have reached a pressure exactly bal- ancing one atmosphere. This temperature is the boiling- point of the liquid or that point at which bubbles of vapor form in the liquid itself. At higher elevations than sea level, i.e., under reduced atmospheric pressures, this attainment of the external pressure by the vapor of the liquid (the boiling-point) will take place at lower tempera- tures. As the temperature falls the aqueous vapor above the mercury in the barometric tube just mentioned will decrease in concentration and exert less pressure; conse- quently mercury will be forced into the tube by the pres- sure of atmosphere from without. The difference in height of the mercury in this tube and that in a barometer, where no moisture is present, will give the height of mercury, or pressure, corresponding to that of the aqueous vapor at the recorded temperature. When any gas is measured over water at atmospheric pressure, and a temperature constant for both gas and liquid, we obtain the partial pressure of the dry gas in the mixture by subtracting from this barometric reading the pressure which aqueous vapor exerts at the observed tem- perature. (The values for a range of temperatures are found in Appendix II.) Example 12. 100 c.c. of a gas at 10 and 750 mm., measured over water, will have what volume under standard conditions? At 10 the pressure of aqueous vapor is 9.2 mm. (see Appendix II), therefore the barometric reading, 750 mm., represents the sum of the pressures of the aqueous vapor and that of the dry gas, and, as standard conditions are desired, the volume of the gas should be determined in the dry state. Accordingly 750 mm. 9.2 mm. = 740.8 mm., the pressure of the dry gas. The problem then resolves 30 CHEMICAL CALCULATIONS itself into one exactly as in Ex. 8, page 24. The volume of a gas at 740.8 mm., when calculated at a pressure of 760 mm., or a greater pressure, must be necessarily diminished; therefore 740.8/760 is the pressure-fraction. The temperature-fraction is 273/283, consequently the new volume will be derived from the equation V = 100 X 740.8/760 X 273/283, or V = 94.03 c.c. PROBLEMS. 18. 500 c.c. of hydrogen at 25 and 745 mm. pressure will have what volume at 15 and 755 mm. pressure? Ans. 476.8 c.c. 19. A vessel of 2000 c.c. capacity held 5 grams of a vapor at the standard conditions of temperature and pressure. What weight of this vapor at 10 and 750 mm. pressure can be held in this vessel, the capacity considered constant? Ans. 4.76 grams. Note. This volume of vapor (x), at and 760 mm., with the weight (10), when raised to the new volume (?/), will still have the same weight. Consequently the vessel can be made to contain only that part of the original weight which is denoted by the fraction x/y. 20. 10 liters of a gas measured at 20 and 750 mm. pressure weighed 14 grams. What weight of this gas could be contained in a smaller vessel holding 4 liters at 10 and 760 mm. pressure? Ans. 5.87 grams. 21. What volume will 1000 c.c. of gas at 30 and 740 mm. pressure occupy when reduced to standard conditions? Ans. 877.3 c.c. 22. 1000 c.c. of a gas measured at 10 and 750 mm. pressure was increased to 1120 c.c. by warming. The final pressure read 740 mm. What was the final tejnperature of the gas? Ans. 39.7. 23. A volume of gas measured at 10 and 750 mm. pres- sure will have what pressure at 20, the volume a constant? Ans. 776.5 mm. 24. A volume of gas measuring 1000 c.c. at 15 and 745 mm. pressure was warmed to 32. What increase in the pressure of the gas would be recorded, the volume a constant? Ans. 44 mm. TEMPERATURE AND PRESSURE UPON GASES 31 25. What decrease in temperature will be necessary to reduce 400 c.c. of a gas, at 20 and 765 mm., to a volume of 300 c.c. at 750mm.? Ans. 77.56. 26. What increase in temperature will be necessary to bring a volume of gas, measuring 560 c.c. at 10 and 745 mm. pres- sure, to a volume of 600 c.c. at this same pressure? Ans. 20.2. 27. A volume of hydrogen measuring 500 c.c. at 25 and 730 mm. was reduced in volume to 400 c.c at 0. What was the final pressure of the gas? Ans. 836 mm. 28. What increase in pressure is necessary to force 100 c.c. of hydrogen, at 10 and 740 mm., into a vessel of 80 c.c. capacity, when the temperature of this vesselis constant at 0? Ans. 152.3 mm. 29. What is the absolute density of hydrogen at 20 and 740 mm. pressure? Ans. 0.08153. 30. What is the absolute density of air at 10 and 750 mm. pressure? The absolute density at the standard conditions is 1.293. Ans. 1.231. 31. 400 c.c. of gas, with the absolute density 6 and measured at 25 and 750 mm. pressure, is to be brought to a temperature of 10 and a pressure of 760 mm. What weight of this final gas can be contained in a vessel of 100 c.c. capacity? Ans. 0.64 gram. 32. 10 grams of a gas, measured at 48 and 600 mm. pres- sure, was expanded by heating to 177 and reducing the pressure to 480 mm. Of this rarefied gas 250 c.c. weighed 0.5 gram. What was the original volume of gas and what was the density of the gas at its original and final volume? Ans. 2000 c.c. Original density = 5. Final density = 2. 33. 150 c.c. of air measured over water at 18 and 746 mm. pressure will have what volume at standard conditions? Ans. 135.3 c.c. 34. 1 liter of oxygen at standard conditions weighs 1.429 grams. 440 c.c. of this gas measured over water at 24 and 742 mm. pressure will contain what weight of the dry gas? Ans. 0.547 gram. 32 CHEMICAL CALCULATIONS 35. 545 c.c. of nitrogen as measured over water at 22 and 748 mm. pressure contain what weight of the dry gas (density = 1.2507)? Ans. 0.6045 gram. 36. 2.2 grams of oxygen will occupy what volume at 20 and 770 mm. pressure when transferred to a vessel inverted over water? Ans. 1667 c.c. 37. 1 gram of hydrogen (density = 0.08987) measured at standard conditions will occupy what volume when transferred to a vessel over water at and 760 mm. pressure? Ans. 11,195 c.c. 38. 400 c.c. of oxygen at standard conditions will have what volume when measured over water at 20 and 755 mm. pressure? Ans. 442.3 c.c. 39. 100 c.c. of a gas, measured over water at 25 and 745 mm. pressure, will have what volume when deprived of moisture? The atmospheric conditions constant. Ans. 96.8 c.c. 40. 1000 c.c. of oxygen, measured over water at 10 and 750 mm. pressure, will have what volume at 10 and 770 mm. pressure when deprived of its moisture? Ans. 894.1 c.c. 41. 500 c.c. of a gas contained in a tube inverted over water, and measured at 10 and 765 mm. pressure, will have what volume under a change in the atmospheric conditions to 20 and 745 mm. pressure? Ans. 537.7 c.c. 42. What decrease in pressure will be necessary to raise a volume of gas measuring over water 400 c.c., at 22.5 and 748 mm. pressure, to a volume of 440 c.c. under the same conditions? Ans. 66.2 mm. Note. The actual atmospheric pressure of the moist gas required for the new conditions will be obtained by adding the tension of aque- ous vapor to the value for P'. 43. What increase in atmospheric pressure will be necessary to reduce 200 c.c. of a gas, measured in a tube over water at 10 and 720 mm., to a volume of 100 c.c. at 20 in this same tube? Ans. 769.2mm. Note. The calculated pressure for the dry gas must be increased by the tension of aqueous vapor at the new conditions in order to obtain the actual pressure that would be recorded (cf. Prob. 42). CHAPTER VI. AVOGADRO'S HYPOTHESIS AND SOME OF ITS APPLICATIONS. THE hypothesis of Avogadro is the outcome of the Law of Combining Volumes (Gay-Lussac), and assumes that, if the Laws of Boyle and Charles are strictly true, there must be a uniform distribution of molecules in all volumes of gases at the same conditions of temperature and pres- sure. More generally the hypothesis takes the following form: Under the same conditions of temperature and pressure equal volumes of all substances in the state of vapor contain an equal number of molecules. By mole- cules are meant the smallest parts in which a substance maintains its identity. The complete chemical combina- tion between the molecules of equal or multiple volumes of gases to form definitely related volumes of gaseous products, with no molecules of either constituent remain- ing free or uncombined, presupposes this even distribution of molecules. Now the weight of a given volume of gas when compared with the weight of an equal volume of another gas, under the same conditions, will represent not alone the ratio between the sum of the weights of all the molecules in one volume and that of the other, but also the ratio between the weight of an individual molecule of one gas and that of the other. This follows, of course, from the fact that the numerical ratio between the weights of each equal volume is not altered through division by a common factor, the unknown yet equal number of molecules in both 33 4 CHEMICAL CALCULATION^ volumes. In determining the relative weights of mole- cules it is therefore of no importance as to what particular volume of gas is weighed, so long as the weight is to be compared with the weight of an exactly equal volume of some other gas at the same conditions of temperature and pressure. Any known volumes of gases upon which accu- rate data are at hand may serve for this purpose when they are reduced to the same conditions (cf. Chap. V) and comparison made between equal volumes of each. The Relation of Density to Molecular Weight. This comparison in the weights of equal volumes of gases is, after all, nothing more or less than the comparison in gas densities. Thus the absolute density of hydrogen, 0.08987, when compared with that of oxygen, 1.429, is found to be about 1/16 of the latter, or, if the density of hydrogen is considered as unity, the density of oxygen becomes 15.90: 0.08987 : 1.429 = 1 : 15.90. The relative weight of a molecule of oxygen, its molecu- lar weight, would be, accordingly, 15.90 times the molecular weight of hydrogen. But from accurate deter- minations of the atomic weights of these elements the molecular weight of oxygen is found to be only 15.88 times that of hydrogen.* * The determination of the atomic weights of all the elements gives us the most accurate means of obtaining the true molecular weights. Density determinations involve numerous complications, and are there- fore liable to errors, but nevertheless they give us a very fair approach to the true values. The various degrees of cohesion between the molecules of gases bring into existence a greater or less deviation from the uniform packing of these molecules as assumed by Avogadro's hypothesis for a perfect gas. Thus with oxygen we have a slightly greater packing than with hydrogen, as is shown by the increase of the relative density 15.90 over the calculated value 15.88. In order that our calculations may be free of these slight discrepancies, densi- ties may be given as calculated back from the correct molecular weights, AVOGADRO'S HYPOTHESIS 35 The relation in the absolute densities applies equally well to the relative densities of gases providing, of course, that they are referred to the same standard. Thus the relative densities of hydrogen and oxygen upon air as a standard are 0.0696 and 1.105 respectively. The com- parison of these values gives the same relative weights of the molecules of hydrogen and oxygen as shown above. We have several cases which show that a volume of hydrogen may enter into combination with an exactly equal volume of another gas, and the resulting gaseous compound occupy the volume originally held by the two gases severally, i.e., the sum of the two gaseous volumes now makes up the volume of the compound. Since temperature and pressure remain the same throughout, the number of molecules in the volume of the resulting compound must be twice the number of molecules in either volume of gaseous constituent. As each of the molecules in the compound contains hydrogen in chemical combination, we clearly see that there are now twice as many hydrogen parts present as there were in the original volume of hydrogen. Since weight for weight there can be no change in the amount of hydrogen present in the two cases, we naturally infer that the original molecule of hydrogen contained two smaller parts or units, atoms, and that it was these atoms which were concerned in the chemical combination. Never have we been able to get the element hydrogen to spread out over more than twice its original volume when entering into chemical combination, therefore we assume that the hydrogen as we know it contains at least two atoms to its molecule. and consequently the value for oxygen will henceforth be taken as 15.88 instead of 15.90, when referred to hydrogen as unity. Unless otherwise stated, the densities hereafter considered are the calculated weights of 1 liter of gas at standard conditions. 36 CHEMICAL CALCULATIOKS We have a number of elements the molecules of which under certain conditions are shown in like manner to possess two atoms (i.e., diatomic), e.g., oxygen (O 2 ), chlorine (C1 2 ), bromine (Br 2 ), iodine" (I 2 ), and nitrogen (N 2 ), whereas the molecule of mercury in the vapor state is monatomic (Hg). As already stated, from the comparison in the weights of equal volumes of gases, we may derive the relative weights of the several kinds of molecules. With the hypothesis now that the molecules themselves are com- posed of atoms, we must interpret these relative values as those which stand for the ratios between the sums of the weights of the atoms in the molecules, i.e., the molecular weights. A comparison between the relative weight of a molecule of hydrogen and a molecule of oxygen gives us the ratio 1 : 15.88. Since both hydrogen and oxygen molecules contain two atoms each, this same ratio stands likewise for that between the weights of the hydrogen and oxygen atoms. In order, therefore, to include the relative weights of the atoms in the molecular weights and thus be able to compare the relative weights of the molecules upon their smallest units as a basis, it is only natural that we regard the ratio 1 : 15.88 as the ratio between the relative weights of the hydrogen and oxygen atoms. The molecu- lar weights of hydrogen and oxygen then become 2 and 31.76 respectively, values which indicate, of course, that they are made up of the sums of the respective atomic weights in each molecule. All of this presupposes that hydrogen, with the smallest atomic weight, should be re- garded as unity for the convenience of these comparisons. The importance of the element oxygen with its great number of well-defined compounds and the comparative ease with which these compounds may be utilized, through analyses, for the determination of the relative weights of AVOGADRO'S HYPOTHESIS 37 molecules of other elements, has led to its universal adop- tion as the standard of molecular weights. In order that the values upon this scale may not depart far from those already determined upon the basis of hydrogen as unity, the former value for oxygen, 31.76, was increased to 32, a whole number, and thus the atomic weights of all the elements allowed to retain a value greater than unity. With oxygen changed from 31.76 to 32, we may readily derive the corresponding value for the molecular weight of hydrogen from the simple proportion: 31.76 : 2 = 32 : x, which gives the value 2.016. From the molecular weight of oxygen (32) the relative weight of the atom must be- come 16, while the relative weight of the hydrogen atom will be 1.008. An imaginary gas with the atomic weight 1/16 of the atomic weight of oxygen (16) may be regarded, therefore, as the basis or unit weight of these values. These numerical values refer to no particular standard of weights, but when expressed in grams it is customary to call them gram-molecular weights; thus 32 grams is the G.M.W. of oxygen. As already stated, the comparison of the densities of all substances in the state of vapor gives at once the relative weights of their molecules. In order, then, to obtain the molecular weight of any substance upon the basis of oxy- gen as 32 it becomes necessary to effect a comparison between its absolute density and that of oxygen on this basis. The calculated absolute density of chlorine is 3.166 (actual value = 3.22). The absolute density of oxygen, 1.429, will bear the same relation to this value for chlorine as the weight 32 bears to the molecular weight of chlorine, 70.92: 1.429 : 3.166 = 32 : x, or x = 70.92. 38 CHEMICAL CALCULATIONS When the relative density of a gaseous substance is given, i.e., the value obtained by referring the absolute density to that of oxygen as unity, the proportion above will of course take the form: 1 : Rel. density of gas = 32 : x. This proportion shows that the molecular weight (x) is merely the product of the relative density by 32. This is the logical outcome of making oxygen the standard both of gas densities and molecular weights. Since as the basis of relative densities the weight of a unit volume of oxygen is considered unity and as the basis of molecular weights it is considered 32, we naturally need only multiply any particular relative density by 32 in order to obtain the molecular weight of this substance. If, on the other hand, the relative density of a gas is given in terms of some other density besides oxygen as a standard, the molecular weight of this standard substance must of course replace 32 in the proportion above. For- merly the relative densities were determined upon the basis of hydrogen as unity. The proportion indicating the molecular weight upon this basis is as follows: 1 : Rel. density (H = l) = 2.016 : x. Consequently the molecular weight may be derived from the relative density (H = l) by multiplying this value by 2.016. The Calculation of Relative Densities upon Different Standards. To convert the relative density of a gas upon the hydrogen standard over to the standard of oxy- gen, or vice versa, we shall need to refer this value to the ratio between the absolute densities of the two standards, namely 0.08987 and 1.429, in an inverse order. As the means and extremes of the proportion we must have the product of the relative density of a substance by the AVOGADRO'S HYPOTHESIS 39 absolute density of the corresponding standard, a product always equal to the absolute density of the substance in question; and hence a constant for the proportion. Thus: Abs. density 1 oxygen (1.429) Abs. density hydrogen (0.08987) Rel. density substance (H = l) Rel. density substance (O = 1). Example 13. The calculated relative density of chlorine (H = 1) is 35.23. What is the relative density of this gas referred to oxygen? According to the proportion : 1.429:0.08987 = 35.23 :x, the value of x, or relative density, (O = 1), is calculated as 2.216. When this is multiplied by 32 we obtain the molecular weight of chlorine, 70.92. Quite often the vapor density of a substance is deter- mined upon air as a standard. The ratio 1.429 : 1.293 is the ratio between the absolute densities of oxygen and air respectively, hence the calculated molecular weight of air referred to oxygen as 32 will be derived from the propor- tion: 1.429 : 1.293 = 32 : x. From this the value 28.955 is found; a value which signifies that if air were a com- pound its molecular weight would be 28.955. The molecular weight of a substance may be calculated, therefore, from its vapor density referred to air if we substitute this value, 28.955, for 32 in the proportion on page 38, and the relative vapor densities are made one member of the proportion. Example 14. The calculated vapor density of mercury is 6.908 referred to air. What is its molecular weight? The proportion 1 : 6.908 = 28.955 : x gives 200 as the molecular weight of mercury. The conversion of any vapor density (air = 1) to the basis (O = 1) is exactly analogous to the case of hydrogen and oxygen above. The absolute density of air is 1.293. 40 CHEMICAL CALCULATIONS The Gram-Molecular Volume. If the actual weight of 1 c.c. of a gas is known in grams and this value is divided into the molecular weight also expressed in grams, the quotient will indicate the exact number of cubic centi- meters of the gas in question that will be required to give this weight or the gram-molecular weight (G.M.W.). In other words, the quotient expresses the volume in cubic centimeters which contains the gram-molecular weight. For example, 32 = G.M.W. of oxygen; 0.001429 gram is the weight of 1 c.c. of oxygen, therefore qo 22,390 (approximately), 0.001429 the number of cubic centimeters of oxygen, at and 760 mm., necessary to give its G.M.W. Again, the G.M.W. of hydrogen is 2.016 and the weight of 1 c.c. is 0.00008987 gram, therefore 2>016 22,400 (approximately), 0.00008987 the number of cubic centimeters of hydrogen at and 760 mm. necessary to give its G.M.W. This volume occupied by the G.M.W. of a substance is known as the gram-molecular volume (G.M.V.). Its calculation from the density of a gas gives values which diverge slightly from the average obtained for the more nearly perfect gases (i.e., 22,400 c.c.) according as the degree of packing of the molecules in each diverges from that in the latter. We deduce, therefore, the simple statement that the weight of 22,400 c.c. of a gaseous sub- stance, at and 760 mm., will give, when expressed in grams, the gram-molecular weight of that substance. From the weight of any given volume of a gas the weight of 22,400 c.c. may be calculated by simple proportion; AVOGADKO'S HYPOTHESIS 41 hence a ready means is given for determining molecular weights in general. Example 15. What is the molecular weight of a gas, 5600 c.c. of which at standard conditions weigh 5 grams? The value 22,400 is a standard, and the volume here to be compared with it (5600) must form the same ratio thereto as the relative weights of the two volumes : 5600:22,400 = 5 : *, or Jgl . |. From which the molecular weight of this gas (x) is found to be 20. With this definite relation established between the volume 22,400 c.c., calculated at and 760 mm., and that weight in grams which stands for the molecular weight of an element or compound, we may derive the fractional volume of this G.M.V. which any corresponding fractional part of the G.M.W. of a substance may occupy, and, vice versa, any fractional weight of the G.M.W. which any fractional part of this G.M.V. of the substance in state of vapor may have. In other words, the same fractional parts of the G.M.V. and the G.M.W. are directly pro- portional to each other. Example 16. 16 grams of oxygen will occupy what vol- ume at and 760 mm. pressure? Since 32 grams occupy 22,400 c.c. at these conditions, the volume occupied by 16 grams will be 16/32 of 22,400 or 11,200 c.c. , or, by simple proportion, 32 : 16 = 22,400 : x, or x = 11,200 c.c. To apply the effect of temperature and pressure changes let it be desired to find the volume which 2 grams of nitrogen will occupy at 20 and 750 mm. The molecular weight of nitrogen is 28. 22,400 : x = 28: 2, or x = 1600 c.c. This volume corrected for temperature and pressure be- 42 CHEMICAL CALCULATIONS comes increased at 20 by 293/273 of itself, and, at 750 mm., by 760/750 of itself, and hence the corrected volume will be 1600 (|fx), or 1739.2 c.c. Vice versa, the weight of any volume of a substance in the state of vapor may be determined if the gram-molecu- lar weight of the compound is known. Example 17. What is the weight of 5600 c.c. of hydrogen chloride measured at and 760 mm.? The molecular weight of hydrogen chloride is 35.46 + 1.01, or 36.47. Therefore 22,400 : 5600 = 36.47 : x, or x = 9.118 grams. Calculation of Densities of Gases from the Molecular Weights. From these relations the determination of the density of any volume of vapor is reduced to the simple task of ascertaining from the molecular weight of the substance and this volume 22,400 c.c., what the weight of a unit volume (1 liter for gases) will be at standard con- ditions. The determination of the relative density of one gaseous substance upon any other as a standard resolves itself, therefore, into a comparison of the gram-molecular weights of the two gases, i.e., the comparison of the weights of these equal volumes, the G.M.V. With air, the weight of 22,400 c.c. has been given as 28.955 grams (this may be construed as the hypothetical gram-molecular weight of air). Example 18. What is the relative density of chlorine re- ferred to (a) air? to (6) oxygen? (a) The gram-molecular weight of chlorine is 70.92, the weight of 22,400 c.c. The weight of 22,400 c.c. of air is 28.955 grams. Therefore the ratio 70.92/28.955 when made equal to the ratio x/l (where the weight of an equal volume AVOGADRO'S HYPOTHESIS 43 of air is unity) gives us the proportion : 70.92: 28. 955 = x : 1, or the relative density of chlorine as 2.449 (air =1). (6) When referred to oxygen the relative density of chlorine is determined in an exactly similar manner. The G.M.W. of oxygen, 32, replaces the weight of 22,400 c.c. of air in the example above. The process is of course the reverse of that in which the molecular weight of a sub- stance is derived by multiplying its relative density (O = 1) by the value 32, the basis of molecular weights (O = 32) (cf. p. 38). PROBLEMS. 44. The relative density of carbon dioxide (H=l) is 21.83. What is the relative density of this gas upon the oxygen stand- ard? Ans. 1.373. 45. The relative density of chlorine (air = 1) is 2.449. What is its relative density upon the oxygen standard? Ans. 2.216. 46. The vapor density of phosphorus trichloride (air = l) is 4.745. What is its molecular weight? Ans. 137.38. Note. Though Ex. 14 may be followed it is much more simple to calculate the weight of 1 liter of the vapor, which is 4.745 times the weight of 1 liter of air (1.293 grams), and with this weight of 1 liter to apply Ex. 15. 47. What is the molecular weight of mercuric chloride, the vapor density of which is 9.354 referred to air? Ans. 270.84. 48. The vapor density of water is 0.622 (air = l). What is its molecular weight? Ans. 18.016. 49. What is the molecular weight of sulphur dioxide, the relative density of which is 2.002 (0 = 1)? Ans, 64.07. 50. What is the molecular weight of a gas, 2 liters of which at standard conditions weigh 12 grams? Ans. 134.4. 51. The weight of 3840 c.c. of a certain vapor, at standard conditions, is 24 grams. What is the molecular weight of the substance? Ans. 140. 44 CHEMICAL CALCULATIONS 52. 3180 c.c. of a gas measured at 24 and 750.2 mm. pressure weighed 6 grams. What is the molecular weight? Ans. 48. 53. 8019 c.c. of a gas measured over water at 20 and 742.4 mm. pressure weighed 14 grams when deprived of aque- ous vapor. Calculate the molecular weight. Ans. 44. 54. 5647 c.c. of a gas measured over water at 24 and 754.2 mm. pressure weighed 6.254 grams when deprived of aque- ous vapor. Calculate its molecular weight. Ans. 28.02. 55. What volume will 49.63 grams of chlorine (mol. wt. = 70.92) occupy at standard conditions? Ans. 15,680 c.c. 56. What is the volume occupied by 8.8 grams of carbon dioxide (moL wt. = 44) at 12 and 752 mm. pressure? Ans. 4726 c.c. 57. 10 grams of carbon dioxide (mol. wt. = 44) will occupy what volume when contained in a vessel over water at 20 and 742.4 mm. pressure? Ans. 5728 c.c. 58. 1 gram of oxygen will occupy what volume over water at 30 and 756.5 mm. pressure? Ans. 814.3 c.c. 59. 8 grams of hydrogen (mol. wt. = 2.016) are to be ad- mitted into a balloon at a temperature of 20 and a pressure of 740 mm. What must be the capacity of the balloon? Ans. 97,980 c.c. 60. Of what capacity is that vessel which contains 4 grams of oxygen (mol. wt. = 32) at the temperature of 18 and pres- sure of 748.4 mm.? Ans. 3031 c.c. 61. 10.08 grams of hydrogen (mol. wt. = 2.016) at and 760 mm. pressure are to be forced into a vessel of 11.2 liters capacity. Under what pressure will the hydrogen be at this same temperature? Ans. 10 atmospheres. 62. At what temperature will 8 grams of oxygen (mol. wt. = 32) under a pressure of 760 mm. occupy a volume of 11.2 liters at this same pressure? Ans. 273. 63. A vessel holding 30.8 grams of carbon dioxide (mol. wt. = 44) at 10 and 740 mm. pressure is to be brought to a temperature of 50 and a pressure of 750 mm. What weight and volume of carbon dioxide will be lost? Ans. 2105 c.c. gas at latter conditions; 3.449 grams. AVOGADRO'S HYPOTHESIS 45 64. What is the weight of 6.594 liters of oxygen (mol. wt. = 32) at 40 and 740 mm. pressure? Ans. 8 grams. 65. What is the absolute density of hydrogen sulphide (mol. wt. = 34.09). Ans. 1.52. 66. What is the relative density of hydrogen sulphide (mol. wt. = 34.09) referred to air and also to oxygen? Ans. 1.177 (air = 1). 1.065 (0 = 1). 67. What is the relative density of hydrogen chloride (mol. wt. = 36.47) referred to hydrogen, air, oxygen and chlorine? Ans. 18.05 (H = 1). 1.259 (air = 1). 1.139 (0 = 1). 0.514 (Cl = 1). 68. 8 grams of oxygen were mixed with 10.08 grams of hydrogen (mol. wt. = 2.016). Both gases were measured at the standard conditions. What was the relative density of this mixture (O = 1)? Ans. 0.1076. 69. A gas with the relative density 1.5 (O = l) was reduced from standard conditions to 20 and 740 mm. pressure so that it might have a volume measuring 6172.3 c.c. What is the weight of this volume of the gas? Ans. 12 grams. 70. A volume of gas, with the relative density 0.8757 (0 = 1), was found to measure 1560 c.c. when transferred to a vessel over water at 18 and 742.4 mm. What is the weight of the dry gas here concerned? Ans. 1.75 grams. CHAPTER VII. THE LAW OF DEFINITE PROPORTIONS. WHEN two or more elements unite in chemical combi- nation, a definite ratio is found to exist between the quan- tities of each element present and also between each of these several quantities and that of the molecular weight of the compound itself. In other words, upon the basis of 100 as the molecular weight, the amount by weight, here the percentage, of each element in the compound is constant and must bear a constant and unalterable ratio to the percentage by weight of any other element present. The molecular weight of a compound represents the sum of the atomic weights that go to form it. The atomic or unit weight, i.e., the symbol weight, is regarded as the smallest weight in which an element can be present in a molecular weight of any of its volatile compounds. A formula, therefore, is merely the expression in symbols which, under the existing conditions, denotes the number of atomic weights of the several elements in one molecu- lar weight; this is better called the molecular formula (cf. Chap. VIII). The multiples of these atomic weights which occur in the molecular weight of a compound are always expressed by integers which are usually small. The Percentage Composition of a Compound. The percentage composition of a compound may be deter- mined by analytical or synthetical means. Thus, for example, 10 grams of magnesium unite with 6.58 grams of oxygen (when heated in an atmosphere of oxygen), and the resulting magnesium oxide weighs 16.58 grams. 46 THE LAW OF DEFINITE PROPORTIONS 47 Of the 16.58 grams magnesium oxide, 10 grams are of course magnesium, or the fractional quantity 10/16.58 rep- resents the weightxof magnesium in magnesium oxide; the fraction 6. 5$/ 16.58 will represent, then, the propor- tional amount of oxygen present in the compound. These same definite ratios hold for all quantities of magnesium oxide; hence, if 100 grams are taken as a basis, 10/16.58 of 100, or 60.3, will give the amount of magnesium in the 100 grams, i.e., the percentage of this element in the compound. Similarly, 6.58/16.58 of 100, or 39.7, is the percentage of oxygen present. By reference to simple proportion upon the basis of 100 as the total weight, these same percentage amounts are readily determined: MgO : Mg = 100 : x 16.58 : 10 = 100 : x, or 60.3 per cent MgO : O = 100 : x 16.58 : 6.58 = 100 : x, or 39.7 per cent. Now the formula of magnesium oxide is MgO, in which one atomic weight of magnesium (24.32) is combined with one atomic weight of oxygen (16). From the Law of Definite Proportions we know the ratios Mg/MgO and O/MgO are constant and, when expressed in numerical values of the several atomic weights concerned, become respectively: Mg or 24.32 24.32 and Q 16 MgO 24.32 + 16 The constant ratio 24.32/40.32 stands for the amount of magnesium in magnesium oxide in all amounts of the latter, and upon a percentage basis will be represented by this same fractional part of 100, i.e., 24.32/40.32 of 100, or 60.3 per cent. Similarly, for the amount of oxygen in 48 CHEMICAL CALCULATIONS magnesium oxide when expressed on a percentage basis, 16/40.32 of 100, or 39.7 per cent, will be obtained. These values are identical (cf. preceding paragraph) with those obtained in the synthesis of this compound where a defi- nite weight of magnesium was converted into the oxide. In accordance with the Law of Definite .Proportions, the same relations must hold for all determinations in this compound. The general expression for problems of this type when the relative molecular weights are referred through simple proportion to a basis of 100 is as follows: MgO : Mg = 100 : x 40.32 : 24.32 = 100 : (60.3) and MgO : O = 100 : x 40.32 : 16 = 100 : (39.7). In the simple compound water, with the formula H 2 0, the quantity of hydrogen in one molecular weight is rep- resented by two atomic weights of this element. This aggregate of atomic weights, which represents the quan- tity in which any element, or group of elements, is present in a molecular weight of some compound, may be regarded as the formula-quantity for the particular element, or ele- ments, concerned in the total molecular weight. Thus 2 H stands for the formula-quantity of hydrogen (2) in water (18), and the ratio 2 H/H 2 O or 2/ 18 or 1 /9 represents this constant proportion between the formula-quantity of hydrogen and the molecular weight of water when con- cerned, of course, in this compound.* As there is but one * The molecular weight of a substance is, after all, a formula- quantity for that particular group of elements when associated in this compound. The ratio between the two is always expressed by unity. Thus 18 is the formula-quantity of water in its molecular weight as steam, whereas in the form of ice the formula-quantity, just as the molecular weight, must be increased to 3 (?) times this number if the molecule is to be regarded as (H 2 0) 3 . THE LAW OF DEFINITE PROPORTIONS 49 atomic weight of oxygen in the molecular weight of water, the second ratio is simply O/H 2 O or 16/18 or 8/9. Hence in water 1 part by weight of hydrogen (1/9) is combined with 8 parts by weight of oxygen (8/9). Ex- pressed in percentage composition we have: H 2 18.02 2H 2.02 100 : x 100 : 11.2 per cent hydrogen and H 9 : = 100 : 88.8 per cent oxygen, the same ratio, 1 : 8, holding true, of course, for the percentages. In the compound sodium sulphate, Na 2 S0 4 , we may express the constant ratios: 2Na/Na 2 SO 4 , S/Na 2 SO 4 , and 4O/Na 2 SO 4 , by percentages of the molecular weight, 142.07, or (46.0 + 32.07 + 64), as follows: Na 2 S0 4 142.07 2Na = 46.0 = 100 100 Na 2 S0 4 142.07 S 32.07 = 100 100 Na 2 S0 4 142.07 4O = 64 = 100 100 1UU X 100 32.38 per cent 100 X 100 22.57 per cent 100 X 100 45.05 per cent The sum of these = 100 per cent. The Relation of Formula-Quantity to Molecular Weight. Whatever the weight of a substance under considera- tion may be, the ratio between any one formula-quantity and the total molecular weight of the compound is always equal to the ratio between the actual weights which here correspond to these respective quantities. Example 19. Calculate the weight of chlorine present in 50 grams of sodium chloride. 50 CHEMICAL CALCULATIONS The formula of sodium chloride, Nad, indicates that 1 atomic weight of sodium (23.00) is combined with 1 atomic weight of chlorine (35.46) to give a molecular weight of 58.46. The ratio Cl/NaCl or 35.46/58.46 expresses the relation between the formula-quantity of chlorine and the molecular weight of the salt, and all weights of these substances when jointly involved in the compound salt must be in accord with this ratio. When this ratio is converted to a percentage basis we obtain the value 60.6 as the percentage of chlorine in all amounts of salt: 58.46 : 35.46 = 100 : 60.6. If, then, 60.6 per cent of any weight of salt is chlorine, 50 grams X 60.6 per cent or 30.3 grams is the amount of chlorine present in the known weight of salt. Without resort to the percentage composition of a com- pound the constant ratio between the formula-quantity of any element present and the total molecular weight may be used directly for the determination of the amount of this element in any weight of compound. In the example just given, 35.46/58.46 represents the propor- tional amount of chlorine in salt. Without reference, then, to 100 as a basis, this same proportional part of the weight of salt taken must give the required value foi chlorine; thus 50 grams X 35.46/58.46 = 30.3 grams or weight of chlorine present; or, as is more customary, we may express the same by simple proportion: Since 58.46 grams of salt contain 35.46 grams of chlorine, 50 grams of salt can contain only that proportional part of its own weight which 35.46 is of 58.46; the ratio 58.46 : 35.46 at once becomes equal to the ratio 50 : x; thus 58.46 : 35.46 = 50 : x, or x = 30.3 grams. These proportions are most easily formed when we bear in mind that only one set of related terms, e.g., the THE LAW OF DEFINITE PROPORTIONS 51 formula-quantities, constitute one member of the propor- tion, whereas the other set of related terms, the actual weights (here with one unknown factor), constitute the other member. Example 20. Calculate the weight of sodium chloride that can be prepared from 30.3 grams of chlorine. The amount of salt that can be prepared from this definite weight of chlorine (30.3 grams) will be represented by the same ratio, Cl/NaCl, as under the preceding example (Ex. 19); but, for the purposes of multiplication into known values, it is necessary to use the inverted form NaCl/Cl or 58.46/35.46, with the large value in the numerator so that a proportional increase in the values affected by this ratio may be obtained. The same end may be reached by the use of simple proportion, where the unknown term increases over the known in direct ratio to the corresponding formula-quantities: 35.46 : 58.46 = 30.3 : x, or x >= 50 grams. The Relation of Formula-Quantities to Each Other. The ratio between any two formula-quantities occurring in one molecular weight may be used, of course, inde- pendently of the molecular weight, and when the ratio between certain groups of elements in a formula is to be determined the same principle will be found to apply as between single elements. The symbols representing these groups serve as usual for the calculation of the correct formula-quantities. In analyses of minerals the determination of these formula-quantities for various groups of oxides, etc., aids materially in the general classi- fication; thus in the mineral feldspar, A1K (Si 3 O 8 ), or Al 2 K 2 (Si 3 O 8 ) 2 , when intended for resolution into the con- stituent groups, becomes A1 2 3 . K 2 O . 6 Si0 2 , from which 52 CHEMICAL CALCULATIONS the percentage of any one of the oxides present may be calculated from the corresponding formula-quantity. Example 21 . What weight of aluminium is present in a sample of pure cryolite, A1 2 F 6 . 6 NaF, which analyzed for 10 grams of sodium? The ratio between the formula-quantity of each of the two substances concerned is 2 Al/6 Na, or 2 (27.1)/6 (23), or 54.2/138. As this is a constant we need only refer the ratio directly to the equivalent ratio when based upon sodium as 10 grams, 54.2/138 = x/ 10, or, as has been the practice, 138 : 54.2 = 10 : x, from which we ob- tain the value 3.93 grams for x, the weight of aluminium present. The Relation of Formula-Quantities to Percentage Com- position. When data are given in percentages and it is desired to ascertain the percentage amount of some other element or group of elements present in the same com- pound, the calculations are made directly upon these per- centages. They occupy, of course, the identical relation to each other as do the actual weights themselves, but with the condition that a constant value of 100 serves as a basis for the entire molecular weight. Example 22. What percentage of potassium oxide, K 2 0, is present in a sample of feldspar, AlK(Si 3 O 8 ), which analyzed for 12.2 per cent potassium? The ratio between the potassium and its corresponding amount of potassium oxide possible of existence in this same molecule is expressed by 2 K/K 2 O. With this defi- nite quantity of oxygen in the molecule now to be asso- ciated with the potassium, there naturally will be an increase in the percentage amount of the group K 2 over that of the potassium alone. The ratio between these percentage amounts must be in accordance with the corre- THE LAW OF DEFINITE PROPORTIONS 53 spending formula-quantities, 2 K and K 2 0, upon which their weight and consequent percentage is dependent; therefore the ratio 2 K/K 2 O must be equal to the ratio 12.2% /x%. From the equation 2 K/K 2 or 78.2/94.2 = 12.2 / x we obtain, as the value of x, 14.7 per cent, i.e., 14.7 per cent of a compound of the composition noted, and analyzing for 12.2 per cent potassium, may be con- sidered as potassium oxide. Percentage Composition as a Basis for Percentage Purity. Upon this method of comparison it is a very simple task to calculate the percentage purity or per- centage amount of some compound in one of its samples submitted to an analysis; and this too upon the data secured in the determination of only one of its constituent elements. Example 23. A sample of salt analyzed for 50 per cent chlorine. Assuming all of the chlorine to have been present as sodium chloride, what was the percentage of salt in the sample? The ratio Cl/NaCl must be equal to the ratio between the corresponding percentage amounts of these quantities, 50%/z%. Cl/NaCl = 35.46 : 58.46 = 50/z or, by simple proportion, 35.46 : 58.46 = 50 : x. From this, x is found to be 82.4 in place of the theoretical 100. This is, there- fore, the percentage purity of the sample. When analyti- cal data are not given in percentages, the weight of both sample and final product must be known in order to deter- mine the percentage amount of any constituent present. The percentage purity of a compound may be deter- mined also by calculating the weight of substance theo- retically possible from the results at hand (cf. Ex. 20) and then determining what percentage the weight of sample bears to this calculated value. The Interrelationship of two Formula-Quantities through a Common Quantity. When the results of an analysis 54 CHEMICAL CALCULATIONS are given in terms of some compound other than that involved in the investigation it is necessary to calculate these results through the element common to both com- pounds. Example 24. A sample of feldspar, KAlSi 3 8 , weighing 0.2507 gram gave upon analysis 0.0658 gram of potassium sulphate, K 2 S0 4 . What was the percentage of potassium oxide present in the sample? We have here two distinct ratios: K 2 S0 4 /2 K, which answers for the definite quantity of potassium present in this weight of potassium sulphate, and also 2 K/K 2 for the quantity of oxide possible from a definite weight of potassium (cf. Ex. 22). Since the ratios involve the same definite quantities of potassium throughout, they may be considered as equivalent in respect to this ele- ment, and when brought together may be cleared of this common term, 2 K, as indicated when K 2 S0 4 /2 K is involved with 2 K/K 2 0. By multiplication we should have K 2 S0 4 /K 2 0, or that ratio with which the corre- sponding ratio between the actual weights of these substances, involving the same weight of potassium, must accord. From the known weight of potassium sul- phate, the actual weight of potassium oxide is easily calculated: K 2 S0 4 /K 2 O or 174.27/94.27 = 0.0658/z, i.e., 174.27 : 94.27 = 0.0658 : x, or x, the weight of potassium oxide in the sample, is equal to 0.0356 gram. From this weight of oxide and the original weight of the sample the percentage amount of potassium oxide in the compound may be calculated: 0.2507 : 0.0356 = 100 : x, orx = 14.2 per cent. Without the simplification noted it would have been necessary first to determine the actual weight of potassium present in the potassium sulphate, and second, from the weight of potassium, the consequent weight of oxide. THE LAW OF DEFINITE PROPORTIONS 55 This cancellation or elimination of a common term between two ratios may be illustrated by a second example somewhat more complicated. Example 25, What percentage of potassium chloride can be regarded as present in a sample of potassium chlorplatin- ate, K 2 PtCl 6 , which analyzed for 43.6 per cent chlorine? The ratio of the formula-quantity of chlorine to that of potassium is 6 Cl/2 K. Now the ratio between potassium and potassium chloride in one molecular weight of potas- sium chloride is K/KC1. By examination of the formula of potassium chlorplatinate it is seen that two molecules of potassium chloride are involved in each molecule of chlorplatinate. This latter ratio then becomes 2 K/2 KC1 when considered in the molecule of chlorplatinate. By this manner the same formula-quantity of potassium is considered in both ratios (6 Cl/2 K and 2 K/2 KC1); consequently this equivalent quantity may be eliminated from each (cf. Ex. 24), and the final ratio 6 Cl/2 KC1 ob- tained for the corresponding weights of chlorine and potas- sium chloride possible in a molecule of chlorplatinate. Though a fractional quantity of the total chlorine is brought into consideration as potassium chloride, this in no way disturbs the definite relation between the total chlorine and the total potassium even when considered as chloride. The ratio may be simplified, if desired, to 3 C1/KC1, and is equal to the ratio between the cor- responding percentage amounts of these substances, 43.6% \x%. From the equation 3 C1/KC1 = 106.38 : 74.56 = 43.6/z we obtain 30.56 as the percentage amount of potassium chloride possible in the sample. 56 CHEMICAL CALCULATIONS PROBLEMS. 71. Determine the percentage composition of calcium car- bonate, CaC0 3 . Am. Ca, 40.05% ; C, 11.99% ; O, 47.96%. 72. Determine the percentage composition of sodium nitrite, NaN0 2 . Ans, Na, 33.33%; N, 20.30%; 0, 46.37%. 73. Determine the percentage composition of potassium chlorate, KC1O 3 . Ans. K, 31.90%; Cl, 28.93%; O, 39.17%. 74. Determine the percentage composition of crystallized sodium sulphate, Na 2 SO 4 . 10 H 2 O. Ans. Na, 14.28%; S, 9.95%; 0, 69.51%; H, 6.26%. 75. Determine the percentage composition of sulphuric acid, H 2 SO 4 . Ans. H, 2.05%; 8,32.70%; O, 65.25%. 76. Determine the percentage of sulphur trioxide, S0 3 , in sulphuric acid. Ans. 81.64 per cent. 77* Determine the percentage of water in crystallized sodium carbonate, Na 3 C0 3 . 10 H 2 0. Ans. 62.95 per cent. 78. Determine the percentage composition of ordinary alum, K,Al t (S0 4 ) 4 .24H.O. Ans. K, 8.24%; Al, 5.71%; S, 13.52%; 0, 67.43%; H, 5.10%. 79. What weight of sodium is present in 50 grams of sodium nitrate, NaNO s ? Ans. 13.53 grams. 80. What weight of sodium is present in 100 grams of sodium hydrogen carbonate, NaHCO 3 ? Ans. 27.38 grams. 81. What weight of oxygen is present in 200 grams of potas- sium chlorate, KC10 3 ? Ans. 78.34 grams. 82. What weight of oxygen is present in 200 grams of mer- curic oxide, HgO? Ans. 14.815 grams. 83. What weight of mercuric oxide, HgO, will contain 30 grams of oxygen? Ans. 405 grams. 84. What weight of potassium chlorate will contain 30 grams of oxygen? Ans. 76.6 grams. 85. What weight of sulphuric acid can be prepared from 100 grams of sulphur? Ans. 305.9 grams. 86. Calculate the weight of potassium in a sample of pure sylvite, KC1, which analyzed for 2.230 grams of chlorine. Ans. 2.459 grams. THE LAW OF DEFINITE PEOPOBTIONS 57 87. What weight of copper is present in a sample of pure copper sulphate, CuS0 4 . 5 H 2 0, which analyzed for 30.2 grams of sulphur trioxide, S0 3 ? Ans. 23.97 grams. 88. Calculate the percentage purity of a sample of horn- silver, AgCl, which analyzed for 74.2 per cent silver. Ans, 98.61 per cent. 89. Calculate the percentage purity of a sample of marble, CaC0 3 , which analyzed for 39.6 per cent calcium. Ans. 98.88 per cent. 90. Calculate the percentage of potassium chloride in a sample of carnallite, KC1 . MgCl 2 . 6 H 2 0, which analyzed for 37.72 per cent chlorine. Ans. 26.44 per cent. Note. Theory calls for 26.83 per cent KC1. Therefore the devia- tion, 0.39, between the percentage values means an error of 1.5 per cent from the theory (26.83). Halogen determinations are usually quite accurate. If the analysis is at all good, the salt falls a little short of purity. 91. Calculate the percentage of calcium oxide, CaO, present in a sample of marble, CaC0 3 , which analyzed for 43.8 per cent carbon dioxide. Ans. 55.84 per cent. Sample practically pure; theory, 56.04 per cent. 92. A sample of sodium chromate, Na 2 CrO 4 , weighing 1.6780 grams, gave upon analysis 1.4620 grams of sodium sulphate, Na 2 SO 4 . What was the percentage of sodium oxide, Na 2 0, in the sample? Ans. 38.01 per cent. Theory, 38.25 per cent. 93. Calculate the percentage purity of a quantity of potas- sium ferrocyanide, K 4 Fe(CN) 6 , 0.5793 gram of which gave upon analysis 0.4650 gram of potassium sulphate, K 2 SO 4 . Ans. 8^.8 per cent. Note. From this weight of K 2 SO< determine the actual weight, and finally the percentage, of potassium in the sample and then apply Ex. 23. 94. What is the percentage of potassium sulphate, KzS0 4 , in a sample of common alum, K 2 SO 4 . A1 2 (S0 4 ) 3 . 24 H 2 0, which analyzed for 33.51 per cent sulphur trioxide, S0 3 ? Ans. 18.23 per cent K 2 S0 4 . Sample practically pure ; theory, 18.36 per cent K 3 S0 4 . 58 CHEMICAL CALCULATIONS 95. What is the percentage of copper carbonate, CuG0 3 , in a sample of malachite, Cu(OH) 2 . CuC0 3 , which analyzed for 57.1 per cent copper? Ans. 55.49 per cent CuCO 3 . The salt is probably pure ; theory, 55.87 per cent CuC0 3 . 96. A sample of carnallite, KC1 . MgCl 2 . 6 H 2 0, analyzed for 35.34 per cent chlorine. What is the percentage of magnesium chloride present? Ans. 31.64 per cent MgCl 2 . Note. Theory: 34.27 per cent MgCl 2 . Therefore 92.31 per cent pure. 97. A sample of crystallized ferrous-ammonium sulphate, FeSO 4 . (NH 4 )2S0 4 . 6H 2 O, analyzed for 8.66 per cent ammonia, NH 3 . What is the percentage of ferrous sulphate, FeS0 4 , present? Ans. 38.60 per cent FeSO 4 . Theory, 38.74 per cent FeS0 4 . 98. A sample of gahnite, Zn(A10 2 ) 2 or ZnO . A1 2 O 3 , weighing 0.1909 gram, gave on analysis 0.1664 gram of zinc sulphate, ZnSO 4 . What was the percentage of alumina, A1 2 S , in the sample? Ans. 55.18 per cent. If analysis is correct, the sample is slightly impure. Theory, 55.67 per cent. CHAPTER VIII. THE DERIVATION OF CHEMICAL FORMULA. Molecular Formulae from Molecular Weight and Per- centage Composition. The formula of a compound, designated by symbols which stand individually for one atomic weight of an element, is derived from data afforded by analysis or synthesis. In reverse to the determination of the percentage composition, as described in the preced- ing chapter, we can derive the true molecular formula when given the correct molecular weight of a compound and the percentage amount of each element present. The proportional parts of the molecular weight indicated by the several percentages are of course the quantities (formula-quantities) of the respective elements in one molecular weight of the compound, and necessarily must be multiples of the corresponding atomic weights by unity or a small integer. Example 26. The molecular weight of water is 18.016. The' amount of hydrogen present is 11.2 per cent and that of oxygen 88.8 per cent. What is the molecular formula? Proportional Unit or Molecular weight. Percentage indicated. part of molec- ular weight corresponding atomic weight of element. Number of units. to element. 18.016 H= 11.2% 2.016 1.008 2 18.016 O= 88.8% 16.00 16.00 1 59 60 CHEMICAL CALCULATIONS The values 2.016 and 16 represent respectively the sum of the atomic weights of hydrogen and oxygen which are present in the molecular weight of water. They are, therefore, the formula-quantities of these elements in this compound. With 1.008 and 16 as the atomic weights of hydrogen and oxygen respectively, it is only a simple step to determine the number of hydrogen units (2) and oxygen units (1) which are necessary to make up the formula-quantity of each element in this molecule and consequently, together, the molecular weight of water; hence the formula H 2 0. The correct molecular weight of a compound is rarely ever at hand for the determination of chemical formulae. The usual procedure lies, then, in the determination of the formulae from the percentage composition or other data, and in the final adjustment of these to accord with the molecular weights which may have been determined only with approximate exactness. Empirical Formulae from Percentage Composition. Upon the percentage basis the molecular weight is re- garded as brought over to the value 100; this number, however, gives in itself no clue to the probable molecular weight. In Example 26 we were given the correct molec- ular weight; consequently, the proportional parts of this total weight, as indicated by the percentage amounts, agree always with the exact quantity of each element present in one molecular weight of the compound. Now, upon the adoption of 100 as the molecular weight, each percentage amount becomes the accepted value for the formula-quantity of that particular element in the molec- ular weight. In dividing each of these values through by the atomic weight of the corresponding element, we obtain not whole numbers, indicating the number of atomic weights of the respective elements present, but fractions DERIVATION OF CHEMICAL FORMULAE 61 or factors of these whole numbers, all of which are related to the true numbers in the same ratio as the adopted value, 100, is related to the true molecular weight. Since a chemical formula calls for simple multiples of the atomic weights concerned, we need only raise the entire range of factors by some term which will bring each and all into whole numbers. When the smallest factor is made the divisor, then all of the other factors divided through by it must necessarily give quotients which are equal to or greater than this smallest factor as unity. In bringing the correct molecular weight over to 100 as a basis, this smallest factor was of course reduced from unity, or a multiple of it, in the same proportion as were all of the other factors. The quotients, then, upon the basis of this small factor as unity, will possess values close to their former and correct numbers. If the smallest factor had been reduced from unity itself, then the quo- tients will represent the correct formula, a molecular formula, of the compound. If, however, the original value of this smallest factor was a simple multiple of unity (e.g., 3), then the quotients will vary from their true values in a molecular formula by just this same fractional amount that unity is of the simple multiple. (If 3 were the mul- tiple, then the quotients would stand at one-third of their original values.) This simplest expression of a formula in symbols is known as an empirical formula, and the sum of the atomic weights therein represented, though a formula weight, is not necessarily the molecular weight. From the empirical formula the true percentage com- position of a compound is always derivable. Its formula weight, however, may not be coincident with the molecular weight, and, if so, the correct molecular formula can be derived only when the molecular weight, or at least a fair 62 CHEMICAL CALCULATIONS approach to this, is known. The molecular formula, there- fore, is always a multiple of the empirical formula by some small integer. Example 27. A substance by analysis was found to contain 32.32 per cent sodium, 22.44 per cent sulphur, and 45.24 per cent oxygen. What is the formula of the compound? The method of solution will be made most apparent when we set down the percentages and refer these to the respective atomic weights of the elements present, as indicated below: Percentage Substance. Percentage. referred to corresponding Factor. Factor to unit value. at. wt. as basis Sodium. . 32.32 32.32/23.00 1.406 ] Divide (2.01 Sulphur. . 22.44 22.44/32.07 0.699 1 through \ 1.00 Oxygen. . 45.24 45.24/16 2.827 J by 0.699 I 4.04 100.00 These unit values approach very closely to the integers 2, 1 and 4, and designate the number of unit or atomic weights of sodium, sulphur and oxygen, respectively, in their corresponding formula-quantities in a formula weight of the compound, Na 2 S0 4 . In Chapter VII we have noted the means for calculating the theoretical percentage composition of a compound. In the case of a derived formula it is always well to calcu- late from it the percentage amount of each element present and note whether or not these values check with those derived by analysis. The derived formula Na 2 S0 4 presents the following percentage composition: DERIVATION OF CHEMICAL FORMULAE 63 Na 2 S0 4 142.07 Na 2 SO 4 142.07 Na 2 S0 4 142.07 2Na 46 4O > = 100 : x, or 32.38 per cent sodium. : x, or 22.57 per cent sulphur. = 100 : x, or 45.05 per cent oxygen. 100 per cent. The percentage amounts actually found by analysis approach very closely to these theoretical values based upon the formula Na 2 SO 4 ; consequently we may conclude that this formula represents the constitution of the com- pound. The molecular weight of sodium sulphate is found to be close to 142; hence the formula Na 2 S0 4 (with the formula' weight 2 (23) + 32.07 + 4 (16) or 142.07) satisfies also the requirements for a molecular formula. Accuracy in analytical data is dependent upon purity of material and method of operation. In all of our results we may expect a certain degree of variation from the theoretical values, but the limits of error in the deter- mination of chemical formula? should not greatly exceed two-tenths of 1 per cent. Much depends, however, upon the particular element considered. For example, in the case of hydrogen the error often may be as much as four- tenths of 1 per cent above the theoretical, due to insuffi- cient removal of moisture from the sample or to a faulty combustion. In the case of oxygen the data are rarely ever determined directly, but by difference; thus in Example 27 the sum of the percentage amounts for sodium and sulphur was subtracted from the total 100 per cent, and the difference considered as the value for oxygen, a value altogether dependent upon the accuracy 64 CHEMICAL CALCULATIONS of the other data. From these considerations the chemist finds it greatly expedient, in problems of this nature, to consider each analysis separately and to determine what one is likely to be most free of^error. The factor, there- fore, which corresponds to the percentage amount of this one element is the best to select as a basis for the reduction of all the other factors to unit values. Let us suppose, in Example 27, that the value for sodium was known to be more accurately determined. The factor here is 1.406, and, if we base all the other data upon this one, i.e., by dividing each factor by 1 .406, we shall obtain the values, 1, 0.498 and 2.010, for sodium, sulphur and oxygen respec- tively. These may be brought to unit values by multi- plication by a small integer, here by 2, and we obtain finally 2, 0.996 and 4.02, values v/hich approach the respective integers 2, 1 and 4 somewhat more closely than in the previous calculation. This, of course, is due to the close agreement between the theoretical and found per- centage values for sodium. It is now seen why the factor for oxygen is rarely selected as a basis for formula deter- minations. Example 28. Determine the formula for that substance which presented, by analysis, the following percentage com- position: carbon, 39.78 per cent, hydrogen, 6.97 per cent and oxygen, 53.25 per cent. Above 230 the vapor of this sub- stance gave a constant relative density, 2.09 (air = 1), calcu- lated to standard conditions. From the relative density it is only a simple step to calculate the molecular weight (cf. Ex. 14) : 1 : 2.09 = 28.955 : x, where x, the molecular weight, is found to be 60.52. The following table of percentages and corresponding atomic weights is easily constructed as under Example 27: DERIVATION OF CHEMICAL FORMULAE 65 Substance. Percentage. Percentage referred to corresponding at. wt.as basis. Factor. Unit value. Carbon Hydrogen . Oxygen 39.78 6.97 53.25 39.78/12 6.97/1.01 53.25/16 3.315 ) Divide f 1.00 6.90 [ through {2.08 3.33 J by 3.315 I 1.01 These unit values approach well the integers 1, 2 and 1, which represent now the number of atomic weights of these respective elements in a formula weight of the compound, CH 2 O. The sum of the formula-quantities thus repre- sented in the formula weight is 30.02 (i.e., 12 + 2.02 + 16), a value which does not coincide with the molecular weight calculated from the vapor-density determination above (60.52). The molecular formula as previously stated is always a multiple of this simplest or empirical formula, and by just that integer which brings the empirical formula weight up to the molecular weight or an approximation to the same. The integer in the present case is 60.52/30.02 or 2.02. This is well within the limits of error, which in some cases may exceed a variation of 0.2 from an integral value. The empirical formula CH 2 must be multiplied now by the integer 2, when we shall obtain C 2 H 4 2 , the correct molecular formula, or that formula in which the sum of the atomic weights involved gives the molecular weight. The calculation of the theoretical percentage composition for this substance, acetic acid, gives the following: carbon, 39.97 per cent, hydrogen, 6.73 per cent and oxygen, 53.50 per cent. The analytical data, there- fore, are closely in accord with the theoretical. Formulae from Percentage Composition involving Radi- cals. As a more complex example it will be well to 66 CHEMICAL CALCULATIONS examine the analytical data upon some mineral on which the results are given in percentage amounts of the several groups present. Feldspar belongs to this class of sub- stances, and in its analyses certain values will be found for silicon dioxide (Si0 2 ), aluminium oxide (A1 2 O 3 ) and potassium oxide (K 2 O); but associated with these there may occur other oxides which are isormorphous with one or more of the above. Consequently the factor for these isomorphous substances, found, of course, as heretofore described, by the relation of percentage composition to atomic weight or groups of atomic weights representing the substance, must be reckoned together as one factor and hence as one group, in order to determine the extent to which this particular group of isomorphous substances may be present in the complete formula. If, perchance, a portion of the data were not given in terms of the proper isomorphous compound, but in some part of it, for instance, the percentage of potassium instead of potassium oxide, we should require only a single calculation to con- vert the data into the correct form (cf. Ex. 22) : 2 K : K 2 = per cent given : per cent sought. Example 29. A sample of the mineral orthoclase (a feld- spar) gave by analysis the following percentage composition: Si0 2 , 65.69 per cent, A1 2 3 , 17.97 per cent, (CaO, 1.34 per cent, K 2 0, 13.99 per cent, Na 2 0, 1.01 per cent, isomor- phous). What is its formula? By drawing up a table of these percentages and referring each to the formula weight of the respective groups, we may expect to obtain factors which in this case do not refer to the simplest relation between the elements, but rather between the groups of elements. The factors for K 2 O, Na 2 O and CaO, since these oxides are isomorphous and mutually may replace each other in DERIVATION OF CHEMICAL FORMULAE 67 the molecule, must be added together and considered as some generic formula such as R 2 O, when in the final adjustment of the number of each group present in the formula they will figure as an individual group of uniform composition. Percentage re- referred to Sub- stance. Per- centage. formula wt. of group Factor. Factor of isomor- phous groups. Unit value. as basis. SiO 2 . . 65.69 65.69/60.3 1.0894 1.0894 6.258 A1 2 3 - 17.79 17.79/102.2 0.1741 0.1741 Divide 1.000 CaO.. 1.34 1.34/56.09 0.0239] through- K 2 0.. 13.99 13.99/94.2 0.1485[ 0.1887 by 1.084 Na 2 O. 1.01 1.01/62.2 0.0163J 0.1741 As has been already noted in selecting the factor of those equally small, for division into the entire range of factors, it is always better to select that one which is likely to have been based upon more accurate analytical results, in this case, the aluminium oxide and not the composite oxide (into the determination of which three analyses must have entered). The resulting integers are here found to be 6, 1 and 1, and consequently the formula for the feldspar is (K 2 O . CaO . Na 2 0) (A1 2 3 ) 6 (SiO 2 ), in which the oxides of the metals K, Na and Ca make up one single group of the general formula of a feldspar, (K 2 0) . (A1 2 O 3 ) . 6 (SiO 2 ), or KAlSi 3 O 8 . The extent of these isomorphous replacements within any group may vary considerably. This empirical formula is accepted as the true formula in the absence of any conflicting statements in regard to the true molecular weight. Formulae from the Relation of Formula-Quantities to Each Other. Without recourse to percentage composi- 68 CHEMICAL CALCULATIONS tion, the actual weight of any element found by analysis to be present in a known weight of compound may be referred directly to the formula-quantity of this element in the total molecular weight of that compound. Since these analytical data are dependent upon the number of atomic weights of each element present in one molecular weight, the ratio between the weight of each element in a known weight of compound containing, for example, two constituents, must be equal to the ratio between a certain unknown number of atomic weights of the one element and that of the other, i.e., to the ratio between the cor- responding formula-quantities. Example 30. Determine the formula of an iron oxide pro- duced in the oxidation of 22.4 grams of iron to a final weight of 32 grams. Here 32 22.4 = 9.6 grams of oxygen taken up. The ratio between the weight of each of these two constituents in this sample of oxide, i.e., 22.4/9.6, must be equal to the ratio between the formula-quantity of each corresponding element in the molecular weight. Since the number of atomic weights of each element, grouped in its formula-quantity, is here unknown, only the algebraic expression Fe^Oj/ can be written for the complete formula; x and y representing, of course, these unknown integers. Given, then, the atomic weights of these two elements, iron (55.85) and oxygen (16), we may easily draw up the ratio between the formula-quantity of each in the compound. This ratio involves an unknown factor (an integer) in each term, but as a ratio it must be always equal to any other ratio that can be drawn up from data on the weights of these same particular sub- stances in a definite weight of the compound. Thus the ratio Fe x /O y , as determined in the analysis of the oxide given above, is represented by 22.4/9.6, and consequently, DERIVATION OF CHEMICAL FORMULAE 69 if the analyses are correct, these two ratios must be identical: Fe^ = 22.4 55.85 x = 22.4 O y '' '' 9.6 ' 16 y '' ~~ 9.6 " The exact numerical relation between x and y may be readily calculated by bringing the numerical values to one side of the equation (i.e., by multiplying the equation through by 16/55.85), when we obtain x_ 22.4 X 16 358.4 y " 9.6 X 55.85 " 536.4 ' a result likewise obtained through the simple proportion: 55.85 x : 16 y = 22.4 : 9.6 (55.85 X 9.6) x = (16 X 22.4) y x/y = 358.4/536.2. In one equation involving two unknown terms we can only expect to determine the simplest ratio between them. The ratio 358.4/536.4 when reduced by division of each term by the smaller (358.4), gives us 1/1.49, and this in turn, through multiplication of each term by 2, is brought very close to 2/3, which, as the ratio between the smallest integers, indicates the value of x and y respectively; hence the formula of the compound, Fe 2 3 . Whether the molecular formula calls for this simplest form or some multiple of it by a simple integer cannot be determined without knowledge of the molecular weight of the compound. The application of this method to compounds contain- ing three or more elements introduces, of course, this corresponding number of unknown terms (the number of atomic weights of each element present), and lends itself less readily to a simple solution. When, however, two or more groups of elements, as radicals, etc., are present, the method may be applied to the determination of the number of each of the two groups in the molecule. 70 CHEMICAL CALCULATIONS Example 31 . What is the formula of that oxy-halogen salt of potassium, (~KCl)yO X) 13.9 grams of which lost 6.4 grams of oxygen upon heating, and gave a residue of 7.5 grams (13.96.4) of potassium chloride, KC1? The ratio between the weight of oxygen driven off and the residue of potassium chloride left is 6.4/7.5, which, of course, is equal to the ratio between the formula-quantity of each: Ox = 6.4 or (16) s = 6.4 (KCl) y 7.5' (39.1 + 35.46) y 7.5* or x_ (6.4) (74.56) 476.2 y = (7.5) (16) 120 ' From this we obtain x/y = 3.97/1 or, by simplest integers, z = 4 and y= 1. The formula of the compound (KC\) y O x . now becomes (KC1) 1 O 4; i.e., KC1O 4 . Formulae Containing Water of Crystallization. Again we have a good illustration of examples of this kind in the consideration of those salts which contain water of crystallization, definite in amount under certain defi- nite and fixed conditions. Example 32. 7.15 grams of crystallized sodium carbonate, NaaCOg . x H 2 O, lost all of its water of crystallization when gently heated. The weight of the residue was 2.655 grams. What is the formula of the crystallized salt? Here 7.15 2.655 = 4.495 grams, or the weight of water in this weight of salt. The ratio between the weight of water actually associated with the weight of anhydrous salt (that portion of the crystallized salt left after the elimination of the water) is 4.495/2.655. This ratio there- fore must be equal to the ratio between the formula-quan- tity of water and the formula-quantity of the anhydrous salt in the molecular weight of the crystallized salt. In that the water of crystallization is usually determined DERIVATION OF CHEMICAL FORMULAE 71 with reference to one molecule of the anhydrous salt, one of the unknown terms in the ratio between the formula- quantities drops out (i.e., equals unity) and we have only the determination of the unknown number of molecular weights (x) of water present. Thus: x (H 2 0) = 4.495 Na 2 CO 3 2.655 ' By substituting the atomic weights indicated, we obtain x (18.02) 4.495 106 2.655' which, when solved for x, gives the value 10 as the number of molecules of water associated with one molecule of the anhydrous salt to form a molecule of crystallized sodium carbonate, Na 2 CO 3 . 10 H 2 O. The calculation may be conducted by simple proportion as follows: Na 2 CO 3 : x (H 2 0) = wt. anhydrous salt : wt. H 2 O. 106 : x (18.02) = 2.655 : 4.495. The expression x (18.02) is found equal to 179.5, hence the integral value for x will be 10 and accord with the formula above. Formulae from the Relation between Formula-Quan- tities and Their Corresponding Weights. This constancy in the ratios between the quantities of all the elements either singly or collectively in a chemical compound was comprehended, of course, in the Law of Definite Pro- portions; and the proportions in which the elements enter into chemical combination are seen, accordingly, to be functions of the corresponding atomic weights. In any known amount of substance there is always a definite ratio between the actual weight of some one element, or group of elements, present and the corresponding formula- 72 CHEMICAL CALCULATIONS quantity. Furthermore, this ratio is always identical with every other ratio between the weight of any other element in this same sample and its corresponding formula-quan- tity.* The ratio can be unity only when we are dealing with a gram-molecular weight of the substance. Thus, in 18 grams of water, the weight of oxygen (16 grams) bears the ratio of unity to the formula-quantity of this element in the molecular weight, and also the same ratio, unity, exists between the weight of hydrogen (2 grams) and its formula-quantity. As is often the case, no one correct formula-quantity is at hand. Under such circumstances it is only possible to refer any and all of the known weights of the elements in a given sample of compound directly to their corre- sponding atomic weights. In order to maintain through- out the same constant ratio between these values actual weights and atomic weights it is necessary to select the ratio between the actual weight of some one element in the sample and the atomic weight correspond- ing thereto as the standard ratio. By means of this ratio the actual weight of any other element present in the sample may be brought over immediately (simple pro- portion) to a value which holds the same relation to its true formula-quantity as the single atomic weight of the first element will be found to have toward its formula- * The equivalence in the ratios between any two formula-quantities, as x and y, and the corresponding weights, as w^ and w 2t which repre- sent them respectively in any given sample, is usually expressed as x /y = wjw v By multiplying through by y/w l we immediately obtain the expression x/w l = y/w 2 for this equivalence in the ratios between each formula-quantity and the weight representing it in some sample. The two expressions take the following forms through simple proportion: x : y = w l : w 2 and x : w l = y : w v each of which follows algebraically from the other, i.e., the ratio between the first and second terms of a proportion, when equal to the ratio between the third and fourth terms, signifies also this equivalence in the ratios between the first and third terms and the second and fourth terms. DERIVATION OF CHEMICAL FORMULA 73 quantity. In this latter case, with the formula-quantity always a multiple of the atomic weight by a small integer, the process of determining the true formula-quantity is comparatively a simple one. The integer selected, how- ever, must also bring each and all of the other values, similarly found, to multiples of their individual atomic weights. In other words, we determine the least common multiple for the entire range of values, such that each in order will be raised to a multiple, by unity or a small integer, of its atomic weight. Finally, the quotients of these so derived formula-quantities by the corresponding atomic weights will give the units of each element neces- sary in the simplest formula; one which can be raised later to any desired multiple, depending, of course, upon the molecular weight in question. Example 33. Derive the formula of iron oxide from the data in Example 30 : 22.4 grams of iron gave 32 grams of oxide. The weight of oxygen entering into the compound is 9.6 grams. Accordingly, the ratio 9.6/16, between weight of oxygen present in the sample and the atomic weight of this element, may be taken as the standard ratio. By means of this ratio the weight of iron present is now brought over to a value always equivalent to oxygen as 16 in this compound.* Thus, 9.6 : 16 = 22.4 :x, where * In the combinations between hydrogen and oxygen no com- pound has been found to contain more than two atomic weights of hydrogen with one of oxygen in a single molecular weight. Con- sequently the lowest weight of oxygen corresponding to the lowest weight of hydrogen (1.008, its atomic weight) will be one-half of 16, or 8. This weight of oxygen, the smallest equivalent to one atomic weight of hydrogen, is often used as a basis for the conver- sion of known weights or percentages of other elements (associated with oxygen) to the scale of atomic weights. Though it brings into the calculations numbers somewhat smaller (by one-half) than other- wise obtained through the use of the actual atomic weight, 16, the process of determining the least common multiple is in no way simplified. 74 CHEMICAL CALCULATIONS this value for iron is found to be 37.2. By referring these values now to the respective atomic weights we have: Relative Value Sub- stance. amount in known weight of Value con- structed upon oxy- Multiple raised to multiple of correspond- Atomic weight. Unit value. compound. gen as 16. ing at. wt. Iron. . . 22.4 37.2 111.6 55.85 2 Oxygen 9.6 16 48 16 3 From this the formula of the compound is found to be Fe 2 O 3 . Formulae from the Relation between Formula-Quantity and Percentage Composition. In extending this method to compounds which contain a number of elements, it serves equally well if we employ, in place of actual weights, the percentage amounts of the elements, or groups of ele- ments, present. These values, of course, are as definitely related to each other as any other fractions of the total weight of a substance. The ratio between the percentage of any one element and its atomic weight suffices, through simple proportion, to bring over each of the other per- centages to the corresponding values for these elements in the compound. Example 34- Derive the formula of sodium sulphate from the data given in Example 27. Sub- stance. Percent- age. Value con- structed upon oxy- Multiple Value raised to multiple of correspond- Atomic weight. Unit value. gen as 16. ing at. wt. Sodium . . . Sulphur.. . Oxygen. . . 32.32 22.44 45.24 11.43 7.94 16 4 45.72 31.76 64 23 32.07 16 1.99 0.99 4 100.00 DERIVATION OF CHEMICAL FORMULAE 75 The value of oxygen is made equal to the atomic weight of this element, and the ratio 45.06 : 16 is used to reduce all of the other values proportionately, e.g.: with sulphur 45.24 : 16 = 22.44 : x, or x = 7.94 and with sodium 45.24 : 16 = 32.32 : x, or x = 11.43. These final values represent correctly the relative amounts of each element. One, oxygen, was made to coincide with its corresponding atomic weight, hence each of the other values will represent a weight of that respective ele- ment in this compound equivalent to oxygen as 16; and, as this is the unit value for oxygen, the entire range of values can be raised only through multiplication by simple integers until the lowest possible formula-quantity for each is obtained. The integer 4, found by trial, here raises all to values that are multiples of the corresponding atomic weights. These formula-quantities when divided by the respective atomic weights give at once the number of atomic weights of each element in the molecule, hence the formula Na 2 S0 4 . The Derivation of Formulae of a Known Type from a Single Analytical Value. If it is known to what general class (oxides, chlorides, sulphates, etc.) a particular sub- stance of simple type (not involving replacements) belongs, its complete analysis is not necessary for the determination of the formula. For instance, the percentage of copper in a sample known to be a copper chloride will be suffi- cient to this end. This follows by reason of the fact that the percentage amount of any element found comes from the ratio between its formula-quantity in the molecule and the total molecular weight, a ratio represented, we shall say, by the expression, mol. wi./x (at. wt.). Upon analysis certain definite numerical values fall to this ratio j in percentages, for example, we have the 76 CHEMICAL CALCULATIONS ratio 100 /per cent element found. We thus form the equation: 100 _ mol. wt. % element found x (at. wt. element) Example 85. A sample of a chloride of copper gave upon analysis 47. 9 per cent copper. What is the formula of the com- pound? The ratio between the known and theoretical values now becomes 100/47.9 = mol. wt./z (63.57), or 100 : 47.9 = mol. wt. : x (63.57). As x is always a small integer it may be neglected for the first calculation: 100 : 47.9 = mol. wt. : 63.57. From this we obtain a value 132.71, as the molecular weight, in which one atomic weight of copper must be present to the extent of 47.9 per cent. The difference, therefore, or 69.14 (i.e., 132.71 - 63.57 = 69.14), must represent the atomic weights of chlorine associated with this one atomic weight of copper. The quotient of 69.14 by 35.46, the atomic weight of chlorine, gives 1.95, or practically 2, and we at once draw up the formula CuCl 2 (63.57 + 2 (35.46) = 134.49), with the molecular weight of which the an- alytical data are well in agreement; hence the correct formula. If there had been obtained some fractional quantity as a quotient (in place of 2 above), this quantity, together with the value for copper as unity, may be raised by multiplica- tion to integral values which indicate the probable formula. This, of course, is identical with multiplying the molecular weight found by x (the smallest integrals in their order) until a value is obtained that can be made up of atomic DERIVATION OF CHEMICAL FORMULA 77 weights without fractional parts. This try-out of a for- mula is nothing more than the reverse of the method outlined in Example 23, wherein was shown the means of determining the amount of any salt present in any one of its samples when given the percentage of some one element in the salt. In place of the molecular weight there given we have here only to set down, in order, the trial values for this molecular weight which correspond to the possible formula. Thus : ( Percentage amount ) ( i nn | _ ( Formula-quantity ) f ( Molecular weight ) ( of copper } ' \ ) ( copper ) ' j calculated. J 47.9 : x = 63.57 : When the value calculated for x, from any trial molecular weight, approaches closely to 100 we are assured of the correct value for this molecular weight and consequently of the correct formula. PROBLEMS. 99. What is the formula of that substance which gave; by analysis, 26.9 per cent sodium, 16.58 per cent nitrogen, 56.52 per cent oxygen? Ans. NaNO 3 . 100. What is the formula of that substance which gave, by analysis, 26.6 per cent potassium, 35.22 per cent chromium, 38.18 per cent oxygen? Ans. K 2 Cr 2 7 . 101. What is the formula of that substance which gave, by analysis, 24.65 per cent potassium, 34.85 per cent manganese, 40.5 per cent oxygen? Ans. KMnO 4 . 102. Derive the formula of that substance with the observed relative density 0.8853(0= 1) and, by analysis, the composition: 85.41 per cent carbon and 14.64 per cent hydrogen. Ans. C 2 H 4 . 103. Derive the formula of that substance with the observed density 1.189 and, by analysis, the composition: 92.1 per cent carbon, 7.85 per cent hydrogen. Ans. C 2 H 2 . 78 CHEMICAL CALCULATIONS 104. Derive the formula of that compound of hydrogen and oxygen which gave, by analysis, 5.93 per cent hydrogen. A determination of its molecular weight gave the value 31.8. Ans. H 2 O 2 . 105. Derive the formulae of that oxide of nitrogen which gave, by analysis, 30.4 per cent nitrogen. In the solid state it was found to have a molecular weight of 92.4, whereas the actual density of its vapor above 140 was only 2.013. ( Solid, N 2 4 . lGas,N0 2 . 106. Derive the formula of that acid which gave, by analysis, 26.5 per cent carbon, 2.2 per cent hydrogen and the rest oxygen. The molecular weight was determined approximately as 90.4. Ans. H 2 C 2 4 . 107. Derive the formula of the mineral chalcopyrite, a speci- men of which gave, by analysis, the following percentage compo- sition: 34.40 per cent copper, 30.47 per cent iron and 35.87 per cent sulphur. Ans. CuFeS 2 . 108. Derive the formula of the mineral dolomite which gave, by analysis, the following percentage composition : CaO 31.37 per cent ) . MgO 21.23 per cent j Amorphous CO 2 47.67 per cent 100.27 Ans. (CaO,MgO)(C0 2 ) or (Ca,Mg) C0 3 . 109. By analysis a specimen of melanterite gave the fol- lowing: FeO 20.37 per cent ) . MgO 4.60 per cent j Amorphous S0 3 29.80 per cent H 2 45.07 per cent 99.84 Derive the formula. Ans. (FeO,MgO)(S0 3 ).7H 2 or (Fe,Mg)S0 4 .7H 2 0. DERIVATION OF CHEMICAL FORMULAE 79 110. By analysis a specimen of xenotime gave the follow- ing: PA Y 2 3 Ce 2 3 Fe 2 3 32.45 per cent 54.13 per cent 11.03 per cent 2.06 per cent isomorphous 99.67 Derive the formula. Ans. (Y 2 3 ,Ce 2 O 3 ,Fe 2 3 ) - (P 2 O 5 ) or ([Y,Ce,Fe]A) (P 2 O 5 ) 111. By analysis a specimen of columbite gave the following: Nb 2 5 Ta 2 5 47.05 per cent 34.04 per cent isomorphous SnO 2 0.30 per cent ] FeO 11.15 per cent isomorphous MnO 7.80 per cent 100.34 Derive the formula. Ans. ([Fe,Mn,Sn]O) . ([Nb,Ta]A). By analysis a specimen of garnet gave the following: Si0 2 A1A Fe 2 3 FeO MnO MgO CaO H 2 39.09 per cent 23.05 per cent 0.53 per cent 0.11 per cent 0.35 per cent 1.01 per cent 35.75 per cent 0.15 per cent (neglected) isomorphous - isomorphous 100.04 Derive the formula. Ans. 3 ([Ca,Mg,Mn,Fe]0) . ([Fe,Al]A) . 3 (Si0 2 ) General type, 3 RO . R 2 3 . 3 Si0 2 or R" 3 R"' 2 (SiO 4 ) 3 . 113. Derive the formula of the oxide produced when 6.87 grams of barium unite with 1.6 grams of oxygen. Ans. Ba0 2 . 80 CHEMICAL CALCULATIONS 114. Derive the formula of the oxide formed in the com- bustion of 2.61 grams of aluminium with oxygen to a final weight of 5.01 grams. Ans. A1 2 3 . 115. Derive the formula of the oxide produced by the com- bustion of 43.45 grams of lead with 4.48 grams of oxygen. Ans. Pb 3 4 . 116. Derive the formula of the oxide produced by the burn- ing of 2.5 grams of phosphorus in oxygen to a final weight of 5.7 grams. Ans. P 2 5 . 117. Derive the formula of the nitrate, 19.7 grams of which were prepared from 10.4 grams of bismuth. Ans. Bi(N0 3 ) 3 . 118. Derive the formula of the nitrate, 17 grams of which gave a residue of 13.8 grams of sodium nitrite, NaNO 2 , upon heating. Ans. NaN0 3 . 119. Derive the formula of that chlorate, 4.165 grams of which lost 1.3 grams of oxygen upon heating and gave a resi- due of barium chloride, BaCl 2 . Ans. Ba(C10 3 ) 2 . 120. Derive the formula of mercuric cyanide, 5.4 grams of which lost 1.1 grams of cyanogen upon heating. Ans. HgC 2 N 2 . 121. Derive the formula of the double salt of ammonium sulphate and copper sulphate, 4.12 grams of which lost 1.81 grams of ammonium sulphate upon heating. Ans. (NH 4 ) 2 S0 4 . CuSO 4 . 122. Derive the formula of crystallized sodium sulphate, 8.16 grams of which lost 4.51 grams of water upon dehydration. Ans. Na 2 S0 4 . 10 H 2 0. 123. Derive the formula of crystallized copper sulphate, 7.84 grams of which lost 2.79 grams of water upon dehydration. Ans. CuS0 4 . 5 H 2 0. 124. Derive the formula of crystallized aluminium sulphate, 9.54 grams of which lost 4.61 grams of water upon dehydration. Ans. A1 2 (S0 4 ) 3 . 18 H 2 0. 125. Derive the formula of aluminium hydroxide, 4.75 grams of which lost 1.64 grams of water and left a residue of A1 2 O 3 . Ans. A1(OH) 8 . DERIVATION OF CHEMICAL FORMULAE 81 126. Derive the formula of that nitrate which gave, by analysis, 62.45 per cent lead, 8.68 per cent nitrogen, 28.85 per cent oxygen. Ans. Pb(NO 3 ) 2 . 127. Derive the formula of that substance which gave, by analysis, 52.02 per cent carbon, 13.2 per cent hydrogen, 34.68 per cent oxygen. Ans. C 2 H 6 0. 128. Derive the formula of that acetate which gave, by analy- sis, 63.61 per cent lead, 14.62 per cent carbon, 1.98 per cent hydrogen, 19.79 per cent oxygen. Ans. Pb(C 2 H 3 2 ) 2 . 129. A salt of mercury known to be a chloride analyzed for 26.15 per cent chlorine. What is the formula? Ans. HgCl 2 . 130. A salt known to be a nitrate analyzed for 62.4 per cent lead. What is the formula? Ans. Pb(N0 3 ) 2 . 131. An oxide of iron gave, by analysis, 69.80 per cent iron. What is the formula? Ans. Fe 2 3 . 132. An oxide of barium gave, by analysis, 81 per cent barium. What is the formula? Ans. BaO 2 . 133. Derive the formula of the crystallized salt which, by analysis, gave the percentage composition: 19.98 per cent iron, 11.47 per cent sulphur, 5.24 per cent hydrogen, 63.31 per cent oxygen. 10 grams of this crystallized salt lost 4.5 grams of water upon dehydration. Ans. FeS0 4 . 7 H 2 0. 134. The percentage composition of a certain salt is: 15.6 per cent chromium, 14.41 per cent sulphur, 4.79 per cent hydro- gen, 65.2 per cent oxygen. 10 grams of the crystallized salt lost 4 grams of water upon dehydration and gave a residue of Cr 2 (S0 4 ) 3 . What is the formula? Ans. Cr 2 (S0 4 ) 3 . 15 H 2 O 135. The percentage composition of a certain salt is: 27.51 per cent calcium, 22.15 per cent sulphur, 1.02 per cent hydrogen, 49.32 per cent oxygen. 10 grams of this crystallized salt lost 0.6 gram of water upon dehydration. What is the formula? Ans. (CaS0 4 ) 2 . H 2 O. 136. 2.5 grams of a crystallized salt known to be a sulphate of iron, and containing 20 per cent of iron, lost 1.13 grams of water upon dehydration. What is the formula? Ans. FeSO 4 . 7 H a O. CHAPTER IX. CALCULATIONS DEPENDING UPON CHEMICAL EQUATIONS. IN calculations which depend upon chemical reactions, the equations representing these reactions must first be constructed. In all chemical equations the number of t atomic weights of any one element concerned remains a constant; the relative amount of each element, therefore, will be alike for both sides of the equation. Naturally the valence of each element in the reaction under consid- eration must be known, as upon these factors the balancing of equations is dependent. This valence or measure by which each atom of any element can enter into combina- tion determines, accordingly, the number of other atoms or groups of atoms necessary for consideration in any reac- tion. When, however, an element undergoes a change of valence the arrangement of these atom groupings must be made to accord with this change (cf. Chap. XII). The representation of all substances in the molecular form constituting here a molecular equation has the great advantage of indicating the corresponding volume rela- tions between the gaseous substances by reason of the like molecular volumes of all substances in the state of vapor. Reaction-Quantities. The action of metallic sodium upon water is shown in the following equation : 2 Na + 2 H 2 O = 2 NaOH + H 2 2 mol. + 2 mol. = 2 mol. + 1 mol. Relative parts by weight. 2 (23.00)+ 2 (18.02) = 2 (40.01)+ 2.02 46.00 + 36.04 = 80.02 + 2.02 82.04 = 82.04 82 CHEMICAL EQUATIONS 83 In this simple equation each of the quantities repre- sented is definite and bears that relation to every other quantity as is indicated by the corresponding group of atomic weights present in each. These quantities may be considered as the reaction-quantities, atomic or molecular quantities definite for any given reaction and always proportional to each other, such that, when in- volved together, the ratio between them will be exactly equal to the ratio between the actual weights, of whatever denomination, which may represent them. This propor- tionality, relative, of course, to the construction of the equation itself, follows naturally from the Law of Definite Proportions. The reaction equation only indicates the apportionment of the various atomic or molecular group- ings for the several compounds possible under the observed conditions. The formula-quantities going to make up any number of molecules of a compound indicated, still bear a definite ratio to every other formula-quantity, with the result, of course, that the weight which represents any of these reaction-quantities in a given equation must also bear a similar and definite proportion to each other, no matter whether they are upon the same or opposite sides of the equation. The reaction-quantity of sodium in the equation above is represented by two molecular weights; the reaction-quantity of sodium hydroxide, corresponding to this value for sodium, is also represented by two molec- ular weights. As both involve the same quantity of sodium they may be regarded also as equivalent quantities. The ratio between them, 2 Na/2 NaOH, is more simply expressed in the unimolecular form, Na/NaOH. The reaction-quantity of hydrogen directly proportional to the reaction-quantity of sodium in this equation is rep- resented by one molecular weight, or H 2 . The ratio 84 CHEMICAL CALCULATIONS 2 Na/H 2 is definite, therefore, for all possible values for these substances in this reaction. Example 36. What weight of sodium hydroxide and of hydrogen can be procured by the action of 50 grams of sodium upon water? From the equation just studied, the ratio 2 Na/2 NaOH, or Na/NaOH, i.e., 23/40.01, is constant for all proportional amounts of these two substances. Therefore the ratio 50 /x, between this known weight of the metal (50 grams) and its proportional value in hydroxide (x), must be equal to the ratio above (the related terms in these ratios are of course placed similarly), and we shall have 23/40.01 = 5Q/x. By simple proportion this may be expressed as 23 : 40.01 = 50 : x. From these expressions the value of x (the weight of sodium hydroxide) is found to be 86.98 grams. In an exactly similar manner the ratio between the reaction-quantities of sodium and hydrogen, 2 Na/H 2 , or Na/H, or 23/1.01 may be made equal to the ratio between the known or unknown weights of these substances con- cerned in the reaction. Between the weight of sodium (50 grams) and its unknown proportional weight of hydrogen (x) we have the ratio 50 /a;; this is therefore immediately referred to the ratio between the correspond- ing proportional quantities above, Na/H or 23/1.01. Thus, 23/1.01 = 50/z, or 23 : 1.01 = 50 : x, from which the value for x is found to be 2.2 grams. In general we may state that when any substance is concerned in a chemical reaction, the amount of any other substance, resulting directly or indirectly through this reaction, bears to the former a definite ratio and one always CHEMICAL EQUATIONS 85 equal to the ratio between the corresponding reaction- quantities of the substances in the equation. Example 37. What weight of magnesium will be required for the liberation of 10 grams of hydrogen from water or an acid? We may first consider the decomposition of water by magnesium as shown in the following: Mg +H 2 =MgO +H 2 1 mol. + 1 mol.= 1 mol.+ 1 mol. Parts by weight. {24.32 +18.02 = 40.32 + 2.02 and again, the reaction of magnesium upon an acid such as hydrochloric acid: Mg +2 HC1 = MgCl 2 + H 2 1 mol. + 2 mol. =* I mol. + 1 mol. Parts by weight. {24.32 + 2 (36.47) = 95.24 + 2.02 In each case we find that the molecule of magnesium containing one atomic weight displaces, and is equi- valent to, one molecule of hydrogen containing two atomic weights. The quantities Mg and H 2 are there- fore directly proportional to each other in both equations, and the constant ratio Mg/H 2 , or 24.32/2.02, must be equal to the ratio between the corresponding weights of these elements here concerned, i.e., x/10. From the equation 24.32/2.02 = x/ 10, or the simple proportion 24.32 : 2.02 = x : 10, we calculate the value of x to be 120.4 grams (magnesium). In the more complicated reactions the relations between the several quantities on the two sides of an equation are often difficult to ascertain. To say nothing of reversible actions and the possibilities for dependent or secondary reactions to take place between certain of the quantities, we have assumed and must continue to assume that the reaction in question takes place along the lines indicated 86 CHEMICAL CALCULATIONS in that equation best substantiated by the facts under the observed conditions. Example 88. What weight of sulphur dioxide, S0 2 , can be obtained by the action of 10 grams of copper upon concentrated sulphuric acid? Copper acts upon sulphuric acid (cone.), to give sulphur dioxide, water and copper sulphate. Notwithstanding the slight amount of cuprous sulphide formed here through a secondary reaction, we shall represent the action by a single equation: Cu + 2 H 2 S0 4 = CuS0 4 + 2 H 2 + S0 2 63.57 64.07 The reaction-quantities, Cu and S0 2 , are directly pro- portional; the ratio between them, Cu/S0 2 , or 63.57/64.07, is constant for all amounts here involved, and conse- quently equal to the ratio, 10 /x, representing these sub- stances respectively in this example. From the equation 63.57/64.07 = 10/z the value of x is calculated as 10.08, the weight in grams of sulphur dioxide. Calculation of Volume Relations Introduced by Chem- ical Equations. In the consideration of gaseous sub- stances formed in these reactions, their gram-molecular quantities may be referred at once to the gram-molec- ular volumes as a basis (cf. Ex. 16). For example, to recall the illustrations at the beginning of this chapter, two gram-molecular weights of sodium displace one gram- molecular weight of hydrogen, which under standard conditions occupies 22,400 c.c. Whatever be the weight of hydrogen evolved, the ratio between the G.M.W. and this weight will be identical with that between the G.M.V. and the volume of hydrogen corresponding to the known weight above (cf. Ex. 15). CHEMICAL EQUATIONS 87 Example 39. Calculate the volume of hydrogen, at 10 and 750 mm., that will be evolved by the action of 100 grams of zinc upon an acid. The nature or strength of the acid here is of no concern, for, so long as the zinc is used up, the proportional amount of hydrogen must be displaced, and from this weight of hydrogen displaced its volume may be easily calculated. With sulphuric acid the reaction may be represented as follows : Zn + H 2 S0 4 = ZnSO 4 + H 2 65.37 + 98.09 = 161.44 + 2.02. The reaction-quantities Zn and H 2 establish the ratio Zn/H 2 , or 65.37/2.02, as constant for all proportional amounts of these elements ; hence the equation 65.37/2.02 = 100/z. From this the value of x is calculated as 3.09 grams. A gram-molecular weight of hydrogen, 2.02, occupies 22,400 c.c. at standard conditions. As previously stated the volume occupied by any other weight of the gas will stand in the same proportion to this gram-molecular volume as does its weight to the gram-molecular weight. Thus the ratio 2.02/3.09, between the known weights, will be exactly equal to the ratio between the corresponding volumes, 2.02/3.09 = 22,400/z. By simple proportion we should have 2.02: 3.09 = 22,400: x; or, with reference to the fractional part of the volume occupied by 2.02 grams of the gas, we know that 3.09 grams must occupy 3.09/2.02 times this known volume, i.e., 3>09 * 22 ' 400 or 34 261 c.c. 2.02 The value of x, as here calculated, must be adjusted now to the observed conditions of temperature and pressure. For this we need but a moment's glance to see that the corrected volume, x', will be expressed by the equation 88 CHEMICAL CALCULATIONS The reverse process that of determining the amount of any substance which will give a definite volume of gas measured at certain definite conditions is made clear when the weight of this gas is ascertained (cf. Ex. 17). So also the consideration of aqueous vapor, when present, brings into these calculations only those methods already outlined in previous examples. Calculations with Reference to Degree of Purity. Oftentimes the purity of materials employed in chemical operations does not come up to the standard or 100 per cent. In such cases, where the deviation from the theo- retical is known, the results found must be recalculated with reference to the standard purity. For instance, if the magnesium considered in Example 37 had been of only 80 per cent purity (contaminated, we shall say, with 20 per cent of inert or extraneous matter), then our results, wrongly based upon a valuation of 100 instead of the actual 80, must be raised in accordance with the ratio existing between the actual and theoretical value, i.e., according to the ratio 80/100. The result in the exam- ple cited would be changed through the simple pro- portion: 120.4 : x = 80 : 100; or 150.5 grams would be the weight of magnesium (80 per cent) necessary to give 10 grams of hydrogen. Conversely, if it were desired to ascertain what weight of hydrogen could be obtained from this weight (150.5 grams) of magnesium (80 per cent pure), then the result upon the basis of 150.5 grams of pure magnesium would need to be lowered in the same proportion: 100 : 80. The same end is attained by bringing into the original calculation just 80 per cent of the weight of magnesium (80 per cent pure), i.e., 120.4 grams in this example. These calculations afford, therefore, a simple means for estimating the degree of purity of a definite weight of a CHEMICAL EQUATIONS 89 substance when the amount of some constituent con- cerned in one of its reactions is referred to the correspond- ing theoretical value. Example 40. A specimen of marble weighing 5 grams evolved 2.1 grams of carbon dioxide (corresponding to a theo- retical volume of 1162.5 c.c. at 20 and 750 mm.) when acted upon by an acid. Estimate the degree of purity, assuming that all of the carbonate was present as calcium carbonate. The reaction with hydrochloric acid is here given: CaC0 3 + 2 HC1 = CaCl 2 + H 2 O + CO 2 . 100.09 44 From this equation the proportionality between the reaction-quantities CaC0 3 (100.09) and C0 2 (44) leads to the equation j^~ = ~~' wnere x ) 2-2 grams, is the weight of carbon dioxide theoretically possible from this weight of pure calcium carbonate. The specimen is accordingly only ^2 > or 95.45 per cent, pure (2.2 : 2.1 = 100 : x). Calcu- lating from the standpoint of the carbon dioxide, we natu- 44 21 rally reach the same value : nn AQ = ; or x, 4.772 luu.uy x grams, is the theoretical amount of calcium carbonate re- quired for the production of 2.1 grams of carbon dioxide. 4772 This weight is , or 95.45 per cent, of the weight of the oUUU specimen; consequently, just this fraction of the specimen can be considered calcium carbonate, i.e., its degree of purity is 95.45 per cent. Calculation of Products Resulting from Mixtures. The definite weights of various substances brought together for chemical combination are rarely ever present in the exact proportions indicated by the equation representing the particular reaction. 90 CHEMICAL CALCULATIONS Thus, in the interaction between a metal and an acid, as already noted, the acid is taken in excess of the theo- retical amount proportional to the definite weight of metal concerned. Such an excess is easily removed from the final product by volatilization. In cases where this cannot be accomplished without decomposition of some product sought, other means (crystallization, etc.) are employed. In order to ascertain the exact amount of any substance formed through one of these inter- actions it is necessary first to determine what particular component or components can be acted upon to com- pletion by the others present. This, of course, presupposes the tendency for the reaction to run to completeness in some one direction, and implies the elimination of those products which may lead to a reversal of these conditions. Example 4! What weight of sodium sulphate may be expected to result from the interaction of 10 grams of sodium, 11 grams of sulphur and 40 grams of oxygen? The equation representing the possible combination of these elements to this end is as follows : 2 Na + S + 2 O 2 = Na 2 S0 4 46 + 32.07 + 64 = 142.07 Weights given. {10 1 1 40. The actual weights of the several substances here entering into combination as sodium sulphate must stand in the same relation to each other as do the cor- responding reaction-quantities. There is then an equiva- lence in the ratios between each reaction-quantity and the corresponding weight that represents it in the reaction (see footnote, page 72). If these conditions are fulfilled in the example, we shall have an equality in the ratios, 46/10, 32.07/11 and 64/40. A single glance, however, disproves this CHEMICAL EQUATIONS 91 point. It then becomes necessary to determine by trial which one of these ratios, between a reaction-quantity and its corresponding weight as given in the example, is the correct ratio to select as the basis of calculation. For example, upon the basis of 64/40, we must have: 64 46 32.07 40 a; y spective values for sodium and sulphur, actually exceed the amounts at hand. Upon the basis 32.07/11, we have: 32J07 = 46 = 64^ where (153) and y, (22), are the 11 x y respective values for sodium and oxygen. In this latter case only the sodium is higher than the weight stipu- lated. Finally, upon the basis 46/10, we have the equa- tion = 5^ = , and from this both x, (6.97), the 10 x y weight of sulphur, and y, (13.9), the weight of oxygen, fall below the amounts given; hence 46/10 is the proper ratio upon which to base the calculation. The sum of the respective weights thus obtained, 30.87 grams, (10 + 6.97 + 13.97), gives the highest weight of sodium sulphate possible from the data in the problem. In general we select as the basis of calculation for any given reaction only that ratio which brings into consid- eration those values for the various substances involved as do not exceed the amounts present. More simply stated, perhaps, we base our calculations upon that weight of a particular substance at hand which gives, with its corresponding reaction-quantity as denominator, the smallest numerical factor. Thus the total weight of this particular substance determines the theoretically propor- tional weights of the other substances. Under conditions where reversible reactions are likely, for example the preparation of sodium sulphate by the 92 CHEMICAL CALCULATIONS action of sulphuric acid upon common salt, as shown in the equation: 2 NaCl + H 2 S0 4 = Na 2 S0 4 + 2 HC1, we determine in similar manner the weight of sodium sulphate possible from any known weight of salt (if the sul- phuric acid is in excess) or from the weight of sulphuric acid (if the salt is in excess). The presence of hydro- chloric acid as the reversing agent must be removed in either case if we wish to obtain the calculated results. Calculation of Ratios between Reaction-Quantities in Dependent Equations. When an element or group of elements enters into a series of successive reactions and the equation for each reaction can be constructed, we may draw up a proportionality between the reaction- quantities in any two of the equations providing that some quantity is common to both. In like manner this proportionality may be extended step by step over any number of dependent equations (cf. Ex. 24). Example 4> What weight of bromine can be liberated from a concentrated solution of potassium bromide (excess) by the addition of 12 grams of hydrochloric acid (containing 39.1 per cent HC1) and an excess of manganese dioxide? The two equations representing the action are as fol- lows: (a) MnO 2 + 4 HC1 = MnCl 2 + C1 2 + 2 H 2 O (b) 2 KBr + C1 2 =2 KC1 + Br 2 . It is observed that the reaction-quantity C1 2 is common to both (a) and (6), consequently the reaction-quantities directly proportional to this quantity in either equation will be also directly proportional to each other. The ratio 4 HC1/C1 2 from equation (a) and the ratio Cl 2 /Br 2 from equation (b) give us, accordingly, the ratio 4 HCl/Br 2 for the proportionality between the reaction-quantities CHEMICAL EQUATIONS 93 required for this example. From the equivalence between this ratio and the ratio of the weights corresponding thereto, we have 4 HCl/Br 2 , or 4(36.47) /2(79.92), or 145.88/159.84 = 12 /x, which, solved for x, gives a value of 13.1 grams. This weight of bromine, 13.1 grams, is based upon the hydrochloric acid as 100 per cent hydro- gen chloride. The concentrated acid at our command, the acid stated in the example, contained only 39.1 per cent hydrogen chloride. It remains then to calculate the weight of bromine which a 39.1 per cent acid can give. By reference to Example 40, the method is outlined to be simply one of proportion, according to which we shall have 100 : 39.1 = 13.1 : x, or z = 5.13 grams, the weight of bromine evolved by 12 grams of 39.1 per cent hydrochloric acid. This same result is easily obtained by determining first the weight of hydrogen chloride in the 12 grams of 39.1 per cent acid, (12 X 39.1 per cent = 4.69), and working the example with this value for the hydrogen chloride. The elimination of a quantity occurring in two depend- ent equations may necessitate a readjustment of one or both of the equations containing it before the quantity becomes alike in each. Example 48. What weight of iron sulphide will be required to furnish sufficient hydrogen sulphide for the reduction of 10 grams of sulphur dioxide to sulphur? The two equations here required are as follows : (a) FeS + 2 HC1 = FeCl 2 + H 2 S (6) 2 H 2 S + S0 2 =2 H 2 O + 3 S. The reaction in equation (6) depends upon the hydro- gen sulphide that is evolved in (a) ; consequently this sub- stance must constitute the common reaction-quantity. In order to make this quantity alike for the two equations 94 CHEMICAL CALCULATIONS and thus eliminate it from the calculations, it is only neces- sary to multiply equation (a) by 2, when we obtain (a') : (a') 2 FeS + 4 HC1 = 2 FeCl 2 + 2 H 2 S. The comparison of any of the reaction-quantities in the two equations (a') and (6) with reference to hydrogen sul- phide is now made simple. In the example given, iron sulphide and sulphur dioxide are found to be related through the ratios 2 FeS/ 2 H 2 S and 2 H 2 S/S0 2 , or directly as 2 FeS/S0 2 . The calculation is conducted, therefore, as follows : 2 FeS/S0 2 , or 2 (87.92) /64.07, or 175.84/64.07 = z/10. From which x, the weight of iron sulphide, is found to be 27.44 grams. Without this elimination of the quantity common to both equations, the calculation, of course, can be made directly toward ascertaining the weight representing this quan- tity in the first equation, and then finally, from this weight, the weight representing any other reaction-quantity pro- portional to it in the second equation. Such calculations, here involving the weight of hydrogen sulphide, are indeed roundabout and entirely unnecessary. Calculation of Ratios between Reaction-Quantities in Independent Equations. In the study of dependent equations the reaction-quantities may be regarded as related to each other, through this common reaction-quan- tity, as are the members of a single equation. The reac- tion-quantities of the second equation are, so to speak, brought into existence through the agency of this com- mon quantity. In the study of independent equations, where there is present no one quantity which has a direct bearing upon any other equation, we have simply a fur- ther application of this same principle; namely, the com- parison of all the possible reaction-quantities in the CHEMICAL EQUATIONS 95 separate equations so long as some quantity can be made common to each. Quantities so compared will be directly proportional to each other, but only in respect to this common reaction-quantity. As an illustration, the fol- lowing equations are cited: (a) Na 2 C0 3 + 2 HC1 = 2 Nad + H 2 + C0 2 (6) NaHC0 3 + HC1 = Nad + H 2 O + CO 2 . In these independent equations there are a number of substances represented for which the reaction-quantities could be adjusted alike for both; thus the reaction-quan- tity C0 2 is a common one. Upon this fact we may draw up the ratio Na 2 C0 3 /NaHC0 3 to express the proportion- ality between the relative amounts of normal carbonate and primary carbonate necessary to give an equal amount of carbon dioxide, i.e., the relation is based upon the carbon- dioxide content. In order to obtain a comparison with reference to the salt, equation (6) must be doubled to (&'): (&') 2 NaHC0 3 + 2 HC1 = 2 Nad + 2 H 2 + 2 C0 2 . The ratio Na 2 CO 3 /2 NaHC0 3 then expresses the relation between the relative amounts of each carbonate neces- sary to give equal amounts of salt with hydrochloric acid. Or, since sodium is always a constant quantity in salt, we may say that the ratio above is that based upon a like content of sodium in each carbonate. In addition to the carbonate discussed in the preceding paragraph, we may also compare the reaction-quantities for the carbon dioxide present. Between equations (a) and (6), where this is a common quantity, we have the ratio 2 HC1/HC1 representing the relative amounts of acid necessary in the respective cases to give equal amounts of carbon dioxide. Between equations (a) and (&') we have the ratio C0 2 /2 CO 2 representing the relative 96 CHEMICAL CALCULATIONS amounts of carbon dioxide evolved from equal amounts of sodium when contained respectively in normal car- bonate or primary carbonate, as indicated by the ratio Na 2 CO 3 /2 NaHCO 3 , or the ratio 2 NaCl/2 NaCl, in each of which the sodium content is the same for both terms of the ratio. This latter point may be illustrated by the equation : 2 NaHC0 3 = Na 2 C0 3 + CO 2 + H 2 0. Here it is shown that two molecules of the primary car- bonate break down into one molecule of the normal car- bonate with the loss of one molecule of carbon dioxide. The ratio 2 NaHC0 3 /CO 2 represents the relative amounts of these substances concerned in the action of heat upon the primary carbonate. This, in fact, is a case where the entire quantity of a substance, (CO 2 ), available in a com- pound need not be concerned in a ratio for the study of that compound, but only that portion of it separately in- volved as a reaction-quantity and made directly propor- tional, therefore, to some other reaction-quantity. Example 44. What relative weights of mercuric oxide and barium peroxide are required in the preparation of equal amounts of oxygen? The molecular equations with oxygen as the common quantity adjusted alike in both are as follows: 2 HgO = 2 Hg + 2 2 BaO 2 = 2 BaO + O 2 . The ratio 2 HgO/2 Ba0 2 , or HgO/Ba0 2 , or 216/169.37, determines accordingly the relation between the corres- ponding weights of these two substances in this problem. If we take one, e.g., the mercuric oxide, as 100 we re- duce the second to a comparatively simple value: 216 = 100 a 169.37 x CHEMICAL EQUATIONS 97 This gives 78.4 as the value of x, or that weight in grams of barium peroxide equal to 100 grams of mercuric oxide in the preparation of oxygen. Calculation of Reaction-Quantities from the Weights of Substances Involved. The direct proportionality which exists between the reaction-quantities of a given equation requires, as we have seen, the same proportion- ality between the corresponding weights which represent them in this particular reaction. This permits of an equivalence in the ratios between each reaction-quantity and its corresponding weight (cf. Ex. 41 and footnote, page 72). In the construction of chemical equations from the actual weights of substances therein concerned, we must bear in mind the possibility of deviations in these weights from those demanded in the reactions. These deviations may be due to any number of causes and rapidly increase with the instability of the compounds considered as well as with the tendencies for secondary reactions. Conse- quently the determination of the correct reaction-quan- tities for these equations must be made a special study in each individual case. Example 46. A solution of 10 grams of crystallized sodium thiosulphate, Na 2 S 2 3 . 5 H 2 O decolorized 5.1 grams of iodine. What molecular quantity of this salt was associated with one molecule of iodine in this reaction? From the equivalence in ratios between the reaction- quantities and the actual weights involved we have : JL Jl . JL_ 253.84 10" 5.1 ' Cr 10" 5.1 This gives to x the value 497.7, the reaction-quantity of the thiosulphate associated with one molecular weight of 98 CHEMICAL CALCULATIONS iodine. The molecular weight of sodium thiosulphate, Na 2 S 2 O 3 . 5 H 2 0, is 248.2, consequently we have here 497.7/248.2 or approximately 2 molecules of this salt in its reaction-quantity, i.e., the equation constructed upon the iodine involved as just one molecule will be 2 Na 2 S 2 3 . 5 H 2 O + I 2 = (2 Nal + Na 2 S 4 6 ). Complex Reaction-Quantities. There are a number of equations in which the reaction-quantities upon one side are incorporated into one single reaction-quantity upon the other side. The same principles hold here as in the cases just discussed, but the study of this larger quantity is nothing more or less than the study of the formula- quantities present in it. The consideration of the so-called molecular compounds, as are the double salts and salts containing water of crystallization, illustrates this point. Example 46. 100 grams of copper sulphate will give what weight of blue vitriol (CuS0 4 . 5 H 2 0)? From the equation, CuS0 4 + 5 H 2 = CuSO 4 . 5 H 2 159.64 + 90.1 = 249.74, the ratio GuSO 4 159.64 * f\f CuS0 4 .5H 2 O' 249.74 is a constant, and denotes the direct proportionality between these two reaction-quantities. The ratio between the actual weights, 100 and x, when put equal to the ratio between the molecular quantities, gives us 159.64 100 249.74 x ' where x, with the value 156.4 grams, is the weight of blue vitriol. As the percentage composition of the anhydrous copper sulphate is a constant, we may just as well calcu- late what weight of pure copper, Cu, or sulphur, S, or even CHEMICAL EQUATIONS 99 oxygen, 4 O, is necessary to give any definite weight of anhydrous copper sulphate and finally blue vitriol; or, vice versa, what weight of blue vitriol is obtainable from any definite weight of one of these constituents. As another illustration of this point we may take the formation of ordinary alum: K 2 SO 4 . 6 HP + A1 2 (SO 4 ) 3 . 18H 2 O = I^SO, . A1 2 (S0 4 ) 3 . 24H,O 1 mol. + 1 mol. = 1 mol. A molecular quantity of one sulphate unites with one of another to form one molecular quantity of the double sulphate. These quantities are all directly proportional to each other, and from the equation we write the following ratios: K 2 S0 4 .6H 2 (W K 2 SO 4 . A1 2 (S0 4 ) 3 . 24 H 2 O A1 2 (SO 4 ) 3 . 18 H 2 O K 2 S0 4 . A1 2 (S0 4 ) 3 . 24 H 2 O ( , K 2 S0 4 . 6 H 2 \ c ) A1 2 (SO 4 ) 3 . 18 H 2 O These serve as the ratios for calculating, upon the molecu- lar quantities involved, the amount of alum obtainable from certain known amounts of (a) crystallized potassium sulphate; (6) crystallized aluminium sulphate; and also (c) the amount of crystallized aluminium sulphate neces- sary for combination with one molecule of the crystallized potassium sulphate, or vice versa. Though somewhat more complicated molecular aggregates may be present, the ratios between the several factors are always constant. Example 47. How much alum can be prepared from 100 grams of anhydrous potassium sulphate? The ratio K 2 S0 4 174.27 K 2 S0 4 . A1 2 (S0 4 ) 3 . 24 H 2 ' r 949.16 100 CHEMICAL CALCULATIONS is placed equal to the ratio 100/z, and the equation solved in the usual manner. The value of x, the weight of alum, is found to be 544.6 grams. PROBLEMS. 137. What weight of potassium hydroxide may be prepared by the action of 100 grams of potassium upon water? Ans. 143.5 grams. 138. What weight of potassium will be required in the prep- aration of 20 grams of potassium carbonate, K 2 CO 3 ? Ans. 11.3 grams. 139. What weight of magnesium chloride, MgCl 2 , may be obtained by the action of hydrochloric acid upon 10 grams of magnesium carbonate, MgC0 3 ? What weight of carbon dioxide will be liberated? Ans. 11.3 grams MgCl,. 5.2 grams C0 3 . 140. Recalculate Problem 139 on the supposition that the magnesium carbonate contained 10 per cent of insoluble matter. Ans. 10.16 grams MgCl a . 4.7 grams CO,. 141. What weight of sulphur dioxide, SO 2 , can be obtained by the action of an acid upon 250 grams of sodium sulphite, Na 2 SO 3 ? Ans. 127 grams. 142. Recalculate Problem 141 on the supposition that 20 per cent of the sulphite had become oxidized to sulphate. Ans. 101.6 grams. 143. Calculate the volume of carbon dioxide, at 22 and 740 mm. pressure, that will be liberated by the action of acid upon 20 grams of calcium carbonate, CaCO 3 . Ans. 4967 c.c. 144. What weight of magnesium will be required for the liberation of 500 c.c. of hydrogen, at 20 and 740 mm. pressure, when acted upon by an acid? Ans. 0.49 gram. 145. Calculate the volume of hydrogen, measured over water at 17 and 742 mm. pressure, that can be liberated by the action of 10 grams of sodium upon water. Ans. 5403 c.c. 146. What weight of aluminium will be required for the liberation of 420 c.c. of hydrogen, measured over water at 18 CHEMICAL EQUATJONS ' 101 and 746.4 mm. pressure, when acted' upbri J 16y hydrodHlofic' acid? Ans. 0.306 gram. 147. What weight of ammonium nitrite, NH 4 NO 2 , will evolve, when heated, 480 c.c. of nitrogen, measured over water at 21 and 747.5 mm. pressure? Ans. 1.22 grams. 148. Determine the purity of a sample of anhydrous sodium carbonate, Na 2 C0 3 , 6 grams of which gave, when acted upon by an acid, 1310 c.c. of carbon dioxide, at 10 and 750 mm. pressure. Ans. 98.35 per cent pure. 149. Determine the purity of a sample of anhydrous sodium carbonate, 3 grams of which gave, by treatment with sulphuric acid and final ignition, 3.99 grams of sodium sulphate, Na 2 S0 4 . Ans. 99.25 per cent pure. 150. A sample of iron wire weighing 2.4 grams was found to give, when acted upon by excess of acid, a volume of hydrogen measuring 1015.1 c.c., at 10 and 745 mm. pressure. What was its degree of purity? The presence of any other substance capable of liberating hydrogen from an acid is here disregarded. Ans. 99.72 per cent pure. 151. A sample of silver nitrate weighing 2.40 grams was brought into solution and treated with a soluble chloride (excess). The weight of silver chloride, AgCl, precipitated was 2.01 grams. What was the purity of the sample? Ans. 99.26 per cent pure. 152. What weight of zinc (98 per cent pure) will be required for the liberation of the hydrogen from 10 grams of hydro- chloric acid containing 39.1 per cent HC1? Ans. 3.576 grams. 153. What weight of sulphuric acid containing 27.32 per cent H 2 S0 4 will be required for interaction with 2.17 grams of iron wire (99 per cent pure)? Ans. 13.82 grams. 154. What volume of hydrogen, measured over water at 18 and 746.4 mm. pressure, will be liberated by the action of aluminium upon 20 grams of sulphuric acid containing 41.5 per cent H 2 S0 4 ? Ans. 2100 c.c. 155. What weight of sulphuric acid (27.32 per cent H 2 S0 4 ) will be required for interaction with a metal (Zn, Mg, etc.) in order to give a volume of hydrogen measuring over water 103 2 - c.c., at 16 and 742.5 mm. pressure? Ans. 15 grams. 102 CHEMICAL CALCULATIONS 156. What weight 'of hydrochloric acid (23.82 per cent HC1) will be required for interaction with iron sulphide, FeS, in order to give a volume of hydrogen sulphide, measuring 613.1 c.c., at 20 and 745 mm. pressure? Calculate also the weight of iron sulphide (98 per cent pure) here consumed. Ans. 7.655 grams acid. 2.243 grams FeS (98 per cent). 157. When 100 grams of mercury and 20 grams of sulphur are rubbed together what weight of mercuric sulphide, HgS, may be formed? Ans. 116 grams. Note. From the reaction-quantities involved it is seen that the sulphur is in excess. This excess is easily removed by solution in carbon disulphide. 158. A mixture of 10 grams of zinc dust and 2 grams of sulphur was gently heated to point of reaction. What weight of zinc sulphide, ZnS, was formed? Ans. 6.053 grams. 159. A mixture of 10 grams of iron and 8 grams of sulphur was gently heated to point of reaction. What weight and volume (at standard conditions) of hydrogen sulphide could be obtained from the final product, FeS, by the action of an acid? Ans. 6.104 grams, or 4010.5 c.c. H 2 S. 160. A mixture of 4 grams of sodium oxide, Na 2 0, and 6 grams of sulphur trioxide will give what weight of sodium sulphate? Ans. 9.166 grams. 161. 3 grams of silver nitrate, AgN0 3 , and 1 gram of potas- sium chloride, KC1, were brought together in aqueous solution. What weight of silver chloride, AgCl, was precipitated? Ans. 1.923 grams. 162. 8.2 grams of crystallized barium chloride, BaCl 2 . 6 H 2 0, and 7 grams of sulphuric acid (70 per cent H 2 S0 4 ) were brought together in aqueous solution. What weight of barium sulphate, BaS0 4 , was precipitated? Ans. 6.05 grams. 163. 10 grams of a mixture of marble (CaC0 3 ), magnesium, and an inert substance were acted upon by excess of acid. The carbon dioxide evolved (taken up in a solution of potassium hydroxide) was found to weigh 1.318 grams. The volume of the other gas, hydrogen, measured over water at 10 and CHEMICAL EQUATIONS 103 750.2 mm. pressure, was 5874 c.c. Calculate the percentage amount of each component in the mixture. Ans. 30 per cent marble. 60 per cent magnesium. 10 per cent insoluble matter. 164. 15 grams of an alloy of zinc and copper (containing 10 per cent of copper) were placed in a vessel containing 100 grams of sulphuric acid (25 per cent H 2 S0 4 ). What weight of hydrogen was liberated? Ans. 0.417 gram. 165. 12 grams of an alloy of aluminium and zinc (containing 33J per cent of zinc) were placed in a vessel containing 180 grams of hydrochloric acid (35 per cent HC1). What volume of hydrogen, at standard conditions, was liberated? Ans. 11,290 c.c. 166. A specimen of silver, containing 3 per cent copper, weighed 9.8 grams. After solution in nitric acid an excess of sodium chloride was added to it. Calculate the weight of the silver chloride precipitated. Ans. 12.632 grams. 167. 10 grams of phosphorus tribromide, PBr 3 , were mixed with an excess of water and an excess of silver nitrate added to, this solution. What weight of silver bromide was formed? Calculate also the exact weight of silver nitrate required for this removal of bromide. Ans. 20.81 grams AgBr. 18.82 grams AgN0 3 . 168. What volume of hydrogen sulphide, at standard con- ditions, would be required for interaction with an excess of iodine, in aqueous suspension, in order to furnish an amount of hydriodic acid sufficient for the precipitation of 10 grams of silver iodide from a solution of silver nitrate? Ans. 477 c.c. 169. What weight of fluorspar, CaF 2 , would be required to furnish sufficient hydrogen fluoride (by interaction with sul- phuric acid) to convert 5 grams of quartz, SiO 2 , into silicon fluoride, SiF 4 ? Ans. 12.95 grams. 170. Calculate the volume of chlorine, at standard condi- tions, necessary to give, by interaction with water, an amount of oxygen which will just suffice for the oxidation of 10 grams of mercury to mercuric oxide, HgO. Ans. 1120 c.c. 104 CHEMICAL CALCULATIONS 171. Calculate the volume of chlorine, at standard condi- tions, necessary to give an amount of potassium chlorate (by interaction with a hot solution of potassium hydroxide) which would just suffice, in its decomposition into oxygen and potas- sium chloride, for the oxidation of 5 grams of hydrogen to water. Ans. 55,446 c.c. 172. 100 grams of iron pyrites, FeS 2 , were roasted to ferric oxide, Fe 2 O 3 , and sulphur dioxide. The sulphur dioxide was then taken up by sodium peroxide, Na 2 O 2 , to form sodium sulphate, and this product treated with a solution of barium chloride, BaCl 2 . What weight of barium sulphate was pre- cipitated? Ans. 389.1 grams. 173. 20 grams of nitrogen were carried through the follow- ing series of reactions. Calculate the resulting volume of nitrous oxide, N 2 0, at standard conditions. 3 Mg + N 2 = Mg 3 N 2 Mg 3 N 2 + 6 H 2 O = 3 Mg (OH), + 2 NH 3 NH 3 +HNO 3 = NH 4 N0 3 NH 4 N0 3 = N 2 + 2 H 2 Ans. 50,237 c.c. 174. Calculate the relative weights of sodium chlorate and potassium chlorate necessary to give the same volume of oxygen. Ans. 100 (NaC10 3 ): 115 (KC10 3 ). 175. Calculate the relative weights of potassium chlorate and perchlorate, KC10 4 , necessary to give the same volume of oxygen. Ans. 100 (KC10 3 ) : 84.8 (KC10 4 ). 176. Compare the weights of aluminium and zinc neces- sary for the production of equal weights of hydrogen by inter- action with an acid. Ans. 100 (Al) : 361.8 (Zn). 177. Compare the weight of calcium nitride, Ca 3 N 2 (in its interaction with water), and the weight of ammonium chloride (in its interaction with a base), necessary to give the same weight of ammonia. Ans. 100 (Ca 3 N 2 ) : 72.2 (NH 4 C1). 178. What relative weights of cupric oxide, CuO, and cuprous oxide, Cu 2 0, are procurable from the same weight of copper? Ans. 100 (CuO) : 89.95 (Cu 2 0). CHEMICAL EQUATIONS 105 179. In the interaction of methane, CH 4 , and chlorine, 2 grams of the former required 17.7 grams of the latter. Calcu- late the reaction-quantity of chlorine per molecule of methane. Ans. 2 C1 2 . 180. 2 grams of hydrogen sulphide, H 2 S, decolorized 5.8 grams of potassium dichromate, K 2 Cr 2 7 (in acid solution). Calculate the reaction-quantity of the former required per molecule of the latter. Ans. 3 H 2 S. 181. 2.4 grams of ammonia, NH 3 , reduced 17 grams of hot cupric oxide, CuO, to copper. Calculate the reaction-quantity of cupric oxide required per molecule of ammonia. Ans. 1$ CuO. NOTE. In order to avoid fractional quantities we may here mul- tiply by two and obtain 3 CuO per 2 NH 3 . 182. What weight of chrome-alum, K 2 S0 4 . Cr 2 (S0 4 ) 3 . 24 H 2 O, may be obtained from 20 grams of crystallized potas- sium sulphate, K2S0 4 . 6 H 2 O, and an excess of chromium sulphate? Ans. 70.76 grams. 183. What weight of ammonium-magnesium phosphate, NH 4 MgP0 4 . 6 H 2 0, could be formed from a solution containing 50 grams of crystallized magnesium sulphate, MgS0 4 . 7 H 2 0, and an excess of ammonia and sodium phosphate? Ans. 49.79 grams. 184. What weight of iron-ammonium alum, (NH 4 ) 2 SO 4 . Fe 2 (S0 4 ) 3 . 24 H 2 0, may be formed when 12 grams of ammo- nium sulphate, (NH 4 ) 2 SO 4 , and 30 grams of ferric sulphate, Fe 2 (S0 4 ) 3 , are brought together in concentrated aqueous solution? Ans. 72.3 grams. CHAPTER X. ' ':V NORMAL SOLUTIONS. IN reactions between substances in solution no atten- tion thus far has been given to the actual amount of substance contained in a definite volume of solvent. In the action between an acid, furnishing hydrogen- ion, and a base, furnishing hydroxide-ion, the point of neutralization is reached when equal quantities of these two kinds of ions chemically equivalent are present, and the complete removal of both in the form of the compound water as a slightly ionized substance is effected. The detection of this point in solution is readily accomplished through the use of an indicator, or some substance which shows a change in color by the merest trace of either the one or the other of these two ions. The operation of ascertaining this end-point is called titration. Naturally it may be applied to the determination of other end-points, or points of comple- tion of definite chemical reactions in solution, as well as to this process of neutralization. The neutralization of hydrochloric acid by sodium hy- droxide is shown by the following equation: Na* + OH* + H* + C1' = Na*+Cl' + H 2 O 23 + 17.01 1.01 +35.46 = 23 + 35.46 ,18^?. 40.01 36.47 58.46 18.02* A solution which contains 40.01 parts by weight of sodium hydroxide will exactly neutralize one which con- tains 36.47 parts by weight of hydrogen chloride. This 106 NORMAL SOLUTIONS 107 follows from the fact that in the former there are 17.01 parts of ionizable hydroxyl and in the latter 1.01 parts of ionizable hydrogen, the exact proportions of these two substances necessary for the formation of water. When equal volumes of these solutions neutralize each other, then the concentration of ionizable hydrogen in the acid solution is equal to the concentration of ionizable hydroxyl in the solution of the base; i.e., the relative amounts of each are directly proportional to 1.01 and 17.01 respectively. The liter has been adopted as the standard volume for reactions in solution; when a gram- molecular weight of a substance is contained in this vol- ume we have what is called a gram-molecular solution or more commonly a Molar Solution. Some definite tem- perature, circa 20, is usually understood. From the reaction between sodium hydroxide and hydrochloric acid, and from the definitions just given, we are aware that one liter of a molar solution of the former will exactly neutralize one liter of a molar solution of the latter; consequently any fractional part of the one solution will neutralize this same fractional part of the second solution. In the neutralization of this same base by a dibasic acid, such as suj^^c acid, the following equation comes into consideral 2 Na*+ 2 OH' *2 H' + SO/ =2 Na'+ SO/ + 2 H 2 O 2(23) +2(17.01) +2(1.01) +96.07 = 2(23) +96.07 + 2(18.02) From this it is evident that a molar solution of sulphuric acid, with 98.09 grams of the acid per liter, will contain 2.02 grams of ionizable hydrogen, a quantity that re- quires 34.02 grams (2 X 17.01) of ionizable hydroxyl for its complete neutralization. This quantity of hydroxyl is furnished, as the equation indicates, through the use 108 CHEMICAL CALCULATIONS of two gram-molecular weights of sodium hydroxide. If we were dealing with molar solutions of these substances, two volumes of the sodium hydroxide solution would be required for the neutralization of one volume of a molar sulphuric acid solution. By reason of this variation in the^ number of ionizable hydroxyl and hydrogen groups in the various substances, it is found advisable to base our standard solutions upon the exact number of these univalent groups which they contain, rather than upon the entire molecular weight of the substance itself. A solution which contains in one liter exactly 1.01 grams of ionizable hydrogen is taken as the standard for acids, while that which contains in one liter exactly 17.01 grams of ionizable hydroxyl is taken as the standard for bases. These two solutions are, volume for volume, always equivalent and may be termed Equivalent Normal Solutions or, as is more gen- erally the case, Normal Solutions. The molar solution of sodium hydroxide is identical, of course, with its normal solution. The molar solution of sulphuric acid contains twice as much ionizable hydrogen as is required for its normal solution, a fact indicated by the use of two volumes of the molar sodium hydroxide solution above to neutralize only one ^fkiie of this acid. We are therefore required to disscH Be-half of the gram-molecular weight (98.09) of TB^uric acid, or 49.04 grams, in water and bring this to one liter in order to obtain its true normal solution. In the same manner a base such as barium hydroxide, Ba(OH) 2 , with a molecular weight of 171.39, will con- tain in its molar solution 171.39 grams of substance of which 34.02 grams is ionizable hydroxyl. A definite volume of this molar solution would neutralize two volumes of a normal hydrochloric acid solution; conse- NORMAL SOLUTIONS 109 quently to obtain a solution of 17.01 grams of hydroxyl to the liter (a normal solution) we should need to dis- solve 171.39/2 grams of the substance in a liter of solu- tion. A solution of this concentration is here unattain- able, as the solubility falls below the value required. In such cases various degrees of dilution are used, as will be indicated below. Solutions that furnish neither hydrogen- nor hydrox- ide-ion are considered normal when they contain, per liter, an equivalent of 1.01 grams of hydrogen or 17.01 grams of hydroxyl. This signifies that in the reactions in which they are concerned they will have, per liter, the power of combining with or displacing, either directly or indirectly, these proportional amounts of hydrogen or hydroxyl. Thus, in the action of hydrochloric acid upon sodium carbonate: Na 2 CO 3 + 2 HC1 = 2 Nad + H 2 O + C0 2 , it is observed that one gram-molecular weight of the car- bonate brings into the reaction 2.02 grams of hydrogen; consequently one-half of its gram-molecular weight (106), or 53 grams of sodium carbonate, is required in 1 liter of its normal solution. Normality Factors. For chemical purposes it is not necessary to bri^^very solution to the same standard of concentration, ^Tne normal solution. The variations from the true normality may be readily expressed by fractions, or factors, which designate at once the actual concentration of the solutions in terms of the normal. Thus a molar^solution of sulphuric acid contains ^twiee what a normal solution should contain. Its normality, therefore, is 2, and is expressed as 2 N. A solution con- taining 1 0.365 gram of hydrogen chloride per 100 c.c. would contain 3.65 grams per liter. This is 1/10 of what a 110 CHEMICAL CALCULATIONS normal solution contains; hence the solution is N/10 or 0.1 N, (deci-normal). By another method of procedure, this solution, contain- ing 0.365 gram of hydrogen chloride per 100 c.c., may be compared directly with the amount required in a liter of a normal solution, namely, 36.5 grams: 36.5: 0.365 1 : x. The factor (x) is here 0.01, hence this weight (0.365 gram) of hydrogen chloride would be contained in 1/100 of 1000 c.c., or 10 c.c. of the normal acid. As it actually occurs in 100 c.c., then the solution in question is 10/100 N or N/10; that is, 100 c.c. of this solution is necessarily equivalent to 10 c.c. of the normal solution. By reason of the equivalence between equal volumes of normal solutions we can readily calculate the normality of any solution if we are given the normality of that solution which is to be titrated against it. Example 48. 100 c.c. of N/2 sodium hydroxide solution were required in the neutralization of 400 c.c. of an unknown acid solution. Calculate the normality of this acid? Here, of course, 100 c.c. of N/2 solution is the equivalent of 50 c.c. of a normal solution; i.e., in 100 c.c. of N/2 sodium hydroxide solution we have 100/1000 or 1/10 of 17.01/2 grams, or 17.01/20 gram, of hydroxyl, which is exactly the amount contained in 1/20 of a liter (50 c.c.) of abnormal solution containing 17.01 grams per liter. In order to find the normality of the unknown solution it will be necessary to get some expression for it in terms of the known or normal solution. For example, 400 c.c. of this unknown solution neutralize 50 c.c. of the normal. By reduction, 1 c.c. neutralizes 1/8 c.c. of the normal. The relative volumes are equivalent; hence 1 liter of the unknown solution must neutralize, and possess an equiva- lent amount of substance to, that contained in one-eighth NORMAL SOLUTIONS 111 of a liter of normal sodium hydroxide solution. It is, therefore, but 1/8 N, a fact indicated at once by the factor obtained for the equivalent of that unit value 1 c.c. above. If we had taken as the unit value 1 c.c. of the known solution, we should have obtained 8 as a factor denoting the number of cubic centimeters of the unknown acid solution equivalent in value to 1 c.c. of the known. This comes to the same end and indicates the strength of 1 c.c. of the unknown solution as 1/8 that of the normal. Calculation of Normality by Simple Proportion. A very simple method for calculations of this sort rests upon the consideration of the proportionality which exists between the normality factors and the volumes for these equivalent solutions. Example 49. 400 c.c. of N/8 acid solution neutralized 100 c.c. of an unknown alkaline solution. Calculate the normality of the alkali. (jNbw the volume of a solution when multiplied by its normality factor gives, as we have seen, the 'equivalent volume in terms of its normal solution)- As all of these solutions are balanced or titrated to an end-point which signifies that equivalent quantities of the various sub- stances are present, we may at once place the two expres- sions for the two solutions as equal to each other. Thus : 400 X 1/8 = 100 X x. All of this is in exact accord with our premises which make it necessary for equal volumes to neutralize equal volumes when an equivalent amount of substance is present in each. By separating these terms of the equation into means and extremes of a simple proportion (for example by dividing through by the quantity 100 X 1/8) we obtain 400 x_ 100" 1/8' 112 CHEMICAL CALCULATIONS This is synonymous with saying that the normality factors of the two equivalent solutions stand to each other in an inverse proportion to the corresponding volumes required in the titration: 100 : 400 = 1/8 : x. By calculation x is found to be 1/2, i.e., the alkaline solu- tion is 0.5 normal. Calculation of the Weights of Substances Present, in Standard Solutions. When the normality of an unknown solution has been determined it is often desirable to find the exact amount of substance in a given volume of this solution. For example, in the last paragraph the normal- ity of the alkaline solution was found to be N/ 2. If it is now desired to learn the amount of sodium hydroxide actually present in the 100 c.c. of solution, we need only take the proportional amount of podium hydroxide in a liter of N/2 solution (20 grams) as is indicated by the fractional part which this volume is of 1000, 100/1000 or 1/10, i.e., 1/10 of 20 grams, or 2 grams. By simple proportion a comparison of the volume relations with the corresponding weights of the substances gives the follow- ing: 1000 : 100 = 20 : x. Standardization of Solutions by Gravimetric Means. For our standard solutions, it is usually customary to dissolve a certain calculated amount of substance in water and bring the volume up to the desired mark by the gradual addition of more water. As these solutions may vary slightly from the true values, it is always desirable to standardize them by purely chemical means, such as titration against certain accurately prepared solutions, or if possible by the formation of precipitates in a known volume of their solution. These precipitates when dried NORMAL SOLUTIONS 113 and weighed serve as a means for the calculation of the exact normality. Example 50. 100 c.c. of a hydrochloric acid solution; made up approximately to 1/5 N, gave, when treated with a slight excess of a silver nitrate solution, 3.22 grams of silver chloride. What is the normality of the acid? ^ From the equation: AgN0 3 + HC1 = AgCl + HN0 3 , the proportionality between all amounts of hydrogen chloride and silver chloride are indicated by the ratio: AgCl : HC1 (107.88 + 35.46) : 36.47. From this ratio the weight of hydrogen chloride, 0.8195 gram, corresponding to the weight of silver chloride, 3.22 grams, is easily calculated : AgCl : HC1 = 3.22 : x 143.34 : 36.47 = 3.22 : 0.8195. This weight of hydrogen chloride is found present in 100 c.c. of solution. In one liter we shall have 8.19 grams, whereas we should have 36.47 grams if it were a normal solution. The fraction S. 195/36. 47 represents then the normality, expressed decimally as 0.2247 N, and gives to the solution a value somewhat higher than that estimated, (N/5). Standardization of Solutions by Volumetric Means. In place of the method of precipitates another very instruct- ive method is applicable in standardization; chiefly with acids. This consists in measuring the volume of a gas evolved by the action of some substance upon a known volume of the acid solution. Example 51. 250 c.c. of an acid solution gave, when acted upon by zinc, 560 c.c. of hydrogen (calculated at standard con- ditions of temperature and pressure). What is the normality of the acid. 114 CHEMICAL CALCULATIONS 560 c.c. of hydrogen from 250 c.c. of solution would mean, of course, 2240 c.c. from one liter of the solution. By definition a normal solution of an acid is one that contains 1.01 grams of ionizable hydrogen per liter. The gram-molecular weight (2.02) of hydrogen occupies a volume of 22,400 c.c. at standard conditions. The volume occupied by 1.01 grams therefore will be just one-half this, or 11,200 c.c. Accordingly every liter of a normal acid solution must contain that weight of hydro- gen which when set free will occupy 11,200 c.c. under the standard conditions of temperature and pressure. It is only a simple step to calculate the volume of hydrogen per liter when we have given the volume for any fraction of a liter. In the problem above, 250 c.c. of solution evolved 560 c.c. of hydrogen; consequently 1 liter will evolve 2240 c.c. of this gas: 250 : 1000 = 560 : x. As a liter of normal acid should give 11,200 c.c., the solution in question is less than normal, and in accord- ance with the ratio 11,200 :2240 = 1 : x, or x = 1/5, i.e., the acid is 0.2 N. In other words, the normality is expressed by the fraction, or factor, which the volume of hydrogen evolved per liter makes with the total volume of hydrogen that can be evolved from a liter of the normal acid. If the hydrogen is measured at room temperature and pressure, it is only necessary to calculate the volume it would occupy at the standard conditions and thus make possible the comparison between this and the standard volume per liter. If this is not done, then the standard volume per liter (11,200 c.c.) must be calculated to the conditions of the experiment under which the gas is measured. NOKMAL SOLUTIONS 115 Similarly, the evolution of other gases by chemical action, from known volumes of solutions, may serve for the estimation of the normalities of these solutions. For example, a solution of sodium carbonate may be treated with an acid and the carbon dioxide set free measured. The following equation is here involved: Na 2 CO 3 + H 2 SO 4 = Na 2 S0 4 + C0 2 + H 2 O. The direct proportionality between the reaction-quan- tities concerned is expressed by the ratio C0 2 /Na 2 CO 3 or 44/106. This shows that for every gram-molecular weight of carbon dioxide (44) we must estimate the pres- ence of one gram-molecular weight of sodium carbon- ate (106). Now a normal solution of sodium carbonate contains only one-half of the gram-molecular weight in 1 liter; a fact readily determined by its titration with a normal solution of any acid. This is indicated in the equation above, wherein we note that 2.02 grams of ionizable hydrogen (2 H* + SO 4 ") are required for the complete action, and consequently only one-half of the gram-molecular weight of the carbonate, 53 grams, can be equivalent to 1.01 grams of hydrogen. The amount of carbon dioxide evolved from one gram-molecular weight of the carbonate is one gram-molecular weight, or a volume of 22,400 c.c.; hence from a normal solu- tion with one-half the gram-molecular weight of car- bonate, we should have, per liter, just 11,200 c.c. of this gas. When the volume of carbon dioxide evolved from a definite volume of solution is known we need only to calculate the volume of gas evolved, per liter, and com- pare this volume at standard conditions with the stand- ard volume, 11,200 c.c. The ratio to this value gives the ratio to unit normality. 116 CHEMICAL CALCULATIONS Example 52. What is the normality of a sodium carbonate solution, 125 c.c. of which evolved 350 c.c. of carbon dioxide (at standard conditions) when treated with an excess of acid? Since 125 c.c. of the solution gave 350 c.c. of the gas, 1000 c.c. will give 2800 c.c. as determined from the pro- portion 125 : 1000 = 350 : x. A normal solution should evolve 11,200 c.c. of this gas; consequently the solution in question is only 2800/11,200 or 1/4 N. The evolution of carbon dioxide in the equation above may serve equally well in determining the normality of the sulphuric acid used. Each liter of normal sulphuric acid will liberate 11,200 c.c. of the gas. The ratio of comparison, therefore, is carried out just as described in the previous paragraph. Comparison of Solutions with Standard. When a solution is once standardized other solutions may be standardized by comparison with it either directly or indirectly. In the case of a second acid solution it is necessary to ascertain what volumes of both this acid and our standard acid are required for the neutralization of equal volumes of some alkaline solution. These two equivalent volumes are then compared just as in the preceding examples. Example 63. 10 c.c. of 1^> N hydrochloric acid neutralized 40 c.c. of an alkaline solution. 50 c.c. of an unknown sulphuric acid solution neutralized 80 c.c. of this same alkaline solution. Calculate the normality of the sulphuric acid. 50 c.c. of the sulphuric acid neutralized 80 c.c. of the alkali; 25 c.c. of the acid would neutralize 40 c.c. of the alkali. This is the same volume of alkali neutralized by 10 c.c. of 1.5 N hydrochloric acid; hence these two volumes of the acid solutions must be equivalent: 10 c.c. X 1.5 N HC1 = 25 c.c. X (x) N H 2 S0 4 . NORMAL SOLUTIONS 117 By proportion : 25 : 10 = 1.5 : x, or x = 3/5. Hence the normality of the sulphuric acid is found to be 3/5, i.e., 0.6 N. Adjustment of Solutions to a Desired Standard. When a solution has been standardized and found to vary somewhat from the estimated normality, it is cus- tomary to calculate the amount of substance (if too dilute) or wa^er (if too concentrated) that will bring the solution to the desired normality. The latter, which con- stitutes the more simple case, is illustrated in the follow- ing example: Example 64. A solution of hydrochloric acid was found to have a normality of 1.05. What volume of water must be added to 400 c.c. of this solution to make it exactly normal? 1 c.c. of 1.05 N hydrochloric acid is equivalent by definition to 1.05 c.c. of a normal hydrochloric acid solution. Therefore, to make this more concentrated acid normal, we need only add water until 1.05 c.c. of the diluted solution will be exactly equivalent to 1.05 c.c. of a normal solution; in other words, 1.05 c.c. 1 c.c. = 0.05 c.c., or that volume of water required per cubic centimeter of the 1.05 N acid. 400 c.c. will require 400 X 0.05 c.c., or 20 c.c. Therefore, when 400 c.c. of this 1.05 N acid are diluted with 20 c.c. of water, the final 420 c.c. will be just normal. By comparison of the actual weights of hydrogen chlo- ride in equal volumes of these acids and the direct pro- portionality between these weights and the corresponding volumes which are equivalent, volume for volume, we may calculate the amount of dilution necessary to bring any 118 CHEMICAL CALCULATIONS known solution to a desired normality. Thus in 1 c.c. of normal hydrochloric acid there is 0.0365 gram of hydro- gen chloride, and in 1 c.c. of 1.05 N hydrochloric acid there is 1.05 X 0.0365 gram of hydrogen chloride. There- fore 0.0365 : (1.05 X 0.0365) = 400 : x, or x = 420. This gives the volume to which 400 c.c. of 1.05 N acid must be diluted. When a solution is found too dilute, a similar calcula- tion will give the amount of water that should be removed from the volume in question. Without resorting to this procedure, it is found better to add to the entire volume that weight of substance (usually a well-defined salt) necessary to make with this excess of water a solution of the desired normality. Any change in volume due to process of solution of the salt may be neglected. The calculation of results through reactions which in- volve solutions of definite concentration follows the general outlines presented in Chapter IX. The amount of any substance present in a required volume of a standard solution is to be considered, of course, with reference to its corresponding reaction-quantity. The method of determining the amount of a substance necessary to complete a given reaction with a known amount of some other substance contained in a definite volume of solution is called " Volumetric Analysis." When no reference is made to volume relations, but calculations are made upon the weighed quantities which come under consideration, we have the more common " Gravimetric Analysis." These two form the basis of work in quanti- tative chemical analysis. 119 PROBLEMS. 185. 400 c.c. of N/4 potassium hydroxide solution were required for the neutralization of 600 c.c. of an unknown acid solution. Calculate the normality of this acid solution. Ans. N/6. 186. 500 c.c. of N/10 acid solution were required for the neutralization of 25 c.c. of a solution of sodium hydroxide. Calculate the normality of this latter solution. Ans. 2 N. 187. 2^0 c.c. of N/20 acid solution were required for titra- tion with 1^4 c.c. of a solution of barium hydroxide. Calcu- late the normality factor of the barium hydroxide solution. Ans. .0887 N. 188. What volume of N/10 acid solution will be required in the titration of 440 c.c. of N/4 sodium hydroxide solution? Ans. 1100 c.c. NOTE: The method by simple proportion, with one of the volumes as the unknown term, will be found to serve well in such examples. 189. What volume of N/6 alkaline solution will be required in the titration of 254 c.c. of N/10 acid solution? Ans. 152.4 c.c. 190. Calculate the weight of hydrogen chloride present in 400 c.c. of a hydrochloric acid solution which required 320 c.c. of N/4 alkaline solution for titration. Ans. 2.918 grams. 191. Calculate the weight of sulphuric acid present in 150 c.c. of a solution which required 48.1 c.c. of 0.78 N alkali for titration. Ans. 1.84 grams. 192. An excess of silver nitrate solution was added to 350 c.c. of a solution of hydrochloric acid. The precipitate of silver chloride weighed 7.54 grams. Calculate the normality of the acid. Ans. 0.150%N. 193. A slight excess of barium chloride solution was added to 400 c.c. of a solution of sulphuric acid. From the weight of barium sulphate, BaSO 4 , precipitated, 4.12 grams, calculate the normality of the acid. Ans. 0.0882 N. 120 CHEMICAL CALCULATIONS 194. 600 c.c. of a sulphuric acid solution, when acted upon by an excess of zinc, evolved 1242 c.c. of hydrogen (at standard conditions). Calculate the normality of the acid. Ans. 0.1848 N. 195. 440 c.c. of an acid solution, when acted upon by an excess of zinc, evolved 2430 c.c. of hydrogen, measured over water at 21 and 747.5 mm. pressure. Calculate the nor- / mality of the acid. Ans. 0.4408 N. 196. Calculate the normality of an acid solution, 600 c.c. of which, when acted upon by an excess of sodium carbonate, evolved 2100 c.c. of carbon dioxide (calculated to standard conditions). Ans. 0.3125 N. 197. Calculate the normality of a solution of potassium carbonate, 200 c.c. of which, when treated with an excess of /acid, evolved 4502 c.c. of carbon dioxide (calculated to standard conditions). Ans. 2.01 N. I 198. An excess of iron sulphide, FeS, was added to 500 c.c. of a solution of sulphuric acid. The volume of hydrogen sul- phide set free measured 4640 c.c. (at standard conditions). Calculate the normality of the acid. Ans. 0.8285 N. 199. An excess of sodium sulphite, Na 2 S0 3 ; was added to 400 c.c. of a solution of hydrochloric acid. The volume of / sulphur dioxide set free measured 5600 c.c. (at standard con- ' ditions). Calculate the normality of the acid. Ans. 1.25 N. 200. 1400 c.c. of ammonia (calculated to standard condi- tions) were passed into 500 c.c. of N/2 hydrochloric acid solu- tion. Calculate the normality of the hydrochloric acid still present. Ans. 3/8 N. 201. 210 c.c. of carbon dioxide (at standard conditions) were passed into 250 c.c. of N/10 barium hydroxide solution. / Calculate the normality of the barium hydroxide solution still present. Ans. N/40. 202. 50 c.c. of N/5 hydrochloric acid solution neutralized 40 c.c. of an unknown alkaline solution. 300 c.c. of a sul- phuric acid solution neutralized 60 c.c. of this same alkaline solution. Calculate the normality of the sulphuric acid. Ans. 0.05 N. NORMAL SOLUTIONS 121 203. 200 c.c. of a barium hydroxide solution were required in the titration of 40 c.c. of an acid solution. 100 c.c. of this acid solution exactly neutralized 80 c.c. of N/2 alkaline solu- tion. Calculate the normality of the barium hydroxide solu- tion. Ans. 0.08 N. 204. A solution of hydrochloric acid is desired to be made exactly normal. 40 c.c. of the solution neutralized 50 c.c. 'of 0.84 N sodium hydroxide solution. Calculate the volume of water that must be added per 100 c.c. of the acid solution. Ans. 5 c.c. 205. A solution of sodium hydroxide is desired to be made exactly 0.5 N. 32 c.c. of the solution at hand were required for the titration of 28 c.c. of 0.8 N hydrochloric acid. Calcu- late the volume of water that must be added per 100 c.c. of the alkaline solution. Ans. 40 c.c. 206. A solution of sodium carbonate is desired to be made exactly 0.05 N. 24 c.c. of the solution at hand neutralized /9.6 c.c. of 0.12 N hydrochloric acid solution. Calculate the weight of anhydrous salt, Na 2 C0 3 , that must be added per 100 c.c. of solution. Ans. 0.0106 gram. 207. What weight of iron will be required for interaction with 400 c.c. of N/5 hydrochloric acid? Ans. 2.234 grams. 208. What weight of sodium carbonate will be required for interaction with 600 c.c. of N/8 sulphuric acid? What volume ' of carbon dioxide (at standard conditions) will be evolved? Ans. 3.975 grams. 840 c.c. C0 2 . 209. What weight of sodium hydrogen carbonate, NaHC0 8 , will be required for interaction with 600 c.c. of N/8 sulphuric acid? What volume of carbon dioxide (at standard conditions) will be evolved? Ans. 6.3 grams. 1680 c.c. CO a . 210. Calculate the weight of crystallized oxalic acid, /C 2 H 2 4 . 2 H 2 O, required for a solution which is to be made up to 500 c.c. in volume at N/2. Ans. 15.75 grams. 211. Calculate the volume of nitric oxide, NO (at standard conditions) that could be evolved by the action of copper upon 1000 c.c. of a 7 N nitric acid solution. Ans. 39,200 c.c. CHAPTER XI. COMBINATIONS BETWEEN GASES BY VOLUME. WHEN expressed in grams the molecular weights of all substances in the state of vapor occupy a volume of 22,400 c.c. at the standard conditions of temperature and pressure. In Chapter IX it was observed that a molecular equation offered for this reason an insight into the volume relations of the various substances concerned in a given reaction. Molecular Volumes. These volumes comply naturally with the laws relating to gases, and, further, are subject to the operation of those properties, characteristic of each substance, which may here be brought into consideration through the conditions of the experiment. Thus the con- densation of a gas or vapor to the liquid or solid state, or its solution in, or combination with, various substances which may be present, will remove it completely from further considerations. As a simple illustration of these facts an excellent ex- ample is found in the combination of hydrogen with oxygen: 2 H 2 + 2 = 2 H 2 0. By molecules: 2 +1=2. By gram-molecular volumes: 2 (22,400) c.c. + 1 (22,400)c.c. = 2 (22,400) c.c. + (22,400 c.c. contraction). 122 COMBINATIONS BETWEEN GASES BY VOLUME 123 Reduced through the common term, 22,400: 2 vol. + 1 vol. = 2 vol. + (1 vol. contr.). Each gram-molecular weight may be represented by one volume, i.e., its gram-molecular volume. The coeffi- cients or integers which designate the number of gram- molecular weights will stand likewise for the number of gram-molecular volumes involved in the reaction. The combination of gases by volume expressed thus by simple numbers has been developed experimentally and is com- prehended in Gay-Lussac's well-known Law of Combining Volumes. In the reaction above we observe that 2 volumes of hydrogen and 1 volume of oxygen unite to form 2 volumes of aqueous vapor, the temperature of 100 or above, and the observed pressure, remaining constant through- out. This diminution in the total volume of the gaseous components when transformed into aqueous vapor has been considered in the study of Avogadro's hy- pothesis. If the temperature falls below 100 a further contraction in volume occurs. Since this is due to the condensation of aqueous vapor to the liquid state, and increases accord- ingly with a lowering of temperature, there is left, eventu- ally, no volume of gaseous product. The water formed becomes associated with the liquid over which the gases are measured, and consequently drops out of further consideration through the equalization or adjustment of the levels within and without the vessel to bring all to uniform pressure. Of course a small amount of water remains in the vapor state, even at a low temperature; corrections for this are made by reference to a table of aqueous vapor tensions (Appendix II). 124 CHEMICAL CALCULATIONS The volume of the dry gaseous product, which in this case is theoretically nil, may be expressed in this manner : 2 H 2 + O 2 = 2 H 2 O. Molecular volumes: 2 +1 = 2 + (1 vol.'contr.). Molecular volumes below 100: 2 +1 = + (3 vol. contr.). The Relation of Molecular Volume to a Definite Volume- Unit. In the study of these volume-changes in known reactions it is always necessary to determine the exact relation which any measured volume of vapor, under consideration, bears to the corresponding molecular vol- ume representing it in the molecular equation governing the reaction. The coefficients of the quantities in a molecular equation determine the ratios between the volumes of these sub- stances when in the state of vapor. The coefficient unity stands, of course, for the unit of molecular volume (22,400 c.c.), a volume corresponding to a gram-molecular weight of substance. Any fractional part of the gram-molecular weight of a substance will occupy, therefore, under stand- ard conditions, a volume denoted by this same fractional part of the gram-molecular volume, 22,400 c.c. What- ever may be this fractional part of the molecular quantity of a substance under consideration in a reaction, all other substances possible of interaction with this one must be correspondingly reduced from their own molecular quan- tities. These coefficients, therefore, are constant for any known molecular equation and stand for the ratio between the volumes of the substances in state of vapor, be they gram-molecular volumes or any definite fractional parts of these. That volume which is represented by the co- efficient unity in a molecular equation is indeed the unit ) COMBINATIONS BETWEEN GASES BY VOLUME 125 upon which all of the other substances, in the state of vapor, enter into the reaction. It may be regarded as the Volume-Unit for the reaction given. If the coefficient representing a certain substance in a molecular equation is greater than unity, then, of course, from the actual volume of vapor of the substance here concerned, the true volume-unit for the equation can be derived only by dividing this known volume by the corresponding coefficient. Calculation of the Volume-Unit from a Single Known Volume. When once the volume-unit for a reaction is established, the actual volumes of the several substances here entering into combination are easily determined from the product of the several coefficients in the molecu- lar equation by the volume-unit. Example 55. What volume of oxygen will be required to burn 300 c.c. of hydrogen and what volume of aqueous vapor will result, the temperature of 100 and the atmospheric pressure remaining constant throughout? 2 H 2 + 2 = 2 H 2 0. Molecular volumes: 2 +1 = 2 + (1 vol. contr.). By placing the molecular volume coefficients out by themselves, the relative volumes of the several sub- stances are indicated. Since we can make our calcula- tions only from the volumes actually given as data, we refer the volume of hydrogen, 300 c.c., directly to the coefficient 2. From this we derive the volume 150 c.c. (300 -T- 2) as the value for the coefficient of unity, i.e., the volume-unit of the reaction. There remains now only to substitute this value for each coefficient through- out the entire molecular equation. 126 CHEMICAL CALCULATIONS 2 H 2 + O 2 = 2 H 2 0. Molecular volumes: 2+1 = 2 + (1 vol. contr.). Volumes upon the unit 150 c.c.: 300 c.c. + 150 c.c. = 300 c.c + (150 c.c. contr.). The solution of the problem is clearly seen. 150 c.c. of oxygen are required for the combustion of the 300 c.c. of hydrogen, while the volume of aqueous vapor that results will measure 300 c.c. Calculation of Volume-Unit from Known Mixtures of Gases. Since the coefficients or integers represent the ratios between the several volumes which enter into chemical combination, it must be understood that these combinations take place only in accordance with these ratios, and that, if any substance is present in amount to exceed the stipulated volume, this excess must remain unaffected in the reaction; if present in amount less than the stipulated volume, the entire reaction must run upon a somewhat smaller scale, or volume-unit, for this particular substance; with the result that the other sub- stance or substances concerned will be in excess of the corresponding volumes stipulated by the ratios, and consequently a fraction of each of these will remain unaffected. Example 56. A mixture of 250 c.c. of hydrogen and 150 c.c. of oxygen was submitted to the action of an electric spark. What was the volume of the product after the explosion, the temperature 100 and atmospheric pressure constant? 2 H 2 + 2 = 2 H 2 O. Molecular volumes: 2 +1 = 2 +(1 vol. contr.). In making hydrogen the basis for our calculations, we derive from the volume 250 c.c. the value 125 c.c. as the volume-unit. Substituting this value for unity in the equation above we obtain the following: COMBINATIONS BETWEEN GASES BY VOLUME 127 Volume 250 c.c. + 125 c.c. = 250 c.c.+ (125 c.c. contr.). relations, Volumes 250 c.c. + 150 c.c. given, Excess in f 25 c.c. oxygen, { The volume of aqueous vapor formed is 250 c.c. The volume of oxygen, however, is seen to be in excess by 25 c.c. of the volume actually required in the reaction; hence the apparent volume of product will be increased by this same amount: 250 c.c. + 25 c.c. = 275 c.c., which is the volume of gaseous mixture after the explo- sion. In place of hydrogen as the basis for this calculation of the volume-unit, we may now choose oxygen. The volume given, 150 c.c., will then correspond to a volume- unit in the equation; hence by substituting this value throughout we obtain 2 H 2 + O 2 =2 H 2 O 2+1 =2. Volume relations: 300 c.c. + 150 c.c. = 300 c.c. Volumes given: 250 c.c. + 150 c.c. Here the volume of hydrogen demanded by the equa- tion for combination with 150 c.c. of oxygen actually exceeds the amount of hydrogen at our command. Our volume-unit, therefore, has been placed too high. It is necessary 5 then, to reduce these high values until the calculated volumes of all the separate components fall equal to or under the volumes of substances actually present. Such, of course, was true in the selection of hydrogen, as above, for the basis of the calculation. 128 CHEMICAL CALCULATIONS Calculation of True Volume-Unit for a Reaction. In general, we determine a volume-unit for a reaction from each of the known volumes of substances present; this of course by dividing each volume by the corresponding coefficient which represents it in the molecular equation. Upon comparison of the volume-units thus derived, the smallest will stand for the true unit. In the example above the volume-unit upon the hydrogen basis is 125 c.c., upon the oxygen basis it is 150 c.c; consequently the value 125 c.c. alone fulfills the requirements of the reaction. Example 57. 100 c.c. of ammonia and 90 c.c. of oxygen were exploded. What was the final volume of the product? A temperature of 100 and atmospheric pressure constant. The reaction proceeds in accordance with the equation: 4 NH 3 + 3 2 = 2 N 2 + 6 H 2 0. Molecular f . {4+3=2+ 6 (1 vol. expan.). volumes : [ From the volume of ammonia: 100/4 = 25 c.c. = volume-unit. From the volume of oxygen: 90/3 = 30 c.c. = volume-unit. The smaller value, 25 c.c., must be, therefore, the volume-unit for this reaction, while the oxygen will be found slightly in excess of the required amount. Substituting the value 25 c.c. for unity throughout, we obtain 4 NH 3 + 3 2 = 2 N 2 + 6 H 2 O 4+3 =2 + 6- (1vol. expan.). Volume relations : 100 c.c. - h 75c.c.= Volumes given: Excess: 100 c.c. H -90c.c. 15 c.c. COMBINATIONS BETWEEN GASES BY VOLUME 129 Therefore the volume of product will be equal to 50 c.c. + 150 c.c., or 200 c.c., plus the excess of oxygen, 15 c.c., remaining unacted upon, or 215 c.c. in all. The determination of this volume-unit in combinations between gases is at the basis of all these calculations. In examples where the volume of only one substance is given it is often difficult to ascertain the volumes of other sub- stances that may enter into the specified reaction. Calculation of Volume-Unit from Measured Volume- Changes. Where calculations are based alone upon the volumes of gases that enter into combination, the exact proportion of each and every component of the mixture must be known; otherwise the volume-units for the several reactions cannot be determined. If one or more of these factors are unknown, then a study of the various con- tractions and expansions in the total volume of product over that of the original mixture offers a direct method for the solution. Heretofore we have concerned ourselves only with the components that entered into a reaction. The study of the products, however, offers far greater possibilities for the reason that the numerous condensa- tions and absorptions serve to estimate the volumes of the many substances possible of formation. Through these observations, and a study of the equations involved, we may deduce all of the relations between the various components that entered into a reaction. These conclusions are made possible by the conditions of a molecular equation, wherein only molecular volumes and multiples of these are concerned. An expansion or contraction in total volume of product over the original volume must be represented, therefore, by a gain or loss in a definite number of molecular volumes. This number will be denoted by the difference between the sum of the molecular volumes on one side of the equation and the 130 CHEMICAL CALCULATIONS sum of the molecular volumes on the other. If this difference corresponds to a single molecular volume, i.e., a volume with coefficient unity, then the expansion or contraction observed is in reality the exact measure of the volume-unit for the reaction; whereas, if this difference is greater or less than unity, the change in volume observed must be reduced, through division by the coefficient for this difference, when a volume is obtained corresponding to a coefficient of unity, i.e., the volume-unit. Example 58. 100 c.c. of a sample of air were mixed with 100 c.c. of hydrogen (an excess) and exploded. After the removal of the aqueous vapor by absorption, the volume of dry gaseous product read 140 c.c. Calculate the percentage of oxygen in this sample of air, the room temperature and pressure remain- ing a constant throughout. 2 H 2 + 2 = 2 H 2 0. Molecular volumes , , at 100: =2+(lvol.contr.). Molecular volumes at room tempera- 2+1=0 +(3vol.contr.). ture: If the conditions were such that all of the water remained in state of vapor (100), the contraction observed, and rep- resented by a coefficient of unity, would stand for the con- traction of 1 volume-unit in the equation, and from this value the volume of oxygen present 1 volume-unit would be found equal in value to the contraction itself. With the conditions otherwise and the loss of aqueous vapor increasing the molecular volume contraction by 2 molecular volumes (2 volume-units), we note that the observed contraction must be due to the loss of 3 volume- units from the side of the products. The original mixture, 100 c.c. + 100 c.c., or 200 c.c., lost 60 c.c. in this reaction (200 c.c. 140 c.c.); hence 60 c.c. represents the 3 volume- COMBINATIONS BETWEEN GASES BY VOLUME 181 units, or 1 volume-unit will be equal to 20 c.c. From the equation, the oxygen consumed is exactly 1 volume-unit and hence 20 c.c. This indicates that 20 per cent of the sample of air was oxygen. Example 69. A volume of ammonia was mixed with a large excess of oxygen and exploded. The expansion in volume of product over the original mixture was 50 c.c. What was the volume of ammonja? Temperature of 100 and atmospheric pressure a constant. The equation for this action has been given before : 4 NH 3 + 3 2 = 2 N 2 + 6 H 2 O. Molecular f 4 +3 = 2+ 6. volumes: } 7 vol. = 8 vol. (1 vol. expan.). In this case the expansion is seen to be exactly equal to a unit coefficient in volume; consequently 50 c.c. is the volume-unit for the reaction. By substituting this value in the molecular equation, the volume of ammonia (4 vol- ume-units), which in its combustion can produce this expansion of 50 c.c., is found equal to 4 X 50 c.c. or 200 c.c. From a study of the volume of products, and the alter- ations in this volume through elimination of certain of the substances, we are able to determine the actual com- position of unknown mixtures of gases. Example 60. 200 c.c. of a mixture of nitrogen and methane were exploded with 400 c.c. of oxygen. The volume of the dry gaseous product measured 500 c.c. What was the percentage of methane in the original mixture? The room temperature and pressure a constant. CH 4 + 20 2 = C0 2 + 2 H 2 0. Molecular volumes : 1+2=1+ 2 Molecular volumes, aqueous vapor 1 +2 =1 + (2 vol. contr.). removed: 132 CHEMICAL CALCULATIONS Here no change in volume results from the explosion when the temperature is 100 or over. When the temperature falls and the aqueous vapor is condensed we have a new relation between the volumes of product and original mixture such that a contraction corresponding to 2 volume- units occurs. In the example the actual contraction due to this removal of the water is 100 c.c., (200 +400 - 500). From this we derive the value of a single volume-unit as equal to 50 c.c. By reference then to the volume equation above, 1 volume-unit, i.e., 50 c.c., represents the amount of methane present in the original mixture, of which it constitutes 25 per cent (by volume). The nitrogen is considered as without action. In practice the removal of aqueous vapor is usually not carried out. Since at room temperatures the greater part of this vapor condenses to the liquid state, the partial pressure of this vapor at the observed conditions gives at once a means of calculating the volume of dry gaseous product. Some definite conditions of temperature and pressure are of course taken as a basis. In this manner a direct comparison between the volume of product and that of the original mixture can be readily made. Example 61. A mixture of 200 c.c. of ethylene and 800 c.c. of oxygen (an excess), at 24 and 756.5 mm. pressure, was exploded. After the explosion the volume of product read 649.1 c.c.; at 27 and this same pressure. Calculate the per- centage purity of the ethylene by volume. The reaction takes place according to the equation : C 2 H 4 + 3 2 = 2 C0 2 + 2 H 2 0. Molecular volumes: 1+3 = 2+ 2 Molecular volumes, aqueous vapor removed: 1+3 = 2 + (2 vol.contr.) COMBINATIONS BETWEEN GASES BY VOLUME 133 When allowance is made in this final volume of product for the presence of aqueous vapor, which at 27 has a tension of 26.5 mm., we have only to calculate the volume which the gaseous product will occupy at the original con- ditions, 24 and 756.5 mm. pressure, the basis thus selected for comparison. The expression is V = 620 c.c. The original mixture, 200 c.c. + 800 c.c., or 1000 c.c., suffered therefore a contraction of 380 c.c. (1000 c.c. 620 c.c.). This contraction, due to the removal of the aqueous vapor from the side of the products, corresponds to 2 volume-units; consequently 1 volume-unit for this reaction is equal to 190 c.c., and the volume of ethylene concerned (1 volume-unit) is also 190 c.c. We may as- sume, accordingly, the presence of 10 c.c. of some inert gas (e.g. nitrogen) in the original volume (200 c.c.) of ethylene taken. This signifies that the sample of ethylene was only 95 per cent pure. Considerations into which Different Volume-Units may Enter. When a contraction arises from the combined effect of two or more reactions it may be impossible to derive any one of the volume-units, for, unless the rela- tive volumes of the several gases going to produce the contraction is known, we cannot properly apportion this volume of contraction between the several equations. In other words, the volume-unit may be different for every equation that is brought into consideration. In such cases as these it is necessary to bring the final products under new conditions, whereby further conden- sations or absorptions can take place, and the possibility of involving separately only one product from one of the 134 CHEMICAL CALCULATIONS reactions at a time made likely. An individual reaction thus concerned yields itself at once to the determination of its specific volume-unit. By reference of this volume- unit to the several reactions of a problem it is often pos- sible to determine other volume-units. Still further con- densations or absorptions through new conditions may be necessary, however, to aid in determining the volume-units of certain equations when a large number of substances are present in the original mixture. When all of these values are found the solution of the problem is compara- tively simple. Example 62. A mixture of nitrogen, hydrogen and carbon monoxide, 450 c.c. in volume, was exploded with an excess of oxygen, 250 c.c. After the explosion the volume of gaseous product measured 500 c.c. With the removal of the aqueous vapor the volume of product measured 400 c.c. What was the volume of each component in the original mixture? A tem- perature of 100 and atmospheric pressure considered constant. The two equations are as follows: 2 H 2 + 2 = 2 H 2 0. Molecular volumes: 2 +1=2 + (1vol. contr.). 2 CO + 2 = 2 CO 2 . Molecular volumes: 2 +1=2 + (1 vol. contr.). In this problem the dry gaseous product measured 400 c.c., i.e., by loss of the aqueous vapor, a contraction of 100 c.c. was recorded. This contraction is due alone to the reaction of hydrogen with oxygen, and stands for the loss of 2 volume-units of aqueous vapor in the first volume equation. Therefore 1 volume-unit in this equation is equivalent to 50 c.c., and 2 volume-units, representing the actual amount of hydrogen concerned in the reaction, correspond to 100 c.c. COMBINATIONS BETWEEN GASES BY VOLUME 135 In the combustion of hydrogen the contraction actually possible corresponds to 1 volume-unit of the reaction. This volume-unit has just been determined as 50 c.c. The total contraction in the problem, due to the two reactions and without elimination of aqueous vapor, is recorded as 200 c.c.; hence the difference, 200 c.c. 50 c.c., or 150 c.c., represents the contraction due to the reaction between carbon monoxide and oxygen. This contraction, as shown in the second volume equation above, is equiva- lent to 1 volume-unit. Accordingly the volume of the carbon monoxide present 2 volume-units must have been 2 X 150 c.c. or- 300 c.c. All told, we derive the following composition for the original mixture: 100 c.c. of hydrogen, 300 c.c. of carbon monoxide and 50 c.c. of nitrogen (considered in these problems as an inert gas). The Algebraic Method for the Calculation of Volume- Units. More complicated examples may be given, but throughout all the same principles hold. We must first determine the volume-unit for each of the reactions under consideration before we can calculate the several volumes concerned in the problem. When a change in volume of product is found to result, not from one single reaction, but from a combination of several, and when also, through new conditions presented, still further condensations occur, which in turn are found to be due to a combination of reactions, we find it necessary to give the volume-unit for each reaction an algebraic symbol, and solve for the value of each unit from the several equations that may be constructed. Example 63. A mixture of nitrogen; carbon monoxide, methane and ammonia, amounting to 1250 c.c., was exploded with an excess of oxygen (1100 c.c.), and the volume of product found to measure 2450 c.c. A temperature of 100 and 760 mm. constant throughout. After withdrawal of aqueous vapor by 136 CHEMICAL CALCULATIONS absorption, the volume of the dry gases at the recorded constant conditions was 950 c.c. After removal of carbon dioxide (by passing gases over lime) the volume read 600 c.c. What was the volume of each component in the original mixture? The equations for the three reactions are as follows: 2 CO + 2 = 2 CO 2 Molecular volumes : 2 +1 = 2 + (1 vol. contr.). CH 4 + 20 2 =C0 2 + 2H 2 O Molecular volumes : 1 +2 =1 +2. 4NH 3 +3O 2 =2N 2 + 6H 2 Molecular volumes: 4 +3 =2 + 6 (1 vol. expan.). An expansion follows the combustion of ammonia with oxygen. A contraction due to the combustion of carbon monoxide is not sufficient to make up for expan- sion in the former case, since, the final volume, 2450 c.c., is 100 c.c. larger than the volume of the original mixture of gases. As the contraction and expansion are each represented by a volume-unit in their respective reac- tions, we draw the conclusion that the volume-unit in the ammonia reaction is larger by 100 c.c. than the volume- unit in the carbon monoxide reaction. The withdrawal of aqueous vapor is similarly distributed over two reac- tions, as is also the withdrawal of carbon dioxide. The data from this problem do not yield positive information in regard to any single volume-unit of any reaction. Combinations between the reactions are of course easily intelligible. Thus the withdrawal of the water removes 6 volume-units from the ammonia reac- tion and 2 ^ volume-units from the methane reaction. With all the volume-units unknown we may profitably study the combination of these units with the idea of deriving some one of them and eventually all. For this COMBINATIONS BETWEEN GASES BY VOLUME 137 purpose let x represent the volume-unit of the carbon monoxide reaction, y the volume-unit of the methane reaction, and z the 'volume-unit of the ammonia reac- tion. From the preceding remarks and a study of the reactions themselves we may now draw up the following equations. The expansion is due to the larger value of z compared with x: (a) z - x = 100. The volume of aqueous vapor is represented by 1500 c.c. (2450 c.c. 950 c.c.) ; and is therefore expressed by (b) 6 z + 2 y = 1500. The volume of ca'rbon dioxide corresponds to 2 volume- units of x and 1 volume-unit of y and is represented by 350 c.c. (950 c.c. - 600 c.c.). (c) y + 2 x = 350. By combining equations (6) and (c) to eliminate y } (b) 6 z + 2 y = 1500. Twice (c) or 4 x + 2 y = 700, and subtracting 6 z 4 x = 800, and subtracting ' four times (a) or 4 z 4 x = 400 2~T~ = 400 z = 200. With the volume-unit of the ammonia equation thus derived and equal to' 200 c.c., we have only to derive the value of x from equation (a) as equal to 100 c.c. and then in turn from (b) we derive the value of y as equal to 150 c.c. Substituting these values in their proper equations the volume of carbon monoxide (2 x) is found to be 200 c.c., that of methane (y) 150 c.c., and that of ammonia (4 z) 800 c.c!, making in all 1150 c.c. Therefore the remaining 138 CHEMICAL CALCULATIONS volume of the original 1250 c.c., or 100 c.c., is calculated as nitrogen. Considerations into which the Formation of Non-gase- ous Substances Enter. In the study of those molecular equations in which a number of non-gaseous substances come under consideration we have simply another case of removal of molecular quantities. Example 64. When phosphorus is burned in a vessel con- taining 100 c.c. of nitrous oxide, what will be the volume of nitrogen left? The room temperature and pressure constant throughout. The solid phosphorus pentoxide as well as the phos- phorus fall out of consideration in the volume equation: 5 N 2 O + P 2 = 5 N 2 + P 2 5 . 5 N 2 = 5 N 2 . Molecular volumes: 5 =5. There is, therefore, no change in volume and 100 c.c. will represent also the volume of the nitrogen. Again we may consider the reaction of hydrogen chloride upon a carbonate: Example 65. An excess of sodium hydrogen carbonate was placed in a vessel containing 100 c.c. of hydrogen chloride. Calculate the volume of dry carbon dioxide liberated in the reaction. Room temperature and pressure a constant. NaHC0 3 + HC1 = NaCl + H 2 + C0 2 . Molecular volumes : 1 = 1+1. We have in this volume equation 1 volume of hydrogen chloride liberating, from 1 molecule of sodium hydrogen carbonate, 1 volume of carbon dioxide and 1 volume of aqueous vapor. In the dry state, therefore, the volume COMBINATIONS BETWEEN GASES BY VOLUME 139 relations of carbon dioxide and hydrogen chloride are as 1:1. In the case of the normal sodium carbonate the volume relations of these dry gases are as 1 : 2. Na 2 C0 3 + 2 HC1 = 2 NaCl + H 2 + C0 3 . 2 = 1. The proportional values of carbon dioxide and hydrogen chloride are different in the two cases. This, however, is dependent upon the nature of the substances concerned and is easily explainable by a study of the two reactions. Whatever be the complexity of the examples here pre- sented, they nevertheless may be made to conform to very simple interpretations when once the molecular equations for the reactions are constructed and a study of the volume relations undertaken. PROBLEMS. 212. In the combustion of 400 c.c. of hydrogen with oxygen, what volume of oxygen will be required, and what volume of aqueous vapor will result? A temperature of 100 and a pres- sure of 760 mm. constant throughout. Ans. 200 c.c. oxygen. 400 c.c. vapor. 213. A mixture of 300 c.c. of hydrogen and 200 c.c. of oxygen was exploded by electric spark. What was the volume of product? A temperature of 100 and pressure of 760 mm. constant throughout. Ans. 350 c.c. 214. A mixture of 300 c.c. of hydrogen and 130 c.c. of oxy- gen was exploded. What was the volume of product? Tem- perature of 100 and pressure of 760 mm. constant throughout. Ans. 300 c.c. 215. A mixture of 420 c.c. of hydrogen and 180 c.c. of oxygen was exploded. What was the volume of product after the removal of aqueous vapor (by absorption with phosphorus pentoxide)? Temperature of 100 and pressure of 760 mm. a constant. Ans. 60 c.c. 140 CHEMICAL CALCULATIONS 316. A mixture of 300 c.c. of methane and 150 c.c. of oxy- gen was exploded. What was the volume of the product after the removal of aqueous vapor (by absorption) ? Temperature of 100 and pressure of 760 mm. constant throughout. Ans. 300 c.c. 217. A mixture of 250 c.c. of carbon monoxide and 120 c.c. of oxygen was exploded. Calculate the volume of gaseous product. Temperature and pressure a constant. Ans. 250 c.c. 218. A mixture of 320 c.c. of carbon monoxide and 180 c.c. of oxygen was exploded. Calculate the volume of gaseous product after the removal of the carbon dioxide (absorption by lime). Temperature and pressure a constant. Ans. 20 c.c. 219. A mixture of 200 c.c. of methane and 300 c.c. of carbon monoxide was exploded with 600 c.c. of oxygen (excess). Cal- culate the volume of gaseous product. Calculate, also, the volume of product when deprived of aqueous vapor. Tem- perature of 100 and pressure of 760 mm. a constant throughout. Ans. 950 c.c. (moist). 550 c.c. (dry). NOTE. Since the oxygen is in excess we know the calculation is to be based upon the CH 4 and CO. The oxygen left unconsumed in these two reactions is readily determined by difference. 220. A mixture of 80 c.c. of methane and 200 c.c. of oxygen was exploded over mercury. Calculate the volume of product. A temperature of 20 and barometric pressure of 757.4 mm. constant throughout. ^Lns. 122.83 c.c. NOTE. The volume of dry gaseous product, at 20 and 757.4 mm., is simply to be changed to accord with the observed conditions in the presence of aqueous vapor. These experiments are usually carried out over mercury. The excess of water condenses of course upon the mercury column. If an appreciable quantity were present allow- ance for its pressure upon the column would need to be made. 221. A mixture of 400 c.c. of hydrogen and 300 c.c. of oxygen was exploded. Calculate the volume of product. A temperature of 25 and pressure of 753.6 mm. constant. Ans. 103.23 c.c. COMBINATIONS BETWEEN GASES BY VOLUME 141 222. A mixture of 200 c.c. of carbon monoxide and 300 c.c. of oxygen was exploded over mercury. Calculate the volume of product. A temperature of 27 and pressure of 740 mm. constant throughout. Am. 400 c.c. 223. A mixture of oxygen and hydrogen, measuring 150 c.c. in volume, was exploded by means of an electric spark. The volume of product measured 125 c.c. Temperature of 100 and atmospheric pressure a constant. Calculate the volumes of hydrogen and oxygen concerned in the reaction. The resid- ual gas is of course the excess of either the hydrogen or oxygen present. Ans. 50 c.c. hydrogen. 25 c.c. oxygen. 224:. A mixture of oxygen and hydrogen, measuring 150 c.c. in volume, was exploded. The volume of product measured 76.47 c.c. Temperature of 17 and pressure of 754.4 mm. a constant. Calculate the volumes of hydrogen and oxygen con- cerned in the reaction. Ans. 50 c.c. hydrogen. 25 c.c. oxygen. 225. 250 CoCo of dry air were mixed with 150 c.c. of hydrogen (an excess) and exploded. The volume of product was 350 c.c. Calculate the percentage of oxygen (by volume) in the sample of air. Temperature of 100 and atmospheric pressure a con- stant. Ans. 20 per cent. 226. 300 c.c. of dry air were mixed with 250 c.c. of hydrogen (an excess) and exploded. The volume of the dry gaseous product was 361 c.c. Calculate the percentage of oxygen (by volume) in the sample of air. Temperature of 20 and pressure of 745 mm. a constant. Ans. 21 per cent. 227. A mixture of hydrogen sulphide with an excess of oxygen measured 350 c.c. at 100 and 750 mm. pressure. After explosion (with complete combustion) the volume of dry gase- ous product read 260 c.c. at these same conditions. Calculate the volume of hydrogen sulphide in the mixture. Ans. 60 c.c. 228. A mixture of hydrogen sulphide with an excess of oxygen measured 200 c.c. at 20 and 747.4 mm. pressure. After explosion (with complete combustion) the volume of product read 143.3 c.c. at these same conditions. Calculate the volume of hydrogen sulphide in the mixture. Ans. 40 c.c. 142 CHEMICAL CALCULATIONS 229. A mixture of acetylene, C 2 H 2 , with an excess of oxygen measured 350 c.c. at 25 and 745 mm. pressure. After explo- sion the volume of dry gaseous product read 275 c.c. at the same conditions. Calculate the volume of acetylene in the mixture. Am. 50 c.c. 330. A mixture of acetylene, C 2 H 2 , with an excess of oxygen measured 240 c.c. at 24 and 752.4 mm. pressure. After ex- plosion the volume of product read 221.8 c.c. at 28 and 750.1 mm. pressure. Calculate the volume of acetylene in the mix- ture. Ans. 20 c.c. 331. An excess of oxygen, 400 c.c., was admitted into a volume of ammonia (containing an impurity of air, i.e., nitro- gen and oxygen) which measured 240 c.c., and the mixture, at 22 and 745.2 mm. pressure, exploded. The volume of product measured 415.1 c.c. at 24 and the same pressure. Calculate the percentage purity of the sample of ammonia. Ans. 80 per cent. 232. Into a stoppered tube (90 c.c. in capacity) filled with chlorine, a small quantity of concentrated ammonia water (10 c.c. or an excess) was admitted. After shaking, the mouth of the tube was opened under a dilute acid solution contained in a tall cylinder. The excess of ammonia, together with the hydrogen chloride, formed in the reaction, were thus removed. The residual volume of gas may be considered as nitrogen. Calculate the volume of nitrogen in the tube. The room tem- perature and pressure considered a constant throughout. Ans. 30 c.c. 233. A mixture of carbon monoxide and methane contain- ing also nitrogen measured 260 c.c. Into this mixture was admitted an excess of oxygen, 300 c.c., and the whole exploded. After the explosion the volume of product measured 520 c.c., but after the removal of aqueous vapor (by absorption) from the gaseous product the volume measured 320 c.c. Calculate the volume of each component in the original mixture. A tem- perature of 100 and atmospheric pressure constant. Ans. 80 c.c. CO. 100 c.c. CH 4 . 80 c.c. N. COMBINATIONS BETWEEN GASES BY VOLUME 143 234. A sample of water gas containing an impurity of air measured 240 c.c. in volume. Into this was admitted an excess of oxygen, 400 c.c., and the mixture exploded. The volume of product measured 540 c.c. After removal of aque- ous vapor (by absorption) the volume of product measured 440 c.c. Calculate the percentage of hydrogen and carbon monoxide in the sample. A temperature of 100 and atmos- pheric pressure constant throughout. Ans. 41.25 per cent H. 41.25 per cent CO. 235. 500 c.c. of a mixture of ammonia, methane and nitrogen were mixed with 500 c.c. of oxygen and the mixture exploded. The volume of the product measured 1050 c.c. A constant temperature of 100 and pressure of 760 mm. maintained throughout. Upon cooling the gaseous product to remove the aqueous vapor and recalculation of the volume of dry gas to the original conditions (100), the volume measured 550 c.c. What was the composition of the original mixture? Ans. 100 c.c. CH 4 . 200 c.c. NH 3 . 200 c.c. N. 236. A mixture of methane, hydrogen sulphide and air measured 300 c.c. When mixed with an excess of oxygen, 500 c.c., and exploded the volume of product measured 720 c.c. A temperature of 100 and atmospheric pressure constant throughout. Upon cooling the gaseous product to the room temperature and recalculation of the volume of dry gas to the original conditions, the volume measured 320 c.c. Calculate the composition of the original mixture. Ans. 120 c.c. CH 4 . 160 c.c. H 2 S. 20 c.c. air. 237. A mixture of methane and acetylene measuring 500 c.c. was mixed with 1500 c.c. of oxygen (an excess) and ex- ploded. The volume of product measured 1850 c.c. A tem- perature of 100 and atmospheric pressure constant. Upon removal of aqueous vapor (by condensation and recalcula- tion to the original conditions) the volume measured 1150 c.c. Calculate the composition of the original mixture. Ans. 200 c.c. CH 4 . 300 c.c. C,H 2 . 144 CHEMICAL CALCULATIONS 238. A mixture of hydrogen, methane and nitrogen measur- ing 350 c.c. was exploded with an excess of oxygen, 500 c.c. After the explosion the volume of dry gaseous product meas- ured 475 c.c. Upon removal of carbon dioxide the volume of gas measured 400 c.c. Calculate the composition of the original mixture. Room temperature and pressure a constant. Ans. 150 c.c. H. 75 c.c. CH 4 . 125 c.c. N. 239. A mixture of cyanogen, methane and nitrogen meas- uring 200 c.c. in volume was exploded with an excess of oxygen, 500 c.c. After the explosion the volume of product meas- ured 619.6 c.c. Upon removal of carbon dioxide (by lime) the volume of dry gas measured 430 c.c. Calculate the com- position of the original mixture. A temperature of 25 and pressure of 746.6 mm. constant throughout. Ans. 50 c.c. CH 4 . 60 c.c. C 2 N 2 . 90 c.c. N. 240. A liter of methane contaminated with the vapor of carbon disulphide was mixed with an excess of oxygen, 3000 c.c., and exploded. The volume of product measured 3840 c.c. Calculate the percentage impurity of the methane. A tempera- ture of 100 and atmospheric pressure constant throughout. Ans. 16 per cent. 241. An excess of oxygen was admitted to a vessel contain- ing hydrogen and ammonia and the mixture exploded. After the explosion there was observed no change in volume of product from that of the mixture. Calculate the percentage composition of the hydrogen and ammonia mixture. A tem- perature of 100 and atmospheric pressure constant. Ans. 33^ per cent H. 66f per cent NH 3 . 242. A mixture of ethylene and ammonia, contaminated with air measured 600 c.c. When this volume of gas was mixed with an excess of oxygen, 2000 c.c., and exploded, the volume of product measured 2640 c.c. Upon removal of aqueous vapor (by cooling and recalculation) the volume of dry gas COMBINATIONS BETWEEN GASES BY VOLUME 145 measured 2120 c.c. Calculate the composition of the original mixture. A temperature of 100 and atmospheric pressure constant. Am. 160 c.c. NH 3 . 140 c.c. C 2 H 4 . 300 c.c. air. 343. A mixture of ethylene and ammonia contaminated with air measured 400 c.c. When this volume of gas was mixed with an excess of oxygen, 1200 c.c., and exploded, the volume of product measured 1293.6 c.c. Upon removal of carbon dioxide (by lime) the volume of dry gas measured 1050 c.c. Calculate the composition of the original mixture. A temperature of 26 and pressure of 745.6 mm. constant throughout. Ans. 100 c.c. C 2 H 4 . 120 c.c. NH 3 . 180 c.c. air. 244. Into a mixture of cyanogen, ethylene and nitrogen was admitted a large excess of oxygen and the mixture, which measured 1600 c.c. at 22 and 753 mm. pressure, exploded. After the explosion the volume of product measured 1257 c.c. at 24 and 746 mm. pressure. Upon removal of carbon dioxide and moisture the volume of dry gas (still at 24 and 746 mm.) measured 609.7 c.c. Calculate the amount of cyanogen and ethylene in the original mixture. Ans. 100 c.c. C 2 N 2 . 200 c.c. C 2 H 4 . 245. A mixture of hydrogen and ammonia contaminated with air measured 400 c.c. Into this was admitted an excess of oxygen, 600 c.c., and the final mixture exploded. After the explosion the volume of product measured 1025 c.c. Upon removal of aqueous vapor (by absorption) the volume read 675 c.c. Calculate the composition of the original mixture. A temperature of 100 and atmospheric pressure constant through- out. (Cf. Ex. 63.) Ans. 50 c.c. H. 200 c.c. NH 3 . 150 c.c. air. 346. A mixture of hydrogen, methane, carbon monoxide and nitrogen measured 500 c.c. Into this was admitted an excess of oxygen, 900 c.c., and the final mixture exploded. After the explosion the volume of product measured 1250 c.c. Upon removal of aqueous vapor (by absorption) the volume measured 1000 c.c. Upon removal of carbon dioxide (by 146 CHEMICAL CALCULATION^ lime) the volume measured 725 c.c. Calculate the composition of the original mixture. A temperature of 100 and atmos- pheric pressure constant. (Cf. Ex. 63.) Ans. 100 c.c. H. 200 c.c. CO. 75 c.c. CH 4 . 125 c.c. N. 247. A mixture of carbon monoxide and methane contami- nated with air and measuring 400 c.c. was exploded with an excess of oxygen, 1000 c.c. The volume of product measured 1025.4 c.c. Upon removal of the carbon dioxide (by lime) the volume of dry gas measured 725 c.c. Calculate the composi- tion of the original mixture. A temperature of 21 and pres- sure of 745.5 mm. constant throughout. Ans. 100 c.c. CO. (Cf. Ex. 63.) 175 c.c. CH 4 . 125 c.c. air. 348. A mixture of cyanogen, methane and hydrogen con- taminated with air measured 800 c.c. An excess of oxygen, 2000 c.c., was admitted into the vessel containing this mixture and the final mixture exploded. After the explosion the vol- ume of product measured 2750 c.c. After removal of aqueous vapor (by cooling and recalculation) the volume measured 2250 c.c. After removal of carbon dioxide (by lime) the final vol- ume measured 1450 c.c. Calculate the composition of the original mixture. A temperature of 100 and atmospheric pressure constant. Ans. 100 c.c. H. (Cf. Ex. 63.) 200 c.c. CH 4 . 300 c.c. C 2 N,. 100 c.c. air. 249. An excess of oxygen, 800 c.c., was added to a mixture of ammonia, ethylene and hydrogen, 310 c.c. in volume, and the final mixture exploded. After the explosion the volume of product measured 648.35 c.c. Upon removal of carbon dioxide (by lime) the volume of dry gas measured 435 c.c. Calculate the composition of the original mixture. A tempera- ture of 18 and pressure of 746 mm. constant throughout. (Cf. Ex. 63.) Ana. 50 c.c. H. 100 c.c. C 2 H 4 . 160 c.c. NH 3 . NOTE. The contraction due to the reactions is here masked in the contraction from the loss of aqueous vapor. Construct the COMBINATIONS BETWEEN GASES BY VOLUME 147 third equation from the volume of oxygen required in the several combustions, bearing in mind that the volume, 365 c.c. (800 c.c. 435 c.c.), which nominally represents this quantity in the problem, is in fact less than the actual volume by that volume of nitrogen left free. 250. A mixture of methane and ethylene measuring 300 c.c. was exploded with an excess of oxygen, 1500 c.c. After the explosion the volume of dry gaseous product (calculated to observed conditions of temperature and pressure) measured 1200 c.c. In order to determine the volume of oxygen actually consumed in this combustion and thus present data upon which a second equation may be constructed, an excess of hydrogen, 900 c.c., was admitted to the vessel containing the dry gaseous product. After an explosion the volume of this second product measured 1400 c.c. (calculated as above, with absence of aqueous vapor). Calculate the amount of methane and ethyl- ene in the original mixture. The room temperature and pres- sure remained constant throughout. Ans. 200 c.c. CH 4 . (Cf. Ex. 63.) 100 c.c. C 2 H 4 . 251. A volume of nitric oxide, NO, measuring 400 c.c. was required for the combustion of a definite weight of phosphorus. What volume of nitrogen remained free? The room tempera- ture and pressure a constant. Ans. 200 c.c. 252. A quantity of phosphorus was burned in a vessel over water holding 222.7 c.c. of nitrous oxide, N 2 O, meas- ured at 20 and 750 mm. pressure. What volume of nitrogen remained, the temperature and pressure constant? Ans. 222.7 c.c. 253. What volume of water gas is theoretically possible from the action of 1 liter of steam upon heated coke? Tempera- ture and pressure a constant. Ans. 2000 c.c. 254. In the decomposition of methane by chlorine what volume of hydrogen chloride corresponds to 1 volume of meth- ane? Temperature and pressure constant. Ans. 4 volumes. 255. A mixture of 400 c.c. of methane and 1000 c.c. of chlo- rine was exploded. Calculate the volume of gaseous product, temperature and pressure a constant. Ans. 1800 c.c. 256. What volume of gaseous product may be obtained in the decomposition of 100 c.c. of ammonia by heated cupric oxide? Temperature and pressure a constant. Ans. 200 c.c. (Steam + N). CHAPTER XII. COMPLEX EQUATIONS. THE study of chemical reactions as outlined in Chapter IX was not inclusive of those examples where a change in the valence of an element occurs. Although these changes may complicate the matter of drawing up the equations, the same principles as already discussed will be found to apply. The definite quantity of each element concerned, whether OT not it exerts the same combin- ing capacity for other elements on the two sides of an equation, is nevertheless a constant for the particular equation in question. The exact application of these principles will rest, of course, upon the construction of the equations in their molecular form. Oxidation and Reduction. The increase in valence or combining capacity* which accompanies an element when, for instance, it passes from a lower to a higher oxide, has been taken as the measure of the degree of oxidation. The position of oxygen in this connection might be filled by almost any non-metallic element and still we may have the means of measuring the increase in com- bining capacity of a metallic element or radical acting as such toward a non-metallic element, i.e., oxidation. The unit of measure, rather than oxygen with its 2 com- bining capacities, is referred to an element with but 1 * The author is indebted to his colleague, Professor S. Laurence Bigelow, for this interpretation of valence as the capacity factor of chemical energy, i.e., the combining capacity of a unit quantity of an element or radical. 148 COMPLEX EQUATIONS 149 capacity, e.g. hydrogen. Thus 1 molecule of chromic oxide, Cr 2 III O 3 ", undergoes oxidation to 2 molecules of chromium trioxide, 2 Cr VI 3 ", and the oxygen intake for this definite quantity of chromium is 3 oxygen atoms, i.e., the equivalent of 6 hydrogen atoms in combining capacity, or 6 combining capacities. This amount for 2 atoms of chromium is equivalent to 3 combining capac- ities for 1 atom of chromium and thus this increase will be denoted in the second compound Cr VI O 3 n . The reverse process, or a decrease in this combining capacity toward a non-metallic element, is known as reduction. For example, ferric iron as Fe 111 ^ 1 is reduced to ferrous iron, Fe 11 ^ 1 . Analogous to this the increase in combining capacity of a non-metallic element, or a radical acting as such, for oxygen or some other non-metal will be a measure of the oxidation, e.g. CyO" -> C1 2 VII O 7 ". When, however, this capacity of a non-metal is considered in relation to its combination with metallic groups the increase in its com- bining capacity is in fact a measure of its reduction. Thus potassium ferricyanide, K 3 I [Fe m (CN) fl I ] m , is reduced to potassium ferrocyanide, K 4 I [Fe"(CN) 6 I ] IV , when the ferri- cyanide radical increases its combining capacity toward the metallic element potassium. This. result is of course due to the reduction of the ferric iron in the former radical to ferrous iron in the latter. In general the increase in the combining capacity of a non-metallic radical for a metallic one signifies a demand" for a lower number of these capacities required of an intra-radical metal for complete combination, or what was previously interpreted as a reduction. Reaction-Quantities Involved in Oxidation and Reduc- tion. The change of an element from the non-metallic to the metallic groupings, or vice versa, may be accom- 150 CHEMICAL CALCULATIONS panied by a loss or gain in its combining capacity, and consequently we speak of it as undergoing a reduction or oxidation. Thus potassium dichromate, K 2 Cr 2 7 , con- tains the chromium associated directly with oxygen in the form of the acid H 2 CrO 4 (or oxide Cr VI 3 "), but by the action of a strong acid this chromium, with a combining capacity of 6, is completely transformed into a chromic compound (where chromium is the basic element) and then has a combining capacity of only 3, Cr 2 in 3 n . The action therefore is described as one of reduction. In illustration of these points a few of the more com- mon examples will be given. First we may mention the action of hydrogen sulphide upon a solution of ferric chloride: 2 FeCl 3 + H 2 S = 2 FeCl 2 + 2 HC1 + S. Expressed by ions: 2 Fe"* + 6 CV + 2 H' + S" = 2 Fe" + 4 CV +2 H* + 2 Cl' + S, where the combining capacities are represented in solu- tion as ionic charges. From the reaction-quantities noted we may draw up the several ratios of proportionality here existent. For example the amount of ferrous chloride resulting from a definite weight of ferric chloride is expressed by the ratio FeCl 2 /FeCl 3 , and then again the weight (as well as the volume eventually) of hydrogen sulphide necessary for this reduction of the ferric chloride is in accord with the ratio H 2 S/2 FeCl 3 . The hydrogen sulphide molecule is equivalent therefore in its reducing action to 2 units of hydrogen, whereas 2 molecules of ferric chloride are equivalent in oxidizing action to 1 oxygen atom. Oxidizing Action of Potassium Dichromate. As has been stated, the oxidizing action of potassium dichro- COMPLEX EQUATIONS 151 mate, K 2 O 2 O 7 , in the presence of a strong acid and sub- stance capable of oxidation, is due to the transformation of the dichromate (Cr VI ) into a chromic compound (Cr 111 ). The oxidizing power of this substance is measured by the liberation of 3 atoms of oxygen per molecule of dichromate, as may be shown by the following hypotheti- cal equation for the action with sulphuric acid: K 2 Cr 2 7 + 4 H 2 S0 4 = K 2 SO 4 + O 2 (S0 4 ) 3 + 4 H 2 + 3 O. When an active reducing agent, e.g. ammonium sul- phide, is present the reduction of the dichromate may be accomplished without the presence of an acid: K 2 Cr 2 7 + 3 (NH 4 ) 2 S + 7 H 2 O = 2 Cr(OH) 3 + 2 KOH + 6 NH 4 OH + 3 S. In other cases some oxidizable substance must be added to the acid mixture in order that the oxygen capable of liberation may be actually consumed and the reaction proceed. A number of substances, such as hydrogen sulphide, sulphur dioxide, carbon monoxide and alcohol, are readily oxidized in this manner to sulphur, sulphur trioxide, carbon dioxide and aldehyde, respectively. Thus with hydrogen sulphide: K 2 Cr 2 7 + 4 H 2 S0 4 + 3 H 2 S = K 2 S0 4 + Cr 2 (S0 4 ) 3 + 4 H 2 + 3 H 2 O + 3 S. The presence of 3 molecules of hydrogen sulphide is neces- sary for interaction with the 3 atoms of oxygen liberated per molecule of dichromate; the products by this oxidation are indicated in the equation. 1 molecular weight of potassium dichromate is here directly proportional to 3 molecular weights of hydrogen sulphide, hence in this 152 CHEMICAL CALCULATIONS reaction all values for these substances must accord with the ratio 3H 2 S 3 (34.09) K 2 Cr 2 CV C 294.2 ' From the relation between the molecular weight of a substance and its volume as a vapor we are able to calcu- late, for example, the volume of hydrogen sulphide that can be oxidized by a corresponding proportional weight of the dichromate. Example 66. What volume of hydrogen sulphide, at stand- ard conditions, can be oxidized by 7.36 grams of potassium dichromate in an acid solution? The volume of hydrogen sulphide oxidized by 1 gram- molecular weight of potassium dichromate (294.2 grams) is that volume corresponding to 3 gram-molecular weights of the gas, or 67,200 c.c. (i.e., 3 X 22,400 c.c.) at standard conditions. The direct proportionality between the gram- molecular weight and the actual weight of dichromate here given, determines also the proportionality between the respective volumes of hydrogen sulphide correspond- ing to these values: 294.2 : 7.36 = 67,200 : x 9 in which x represents the volume (1681 c.c.) of gas oxi- dized. The weight of sulphur precipitated, as well as any other substance present, may be calculated from the standpoint of these proportional values expressed in the equation. For example, the relation between the sulphur and potas- sium dichromate is expressed by the ratio: 3 S 96.21 K 2 Cr 2 7 ' 294.2' COMPLEX EQUATIONS 153 When hydrochloric acid is used the excess of acid is itself oxidized by the liberated oxygen with the produc- tion of free chlorine and water: K 2 2 7 + 14 HC1 = 2 KC1 + 2 CrCl 3 + 7 H 2 O + 3 C1 2 . Standard Oxidizing Solutions with Potassium Dichro- mate. From reactions of oxidation with a potassium dichromate solution we are able to draw up the value of such a solution in terms of the hydrogen equivalent. 1 atom of oxygen will oxidize and combine with 2 atoms of hydrogen, hence the equivalence of 3 atoms of oxygen in terms of hydrogen is necessarily 6. Since 1.01 grams of hydrogen or this unit equivalent are required per liter of a normal solution, we shall require here in a normal solu- tion of potassium dichromate that solution which per liter can furnish just sufficient oxygen to unite with 1.01 grams of hydrogen exactly one-sixth of the gram-molecular weight of this salt, i.e., 294.2/6 or 49.03 grams. In the work of titration with standard solutions of this salt the change in color from red to green indicates complete reduction, but for accurate determination of this point it is customary to involve small portions of the solution in further reactions. Example 67. What is the normality of that potassium dichromate solution, 40 c.c. of which oxidized 0.590 gram of ferrous sulphate to the ferric state? The more commonly used ferrous ammonium sulphate, FeS0 4 (NH 4 ) 2 S0 4 .6H 2 0, is here replaced by ferrous sulphate, FeS0 4 , for the sake of simplicity. The reaction for the oxidation of this latter to ferric sulphate involves 1 oxygen atom, with 1 molecule of sulphuric acid, per 2 molecules of the ferrous salt: 2 FeS0 4 + H 2 SO 4 + O = Fe 2 (S0 4 ) 3 + H 2 0. 154 CHEMICAL CALCULATIONS Consequently to include this action in the equation which represents the oxidation by means of 1 molecule of potas- sium dichromate, with its 3 possible oxygen atoms, we shall need to multiply the equation above by 3 and thus obtain 3 O as a reaction-quantity common to both equa- tions : K 2 Cr 2 7 + 4 H 2 S0 4 = K 2 SO 4 + Cr 2 (S0 4 ) 3 + 4 H 2 + 3 6 FeSO 4 + 3 H 2 SO 4 + 30 = 3 Fe 2 (S0 4 ) 3 + 3 H 2 0. It comes to the same end if we simply combine these equations and let this common quantity (30) drop out, when we shall have the final expression: K 2 Cr 2 7 + 6 FeS0 4 + 7 H 2 S0 4 = K 2 S0 4 + Cr 2 (S0 4 ) 3 + 3 Fe 2 (SO 4 ) 3 + 7 H 2 0. In either case the reaction-quantities 6 FeS0 4 and K 2 Cr 2 O 7 are directly proportional to each other, and the ratio 6 FeS0 4 /K 2 Cr 2 O 7 , or 911.52/294.2, determines the ratio between all weights which severally may represent these quantities. In the example 0.590 gram of FeS0 4 was oxidized. Therefore the following equivalence in the ratios: 911.52 _ 0.590 294.2 = x The value for x is found to be 0.1904 gram. This weight, however, was present in 40 c.c. of solvent, consequently in 1 liter we should have 25 X 0.1904 or 4.76 grams of potassium dichromate. A normal solution of this salt contains 49.03 grams per liter (p. 153), hence we have a solution of 4.76/49.03 normality, i.e., 0.0971 normal, a trifle less than N/ 10. Oxidizing Action of Potassium Permanganate. As a second illustration the oxidizing action of potassium per- COMPLEX EQUATIONS 155 manganate is of much importance. The action in an acid solution is analogous to that of potassium dichromate, only the reduction of Mn vn to Mn" is accompanied here with the liberation of 5 atoms of oxygen for every 2 mole- cules of permanganate : 2 KMn0 4 + 3 H 2 S0 4 = K 2 S0 4 + 2 MnSO 4 + 3 H 2 + 5 O. The presence of some substance capable of oxidation is necessary in order to bring about this reaction in dilute solutions. We may take again the example of hydrogen sulphide as affected by an oxidizing agent: 2 KMnO, + 3 H 2 S0 4 + 5 H 2 S = K 2 S0 4 + 2 MnS0 4 + 3 H 2 + 5 H 2 + 5 S. From this equation the relative amounts of hydrogen sul- phide and potassium permanganate involved will always accord with the ratio between their respective reaction- quantities : 5H 2 S 5 (34.09) 2KMn0 4 ' 2(158.03)' When expressed by molecular equations the substances here concerned may be studied also in regard to their volume relations. Example 68. What weight of potassium permanganate will be reduced by 2800 c.c. of hydrogen sulphide, at standard condi- tions of temperature and pressure? The reaction already discussed supplies the necessary data: 2 gram-molecular weights (316.06 grams) of the permanganate are reduced by 5 X 22,400 c.c. of hydrogen sulphide. By simple proportion the weight of perman- ganate that can be reduced by 2800 c.c. is derived as follows : (5 X 22.400) : 2800 = 316.06 : x, where x, or 7.9 grams, is this weight of permanganate. 156 CHEMICAL CALCULATIONS The reverse operation, or that of determining the weight (or volume) of hydrogen sulphide oxidized by a known weight of potassium permanganate, follows the same principle. Standard Oxidizing Solutions with Potassium Perman- ganate. The liberation of 5 atoms of oxygen per 2 molecules of permanganate also determines the amount of this salt that must be present in one liter of its normal solution for use in the presence of acid. By calculation, in a manner similar to that described under the standard dichromate solution, it is found necessary to have one- tenth of the 2 gram-molecular weights of potassium permanganate (316.06 grams), or 31.606 grams, in each liter of solution; whereby this volume of solution will be capable of furnishing just sufficient oxygen to oxidize 1.01 grams of hydrogen. The change from the deep purple color of the permanganate to the almost colorless manganous salt serves admirably to mark the point when the reduction is complete and hence to fix the end-point in titration with a permanganate solution. Example 69. 100 c.c. of a hydrogen peroxide solution were required to decolorize 500 c.c. of N/10 potassium permanganate solution (acidulated). What was the percentage concentration of the hydrogen peroxide solution? 2 KMn0 4 + 3 H 2 S0 4 + 5 H 2 2 = K 2 S0 4 + 2 MnS0 4 + 3 H 2 O + 5 H 2 + 5 2 . As the equation indicates, the 5 atoms of oxygen liberated from the permanganate withdraw 5 atoms of oxygen from 5 molecules of hydrogen peroxide and are together removed from the reaction in the molecular form (5 molecules). 500 c.c. of N/10 potassium permanganate solution are equivalent to 50 c.c. of the normal solution. One liter of COMPLEX EQUATIONS 157 a normal potassium permanganate solution, containing 31,606 grams (1/10 of 316.06 grams as represented by the reaction-quantity), can only react with 1/10 of the quan- tity 5 H 2 O 2 required by the reaction, or with 1/2 of one molecule of hydrogen peroxide. Here 50 c.c. (1/20 liter) can act upon just 1/20 of the amount of hydrogen per- oxide that a liter of the normal solution can oxidize (namely, 1/2 of a gram-molecular weight ; or 17 grams), i.e., 1/20 of 17 grams or 0.85 gram. Consequently in 100 c.c. of the hydrogen peroxide solution there is only this amount, 0.85 gram, of the compound, and if the weight of the solution is assumed to be 100 grams, then the percentage of hydrogen peroxide present is 85/100 of 1 per cent. From the conditions of this reaction the amount of any one substance present can be calculated also from the volume of oxygen evolved. Just as potassium dichromate so also an acidulated solution of potassium permanganate readily oxidizes solu- tions of ferrous salts to the ferric state. The equation indicates, of course, this liberation of 5 oxygen atoms per 2 molecules of permanganate: 2 KMn0 4 + 10 FeS0 4 + 8 H 2 S0 4 = K 2 S0 4 + 2 MnS0 4 + 5 Fe 2 (S0 4 ) 3 + 8 H 2 0. Again the action of an acidulated potassium perman- ganate solution upon a hot solution of oxalic acid (weighed out as C 2 H 2 4 . 2 H 2 0), with the complete oxidation of the latter into carbon dioxide and water, affords an admirable means for the titration and standardization of permanganate solutions : 2 KMnO 4 + 5 C 2 H 2 4 . 2 H 2 + 3 H 2 S0 4 = K 2 S0 4 + 2 MnSO 4 + 10 C0 2 + 8 H 2 O + (10 H 2 0). 158 CHEMICAL CALCULATIONS Hydrochloric acid, except in the cold and under proper precautions, is rarely ever used for the acidulation of permanganate solutions owing to its ready oxidation, similar to that by potassium dichromate, into chlorine and water: 2 KMn0 4 + 16 HC1 = 2 KC1 + 2 MnCl 2 + 5 C1 2 + 8 H 2 0. In neutral or alkaline solutions potassium permanganate is reduced by the presence of oxidizable substances to manganese dioxide (precipitated), and somewhat less oxygen per molecule of permanganate is thereby liberated. The action may be represented in part as follows: 2 KMn0 4 + H 2 O = 2 KOH + 2 Mn0 2 + 3 O. The 3 oxygen atoms liberated from 2 molecules of potas- sium permanganate determine the amount of this salt in its normal solution to be 1/6 of 316.06 grams (repre- sented by the reaction-quantity 2 KMn0 4 ), or 52.67 grams per liter. lodimetry. The action of iodine in the presence of water upon a number of compounds is interesting to show one of the indirect methods of estimating substances in solution. Iodine dissolves readily in water containing potassium iodide. This solution acts as a mild oxidizing agent as may be seen from the following equation: S0 2 + 2 H 2 + I 2 = H 2 S0 4 + 2 HI. In the presence of an oxidizable substance the one mole- cule of iodine, through the decomposition of water present, furnishes sufficient oxygen to oxidize one molecule of sulphur dioxide to the trioxide, or an amount capable also of oxidizing two atoms of hydrogen. The normal value, therefore, of such an iodine solution must be fixed COMPLEX EQUATIONS 159 at one-half the gram-molecular weight (253.84 grams), or 126.92 grams per liter. The employment, however, of a more dilute solution as N/10, etc., is more common in practice. The end-point in titrations with iodine solu- tions is detected by the addition of a slight amount of a starch solution, with which substance the iodine, when occurring in slight excess, forms a deep blue mixture. An iodine solution may also exert an oxidizing action by withdrawing a number of positively charged ions from a salt in solution: 2 Na 2 S 2 O 3 + I 2 = Na 2 S 4 8 + 2 Nal, or 4 Na* + 2 S 2 O 3 " + I 2 = 2 Na + S 4 O 6 " + 2 Na + 2 I'. A solution of sodium thiosulphate can be easily made up from the well crystallized salt, Na 2 S 2 O 3 . 5 H 2 O, and this solution titrated against an iodine solution, using starch as an indicator. One molecule of iodine reacts with two molecules of the thiosulphate; consequently a normal solution of the latter will contain (with reference to a normal solution of iodine) just one-half of this reaction- quantity or one gram-molecular weight per liter. Substances that are capable of liberating iodine from potassium iodide, e.g. chlorine, bromine, etc., may be brought in this way into estimations by volumetric analysis : 2 KI + C1 2 = 2 KC1 + I 2 . The amount of iodine liberated is in exact accord with the proportional amount of chlorine under investigation, and the iodine in turn is open to determination from the amount of thiosulphate required for titration. This in- teresting application of volumetric analysis is known as lodimetry. Example 70. An unknown weight of sodium dichromate was warmed with an excess of hydrochloric acid, and the chlorine, 160 CHEMICAL CALCULATIONS set free, passed into a solution of potassium iodide. 400 c.c. of N/10 sodium thiosulphate solution were required for the titra- tion of the iodine thereby liberated. Determine the amount of chlorine evolved, and also the weight of dichromate reduced. 400 c.c. of N/10 sodium thiosulphate solution are equivalent to 40 c.c. of a normal solution. 1 liter of a normal solution of this salt exactly decolorizes 1 liter of a normal solution of iodine, or this amount of iodine (126.92 grams) in whatever volume of solution it is con- tained. Consequently 40 c.c. of the normal thiosulphate solution will act upon that amount of iodine as indi- cated in the proportion : 1000 : 40 = 126.92 : x, or x = 5.177 grams. The amount of chlorine proportional to this weight of iodine is in accord with the ratio Cl/I, or 35.46/126.92, as indicated by the reaction above. By proportion we have: 126.92 : 35.46 = 5.177 : x, where x, or 1.45 grams, is the weight of chlorine set free and thus made capable of liberating the amount of iodine above calculated. The determination of the amount of sodium dichromate necessary to liberate this weight of chlorine rests upon an equation exactly analogous to the action of hydrochloric acid upon potassium dichromate (see p. 153), and is indicated by the ratio: 3C1 2 212.76 Na 2 Cr 2 7 ' 262 Accordingly, 212.76 : 262 = 1.45 : x. From this we cal- culate the weight of dichromate as 1.786 grams. No matter how complex the equations may become, the true relationship between all of the substances concerned is found to be in direct proportionality to their respec- tive reaction-quantities. COMPLEX EQUATIONS 161 PROBLEMS. 257. What weight of ferric chloride, FeCl 3 , can be reduced to ferrous chloride, FeCl 2 , by 8.52 grams of hydrogen sulphide? Ans. 81.10 grams. 258. What volume of hydrogen sulphide (at standard con- ditions) will be required for the reduction of 100 grams of ferric chloride, FeCl 3 , to the ferrous salt? Ans. 6903.8 c.c. 259. Calculate the weight of sulphur precipitated in the reduction of 100 grams of ferric chloride, FeCl 3 , to the ferrous salt by the action of hydrogen sulphide. Ans. 9.884 grams. 260. 1000 c.c. of hydrogen sulphide (at standard conditions) will reduce what weight of potassium dichromate in acid solu- tion? Ans. 4.378 grams. 261. What volume of hydrogen sulphide (at standard con- ditions) will be required for the reduction of 100 grams of potas- sium dichromate in acid solution? Ans. 22,842 c.c. 262. What weight of hydrogen sulphide can be oxidized by 400 grams of potassium dichromate, in acid solution? What weight of sulphur will be precipitated? Ans. 139.05 grams H 2 S. 130.81 grams S. 263. What volume of sulphur dioxide (at standard condi- tions) can be oxidized by 30 grams of potassium dichromate in acid solution? Ans. 6852 c.c. 264. What volume of hydrogen sulphide, at 24 and 750 mm. pressure, can be oxidized by 10 grams of potassium dichromate in acid solution? Ans. 2518 c.c. 265. What weight of potassium dichromate, in acid solution, will be reduced by 1653.5 c.c. of sulphur dioxide at 22 and 745 mm. pressure? Ans. 6.567 grams. 266. Calculate the volume of chlorine liberated in the action of 40 grams of potassium dichromate upon a hydrochloric acid solution. Ans. 9137 c.c. 267. What weight of sodium dichromate must enter into reaction with a hydrochloric acid solution in order to liberate 100 grams of chlorine? Ans. 123.14 grams. 162 CHEMICAL CALCULATIONS 268. Determine the normality of a potassium dichromate solution, 25 c.c. of which oxidized 1.24 grams of ferrous sulphate to the ferric salt. Ans. 0.3265 N. 269. Determine the normality of a sodium dichromate solu- tion, 50 c.c. of which oxidized 3.85 grams of ferrous ammonium sulphate, FeS0 4 . (NH 4 ) 2 S0 4 . 6 H 2 to the ferric salt. 0.1963 N. 270. What volume of hydrogen sulphide, at standard con- ditions, will be required to reduce 200 c.c. of N/10 potassium dichromate solution (acidulated)? Ans. 224 c.c. 271. What weight of potassium permanganate, in acid solu- tion, will be reduced by 5000 c.c. of hydrogen sulphide at standard conditions? Ans. 14.11 grams. 272. Calculate the volume of hydrogen sulphide, at stand- ard conditions, that can be oxidized by 4 grams of potassium permanganate in acid solution. Ans. 1417 A c.c. 273. What weight of sulphur dioxide can be oxidized by 200 grams of potassium permanganate in acid solution? Ans. 202.7 grams. 274. 50 c.c. of a solution of hydrogen peroxide were required to decolorize 400 c.c. of N/5 potassium permanganate solution (acidulated). Calculate the percentage concentration of the hydrogen peroxide solution. Ans. 2.72 per cent. 275. Calculate the weight of crystallized oxalic acid, C 2 H 2 4 . 2 H 2 O, required for the reduction of 100 grams of potassium permanganate in acid solution. What volume of carbon diox- ide, at standard conditions, would be liberated? Ans. 199.4 grams. 70,873 c.c. C0 2 . 276. 50 c.c. of an acidulated potassium permanganate solu- tion were reduced by 2.4 grams of anhydrous oxalic acid, C 2 H 2 4 . Calculate the normality of the permanganate solution. Ans. 1.066 N. 277. What weight of sulphur dioxide will be oxidized by 50 grams of potassium permanganate in alkaline solution? Ans. 3.04 grams. COMPLEX EQUATIONS 163 278. The bromine set free by the action of manganese dioxide upon a hydrobromic acid solution was passed into a solution of potassium iodide. 200 c.c. of N/10 sodium thio- sulphate solution were required for the titration of the free iodine. Calculate the weight of bromine evolved. Ans. 1.6 grams. 279. Determine the purity of a sample of manganese dioxide, 2.2 grams of which, with excess of hydrochloric acid, set free sufficient chlorine to liberate a quantity of iodine that required 250 c.c. of N/5 sodium thiosulphate solution for titration. Ans. 98.79 per cent pure. 280. 0.2452 gram of potassium dichromate, acting upon an excess of hydrochloric acid, set free an amount of chlorine which, when passed into a solution of potassium iodide, liber- ated iodine sufficient for the titration of 50 c.c. of an unknown sodium thiosulphate solution. Calculate the normality of this latter solution. Ans. N/10. APPENDIX. APPENDIX I. Correction of Barometic Readings. A barometric reading at room temperature is reduced to the corresponding height of a column of mercury at by subtracting from the actual reading in millimeters that number in the second column below (Correction) set opposite the observed temperature. The barometric read- ings throughout these problems are given in corrected form. Tempera- Correc- Tempera- Correc- Tempera- Correc- ture. tion. ture. tion. ture. tion. 12 1.6 17 2.2 23 3.0 13 1.7 18.5 2.4 24.5 3.2 14 1.8 20 2.6 25 3.3 15 2.0 21.5 2.8 26 3.4 APPENDIX II. Tension of Aqueous Vapor in Millimeters. Tempera- ture. Pressure. Tempera- ture. Pressure. Tempera- ture. Pressure. 4.6 16 13.5 26 25.1 5 6.5 17 14.4 27 26.5 8 8.0 18 15.4 28 28.1 9 8.6 19 16.3 29 29.8 10 9.2 20 17.4 30 31.5 11 9.8 21 18.5 31 33.4 12 10.5 22 19.7 32 35.4 13 11.2 23 20.9 33 37.4 14 11.9 24 22.2 34 39.6 15 12.7 25 23.6 100 760.0 167 168 CHEMICAL CALCULATIONS APPENDIX III. Four-Figure Logarithms. irst two igures. Third figure. Fourth-figure differences. In O 4e c 70 A fe^ 0000 . A O O o 9 10 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 8 12 17 21 25 29 33 37 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 8 11 15 19 23 26 30 34 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 3 7 10 14 17 21 24 28 31 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 3 6 10 13 16 19 23 26 29 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 369 12 15 18 21 24 27 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 368 11 14 17 20 22 25 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 358 11 13 16 18 21 24 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 257 10 12 15 17 20 22 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 1257 9 12 14 16 19 21 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 247 9 11 13 16 18 20 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 246 8 11 13 15 17 19 21 3222 3243 3263 3284 3304 3224 3345 3365 3385 3404 246 8 10 12 14 16 18 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 246 8 10 12 14 15 17 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 246 7 9 11 13 15 17 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 245 7 9 11 12 14 16 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 235 7 9 10 12 14 15 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 235 7 8 10 11 13 15 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 235 689 11 13 14 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 235 689 11 12 14 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 1 3 4 679 10 12 13 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 1 3 4 679 10 11 13 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 1 3 4 678 10 11 12 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 134 578 9 11 12 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 1 3 4 568 9 10 12 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 1 3 4 568 9 10 11 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 2 4 567 9 10 11 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 2 4 567 8 10 11 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 2 3 567 8 9 10 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 2 3 567 8 9 10 39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 2 3 457 8 9 10 40 6021 6031 6042 6053. 6064 6075 6085 6096 6107 6117 2 3 456 8 9 10 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 2 3 4 5 6 789 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 2 3 5 6 789 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 2 3 5 6 789 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 2 3 5 6 789 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 2 3 5 6 789 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 2 3 5 6 778 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 2 3 5 5 678 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 2 3 * 678 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 2 3 678 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 2 3 3 5 678 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 2 3 3 5 678 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 1 2 2 3 5 677 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 1 2 2 3 5 667 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 122 3 5 667 APPENDIX 169 Four-Figure Logarithms. irst two igures. Third figure. Fourth-figure differences. m 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 1 2 2 345 567 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 1 2 2 345 567 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 122 345 567 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 1 1 2 344 567 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 1 1 2 344 567 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 1 2 344 566 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 1 2 344 566 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 1 2 334 566 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 1 2 334 556 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 1 2 334 556 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 1 2 334 556 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 1 2 334 556 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 1 2 334 556 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 1 2 334 456 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 1 2 234 4 5 6 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 1 2 234 456 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 1 2 234 455 72 8573 8579 8585 8591 8597 8603 8609' 8615 8621 8627 1 2 234 455 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 1 2 234 455 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 1 2 234 455 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1 2 233 4 5 5 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 1 2 233 455 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 1 2 2 3 445 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 1 2 2 3 445 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 1 2 2 3 445 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 1 2 2 3 445 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 1 2 2 3 445 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 1 2 2 3 445 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 1 2 2 3 445 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 1 2 2 3 445 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 1 2 2 3 445 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 1 2 2 3 445 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 1 2 2 344 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 1 2 2 344 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 1 2 2 344 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 1 2 2 344 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 1 2 2 344 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 1 2 2 344 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 1 2 2 3 4-4 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 1 2 2 344 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 1 2 2 344 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 1 2 2 344 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 1 2 2 344 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 1 2 2 344 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 1 223 334 170 CHEMICAL CALCULATIONS APPENDIX IV. Antilogarithms. First two figures of mantissa. Third figure. Fourth-figure differences. 1 2 3 4 5 .00 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 1 1 1 222 .01 1023 1026 1028 1030 1033 1035 1038 1040 1042 1045 1 1 222 .02 1047 1050 1052 1054 1057 1059 1062 1064 1067 1069 1 1 222 .03 1072 1074 1076 1079 1081 1084 1086 1089 1091 1094 1 1 222 .04 1096 1099 1102 1104 1107 1109 1112 1114 1117 1119 1 1 222 .05 1122 1125 1127 1130 1132 1135 1138 1140 1143 1146 1 1 222 .06 1148 1151 1153 1156 1159 1161 1164 1167 1169 1172 1 1 222 .07 1175 1178 1180 1183 1186 1189 1191 1194 1197 1199 1 1 222 .08 1202 1205 1208 1211 1213 1216 1219 1222 1225 1227 1 223 .09 1230 1233 1236 1239 1242 1245 1247 1250 1253 1256 1 223 .10 1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 1 1 223 .11 1288 1291 1294 1297 1300 1303 1306 1309 1312 1315 1 2 223 .12 1318 1321 1324 1327 1330 1334 1337 1340 1343 1346 o ; 2 2 223 .13 1349 1352 1355 1358 1361 1365 1368 1371 1374 1377 1 2 2 2 3 3 .14 1380 1384 1387 1390 1393 1396 1400 1403 1406 1409 1 2 2 233 .15 1413 1416 1419 1422 1426 1429 1432 1435 1439 1442 1 2 2 233 .16 1445 1449 1452 1455 1459 1462 1466 1469 1472 1476 1 2 2 233 .17 1479 1483 1486 1489 1493 1496 1500 1503 1507 1510 2 2 233 .18 1514 1517 1521 1524 1528 1531 1535 1538 1542 1545 2 2 233 .19 1549 1552 1556 1560 1563 1567 1570 1574 1578 1581 2 2 333 .20 1585 1589 1592 1596 1600 1603 1607 1611 1614 1618 1 2 2 333 .21 1622 1626 1629 1633 1637 1641 1644 1648 1652 1656 222 333 .22 1660 1663 1667 1671 1675 1679 1683 1687 1690 1694 222 333 .23 1698 1702 1706 1710 1714 1718 1722 1726 1730 1734 222 334 .24 1738 1742 1746 1750 1754 1758 1762 1766 1770 1774 222 334 .25 1778 1782 1786 1791 1795 1799 1803 1807 1811 1816 222 334 .26 1820 1824 1828 1832 1837 1841 1845 1849 1854 1858 2 2 3 334 .27 1862 1866 1871 1875 1879 1884 1888 1892 1897 1901 223 334 .28 1905 1910 1914 1919 1923 1928 1932 1936 1941 1945 1 223 344 .29 1950 1954 1959 1963 1968 1972 1977 1982 1986 1991 1 223 344 .30 1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 1 223 3 4 4 .31 2042 2046 2051 2056 2061 2065 2070 2075 2080 2084 1 223 344 .32 2089 2094 2099 2104 2109 2113 2118 2123 2128 2133 1 223 344 .33 2138 2143 2148 2153 2158 2163 2168 2173 2178 2183 1 223 344 .34 2188 2193 2198 2203 2208 2213 2218 2223 2228 2234 1 2 233 445 .35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 1 2 233 445 .36 2291 2296 2301 2307 2312 2317 2323 2328 2333 2339 1 2 233 445 .37 2344 2350 2355 2360 2366 2371 2377 2382 2388 2393 1 2 233 445 .38 2399 2404 2410 2415 2421 2427 2432 2438 2443 2449 1 2 233 445 .39 2455 2460 2466 2472 2477 2483 2489 2495 2500 2506 1 2 2 3 3 455 .40 2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 1 2 234 455 .41 2570 '2576 2582 2588 2594 2600 2606 2612 2618 2624 1 2 234 455 .42 2630 2636 2642 2649 2655 2661 2667 2673 2679 2685 1 2 234 456 .43 2692 2698 2704 2710 2716 2723 2729 2735 2742 2748 1 2 334 456 .44 2754 2761 2767 2773 2780 2786 2793 2799 2805 2812 1 2 334 456 .45 2818 2825 2831 2838 2844 2851 2858 2864 2871 2877 1 2 334 556 .46 2884 2891 2897 2904 2911 2917 2924 2931 2938 2944 1 2 334 556 .47 2951 2958 2965 2972 2979 2985 2992 2999 3006 3013 1 2 334 556 48 3020 3027 3034 3041 3048 3055 3062 3069 3076 3083 1 2 344 566 .49 3090 3097 3105 3112 3119 3126 3133 3141 3148 3155 1 2 3 4 4 566 APPENDIX 171 Antilogarithms. First two figures of mantissa. Third figure. Fourth-figure differences. 8 .50 3162 3170 3177 3184 3192 3199 3206 3214 3221 3228 1 1 2 344 567 .51 3236 3243 3251 3258 3266 3273 3281 3289 3296 3304 122 345 567 .52 3311 3319 3327 3334 3342 3350 3357 3365 3373 3381 1 2 2 345 567 .53 3388 3396 3404 3412 3420 3428 3436 3443 3451 3459 1 2 2 345 667 .54 3467 3475 3483 3491 3499 3508 3516 3524 3532 3540 122 345 667 .55 3548 3556 3565 3573 3581 3589 3597 3606 3614 3622 1 2 2 345 677 .56 3631 3639 3648 3656 3664 3673 3681 3690 3698 3707 1 2 3 345 678 .57 3715 3724 3733 3741 3750 3758 3767 3776 3784 3793 1 2 3 345 678 .58 3802 3811 3819 3828 3837 3846 3855 3864 3873 3882 1 2 3 445 678 .59 3890 3899 3908 3917 3926 3936 3945 3954 3963 3972 123 455 678 .60 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 1 2 3 456 678 .61 4074 4083 4093 4102 4111 4121 4130 4140 4150 4159 1 2 3 456 789 .62 4169 4178 4188 4198 4207 4217 4227 4236 4246 4256 1 2 3 456 789 .63 4266 4276 4285 4295 4305 4315 4325 4335 4345 4355 1 2 3 456 789 .64 4365 4375 4385 4395 4406 4416 4426 4436 4446 4457 1 2 3 456 789 .65 4467 4477 4487 4498 4508 4519 4529 4539 4550 4560 1 2 3 456 789 .66 4571 15S1 4592 4603 4613 4624 4634 4645 4656 4667 1 2 3 456 7 9 10 .67 4677 K;SS 4699 4710 4721 4732 4742 4753 4764 4775 1 2 3 457 8 9 10 .68 4786 4797 4808 4819 4831 4842 4853 4864 4875 4887 123 467 8 9 10 .69 4898 4909 4920 4932 4943 4955 4966 4977 4989 5000 1 2 3 567 8 9 10 .70 5012 5023 5035 5047 5058 5070 5082 5093 5105 5117 1 2 5 6 7 8 9 11 .71 5129 5140 5152 5164 5176 5188 5200 5212 5224 5236 1 2 567 8 10 11 .72 5248 5260 5272 5284 5297 5309 5321 5333 5346 ->::,-s 1 2 567 9 10 11 .73 5370 S383 5395 5408 5420 5433 5445 5458 5470 5483 1 3 568 9 10 11 .74 5495 5508 5521 5534 5546 5559 5572 5585 5598 5610 1 3 568 9 10 12 .75 5623 5636 5649 5662 5675 5689 5702 5715 5728 5741 1 3 578 9 10 12 .76 5754 5768 5781 5794 5808 5821 5834 584 S 5861 5875 1 3 578 9 11 12 .77 5888 6902 5916 5929 5943 5957 5970 5984 5998 6012 1 3 578 10 11 12 .78 6026 6039 6053 6067 6081 6095 6109 6124 6138 6152 1 3 678 10 11 13 .79 6166 6180 6194 6209 6223 6237 6252 6266 6281 6295 1 3 679 10 11 13 .80 6310 6324 6339 6353 6368 6383 6397 6412 6427 6442 1 3 4 679 10 12 13 .81 6457 6471 6486 6501 6516 6531 6546 6561 6577 6592 235 689 11 12 14 .82 6607 6622 6637 6653 6668 6683 6699 6714 6730 6745 235 689 11 12 14 .83 6761 6776 6792 6808 6823 6839 6855 6871 6887 6902 235 689 11 13 14 .84 6918 6934 6950 6966 6982 6998 7015 7031 7047 7063 235 6 8 10 11 13 15 .85 7079 7096 7112 7129 7145 7161 7178 7194 7211 7228 235 7 8 10 12 13 15 .86 7244 7261 7278 7295 7311 7328 7345 7362 7379 7396 235 7 8 10 12 13 15 .87 7413 7430 7447 7464 7482 7499 7516 7534 7551 7568 235 7 9 10 12 14 16 .88 7586 7603 7621 7638 7656 7674 7691 7709 7727 7745 245 7 9 11 12 14 16 .89 7762 7780 7798 7816 7834 7852 7870 7889 7907 7925 245 7 9 11 13 14 16 .90 7943 7962 7980 7998 8017 8035 8054 8072 8091 8110 246 7 9 11 13 15 17 .91 8128 8147 8166 8185 8204 8222 8241 SL'lid 8279 8299 246 8 9 11 13 15 17 .92 8318 8337 8356 8375 8395 8414 8433 8453 8472 8492 246 8 10 12 14 15 17 .93 8511 8531 8551 8570 8590 8610 8630 8650 8670 8690 246 8 10 12 14 16 18 .94 8710 8730 8750 8770 8790 8810 8831 8851 8872 8892 246 8 10 12 14 16 18 .95 8913 8933 8954 8974 8995 9016 9036 9057 9078 9099 246 8 10 12 15 17 19 .96 9120 9141 9162 9183 9204 )22ti 9247 9268 9290 9311 246 8 11 13 15 17 19 .97 9333 9354 9376 9397 9419 9441 9462 9484 9506 9528 247 9 11 13 15 17 20 .98 9550 9572 9594 9616 9638 9661 9683 9705 9727 9750 247 9 11 13 16 18 20 .99 9772 9795 9817 9840 9863 9886 9908 9931 9954 9977 2 5 7 9 11 14 16 18 20 INDEX PAGE Analysis, volumetric and gravimetric 118 Aqueous vapor, tension of. . . 28 Atom, definition of 35 Atomic weight, definition of, 36, 46 Avogadro's hypothesis 33 Boiling-point 29 Boyle, law of 9 Charles, law of Coefficient of expansion of Cohesion in 15 15 34 Dalton, law of 27 Definite proportions, law of 46 Density, absolute, definition of 5 relative, calculation of, from molecular weight 42 definition of 5 standard of, for gases. . 6 upon different standards 38 relation of, to molecular weight 34 to pressure of gases. ... 12 to temperature of gases 18 to pressure and tempera- ture of gases 26 Dyne, definition of 2 Empirical formulae, definition of 61 derivation of, from percent- age composition 60 End-point in titrations 106 Erg, definition of 3 Equations, dependent 92 independent 94 study of 82 volume relations in. . 86 PAGE Force, unit of 2 Formulas, derivation of 59 for salts containing water of crystallization. ... 70 from percentage composi- tion 60,65 from relation of formula- quantities to each other 67 from relation between for- mula-quantities and their corresponding weights 71 from relation between for- mula-quantities and percentage composi- tion 74 from single analytical val- ues 75 Formula-quantity, definition of 48 relation of, to molecular weight 49 to other formula-quan- tities 51,53 to percentage composi- tion. . . 52 Gay-Lussac, law of 33 Gram, definition of 2 Gram-molecular volume, defi- nition of 40 in chemical reactions 122 Gram-molecular weight, defi- nition of 37 Gravimetric analysis 118 Indicators in titrations 106 lodimetry 158 Ions, hydrogen and hydrox- ide 106 173 174 INDEX Length, unit of. . . . Liter, definition of. PAGE 1 2 Mass, definition of ......... 3 unit of ................. 2 Meter, definition of ........ 1 Molar solution, definition of 107 Molecule, definition of ...... 33 Molecular formulae, definition of ................ 46,61 derivation of, from percent- age composition ..... 59 Molecular volumes of gases, combinations between 122 relation of, to volume-unit 124 Molecular weight, definition of ................. 34,46 Neutralization ............ 106 Normality factor .......... 109 calculation of, by simple proportion .......... Ill Normal solution, definition of 108 Oxidation, definition of. ... Oxidizing solutions, with 148 150 withKMn0 4 ............ 154 Oxygen, as standard of rela- tive densities of gases . 6 as standard of molecular weights ............ 6 Partial pressures of gases ... 27 Percentage composition of compounds .......... 46 Percentage composition, re- lation of, to formula- quantities .......... 52 as basis for percentage purity .............. 53 Percentage purity .......... 53 calculation of ........... 88 Pressure, normal .......... 3 relation of, to temperature of gases ............ 23 to volume of gases. ... 11 to volume and tempera- ture of gases ........ 26 Pressure-fraction .......... 11 Products resulting from mix- tures, calculation of . . 89 PAGE Reaction-quantities 82 calculation of ratios be- tween 92, 94 calculation of, from weights of substances involved 97 complex 98 Reduction, definition of .... 148 Second, definition of 1 Solutions, adjustment of, to desired standard 117 comparison of, with stand- ard 116 standard 108 standard oxidizing 153, 156 Specific gravity, definition of 5 Standardization of solutions, 112,113 Temperature, normal 3 relation of, to density of gases 18 to pressure of gases .... 23 to volume of gases 18 to volume and pressure of gases 25 Temperature-fraction 18 Time, unit of 1 Titration of solutions 106 Valence, definition of 148 Vapor density 6 Volume relations in chemical equations 86 Volume, relation of, to pres- sure of gases 9 to temperature of gases 17 to temperature and pres- sure of gases 21, 24 unit of 2 Volume-fraction 11 Volume-unit, for reactions between gases 124 calculation of, from known volumes of gases ; 125,126, 128 calculation of, from ob- served volume-changes 129 calculation of, from sev- eral simultaneous re- actions 133, 135 Volumetric analysis 118 Work, unit of 3 D. VAN NOSTRAND COMPANY 25 PARK PLACE NEW YORK SHORT-TITLE CATALOG OF OP SCIENTIFIC AND ENGINEERING BOOKS This list includes the technical publications of the following English publishers : SCOTT, GREENWOOD & CO. CROSBY LOCKWOOD & SON CONSTABLE & COMPANY, Ltd. TECHNICAL PUBLISHING CO. ELECTRICIAN PRINTING & PUBLISHING CO., for whom D. Van Nostrand Company are American agents. Descriptive Circulars sent on request. 4 D. VAN NOSTHAND COMPANY'S SHORT-TITLK CATALOG Atkinson, A. A. Electrical and Magnetic Calculations 8vo, *i 50 Atkinson, J. J. Friction of Air in Mines. (Science Series No. 14.) : i6mo, o 50 Atkinson, J. J., and Williams, E. H., Jr. Gases Met with in Coal Mines. (Science Series No. 13.) i6mo, o 50 Atkinson, P. The Elements of Electric Lighting i2mo, i 50 The Elements of Dynamic Electricity and Magnetism. i2mo, 2 oo Power Transmitted by Electricity I2mo, 2 oo Auchincloss, W. S. Link and Valve Motions Simplified. . . .8vo, *i 50 Ayrton, H. The Electric Arc 8vo, *5 oo Bacon, F. 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