GIFT OF Pony A SERIES OF CHEMICAL PROBLEMS A SERIES OF CHEMICAL PROBLEMS FOR USE IN COLLEGES AND SCHOOLS T. E. THORPE, B.Sc. (Vicr.), PH.D. F.R.S. REVISED AND ENLARGED BY W. TATE, Assoc. N.S.S. (HONOURS),- F.C.S. WITH A PREFACE BY SIR H. E. ROSCOE, B.A. PH.D. F.R.S. M.P. N%W EDITION o , pontoon MACMILLAN AND CO. AND NEW YORK 1891 The Right of Translation and Reproduction is Reserved T RICHARD CLAY AND SONS, LIMITED, LONDON AND BUNGAY. Copies Received 1870-1874. /?*>* Edition Printed for MACMILLAN & Co., 1874. Reprinted 1876, 1877. 7F#A AVy 1878. Reprinted 1880, 1881, February and December 1882, 18 April and December 1888, 1890. .AVw Edition 1891. PREFACE THE following complete series of Questions and Problems in Elementary Chemistry will prove a great boon to those engaged either in teaching or in learning the science. They were prepared by my friend DR. THORPE, with special reference to our junior classes in Owens College, in which my " Elementary Chemistry" is used ; but they will of course also be available where any other modern text-book is employed. My experience has led me to feel more and more strongly that by no method can accuracy in a knowledge of Chemistry be more surely secured than by attention to the working of well-selected problems, and Dr. Thorpe's thorough acquaintance with the wants of the student is a sufficient guarantee that this selection has been care- fully made. I intend largely to use these Questions in my own classes, and I can confidently recommend them to all teachers and students of the science. HENRY E. ROSCOE. Oct. 15, 1870. 237344 TABLE OF CONTENTS PAGE vletric System of Weights and Measures i Conversion of Thermometric Scales 6 Correction of Thermometer Readings 8 Correction of Barometer Readings 10 Gay Lussac's Law ; Boyle's Law ; Correction of the Volumes of Gases for Temperature and Pressure ; Law of Partial Pressures 13 Relative Density of Solids, Liquids, and Gases ; Vapour Density 17 Avogadro's Hypothesis ; Density and Molecular Weight ; Valency, Equivalents, and Atomic Weights 33 Deduction of Empirical Formula from Percentage Compo- sition. Formulae of Minerals 36 To Calculate the Percentage Composition of a Compound from its Formula 42 Chemical Equations ; to Calculate the Amount of Material Required to Produce a Given Weight of any Substance, or the Quantity of the Substance Produced by the De- composition of a known Weight of the Material ... 44 Combination and Decomposition of Gaseous Bodies . . 52 Gas Analysis Calculations 54 Calculation of the Results of Atomic Weight Determina- tions 67 Calculations Involved in Indirect Analysis 70 General Analytical Questions 74 Vlll TABLE OF CONTENTS. PAGE Solution of Gases in Liquids 78 Solubility of Solids in Liquids ; Molecular Weight and the Lowering of the Freezing-Point of Solutions 82 Exercises on the Specific Heat, Latent Heat, and Atomic Heat of Substances 86 I lent of Solution ; Heat of Combination ; Calorific Power and Calorific Intensity 91 APPENDIX. Table I. Atomic Weights and Symbols of the Elements . 97 Table II. Weight of One Cubic Centimetre of Atmo- spheric Air at Different Temperatures from o to 300, at 760 mm. Pressure 99 Table III. The Weight of 1000 c.c. of Water at tC., when Determined by Means of Brass Weights in Air of oC., and at a Tension 0*76 m., is equal to IOOCKA; grams ~. . . . 100 Table IV. Volume and Density of Water at Different Temperatures . . 101 Table V. For the Calculation of i o . ocn 6 7 r r IO2 Table VI. Vapour Pressure (Tension) of Water .... 103 Table VII. For the Conversion of Degrees (T') of a Mercurial Thermometer into the Corresponding Values of an Air Thermometer 104 Table VIII. Table for Correction of Thermometer Read- ings 105 Table IX. Table of Useful Constants ic6 Table X. Logarithms and Anti-Logarithms 107 KEY TO CHEMICAL PROBLEMS 117 CHEMICAL CALCULATIONS. METRIC SYSTEM OF WEIGHTS AND MEASURES. MEASURES OF LENGTH. Standard : i metre, the length at oC. of a platinum bar preserved in Paris. Originally constructed to repre- sent the 4^0^0000 P art f tne earth's circumference, measured along the Paris meridian ; according to modern measurements it does not fulfil this intention and can therefore only be viewed as an empirical standard. 10 decimetres (dcm.) . . . I metre (m.). 100 centimetres (cm.) ... 1000 millimetres (mm.) ... looo metres . . . . I kilometre. i inch = 2-539954 centimetres. i foot = 3-047945 decimetres. I yard = 0-914383 metre. i mile = 1*609315 kilometres. MEASURES OF SURFACE. I oo square decimetres (sq. dcm.) i square metre or centiare (sq. m.). 10,000 centimetres (sq. cm.) 1,000,000 millimetres (sq. mm.) 100 square metres i are. 10,000 .... i hectare. I square inch = 6*4513669 sq. cm. i foot = 9*2899683 sq, dcm. i yard = 0*83609715 sq. m. i acre = 0*404671021 hectare. B B 2 METI-J/C SYSTEM OF MEASURES OF CAPACITY. 1000 cubic decimetres (c.dcm.) I cubic metre or stere. 1,000,000 centimetres (c.c.) 1,000,000,000 millimetres (c.mm.) I cubic decimetre ... I litre. I cubic inch = 16-386176 c.c. I foot = 28*315312 c.dcm. I gallon = 4-54345797 litres. MEASURES OF WEIGHT. Standard : I kilogram, the mass of a piece of platinum weighed in vacuo ; preserved in Paris. Theoretically the gram equals the mass of ic.c. and the kilogram the mass of 1000 c.c. of distilled water at 4C, at the latitude of Paris ; the actual standard differs slightly from its value as thus defined. 1000 grams (grms.) . . . i kilogram 10,000 decigrams. .... 100,000 centigrams .... 1,000000 milligrams .... I grain = 0*06479895 gram. I troy oz. = 31*103496 grms. i Ib. avd. = 0*45359^65 kilo, i cwt. = 50*80237689 kilos. To reduce grams to grains. Log. grams + 1*188432 = log. grains. To reduce cubic centimetres to cubic inches. Log. c.c. + ( 27855224) = log. cubic inches. To reduce millimetres to inches. Log. mm. + ( 2*5951742) = log. inches. To convert grains into grams. Log. grains -j- (-2*8115680) = log. grams. To convert ctibic inches into cubic centimetres. Log. cb. in. -+- 1*2144776 = log. c.c. To convert inches into millimetres. Log. inches + 1*4048258 = log. mm. WEIGHTS AND MEASURES. 3 SURFACES AND CAPACITIES. Area of a square = side 2 . Area of a rectangle, rhombus, or rhomboid = side X perpendicular height. Area of a triangle = half the side X perpendicular height. Area of a circle = 3*141593 X radius 2 . Area of an ellipse = 3*141593 X major semi-axis X minor semi-axis. Surface of a cube = 6 X edge 2 . Surface of a sphere = 12*566370 X radius 2 . Surface of a cylinder = 6*283185 X radius of base X sum of height and radius of base. Surface of a spherical segment = 6*283185 X height X radius of circular base. Volume of a cube = edge 3 . Volume of a sphere = f X 3*141593 X radius 3 . Volume of a cylinder = 3*141593 X height X radius 2 . Volume of a prism = base area X height. Volume of a cone or pyramid = J X base area X height. EXAMPLES. 1. How many millimetres are contained in 5 metres ? From the table we find that looo millimetres = i metre. .*. 5 metres = 1000 X 5, or 5000 millimetres. 2. How many decimetres are equivalent to 106725 millimetres ? i decimetre = 100 millimetres. .'. 106725 millimetres = 1067*25 decimetres. 3. How many litres of air are contained in a room measuring 4X3X2 metres ? 4 X 3 X 2 = 24 cubic metres. i cubic metre = 1000 litres. .'. looo X 24 = 24,000 litres. 4. Required the number of milligrams in 15 c.c. of water measured at 4 C. i c.c. of water at 4 C. = I gram. looo milligrams = I gram. .'. 15 x looo 15,000 milligrams. B 2 METRIC SYSTEM OF QUESTIONS. 1. How many millimetres, centimetres, and decimetres are respectively contained in 0*437 of a decimetre ? 2. Required the number of (a] centimetres, (b) deci- metres, (c) metres, in 1098765421 millimetres. 3. Required the number of square millimetres, square centimetres, square decimetres contained in the top of a table measuring i metre by 70 centimetres. 4. Define the gram and the litre How many grams are contained in 1725 kilograms ? 5. How many centigrams are contained in 2*567 kilograms ? 6. Required the number of milligrams contained in 5 cubic centimetres of water measured at 4C. 7. Calculate the number of square centimetres con- tained on the surface of a paper filter possessing a radius of 5 centimetres. 8. In an English inch are contained 25*3995 millimetres. How many kilometres are there in a mile ? 9. The mean height of the barometer in the latitude; of Greenwich at the level of the sea is 30 inches. How many millimetres is this equivalent to ? 10. A gallon is equal to 4*543 litres. How many cubic centimetres are contained in one pint ? 11. A spherical glass bulb possesses a radius of 3 centi- metres ; calculate its capacity. 12. Calculate the length of the polar axis on the suppo- sition that the earth is spherical, and that the actual metre is identical with the metre as defined. 13. A fine wire 255 millimetres in length weighs 0*172 grams. What length of such wire would it take to make a centigram ' rider ' ? WEIGHTS AND MEASURES. 5 14. A piece of platinum foil measuring 10*5 centimetres by 1*5 centimetres weighs 0723 gram. Into how many pieces, each weighing one decigram, may it be divided ? 1 5. A piece of Swedish filter paper measuring 60 centi- metres square, leaves on burning 0*1062 gram of ash. Calculate the amount of ash left on burning filters pos- sessing the following radii : (a) 3 centimetres, (b) 4 centi- metres, (c] 5 centimetres, (d) 6 centimetres, (e) 8 centi- metres, (/) 10 centimetres. 1 6. A perfectly cylindrical tube 90 millimetres in length holds exactly one gram of water at 4 C. Calculate its internal diameter. 17. I require a perfectly spherical bulb to hold 300 cubic centimetres. What internal diameter must it have ? 1 8. A cylindrical graduated tube 15 millimetres in diameter holds 25 cb. centimetres up to the hundredth division. Calculate the value in grams of water at 4 C. of one division. What will be the length between each division ? 19. A flask, nominally of I litre capacity, is found to contain 997 grams of distilled water at 15 C., when the lowest point of the meniscus coincides with the graduation mark ; (see table III) by how much do its contents fall short of their supposed value ? 20. A 5OC.C. pipette delivers 50 grams of water at 16 C.; what is the true volume of the space up to the graduation mark ? 21. In the verification of a Monies burette successive volumes of 10 c.c.'s of water, at 15 C., weighed 9-94 grm., 9*99 grin., io'o6grm., IO'I3 grm., 10*19. grm. What correc- tions should be applied to obtain the volumes delivered from the zero-mark to the 20, 40, and 50 c.c. marks respec- tively ? CONVERSION OF CONVERSION OF THERMOMETRIC SCALES. (C. Centigrade ; F. Fahrenheit ; R. Reaumur.) Fahrenheit to Centigrade f (F. 32) = C. Centigrade to Fahrenheit j C.+32 = F. Reaumur to Fahrenheit f R.+32 = F. Fahrenheit to Reaumur |(F. ~32) = R. Centigrade to Reaumur | C. = R. Reaumur to Centigrade f R. = C. EXAMPLES. I. What temperature on the Centigrade scale is equal to 212 F.? 212 - 32 = I08 180 x 5 900 2. What temperature on the Centigrade scale is equal to o F. ? o - 32 = - 32 - 32 X 5 - 160 -JT - - 17 77 C 3. Express 60 C. on Fahrenheit's scale. 60 X 9 540 5 T = Io8 Ans. 108 -f- 32 = 140 F. 4. Express 1 5 C. on Fahrenheit's scale. - 15 X 9 = ~ 135 5 5 Ans. - 27 + 32 = + 5 F. THERMOMETRIC SCALES. QUESTIONS. 22. Express the following temperatures in C. 60 F. o R. -10 R. 80 R. -I5 C '5 F. 25 C 'S R. o F. 500 F. o c '2 F. 23. Express the following temperatures in R. I5'5C. i2ooF. o-25C. 60 F. -40 C. o-o F. 32 F. 240 C. i5 c 'oF. 24. Express the following temperatures in C F. 4-o C. 12 R. 212 C. I5 C '5 C. o'i C. 360 R. 60 C 80 R. i-oC. 25. Bromine boils at I387 F. What is the correspond- ing temperature on the Centigrade scale ? 26. Ether boils at 34'9 C., alcohol at 78'4 C., and sulphuric acid at 338 C. What are the corresponding boiling-points on Fahrenheit's scale ? 27. Mercury freezes at -38'8 C. and boils at 3S7' 2 S C. Calculate the corresponding temperatures on the scales of Reaumur and Fahrenheit. 28. Cast-iron melts at 1550 C., zinc at 423 C., lead at 334 C., tin at 235 C. Calculate these temperatures in 29. At what point are the numbers on the scales of Fahrenheit and Celsius identical ? To what temperature does this point correspond on the scale of Reaumur ? CORRECTION OF CORRECTION OF THERMOMETER READINGS. Thermometer readings require, in general, to be corrected : (1) For change in the zero-point. (2) For the cooling of that portion of the thread not immersed in the heating medium (the emergent thread). For the most accurate work a correction is also applied for the want of uniformity of the bore of the capillary stem, generally by means of a table formed by calibration of the tube or by comparison of the individual thermo- meter with a standard instrument. Finally, the mercurial- thermometer readings, to be comparable, should all be reduced to air-thermometer readings by means of Table VII. (Appendix). An approximate correction for the emergent thread may be obtained by means of the expression 8 (t /')#, where S the co-efficient of apparent expansion of mercury in glass = '000154, / = the observed temperature on the thermometer, /' = the mean temperature of the cooled column as determined by an attached thermometer, and n the length in degrees of the cooled column. In ordinary cases 8 = '000145 gives a result nearer the truth than the actual coefficient ; a useful table calculated by means of this constant is appended (Table VIII).* EXAMPLES. I. In the determination of the boiling-point of a sample of ethyl formate a thermometer is used indicating o'8 when plunged into melting ice. When the stem is en- tirely surrounded by vapour the reading taken is 55. What is the corrected boiling-point ? * Several other formulae have been proposed to calculate the correction required, but neither these nor the one given above give the true correction in all cases ; for moderate values of n Table VIII. is fairly true if thermo- meters of the ordinary type are employed. See the article "Zur Korrection der Thermometerablesungen fur den herausragenden Faden," by Dr. E. Rimbach, in the Zeitschrift fur Instrumentenkunde, May 1890. THERMOMETER READINGS. 9 The zero of this thermometer has risen o*8, hence all readings must be diminished by this amount and we have 55-o-8 = 54*2 BP required. 2. With a thermometer reading o*2 in melting ice, the BP of propionic acid is indicated as I39*2 ; the column is immersed in the vapour to the 70 mark and the tem- perature of the exposed part is 26 ; what is the corrected BP? What is the corresponding H - thermometer tem- perature ? Correction for change of zero-point = -o*2. Correction for cooled column = + I*I3 for n = 139*2 - 70 = 69*2 (70 nearly), and /-/' = 139*2- 26 = 113*2, while from Table VIII. for n = 70 and /-/' = no and 1 20 the corrections are + i*io and + i*2o respectively. The required BP is therefore 139*2 -o'2 + i*i3 = I4o*i3. From Table VII. 139*85 mercury thermr. = 140 hydrogen thermr. ; hence 140* 13 = (140 - 139*85) = 140*! 3 + o*i5 = I4o*28 is the corresponding hydrogen- thermometer temperature. QUESTIONS. 30. The indicated temperature of the vapour of boiling methyl acetate is 57*9 when the thermometer is immersed in the vapour to the 100 mark ; the same instrument reads o'5 in melting snow ; what is the corrected boiling point ? 31. During a fractionation of isobutyric acid 4 portions are collected (a) from I49*5 to i5o*3 (b) from I5o c *3 to I5o c *9 (c) from I5o c '9 to I5i*i (d) I5i c *i to I5i*5 as in- dicated by a thermometer (standing at o c *4 in clean melting ice), having 55 immersed in the vapour, temperature of outside column as given by a second thermometer being 27. What are the corrected temperatures between which the four fractions pass over ? 32. A sample of ethene dibromide distils over mostly between I3o c *6 and J3O G *7, n = 55 and t' = n c *3. What is the corrected BP of this fraction? What is the corre- sponding temperature on the hydrogen thermometer scale ? IO CORRECTION OF 33. Reduce the indicated BP's given below to tem- peratures on the hydrogen-thermometer scale. N 2 4 2i-8 5 (n = 22, /' = i 3 - 5 ) SiCl 4 58'2o (n = o) PC1 3 76 c 'i5(>* = i8-2, /'= 2 6 C ) POC1 3 io7'3o (n = o). VOC1 3 I27 c 'i5 (n = 10, /' - 30) CORRECTION OF BAROMETER READINGS. Observed barometric heights are reduced to the height of a column of mercury at o and at the level of the sea in latitude 45, which would produce the same pressure. The corrections needed for this reduction are : I. Correction for temperature, due to expansion of the mercury and the material of the scale. Let m = coefficient of expansion of Hg, S = co- efficient of linear expansion of the substance of the scale, h = the observed height of the barometer, and H the corrected height ; then H = h ( l * I mt' or approximately H = h {i - (m -s]t\ Hence the required correction is - (m - s)t.h. With a brass scale we have H = h (i o*oooi62/) and the correction is -o'oooi62././ ; and with a glass scale similarly H = h (i - 0*0001 57/) and the correction is 0*000157. t.h. II. Correction for height above the sea-level and lati- tude, obtained as follows : If X be the latitude, and x the height expressed in metres of the place of observation above the sea-level, the length of the column of mercury which produces the standard pressure is BAROMETER READINGS. II 760*000= 1*946 cos 2 X + 0*000 1 49-r. mm. therefore if the observed height of the barometer be h the true or re- duced height jj = h X 760 760 + 1*946 cos 2\ -f- 0*000149 x- If x be less than loom, the term 0*000149^ may be omitted. As these corrections are small, a result approximately true is obtained by taking H = h 1*946 cos 2\ 0*000149 *" the greatest error so made being 0*04 mm. A further correction is required for the combined effects of error in adjustment of the scale and the depression of the column produced by capillarity, this correction should be given by the maker from comparison with a standard instru- ment. EXAMPLE. A Fortin barometer at latitude 52 and 2im. above sea-level stands at 754*34 mm. The instrument has a brass scale, the correction for index-error and capillarity is + o ' 2 6 mm., and the temperature at time of observation is 1 6 C. ; what is the reduced reading? Neglecting quantities beyond the second place of decimals. I. H = 754*34 (i -0*000162. /) mm. = 754'34~ 754*34 X 0*000162 X 16 mm. = 7 54*34 -i -96 mm. 75 2 *38 mm - II. H= H X7 6o 760 + I'946 COS 2 X + O*OOOI49. *"* 752*38 x 760 760 + 1*946 cos 104 + 0*000149. 2I - 752*38 x 760 760+ 1*946 (-0*2419) 752-38X760 =752*84 mm. 759*53 12 CORRECTION OF THE Hence true height of barometer, reduced to oC, sea- level, and latitude 45 = 752*84 + "26 mm. = 753*10 mm. QUESTIONS. 34. The barometer-readings taken during a gas-analysis with a Fortin barometer, of which the maker's correction was + 0*19 mm., 'were 754*83 mm., 756*21 mm., 757*39 mm., and 758*27 mm. The temperature of the instrument was throughout 18 C., what were the barometer-heights re- duced to oC. ? 35. Supposing the barometer to stand at 753*82 mm. at London (Lat. 51 30' N. T = 8 C C)., Liverpool (Lat. 53 25' N. T = I4 C C), Madeira I. (Lat. 32 45' N. T = 2O C C), and Bombay (Lat. 19 8' N.T = I9C.); what is the true pressure of the atmosphere at each of these places ? 36. Fill in the reduced barometric heights, in column VI. of the annexed table. All the barometers have brass scales. I. Temperature of barometer at the time of observa- tion. II. Height of place of observation above the sea-level. III. Latitude of IV. Maker's correction. V. Observed height of mercury column. I. II. III. IV. V. VI. A. 8 C. 20 m. 45 N. o'i2 mm. 753-72 mm. B. 15 C. 357m. 52 S. o'35 mm. 761 '08 mm. C. 13-5 C. 75 m. 63 N. + o'o8 mm. 755 -59 mm. D. 17 -2 C. 28 m. 25 S. - o'2i mm. 765-34 mm. E. i6'8 C. 23401. + o'i7 mm. 759-37 mm. F. 20 C. 998 m. 72 N. + o'i5 mm. 76o'i8 mm. G. xi C. 105 m. 58 N. - 0*23 mm. 762*15 mm. VOLUMES OF GASES. 13 CORRECTION OF THE VOLUMES OF GASES FOR TEMPERATURE AND PRESSURE. % Law of Charles : Gases under constant pressure expand ^73 part of their volume at o C. for every increase in temperature of i c C. Thus 273 volumes of any gas at o C. become 274 i C. 275 2 C. 276 3 C. 273 +* t This fraction 2*3, or the slightly more accurate decimal fraction 0*003665, is termed the coefficient of expansion of gases. EXAMPLES. 1. 10 c.c. of a gas are measured at 15 C. What volume will the gas occupy at 150 C. ? 273 + 15 : 273 + 150 :: 10 : x 423 X io -^s~ = I4 ' 69 cc - 2. ioo c.c. of hydrogen are measured at 100 C. How many cubic centimetres will the gas occupy at 100 C. ? 2 73 + 10 ' 273 - ioo :: ioo : x 173 X ioo - = 6- c.c. 3. What will be the volume at 25 C. of 252 c.c. of oxygen measured at 1 5 C. ? i + (15 X 0-003665) : i + (25 X 0-003665) :: 252 : x 1*054975 : 1-091625 :: 252 \x 1-091625 X 252 *"- 1-054975 14 CORRECTION OF THE By increasing the pressure upon a gas its volume is diminished : by diminishing the pressure the volume is increased. Boyle's Law : The volume of a gas at constant temperature is inversely proportional to the pressure to which it is subjected. i. 1000 c.c. of hydrogen are measured under a baro- metric pressure of 740 mm. What will the volume become under the standard pressure of 760 mm. ? 760 : 740 :: 1000 : x 740 X 1000 ~~~' 2. A litre of air is measured at o C. and 760 mm. What volume will it occupy at 740 mm. and 15' 5 C. ? 288-5 X 760 x 1000 273 x 740 ='085-340.0. Daltoris Law of Partial Pressures : The pressure of a mixture of chemically indifferent gases and vapours is equal to the sum of the pressures which each would exert if it alone occupied the whole space. EXAMPLES. I. If loo vols. of air consist of 20*66 O, 77*9 N, 0*03 CO 2 , and 1*40 of aqueous vapour, and the barometer stands at 760 mm. ; what is the partial pressure of each of these vapours and gases ? 760 X 20-66 Partial pressure of O = - -- = I57'oi mm. 760 x 77'9i N = --- ^ - = 592-11 mm. ~^ 760 x 0*03 CO 2 = i = 0*23 mm 100 O =7 6o x 1-40 = J0 . 64mnl 100 VOLUMES OF GASES. 1 5 The pressure of a vapour in contact with its own liquid is always the same at the same temperature. 2. An eudiometer, graduated in mm. from the closed end, contains coal gas ; pure boiled water is introduced until the mercury meniscus remains wet ; the upper level of the mercury with gas dry stands at 400 mm., the lower level at 800 mm. ; the average volume of i cm. of the tube = 1*66 c.c., barometer stands at 756 mm., temp, of air is i8C. What are the volumes of the dry coal gas and of the vapour of water in the moist gas respectively, reduced to standard conditions ? Vol. of dry coal gas = 1*66 x 40*0= 66' 4 c.c. at 756 - (800-400) or 356 mm. and i8C. Hence vol. at o c C. and 760 mm. = *&x 35^ x 273 760 x 291 = 29-18 c.c. Tension of aqueous vapour at i8 c C. = 15*33 mm - ; hence upper level of mercury is depressed to (400 X 15*33) mm. and vol. of aqueous vapour = i'66 X 41*536 = 68*95 c.c. at 15*33 mm - an d i8C., and this volume of a gas m 68-95 x 15*33 X 273 = cc at 6o mm> and CQ 760 X 291 QUESTIONS. 37. 170 volumes of oxygen are measured at ioC. What will the volume be on the temperature sinking to oC. ? 38. A certain weight of air measures a litre at o c C. How much will the air expand on being heated to iooC. ? 39. A gas has its temperature raised from 15 C. to 50 C. ; at the latter temperature it measures 15 litres. What was the initial volume ? 40. A volume of hydrogen measures 1500 cubic centi- metres at o C. How many cubic centimetres .will it measure at (a) I5 ? '5 C., (b) at 50 C., (c) at 100 C., and (d) at 300 C. ? At what temperature will it measure exactly 1000 cubic centimetres? 1 6 CORRECTION OF THE 41. The coefficient of expansion for gases for i C. is 0*003665. What is the coefficient for (a) i F., and (b) i R. ? 42. A certain volume of air preserved at a constant temperature measures 150 cubic centimetres when the barometer stands at 760 millimetres. On the following day its volume is found to have decreased 1*52 cubic centimetres. Calculate the alteration in the height of the barometer which must have ensued. 43. A balloon containing 1 200 cubic metres of coal gas under a pressure of 770 millimetres of mercury ascends until the barometer stands at 530 millimetres. What volume would the gas in the balloon now occupy suppos- ing none to have escaped ? 44. A certain quantity of nitrogen measures 155 cubic centimetres at 10 C., and under a pressure of 530 milli- metres of mercury. What will the volume become at i87 C, and under a pressure of 590 millimetres of mer- cury. 45. Calculate the temperature at which air possesses a density equal to that of hydrogen at o C. 46. Oxygen is passed into an eudiometer till the upper level of the mercury stands at 250 mm., level in trough at 780 mm. It is required to add double the quantity of hydrogen ; if the tube be adjusted till the trough level becomes again 780 mm. and the average volume of 10 mm. of tube is 1 74 c.c., what will be the reduced volume of the added hydrogen, and what the reading of the upper level of the mercury in the eudiometer after the addition (Bar. 754mm., T. 17-5 C.)? 47. Supposing the atmosphere to be saturated at 15 C., barometer 754 mm., what is the per centage volume of aqueous vapour contained in the air ? 48. Defining the dew-point as the temperature at which the air would be saturated when containing the amount of vapour of water present at the time of observation, give the per centage volume of aqueous vapour present in the air when the dew-point is 15 C. and 5 C. respectively (Bar. 765 mm.). VOLUMES OF GASES. 17 49. A gas burette contains 53*2 c.c. of moist coal gas at 760 mm. pressure and 18*5 C: find the volume of the dry gas under standard conditions. 50. 2370 c.c. of dry nitrogen are contained in an ab- sorptiometer, 5 c.c. of water are introduced and by agita- tion 0*07 c.c. of the gas is absorbed. The pressure is brought to that of the atmosphere, 757*3 mm., when reading the volumes, and the temperature remains at 16 C. : what is the volume of moist nitrogen indicated at the end of the experiment ? RELATIVE DENSITY OF SOLIDS, LIQUIDS, AND GASES ; VAPOUR DENSITY. THE specific gravity (sp. gr.) or relative density of a solid or liquid substance is the ratio of its mass to the mass of an equal volume of some liquid taken as unity. The standard universally adopted is pure water at its maximum density. The number which expresses the relative density of a solid or liquid substance denotes, therefore, how much heavier or lighter the substance is than an equal bulk of water at 4 C. I. The relative density of a solid is generally ascertained by: I. The hydrostatic balance method. Weight of substance (W) W nS * "" Weight of equal vol. water at 4 C. ^ W - _ w . 760 . (i + 0-00367 T) ~ v . 0-00008958 .(H-h).(i+ kT)> = (25*0122 - 24*4722) + (169*5 X 0-001252-^8) (169*5 - I2*5).0*000722'f| = 0*5400 + 0*2128 iS7.o-ooo722.m-* 4 ' 43 = 07528.14-436. 1 57.0-000722. ?-u. = From the more accurate formula '' + Q-Q0367 T). 760. 7- ^1 ^'.0-00008958 / I +0-00367. /J 76o.(i +0-003677;) Taking k for glass equal to 0*000023, we have SOLIDS, LIQUIDS, AND GASES. 25 , o*ooi2932.(i69*5-i2*5).762. 25-0122 -24'4722+ / - N ^ J _ * ^ ^( 1 4-Q-OQ367 X 2 1 5)760 762.0*00008958 760. (1+0-00367x21 5). _ [170-3- 2r6] 762.0-00008958. J 76o 1789 = 737i XI 789* 76o 1487x0-00008958x762 ~ 9 73 * QUESTIONS. 51. Calculate the relative density of a solid substance from the following numbers : Weight of substance in air ... 2*4554 grams. water . . . 2*0778 grams. 52. Determine the relative density of "gold from the following data : Weight of gold in air . . . . 4*6764 grams. Loss of weight in water . . . 0*2447 grams. 53. Find the relative densities of the following sub- stances : Granite Marble Hamatite Weight in air. 409*82 grams. 53*2841 grams. 13.6287 grms. water. 259*31 . 33*4020 . 10*9406 . 54. Determine the relative density of wood from the following numbers : Weights of : wood in air, 4 grams, ; a silver sinker, 10 grams, ; wood and sinker under water, 8*5 grams. Relative density of silver = 10*5. 55. A platinum wire 10 cm. in length weighs 0*12 gram. The relative density of platinum being 21*275, calculate the diameter of the wire. 26 RELATIVE DENSITY OF 56. A glass bulb of 120 c.c. capacity weighs 14*5 grams. The relative density of the glass being 2*48, calculate the thickness of the walls of the bulb. 57. Faraday calculated that the gold contained in 4 sovereigns could be drawn into a wire long enough to surround the earth. The weight of a sovereign is 7 '988 grams, and 91*66% of this is pure gold, possessing a relative density of 19*3. Assuming that the length of a quadrant of the earth is 10,000,857 metres, find the thick- ness of the wire. 58. A piece of bell-metal, weighing 7*550 grams in air, weighs 6*6635 grams when weighed under water. Determine the percentage amount of copper and of tin in the piece on the assumption that no alteration in bulk has occurred in alloying the two metals. Relative density of copper = 8*93 ; of tin = 7*292. 59. A piece of cast iron, the relative density of which is known to be 7 *2, is suspected to have internal cavities. Its weight is found to be 203*04 grams, and when sus- pended in water at 4 C. it is found to weigh 171 '84 grams. What is the total volume of these cavities? 60. Find the relative density of calcium from the fol- lowing determination by Bunsen : Weight of empty bottle .... 13*640 grams. bottle filled with naphtha 20*275 partly . 16*650 and calcium 19*150 full of naphtha and 21*576 Relative density of naphtha . . . 0*758 61. From the given data, calculate the relative densities of: Weights of Substance in air. Bottle -f water. Bottle + water and substance. Quartz . . Heavy-spar . Calcite ^ . . Iron pyrites . 3-445 grams. Il'2l6 ,, 34'I5I n o'795 ,, 75 '103 grams. 69*002 ,, 8o'20I ,, 5*337 n 77-248 grams. 77*7i8 ioi'824 ,, 5*973 ,, SOLIDS. LIQUIDS, AND GASES. 62. Calculate the relative density of sea water from the following data : Weight of bottle empty 3'535 grams. filled with sea water, 77849 filled with water 7*6722 63. From the given data, calculate the relative densities of the liquids given. Weights of Bottle empty. Bottle filled with water. Bottle filled with substance. A. A mineral water . . B. Vanadyl trichloride . C. Chromyl dichloride . D Alcohol . . 14*1256 gm. i'74*3 3'53i2 111*1370 gm. 6*8880 7-6649 20*2051 111*7050 gm. 11*2190 11*4692 ,, 17*0876 ,, 64. A specific gravity flask holds 2,545 milligrams of alcohol, 42,740 of mercury, and 5,829 of sulphuric acid. Calculate the relative density of the sulphuric acid and mercury, the density of the alcohol being 0-8095. 65. A solid weighs in vacua 100 grams ; in water 85 grams ; and in another liquid 88 grams. What is the relative density of this liquid ? 66. A piece of a certain metal weighs 37395 grams in air ; 2*3545 grams in water ; and 2*0896 grams in another liquid. Calculate the relative density of the metal and of the liquid. 67. A column of distilled water 100 mm. high is found to balance columns of (a) glycerine 79*3 mm. high, (b) olive oil 109*3 mm. high, (c) turpentine 114*9 rnm. high. What are the relative densities of these three liquids ? 68. When a mercury barometer registers 760 mm., a glycerine barometer indicates 8203 mm., and a water barometer 10,162 mm. ; the vapour tension of water at the temperature of the air, 15 C., is 127 mm. of mercury. What are the relative densities of the mercury and glycer- ine? 69. Regnault obtained, by his method of counter- balanced glass globes, the following data, from which it is required to calculate the relative density of nitrogen (air = i). 2$ RELATIVE DENSITY OF I. Balloon filled with air while im- mersed in melting ice. Height of baro- meter, reduced to o C., at the moment of closing the tap H Q = 761-19 mm. Weight added to balloon p = i -487 gram. II. Balloon evacuated. Pressure of residual air, when balloon is immersed in ice h Q = 843 mm. Weight added to balloon . . . . P = 14* 141 gram. Air being replaced by nitrogen, same conditions. I. HQ = 758'55 mm. p = 1-8725 gram. II. /z = 2*18 mm. P =14-227 gram. 70. With the same apparatus, the figures given for hy- drogen were : I. //o = 756-16 mm. p =13-301 gram. II. k$ = 3-40 mm. P = 14*1785 gram. With air = I, find the relative density of hydrogen. 71. Calculate the density of thiophosphoryl fluoride from the following data : I. Capacity of glass bulb at I5*8C. and 776 mm. 239*86 c.c. Weight of bulb in air at 17*1 C. and 771 mm. . . . 597543 grams. Weight of bulb in air at 17 C. and 771 mm. filled with gas and re- sidual nitrogen at 10*7 C. and 771 mm 60*3954 grams. Volume of residual nitrogen at 1 1 *3C. and 771 mm .. 78*9 c.c. II. Capacity of glass bulb at 15-8 C. and 776 mm 239*86 c.c. Weight of bulb in air at 15*8 C. and 776 mm 59*7282 grams. Weight of bulb in air at 15*8 C. and SOLIDS, LIQUIDS, AND GASES. 29 778 mm. filled with gas and residual nitrogen at 77 C. and 778 mm. 60*5274 grams. Volume of residual nitrogen at 9*6 C. and 777 mm. 43*4 c.c. 72. An experimental determination of the density of hydrofluoric acid under diminished pressure gave : Weight of bulb and caps in air at 197 C. and 772mm 244 0120 grains. Weight of bulb and caps and vapour in air at 17*5 C. and 772 mm. . 244*0025 Cocks turned off at 31*8 C. Bar 772*3 mm., manometer reading 106*2 mm. Weight of bulb and caps + water at 24*3 C 528*18 grams. Residual air, reduced to o C. and 760 mm 3*58 c.c. Weight of I c.c. of air at o C. and 760 mm 0*0012928 Coefficient of cubical expansion of platinum for i C. 0*000027 Sp.gr. of air, H = i 14*435 . Hence find the rel. dens, of the vapour of hydrofluoric acid, under the conditions at closing the cocks, as com- pared with air, and with hydrogen. 73. From the tabulated data, calculate the correspond- ing relative densities of hydrofluoric acid at the given temperatures. Height of Percentage vol. of Pressure of Bar. residual air. vapour. 26*4 756 mm. 1*51 745 mm. 27-8 763 2*23 746 29*2 762 32*0 754*5 33'i 77o'5 767 36-3 751 38*7 764 i'55 75o i'52 743 2-64 750 1*14 758 i*58 739 1*69 751 30 RELATIVE DENSITY OF 74. A determination of the density of methyl bromide made by Bunsen according to his method yielded the following data. Calculate the relative density of the compound (H = i). Vol. of gas, at 0*7464 m. and 16*8 C. . . 42*19 c.c. Height of mercury above the level of the metal in trough 0*0243 m. Weight of the flask and gas 7 '9465 grams. Weight of the flask and air 7*8397 Temperature in balance case 6*2 C C. Height of barometer 0*742 1 m. 75. Bunsen determined the times of effusion through a minute aperture of the same volume of gases as follows : Carbon Air. Oxygen. Air. dioxide. 102*5 108*5 1027 127*0. 103 109 1027 127*5, Air. Hydrogen. (1) 105*5 297 (2) 105*6 29*3 Find the rel. dens, of each gas compared with air. 76. The data below were obtained in three determina- tions of the vapour density of P 4 O 6 according to Hermann's method ; calculate the V.D. in each case, (A) Air = i. (B) I. II. III. Weight of substance taken Temperature of tube and 0*1887 grams. 0-1887 grams. TCQ C o'i887 grams. 184 C Temperature of air . . . Barometer Height of mercury column Volume of vapour at T C. Tension of mercury at T C 20 C. 770 mm. 625 mm. 137-5 c.c. i "5 mm 20 C. 769 mm. 616 mm. 1 39 '6 c.c. 5 '6 mm. 20 C. 769 mm. 605 mm. 141 -45 c.c. 12 mm. 77. A series of experiments by Ramsay and Young, under varying conditions, yielded the following data, from which the vapour-density of acetic acid is to be found in each case (Hermann's method). (H ' = i). SOLIDS, LIQUIDS, AND GASES. Weight taken. Reduced Temp. Corrected pressure. Volume. A. 0*01126 gram. 50 C. 13-4 mm. 170*06 c.c. B. 0*03565 ,, 78-4 63'2 ,, 137-8 C. 0*01126 ,, Il8*2 ,, 84 "05 45'i5 D. 0*03565 ,, 162*5 ,, 373'i 37'o E. 0*03565 ,, 184-1 79*'6 18 78. In the determination of a vapour density by Gay- Lussac's method the following data were obtained : Weight of substance taken, 0*1163 grams. Temp, of bath, 2i5C. Observed volume of vapour, 50770.0. Bar- ometer, 755*5 mm. Difference of height of mercury inside and outside of tube, 80*0 mm. Height of spermaceti column reduced to mm. of mercury, 16*9 mm. Required the sp. gr. of the vapour (H i) ; also the weight of one litre at o C. and 760 mm. barometric pressure, assuming the vapour to behave like a perfect gas. 79. Using a porcelain vessel, heated in a Perrot's fur- nace to 1027 C., Victor Meyer obtained the data given below, from which the vapour density is to be calculated. I. II. Weight of substance taken (Iodine) Barometer . . 0*0874 grams. 722*8 mm 0*0847 grams. 722*8 mm. Temperature of the room .... Volume of air displaced .... 2i*5C. i3'7 c.c. K-aT'C i3'3 c.c. 80. From the tabulated data, obtained by Victor Meyer's method, find the sp. gr. of the vapour of each substance, referred (i) to air, (2) to hydrogen, as unity. Substance. Weight taken. Temp, of room. Bar. Volume of air displaced. A. CHC1 3 B. CS 2 C. OH 2 D. Iodine 0*1008 gram. 'o495 >, O*OIO2 ,, 0*II05 ,, 16*5 C. 16*5 C. 16*1 C. 16*0 C. 707*5 mm. 717*8 mm. 723*3 mm. 714*8 mm. 22 C.C. 16*4 c.c. 14*6 c.c. II'I C.C. 3 2 RELATIVE DENSITY OF 81. The specific gravity of coal-gas is about 0*5 referred to air as unity, and air is 14*435 times heavier than hydrogen. Calculate from these data the weight of coal- gas, at 10 C. and 530 mm. pressure, required to fill a balloon having a capacity of 800 cu. m. i) of a hydride 82. Calculate the vapour density (H - from the following numbers : Balloon with air, 407955 grams. Temperature, 11 C. Balloon with vapour, 41*0960 grams. Temperature of sealing, 173 C. Capacity of balloon, 189 c.c. 83. Calculate the vapour density of camphor vapour (H ' = i) from the following numbers, obtained by Dumas according to his method : Temp, of air, 1 3*5 C. Bar. 742 mm. Temp, of bath at the moment of sealing the globe, 244C. Increase of the weight of globe, 0708 grams. Volume of mercury re- quired to fill the globe, 295 c.c. Residual air nil. 84. In a series of determinations by Deville and Troost of the vapour density of sulphur, the following figures were obtained : I. II. 111. Temperature of the air 10 C. 15 C. 10 C. Height of barometer . . Difference of weighings . Volume of porcelain bulb . 769*5 mm. 0*190 gram. 281 c.c. 760 mm. 0*176 gram. 280 c.c. 769*4 mm. -0*203 gram. 303*5 c.c. Residual air 3 c.c. 6 c.c. a'e c C Temp, of bath (Zn vapour) 1040 C. 1040 C. 1040 C. Find the vapour density in each case, taking air = i. Coefficient of cubical expansion of porcelain = o'ooooioS. (N.B. The second weighing gives a figure here less than the first with the empty bulb in air.) 85. Using a modification of Dumas 7 method, Friedel and Crafts obtained the following data, from which the vapour density of aluminic chloride is to be calculated : Temp, of vapour, 398*2 C. Pressure of vapour, 0*97 SOLIDS, LIQUIDS, AND GASES. 33 atmosphere. Barometer height, 754*8 mm. Volume of globe, 250 c.c. Residual air, 245 c.c. at 10*3 mm. and 1 7 -2 C. Excess of weight of globe, 0*8765 gram. Weight of air displaced 0*3020 gram. AVOGADRO'S HYPOTHESIS ; DENSITY AND MOLECULAR WEIGHT; VALENCY, EQUI- VALENTS AND ATOMIC WEIGHTS. According to Avogadro's Hypothesis equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. Hence it follows that the molecular weights of bodies in the state of gas or vapour are in the same ratio as their specific gravities or vapour densities : e.g. if H = i, the sp. gr. of gaseous HC1. = 18*18. Assuming the molecular weight of H to be 2, the molecular weight of HC1 is 36*36. The molecular weight of a gaseous element or compound may be thus defined : The molecular weight of a gaseous element or compound is a number which expresses how many times greater than two unit masses of hydrogen is the mass of the specified element or compound which occupies (under the same conditions of temperature and pressure) the same volume as is occupied by these two unit masses of hydrogen. The maximum atomic weight of an element is a number which expresses how many times greater is the smallest mass of that element which combines with other elements to form a compound gaseous molecule, than the smallest mass of hydrogen which com- bines with other elements to produce a compound gaseous molecule, such smallest mass of hydrogen being taken as unity. (Watts's Diet. I. p. 341.) That property of an atom which determines the number of other atoms with which it can combine to form a com- pound molecule is termed its valency or atomic value. Valency is usually expressed as the number of hydrogen atoms, or atoms chemically equivalent to hydrogen, which one atom of the specified element can combine with or replace when in the same state as in the compound considered. 34 AVOGADRO'S HYPOTHESIS. The equivalent of an element is the number expressing that weight of the element which will combine with or replace the unit weight of hydrogen. Let E the equivalent of any element, A its atomic weight, and V = its valency in the compound used for determining the equivalent ; the relations between these quantities are expressed by the equation A = E . V. The numerical value of the molecular weight of any substance is obtained by the summation of the atomic weights of the component elements each multiplied by the number denoting how many of such atoms are contained in the molecule, e.g. the molecular weight of alcohol, C 2 H 6 O, is found thus (12 X 2) -f- (i x 6) + 16 = 24+ 6 + 16 = 46. QUESTIONS. 86. If the density of sulphuretted hydrogen = 1791, ammonia = 0*597, nitrous oxide = 1*520, marsh gas = '555> chloroform = 4*20, and stannic chloride = 9*20, as compared with air, what are the molecular weights of these bodies taking the molecular weight of hydrogen as 2 ? 87. Determine the weight of one litre of the following simple gases and vapours at o and 760 mm., on the supposition that they can all exist as perfect gases at the standard temperature and pressure : Oxygen. Sulphur. Sodium. Chlorine. Phosphorus. Arsenic. Iodine. Nitrogen. Mercury. 88. Calculate the volume, at the standard temperature and pressure, of a kilogram of each of the following gases and vapours : Carbon monoxide. Ethylene. Hydrogen sulphide. Oxysulphide of carbon. Marsh gas. Bromine. Water. Hydrochloric acid. 89. What weight of each of the specified substances is represented by the appended formulas, taking one gram as the unit of weight ? Lead chloride, PbCl 2 ; Silicon fluoride, SiF 4 ; Ferric chloride, FeCl 3 ; Methyl bromide, CH 3 Br. AVOGADRO S HYPOTHESIS. 35 90. Find the density (air = i) of : Carbon anonoxide, CO ; Carbon bisulphide, CS. 2 ; Sulphur trioxide, SO 3 ; Boron trifluoride, BF 3 ; Phosphorus pentafluoride, PF 6 . 91. If the electro-chemical equivalent of: Oxygen = 7*98, Silver = 107*66, Chlorine = 35*37, Antimony = 39'86, and Copper = 31*59, and the hydrogen-replacing power of these elements in the compounds used in rinding the equivalents be for one atom of the element taken, Oxygen 2, Silver i, Chlorine I, Antimony 3, and Copper 2 atoms ; what are the atomic weights of these elements ? 92. From the data given deduce the probable atomic weight of oxygen : Gaseous compound. Rel. dens, (air = i). Rel. dens. X 28-87 approximate mol. wt. Mole- cular weight. Analysis stated in parts by wt. per mol. Carbon dioxide **53 44 '2 43'89 31 '920 + n'97C. Sulphur dioxide 2-25 64-9 63-90 31*920 + 31-988. Sulphur trioxide Carbon monoxide Water . . . 2-9 0-97 0*63 SB"? 27-97 l8'2 79-86 28-93 17-96 47-880 + 31-988. 15*960 + 11*97 C. 15-960 + 2 H. Nitric oxide . i 04 30 "o 29-97 15-960 + 14-01 N. 93. Find the most probable atomic weight of carbon from the appended experimental results : Gaseous compound. Rel. dens, (air = i). Analysis in parts by wt. Marsh gas Methyl iodide Chloroform .... Carbon monoxide . . Carbon bisulphide . . o'555 4-883 4*20 0-968 2-645 i'9 7 C. + <.*oH. i '97 C. + 126*53 I + 3*0 H. 1-97 C. + io6'n Cl + 1*0 H. i'9 7 C. + 15-960. 1*97 C. + 63-968. 94. Find the molecular weights of the following bodies : Cadmium at 940, Rel. dens. = 3-94, (air = i). Phosphorus at 500, Rel. dens. = 4-35, Mercury at 446, Rel. dens. = 6*98, D 2 DEDUCTION OF 95. Select the most probable atomic weight for phos- phorus as indicated by the given experimental results : Gaseous compound. Rel. dens, (air = i). Analysis in parts by wt. Phosphorus hydride .... Phosphorus chloride .... Triethyl phosphine oxide . . i'i5 4'88 4'6o 30-96 P. + 3-0 H. 30*96 P. -f io6'n Cl. /3o'96P. + 15-960. \+ 71-820. + 15-0 H. How many atoms are there in the phosphorus molecule ? DEDUCTION OF THE EMPIRICAL FOR- MULA OF A BODY FROM ITS PERCEN- TAGE COMPOSITION; FORMULAE OF MINERALS. In order to calculate the empirical formula of a com- pound, divide the percentage amount of each constituent by its corresponding atomic weight : divide each of the quotients so obtained by the lowest number, and reduce them to their simplest ratios. EXAMPLES. I. A body on analysis yielded the following percentage composition : Carbon .... 27*273 Oxygen. . . . 72727 Calculate its empirical formula. The atomic weight of carbon = 12 ; that of oxygen = 16. EMPIRICAL FORMUL/E. 37 Then c = 27-273 = 2 . 2727 j .; I 0=^1=4-5454 The simplest relation between the carbon and oxygen is at once seen to be as I to 2, since 2-2727 : 4*5454 :: i : 2. Hence the formula of the compound is CO 2 2. A compound of hydrogen and nitrogen was found to possess the following percentage composition. Calculate its formula. Nitrogen . . . 82-353 Hydrogen. . , 17*647 100*000 The atomic weight of N = 14, and of H = i. = 82-353 = . H= 17-647 = The simplest ratio between the nitrogen and hydrogen is as i to 3, since 5-882 : 17*647 :: i -3. Hence the formula of the body is NH 3 . 3. A compound of iron and oxygen possesses the fol- lowing percentage composition. Calculate its formula. Iron 70*01 Oxygen .... 29*99 100*00 At. weight of Fe = 56*0. At. weight of O = 16. 38 DEDUCTION OF Then The ratio between 1*2514 : 1*8744 :: I ' I 'S The simplest ratio, in whole numbers, is therefore 2 : 3. Hence the formula of the body is Fe 2 O 3 . FORMULAE OF MINERALS. There exist many groups of minerals of which the in- dividual members are composed of equal numbers of atoms similarly combined, and which possess the same or nearly the same crystalline form ; such groups are termed isomorphous groups, and the minerals forming each group are said to be isomorphous with each other. Isomorphous minerals may replace each other without affecting the exterior form of the mineral partially re- placed. As the percentage amount of the elements in such minerals is different in each case, mixed minerals do not admit of being directly represented by a simple formula derived as above from the percentage composition. It is the practice to represent minerals of this type by a general formula obtained by considering the particular iso- morphous substance present in greatest quantity as the typical substance and reducing the quantities of the re- placing bodies to equivalent quantities of the typical body, when the formula may be at once obtained by the ordin- ary means. When possible, mineral analyses are ordinarily reported as percentages of oxides, and in the calculation these oxides are considered as the isomorphous components to be reduced to terms of that present in greatest amount. As an example we may consider the case of ankerite : an analysis of which by Schrotter yielded the following results : EMPIRICAL FORMULAE. 39 CaO .... 26-90 MgO. . . . 7-84 FeO .... 19*91 MnO. ... 1-82 C0 2 . . . . 43'OS 99*52 CaO is the oxide present in greatest quantity : we there- fore reduce the other oxides to their equivalents of CaO and adding all together we obtain the typical composition of the mineral considered as if it were a Ca compound ; thus : 7 '4 X 5 __ IQ .g3 _ c a o equivalent to MgO present. 40 19-91 X 56 = g = CaO equivalent to FeO present. 72 1-82 X 56. = r4 6 = CaO equivalent to MnO present. 70 Hence total typical oxide present (M" O) parts. = 26-90 + 10*98 + 15-48 +1*46 = 54*82 combined with CO 2 ... 43*05 By the ordinary method M" O. . . - = 0*98 nearly, and -3- = i 56 0-98 C0 2 . . 43 ' 5 = 0-98 nearly ?2| = I 44 0-98 Hence typical composition is M"O.CO 2 or M"CO 3 , where M" represents Ca, Mg, Fe, and Mn. The formula of ankerite may therefore be written (Ca, Mg, Fe, Mn) CO 3 . DEDUCTION OF QUESTIONS. Deduce the formulae of the following substances : 96. Hydrogen . Oxygen 97. Nitrogen . Oxygen . . ! 98. Iron . . . Oxygen . . 99. Carbon . . Oxygen . . Sulphur loo. Potassium . Hydrogen . Sulphur . . Oxygen . . ! 101. Magnesium Sulphur Oxygen . . Water . . 1 02. Zinc . . . Sulphur Oxygen Water . . . 5-88 . 94-12 1 103. Calcium . . Phosphorus . Oxygen . . 1 104. Sodium . . Aluminium . Fluorine . . 105. Potassium Nitrogen . . Oxygen . . 1 06. Aluminium . Sulphur . . Oxygen . . 107. Copper . . Carbon . . Hydrogen . . Oxygen . . 1 1 08. Hydrogen Oxygen . . Nitrogen . . Sulphur Nickel . . . . 3872 . 20*00 . 41-28 100*00 lOO'OO 30-43 . 69-57 . 3279 . 13-02 . 54-19 100*00 . 70*01 29-99 100*00 lOO'OO . 45-95 . 16-45 - 37-60 . 20-00 . 26-67 53'33 lOO'OO 100*00 8 en K) 1-1 ON 00 ON . 2873 073 . 23*52 . 47-02 lOO'OO . 57-46 5-43 . 0'9I . 36*20 . 976 . 13-01 . 26*01 . 51*22 100*00 100*00 ' 33-88 . 14*84 . 16-95 . 2270 . 11*15 . 22*28 . 43-87 lOO'OO EMPIRICAL FORMULA. 109. Carbon . Hydrogen 1 10. Carbon. Hydrogen Carbon . Hydrogen Sulphur Oxygen . 8571 H'29 lOO'OO . 92*3 77 i oo-o 113. 19-04 476 25-40 50-80 112. Carbon Hydrogen. Nitrogen . Carbon Hydrogen . Nitrogen . Oxygen Platinum . Chlorine . . 74-07 . 8-64 . 17-29 lOO'OO 46-66 4-26 5-20 5-92 18-26 1970 lOO'OO 1114. Show that a mineral having the following per- centage composition is represented by the formula MOCO 2 when M signifies a metal of the magnesium family : Lime Magnesia Iron monoxide . . . . Manganese monoxide . Carbon dioxide . . . 99*3 1 1 5. The mineral kerolite gave the following numbers on analysis. Calculate its formula : SiO 2 MgO H 2 O 46-96 31*26 21*22 = 99-44. 1 1 6. A specimen of cobalt-bloom was found to have the following composition. Determine its formula : As 2 O 5 Co O Fe O H 2 O 38-43 36-52 roi 24-10=100-06 117. Calculate the formula of soda-feldspar from the following analysis : SiO 2 A1 2 O 3 Fe 2 O 3 CaO MgO K 2 O Na 2 O 68-45 i%"Ji 0*27 0-50 0-18 0*65 11-24=100-00. 42 CALCULATION OF 1 1 8. Calculate the formula of labradorite from the fol- lowing analysis : SiO 2 A1 2 O 3 Fe 2 O 3 CaO MgO Na 2 O 52*52 30*03 172 12*58 0*19 4*51 = 101*55. TO CALCULATE THE PERCENTAGE COM- POSITION OF A COMPOUND FROM ITS FORMULA. FIRST calculate the molecular weight of the compound by adding together the sums of its atomic weights. Thus the formula for water is H 2 O ; its molecular weight is therefore 17-96. Atomic weight of H. = i 1x2= 2 Atomic weight of O. = 15*96 15*96 17*96 In order to calculate from the molecular weight of the compound the percentage amount of its constituents that is, to determine how much of each constituent is con- tained in 100 parts of the compound multiply the amounts of the several constituents in the compound by 100, and divide each of the products by the molecular weight of the compound. Thus, to calculate the amounts of hydro- gen and oxygen contained in 100 parts of water : 2 x IPO 17*96 16 x 100 = - = 88 ' 86 1 00 '00 Table I. in the Appendix gives the atomic weights of the elements according to the most trustworthy deter- minations. PERCENTAGE COMPOSITION. 43 QUESTIONS. Calculate the percentage composition of the following compounds : 119. Water H 2 O. 1 20. Potassium chlorate .... K Cl O 3 . 121. Mercury monoxide .... Hg O. 122. Potassium nitrate .... K N O 3 . 123. Sodium nitrate, Na N O 3 . 124. Barium sulphate Ba S O 4 . 125. Calcium carbonate .... Ca C O 3 . . 126. Silver chloride Ag Cl. 127. Magnesium pyrophosphate . Mg 2 P 2 O 7 . 128. Potassium platinum chloride 2 (K Cl) + Pt C1 4 . 129. Sodium thiosulphate. . . . Na 2 S 2 O 3 + 5 H 2 O. 130. Magnetic oxide of iron . . Fe 3 O 4 . 131. Hausmannite Mn 3 O 4 . 132. Copper pyrites Cu 2 S + Fe 2 S 3 . 133. Stromeyerite Ag Cu S. 134. Stilbite Ca O AU O 3 . 6 Si O, + 5 H" 2 O. 135. Idocrase 9 (2 Ca O . Si O 2 ) . 2(2Al 2 3 . 3 Si0 2 ). 136. Spodumene 3 (Li 2 O . Si (X). 4 (A1 2 3 . 3 Si 2 ). 137. Sphene Ca Si O 3 . Ca Ti O 3 . 138. Pyromorphite 3 Pb 3 P 2 O 8 . Pb CL. 139. Ethyl alcohol C 2 H O. 140. Cane sugar C 12 H 22 O n . 141. Xylene C s H 10 . 142. Cymene C 10 H 14 . ( C 2 H 6 . 143. Silicon ethyl Si < * S? J ^2 "6 f C 2 H 5 . 144. Potassium thiacetate ... C 2 H 3 K O S. 145. Potassium ferrocyanide . . K 4 Fe C 6 N 6 . 146. Kreatin C 4 H 9 N 3 O 2 + H 2 O. 147. Rosaniline C 20 H 21 N 3 O. 148. Strychnine C 21 H 22 N 2 O 2 . 44 CALCULATION OF CHEMICAL EQUATIONS; TO CALCULATE THE AMOUNT OF MATERIAL REQUIRED TO PRODUCE A GIVEN WEIGHT OF ANY SUBSTANCE OR THE QUANTITY OF THE SUBSTANCE PRODUCED BY THE DECOM- POSITION OF A KNOWN WEIGHT OF THE MATERIAL. Chemical equations : The results of any chemical actions are represented by equations in which the signs + and = are used in the same sense as in algebra, so far as regards the weights of matter represented by the symbols. The meaning of such an equation will be best shown by examples. 1. H 2 SO 4 + BaCl 2 = BaSO 4 + 2 HC1. This equation indicates not only that sulphuric acid and barium chloride react to produce barium sulphate and hydrochloric acid, but, further, that the molecular weight of sulphuric acid reacts with the molecular weight of barium chloride to produce the molecular weight of barium sulphate and twice the molecular weight of hydrochloric acid ; hence 98 parts by weight of sulphuric acid react with 208 parts by weight of barium chloride to form 233 parts by weight of barium sulphate and 73 parts by weight of hydrochloric acid gas. The number of atoms of each element must be the same on either side of the equation, consequently the total number of atoms must be the same on either side. The numbers of molecules before and after a reaction are not necessarily the same. 2. H 2 +C1 2 =2 HC1. Here, as in (i), it is indicated that 2 parts by weight of hydrogen combine with 7 1 parts by weight of chlorine to produce 73 parts by weight of hydrochloric acid. But the equation signifies more : as we know the molecular weights AMOUNT OF MATERIAL 45 of all the reacting bodies and products it shows that a molecule of hydrogen combines with a molecule of chlorine to form two molecules of hydrochloric acid gas. If we state the molecular weights in grams we know from the equation that 22*4 litres of hydrogen combine with 22*4 litres of chlorine to yield 44*8 litres of hydrochloric acid, for two grams of hydrogen measure 22*4 litres ap- proximately at the standard conditions and, by Avogadro's hypothesis, equal numbers of molecules under the same conditions occupy equal volumes. EXAMPLES. i. How much oxygen can be obtained by the decom- position of loo grams of mercury monoxide ? The symbol for mercury monoxide is HgO : hence its molecular weight is 216. Hg = 200 O = 16 216 When mercury monoxide is heated, it is completely resolved into oxygen and metallic mercury, 2Hg O = 2Hg + 2 , 432 parts of mercury monoxide giving 400 of mercury and 32 of oxygen. If, therefore, 32 grams of oxygen are evolved by the decomposition of 432 grams of mercury monoxide, how many grams of oxygen will be evolved by the decom- position of 100 grams of mercury monoxide? 432 : 100 :: 32 : x. x = 7-407. 7-407 grams of oxygen will therefore be evolved on heating 100 grams of mercury monoxide. 2. I want loo Ibs. of oxygen : how many Ibs. of potas- sium chlorate must I take ? The formula of potassium chlorate is KC1O 3 . Its molecular weight is therefore 46 REQUIRED TO PRODUCE A K = 39'I Cl - 35'5 3 = (16x3) 43-0 This substance on being heated eventually decomposes into potassium chloride and oxygen. 2K Cl 3 - 2K Cl + 3O 2 . That is, 961bs. of oxygen are evolved on heating 245*2 Ibs. of potassium chlorate. Then how many Ibs. of potas- sium chlorate are required to yield 100 Ibs. of oxygen ? 96 : 100 : 245*2 : x. x = 255-4 Ibs. Hence it would require 255*4 Ibs. of potassium chlorate to yield 100 Ibs. of oxygen. 3. How many cubic centimetres of oxygen and hydrogen, measured at ioC. and 770 mm. pressure, can be obtained by the decomposition of I gram of water ? The symbol for water is H 2 O, its molecular weight is 1 8. The weight of hydrogen yielded by a gram of water is 18 : I :: 2 : x x o'liii gram. The weight of oxygen is 18 : i :: 16 : x x = 0-8889 i 'oooo gram. 1000 cubic centimetres of hydrogen at o C. and 760 mm. weigh 0*08936 grams. Then what volume would 0*1111 gram of hydrogen occupy ? 0*08936 : o'lin :: 1000 : x. x 1243*28 c.c. of hydrogen. This at 10 and 770 mm. would measure 273 : 273 + io\ To^-oQ . 770 : 760 J ' I2 4o-S ;,* x = 1272*09 c.c. GIVEN WEIGHT OF SUBSTANCE; 47 1000 cubic centimetres of oxygen at o C. and 760 mm. weigh (16 X 0*08936 gram) = I '42976 grams. Then what volume would 0*8889 grams occupy ? 1*42976 : 0*8889 :: 1000 : x. x = 621713 c.c. of oxygen. This at 10 C. and 770 mm. would measure 273 : 273 770 : 760 273 : 273 + 10 ) .. 67I . 7T , . ] " 62171 ^ ' * x = 636*116 c.c. Therefore, a gram of water on being decomposed would yield at 10 C. and 770 mm. pressure, 1908*206 c.c. of mixed gases consisting of Hydrogen . . 1272*09 c.c. Oxygen . . . 636*116 1908*206 QUESTIONS. 149. How much potassium chlorate is needed to furnish I Ib. of oxygen ? 150. I require 2 kilograms of oxygen : how much (a) mercury monoxide, (6) potassium chlorate, (c) manganese dioxide, (d) sulphuric acid, shall I need ? 151. On completely decomposing by heat a certain weight of potassium chlorate, I obtain 20*246 grams of potassium chloride. What weight of potassium chlorate did I take, and how much oxygen was evolved ? 152. A gas bag has a capacity of 45 litres : how much manganese dioxide containing 70 per cent, of Mn O 2 is required to fill it with oxygen at 15 C. and 760 mm. barometric pressure ? 153. 132*74 kilograms of hydrogen are needed to inflate a balloon. What weight of zinc and sulphuric acid will be required to produce this quantity of gas ? 48 OR THE QUANTITY OF 154. Iron, zinc, and sulphuric acid diluted so as to contain 20 per cent, of real acid (SO 4 H 2 ), are supplied to you. Find the amount of these materials required to produce the above quantity of hydrogen, (i) by the action of sulphuric acid upon iron, (2) by the action of sulphuric acid upon zinc. 155. What weight of potassium chlorate is needed to furnish oxygen sufficient to burn the hydrogen evolved by the action of water upon 200 grams of sodium ? 156. How many cubic centimetres of oxygen and hydrogen measured at 12 C. and under a pressure of 762 mm. of mercury can be obtained by the electrolysis of 10 grams of water ? 157. loo grams of steam is passed over 1000 grams of red-hot iron wire. Required the volume of hydrogen evolved measured at 10 C. and 742 mm. pressure, and the weight of iron oxide produced. I 5^. 77 per cent, of the weight of the air, freed from moisture and carbonic acid, consists of nitrogen. Calcu- late the weight of (a] metallic copper, and (ft) of phos- phorus required to abstract the oxygen from I Ib. of air. 159. Required the weight of ammonia and of chlorine needed to produce a litre of nitrogen. 1 60. What quantities of nitre and Chili saltpetre respectively will be required to obtain the maximum quantity of nitric acid by reaction with 140 kilos, of 97 per cent, sulphuric acid ? ! 161. How much nitre and sulphuric acid shall I need to prepare nitric acid enough to neutralize exactly 5 Ibs. of chalk? 162. Calculate the volume of nitrogen monoxide at 15 C. and 740 mm. produced on heating 30 grams of ammonium nitrate. 163. What weight of copper is required to yield a litre of nitrogen dioxide at o c C. and 760 mm. ? SUBSTANCE PRODUCED 49 164. At the ordinary temperature and pressure water absorbs 50 per cent, of its weight of ammonia. Calcu- late the amount of sal-ammoniac and quicklime needed to produce 10 kilograms of Liquor Ammonice. 165. Coal contains about 2 per cent, of nitrogen. As- suming that 75 per cent, of this amount escapes as ammonia on distillation, calculate the amount of coal required to furnish the sal-ammoniac needed to pro- duce the 10 kilograms of Liquor Ammonia. 1 66. Calculate the weight of air required to burn a ton of coal possessing the following percentage com- position : Carbon, 88*42 ; Hydrogen, 5'6i ; Oxygen, 5*97. 167. How much marble and hydrochloric acid con- taining 22 per cent. HC1 are needed to yield 10 litres of carbon dioxide at 15 C. and 760 mm. barometric pres- sure ? 1 68. According to Boussingault, a square metre of leaf will decompose in sunlight rio8 litres of carbon dioxide in an hour. Calculate in tons the amount of carbon assimilated in an hour by a million trees, each possess- ing 100,000 leaves, and each leaf containing 25 square centimetres. Calculate the volume of the carbon so assimilated on the assumption that it possesses a sp. gr. of I '6. ! 169 How many litres of (i) hydrogen and of (2) car- bonic oxide gases at 10 C. and 750 mm. can be obtained by the decomposition of 100 grams of steam by passing it over red-hot charcoal ? 170. I require 10 litres of carbon monoxide at o C. and 760 mm. pressure. How many grams of (i) oxalic acid, (2) of formic acid, and (3) of potassium ferrocyanide shall I need ? 171. Required the weight in grams of sodium acetate to yield 10 litres of methane at o and 760 mm. 172. Manchester coal-gas contains 35 per cent, by volume of marsh gas. Calculate the weight of this gas in a gasometer holding 100,000 cubic feet of coal-gas. 173. Calculate the weight in kilos, of air needed, and of carbon dioxide and water formed, by the complete com- 50 BY THE DECOMPOSITION bustion of 10,000 litres of Manchester coal-gas possessing the following composition by volume : Hydrogen 45 '58 Marsh gas ........ 34'9 Carbon monoxide . . . . . 6*64 Ethylene 4-08 Butylene 2*38 Sulphuretted hydrogen .... 0*29 Nitrogen 2*46 Carbonic acid 3*67 1 00 '00 174. I oo grams of pure silver cyanide are shaken up with 1 20 grams of hydrochloric acid containing 26*1 per cent, of HC1. Required the amount of silver chloride produced, and the percentage amount of hydrocyanic acid in solution. 175. Required the weight of manganese dioxide con- taining 60 per cent. MnO 2 to liberate all the iodine from 100 grams of potassium iodide. 176. A manufacturer of bleaching powder requires 10 tons of chlorine. How much salt, manganese containing 59 per cent, of the dioxide, and sulphuric acid containing 58 per cent, of real acid, will he need ? 177. What weight of hydrochloric acid gas is produced in the manufacture of 100 tons of salt cake ? What volume of gas escapes supposing the manufacturer to condense only 92 per cent, of the quantity evolved ? J 78- 5*75 grams of silver nitrate are added to 575 grams of a solution of hydrochloric acid containing 10*22 per cent. H Cl. How much silver is precipitated and how much remains in solution ? 179. What weight of bleaching powder can be theo- retically produced from 150 tons of manganese contain- ing 68 per cent, of the dioxide ? 1 80. Required the weight of potassium chlorate yielded by the chlorine evolved from 100 tons of manganese con- taining 60 per cent, of the dioxide. OF A KNOWN WEIGHT OF THE MATERIAL. 51 1 8 1 . What weight of potassium bromate can be obtained by neutralizing 520 grams of bromine with potash ? 182. lodic acid may be obtained by passing a stream of chlorine through water containing iodine in suspension. How much iodine and chlorine will be needed to prepare loo grams of iodic acid ? 183. Required the weight of copper and sulphuric acid needed to yield 3 litres of sulphurous acid at the standard temperature and pressure. ! 184. A vitriol-maker prepares 100 tons of vitriol of specific gravity I '6, containing 70 per cent, of acid : how many tons of pyrites containing 42 per cent, of sulphur must for this purpose be burnt? Supposing that 3 per cent, of the theoretical yield of sulphur remained unburnt in the pyrites, what would be the difference in the pro- duction of sulphuric acid ? 185. What volume of oxygen at 10 C. and 743 mm. of mercury can be obtained by the decomposition of a litre of sulphuric acid possessing a density of 1*854 at o C. ? 1 86. What weight of iron sulphide will be needed to yield a litre of hydrogen sulphide at o C. and 760 mm., and how much air will be required to burn this gas com- pletely to water and sulphur dioxide ? f 187. I gram of phosphorus is to be converted into the pentachloride. How many litres of chlorine at o C. and 760 mm. are required ? 1 88. How much crystallized microcosmic salt must be ignited to furnish a gram of sodium metaphosphate ? ! 189. An alkali-maker consumes 300 tons of salt per week : he divides three-quarters of his yield of soda-ash equally for the preparation of soda crystals, bicarbonate, and solid caustic containing 70 per cent. Na 2 O. What weight of sulphur will he need to use weekly, and what is the theoretical weekly yield of the four products of manufacture ? 190. Required the weight of limestone needed to convert 50 tons of soda crystals into bicarbonate. ! 191. loo grams of pure iron is burnt in excess of (i) oxygen and (2) chlorine. What is the weight of oxide and chloride produced ? E 2 52 COMBINATION AND DECOMPOSITION COMBINATION AND DECOMPOSITION OF GASEOUS BODIES. EXAMPLES. 1. 4 litres of hydrogen are mixed with 5 litres of chlorine, and the mixture exploded. What volume of hydrochloric acid gas is produced? Which gas. and how much of it remains in excess ? From the equation H 2 + C1 2 = 2H Cl 2 volumes + 2 volumes = 4 volumes it is evident that i volume of hydrogen combines with I volume of chlorine to form 2 volumes of hydrochloric acid gas. Therefore 4 litres of hydrogen would require 4 litres of chlorine, and would form 8 litres of hydrochloric acid gas, I litre of chlorine remaining uncombined. 2. 1 50 cubic centimetres of oxygen are mixed with 400 cubic centimetres of hydrogen and the mixture exploded. What volume of steam is produced ? Which gas, and how much of it, remains uncombined ? From the equation 2 H 2 + 2 - 2 H 2 O 4 vols. + 2 vols. = 4 vols. it is evident that 150 cubic centimetres of oxygen would require 300 cubic centimetres of hydrogen, and yield 300 cubic centimetres of steam, 100 cubic centimetres of hy- drogen remaining in excess. 3. loo cubic centimetres of ammonia gas are completely decomposed by a series of electric sparks, yielding 200 cubic centimetres of mixed hydrogen and nitrogen : an excess of oxygen is next added, when the volume of mixed gases is found to amount to 290 cubic centimetres. The mixture is now exploded, when 65 cubic centimetres of gas remain. Show from these data that the symbol for ammonia is NH 3 . OF GASEOUS BODIES. 53 The total volume of the mixed gases before the explo- sion was 290 cubic centimetres ; after the explosion 65 cubic centimetres remain. Hence (290 65) = 225 cubic centimetres of oxygen and hydrogen have disappeared to form water, and two-thirds of this contraction gives the amount of hydrogen in the mixture, and hence in the 200 cubic centimetres resulting from the decomposition of the loo cubic centimetres of ammonia gas . | of 225 = 150 cubic centimetres. Hence in the 200 cubic centimetres of the gas, 150 cubic centimetres were hy- drogen, the remaining 50 being nitrogen. The gases are therefore mixed in the proportion of I volume of nitrogen to 3 volumes of hydrogen, and hence the symbol for ammonia is NH q . QUESTIONS. ! 192. 20 litres of hydrogen are mixed with 10 litres of chlorine. Which gas remains in excess ? How many litres of hydrochloric acid are produced ? f 193. One cubic foot of hydriodic acid is decomposed by an excess of bromine. How many cubic feet of hydro- bromic acid are formed ? f 194. The iodine in 100 volumes of hydriodic acid is liberated in succession by chlorine and by oxygen. How many volumes of chlorine and how many volumes of oxygen are required ? 195. 50 cubic centimetres of hydrogen are exploded with 75 cubic centimetres of oxygen. Required the total volume of the gases after the explosion, measured at 1 50 C. and 760 mm. pressure. 196. How many litres of oxygen are contained in 3 litres of nitrogen peroxide (N 2 O 4 ) ? 197. An unknown volume of hydrogen sulphide re- quired 110-34 cubic centimetres of chlorine for complete decomposition. What was the volume of the hydrogen sulphide ? 54 GAS ANALYSIS CALCULATIONS. 198. How many cubic centimetres of hydrogen and nitrogen are contained in a litre of ammonia gas ? f 199. V volumes of a hydrocarbon C n H 2n are sub- mitted to combustion. How many volumes of oxygen are required for its complete combustion, and how many volumes of carbonic anhydride are generated ? are of 200. How many litres of air at o and 760 mm. required for the complete combustion of 10 litres (i) marsh gas, (2) olefiant gas, (3) acetylene ? 201. 5 litres of chlorine are mixed with 5 litres of carbon monoxide. What volume of phosgene gas is produced, and how much hydrochloric acid and carbon dioxide would be produced by the decomposition of this gas with water ? 202. What volume of arsenic is contained in a litre of oxide of cacodyl vapour measured at 500 C. ? GAS ANALYSIS CALCULATIONS. The processes employed in gas analysis fall naturally under the heads : I. Proximate analysis, effected by the successive ab- sorption of the various constituents of a gaseous mixture by suitable reagents. Ultimate analysis, performed by the combustion of the gaseous mixture and calculation of the quantity of each element present from the observed alteration in volume due to the combustion, and the quantities of CO 2 and residual gas resulting therefrom. The combustion method in most cases allows the proximate composition of the mixture to be obtained indirectly ; e.g. a mixture of x c.c. H,jc.c. CH 4 , and z c.c. C 2 H 2 yields the following data : A = original volume of mixture, B = contraction due to combustion, C = volume of CO 2 produced by the combustion. GAS ANALYSIS CALCULATIONS. 55 From the equations (1) 2 H 2 (2 vols. gas) + O 2 (i vol. gas) = 2 H 2 O (liquid). (2) CH 4 (i vol. gas) + 2 O 2 (2 vols. gas) - CO 2 (i vol. gas) + 2 H 2 O (liquid). (3) 2 C 2 H 2 (2 vols. gas) + 5 O 2 (5. vols. gas) = 4 CO 2 (4 vols. gas) + 2 H 2 O (liquid). we have A = x+y + z. B = \x + zy + f *. C = y + 2z. Hence $A - 28 + C $ A - 2B - C 6 -* -- x,y, and z are thus all expressed in terms of the measured quantities A, B, and C and can therefore be readily found. It is required to measure the comparative quantities of gas present before and after an experiment in both I. and II. Three quantities require to be taken into account in such measurements, namely : (a) the volume of the gas, V ; (ft) its temperature, / (or absolute temperature 273 + / = T) ; (c) its pressure, P. With reference to the measurement of F, T 9 and P four cases occur among common methods of gas analysis : (A). Tand P are kept constant, V alone requires to be measured. Many technical processes, e.g. Orsat's method, follow this plan. (B]. V and T are maintained throughout the same, P is the measured quantity. Regnault's and derived processes use this method. T* (C). The ratio 5, termed the disgregation of the gas, is always brought to the same value, V is the recorded quantity. Doyre's method furnishes an example of this type. 56 GAS ANALYSIS CALCULATIONS. (Z>). Analysis by the eudiometer and absorption tube needs the observation of all these variables, P 9 V, and T. In (A) and (C) the observed volumes are in the direct ratio of the volumes reduced to standard conditions and hence may be directly compared ; in () the measured pressures vary directly as the corrected volumes and can therefore be substituted for the latter in all calculations of the relative composition of the gas sample ; in (D) all volumes must be reduced to some standard conditions before they can be compared. EXAMPLES. i. A heating-gas, made by working coke-gas producers with air and steam, yielded on analysis the following data : (a) Volume of gas employed 977 c.c. (&) Vol. after absorption of CO 2 by potash . . 87*5 c.c. (c) Vol. after absorption of CO by cuprous chloride 68*6 c.c. (d) Vol. taken from (c) for estimation of H . . 59-3 c.c. (e) Vol. after addition of air 98*8 c.c. (/) Vol. after combustion with palladium asbestos in capillary tube 80*5 c.c. Assuming the pressure and temperature to have remained constant during the analysis, calculate the percentage composition of the gas. CO 2 absorbed by potash = 977 - 87*5 = 10*2 c.c. hence % of CO, = '^-^? = 10-45. 977 CO absorbed by cuprous chloride = 87-5 - 68'6 = 18-9 c.c. iS'Q X 100 hence L of CO = = IQ'34- 977 Contraction on combustion = 98^8 8o'5 = i8'3 c.c. GAS ANALYSIS CALCULATIONS. 57 From equation 2 H 2 + O 2 = 2 H 2 O, it is seen that f of observed contraction represents vol. of H present ; hence vol. of H in 59*3 c.c. (d) = = 12*2 c.c. 12-2 X 68-6 X IPO and .*. % of H in gas = 59-3 x 977 = I4 * 44 * By difference, the % of N is 100 - 10-45 ~ 19*34 ~ I4'44 = 5577- The percentage composition is then : Carbon dioxide .... 10*45 P er cent. Carbon monoxide . . . I9'34 Hydrogen 14-44 >, Nitrogen ...... 5577 2. A determination of the composition of air by Frankland and Ward's modification of Regnault's process supplied these figures (V and T are constant in this method) : (a) Air alone : Height of Hg. in barometer .... 673*0 mm. Height of mark on measuring tube. . 383*0 mm. (b) Air + hydrogen : Height bar 945'o mm. mark 383*0 mm. (c) After explosion : Height bar 763*3 mm. mark 383*0 mm. Hence pressures are : (a) 290*0 mm. (b) 562*0 mm. (c) 380*3 mm.; and the corrected volumes are in the same ratio as the observed pressures, therefore the percentage of oxygen present in the sample of air is 58 GAS ANALYSIS CALCULATIONS. 562-0 - 3803 100 290 60-56 X ioo 290 = 20-88. 3. Find the composition of a sample of coal-gas from the data given (Bunsen.) Volume. Pressure. Temperature C. \ Gas taken I 34'3 0*7285 m. I 4'5 After KOH absorption . After pyrogallate absorp-\ tion .... J 131-1 130*6 0-7317 m. 0*7293 m. 17-8 i7'S A portion of the residue was treated in an absorption- tube with H 2 SO 4 containing excess of SO 3 , and then washed with KOH. Volume. Pressure. Temperature c. B Portion taken 97*3 0*6923 m. 18*5 After absorption of ethy-^ lene, propylene, and > benzene- vapour . . .; 93'2 0*6835 m. : 18*0 A part of the residue from B was then exploded in the eudiometer. Volume. Pressure. Temperature C. C Part taken 150*9 0*1774 n ^' 18*4 After addition of O. 281*5 0*3041 m. 18*9 After further addition ofl air / 47 2 '3 0*4942 m. 19*0 After explosion .... 425'5 0*4485 m. 17-8 After absorption of CO2 407-7 0-4375 m. 18*0 On addition of H . . . 561*7 0-5888 m. i9'3 After explosion .... 367-5 0*3965 m. 19*6 The remainder of the gas from A was then similarly analysed. GAS ANALYSIS CALCULATIONS. 59 Volume. Pressure. Temperature C. D. Portion taken for analysis 139-3 0*1623 m I5'2 After addition of O. . . 2?5'8 0*2963 m 15-7 After further addition ofl air / 446-6 0*4704 m I3'3 After explosion .... After absorption of COg . Residue + H 407-0 383-0 592*8 0*4296 m 0*4236 m 0*6296 m 14*1 16*8 17*1 After explosion .... 424-0 0*4629 m 17*2 All the volumes given in this example require to be reduced to some common temperature and pressure ; taking o C. and I m. pressure for standards, we have : A. Gas taken = 92*91 ; after KOH absorption = 90*06 ; (92*91 - 90*06) 100 285 hence % of CO 2 = ^^ r ' = = 3'o7- 92*91 92*91 After absorption by alkaline pyrogallate, remaining vol. = 89*51, hence % of O = (90*06 - 89*51) IOQ 0*55 x 100 92*91 92*91 = 0-59. B. Part taken = 63*08 ; after absorption of ethylene propylene, and benzene-vapour by SO 3 = 59*75 ; hence (63*08 - 59*75) X 89*5 IX ioo c / of these constituents = >- -^K^p 3*33 X 89*51 X ioo = 63*08 X 92*91 C. Portion taken = 25*1 ; after addition of O = 80* I ; after addition of air = 218*2 ; after explosion = 179*2 ; hence contraction = 218*2 - 179*2 = 39*0. (i) After absorption of CO 2 produced = 167*4 ; hence vol. of CO 2 = 179*2 - 167*4 = 11*8 (2) After addition of H = 308*9 ; after explosion = 135*9 ; hence second contraction =173. (3) The incombustible residue of N from (2) is therefore, 167*4 "HP = l &7'4 ~~ 57'66 = 109*74, of which 0*7904 60 GAS ANALYSIS CALCULATIONS. (218*2 80*1) = 109*15 belongs to added air, hence N in 25 'I vols. taken = 10974 109*15 = 0*59 vols.; therefore 0-59 X 5975 X 89*51 X 100 / afNmgas = = 2*15. Combustible gas = 25*1 0*59 = 24*51 ; and this con- sists of x vols. H + y vols. CH 4 + 2 vols. CO. From the equations : 2 H 2 + 2 = 2 H 2 0. CH 4 + 2 O 2 = CO 2 + 2 H 2 O. 2 CO + 2 = 2 C0 2 . we have, contraction 39 = f x + 2 j -f- J # () CO 2 formed 11-8=7 + *. W ; and total volume 24*51 = x -\- y + 2. (c) ; hence ^r \from (^) and (*)/ = 12*71. Solving (a) and (^) for *, we have : 39 = 19*06 + 27 + ^2- 23 = 2y + 2 * 3 -^ hence = 23*6 + 19*06 - 39 = 3*66. 3*66 X 2 and 2 --- = 2*44. From (ff) y + 2 = 1 1 *8 so y = 11*8 z = 11*8 - 2*44 = 9*36. 12*71 X 5975 X 89*51 X IPO Therefore % of H= ,;.. ^ 63-08 X Ui " ^'^ o, of CH _ 9-36X5975X89-5 IX loo _ ' f CH4 - 2 S -ix6 3 -o8X92- 9 i 2-44X5975x89-51 X loo and c / of C0 2 = 2S . IX63 . 08X92 . 9I GAS ANALYSIS CALCULATIONS. 6 1 D. Portion taken containing all the combustible gases = 21*5 ; after addition of O = 77*3 ; after addition of air = 200-4; after explosion = 166*2; hence contraction = 34'2. (i) After absorption of CO 2 produced, vol. = 152-9 ; hence CO 2 = 166-2 - 152-9 = 13-3. (2) After addition of H to residue, vol. = 351*3 ; after ex- plosion 184-8; hence second contraction = 166*5. (3) The contraction on explosion (1) = = 153-2 per cent, of the volume 21-5 X 92-91 of the original gas. The CO 2 produced I.V3 X 89-51 X loo (2) = = 59*60 per cent, of the same. 21-5 X 92*91 Contraction due to CH 4 , CO, and H, from C. (i), 39 X 5975 X 89-51 X 100 * -5sTx-6 3 -o8 x 92-91 = I41 ' 8 Per original vol. ; hence contraction due to ethylene, propylene, and benzene, = 153*2 141*8 =11*4 per cent, of vol. of original gas. Similarly, CO 2 produced from CH 4 and CO, from C (2) ir8 X 5975 X 89-51 X loo 25*1 X 63*08 X 92-91 =42*90 per cent, and so vol. of CO 2 produced from ethylene, propylene, and benzene, = 59*60 42*90 = 16*70 per cent, of vol. of original gas. The percentage of ethylene, propylene, and benzene- vapour, present in the original gas is 5*09. (B). Let x '= percentage of C 2 H 4 , y = percentage of C 3 H 6 , and z = percentage of C 6 H 6 "; then, from the equations : C 2 H* 4 + 3. O 2 = 2CO 2 + 2H 2 0. 2C 3 H 6 + 9. 2 = 6C0 2 + 6H 2 0. 2C 6 H C + 15. O 2 = I2CO 2 + 6H 2 O. 62 GAS ANALYSIS CALCULATIONS. we have contraction 1 1 '4 = ix + f/ + # (4) and vol. of CO 2 1670 = ix + 3j/ + 62-; (S) also vol. of gases 5*09 = x + j + z. (6) From (4) and (6) 11-4 = 2;r + \y + |^\ 1272 = I.T-J- fj + f*/ whence 1*32 = J;r and * = 2-64 % of C 2 H 4 . From (5) and (6) 1670 = 5*28 + 3j/ + 6z 5-09 = 2-64 or 11*42 = 3^ +6,2'\ 2'45 = J + * ) or 11*42 = 3/ + 6^\ 1470 = 6y + 6z ) whence 37 = 3*28 and y = 1*09 % f C 3 H 6 . Again, j + z = 2*45 .*. ar = 2*45 - 1*09 = 1*36 / of C 6 H 6 . Tabulating our results, the percentage composition of coal gas is : Carbon dioxide ............. 3*07. Oxygen ................ 0*59. TEthylene ..... 2-64. Absorbed by SO 3 . 5*09 %? Propylene .... 1*09. (^Benzene-vapour . . 1*36. Nitrogen ............... 2*15. Hydrogen ............... 46*21. Marsh-gas ............... 34*03. Carbon monoxide ............ 8*87. QUESTIONS. 203. Calculate the percentage composition of a sample of atmospheric air from the following numbers : GAS ANALYSIS CALCULATIONS. Volume. Pressure. Temperature c.. Air employed . 863*7 0*5576 m e. *c After addition of H ... After explosion 1006 '7 800-7 o'69ii m. 0-4914 m. 5*5 5'6 204. A gaseous mixture containing oxygen, nitrogen, and carbonic acid, yielded the following numbers on analysis. Calculate the proportion of the constituent gases in 100 volumes of the mixture (Bunsen) : Volume. Pressure. Temperature C. Original gas 171 *2 0-62 o m TO'- After absorption of carbonic \ acid ) 167-3 0*6196 m. i3'5 After absorption of oxygen . 147-0 0*6058 m. 13*9 205. Determine the amount of aqueous vapour in a sample of air which yielded the following numbers on analysis (Bunsen) : Air saturated with moisture. Lower level Hg. stands at. .... 565*9 Upper level in eudiometer . . . . 317*3 Vol. corresponding to 317*3 . . . . 292*7 Meniscus correction 0*4 Air temperature 20*2 C. Height of barometer 07469 m. Same volume of air dried by Ca C1 2 . Lower level Hg. stands at .... 565*9 Upper level in eudiometer ..... 310*7 Vol. corresponding to 310*7 . . . . 286*0 Meniscus correction 0*4 Air temperature 20*2 C. Height of barometer . . . . . . 0*7474 m. 206. Calculate the composition by volume of water from the following eudiometric synthesis : 6 4 GAS ANALYSIS CALCULATIONS. A. Hydrogen saturated with moisture. B. Hydrogen + Oxygen saturated with moisture. C. Residue after explosion saturated with moisture. A. B. C. Observations at lower level of Hg 77i'8 768 '2 771 "9 Observations at upper level in eudiometer 238-0 435 'o 332 '0 Corresponding volume .... 262-5 468-3 360-8 Barometer Thermometer 12*1 1 2 '6 I3'o Corresponding tension of aqueous vapour 10 '5 mm. io"9 mm. 11*2 mm. 207. Find the volume of each constituent contained in one volume of methyl ether from the data below, and deduce its empirical formula : Volume. Pressure. Temperature. A Gas taken . 0-6 0*1419 m i*4C After addition of oxygen . 199-8 0-3112 m. 2 *6C. After explosion .... After absorption of COo . After addition of H . 172-4 132-8 547'3 0*2738 m. 0*2409 m. 0*6955 m. 3* 7 C. 3'9C. 2*6C. After explosion .... 466*6 0*6126 m. x'S'C. B. Gas taken 79 "6 0*3140 m. 4*oC. After addition of oxygen . After explosion .... 327-2 268-7 0*5615 m. 0*4915 m. 5'oC. 4* 9 C. After heating to 99 "5 . . 418-1 0*6752 m. 99'5C. After cooling .... 268-2 0*4914 m. 3'7C. After absorption of CO^ . *93'3 0*4188 m. o* 7 C. 208. Find the formula of nitrous oxide from the analytical results given : Volume. Pressure. Temperature. Gas taken 140*2 0*2175 m. i4'4C. After addition of H . . . . 248-4 191 '4 0*3219 m. 0*2664 ^' *<&. 14 o (^. After addition of O . . . . After explosion 228-8 1 60*6 0*2931 m. 0*2253 m - i2*8C. i 3 * 7 C. GAS ANALYSIS CALCULATIONS. 209. A gaseous mixture collected from Hekla a month after an eruption yielded the following figures : find its percentage composition : Volume. Pressure. Temperature. Gas taken 114*9 0*6944 m. 2o*4C. Afte absorption with MnO2 . Afte absorption with KOH . 112*9 108*1 0*6958 m. 0*7092 m. 2o*4C. 20'6C. Gas decanted into eudiometer 136*7 0*3460 m. 2o-6C. Afte explosion with (H2+O) Afte addition of H . . . . 137-2 190-4 0*3452 m. 0*3980 m. 20- 7 C. 2o'5C. Afte explosion Afte absorption with KOH . 152*7 148-9 0*3585 m. 0*3665 m. 20'3C. i8- 9 C. 210. Find the percentage volume of H,SH 2 , CO 2 , N, CO, and CH 4 present in a fumarolle-gas giving on analysis : Volume. Pressure. Temperature. Gas taken 0*6945 ni irVC Afte absorption of H^S . . Afte absorption of CC*2 . Gas decanted into eudiometer 73'7 46*1 96*8 0*6728 m 0*6502 m 0*3093 m i3'6C. i 3 *6C. i3'iC. Afte addition of air . 243*0 0*4534 m i3'6C. Afte explosion 172*0 0*3839 m X3-7C. Afte treatment with KOH . 168*6 0*3902 m i3'iC. 211. From the tabulated data, calculate the percentage amount of nitrogen present in each body analysed, and give the volumes, reduced to standard conditions, of the nitrogen and nitric oxide respectively in each case. Measured in Frank land and Ward's apparatus. Substance taken. Volume in c.c. Temperature. Pressure N + NO Pressure N. A. 0-1263 gm. 13*79 i8C. 451 mm. 423 mm. B. 0-0832 ,, 22-32 20C. 411 , 378 , C. 0*0827 ,, 22-32 2 4 C. 416 , 373 D. 0-0856 E. 0-0878 22*32 *3*5 29 C. i 9 C. 396 , 597 i 375 , 582 , F. 0*1259 9*02 20C. 408 , 45 , 66 GAS ANALYSIS CALCULATIONS. 212. Find the composition of marsh gas from the figures given below, obtained by Thomas's modification of Frankland and Ward's apparatus. Temperature constant at 15*4 C. Gas taken at 118*5 mm - pressure. After addition of oxygen, pressure = 357*8 mm. Pressure after explosion =121*1 2 mm. Pressure after absorption of C0 2 =2*32 mm. 213. Calculate the percentage of CO 2 by volume con- tained in a sample of air which gave the accompanying data 10 c.c. of baryta solution agitated in a flask, con- taining the air, of 618 c.c. capacity, required 6'o c.c. of normal oxalic acid for neutralization, of which I c.c. equals I c.c. of CO 2 at o C. and 760 mm. ; bar. 726 mm., temp. 21 C. The same amount of baryta solution directly titrated with normal oxalic acid required 8'8 c.c. 214. To a solution of baryta sufficient alcoholic solu- tion of phenolphthalein is added to produce a distinctly pink coloration; I c.c. of the solution =0*104 c.c. carbon dioxide. A volume of 25 c.c. of the coloured solution becomes decolorized on the aspiration of 540 c.c. of air through it, owing to the absorption of CO 2 ; find per- centage of CO 2 present. 215. An analysis of a specimen of coal-gas is made by (i) successive absorption of carbon dioxide, ethylene (propylene, butylene), benzene, oxygen, and carbon monoxide by means of HempePs pipettes ; (2) combus- tion of the hydrogen in a palladium-asbestos tube ; (3) combustion of the methane by red-hot copper oxide with estimation of the carbon dioxide produced by titra- tion. From the subjoined figures find the composition by volume of the sample analysed : Volume of gas employed 99*1 c.c. (i) Volume after absorption by KOH . . . 97*7 c.c. Volume after treatment with Br water . . 93*9 c.c. Volume after treatment with fuming NO 3 H 92*9 c.c. Volume after alkaline pyrogalldl absorption 92*6 c.c. Volume after treatment with cuprous chloride ' . . . . . . 86*5 c.c. ATOMIC WEIGHT DETERMINATIONS. 67 (2) Non-absorbable gas used for H estimation 40*6 c.c. Non-absorbable gas + air 99*0 c.c. Volume after combustion 657 c.c. (3) Residue of 657 c.c. is burned by air and copper oxide, and CO 2 titrated by baryta-water and oxalic acid. Gas measures 657 c.c. at 736 mm. and 20 C. Baryta- water, I c.c. = 1*04 c.c. oxalic acid = 1*04 c.c. methane. Baryta-water employed 50 c.c. Oxalic acid required for re-titration . . 37*4 c.c. 216. 100 volumes of Manchester cannel gas contained 4*98 volumes of defines, which yielded on combustion 13*93 volumes of carbon dioxide. Required the volumes of ethylene (C 2 Hi) and butylene (C 4 H 8 ) contained in the gas, and the value of the illuminating power of the gas expressed in percentages of ethylene. CALCULATION OF THE RESULTS OF ATOMIC WEIGHT DETERMINATIONS. The data required for these calculations , when not other- wise given, may be obtained from Table I. in the Appendix. 217. DUMAS found, on heating copper oxide in a stream of dry hydrogen, that a certain weight of this substance lost 59789 grams of oxygen, and yielded 67*282 grams of water. Calculate from these numbers the atomic weight of hydrogen. 218. Marignac obtained 314*894 grams of silver nitrate from 200 grams of silver. 14*110 grams of silver nitrate required 6*191 grams of potassium chloride for complete precipitation, and 10*339 grams of silver dissolved in nitric acid required 5*1 20 grams of ammonium chloride for precipitation. Calculate the atomic weight of nitro- gen from these data. F 2 68 CALCULATION OF 219. Stas found that 91*462 grams of metallic silver, when heated in a stream of chlorine, yielded 121*4993 grams of pure silver chloride. Calculate from this the atomic weight of chlorine. 220. Stas found, after adding 7*25682 grams of potas- sium chloride to 10*51995 grams of silver dissolved in nitric acid, that 0*0194 grams of silver remained in solu- tion. Calculate from these data the atomic weight of potassium. 221. Erdmann and Marchand obtained 109*6308 grams of mercury from 1 1 8*3938 grams of the red oxide. Calcu- late the atomic weight of mercury. 222. Calculate the atomic weight of carbon from each of the following analyses by Redtenbacher and Liebig of Salt. Silver. 1. Silveracetate C 2 H 3 AgO 2 28*803 grams gave 18*612 2. Silvertartrate C 4 H 4 Ag 2 O 6 16*220 9*6175 3. Silvermalate C 4 H 4 Ag 2 O 5 25*898 16*059 223. Mallet found that 8*2144 grams of ammonia-alum dried by exposure to air at 21 25 C. yielded 0*9258 gram of A1 2 O 3 . Taking the following atomic. weights : O = 15*961; S = 31*966; N = 14*010; find the atomic weight of aluminium. 224. Mallet found that 6*9617 grams Al Br 3 required 8*4429 grams of Ag for precipitation of the bromine ; if Ag = 107*649 and Br = 79754 what is the atomic weight of Al as deduced from this result ? 225. 0*3697 gram of aluminium liberated 0*04106 gram of hydrogen on being dissolved in a strong solution of sodium hydrate ; find atomic weight of aluminium. 226. A determination of the atomic weight of titanium was made by finding the weight of pure silver required to precipitate the chlorine from an aqueous solution in which a bulb containing a known weight of TiCl 4 had been broken. A quantity of the pure silver nearly sufficient was in each experiment first converted into nitrate and then mixed with the titanium solution, the precipitation was finished by addition from a burette of a centinormal ATOMIC WEIGHT DETERMINATIONS. 6 9 solution of AgNOg. Find the atomic weight of Ti from the annexed data in each case. (Take Ag = 67456, Cl = 2-21586, H - 0-06265, O = I.) Corrected weight of TiCl 4 used. Weight of silver weighed out. Weight of silver added from burette. 2 '43 2 75 grams. 5-49288 grams. 0-03509 gram. 5'42332 ,, 12-30669 o"oi59i ,, 3'596oi 8-15960 o'oisoi ,, 3'3*222 7-44102 ,, 0*08619 ,, 4-20093 9'52445 0-02234 ,, 227. A series of experiments made for the purpose of determining the atomic weight of gold consisted in finding the ratio existing between the quantities of KBr and metallic Au remaining on the decomposition of potassium bromo-aurate by heat. Taking the equivalent of H, 0*06265, K, 2*44523, and Br 4*99634, when O = i, calculate the atomic weight of gold from the following data : Weight of Au. 6*19001 4*76957 4*14050 3*60344 3-67963 4*57757 5-36659 5*16406 228. The atomic weight of silicon has been determined by estimating the amount of SiO 2 obtained on treatment of known weights of SiBr 4 with water. From the following table of results, find the ratio Si : H, taking H = 0*06265, and Br = 4*99721, (O = i). Weight of KBr. 373440. 2*87715. 2*49822. 2*17440. 2*21978. 276195. 3*23821. Weight of Si Br 4 in vacuo. 9*63007 grams. 12*36099 12*98336 9*02269 15-38426 Weight of Si O 2 in vacuo. 1*67070 grams. 2*14318 2*25244 1*56542 2-66518 7 CALCULATIONS INVOLVED 229. Taking the atomic weights of O = 15*96, N = 14*02, H = I, find the mean atomic weight of chromium from the given experimental results. Corrected weight of (NH 4 ) 2 Cr 2 7 . 1-01275 ro8i8i 1-29430 1-13966 0-98778 1-14319 Corrected weight of Cr 2 O 3 obtained therefrom. 0*61134 0*65266 0*78090 0*68799 o'59595 0*68987 230. From the following readings given by Ramsay, showing the volume of hydrogen obtained by means of pure zinc, calculate the equivalent of zinc (H = i). Weight, in a vacuum, Reduced volume Reduced Thermometer of the zinc used. of hydrogen. pressure. reading, C. / 2i6'86 cc. 729*18 mm. o'54 219-64 727-78 o'54 ! 22273 725*78 o'54 3-299584 c 225*62 724-08 o'54 233*05 719-72 o'53 23 '54 721*10 0*48 228*02 722-65 0-48 CALCULATIONS INVOLVED IN INDIRECT ANALYSIS. There are three main types of Indirect Analysis, namely : (i) The method of substitution; an equivalent amount of some other radicle (or substance) is substituted for the radicle we wish to determine and is then estimated directly; for instance we may determine free Cl by adding to its solution KI when KC1 is formed and free I, equivalent in amount to the Cl present, remains and can be estimated directly. IN INDIRECT ANALYSIS. 7 1 (2) The residue method; the body in which the radicle occurs undergoes a definite chemical change when acted upon by a known amount in excess of some reagent, and the excess of reagent is determined. (3) The method depending on the numerical differences between molecular weights j with a mixture of two salts, differing considerably in molecular weight, but possessing a common constituent, it is possible from the estimation of the common radicle to calculate the quantities of each of the other radicles present in the mixture. EXAMPLES. (i) To 50 c.c. of a solution of Cl, an excess of potassium iodide solution was added ; the liberated I was then estimated by means of a standard solution of Na 2 S 2 O 3 using starch solution to indicate the end of the reaction. I c.c. Na 2 S 2 O 3 solution = 0*0127 gram I, and 22*5 c.c. of this solution were used ; what was the strength of the Cl solution in grams per c.c. ? Iodine set free = 0*0127 X 22*5 gram. .... 0-0127 X 22*5 X 35'37 ,,- which is equivalent to -7- - gram of Cl. which was the weight of Cl contained in 50 c.c. Hence, weight of Cl per c.c. 0-0127 X 22-5 X 35*37 = , *"* = 0-001597 gram. 50 X 126*54 (2). 0*2815 gram of calcite was dissolved in 30 c.c. of normal HNO 3 , and the excess of acid determined by normal NaOH of which 24*43 c - c - were required ; what percentage of CO 2 did the sample contain ? The HNO 3 neutralized by the CaCO 3 present in the calcite was 30*00 - 24*43 = 5 '57 c - c - 126 grams HNO 3 are equivalent to 44 grams CO 2 . 72 CALCULATIONS INVOLVED Hence, amount of CO 2 present was : 0-063 X 5-57 X 44 -- - gram. (3). Two grams of a mixture of barium carbonate and calcium carbonate evolve 0*67 gram of carbon dioxide ; find the percentages of Ca and Ba present in the mixture. Taking the following atomic weights, Ca, 40*0 ; Ba, 137*2 ; C. I2'o ; O. 16 ; we have BaCO 3 contains -^- of its weight of CO 2 , and CaCO 3 44 100 Let .r be the weight of BaCO 8 in the mixture taken, then (2 x) is the weight of CaCO 3 present. C0 2 present is -^ * + (2 - *) = 0-67. 197-2 ' 100 v 44(197-2 - 100) hence, - ' x = 0*21. loo X 197*2 .-. x = 0-968 gm. BaCO 3 . and 2 x = 2 0*968 1-032 gm. Ca CO 3 . Whence % of Ba in mixture = - - - = 33*67. 1*032 X 40 X ioo and L of Ca = - - - - = 20*64. ioo X 2 QUESTIONS. 231. 0-2151 gram of uric acid was heated with soda- lime and the evolved NH 3 absorbed in HC1. The NH 4 C1 was evaporated to dryness and redissolved, the Cl being then thrown down as AgCl, of which 07390 gram was obtained ; what percentage of nitrogen did the sample of uric acid contain ? IN INDIRECT ANALYSIS. 73 232. 25 c.c. of a solution, containing 0*2105 gram of a copper alloy in the form of sulphates, was neutralized with Na 2 CO 3 and then rendered acid by acetic acid ; to the acidulated solution an excess of freshly prepared potassium iodide solution was added and the liberated iodine titrated by a standard Na 2 S 2 O 3 solution ; 31*29 c.c. were required for the reaction. 25*22 c.c. of the thiosulphate solution were found to be equivalent to 25 c.c. of iodine solution of which I c.c. = 0*006313 gram of copper. Calculate the percentage amount of copper present in the alloy. 233. 0*5005 gram of a sample of calcite, was weighed out, dissolved in dilute HC1 and precipitated as cal- cium oxalate ; the washed precipitate was treated with dilute sulphuric acid and the solution titrated with potassium permanganate solution of which I c.c. = 0*0056 gram of iron ; 100*08 c.c. of the permanganate were decolorized. What was the percentage of calcium in the mineral ? 234. In an estimation of nitrogen by the soda-lime method, 0*2102 gram of benzamide was taken and the N evolved ammonia absorbed in sulphuric acid solution, 25 c.c. being taken ; the residual acid required 21*517 c.c. N of - NaOH for neutralization. What was the percentage amount of nitrogen in the benzamide ? ^235. The subjoined data were obtained in an estimation of NH 3 in (NH 4 ) 2 SO 4 by the method of distillation with soda-lime in the wet way. Weight of ammonium sulphate taken = 0*3200 gram. Amount of H 2 SO 4 solution taken = 60 c.c. i c.c. of acid solution = 0*855 c - c - ' NaOH sol. N 41*7 c.c. of soda just neutralized the residual acid. Calculate from these data the percentage amount of ammonia in ammonium sulphate. 74 GENERAL ANALYTICAL QUESTIONS. 236. One hundred cubic centimetres of a semi-normal solution of sodium carbonate were added to 100 c.c. of a neutral solution containing copper sulphate. After filtra- N tion, the residual Na 2 CO 3 was determined by means of ~ sulphuric acid, of which 27 c.c. were required. What weight of Cu SO 4 in grams was present in each litre of the solution ? 237. The following results, obtained in an analysis of Rochelle salt, give sufficient data for the estimation of K and Na in the sample used ; find the percentage of each of these elements present. Weight of salt taken 0*7012 gram. Weight of mixed alkaline chlorides . . . 0*3305 Weight of Ag Cl obtained 0*7129 ,, 238. From 0*9330 gram of a mixture of sodium chloride and iodide, 0*9066 gram of sodium sulphate is obtained. What weights of the sodium chloride and sodium iodide are there in the weight of mixture taken ? (Take Na = 23, S = 32, O = 16, Cl - 35*5 I, = 127.) GENERAL ANALYTICAL QUESTIONS. 239. CALCULATE the formula of a salt containing sodium and chlorine which yielded the following numbers on analysis : (a) 0*1998 gram of the salt gave 0*4865 gram of silver chloride, and 0*0032 metallic silver. (b) 0*9543 gram of the salt gave 1*1584 grams of sodium sulphate. 240. 0*3951 gram of a substance supposed to be arsenic pentoxide was dissolved in dilute ammonia, and precipi- tated as magnesium-ammonium-arseniate. After drying at 110 the precipitate weighed 0*6544 gram. Was the substance As 2 O 5 ? GENERAL ANALYTICAL QUESTIONS. 75 241. The mineral nontronite gave on analysis the fol- lowing numbers. Calculate its formula. 1*4155 grams of the mineral gave 0*5711 grams of silica, and 0*5157 ferric oxide. 0*0380 lime. ri2O5 lost on drying 0*2311 water. 24.2. loo grams of a mixture of potassium and sodium chlorides have furnished 164*1 grams of potassium-plati- num chloride. What is the composition of the mixture ? 243. Calculate the formula of thallium perchlorate from the following determination : 0*1831 gram of salt gave 0*2476 gram of the double chloride of platinum and thal- lium. 244. Calculate the percentage amount of chlorine con- tained in vanadyl trichloride of which 3*9490 grams added to 7*383 grams of silver dissolved in nitric acid required 3 cubic centimetres of centesimal HC1 solution for com- plete precipitation according to Gay-Lussac's method. ! 245. Two equal volumes of a liquid found to contain an oxygen compound of bromine were completely reduced with sulphurous acid, the excess of which was removed by boiling : in one portion the sulphuric acid formed was estimated as barium salt, and in the other the bromine was weighed as silver bromide. Analysis gave : weight of Ba SO 4 = 0*402 gram ; weight of silver bromide, 0*3240 gram. Required the composition of the oxide of bromine. 246. 1*2185 grams of perchloric acid yielded 1*6785 grams of dry potassium salt : of this salt 0*9660 gram lost 0*444 gram on heating, and the residue required 0*744 gram of pure silver for complete precipitation. Required the percentage amount of acid contained in the quantity taken. (Roscoe.) ! 247. Required the percentage of real acid contained in J *483 grams of aqueous acetic acid, which evolved 0*427 gram of carbonic acid on adding it to bicarbonate of sodium. 248. 2*122 grams of aqueous formic acid required for neutralization 36*3 cubic centimetres of soda solution, of 7 6 GENERAL ANALYTICAL QUESTIONS. which each cubic centimetre contained (0*9025 X 31) milligrams of soda. Required the percentage of pure acid contained in the liquid. 249. Calculate the volume of carbonic acid in 10,000 volumes of sea-air, from the following numbers, obtained by Pettenkofer's method : i cubic centimetre of oxalic acid solution = I mgrm. COg. 50 cubic centimetres of baryta water = 55*18 cubic centi- metres of oxalic acid solution. Capacity of flask used, 4815 cubic centimetres. Temp, of air 13- 9. Bar. 753*1 mm. 50 cubic centimetres of baryta solution taken : of this 25 cubic centimetres after the experiment required 26*29 cubic centimetres of oxalic acid solution for neutralization. 250. The following numbers express the amount of the constituents in 1,000 grams of the water of the Irish Channel : 1. Chlorine 1 8 '62650 2. Bromine "06133 3. Sulphuric acid (SO 4 ) . . 2*59280 4. Lime (total) 0*57512 5. Calcium carbonate . . . 0*04754 6. Magnesia 2*03233 7. Potassium 0*39131 8. Sodium 10*40200 9. Ferric oxide "00465 10. Ammonia . . . . . . *ooon 11. Nitric acid '00156 Total fixed constituents 33*83855 Calculate the composition of the saline matter of the water on the following assumption : That the sodium, potassium, and ammonium are combined with chlorine, the excess of chlorine being united to magnesium. The bro- mine and nitric acid are also to be united to the magnesium. The sulphuric acid exists in combination with lime, the excess being united to magnesia. The ferric oxide exists as ferrous carbonate. ! 251. Calculate the illuminating power of a gas flame burning 5 cubic feet per hour, expressed in candles burn- GENERAL ANALYTICAL QUESTIONS. 77 ing 1 20 grains (779 grams) of sperm per hour, from the following data : Rate at which gas issues = 4*8 cubic feet per hour. Consumption of standard candle in 10 minutes = 1*38 grams. Readings on the photometer scale for each minute of observation : (i), 27. (2), 22-5. (3), 17. (4), 21. (5), 18. (6), 19. (7), 19. (8), 20. (9), 19. (10), 21. 252. 0*5637 grams of the barium salt of an organic acid, obtained by the action of carbonic acid on C 6 H 4 BrC 2 H 5 in presence of sodium, lost, on drying at 120 C., 0*0430 gram of water, and gave 0*2733 gram barium sulphate. Required the formula of the salt. ! 253. 4*826 grams of a nitrogenous organic body yielded, after heating with soda-lime, 1*532 grams of the double chloride of platinum and ammonium. Required the per- centage of nitrogen contained in the organic substance. ! 254. Required the percentage composition, observed and calculated vapour density, and formula of a body which gave the following analytical results : (a) 0*4245 gram yielded 0*5670 gram of carbonic acid, and 0*3025 water. (ff) 0*1 8 10 gram yielded. 0*3855 gram of silver chloride, and 0*0165 metallic silver. (c) Determination of vapour density according to Gay-Lussac's method : Weight of substance employed, 0*0893 gram. Temperature of the air, 8 C. Height of barometer, 739 mm. Temperature of vapour, 50 C. Volume, 46*2 cubic centimetres. 7 8 SOLUTION OF GASES IN LIQUIDS. Difference of level, 140*5 mm. Coefficient of expan- sion of mercury = 0*0001815. Coefficient of cubical expansion of glass, o*oooo262. 255. 1*5055 grams of a mixture of sodium and potas- sium chlorides gave 3*4222 grams of silver chloride. Cal- culate the relative amounts of the two chlorides. 256. 1*2060 grams of a mixture of sodium bromide and sodium chloride gave, on complete precipitation by silver nitrate, 2*6554 grams of mixed silver chloride and bromide, which, on reduction, yielded 1*8418 grams of metallic silver. Calculate the proportion of chloride and bromide. SOLUTION OF GASES IN LIQUIDS. GASES fall into two groups when considered in relation to their solubility in liquids : (i). Those gases completely expelled from the solvent by raising its temperature or reducing the pressure to which it is subjected. (2) Other gases. I. In group (i), the amount of gas absorbed depends upon : I. The nature of the gas and of the absorbing liquid. II. The actual pressure of the gas considered. Henry's Law : The quantity of any gas absorbed by a given quantity of a liquid is proportional to the pressure. III. The temperature. Volume absorbed decreases with the increase of temperature ; it may be expressed by means of an empirical formula of the form V=a-&t-\-ct*, where a, , c are constants for each gas and / is the temperature. The absorption-coefficient of a gas is a number which expresses the ratio of the volume of the gas, measured at o C. and the pressure of absorption, absorbed by the liquid at the temperature of observation to the volume of SOLUTION OF GASES IN LIQUIDS. 79 the absorbing liquid. In a mixture of gases, each com- ponent behaves as if it alone were present under the partial pressure which is exerted by itself. EXAMPLE. It is required to calculate the absorption coefficient of nitrogen dissolved in water at 19 C. from the appended data (Bunsen). I. Observations before Absorption. Lower level of mercury in outer cylinder a = 423*6 mm. Upper level of mercury in absorption tube b = 124*1 Height of barometer p 746*9 ,, Temperature of absorptiometer . . . /= 19*2 C. Temperature of barometer . . . . . T= 19*0 C. II. Observations after Absorption. Lower level of mercury in outer cylinder a = 352*2 mm. Upper level of mercury in absorption tube b l = 3507 Water level in absorption tube . . . . c 65*5 Water level in outer cylinder . . . . d^ 8'o Height of barometer P\ 74&3 Temperature of absorptiometer . . . / x = 19*0 C. Temperature of barometer 7^ = 18*9 C. III. Tabular Data. Volumes, according to the calibration table, of: Gas taken, mark b 124*1 mm. ... V = 34*90. Tube above b l = 350*7 mm 200*04. Tube above c^ = 65*5 mm., remaining gas. V = 17*67. Hence water used for absorption. . . . (h) = 182*37. Coefficient of expansion of mercury for i C. . =0*00018. i (VP The absorption coefficient a = 7 ( 5- - tl \ Jr j 8o SOLUTION OF GASES IN LIQUIDS. P and P l being the pressures of the dry gas before and after absorption respectively, V and V^ being the corre- sponding volumes of gas reduced to o C. P = p reduced to o C. (a - b) reduced to o C. - vapour pressure of water at 19*2 C. = (744*4 298*5 - 1 6*6) mm. = 429*3 mm. P 1 = pi reduced to o C. - (a -- b^ reduced to o C. + {(<% b^) -f- (q dj)} reduced to equivalent mercury column at o C. vapour pressure of water at 19 C. - (743*8 - 1*5 + 4*4 16*3) mm. = 730*4 mm. V = it . , N = 34'Qo . ; ^ (i + 0*00367 . 19*2) i + 0*00367 . 19*2 = 32*608. ^ = p i- (I +0-00367. 19) ' = 16*52. T / -2-2'f Hence a = 102-37 \ QUESTIONS. 257. The absorption coefficient of nitrogen dissolved in water is 0*0152 at 12*6 C.; what volume of the gas measured at o C. and 760 mm. pressure is absorbed by one litre of water at 12*6 C. at each of the pressures : looo mm., 748*2 mm., 391 mm., and 14*3 mm. ? 258. What is the volume of nitrogen absorbed in each case in question 257, measured at the temperature and pressure of the experiment ? What is the weight of the gas absorbed in each case ? 259. The absorption coefficient of hydrogen dissolved in water is represented by the interpolation formula a = 0*0215286 0*0001 92 1 6t + 0*000001 7228t 2 : what quantities of the gas measured (a) under the ex- perimental conditions, (b) under standard conditions, SOLUTION OF GASES IN LIQUIDS. 8 1 will be given off on boiling 325 c.c. of the solution made by agitating water in hydrogen (A) at 10 C. and 750 mm., (B) at 14 C. and 767 mm., (C) at 18 C. and 732 mm. ? 260. The coefficient of absorption of carbon dioxide in water is found to be a = 17967 - 0-07761 / + 0*00 16424^; find the coefficients for / = 4-4 C., 8*4 C., 13-8 C, 1 6-6 C, 19-1 C., and 22-4 C. 261. The coefficient of absorption of hydrogen in alcohol is given by the interpolation formula 0=0*06925 0*0001487^ + o'oooooi/ 2 ; find the values of the co- efficient at i, 5, ii'4, 14*4, and 19*9 C., and compare each value with the corresponding value of the coefficient for water as given in question 259. 262. Find an interpolation formula expressing the variation of the coefficient of absorption of carbonic oxide in water with alteration of temperature, the coefficients found by experiment being : at 5*8, 0*028636 ; at 8*6, 0*027125; at 17*4, 0*023854; at 18*4, 0*023147 ; at 9, 0*026855 > an d at 22, 0*022907. 263. At 23 C. the absorption-coefficient of oxygen in water is 0*03402, that of nitrogen is given by the expres- sion a = 0*020346 - o*ooo53887/ + 0*00001 1156^ ; what will be the percentage composition by volume of the gas mixture obtained by boiling water previously saturated with air at 23 C. ? (Air : N, 79*1 ; O, 20*9.) 264. A litre of water saturated with carbon dioxide at 4*4 C. and 748 mm. is shaken up with a litre of nitrogen at 23 C. and 760 mm. ; the temperature of the whole being now 23 C., what is the composition of the gas remaining over the solution, a for nitrogen being taken as in question 263, and for CO 2 being equal to 1*7967 0*07761^ + 0*0016424/2 ? 265. A mixture of hydrogen and carbon dioxide is agitated with 356*4 c.c. of water at 5*5 C., volume of gas before absorption reduced to o C. = 171*29 c.c. and pressure = 0*5368 m. ; vol. of gas after absorption at o C = 119*61 c.c., and its pressure = 0*6809 m. ; a for H = 0*0193, for CO 2 = i'4i99 ; what is the percentage com- position of the original mixture ? 82 SOLUBILITY OF SOLIDS IN LIQUIDS. SOLUBILITY OF SOLIDS IN LIQUIDS. MOLECULAR WEIGHT AND THE LOWERING OF THE FREEZING-POINT OF SOLUTIONS. THE solubility of solids in liquids does not admit of being represented by any simple law ; the alteration of the solubility with temperature may generally be expressed by an equation of the form x a + bt + eft + dfi, where a, b, c, d are constants for each set of substances con- sidered, and / is the temperature. As these constants have only been determined in a few cases, the consideration of problems arising in connection therewith does not fall within the scope of this book. It has been found that substances dissolved in any solvents which solidify at attainable temperatures cause a lowering of the freezing-point according to the following law: If the molecular weight (in terms of any unit) of any substance be dissolved in 100 times the molecular weight (in terms of the same unit) of any liquid, the freezing- point of the latter is lowered by an amount always very near to 0-63 C. Let A be the coefficient of lowering of the temperature of solidification (lowering produced by one gram dissolved in 100 grams of solvent), M be the molecular weight of the dissolved compound, T be the molecular lowering of the freezing-point (that produced by the molecular weight in grams of the substance dissolved in 100 grams of the solvent) ; then MA = T, and M = -. ^TL If P be the weight of the solvent, P 1 the weight of the dissolved body, K the lowering of the freezing-point given by experiment, we have P'Xioo And hence M= PJ\. SOLUBILITY OF SOLIDS IN LIQUIDS. 83 If M' be the molecular weight of the solvent employed, then T= M' X 0*63 very nearly, with the exception of the solvent water. T has been determined by numerous experiments for the commoner solvents ; its mean value for : Acetic acid = (18*0 in a few cases), 38*6. Formic acid = 277. Benzene = 50*0. Nitrobenzene = 707. ( 1 8*5 (for organic substances, some salts of vy , _ J diad metals, all the feeble bases and acids). ~ j 37 (f r alkaline and alkaline earthy salts, and ( all the strong acids and bases). EXAMPLE. The freezing-point of a sample of acetic acid was found to be 16*490 C. ; taking 62*014 grams of this acid and . adding thereto 0*2540 gram of pure propionic acid, the solidifying point of the mixture is found to be 16*277 C. What is the molecular weight of propionic acid ? The observed lowering of the freezing-point is 16*490 16*277 = 0*213 = A", the weight of solvent P =62*014 g ram > weight of dissolved body P 1 = 0*2540 gram, and T = 38*6 for acetic acid, hence M = 38-6(0-2540X100) = 62*014 X 0*213 (Calculated njolecular weight of C 2 H 5 COOH = 74) QUESTIONS. 266. If the molecular lowering of the freezing-point of acetic acid, 16*75 C., be 43 in the case of benzoic acid, what is the temperature of solidification of a 4 per cent, solution ? G 2 84 SOLUBILITY OF SOLIDS IN LIQUIDS. 267. Plot out a curve showing the relation between concentration and K in the case of ethyl formate dissolved in acetic acid, Percentage of Ethyl Formate. Observed K. 0-1977. 0*101 C. 0-3971. 0-212 C. 0-5935. 0-31 5 C. 0-9692. o*5iiC. 1-3907. 0758 C, and find the mean molecular weight of the ethyl formate as given by this series of experiments. 268. Find the coefficient of lowering of the freezing- point of acetic acid by ethyl formate from the data of the preceding question. 269. What percentage of methyl acetate has been added to a sample of acetic acid freezing originally at 16-52 C. and after addition of the ester at 15*619 C. ? 270. By what amounts should the freezing-point of benzene be lowered in 5 per cent, solutions of each of the following bodies toluene, xylene, and anthracene ? 271. Taking 567 gram of acetic acid freezing at 16*52 C. and adding methyl acetate in successive quan- tities as follows : (i) 0*0725 gram, (2) 0*0970 gram, (3) 0*0821 gram, (4) 0-1192 gram, the freezing-points after each addition are observed to be (i) i6*452C.,(2) i6'363 c C, (3) 16*282 C, and (4) 16*177 C. respectively ; what is the molecular weight of methyl acetate found from each observation ? 272. What should be the normal molecular lowering of the freezing-point in the case of (a) anthracene, (b) sodium, (c) tin ? 273. 6 grams of anhydrous magnesium sulphate are dissolved in 100 grams of water ; the observed depression of the freezing-point is 0*958 C. Taking the value of T to be 37, which of the following formulae most probably represents the state of the dissolved salt, (a) MgSO 4 : (b) MgS0 4 , 7H 2 ? SOLUBILITY OF SOLIDS IN LIQUIDS. 274. It has been found that the latent heat of fusion of a solvent is connected with the molecular lowering of its freezing-point as shown in the formula, (absolute temperature) 2 latent heat of fusion T = 0'02. The freezing-point of ethylene dibromide being 7-9 C, and the molecular depression of its freezing-point 117*9, find its latent heat of fusion. 275. The value of Tfor water is 18-5, for formic acid 277, and for acetic acid 38*6, when organic substances are dissolved in each solvent. What inference may be drawn as to the relative complexity of the molecules of each of these solvents ? 276. What conclusion would you draw as to the molecular weight of nitrogen tetroxide from the following data ? (a) 0-5269 gram of CHC1 3 introduced into 17*3046 grams of nitrogen tetroxide lowers the freezing-point of the latter by I '06. (b} Similarly 0*4931 gram of C 6 H 5 C1 mixed with 167650 grams of nitrogen tetroxide lowers its freezing-point 1-09. 277. Find the molecular weight of gold, from the figures below showing the effect of Au in lowering the freezing-point of Na. (Take T= 104). Weights of Na. Weights of Au added in succession. Freezing-point. 20-425 (*) (*> 2'I i'3i6 97*44 91-99 88-59 278. Plot out a curve to show the variation of the molecular lowering of the freezing-point with concentra- tion, using the data obtained by Haycock for mercury dissolved in sodium. 86 EXERCISES ON THE Weight of Na. Weights of Hg added in Freezing-point. succession. 32 '47 97 '47 0-5605 96-6 0-7410 95'38 0-574 94-46 0-5170 93'64 1-771 9o'93 4'95o 83'35 EXERCISES ON THE SPECIFIC HEAT, LATENT HEAT, AND ATOMIC HEAT OF SUBSTANCES. THE capacity of a body for heat is measured by determining the number of units of heat required to raise that body one degree of temperature. The Specific Heat of a body is the ratio of the quantity of heat required to raise that body one degree to the quantity required to raise an equal weight of water one degree. Latent Heat is the quantity of heat which must be communicated to a body in a given state in order to con- vert it into another state without changing its temperature. The Atomic Heat of an element is the product of its atomic weight into its specific heat ; for the greater num- ber of the elements, the mean value of this quantity is 6*40. This is expressed by Dulong and Petit in the form of a law, thus : The atoms of all elementary bodies have exactly the same capacity for heat. The following table by Regnault shows the specific heat of the more important elements and compound gases between o 100 C. Iron .... SPECIFIC O'II37Q HEAT, Sulphur .... 0*20259 Zinc Selenium . . . 0*08370 Copper .... Mercury . . . (solid) . Cadmium . . . Silver .... Arsenic .... Lead .... 0*09515 0*03332 0*03241 0*05669 0*05701 0*08140 Tellurium . . . Potassium . . . Bromine (liquid) . (solid-28 Iodine . . . . Carbon . . . . Phosphorus 0*05155 0*16956 0*11094 c ) 0*08432 0*05412 0-24111 0*18870 Bismuth. . . . Antimony . Tin 0*03084 0*05077 Atmospheric Air . Oxygen . . . . Nitrogen 0-2375 0-2175 0*24.38 Nickel .... Cobalt .... Platinum plate . sponge. Palladium . . . Gold . 0*10863 0*10696 0*03243 0-05927 Hydrogen . . . Carbon monoxide Chlorine . . . Carbon dioxide . Nitric oxide . . Steam 3-4090 0-2479 0*1214 0*2164 0*2238 0*47"; The following table, embodying the results obtained by Person, gives the latent heat of several bodies : Water 79' 2 5 Phosphorus Sulphur . . * Sodium nitrate . Potassium nitrate Sodium chloride . Tin . 5*03 9-37 62-97 47*37 40-70 14-25 Bismuth .... 12*64 Lead . . . . . 5-37 Zinc 28-13 Mercury .... 2*83 Steam 537 Alcohol vapour . . 208 Ether vapour ... 90 The heat required to raise the temperature of a kilogram of water through i C. is able to do work equivalent to the lifting of a kilogram weight through 425 metres. The mechanical equivalent of heat is 425 metre-kilograms. A kilogram of water falling through a height of 425 metres, and having its motion suddenly arrested, would have its temperature raised i C. 88 LATENT HEAT, QUESTIONS. 279. The following quantities of water are mixed together : 1 kilogram at 40 C. 2 kilograms 30 3 20 4 5> j> 5 ?> Calculate the temperature of the mixture. ! 280. If one kilogram of mercury at 20 C. be mixed with one kilogram of water at o, the temperature of the mixture will be 0*634 C. Calculate the specific heat of mercury. ! 281. Determine the specific heat of mercury from the observation that when the same vessel is filled succes- sively with water and mercury, and heated to the same temperature, the water and mercury cool through the same number of degrees in 10 minutes and 270 seconds respectively. The specific gravity of mercury being con- sidered constantly at 13*6. 282. Calculate the specific heat of mercury from the following numbers obtained by Kopp, according to his method : Temperature of mercury bath . . 51*! C. Initial temperature in calorimeter . I3'4i Final . i6*5o Weight of water in calorimeter . . 26*945 grams. Weight of mercury used .... 53*015 ,, Thermal value of apparatus . . . 0*65 1 283. Calculate the specific heat of phosphorus from the following determination by Kopp : Temperature of mercury bath . . 38*8 C. Initial temperature in calorimeter . io c< o5 Final . I3*2O grams. Weight of water in calorimeter . . 26*95 Weight of phosphorus employed . 3*075 Weight of water in tube .... 2*065 AND ATOMIC HEAT, 89 f 284. The specific heat of water is 4 times, and its density 770 times, that of air. Supposing a cubic mile of water to yield up one degree of its heat to a cold atmosphere, what quantity would 1000 cubic miles of the atmosphere be heated ? 285. A bar of platinum weighing 150 grams is heated in a furnace until its temperature becomes constant, when it is thrown into a kilogram of water, the temperature of which it raises from 15 to 20. Required the temperature of the furnace on the assumption that the specific heat of platinum is 0*03308 + o*ooooc>42/ between o and /. (Pouillet.) 286. A determination of the specific heat of iron made in the calorimeter of Lavoisier and Laplace yielded the following data. Calculate the specific heat of iron : Weight of iron taken . . . 100 grams. Weight of ice melted . . . 14*35 Initial temperature of iron . 100 C. Latent heat of water . . . 79*25 287. The mechanical equivalent of heat is 425 metre- kilograms. What is this in foot-pounds ? 288. From what height must a block of ice at o C. fall that the heat generated by its collision with the earth shall be just competent to melt it ? From what height must it fall that the heat generated may be sufficient to convert it into steam ? 289. If W w = the number of scale divisions on a Bunsen's calorimeter equivalent to one gramme-degree unit of heat, T the observed movement of the thread in scale divisions, G the weight of substance taken T and / its temperature, then sp. ht. = -rr? -~ Find the sp. ht. of indium if = 1*1514, t = 99*82 C., W w = 14*657 and T 100*2. Time correction for T = - 3'45-) 290. From the data given, find the specific heats of (A) cast silver, () cast zinc, (C) cast antimony, (D) 9 o OF SUBSTANCES. cast cadmium, (E) roll sulphur, as determined by Bunsen's method. W-w = 14*657. Weight of substance . . Temperature . A. 3-6320 iooC B. 2*5150 QQ-8C C. QQ'8C D. i '8675 QQ-8C E. i "0708 IOOC Scale movement .... 2977 343'8 279-5 1467 268-8 291. One kilogram of steam at 100 C. is condensed in forty-nine kilos of water at 16 C. ; what is the temperature of the mixture ? 292. It is required to distil 2 kilos, of ethyl alcohol (BP. 78*3) per hour. What must be the supply of water at 1 6 C. in order that the temperature of the water round the worm may not average higher than 25 C. ? (Specific heat of ethyl alcohol = 0*615). 293. 75 grams of water are placed in a calorimeter of which the water equivalent is 5 grams ; when at 15 C., 5 grams of steam are passed into and condensed by the water with the result that the temperature of the calori- meter and its contents is raised to 5 1 '6 C. ; from these data calculate the latent heat of vaporization of steam. 294. Using the atomic weights given in Appendix I., calculate the atomic heats of Pb, Ag, Cu, Fe, S, and P. 295. Taking the mean value of the atomic heats of the elements as 6*4, and the specific heats given in the table, find the atomic weights of Ag, Zn, Bi, Sn, and Fe. The stochiometrical quantities of these elements found by analy- sis as equivalent to 35*37 parts of Cl are as follows : Ag 107-66, Zn 32-44, Bi 69*16, Sn 29-339, and Fe 18*626 ; from these numbers deduce the exact atomic weights of the elements in question. 296. It is found that in many compounds the sum of the atomic heats of the atoms of which the molecule is built up equals the product of the molecular weight into the specific heat of the compound (the molecular heat) ; assuming this to be always the case, deduce the atomic heats in the solid state of Cl and O, from the data that the HEAT OF SOLUTION. 91 sp. ht. of PbCl 2 = 0-0644 (Regnault), of Ga 2 O 3 = 0-1062, and of In 2 O 3 = 0*0807. (Specific heats, Ga, 0*079 5 I n ? '57-) 297. Deduce the atomic heat of oxygen from the molecular heat 28-3 of potassium permanganate. (Specific heats, K, 0*166 ; Mn, 0*122.) 298. Find the approximate atomic heat of hydrogen if the molecular heat of NH 4 C1 be taken as 20, and the atomic heats of N and Cl as 5*6 and 6*4 respectively. 299. The atomic heat of lead is deduced from Regnault's value for the specific heat to be about 6*3, the specific heat of cerussite = 0*080, of calc-spar = 0*206, of strontianite = 0*145, an d * witherite = 0*109 5 find tne atomic heats of Ca, Sr, and Ba. HEAT OF SOLUTION ; HEAT OF COMBINA- TION ; CALORIFIC POWER; CALORIFIC INTENSITY. CHEMICAL change is usually accompanied by changes in the distribution of energy in the system considered. By far the larger part of the energy lost to the changing systems during chemical reactions is given out in the form of heat ; reactions in which heat is evolved are said to be exothermic. There exists also a class of reactions requiring heat to be imparted from without the system to the reacting bodies to enable the change to occur ; such reactions are termed endothermic. It is convenient to consider the thermal effects of the solution of substances together with the changes of energy- distribution due to strictly chemical reactions. The thermal unit in general use in connection with problems of this character is defined to be the amount of heat required to raise the temperature of one kilogram of water one degree Centigrade (the Calorie). Generally the heat of combination is only one of a number of factors in the total thermal effect, heat being absorbed in the liquefaction or vaporization and evolved on the solidification of the reacting substances and pro- ducts. 92 HEAT OF COMBINATION. In the phenomena of ordinary combustion the terms calorific power and calorific intensity are used ; by the former we understand the amount of heat produced by the combustion of one unit of weight of the burning sub- stance, whereas the latter indicates the temperature to which the products of combustion can be raised by the heat evolved. TT The calorific intensity 1 = m^i + m 2 s 2 + m 3 s 3 + .... where H represents the calorific power, ^ s 2 s 3 &c., the specific heats, and 1% m 2 m 3 &c., the masses of the pro- ducts of the combustion of one unit weight of substance. The following table expresses the calorific powers of a number of substances burnt in oxygen : 1,144- 602. 2,403- 13,063. 11,858. 6,850. The calorific power of any substance is a constant, being the same whether the body be burnt in oxygen or in air and whether it be burnt rapidly or slowly ; the calorific intensity is modified by the circumstances under which the combustion takes place, any mixture of inert material e.g. nitrogen in air, ash in coal lessening / since heat must be used to raise the temperature of the foreign matters. Again, if radiation be allowed to take place freely the temperature reached during slow com- bustion will not be nearly so high as that attained by a more rapid burning. A special form of notation is used in thermo-chemistry e.g. [H 2 , CP] = 44,000 -f . in words means ' two kilo- grams of hydrogen combine with seventy-one kilograms of chlorine with the evolution (+) of 44,000 Calories. 7 EXAMPLES. I. The calorific power of carbon is 8080, what is its calorific intensity when burnt in oxygen ? Hydrogen . ^4.462. Tin Carbon . . Sulphur . . Phosphorus . Zinc . . . Iron . . . }-(, < T V *' 8,080. 2,220. 5,747- 1,301. 1,576. Copper . . . Carbonic oxide. Marsh gas . Olefiant gas . . Alcohol . . . CALORIFIC POWER. 93 12 parts of carbon give on combustion 44 parts of carbon dioxide, hence I kilo, produces 3-67 kilos. The specific heat of carbon dioxide is 0*2164. Hence 7 = , 8o8 , = io,i74C. 3*67 x 0*2164 2. Find the calorific intensity of hydrogen burning in oxygen from the appended data. Calorific power of hydrogen 34,462. Weight of the product of combustion (H 2 O) yielded by I kilo, of hydrogen, 9 kilos. Specific heat of steam 0*475. Here it must be remembered that at the temperature of combustion the water produced remains in the gaseous state, whereas the calorific power given above includes the heat given out on the condensation of the water vapour produced to water ; hence the latent heat of vaporization of the water must be deducted from the given calorific power in calculating the value for /. Assuming that the initial temperature is oC. the total heat required to raise the temperature of the water produced to T = 9 {100 + 537 + 0*475 (7" 100) J and this is necessarily the same as the calorific power. Hence / = T, and 9 {100+ 537 - 47*5+0*475 ?} = 34,4^2. 589-5+ 0*475 T= 34^62 therefore 1 = T= ft- 589-5) = 682O . 2 c C o'475 3. Determine the heat of formation of CH 2 O 2 from its elements from the data : [C,0 2 ] = 96,960 + ; [H 2 ,0] = 68,360 +; [CH 2 O a ,O] = 65,900 +. The total heat produced by the oxidation of free C. and H 2 to CO 2 and H 2 O must evidently be equivalent to the heat of formation of CH 2 O 2 together with the heat pro- 94 CALORIFIC INTENSITY. duced by the oxidation of CH 2 O 2 , for we obtain the same final products in both cases. [C,0 2 ] + [H 2 ,0] = [C,H 2 ,0 2 ] + [CH 2 2 ,0]. and [C,O 2 ] + ]H 2 ,0] = 96,960 + 68,360 = 165,320 + . .-. [C,H 2 ,0 2 ] = 165,320 - [CH 2 2 ,0] = 165,320 - 65,900 = 99,420+. 4. Find the heat of solution of HBr from the data : (1) [KOH Aq, HC1 Aq] = [KOH Aq, HBr Aq]. (2) [K Br Aq, Cl] = 11,500 +. (3) [Br, Aq] = 500 +. (4) [H, Br] = 8,400. (5) [H, Cl, Aq] = 39,300. From (2). [KBr Aq, Cl] = [K, Cl, Aq] + [Br, Aq] - [K, Br, Aq] - 11,500. Replacement of Br by Cl gives 11500 thermal units. From this (5) and (i) we have [H, Br, Aq] = [H, Cl, Aq] = 11,000 = 28,300 and [HBr, Aq] = [H, Br, Aq] - [H, Br] that is [HBr, Aq] = 28,300 8,400 = 19,900, or the heat of solution of HBr is represented by 19,900 thermal units. QUESTIONS. 300. What weight of water would be heated from o to 1 C. by the combustion of I gram of hydrogen ? 301. One gram of phosphorus is burnt in oxygen. To what temperature would a kilogram of water at o C. be raised by the combustion ? 302. Calculate the amount of water raised 15 by the combustion of i gram of sulphur in oxygen. 303. Calculate the calorific intensity of (i) ethylene, (2) of methane burning in air. CALORIFIC INTENSITY. 95 304. Calculate the calorific intensity of Newcastle Hartley coal possessing the following percentage com- position : Carbon 88*42 Hydrogen .... 5*61 Oxygen 5-97 lOO'OO 305. Determine the amount of heat disengaged by the combination of I gram of carbon with oxygen from the following experiment by Andrews : Weight of substance burnt . . . ro88 grams. Temperature of the air .... io*o6 C. The excess of the final temperature ) of the water above that of the air { ( Increment of temperature found . 2*473 t corrected . 2^464 Weight of water in the calorimeter 318*3 grams. Thermal value of the vessels . . 180 ! 306. Calculate the thermal power of charcoal from the following data : Weight of substance consumed . 10 parts. the. water 8900 the copper vessel . . 1000 Specific heat of copper .... 0*095 1 5 Initial temperature . . . . 11 C. Final .... 20 C. 307. Calculate the height to which a ton weight would be raised by the combustion of a kilogram of this char- coal, supposing that all the heat evolved was utilised in the lifting. 308. Find the heat of formation of H 2 SO 4 from its elements. L / xj _'" ' i. J J s j s - - r u - ~ 7 AJ 5' J^ > [H 2 S0 4 , Aq] = 17,000 : 96 CALORIFIC INTENSITY. 309. Given that [C,O 2 ] = 96,900 and [H 2 ,O] = 68,400, find the heat of formation of CH. 4 , if the complete com- bustion of CH 4 gives 213,500 heat units. 310. From the given data, find the heat of formation of HCN : [C,O 2 ] = 96,900 ; [H 2 ,O] = 68,400 ; [2HCN, 5 0] = 319,000. 311. What is the thermal value of the reaction [N 2 ,O] ? Data : [C,2N 2 O] = 133,900. [C,O 2 ] = 96,900. 312. Calculate the heat of combination of alcohol and acetic acid [C 2 H 6 O, C 2 H 4 O 2 ], having given : [C 2 H6Q,0 6 ] = 330,400 ; [C 2 H*0 2 ,0 4 ] = 210,300; [(C 2 H5)OOC.CH3, 5.0 2 ] - 313. Find the heat of formation of aldehyde from its elements, (a) liquid, (b) gaseous. Data : C 2 H 4 O + 5.0 = 2CO 2 + 2H 2 O. -C'H'O, 5 0] = 275,500; [C,0 2 ] = 96,900; H 2 ,O] liquid = 68,400 ; [H 2 ,O] gaseous = 58,700 ; C 2 H 4 O,5O] gaseous = 266,000. 314. The heat of solution of K 2 SO 4 is 6340 , of CuSO 4 ,5H 2 O is 2430 -, of K 2 SO 4 .CuSO 4 .7H 2 O = 14,360-. What is the heat of formation of the double salt in solu- tion ? 315. [P 2 ,O 5 ] = 369,100 + with ordinary phosphorus, [P 2 ,O 5 ] =326,800 + with amorphous phosphorus. What is the thermal value of the change from the yellow to the red variety ? 316. In the reaction [H 3 PO 3 Aq, x NaOH Aq] the heat of neutralization for x = \ is 7,400 ; for x = I is 14,800 ; for x 2 is 28,500 ; for^r = 3 is 28,900. What deductions may be drawn as to the basicity of this acid ? Questions marked thus (!) are from the Owens College Calendars. Questions marked thus (i) are from the examination papers of the Science and Art Department. APPENDIX. TABLE I. Atomic Weights and Symbols of the Elements. Elements. Symbol. Atomic Weight. Observer. Aluminium Al Mallet Antimony . Sb (Stibium) . . As ii9'6 Cooke. Barium . Ba 136-86 Pelouze ' Marignac Beryllium . Be 9 '08 Nilson & Pettersson. Bismuth Bi 207*5 (Schneider; (208*38: 1 Boron . B . IO'Q \ Classen). Deville. j Bromine Br 79*76 Stas. Cadmium . . Csesium . Cd Cs . . . 111*7 132*7 Lenssen. Johnson & Allen Calcium Ca OQ'QT Carbon . c ii '97 Van der Plaats Cerium . Ce . . . . 129*90 Robinson Chlorine . . Chromium . Cobalt . . . Cl Cr Co SB'S? 52-06 58*6 Stas. Rawson. Russell Copper . . . Didymium . . Erbium Cu (Cuprum) Di. . E 63-18 146*0 1 66 /Hampe; (63*45: \ Richards). Brauner. Cleve Fluorine F . 19*06 Louyet Gallium Ga Germanium Ge . 72*3 Winkler Gold .... Hydrogen Au (Aurum) . 196*85 Thorpe & Laurie. Indium . . In. . . . H3'4 Winkler Iodine Stas Iridium . Ir IQ2'C Iron .... Fe (Ferrum) . . 55-88 /Berzelius ; Erdmann \ & Marchand. H 9 8 APPENDIX. TABLE I. (continued.) Elements. Symbol. Atomic Weight. Observer. Lanthanum La 138*81 Marignac Lead. . . . Lithium . . Pb (Plumbum) . . Li 206 '39 7'oi Stas. Stas Magnesium Mg 2 3'94 Marchand & Scheerer Manganese . Mn 54'8 Mercury Molybdenum . Hg (Hydrargyrum) Mo 199-8 QC'SO Erdmann & Marchand. Von der Pfordten Nickel . . . Ni c8'6 Russell Niobium . . Nitrogen Nb N 937 Marignac. Stas Osmium Os i' Seubert Oxygen . O m'o6 (Dumas ; Erdmann & Palladium Pd io6'35 ( Marchand. Phosphorus p 30-96 SchrStter Platinum . . Potassium . Rhodium Pt K (Kalium). . . . Rh *94'3 39 '03 Seubert. Stas. Rubidium . Rb 85'34 Heycock Ruthenium Ru 103*5 Claus ' Samarium . Sm Cleve Scandium . Sc 43'97 Nilson Selenium Se 78*8? Silicon . . . Silver . . . Sodium . . . Strontium . Si Ag (Argentium) . Na (Natrium) . . Sr 28 -Q 107 '66 22-99 R 7 -Q /Dumas ; Schiel ; (28*33 ' ( Thorpe & Young.) Stas. Stas. Sulphur s 31 '98 Stas Tantalum Ta i82'o Tellurium . Te ... 125 Brauner Thallium Tl Thorium Th 231 "QQ Kriiss & Nilson Tin .... Titanium Sn (Stannum) . . Ti H7'35 48 'o fBerzelius ; (117*8 : Van \ derPlaats). Thorpe Tungsten . . Uranium . . Vanadium W (Wolfram) . . U. . . . . . . v . 183-6 238-9 (Schneider ; Marchand ; \ Persoz. (Zimmermann, Ali- \ begoff & Kriiss. Ytterbium . Yttrium ... . . Zinc .... Zirconium . . Yb Y. ..'.... Zn Zr 172*8 88-9 64*88 90-40 Marignac. Cleve. /Erdmann ; (65*18 : Van \ derPlaats). G. H. Bailey. APPENDIX. 99 TABLE II. Weight of one Cubic Centimetre of Atmospheric Air at different Temperatures from o to 300, at 760 mm. pressure. 0*001293 46 0*001108 92 0*000967 138 0*000858 I 001288 47 001105 93 000964 139 000856 2 001284 48 001 102 94 000962 140 000854 3 001279 49 001098 95 000959 141 000852 4 001275 So 001095 96 000956 142 000850 5 001270 51 001091 97 000953 143 000848 6 001266 52 001088 98 000951 144 000846 7 001261 53 001084 99 000948 145 000844 8 001257 54 001081 00 000946 I 4 6 000842 9 001252 55 001077 01 000943 147 000840 10 001248 56 001074 02 000941 I 4 8 000838 ii 001243 57 001070 03 000938 149 000836 12 001239 58 001067 04 000936 15 000834 13 001234 59 001063 05 000933 000832 14 001230 60 001060 06 000931 152 000830 15 001225 61 001057 07 000928 153 000828 16 OOI22I 62 001053 08 000926 154 000826 17 OOI2I7 63 001050 09 000923 155 000824 18 OOI2I3 64 001047 IO 000921 I 5 6 000822 19 001209 65 001044 II 000919 157 000821 20 001205 66 001041 12 000916 000819 21 OOI2OI 67 001038 13 000914 159 000817 22 OOII97 68 001035 14 OOOgil 160 000815 23 OOII93 69 001032 15 000 9 09 161 000813 24 OOllSg 70 001029 16 000907 162 000811 25 OOIl85 001026 17 000905 163 000809 26 001181 72 001023 18 000 9 03 164 .000807 27 001177 73 OOIO2O 19 000900 165 000806 28 001173 74 OOIOI7 20 000898 166 000804 29 001169 75 OOIOI4 21 000896 167 000802 30 001165 76 OOIOII 22 000894 168 000800 31 001161 77 001008 123 OOoSgi 169 000798 32 001157 78 001005 124 000889 170 000796 33 001154 79 001002 125 000887 171 000794 34 001150 80 001000 126 000884 172 000793 35 001146 81 000997 127 000882 i73 000791 36 001142 82 000994 I2 000880 000789 37 001138 83 000992 129 000878 175 000788 38 001134 84 000989 130 000876 176 000786 39 001131 85 000986 000874 177 000784 40 001128 86 000983 132 000871 178 000782 001124 87 000980 133 000869 179 000781 42 OOII2I 88 000977 134 000867 1 80 000779 43 001118 89' 000974 135 000865 181 000777 44 001114 90 000972 000863 182 000776 45 OOIIII 000969 137 000860 183 000774 H 2 100 APPENDIX. TABLE II. (continued.) 184 0'000772 213 o'ooo725 242 0*000685 271 o'ooo648 185 OOO77O 214 000724 243 000683 272 000647 186 000769 215 000722 244 000682 2 73 000646 187 000767 216 000721 245 000681 274 000645 188 000765 217 000719 246 000679 275 000643 189 000763 218 000718 2 47 000678 276 000642 190 000762 219 000716 248 000677 277 000641 191 000760 220 000715 249 000675 278 000640 192 000758 221 000713 250 000674 279 000639 193 000757 222 000712 2 5i 000673 280 000638 194 000755 223 000710 252 000672 281 000636 195 000754 224 000709 253 000670 282 000635 196 000752 225 000708 2 54 000669 283 000634 197 000751 226 000706 2 55 000668 284 000633 198 000749 227 000705 256 000666 285 000631 199 000748 228 000703 2 57 000665 286 000630 200 000746 229 000702 258 000664 287 000629 201 000744 230 000701 2 59 000663 288 000628 202 000743 231 000699 260 000662 289 000627 203 000740 232 000698 261 000660 290 000626 204 000739 233 000697 262 000659 291 000625 205 000737 234 000695 263 000658 292 000624 206 000736 235 000694 264 000657 293 000623 207 000734 236 000692 265 000655 294 000622 ,208 000733 237 000691 266 000654 295 000621 2O9 000731 238 000690 267 000653 296 000620 210 000730 239 000689 268 000652 297 000619 211 000728 240 000688 269 000651 298 000618 212 000727 241 000686 270 000650 299 000617 300 000616 j TABLE III. The weight of 1000 c.c. of water of t C., when determined by means of brass weights in air of o C., and of a tension o"j6?n. , is equal to 1000 x grms. t* i 2 3 4 5 6 7 8 9 10 ii 12 X 1-26 i '20 1-16 i'i3 1*12 I'I2 1-14 1-17 V22 1-28 i '35 i '44 i '54 t 13 *4 15 16 17 18 19 20 21 22 23 24 25 X 1-65 1-78 1-92 2*07 2'23 2*40 2-58 2- 7 8 2'99 3 '20 3'43 3'66 3 '9 f 26 27 28 29 3 X 4'*7 4 '45 4-72 4 '99 5-26 APPENOIX. TABLE IV. Volume and Density of Water at different Temperatures, (ROSSETTI). Temp. Volume of Water (at o = i). Sp. Gr. of Water (at o = i). Volume of Water (at 4 = i). Sp. Gr. of Water (at 4 = i). I '00000 I '000000 "00013 0*999871 i 0-99994 I '000057 '00007 0-999928 2 0-99990 I "000098 '00003 0-999969 3 '99988 I'OOOI2O 'OOOOI 0-999991 4 0-99987 I '000129 '00000 1*000000 5 0-99988 I '000119 'OOOOI 0-999990 6 0-99990 I '000099 '00003 0-999970 7 0-99994 I '000062 '00007 0-999933 8 0-99999 1*000015 "OOOII 0*999886 9 "00005 o'999953 "00018 0-999824 10 "00012 0-999876 "00025 o"999747 ii "OOO22 0-999784 00034 o'999655 12 "00032 0-999678 00045 0*999549 13 "00044 0*999559 00057 o'99943o 14 "00057 0-999429 00070 0*999299 "00071 0-999289 "00084 0*999160 16 "00087 0-999131 "00100 0*999002 17 "00103 0-998970 00116 0*998841 18 "00122 0*998782 -00135 0-998654 19 'OOI4I 0-998588 -00154 0*998460 20 "OOIOI 0*998388 00174 0-998259 21 "00183 0*998176 '00196 0-998047 22 "00205 o'997956 '00217 0-997828 23 "00228 0-997730 '00240 0*997601 24 'OO25I o*997495 '00264 0-997367 25 '00276 0*997249 '00289 0*997120 26 00301 0-996994 00314 0-996866 27 '00328 0-996732 00341 0^996603 28 00355 0*996460 00368 0-996331 29 00383 0-996179 00396 0-996051 30 'OO4I2 0-99589 00425 0*99577 40 00757 ... 50 'OIl82 ... 60 '01678 ... 70 02243 80 02874 ... ... 9 -03554 ... ... 100 '04299 ... APPENDIX. TABLE v For the Calculation of I + 0*003677* T T T T 1 i 0-99634 38 o'8776i 75 0-78416 112 0*70870 2 99271 39 87479 76 78191 "3 70686 3 98911 40 87199 77 77967 114 70503 4 98553 4i 86921 78 77745 H5 70321 5 98198 42 86645 79 77523 116 70140 6 97845 43 86370 80 77304 117 69960 7 97495 44 86097 81 77085 118 6978! 8 97148 45 85826 82 76867 119 69603 9 96803 46 85556 83 76651 1 20 69425 10 96460 47 85289 84 76436 121 69249 it 96120 48 85022 85 76222 122 69073 12 95782 49 84758 86 76010 123 68899 13 95446 5o 84495 87 75798 I2 4 68725 14 95U3 5i 84234 88 75588 125 68552 15 94782 52 83974 89 75379 x*6 68380 16 94454 53 83716 90 75i7i 127 68209 17 94127 54 83460 9i 74964 128 68038 18 93803 55 83205 92 74758 129 67869 19 93482 56 82952 93 74554 130 67700 20 93162 57 82700 94 74354 131 67532 21 92844 58 82450 95 74148 132 67365 22 92529 59 82201 96 73947 i33 67199 23 92216 60 8i954 97 73747 i34 67034 24 9 I 95 61 81708 98 73548 i35 66870 25 91596 62 8 I4 6 4 99 73350 136 66706 26 91289 63 81221 100 73153 i37 66543 27 90984 64 80979 101 72957 138 66380 28 90682 65 80740 102 72762 i39 66219 29 90381 66 80501 103 72568 140 66059 30 90082 67 80264 104 72376 141 65899 31 89785 68 80028 105 72184 142 65740 32 89490 6 9 79794 106 71993 i43 65582 33 89197 70 7956i 107 71803 144 65424 34 88906 7r 79329 108 71615 i45 65268 35 88617 72 79099 rog 71427 146 65112 36 88330 73 78870 no 71240 i47 64957 37 88044 74 78642 III 7 I0 55 148 64802 APPENDIX. I0 3 TABLE VI. Pressure (tension) of Water Vapour between - 19 and 101 C. t (BROCH). T.C. mm. T. 6 C. mm. T.C. mm. T.C. mm. -19 i '0288 12 10-4322 43 64-3104 74 276-8675 - 18 I'I202 13 11-1370 44 67-7568 75 288-7640 -17 1-2187 14 11-8835 45 71-3619 76 301-0860 -16 1-3248 15 12-6739 46 75'i3i4 77 313-8475 -i5 I '4390 16 13-5103 47 79-0714 78 327'o549 - 14 1*5618 J 7 14-3950 48 83-1883 79 340-7265 ~i3 1-6939 18 15-3304 49 87-4882 80 354'8730 - 12 i'8357 i9 16-3189 50 91-9780 81 369-5075 - II 1-9880 20 17-3632 5i 96-6644 '82 384-6432 - 10 2-1514 21 18-4659 52 101-5541 83 400/2933 g 2-3266 22 19-6297 53 106*6546 84 416-4721 8 2'5i43 23 20-8576 54 111-9730 85 433-1938 -7 2"7i53 24 22-1524 55 117-5162 86 450*4730 -6 2-9304 25 23-5174 56 123-2925 87 468-3240 - 5 3'i6o5 26 24-9556 57 129-3095 88 486-7635 -4 -3 3'4o65 3-6693 27 28 26-4705 28-0654 58 59 135-5750 142-0973 89 90 505-8059 525*4676 2 3-9499 29 29'7439 60 148-8848 9i 545-7650 I 4'2493 30 31-5096 6l 155-9456 92 566-7149 4'5687 31 33-3664 62 163-2889 93 588-3349 + 1 4-9091 32 35-3181 63 170*9236 94 610*6426 2 5"27i9 33 37-3689 64 178-8585 95 633-6567 3 5-6582 34 39-5228 65 187-1028 96 657'3956 4 6*0693 35 41-7842 66 i95 '6663 97 681-8791 5 6-5067 36 44*1577 67 204*5586 98 707-1271 6 6-9718 37 46-6477 68 213-7895 99 733-1602 7 7-4660 38 49-2950 69 223-3691 100 760-0000 8 7-9909 39 51-9965 70 233-3079 101 787*6678 9 8-5484 40 54-8651 7i 243*6163 10 9-1398 4i 57-8700 72 254-3048 ii 9-7671 42 6i'oi67 73 265-3849 JO4 APPENDIX. TABLE VII. For the Conversion of the Degrees (T') of a Mercurial Thermo- meter into the corresponding values (T") of a Hydrogen Thermometer. T' TV/ T' T ,/ T' T" T' TV/ 10 '07 10 "oo 1 00 TOO 189-65 190 280*52 280 o'oo o'oo 109-98 no 199-70 200 290*81 290 + 10 '05 + 10 '00 "9'9S 1 2O 209-75 210 301*08 300 20 '08 20 '00 129-91 130 219*80 220 3"'45 310 30-10 30-00 *39'85 140 229-85 230 321*80 320 40' 1 1 40*00 149*80 ISO 239-90 240 33 2 '4 330 50*10 50 '00 i59'74 1 60 250-05 250 343 'oo 340 60*09 60 'oo 169*68 170 260*20 260 354 'oo 350 70-07 70*00 179*63 1 80 270*38 270 80 '05 80 -oo 90-03 90*00 APPENDIX. '05 TABLE VIII. For the Correction of Thermometer Readings. t-t' n 10 20 30 40 50 60 70 80 90 100 0-14 0-43 IO 20 30 O'OI O'O2 0*04 o'o6 0*09 0*09 0-13 o'o6 O'll 0-17 o'o7 0-14 0'2I '09 0-17 0'26 O'lO O'2O O'll 0'22 0-13 0'26 o'39 40 50 60 0*05 o'o7 o'o8 o'n 0-14 0-17 0-17 0'2I 0*25 0-23 o-35 0'28 o'43 o'34 o'43 o'so o'6o o'6o o'7o 0-52 '64 0-77 o'57 0-71 o'86 70 80 90 O'lO o'n 0-13 0'20 0'23 O'26 0'30 o'34 o'39 o'45 o'Si 0*50 o'57 0*64 o'6o 0-68 0-77 0*70 o'8o o'go o'8o 0*91 i '03 o'go i -03 i'i6 I'OO 1-14 i '30 100 no 1 20 0-14 o'i6 0-17 0'28 0'3I o'34 o'43 0-47 o'57 0*63 o'69 0-71 0-79 0-86 0-85 0-94 103 I'OO I'lO I '20 1*26 i '37 i '29 1-42 i '54 i "43 n t-t' IIO 120 130 140 150 160 170 180 190 200 10 o'i6 0-I 7 o' I9 O'2O 0'2I 0'22 0-24 0*26 0-27 0'2 9 ao 0-31 o'34 o'37 0*40 o'43 0*46 0-49 0*51 o'54 o'57 30 0-47 0*56 o'6o 0*64 o'68 o'73 0-77 0'82 o'86 40 0-63 '69 0-74 o'8o o'86 0*91 o'97 i '03 1-09 1-14 50 0-79 o'86 0^93 I'OO 1-07 1-14 I '22 1-29 1-36 i '43 60 0-94 i "03 I '20 1-29 i '37 1-46 i '54 i'6 3 1-72 70 I'lO I '20 1-30 I'40 i'5o i '60 1-70 i -80 1-90 2 '00 80 '26 i*37 i '49 I '60 1-72 1*83 i '94 2-05 2-17 2'29 90 '42 i '54 i'66 I '80 i '93 2-05 2-17 2-31 2'45 2'54 IOO 58 1-71 1-84 2'OO 2-15 2-29 2'43 2'57 2-72 2'86 no '73 1-89 2-04 2"2O 2-36 2'6 7 2'8 3 2 '99 3'i5 1 20 '89 2 '06 2-23 2'40 274 2 '92 3 '09 3 '43 io6 APPENDIX. TABLE IX. Useful Constants. Number. Logarithm. g, at Greenwich. . 8 Coefficient of expansion of Hg. for i C. . Mass of i c.c. of Hg. at o C Volume of 31*92 gm. of oxygen in litres. . Conventional molecular vol. in litres. . . 0*0001802 13 '596 grams. 22-325 22-4 4' 2 55754 8 1-1334112 1-3487915 1-3502480 Coeff. exp. air, at constant pressure . . . Coeff. elasticity of air at constant volume . One litre of hydrogen at oC and 760 mm. o '00367 o '003665 3-5646661 3-5640740 barometric pressure and at the level of the sea and latitude 45 weighs . . . 0-08958 grms. 2'9522III Do. oxygen 1-4298 1552753 Do nitrogen . Do. air 1-2932 O-IIl6o57 APPENDIX. 107 ON r^ <* M ON r. m ** CM M o ONOO r>. t^vo m m *- ^- co COOOOVO * CM H O ONOO ^vomm^^cococMCM M H H O ON ON ON 0^0 ^- To " TO tlvo vo ONONONON m NNNMH HMHMM M H H H M M M fs. 1/N * CO CM H H OONON 00 00 00 tx t>. ^ txvo VO VO vovo vnmtomvomm^- CO N co co txvo vo vo m m m <* "***3w- COCOCOfOCOCNl IN N ON ** xovo O CM >* ON ON m ON H -^-oo - ON 00 MMMHCMCMCMCM vO VO vO TJ- M ONVO CM t~x CM oo O *-< CM Tf vnvo t~-oo ON vO CO VO *} VO * H MVOVOCO COTh-<*-CM OOO -^-OvO M ts.r^CM COM rJ-VOCMvO t^ OO ON H CM CO VOVO t^-OO ON OO rh CM M CM 10 CM COOO OO COCOON.OOO CM COO mvO VO CM vO ON COVO ON M -i-vO ON M CO VO f^OO O CM CO lOVO ^-00 M VO t^ CM VOVO OO ON M CM CO vovO t-OO ON O ON Th M -i- VOCO VOOOOO f^vO CO t^OO t^ <* O * t-x KSKTsftSs ON ON vo t^vO O M ON CO vo CM VO O COvO ON M CM T*- VO OO ON M CM CO -.o comt^ OOOHHMCMCMCMCM O CM ThVO 00 ON M CO Tl-VO t^ M" vooo M Tj-vOOO ON M si O w CM CO - vovO r->00 ON O M CM CO Th VOVO t^OO Cft M CO TJ- mvo^ .00^ ON io8 APPENDIX. oo t^ ONOO oooooooo t> tx i-x tx CO t^ r*. t- ^ t>. txVO VO VO t^ t^ txvo vo vo VO vo vo VO vovovovovo inminmin ininiomin'ininin-^--^ vo >,(N -^- in CO ON t>OO O\ ON rovo oo oo vo ro ON * t^ in Tj- (N OOOvO -^-n OxvO O M M ro en Ti- xnvo vo t^ ONO Ooovo NOO Nvo ON m ON "ON o" H 'H " {^ ro VO vo vo vo vo VO vo vo vo vo > O O O O O ON OMX> t>.MD O M ro -^- to if)\o r-^oq ON b M S ro% So S ^ ooONONOwHNrofi^ vo m H -^- m ro o -*vo vo m Q" H N S^ ro ^ invo vo" tQ ^H??SI^ COONONOOVO^OOrO^O S^S^tfSSStf R.R.R.g.r,SKR^^ ^^KoScgcocg^coot ^0^-xn^-Hinoooo^ * ro invo in CO ON ^-00 vo d * Tf "8 ^c^^^^S Koo"b^ o H 2" c^ ro ? tn^vo K cSS 0*0 ^'N'S ro^- CO O H w ro <* mMD t^oo ON vo H in t^oo t^ in IN t^. M O H H N ro TJ- u-> mvo t-^ O r-^Tt-HOO -^-MOO -r)- O oooooNOO^cMMroTt- t^ t^ t^oo oo oo oo oo oo oo N N ONfomTj-Hvo ONO O t~sCO^ONO ONt^-^-ONrO O ONt^inTj-w ONt^^j-N O O >-> N ro Tt- -.oo ON O H e4 co <* invo txoo ON vovovovovovovovovovo APPENDIX. 109 ON 00 *" vo 1C * t'-^-^-'tf-Tt-mmmmm corococorococorococo CMCMCMNCMNCMNCMCM NNN(NNN(SCN1NN N(SNN(M t^oo OO ON ON ON oo fHiffffit O M H * o m ONNint^ONOOO ONOO vo ThM r>.moNThONmt^ t*+ iOU->u->T}-N QVO N t~x T m mvo t^ t~-oo oo ON ON Nvo ONM fOri-totoinTi- ^ON^-O mo too mo O t^ThMVO > a O M N fO Tl- irivo t^OO ON O M N c*"> <* tovO t^OO ON O M CM m -4- mvo t-^oo ON ^s no APPENDIX. ON t~^ WNNNN> ON (N 10 10 iO\O t% ON M vO O 10 H oo H Tj-t-~.o r)-r^H -^-oo 00 VO rh T}- Tj-vO 00 H 10 H HI lOONCOtv.H IOO -^"ON vo vo vo t^ t>.OO OO ON ON ON 00 |ff llfflgf N co ^vo ON N tx N oo oo H Tt-r^o cor^o -^-t^ ^NOOOHrJ-t>,MVO HI lOONCOt^H lOON -^-OO vo vo vo C- t^oo 00 00 ON ON tx ^S-vo^^S-^o^ft OOOOHHHHNN Os ON O H co 10 ON cooo TJ- t^ O <* t> O covo O co t^ N COCOCOrJ--^-Tj-iOioiO H 00 tN.VO VO t^ ON N VO N ; H tj-CO (N VO O * ON COOO VO vo VO tx t^OO OO OO ON ON VO rt-00 N VO 0) OO -<1-H ON t> gf "1 2 a?? S3" VO VO tv.OO O N vO O tO O t^ O covo O covD O co i^ (N COCOCO-^-rJ-rflOiOlO t>. ^ CO N N CO IOVO N t^ O *3-ca N vo O -^-oo co f^ VO VO VO tx t^OO OO OO ON ON VO N vo ON * ON vo H ONVO 10 8 So"? 2 7^ 5 3" r> O COVO* ON IN VO ON COvo" N COCOCOCO^--<1-T)-1010 CO HI ONOO 00 ON H rt-00 CO O rj- r>. M to ON rj-oo SN t-> "* ?CO tx H fx N ONVO CO t^OO OO O\ ON CO O CO 10 t^ O CO 1000 H CO U 8" co co co 5- I?^ fo^ ONCOt^H lOONCOt-.HVO HHHHHHHMMH H H H H M H N 1000 NVO t^COOOOVO O N IO t^ O t^oo oo ON ON PJvo O ^-ONIOHOO toco 8(N lOt^ONCS XOt^O CO OOOOHWHWN W M H N -^-vO O^ CO t-N M ONVO CO N N N rj-vo O "* , OO N vO O rj-oo 0) vO HI vo vovo vo ^ t^. t^oO oo ON ON O O COt^NVO NOOION O O N rJ-t^ONO) >4-t^O CO O\00 00 ON O CO 10 ON ^ ON 1OOO H ThOO M Tj- t^ M <* N (N COCOCOTj-Tj-Tj-tOiO VON OOOOOOOONVOO O H N co Th io>O t^oo ON O H N CO Tj- IOVO tvOO ON I HC,CO^vOvOt,OOON j . Ppp op pop p APPENDIX. Ill 0000 vO - NNNNNNNNNN rrrp" co co roco co co co co.. ro HWHMNNNNNN ,...., NNNNNNrOCOCOCO ON tx. . ro ro rt-vo ON ro OsvO rooo rooo rooo ro ON . O OOHHNNCOrOrt-lO . . 1000 N tx, . CO CO 10 IIIlHlltl 00 N O oo oo oo O rooo ro O O C Q H H" N 'N ro'ro . 10 NNNNNNNNNN ONOO ONN 10 H 00 VO vo CO iovo vo" tx.00 OcT ON O H NNNNNNNCOCOCO H VO CO H N .OO . CO CO N ON t>* 10 CO H ONOO tx,vO N N CO * IOVO vo fx,oO ON oo to ro ro co 1000 N oo 10 10 H t^ ro ONVO ro o>vo . .ON.OCO.VOOVO.. NNNNNNNNNN vo rssss&5n * O VO N ON iO N ONVO fO iovo vo t^ t-^OO ON ON O w NNNNNNNNrOCO O OO IO CO H ONOO VO IO . N N CO . IO IOVO tx-OO ON 10 NNNNNNC?NNN H O H COVO H t^ IO lOVO ^-OVO NOO lOHOO ION iovo vo t^ t^oo ON ON O H NNNNNNNNrOCO ON co O oo oo ON rooo vo vo H N co . 10 iovo tx,oo ON - . H ONOO oo ON N vo M tx, OOwnNNroro.. Tj- iovo O <*- H ONOO ON 10 IOVO t^^OO ON ON O H NvONOONH.O twO ONVO . N ONOO vo 10 ro N ro ONVO . ro ro . tx. O ION O>00 ON O COOO . N H N NOO . M t->roO ^.ThH N'N'N c^cTN f?N roco rj-oo . N M rovo H OO tx oo ioroi-1 ONtx.io.N H H N ro . . iovo tx,oo ON N . M ONOO 00 ON H 10 O vo CON N .txH ^lO.lO N oo M- O vo ro ONVO ro O O iovo t-x t~.oo oo ON O w NNNNNNNNrOrO tx, H tx, . ro 1000 ro ONOO H O vo . ro ro .vo O . O OOVOVOOO H 1OMOO t>.bx Mtx.ro ONVO N ON 10 N ON NNNNNNNNCOrO O CO ONVO iovo ON . H .ON H N ro ro . iovo tx.oo oo IO N ONOO OO ON M . ON IO ON .00 rOOO CO ON . ON IO NOON rJ-00 . H O O 10 iovo vo tx,oo oo ON O O NNNNNNNNrOCO N vO H OO tvOO H 10 N O vo cowoovo .rOH O ON H N ro ro . iovo rx.oo oo I O w N CO . iovo tx,oO ON O H N ro -.co O ro looo (N 10 ON ro O H ro <* iovo oo ON O CM CO H CO ONOO H OO ON -<1- CM co iovo oo ON H ro -*J-vo oo ^ oo H t-s iovo oo ro H H ro O H CM ro rt- iovo r>.CO ON oo vo t^. O vo loco co H ro IO t^ ON CM -,Tt-(N ON H O COOO vo 10 tx H oo fx ON ro O O cooo vo oo CM ON O CM ^J-vo OO O rovo O rooo ^- H CM vo -^-vo H O co O CM t^ CM t-^ ro ONVO ro O oo H Tj- ONVO XO h> H t-xVO 00 OO t^vo vo vo vO t-^ t>.oo ON H CM * t^ ON (N IOOO N vo O H CM ro TJ-VO t^.oo O H O t^ t^ H 00 ON -* COVO PJ H IOOVO H t-~,Tt-HtOOVO f OM.CO^-lOVO^OOON O H N CO * IOVO t-sOO ON O H CM ro <* IOVO tv.OO ON, 2 APPENDIX. ON 00 B- & ss- 2 00 vo ON ON O 00 O 00 ^ co ro J J j* 10 10 m VD vo VO H H 0) M M n ro ro ro rf m ON ON O H H M H -* tx 00 co CO 00 CO CO ON. ON ON CO VO vo vo vO VO vo vO - * t - H " I c, . . ON 11 N 1 1 H i 1 ON ON ON 1 00 M ON 8 s N 00 00 1 J. (M 00 ; ON 1 ON 1 I - II ro s 1 S & ON ON i ON ro vo CO OO i 1 ro oo I ON VO Ov S ON ! m m N CO N O N 00 OO oo vo 00 co oo vo & I M ON i 1 - s J 00 OO 3 1 1 1 I I CO I CO oo n g 00 1 i Th 0? ro ON ON VO i i . II 1 00 >n oo ^ C-s oo 1 vo ON VO ro ON s ON 00 ON M II co co oo 00 CO 0? 5 m S 1 1 CO 00 1 S 00 m CO ON 00 ON R ON I 1 i ft * ON CO ON ? ON * .s ON .81 A KEY TO THE SERIES OF CHEMICAL PROBLEMS I 2 KEY TO CHEMICAL PROBLEMS. METRIC SYSTEM OF WEIGHTS AND MEASURES. 1. 437 millimetres : 4*37 centimetres. 2. 109876542-1 centimetres: 1 0987654-2 1 decimetres : 1098765*421 metres. 3. 70 square decimetres : 7,000 square centimetres : 700,000 square millimetres. 4. The weight of a cubic centimetre of pure water weighed at 4 C. in the latitude of Paris, is styled a gram. A litre denotes a volume equivalent to I cubic decimetre. 1,725 grams. 5. 256,700 centigrams. 6. 5,000 milligrams. 7. 78*54 square centimetres. 8. 1*60931 kilometres. 9. 761*985 millimetres. 10. 567*875 cubic centimetres. 11. ii 3*097 cubic centimetres. Il8 KEY TO CHEMICAL PROBLEMS. 12. 12,732,394 metres 13. 14/826 millimetres. 14. 7 pieces and 0*023 gram will remain. , 15. (a) 0-00083 gram, (&) 0-00148 gram, (V) 0*00231 gram, (d) 0*00333 gram, (e) 0*00593 gram, (/) 0*00926 gram. 1 6. 376 millimetres. 17- 8*305 centimetres. 1 8. I *4 millimetres. 0*25 gram. 19. i-ii cubic centimetres. 20. 50*10 cubic centimetres. 21. 0*03 C.C., + 0*20 C.C., -h 0*41 C.C. CONVERSION OF THERMOMETRIC SCALES. 22. I5'5C ioo'o C. i77C. o-o C. ~26'38C. 2 6o-o C. -I2 C *5 C. 3i c '87 C. -if -6 C. 23. i2-5 R. i2*44 R o-o R. 519-! R - 32 c *o R. I92'o R. 0'20 R. - I4*22 R. -7'55R '24. 39'2 F. - 59'9 F. I4o'o F. 59 -o F. 32'i8 F. 2I2'0 F. 4i3'6F. 842'o F. 33'8 F. 25. 59'28 C. 26. 9 4-82 F., i73-i2 F., 640^4 F. 27. -3i'04 R-, -37-84 F. ; 285'8 R., 675^05 F. 28. 2822 F., 793'4 F. 6 3 3-2 F., 455 F. 29. -40., -32R KEY TO CHEMICAL PROBLEMS. 1 19 CORRECTION OF THERMOMETER READINGS. So- 57 '4. 31. (a) i5o76 to I5i'56, (6) I5i'56 to I52*i7, (c) I52 c *i7 to I52*38, 0) I52*38 to iS278. 32. i3i'54to I3i'64; I3i*64 to I3i74- 33. 2i78, 58-! i, 76*22, io 7 '32, i27 - 37 . CORRECTION OF BAROMETER READINGS. 34. 752-82 mm., 754*20 mm., 755'3 8 mm -> 75 6>2 5 mm - 35. 753-29 mm., 752-69 mm., 750-57 mm., 749-97 mm. 36. A. 752-63 mm. E. 755*49 mm. B. 76072 mm. F. 758*87 mm. C. 755*58 mm. G. 762*30 mm. D. 76175 mm. CORRECTION OF THE VOLUMES OF GASES FOR TEMPERATURE AND PRESSURE; LAW OF PARTIAL PRESSURES. 37. 163-992 volumes. 38. 1*3663 litres. 39. 13-3746 litres. 40. (a) 1585-165 cubic centimetres, (b) 1774-725 cubic centimetres, (c) 2049-45 cubic centimetres, (d) 3148-35 cubic centimetres. 91 C. 41. 0-002036 for i F., 0*004581 for i R 42. From 760 mm. to 76777 mm. 43. 1743*4 cubic metres. 120 KEY TO CHEMICAL PROBLEMS. 44. 143-5 1 8 c.c. 45. 3677 c. 46. The volume reduced to o C. and 760 mm. of the added hydrogen will be 24*10 c.c. The reading of the upper level of Hg = 423'! mm. 47. I '68 per cent. 48. I '66 per cent., and 0*85 per cent. 49. 4879 c.c. 50. 24*06 c.c. RELATIVE DENSITY OF SOLIDS, LIQUIDS, AND GASES ; VAPOUR DENSITY. 51- 6-503. 52. 19*11. 53. Granite 2*72 ; marble 2*68 ; haematite 5*07. 54. 0*8793. 55. 0*268 mm. 56. 0*49 mm. 57. 0*0000219 mm. 58. Relative density of alloy is 8-516. Percentage of copper = 74-7. tin = 25*3. 59. 3 cubic centimetres. 60. 1-58. 61. Quartz 2*65 ; heavy-spar 4*48 ; calcite 2*72 ; iron pyrites 5*0. 62. 1*0272. 63. A. 1*00585 ; B. 1*841 ; C. 1*920; D. 0792. KEY TO CHEMICAL PROBLEMS. 121 64. Sulphuric acid 1*84 ; mercury 13*54. 65. 0-8. 66. Metal 27, liquid 1*19. 67. (a) 1*26; (^0-915; Mo-87. 68. Glycerine 1*26 ; mercury 13*598. 69. 0*97176. 70. 0-06935. 71. 59*56. 72. 1-177 ; 16*99. 73. (Air = i). 1*773; 1712; r578 ;.i-377 5 I*32I; I-270; 1*115; 1*021. (H = i). 25-59 ; 24-71 ; 22-77 ; 19*87 ; 19-06; 18-33; 16-10; 1473. 74. 47-29. 75. (i). 0-0792; 1-1205; 1*5292. (2). 0-0770; 1-1199; I*54I4. 76. I. II. III. (A). 7*67 77i. 7*69. (B). 1107 111-29. uro. 77. (A). 49*60 ; (B). 44-70 ; (C). 36-07 ; (D). 34*95 ; () 35*54. 78. 50*29. 4-494 grams. 79. I. 575 ; II. 574. 80. (Air = i). A. 4-036 ; B. 2*621 ; C. o'6oi2 ; D. 8-665. (H = i). A. 58-26 ; B. 37*83; C. 8-678 ; D. 125-1 81. 347*96 kilos. 82. 5177. 83. 76*33. 84. I. 2-21 ; II. 2-30; III. 2-20. 85. 1377 (H-= i). KEY TO CHEMICAL PROBLEMS. DENSITY AND MOLECULAR WEIGHT; VALENCY, EQUIVALENTS, AND ATOMIC WEIGHTS. 86. SH 2 , 34-38 ; NH 3 , 17-23 ; N 2 O, 43-88 ; CH 4 , 16*02 ; CHC1 3 , 121-25 ; SnQ 4 , 265*6. 87. Oxygen 1*4297 gram. Chlorine 3-1666 Iodine n*3355 Sulphur 2-8648 Phosphorus .... 5*546 Nitrogen 1*255 Sodium 2*0598 Arsenic 13*419 Mercury 8*949 88. Carbon monoxide . . 799*3 litres. Hydrogen sulphide . 657*0 Marsh gas .... 1397*7 Water 1243-1 Ethylene 799* i Carbon oxysulphide . 372*6 Bromine I39'9 Hydrochloric acid. . 613*8 89. PbCl 2 , 277*13 gm ; Si F 4 , 104*24 gm ; Fe C1 3 , 161-99 ; CH 3 Br, 94-73 . 90. CO, 0-967 ; CS 2 , 2-630 ; SO 3 , 2766 ; BF 3 , 2-358 ; PF 5 , 4*373- 91. O, 15-96 ; Ag, 107-66 ; Cl, 35*37 ; Sb, 119-58 ; Cu, 63-18. 92. The most probable value for the atomic weight of oxygen is 15*96, for this number represents the smallest quantity by weight of oxygen occurring in the molecular weight of any of the given compounds. 93. The approximate molecular weights, calculated from the relative density in each case, are : 16*0 ; 141*0 ; 121*3 ; 27*96 ; 76*4, hence the parts given in the third column KEY TO CHEMICAL PROBLEMS. 123 refer to parts by weight per molecule and therefore the atomic weight of carbon is 1 1 '97. 94. Cd, 1137 ; P, 125*6 ; Hg, 201-5. 95. Atomic weight of phosphorus 30*96. From mole- cular weight found above, molecule is P 4 . DEDUCTION OF EMPIRICAL FORMULA FROM PERCENTAGE COMPOSITION, FORMULA OF MINERALS. 96. H 2 O 2 . 97. NO 2 . 98. Fe 2 O 3 . 99. COS. loo. KHSO 4 . 101. MgSO 4 + 7H 2 O. 102. ZnSO 4 + 7H 2 O. 103. Ca 3 P 2 O 8 . 104. Na 3 AlF 6 105. KNO 2 . 106. A1 2 (SO 4 ) 3 . 107. CuCO 3 + Cu H 2 O 2 . 108. NiSO 4 2NH 3 . 109. C n H 2n . 1 10. C 2 H 2 . in. C 2 H 6 SO 4 . 112. C 10 H 14 N 2 . 113. (C 21 H 22 N 2 2 ) 2 H 2 PtCl 6 . 114. A comparison of the amount of oxygen contained in the various bases with that contained in the acid, shows that the two quantities are in the ratio of I : 2, in con- formity with the formula MOCO 2 thus Containing oxygen. Ratio. CaO .... 28-4 8*1 1] MgO .... 12-3 4-92 1 6 FeO .... 12-3 273( i MnO .... i -9 0-43] CO 2 .... 44*4 32-3 2 99'3 115. 2(MgO Si0 2 ) + 3H 2 0. 116. Co 3 As 2 O 8 + 8H 2 O. 117. Na 2 O A1 2 O 3 6SiO 2 . 118. 3CaO, Na 2 O, 4A1 2 O 3 , i2SiO 2 . 124 KEY TO CHEMICAL PROBLEMS. TO CALCULATE THE PERCENTAGE COM- POSITION OF A COMPOUND FROM ITS FORMULA. 119. 120. Hydrogen . Oxygen . . Potassium . Chlorine . Oxygen. . 11*14 . 88-86 125. Calcium Carbon . Oxygen . 126. Silver . Chlorine. . . 40*00 . . I2'OI . . 47'99 100*00 1 00*00 . 31*92 . 28-93 39*15 . . 75-27 24-73 T r\f\'nr\ 121. 122. Mercury. Oxygen . . Potassium . Nitrogen . Oxygen. 1UVJ UU 127. Magnesium Phosphorus Oxygen . . 128. Potassium . Platinum . Chlorine . . 21*61 27*95 50*44 . 92-57 . 7-43 100*00 lOO'OO 38'67 . I3-88 47*45 . 16*11 . 40*10 43*79 100*00 100*00 123. Sodium . . Nitrogen . Oxygen . . 124. Barium . . Sulphur . . Oxygen . ' . . 27*09 . 16*50 . 56*41 129. Sodium . . Sulphur . . Oxygen . . Water . . Or Sodium . . Sulphur . . Hydrogen . Oxygen . . . 18*57 25-83 19*33 . 36-27 lOO'OO 100*00 . 58*82 13*74 . 27*44 . 18-57 . 25*83 . 4*04 . 51*56 100*00 KEY TO CHEMICAL PROBLEMS. I2 5 130. Iron . . Oxygen . 131. Manganese. Oxygen . . 72*42 136. Lithia . 27*58 Alumina. . Silica. . 100*00 100*00 6*44 29*21 IOO*OO 72*03 137. Lime. . . . 44*42 27*97 Silicon dioxide 23*82 Titanium 31*76 100*00 132. Copper . Iron . . Sulphur . 133. Silver . Copper . Sulphur . 34^2 34'95 100*00 53*07 3I-I5 15*88 100*00 138. Lead. . . . 76*34 Phosphorus . 6*87 Chlorine. . . 2*62 Oxygen . . . 14*17 139. Carbon . . Hydrogen . Oxygen. 100*00 52*16 13*07 34-77 100*00 134. Lime (CaO) . 9*20 Alumina (A1 2 O 3 ) 16*79 Silica (SiO 2 ) . 59*21 Water . . . 14*80 100*00 140. Carbon . . Hydrogen . Oxygen . . 42*10 6*45 51-45 100*00 135. Lime. . Alumina. Silica. . 43*49 17*64 38-87 100*00 141. Carbon . . Hydrogen . 90*54 9*46 100*00 126 KEY TO CHEMICAL PROBLEMS. 142. Carbon . Hydrogen 143. Silicon . . Carbon . . Hydrogen . 89*53 10-47 lOO'OO 146. Carbon . Hydrogen Nitrogen Oxygen . Water . 19-48 147. Carbon . 66 '6 1 Hydrogen 13*91 Nitrogen Oxygen . 32*18 6*05 28-25 21-45 12*07 lOO'OO 13-20 5-01 lOO'OO 144. Carbon . . . 21*02 148. Carbon . . . 75*41 Hydrogen . . 2*63 Hydrogen . . 6*60 Potassium . . 34*26 Nitrogen . . 8-42 Oxygen . . . 14*01 Oxygen ... 9-57 Sulphur. . . 28-08 lOO'OO 1 00 -00 145. Potassium . . 42*44 Iron .... 15*19 Carbon . . . 19*52 Nitrogen . . 22*85 100*00 KEY TO CHEMICAL PROBLEMS. 127 CALCULATIONS OF THE AMOUNT OF MA- TERIAL REQUIRED TO PRODUCE A GIVEN WEIGHT OF ANY SUBSTANCE, OR OF THE QUANTITY OF THE SUB- STANCE PRODUCED BY THE DECOM- POSITION OF A KNOWN WEIGHT OF MATERIAL. 149. 2-554. Ibs. 150. (a) 27*038 kilos., (b) 5-1088 kilos., (c) 16*307 kilos., (d) 12*208 kilos. 151- 33*275 grams ; 13*029 grams. 152. 709*84 grams. I 53- 4306*0856 kilos, zinc ; 6492*3134 kilos. H 2 SO 4 . T 54- 37o8*8 kilos, iron ; 4306*1 kilos, zinc ; 32461*5 kilos, sulphuric acid. I 5S- 177*256 grams. 156. 12,939*8 c.c. hydrogen ; 6469*9 c.c. oxygen. 157. 322*216 grams Fe 3 O 4 ; 132*029 litres of hydrogen. 158. 0*91048 Ibs. copper ; 0*1784 Ibs. phosphorus. 1 59. 6*096 grams ammonia ; 9*506 grams chlorine. 160. 140*1 kilos. KNO 3 ; 117*8 kilos. NaNO 3 . 161. 10*116 Ibs. nitre ; 9*805 Ibs. sulphuric acid. 162. 9*08 litres. 163. 4*246 grams. 164. 5*477 kilos, lime ; 10*446 kilos, sal-ammoniac. 165. 183*03 kilos, of coal. 128 KEY TO CHEMICAL PROBLEMS. 1 66. Assuming air to contain 23 per cent, by weight of oxygen, 1 1 "94 tons. 167. 42*36 grams marble ; 140*42 grams hydrochloric acid solution. 1 68. 146*215 tons carbon; 3275 cubic feet. 169. (i) 130*53 litres ; (2) 130*53 litres. 170. 40*223 grams oxalic acid ; 20*559 grams formic acid ; 27*444 grams potassium ferrocyanide. 171. 36*67 grams sodium acetate. 172. 707*333 kilograms. 173. 76*395 kilos, air ; 10*73 kilos, water ; 12*353 kilos carbon dioxide. 174. 106*99 grams silver chloride ; the solution will contain 17*79 per cent, of hydrocyanic acid. 175. 43*647 grams. 176. 16*501 tons of salt; 20*778 tons of manganese dioxide ; 47*683 tons of sulphuric acid. 177. 51*294 tons of hydrochloric acid ; 2,558,701 litres will escape. 178. 1*7355 grams precipitated ; 1*907 grams in solution. 179. 192*33 tons of {Ca(OCl)(OH)] 2 , CaCl 2 . 1 80. According to the equation 3C1 2 + 6KOH = KC1O 3 + 5KC1 + 3H 2 O, 28*201 tons of KC1O 3 . 181. i8no grams. 182. 72*135 grams iodine ; 100*815 grams chlorine. 183. 8*49 grams copper ; 26*29 grams sulphuric acid. 1 84. 54*487 tons of pyrites ; if 3 per cent, of the the- retical yield of sulphur remained the dilute acid would be produced. oretical yield of sulphur remained unburnt, 97 tons of ld KEY TO CHEMICAL PROBLEMS. 1 29 185. 224*28 litres. 1 86. 3*935 grams ferrous sulphide ; 9*325 grams air. 187. 1*802 litres. 1 88. 2*0488 grams microcosmic salt. 189. 82*19 tons of sulphur; 68 tons of salt cake; 107742 tons of bicarbonate ; 183*398 tons of soda crystals ; 56*87 tons of solid caustic. 190. 17*474 tons of limestone. 191. 138*14 grams Fe 3 O 4 ; 289*889 grams Fe 2 Cl , COMBINATION AND DECOMPOSITION OF GASEOUS BODIES. 192. 20 litres of hydrochloric acid gas ; 10 litres of hydrogen remain in excess. 193. I cubic foot of hydrobromic acid gas. 194. 50 vols. of chlorine ; 25 vols. of oxygen. 195. 154*97 c.c. 196. 3 litres of oxygen. 197. iio*34c.c. 198. 500 c.c. nitrogen ; 1500 c.c. hydrogen. 199. V. 1*25. n vols. oxygen ; V.n vols. carbon dioxide. 200. (i) 95*42 litres ; (2) 143*13 litres ; (3) 119-27 litres, 201. 5 litres phosgene gas ; 5 litres carbon dioxide, weighing 9*8296 grams, and 10 litres hydrochloric acid gas, weighing 16*2903 grams. 202. i litre, on the assumption that its density is normal. 130 KEY TO CHEMICAL PROBLEMS. GAS ANALYSIS CALCULATIONS. 203. N, 79*060 per cent. ; O, 20*940 per cent. 204. N, 83*37 per cent. ; O, 13*80 per cent. ; CO 2 , 2*83 per cent. 205. 3*48 per cent, of aqueous vapour. 206. 26*46 vols. of oxygen unite with 52*71 vols. of hydrogen. 207. C (hypothetical vapour volume), 1*025 ; H, 3*129 ; O, 0*484. The empirical formula of methyl ether is C 2 H 6 O. 208. N 2 O. 209. N, 81*81 per cent. ; O, 14*21 per cent. ; CO 2 , 2*44 per cent. ; SO 2 , 1*54 per cent. ; H, 45*07 per cent. ; SH 2 , 24*25 per cent. 210. CO 2 , 29-96 per cent. ; N, 0*72 per cent. ; CO & CH 4 , none. 211. The percentages of N are : A, 7*39 ; B, 16*28 ; C, i6'i6 ; D, 15*00 ; ", 13*99 5 F, 4'4& The volumes of gases obtained, reduced to o C. & 760 mm. are : i A. B. C. D. E. F. N. 7*21 c.c. 10-34 c.c. io'c>7 c.c. 9-96 c.c. 9 '66 c.c. 4-48 c.c. NO. 0'47 o'9i ,, 1-16 o'SS 0'25 o'3 j> 212. Total oxygen added = 239*3, f which 236*98 is used. Carbon dioxide formed = 118*8 = oxygen used in burning C. Hence O used to burn H = 118*18. * . Vol. of H would be 236*36 and hypothetical vol. of C. '-! - 59-4- Hence composition of marsh gas is represented by CH 4 . KEY TO CHEMICAL PROBLEMS. 21 3- *53 P er cent, by volume. 214. 0*47 per cent, by volume. 215. The percentage composition of this sample of coal-gas is : CO 2 . 1-41 ; CO. 6-15; C 2 H 4 &c. 3-84 ; H. 4773 ; C 6 H C . 1-04; CH 4 . 35'63; O. 0*30 ; N. 3*90. 216. 3 vols. ethylene ; 1*98 vols. butylene. Equal in illuminating power to a gas containing 6*96 per cent, of ethylene. ATOMIC WEIGHT DETERMINATIONS. 217. 1-00108. 218. From synthesis of silver nitrate 13*967 By precipitation with potassium chloride 14*026 By precipitation with ammonium chloride 13*945 219- 35'356. 220. 39*33- 221. 199*6. 222. (i) 12*014, (2) 12*014, (3) 12*030. Mean 223. 27*029. 224. 27*028. 225. 27*012. 226. 48*06 ; 48*07 ; 47*98 ; 48*04 ; 48*04. 227. 196*88; 196*90; 196*86: 196-85; 196*89 ; 196*86 ; 196*85 ; 196*89. 228. 28*347 ; 28*303 ; 28-347 ; 28*352 ; 28*243. 229. 52*061. 230. 65*506. Mean 13*979 12*019. K 2 132 KEY TO CHEMICAL PROBLEMS. INDIRECT ANALYSIS. 231- 33*51- 232. 93-03. 233- 39*98. 234. 11*61. 235. 25-51. 236. 58-035 grams. 237. K, 13-75 per cent. ; Na, 8*25 per cent. 238. 0*6281 gram NaCl ; and 0*3049 gram Na I. GENERAL ANALYTICAL QUESTIONS. 239. (a) 0*1998 gram of the salt gave 0*4865 gram silver chloride, and 0*0032 reduced silver. 0*0032 gm. Ag = 0*00425 Ag Cl ; 0*4865 + 0*00425 = 0*49075 gm. Ag Cl ; 0-49075 X 3 5 '37 X ioo - = 60-74 per cent, chlorine. 143-03 X 0-1998 (b) 0*9543 gram of the salt gave 1*1584 gram sodium sulphate. 1-1584 X 45-99 M I4r8l 0-37568 X ioo and = 39*36 per cent, sodium. '9543 =17; = 1 7 ; hence the chlorine and sodium 35'37 2 3 in the compound are in the ratio of atom to atom, and the formula of the salt is Na Cl. KEY TO CHEMICAL PROBLEMS. 133 240. 0*3951 gram of substance gave 0*6544 gram of NH 4 MgAsO 4 , H 2 O = 0-3781 gram As ? O 6 : hence the substance contained 9570 per cent, arsenic anhydride. 241. The percentage results of the analysis are : Silica 40*35 Ferric oxide 36*43 Water 20*62 Lime 2*68 100*08 Neglecting the lime as an unessential constituent, the percentage composition of the mineral becomes : Silica 41*46 Ferric oxide . . ... . . 37*43 Water 21*19 100*08 The simplest ratios of the oxygen contained in each of the constituents is 6 : 3 : 5. Hence the formula of the compound is 3 SiO 2 , F 2 O 3 , 5 H 2 O. 242. 164*1 grams of K 2 Pt C1 6 are equivalent to 50*39 grams of potassium chloride. Hence the composition of the mixture is : Potassium chloride 50*39 Sodium chloride 49 *6 1 100*00 243. 0*2476 gram T1 2 Pt C1 R = 0*1239 grams thallium, or 67*67 per cent. Its composition is therefore : Tl 67*67 cio 4 32-33 100*00 134 KEY TO CHEMICAL PROBLEMS. 6767 32*33 2037 ~ 332 ' 99^1 Hence the formula is Tl Cl O 4 . 244. Centesimal hydrochloric acid corresponds to 1*0766 grams of Ag per litre, hence Ag used = 7*383 - 0*0032298 j 7'37977 X 35'37 X ioo grams = 7*37977 grams ; and Ior 66 x 3-949 61*40 per cent, chlorine. 245. 2 BaSO 4 = O. AgBr = Br. 0*402 gram BaSO 4 = 0*01378 O. 0*324 gram Ag Br = 0*1372 Br. 0*01378 0*1372 7- = 0-00086. ,- = 0*00172. 15*96 7970 Hence the formula is Br 2 O. 246. KC1O 4 = HC1O 4 . 1*6785 KC1O 4 = 1*2167 HC1O 4 or 99*85 per cent. HC1O 4 . 0*966 gm. lost 0*444 g m - O, or 45*97 per cent. : KC1O 4 ought to lose 46*18 per cent. 0*966 0*444 = 0*522 KC1 ; this required 0*744 gram of silver for precipitation : theory requires 0*7523 gram. 247. Na H CO 3 + CH 3 . COOH = CH 3 . COONa + CO 2 -f H 2 O or 43*89 parts of CO 2 = 59*86 parts of CH 3 . COOH. Then 0*427 CO 2 = 0*5823 CH 3 . COOH, , 0*5823 X ioo and g- = 39*26 per cent. 248. NaOH = HCOOH. Then 1*01558 NaOH - 1*1664 X ioo 1-1664 HCOOH and = 54*967 per cent. 249. 2*921 vols. CO 2 , in 10,000 vols. of air. KEY TO CHEMICAL PROBLEMS. 135 250. Sodium chloride . . . . . 26*402. Potassium chloride . . . 07459. Magnesium chloride .... 3*0368. Magnesium bromide .... 0*0705. Magnesium sulphate .... 2*0648. Calcium sulphate 1 '2896. Calcium carbonate . $ * k , 0*0475. Ammonium chloride . . ; . 0*00035. Magnesium nitrate . o'- 0*00185. Ferrous carbonate . . . . 0*00674. MgO unaccounted for . . ' . 0*02904, 251. 22*53 candle power. 252. (i) C 6 H 4 Br C 2 H 6 + Na 2 + CO 2 = C 6 H 4 C 2 H 5 COO Na + NaBr. (2) C 6 H 4 C 2 H 5 COO Na + HC1 = C 6 - H 4 C 2 H 5 COOH + NaCL and the acid gave on treating with baryta water the barium compound, of which 0*5637 gram lost 0*0430 gram water, or 7*63 per cent. ; and gave 0*2733 gram barium sulphate, or 28*517 per cent. Ba. Hence the composition of the salt is : C 9 H 9 O 2 . .:-.).. . 63-85 Ba. / . . ..-. . 28*52 H 2 . 7-63 100*00 63*85 28*52 7-63 ' ' ' 17*96 Hence formula is Ba (C 9 H 9 O 2 ) 2 , 2 H 2 O. 253. i'994 per cent, nitrogen. 254. The composition of the body, as determined by experiment, is : Carbon ...... 36*43 Hydrogen ..... 7-91 Chlorine ...... 55'66 136 KEY TO CHEMICAL PROBLEMS. The simplest relation between these numbers is repre- sented by C 2 H 5 Cl, which formula requires : Carbon 37*22 Hydrogen 777 Chlorine 55*01 The vapour density found is 32*44, calculated 32*15. 255. NaCl. 66*39 P er cent - KCI. 33*61 per cent. 256. NaCl. 60*3 per cent. NaBr. 397 per cent. SOLUTION OF GASES. 257. 20 c.c. ; 14*96 c.c. ; 7*82 c.c. ; 0*286 c.c. 258. The volume is the same in each case, 15*90 c.c. The weights are respectively : 0*025124 gram ; 0*018792 gram ; 0*0098237 gram ; 0*0003593 gram. 259. (A) (S) (O (a) 6*664 c.c. ; 6*553 c.c. ; 6*454 c.c. (b) 6*344 c.c. ; 6*291 c.c. ; 5*832 c.c, 260. 1*4584; 1*2607; 1*0385; 0*9610 ; 0*9134 ; 0*8852. 261. 0*06910 ; 0*06853 ; 0*06769 ; 0*06732 ; 0*06669. Water: 0*02134; 0*02099; 0*01956; O'oi9i2 ; 0*01839. 262. 0*032874 - 0*00081632 / + 0*000016421 t-. 263. 39*36 per cent. O. and 60*64 P er cent. N. 264. By volume : 44*7 per cent. CO 2 ; 55*3 per cent. N. 265. 92*07 per cent. H. ; 7*93 per cent. CO 2 . KEY TO CHEMICAL PROBLEMS. 137 MOLECULAR WEIGHT AND LOWERING OF THE FREEZING-POINT. 266. I5'34C. 267. 73-44. 268. 0-5256. 269. i '69 per cent. 270. 2-867; 2-488; 1-482. 271. (i) 72-58 ; (2) 73-50; (3) 71-97 ; (4) 73-60. 272. (a) 111-87 ; (b) for Na, 14-49 5 W for Sn, 73*93- 273. MgS0 4 , 7 H 2 0. O"O2. (28O"Q) 2 274. L - ^ r f- = 13-38. 18-5 277 38-6 275- ~fy = 2 9 5 -^J = 44 ; -^ = 61 ; the numbers here obtained according to Raoult should represent the molecular weight of the solvent in each case ; in the cases of formic and acetic acids this holds very nearly but in the case of water 29 bears no simple relation to 18 the molecular weight of water in the gaseous state, the con- clusion may therefore be drawn that the molecule of water in the liquid state is more complex than the molecules of formic and acetic acids under similar conditions. 276. From (a) the molecular weight deduced is 65*80, from (b) 65-99 > f r N 2 O 4 molecular weight is 92, for NO 2 it is 46 ; hence it is probable that at the temperature of the experiment nitrogen tetroxide consists of a mixture of N 2 O 4 and NO 2 . 277. (a) 196-2 ; (b) 196-5. 138 KEY TO CHEMICAL PROBLEMS. SPECIFIC HEAT, LATENT HEAT, AND ATOMIC HEAT. 279. 1 8 C. 280. 0*0327. 281. 0-03308. 282. 0*0335. 283. 0*208. 284. 3*o8. 285. 893 C. 286. 0*1137. 287. 774*6 for iF. 288. 33681 metres. 289. Sp. Ht. of Indium is 0*0574. 290. (A) 0*0559 ; (B) 0*0935 ; (C) 0*0495 ; (D) 0*0548 ; (E) 0*1712. 291. 28*4 C. 292. 7*284 kilos, per hour. 293- 537*2. 294. Pb 6*48 ; Ag 6-13 ; Cu 6*01 ; Fe 6*36 ; S 6-48 ; P 5*84. 295. Approximate atomic weights : Ag, 112*3 ; Zn, 66*98 ; Bi, 207*5 J Sn, 113*8 ; Fe, 56*23. Exact atomic weights : Ag, 107*66. Zn 64*88 ; Bi, 207*5 ' Sn, 117*35 ' Fe, 55*88. 296. 5*67 = atomic heat of solid chlorine. 2*98 and 3*06 are the atomic heats for O calcu- lated from the respective oxides. 297- 3'8. 298. 2. 299. Ca 5-6 ; Sr, 6*3 ; Ba 6*4. KEY TO CHEMICAL PROBLEMS. 139 HEAT OF SOLUTION AND COMBINATION. 300. 34*462 kilos. 301. 5747 C. 302. 0*148 kilos. 303. (i) ethylene 2714-5 ; (2) methane 2381*3. 304. 2677-6. 305. 7,614 thermal units. 306. 8095-6. 307. 3386-4 metres. 308. 192,700 -f . 309. 20,200 +. 310. 28,400 -. 311. 18,500 - . 312. 13,080 -. 313. (a) 55,ioo + ; 45,200 - 314. 5,590 -. 315. 21,150+- 316. It may be concluded that H 3 PO 3 is a dibasic acid, for the addition of sodic hydrate beyond two equivalents causes but a very slight liberation of heat, and from nothing to two equivalents the heat evolved is considerable and proportional (or nearly so) to the amount of sodic hydrate added. THE END. RICHARD CLAY AND SONS, LIMITED, LONDON AND BUNGAY. 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