t/CSB LIBRARY

 
 NOTES 
 
 CHEMICAL LECTURES 
 
 SECOND-YEAR STUDENTS 
 
 MEDICAL DEPARTMENT OF THE UNIVERSITY 
 OF PENNSYLVANIA. 
 
 PUBLISHED BY AUTHORITY OF PROF. THEO. G. WORMLEY. 
 
 BY 
 
 JOHN MARSHALL. 
 
 PHILADELPHIA: 
 
 AVIL PRINTING COMPANY. 
 
 1894.
 
 Copyright, 1894, by JOHN MARSHALL.
 
 THE following notes, by Dr. Marshall, of my Lectures on 
 Chemistry, for students, have been published with my consent and 
 authority. 
 
 THEODORE G. WORMLEY.
 
 NOTES 
 
 ON 
 
 CHEMICAL LECTURES. 
 
 ORGANIC CHEMISTRY. 
 
 THE production of organic compounds was supposed to be 
 due to the influence of a so-called vital force. This supposi- 
 tion was shown to be fallacious when organic compounds were 
 produced artificially (by synthesis). 
 
 Liebig defined organic chemistry as the chemistry of the 
 compound radicals. This definition is faulty because there 
 are radicals, as NH t , SO 3 , etc., in the domain of inorganic 
 chemistry. 
 
 The names compound radical and radical are synonymous. 
 
 A radical is a chemical combination of two or more ele- 
 ments capable of playing the part of an elementary form of 
 matter. 
 
 Organic chemistry has been defined as the chemistry of 
 the compounds of carbon. It must be remembered that there 
 are two compounds containing carbon in the domain of 
 inorganic chemistry, CO and CO 2 . 
 
 The latest definition of organic chemistry is, the chem- 
 istry of the hydrocarbon compounds and their derivatives. 
 This definition applies to all carbon compounds except HCN 
 (hydrocyanic acid) and CN (cyanogen). 
 
 The number of elements entering into the composition of 
 organic compounds is comparatively small, but the number of 
 atoms in a molecule of a compound may be very large. 
 
 Organic compounds may be composed of simply carbon 
 and hydrogen. Such compounds are termed hydrocarbons, 
 as Ci H 1G , oil of turpentine, CH 4 , methane. 
 
 (5)
 
 6 NOTES ON CHEMICAL LECTURES. 
 
 CN, cyanogen is an organic compound composed of two 
 elements. 
 
 Other organic compounds may be composed of three ele- 
 ments, namely, carbon, hydrogen, and oxygen. Compounds 
 of these three elements containing the hydrogen and oxygen 
 atoms in the proportion to form water are called carbohy- 
 drates, as C 6 H 10 O 5 , starch, C 6 H 12 O 6 , glucose, in the latter there 
 are H 12 O 6 = 6H 2 O. 
 
 There are three principal groups of carbohydrates : 
 The 
 
 Glucose group. 
 
 Saccharose group. 
 
 Amylose group. 
 
 Organic compounds may be composed of four elements, 
 carbon, hydrogen, oxygen, and nitrogen. Compounds 
 containing nitrogen are termed nitrogenous or azotized, as 
 CO(NH 2 ) 2 , urea, C 5 H 4 N 4 O 3 , uric acid. 
 
 Some compounds are composed of only carbon and nitro- 
 gen, as CN, cyanogen, and others of carbon, nitrogen, and 
 hydrogen, as HCN, hydrocyanic acid. 
 
 HCN, hydrocyanic acid, may be looked upon as CH 4 , me- 
 thane, in which three atoms of H have been replaced by an 
 
 atom of triad nitrogen, as C ^TT 
 
 A few organic compounds contain carbon, hydrogen, oxy- 
 gen, nitrogen, and sulphur, as CgoiHg^N^O^Sa, egg-albumen. 
 Some few contain phosphorus, as C^H^NPOg, lecithin. 
 
 A very few organic compounds contain, in addition to 
 carbon, hydrogen, oxygen, nitrogen, and sulphur, a metal, as 
 iron in C 636 H 1025 N 164 O 189 FeS 3 O 2 , oxyhaemoglobin, the molecular 
 weight of which is 14161. 
 
 All organic compounds occurring in nature may be con- 
 sidered as falling under the above classification. By chemical 
 means nearly all of the elements may be introduced into 
 organic chemical compounds. 
 
 The number of possible combinations of these elements to 
 produce new organic compounds is almost infinite. Ordinarily, 
 however, only fourteen or fifteen elements are concerned in 
 chemical combinations in organic chemistry.
 
 
 " -w.
 
 NOTES ON CHEMICAL LETCURES. 7 
 
 Organic compounds containing mercury, copper, gold, etc., 
 and iodine, bromine, chlorine, etc., may be produced artificially 
 by chemical means. These compounds do not occur already 
 formed in nature. 
 
 An organic body is an aggregation of distinct organic 
 compounds called proximate constituents. An organic body 
 may also contain inorganic compounds. Examples, conium 
 maculatum, opium, etc. The percentage proportion of the 
 proximate constituents contained in an organic body is not 
 definite, i. e., two or three samples of opium selected at 
 random will not contain, except by chance, the same percent- 
 age of morphine. 
 
 A proximate principle is an organic compound which is 
 contained in an organic body and to which the physiological 
 action of the organic body is due. Examples, coniine, C 8 H 15 N, 
 the proximate principle of conium maculatum, morphine,. 
 C 17 H 19 NO 3 , codeine, C 18 H 21 NO 3 , etc., proximate principles of 
 opium. 
 
 Oxygen is the predominating element in inorganic chemistry, 
 carbon in organic chemistry. 
 
 In organic chemistry the same laws of combination apply 
 as in inorganic chemistry. 
 
 The belief that a so-called vital force was a necessity in the 
 formation of organic compounds was overthrown in 1828, by 
 Woehler's synthetical production of urea from substances 
 considered inorganic, namely, ammonia and cyanic acid (the 
 two forming NH 4 CNO, ammonium cyanate). 
 
 NH 3 + HCNO = NHjCNO 
 
 By slowly heating NH 4 CNO, ammonium cyanate, on a 
 water-bath at a temperature between 50 C. and 70 C. a re- 
 arrangement of the atoms occurs with the formation of urea. 
 
 NH 4 CNO CO(NH 2 ) 2 
 
 Ammonium cyanate. Urea. 
 
 The next organic compound produced synthetically was 
 CN, cyanogen, by Fownes in 1841. This was produced by 
 passing nitrogen over red-hot charcoal (carbon). 
 
 C 2 + N 2 = C 2 N 2
 
 8 NOTES ON CHEMICAL LECTURES. 
 
 Berthelot followed in 1856 by the synthetical production of 
 HCHO 2 , formic acid. This was produced by passing CO, 
 carbon monoxide, over heated KOH, potassium hydroxide, 
 forming KCHO 2 , potassium formate. The latter compound 
 when treated with HC1 (hydrochloric acid) breaks up into 
 HCHO 2 , formic acid, and KC1, potassium chloride. 
 
 CO + KOH = KCHO 2 
 KCHO 2 + HC1 = HCHO 2 + KC1 
 
 Wurtz in 1862 produced C 2 H 5 OH, ethyl alcohol, syntheti- 
 cally. The alkaloid coniine, C 8 H 15 N, has been produced 
 synthetically. 
 
 Thousands of organic compounds have been produced syn- 
 thetically in the past twenty-five years, some of them of the 
 greatest importance ; for example, the aniline coloring-matters, 
 'indigo, and many of the medicines as antipyrin, phenacetin, 
 etc., now largely used in medical practice. The number of 
 organic compounds produced synthetically is far greater than 
 those existing in nature. 
 
 It was urged that the synthesis of urea was only the pro- 
 duction of a simpler substance from a more complex one. 
 K 4 Fe(CN)6 having been the complex substance used in the 
 production of the NH 4 CNO, ammonium cyanate, from which 
 the urea was finally produced. This was met by the syntheti- 
 cal production of a complex substance, C 21 H 18 N 2 , benzalde- 
 hyde, from a more simple one, C 7 H 6 O, benzamide, oil of bitter 
 almonds. This was effected by passing NH 3 , ammoniacal gas, 
 into C-H 6 O, oil of bitter almonds. 
 
 3C 7 H 6 O + 2NH 3 = C 21 H 18 N 2 + 3H 2 O 
 
 A radical is a chemical combination of two or more elements 
 capable of playing the part of an elementary form of matter. 
 
 Behaving as an element, radicals must have valence and 
 electrical affinities corresponding with elements. They may 
 be monivalent, divalent, etc., and either electro-positive or 
 electro-negative. As they are unsaturated molecules a neutral 
 radical is an impossibility. 
 
 In 1872, Scheele recognized the compound Hg(CN) 2 , 
 mercuric cyanide, and in 1815, Gay-Lussac isolated CN, it 
 being the first radical isolated.
 
 NOTES ON CHEMICAL LECTURES. 9 
 
 CN (or Cy, which is an abbreviation of the name cyanogen,) 
 may be taken as the type of the negative radicals. 
 
 Cyanogen, in combining with positive elements or with 
 positive radicals, may be compared with the negative element 
 Cl, chlorine, for example, 
 
 KC1, also KCN, a simple salt, potassium cyanide. 
 AgCl, also AgCN, a simple salt, argentic cyanide. 
 HC1, also HCN, a hydrogen acid, hydrocyanic acid. 
 HC1O, also HCNO, an oxyacid, cyanic acid. 
 KC1O, also KCNO, an oxyacid salt, potassium cyanate. 
 
 CN therefore unites with 
 
 1. Hydrogen to form a hydrogen acid. 
 
 2. Metals to form simple salts. 
 
 3. Hydrogen and oxygen to form an oxyacid. 
 
 4. Hydrogen, oxygen, and a metal to form an oxyacid salt. 
 
 A radical may contain a metal. When KCN, potassium 
 cyanide, is heated with metallic iron in the presence of air, a 
 radical, called ferrocyanogen, Fe(CN) 6 , is formed. 
 
 6KCN + Fe + O = K 4 Fe(CN) 6 + K 2 O 
 
 The radical Fe(CN) 6 has never been isolated. Sometimes 
 it is expressed FeCy 6 or Cfy. It is a tetrad radical. The K 
 in K 4 Fe(CN) 6 may be replaced by H, forming H 4 Fe(CN) 6 , 
 ferrocyanic acid. 
 
 If chlorine be passed through a solution of K 4 Fe(CN) 6 , one 
 atom of K is withdrawn, and K Fe 2 (CN) J2 , potassium ferri- 
 cyanide, is formed. 
 
 2K 4 Fe(CN) 6 + C1 2 = K Fe 2 (CN) 12 + 2KC1 
 
 The negative radical Fe 2 (CN) 12 is a hexad. It has never 
 been isolated. The K in K 6 Fe 2 (CN) 12 may be replaced by H, 
 forming H c Fe 2 (CN) 12 , ferricyanic acid. 
 
 C 2 H 5 , ethyl, may be taken as the type of the positive organic 
 radicals. Ethyl, in combining with negative elements or with 
 negative radicals, may be compared with the positive element 
 K, potassium, for example,
 
 10 NOTES ON CHEMICAL LECTURES. 
 
 KC1, also C 2 H 3 C1, ethyl chloride. 
 K 2 S, " (C 2 H 5 ) 2 S, ethyl sulphide. 
 
 K 2 O, " (C 2 H 5 ) 2 O, ethyl oxide. 
 KHSO 4 , " C 2 H 5 HSO 4 , ethyl sulphuric acid. 
 KCN, " C 2 H 5 CN, ethyl cyanide. 
 KCNO, " C 2 H 5 CNO, ethyl cyanate. 
 
 C 2 H 5 therefore unites with 
 
 1. Members of the chlorine group to form simple salts. 
 
 2. Oxygen to form an oxide. 
 
 3. An oxyacid to form an oxyacid salt. 
 
 The first positive radical containing a metal isolated was 
 (CH 3 ) 2 As, kakodyl (alkarsin, dimethylarsin). It is sometimes 
 represented by the abbreviation Kd. It is a monad radical. 
 It has an intense affinity for oxygen, combining with it to form 
 ( (CH 3 ) 2 As) 2 O, kakodyl oxide, sometimes represented by the 
 abbreviation Kd 2 O. Kakodyl combines with chlorine to form 
 (CH 3 ) 2 AsCl, kakodyl chloride, and CN to form (CH 3 ) 2 AsCN, 
 kakodyl cyanide. 
 
 Kakodyl alone and in combination (except as kakodylic 
 acid) is very poisonous. 
 
 Kakodyl when exposed to the air in the presence of water 
 forms (CH 3 ) 2 AsOOH, kakodylic acid, sometimes abbreviated 
 to HKdO 2 . 
 
 Kakodylic acid contains 54 per cent of metallic arsenic, 
 equivalent to 71.4 per cent of As 2 O 3 , arsenious oxide, but is 
 not poisonous. It is the only kakodyl compound that is not 
 poisonous. 
 
 Kakodyl, in combining with negative elements or with posi- 
 tive radicals, may be compared with K, for example, 
 
 KC1, also (CH 3 ) 2 AsCl, kakodyl chloride. 
 K 2 O, " ( (CH 3 ) 2 As) 2 O, kakodyl oxide. 
 K 2 SO 4 , " ( (CH 3 ) 2 As) 2 SO 4 , kakodyl sulphate. 
 
 (CH 3 ) 2 As, kakodyl, therefore unites with 
 
 1. Members of the chlorine group to form simple salts. 
 
 2. Oxygen to form an oxide. 
 
 3. An oxyacid to form an oxysalt.
 
 NOTES ON CHEMICAL LECTURES. 11 
 
 There are similar methyl compounds of antimony and of 
 zinc. 
 
 In addition to (CH 3 ) 2 As, dimethylarsin, there also exist, 
 
 CH 3 As, monomethylarsin. 
 (CH 3 ) 3 As, trimethylarsin. 
 
 CH 4 , methane, may be considered a type of the saturated 
 organic compounds. All of the carbon bonds are satisfied, 
 and it is therefore a saturated molecule. One, two, three, or 
 all four atoms of H in CH 4 may be replaced by certain other 
 elements. The radical remaining after the withdrawal of each 
 atom of hydrogen has a valence corresponding to the number 
 of hydrogen atoms withdrawn. 
 
 1. CH 4 H = CH 3 (methyl), univalent. 
 
 2. CH 4 H 2 = CH 2 (methene), bivalent. 
 
 3. CH 4 H 3 CH (formyl), trivalent. 
 
 4. CH 4 H 4 = C (carbon), quadrivalent. 
 
 Thus replacing the H by Cl we have 
 
 1. CH 4 + 2C1 = CH 3 C1 + HC1. 
 
 2. CH 4 + 4C1 = CH 2 C1 2 + 2HC1. 
 
 3. CH 4 + 6C1 == CHC1 3 (chloroform) + 3HC1. 
 
 4. CH 4 + 8C1 == CC1 4 + 4HC1. 
 
 Or replacing the H by iodine we have 
 
 1. CH 4 + 2I = CH 3 I + HI. 
 
 2. CH 4 + 41 = CH 2 I 2 + 2HI. 
 
 3. CH 4 + 61 = CHI 3 (iodoform) + 3HI. 
 
 4. CH 4 + 81 = CI 4 + 4HI. 
 
 A type is a form of chemical combination common to a class 
 of compounds. 
 
 NaCl, CH 3 C1, types of simple salts. 
 K 2 SO 4 , (C 2 H 5 ) 2 SO 4 , types of oxysalts. 
 H 2 O, (C 2 H 5 ) 2 O, the water type. 
 NH 3 , the ammonia type. 
 
 By substitution is meant the replacing of one or more ele- 
 ments or radicals in a compound by one or more elements or 
 radicals without changing the type of the compound.
 
 12 NOTES ON CHEMICAL LECTURES. 
 
 When chemical combination occurs it does not necessarily 
 follow that substitution has taken place. 
 
 Alcohols and ethers may be viewed as substitution com- 
 pounds. 
 
 An alcohol may be considered after the type of water, in 
 which one atom of H in H 2 O has been replaced by an alcohol 
 radical, as 
 
 H OH, water. 
 C 2 H 5 OH, ethyl alcohol. 
 
 An ether may be considered after the type of water, in 
 which both atoms of H in H 2 O have been replaced by alcohol 
 radicals, as 
 
 H 2 O, water. 
 (C 2 H 5 ) 2 O, ethyl ether. 
 
 EXAMPLES OF SUBSTITUTION. 
 
 When HgCl 2 , mercuric chloride, is treated with NH 3 , ammo- 
 niacal gas, one of the atoms of Cl in the HgCl 2 is replaced by 
 the amido-radical NH 2 , and HgNH 3 Cl, amido-mercuric chloride 
 (white precipitate), is formed. 
 
 HgCl 2 + 2NH 3 = HgNH 2 Cl +NH 4 C1 
 
 When Hg 2 Cl 2 , mercurous chloride, is treated with NH 3 , am- 
 monia, one of the atoms of Cl in the Hg 2 Cl 2 is replaced by the 
 amido-radical NH 2 , and Hg 2 NH 2 Cl, amido-mercurous chloride 
 (black precipitate), is formed. 
 
 Hg 2 Cl 2 + 2NH 3 = Hg 2 NH 2 Cl + NH 4 C1 
 
 HgNH 2 Cl and Hg 2 NH 2 Cl are inorganic substitution com- 
 pounds after the type of NH 4 C1, ammonium chloride, in which 
 two hydrogen atoms have been substituted by mercury. 
 
 One, two, or all three atoms of H in NH 3 may be replaced 
 by other elements or radicals. 
 
 An amide may be defined as a substitution compound in 
 which the hydrogen of NH 3 has been partly or wholly replaced 
 by an alcohol radical or a metal. 
 
 Amides may unite with an acid without replacing the hydro- 
 gen of the acid, as in N(CH 3 ) 3 HC1 trimethylamide hydro- 
 chloride.
 
 NOTES ON CHEMICAL LECTURES. 13 
 
 An amide may be a constituent of a double salt, as in 
 N(CH 3 ) 4 CNAgCN tetramethylargentammonium cyanide. 
 
 We may have NH 2 K, potass-amide, produced by the re- 
 placement of one atom of H in NH 3 by K. By the replace- 
 ment of H in NH 3 by radicals we may have 
 
 NH 2 CH 3 , methylamide. 
 
 NHCH 3 C 2 H 5 , methyl-ethylamide. 
 NCH 3 C 2 H-C 3 H 7 , methyl-ethyl-propylamide. 
 NHCH 3 CH 3 , dimethylamide. 
 NCH 3 CH 3 CH 3 , trimethylamide. 
 
 All the atoms of hydrogen in a compound are not neces- 
 sarily replaceable. For example, only one atom of hydrogen 
 in HC 2 H 3 O 2 , acetic acid, is replaceable. 
 
 When C 6 H 10 O 5 , cellulose (cotton), is treated with nitric acid, 
 three atoms of hydrogen in the cellulose are replaced by 
 the NO 2 , nitro radical, forming the explosive compound 
 C 6 H r (NO 2 ) 3 O 5 , trinitrocellulose (gun-cotton). 
 
 C 6 H 10 5 + 3HN0 3 = C 6 H 7 (N0 2 ) 3 5 + 3H 2 O 
 
 When C 3 H 8 O 3 , glycerine, is treated with nitric acid, three 
 atoms of hydrogen in the glycerine are replaced by the NO 2 , 
 nitro radical, forming the explosive compound C 3 H 5 (NO 2 ) 3 O 3 , 
 trinitro glycerine. 
 
 C 3 H 8 O 3 + 3HNO 3 = C 3 H 5 (NO 2 ) 3 O 3 + 3H 2 O 
 
 The hypothetical H 2 CO 3 , carbonic acid, is dibasic ; has two 
 replaceable atoms of hydrogen. 
 
 / OH 
 
 C = O , carbonic acid. 
 \ OH 
 
 By replacing one of the OH hydroxyl groups in carbonic 
 acid by NH 2 , an amido-acid, carbamic acid is obtained. 
 
 / OH 
 
 C = O , carbamic acid. 
 \NH 2
 
 14 NOTES ON CHEMICAL LECTURES. 
 
 Carbamic acid is not known in its free state. It is always 
 in combination as a salt or an ether. The ammonium salt is 
 produced by bringing together dry CO 2 , carbon dioxide, and 
 dry NH 3 , ammoniacal gas. 
 
 CO 2 + 2NH 3 = CO(ONH 4 )(NH 2 ) or NH 4 NH 2 CO 2 , ammo- 
 nium carbamate. 
 
 / ONH 4 
 
 C = O , ammonium carbamate. 
 \ NH 2 
 
 When treated with water it decomposes into ammonium 
 carbonate. 
 
 CO(ONH 4 )(NH 2 ) + H 2 O = (NH 4 ) 2 CO 3 
 
 Ammonium carbamate. Ammonium carbonate. 
 
 The ethers of carbamic acid are called urethans. They 
 are formed by replacing the atom of hydrogen in the remain- 
 ing OH hydroxyl group in CONH 2 OH, carbamic acid, by 
 an alcohol radical. 
 
 / OCH 3 
 
 C O , methyl urethan. 
 \ NH 2 
 
 / OC 2 H 5 
 
 C = O , ethyl urethan. 
 \ NH 2 
 
 Ethyl urethan under the name of " urethan " is used in 
 medicine. 
 
 By replacing both of the hydroxyl groups in H 2 CO 3 , car- 
 bonic acid, by NH 2 , an amide is formed, CO(NH 2 ) 2 , carbamide, 
 urea. 
 
 / NH 2 
 
 C = O , carbamide (urea). 
 \ NH 2 
 
 Antipyrin, phenyldimethyl pyrazolon, C n H 12 N 2 O, or 
 (C 3 H(CH 3 ) 2 N 2 (C 6 H 5 )O), is a substitution product. 
 
 The aniline coloring-matters are also substitution products.
 
 
 
 
 <V- 372 
 
 I lit 67 3' 7 2 - 
 
 
 -.-? A 
 

 
 NOTES ON CHEMICAL LECTURES. 15 
 
 The composition of organic compounds may be ex- 
 pressed by 
 
 1. Empirical formula. 
 
 2. Molecular formula. 
 
 3. Rational formula. 
 
 4. Graphic formula. 
 
 1. Empirical formula : The simplest possible expression 
 by formula of the composition of a compound. It indicates 
 simply the elements that enter into the composition of the 
 compound in their least atomic proportions. Thus CH 2 O 
 for acetic acid. 
 
 An empirical formula may be deduced from the results of a 
 quantitative analysis of a compound. 
 
 2. Molecular formula : A formula that expresses a quantity 
 of a compound by weight twice its specific gravity in the 
 gaseous state compared with hydrogen. 
 
 Or : A quantity of a compound by weight in the gaseous 
 state twice the volume of the atom of hydrogen. Hence all 
 molecules must be of the same size. 
 
 A molecular formula may be determined : 
 
 a. By determining the specific gravity of the compound in 
 the gaseous state. 
 
 Victor Meyer's method; Depends on the vaporization of 
 a weighed quantity of a compound and measuring, by means 
 of a eudiometer, the volume of air the vapor displaces in the 
 apparatus, correcting for temperature and pressure, and com- 
 paring the volume of air displaced with the volume occupied 
 by a quantity of hydrogen equal to the weight of the com- 
 pound volatilized. 
 
 The molecular weight of a compound which undergoes 
 decomposition before its point of vaporization is reached cannot 
 be determined by this method. 
 
 Method : A substance having a fixed boiling-point is placed 
 in the outer large glass tube of the apparatus. For compounds 
 having a low vaporizing-point water may be used. Generally 
 (C 6 H 5 ) 2 NH, diphenylamine, having a boiling-point of 310 C., is 
 employed. Melted lead may be used for compounds having a 
 higher vaporizing-point. The smaller bulbed glass tube having
 
 16 NOTES ON CHEMICAL LECTURES. 
 
 a curved delivery-tube attached is lowered into the larger glass 
 tube until the bulbed end is a few inches above the surface of 
 the substance having a fixed boiling-point. A small quantity 
 of the compound, of which the molecular weight is to be 
 determined, is weighed off in a very small glass tube, which 
 is closed at one end. The tube containing the compound is 
 placed with the opening upward in a perforation in the cork 
 of the smaller bulbed tube, where it is held in place by a wire 
 thumb-screw contrivance, and the cork thus arranged placed 
 in the smaller bulbed tube. Gentle heat is applied to the outer 
 large tube, and gradually increased until the substance with 
 the fixed boiling-point boils. After boiling a few minutes and 
 the vapor of the boiling material reaches and is condensed to 
 a liquid at a point opposite the bulb of the smaller glass tube, 
 the delivery-tube is adjusted so that the exit is below the sur- 
 face of water contained in a suitable vessel. When bubbles 
 of air cease to ascend through the water (at the same time 
 the heating of the material having a fixed boiling point is kept 
 up so that it is boiling), a eudiometer filled with water is 
 inverted over the exit of the delivery-tube under water, and 
 by turning the thumb-screw contrivance the tube containing 
 the compound under examination is allowed to drop to the 
 bottom of the inner bulbed tube. (To break the fall of the 
 tube containing the compound it is customary to previously 
 place pieces of broken glass tubing in the bottom of the 
 bulb of the inner glass tube.) 
 
 Volatilization of the compound immediately occurs, and a 
 volume of air is displaced and collected in the eudiometer 
 exactly equivalent to the gaseous volume furnished by the 
 weight of the compound volatilized. This final part of the 
 operation requires but a few minutes. 
 
 When bubbles of air cease to come over, the cork is taken 
 out of the apparatus and the eudiometer cautiously removed 
 and placed in a vessel containing water. After a time the air 
 in the eudiometer becomes of the same temperature as the 
 water surrounding it. The volume of air is now read off, the 
 temperature of the water surrounding the eudiometer and the 
 barometric pressure are observed, and the necessary calcula- 
 tions performed.
 
 NOTES ON CHEMICAL LECTURES. 17 
 
 Suppose 0.010 gramme acetic acid were volatilized and the 
 volume of air displaced and collected in the eudiometer 
 corrected for temperature (0 C.) and barometric pressure 
 (7GO mm.), measured 3.72 c.c. 
 
 0.0 10 gramme hydrogen at C. temperature and 760 mm. 
 pressure measure 111.6 c.c. 
 
 Then 111.6-^3.72 = 30 
 
 showing the vapor of acetic acid to be 30 times more dense, 
 volume for volume, than hydrogen. 
 
 c.c. H. Gnn. H. c.c. H. Grm.H. 
 
 Or 11160 : 1 :: 3.72 : 0.00^3 
 
 All molecules, elementary or compound, must be equal in 
 volume to the volume occupied by the molecule of hydrogen, 
 /. e., twice the atomic volume. 
 
 If 30 grammes of acetic acid in the form of vapor will 
 occupy the same volume as the atomic volume of hydrogen, 
 namely, 11160 c.c., then 60 grammes of acetic acid will 
 occupy twice the atomic volume, or the molecular volume of 
 hydrogen, namely, 22320 c.c. 
 
 A formula for acetic acid representing 30 as its molecular 
 weight would be CH 2 O = 30, really its empirical formula, and 
 expresses a quantity in the gaseous state equal to its specific 
 gravity compared with hydrogen. By multiplying the 30 by 
 2 = 60, we have a number representing its molecular weight. 
 A formula for the acid constructed to represent this number 
 would be C 2 H 4 O 2 = 60, i.e., a formula that expresses a quan- 
 tity by weight of the compound twice its specific gravity in the 
 gaseous state compared with hydrogen. 
 
 b. The molecular weight of an organic compound, especially 
 those compounds which are not vaporizable, may be deter- 
 mined, 
 
 1. If an acid, by combining it with a metal to form a salt 
 and determining the quantity of the metal in combination in a 
 given quantity of the salt 
 
 2. If a basic substance, by combining it with an acid and 
 determining the quantity of acid in combination in a given 
 quantity of the salt.
 
 18 NOTES ON CHEMICAL LECTURES. 
 
 The molecular formula of many compounds which will not 
 vaporize or form salts is determined by analyzing their sub- 
 stitution products. 
 
 The molecular formula of some compounds which will not 
 vaporize, form salts or substitution products, is not definitely 
 known, as cane-sugar, starch. 
 
 For example, the molecular formula of acetic acid, which 
 forms a salt with silver, may be determined by estimating the 
 quantity of metallic silver in its silver salt. 
 
 Suppose ].0 gramme of argentic acetate be placed in a 
 crucible and heated until all the organic matter is driven off 
 and nothing remains but metallic silver. After cooling, the 
 silver is weighed. 
 
 Quantity of argentic acetate taken 1.0000 grm. 
 
 Weight of resulting metallic silver 0.6468 " 
 
 Quantity of organic compound C, H, and O, which 
 
 must have been in combination with the silver 0.3532 " 
 
 Then 
 
 At. wt. of silver. 
 
 0.6468 : 0.3532 : : 108 : 58.98 = C, H, and O, in combina- 
 tion with one atom of silver. 
 
 Acetic acid is a monobasic acid, i. e., having one replaceable 
 atom of hydrogen. Consequently, as silver is a monad, it 
 would take the place of one atom of hydrogen in the mol- 
 ecule of acid. Add 1 to 58.98 = 59.98, and allowing for-error 
 in analysis, say 60.0. A formula constructed to equal that 
 number would be C 2 H 4 O 2 = 60, acetic acid. 
 
 The formula of the silver salt is AgC 2 H 3 O 2 . 
 
 A recent method for determining molecular weight is based 
 upon the observations of Raoult (1883), namely, that the 
 lowering of tlie freezing point of a solution is proportional to 
 the absolute quantity of substance in solution and inversely as 
 its molecular weight. (Law of Raoult.) 
 
 Beckmann's method of determining the molecular weight 
 of a compound by the depression of the freezing point of
 
 NOTES ON CHEMICAL LECTURES. 19 
 
 the solvent employed, according to the Law of Raoult, is as 
 follows : 
 
 15 to 20 grammes of the solvent, accurately weighed, are 
 placed in a hard glass tube 2 to 3 centimetres in width, having 
 a tube projecting from the side, and closed with a perforated 
 cork, through which are passed an accurately standardized 
 thermometer (Waldferdin Thermometer) and a stout platinum 
 wire which serves as a stirring rod. This tube is then fixed 
 to the depth of the side-projecting tube in a cork fitted to a 
 larger and wider tube. The latter serves as an air jacket. 
 The entire apparatus consisting of one tube inserted in another 
 is fixed in an aperture in the cover of a large glass vessel 
 which contains a freezing mixture. The congealing point of 
 the solvent is first determined by cooling it 1 to 2 below its 
 freezing point and then by agitation with the platinum wire 
 (after having added platinum clippings) inducing the formation 
 of crystals. During this operation the temperature rises and 
 when the mercury in the thermometer is stationary it indicates 
 the freezing point of the solvent. The mass of crystals is 
 permitted to melt and an accurately weighed quantity of the 
 compound, the molecular weight of which is to be determined, 
 is introduced through the side-projecting tube. When the 
 compound introduced is completely dissolved the freezing 
 point of the solution is determined after the manner just 
 described. 
 
 Rule for calculating results in Beckmann's method; 
 
 Multiply the percentage of compound in solution by the con- 
 stant T of the solvent employed and divide by the depression 
 of the freezing point. 
 
 The solvents usually employed in Beckmann's method are 
 water, benzol and glacial acetic acid. 
 
 The constant T of the solvents is as follows : 
 
 Water 19.0 
 
 Benzol 4.9 
 
 Glacial acetic acid . 39.0
 
 20 NOTES ON CHEMICAL LECTURES. 
 
 Example of determination of molecular weight of a 
 compound by Beckmann's method: 
 
 Example : Suppose a definite weight of oxyberberine 
 acetate (C 20 H 17 NO 5 HC 2 H 3 O 2 molec. wt. 411) is employed and 
 glacial acetic acid is used as the solvent. 
 
 1.573 grms. oxyberberine acetate employed. 
 100.400 grms. glacial acetic acid employed. 
 
 Hence 100.4 : 1.573 : : 100 : 1.56 grms. (percentage of 
 oxyberberine acetate in the solvent). 
 
 Freezing point of glacial acetic acid = 16.262 C. 
 
 Freezing point of glacial acetic acid containing 
 
 the oxyberberine acetate = 16.112 C. 
 
 Depression of the freezing point : 0.150 C. 
 
 Therefore, 
 
 39 X 07T50 == 0.15Q = 407 molecular weight 
 of oxyberberine acetates from which the molecular formula 
 is constructed. 
 
 The molecular formula of many substances which before 
 the introduction of Beckmann's method could not be accurately 
 determined, are now determined with accuracy by this method. 
 Example, glucose. 
 
 3. A rational formula attempts to express the arrangement 
 of the atoms or groups of atoms in the molecule of a com- 
 pound. 
 
 The same compound may be represented by various rational 
 formulas, depending on the views of different chemists, e. g,, 
 over twenty rational formulas have been suggested for acetic 
 acid. 
 
 4. A graphic formula attempts to express the arrangement 
 of the atoms or groups of atoms in the molecule of a com- 
 pound by means of lines (pictures). 
 
 Acetic acid. 
 
 Empirical formula, CH 2 O 
 
 Molecular C,H 4 O 2 
 
 Rational " HC 2 H 3 O 2 or CH 3 COOH
 
 , 
 
 
 
 W 
 
 J- 
 
 ' 
 
 rfT d X 

 
 NOTES ON CHEMICAL LECTURES. 21 
 
 H 
 
 I 
 Graphic formula, H C C O H 
 
 H O 
 
 Benzol (benzene). 
 
 Empirical formula, CH 
 Molecular " C 6 H 6 
 
 H 
 
 I 
 C 
 
 / ^ 
 
 Graphic formula, H C C H 
 
 (Kekule's Benzol || | 
 
 ring.) H C C H 
 
 \ // 
 C 
 
 I 
 H 
 
 All of the atoms of hydrogen in benzol are replaceable. 
 The hydrogen atoms in benzol may be replaced by single 
 elements, as chlorine, or iodine, or bromine, or by radicals, or 
 all of the hydrogen may be replaced by elements unlike each 
 other or by unlike radicals. 
 
 By replacing the hydrogen atoms in benzol with chlorine 
 we may have 
 
 C 6 H 5 C1, monochlor benzol. 
 
 C 6 H 4 C1 2 , dichlor 
 
 C 6 H 3 C1 3 , trichlor 
 
 C 6 H 2 C1 4 , tetrachlor 
 
 C 6 HC1 5 , pentachlor " 
 
 C G C1 6 , hexa- or perchlor benzol. 
 
 Amido-benzol, or aniline, C 6 H 5 NH 2 , is a compound formed 
 by replacing one of the atoms of hydrogen in benzol by the 
 NH 2 , amido group.
 
 22 NOTES ON CHEMICAL LECTURES. 
 
 There are three distinct isomeric dinitrobenzols, (dinitro- 
 benzenes). They are 
 
 Orthodinitrobenzol. 
 
 Metadinitrobenzol 
 
 Paradinitrobenzol. 
 
 Kekule's Benzol ring. 
 
 6/\ 2 
 
 \/ 
 
 4 
 
 When the hydrogen atoms at the positions 1 and 2 have 
 been replaced, the compound is termed an Ortho compound. 
 Example : 
 
 N0 2 
 
 I 
 C 
 
 / ^ 
 
 H C C NO 2 
 Orthodinitrobenzol, || | 
 
 H C C H 
 
 \ // 
 C 
 
 I 
 H 
 
 When the hydrogen atoms at the positions 1 and 3 have 
 been replaced, the compound is termed a Meta compound. 
 Example : 
 
 NO, 
 
 I 
 C 
 
 / ^ 
 
 H C C H 
 Metadinitrobenzol, || | 
 
 H - C C - N0 2 
 
 \ // 
 C 
 
 H
 
 NOTES ON CHEMICAL LECTURES. 23 
 
 When the hydrogen atoms at the positions 1 and 4 have 
 been replaced, the compound is termed a Para compound. 
 Example : 
 
 NO 2 
 
 I 
 C 
 
 / ^ 
 
 H C C H 
 Paradinitrobenzol, 
 
 H C C H 
 
 \ // 
 
 C 
 
 I 
 N0 2 
 
 Phenol, or carbolic acid, C 6 H 5 OH, may be produced by 
 replacing one of the hydrogen atoms in benzol by the OH, 
 hydroxyl group. 
 
 Phenol is eliminated in the urine as C 6 H 5 HSO 4 , phenol- 
 sulphuric acid, or in combination as a salt, C 6 H 5 KSO 4 , phenol- 
 potassium sulphate. Phenol (carbolic acid) is poisonous, but 
 the salts of phenol-sulphuric acid are not poisonous. Sodium 
 sulphate, or magnesium sulphate is recommended as an anti- 
 dote in carbolic acid poisoning, converting the acid into the 
 non-poisonous salt phenol-sodium sulphate, or phenol-magne- 
 sium sulphate. 
 
 Precipitation of barium as barium sulphate does not occur 
 on the addition of barium chloride to a solution of the salts 
 of phenol-sulphuric acid. Thus, from the amount of precipitate 
 of barium sulphate obtained on the addition of barium chloride 
 to the urine, after the ingestion of carbolic acid, it appears 
 that the quantity of sulphuric acid is diminished, whereas it 
 really is unchanged or perhaps increased. 
 
 Phenol cannot be detected directly in the urine by any of 
 the tests. It must first be liberated from its combination as 
 phenol-sulphuric acid, distilled, the distillate agitated with ethyl 
 ether, the ethyl ether allowed to separate and then removed 
 from the aqueous solution and allowed to evaporate. The 
 residue is dissolved in a small quantity of water and the tests 
 for phenol applied.
 
 24 NOTES ON CHEMICAL LECTURES. 
 
 Detection of phenol in the urine. 25 c.c. of sulphuric 
 acid are added to 500 c.c. of the urine, the mixture is distilled 
 until bromine water added to a part of the last portion of the 
 distillate fails to produce a turbidity. The distillate is agitated 
 with ethyl ether, after the ether has separated from the water, 
 it is removed with a pipette, and allowed to evaporate. The 
 residue is dissolved in a small quantity of water and the tests 
 for phenol are applied. 
 
 On the addition of a dilute neutral solution of ferric chloride 
 to an aqueous solution of phenol (carbolic acid) an intense 
 purple color is produced. 
 
 On the addition of bromine water to an aqueous solution of 
 phenol until a permanent yellowish coloration is produced, a 
 yellowish white, crystalline precipitate of C 6 H 2 Br 3 OH, tribrom- 
 phenol is formed. If the dilution is 1 to 40,000 the turbid- 
 ity appears immediately, if the dilution is 1 to 50,000 a 
 crystalline precipitate appears only after the solution has stood 
 several hours. 
 
 Isomerism is a term applied to bodies containing the same 
 elements united in the same relative proportions by weight, 
 but differing more or less widely in their physical, physio- 
 logical, and chemical properties. 
 
 The first isomeric compounds discovered were 
 
 CH 4 , methane. 
 CH 4 , attar of roses. 
 
 Isomeric compounds are of two classes. 1. Polymeric : 
 Where the percentage composition is similar, but the molecular 
 composition dissimilar, /. e., the same empirical but different 
 molecular formula. 
 
 The olefines are examples of polymeric compounds, as are 
 also the cyanogen oxyacids. 
 
 HCNO, cyanic acid, monobasic. 
 H 2 C 2 N 2 O 2 , fulminic acid, bibasic. 
 H 3 C 3 N 3 O 3 , fulminuric acid, monobasic. 
 H 3 C 3 N 3 O 3 , cyanuric acid, tribasic. 
 
 2. Metameric: Where both the percentage and the mole- 
 cular compositions are alike.
 
 A '- 'hAjuZs
 
 ' ; 
 
 ". 
 
 l*~*
 
 NOTES ON CHEMICAL LECTURES. 25 
 
 The oils of turpentine, lemons, bergamot, cloves, and pepper 
 are examples of metameric compounds. They all have a 
 similar percentage and molecular composition, C 10 Hi 6 . 
 C 3 H 6 O 3 , lactic acid, ^ 
 
 C 3 H 6 O 3 , paralactic acid, Vare metameric. 
 C 3 H 6 O 3 , hydracrylic acid, J 
 
 An homologous series is a series of chemical compounds 
 made up of the same elements, but having a common differ- 
 ence in their molecular formula. 
 
 Hydrocarbons of equal equivalence, as CH 4 , may exist 
 separately, whilst hydrocarbons of unequal equivalence, as 
 CH 3 , are incapable of existing in the free state, unless, perhaps, 
 
 CH 3 
 as double molecules, /. e., \ = C 2 H 6 . 
 
 CH 3 
 As examples of homologous series we have the 
 
 1. defines. 
 
 2. Alcohol radicals. 
 
 3. Paraffins. 
 
 4. Alcohols. 
 
 5. Aldehydes. 
 
 6. Volatile fat acids. 
 
 7. Ethers. 
 
 In the above homologous series the common difference in 
 the molecular formula is CH 2 . 
 
 1. defines : Diatomic, polymeric, hydrocarbon cormieunds. 
 j/ t- fK~fJi~4' *^&3 * 
 
 >v*f >U ^Wrf-^t<yI 
 
 1. CH 2 , methene (hypothetical) 
 
 2. C 2 H 4 , ethene 14 
 
 3. C 3 H 6 , propene 21 
 
 4. C 4 H 8 , butene 28 
 
 5. C 5 H 10 , pentene . . 35 
 
 6. C 6 H 12 , hexene 42 
 
 7. C 7 H U , heptene 49 
 
 8. C 8 H 16 , octene 56 
 
 9. C 9 H 18 , nonene 63 
 
 10. QoHaj, decene 70 
 
 and continuing to 
 
 30. C^H^, melene, found in wax.
 
 26 NOTES ON CHEMICAL LECTURES. 
 
 2. Alcohol radicals (may exist only as double molecules): 
 Olefines + one atom of H, hydrides of the olefines, mona- 
 tomic. 
 
 I fj+Hz}. CH 3) methyl. 
 
 2. C 2 H 5 , ethyl. 
 
 3. C 3 H 7 , propyl. 
 
 4. QH 9 , butyl. 
 
 5. C 5 H n , amyl. 
 
 Etc. <V-^, 
 
 The alcohol radicals form salts, as CH 3 I, CH 3 C1, etc. 
 
 3. Paraffins : Alcohol radicals -f- one atom of H, hydrides 
 of the alcohol radicals, saturated hydrocarbons. 
 
 1. CH 4 , methane. 
 
 2. C 2 H 6 , ethane. 
 
 3. C 3 H 8 , propane. 
 
 4. C 4 H 10 , butane or quartane. 
 
 5. C 5 H 12 , pentane. 
 
 6. C 6 H 14 , hexane. 
 
 4. Alcohols, monatomic or monohydric, ethylic series : 
 
 Alcohol radicals -f- OH (hydroxyl), may be regarded after 
 
 the type of H 2 O, in which one atom of H in H 2 O has been 
 replaced by an alcohol radical. 
 
 1. 
 
 2. 
 3. 
 4. 
 5. 
 6. 
 
 CH 3 OH, 
 C 2 H 5 OH, 
 C 3 H 7 OH, 
 QH 9 OH, 
 C 5 H U OH, 
 C 6 H 13 OH, 
 Etc. 
 
 Vapor density 
 compared with I 
 
 methylic alcohol (wood spirit) . . 16 
 ethylic " (ordinary alcohol) 23 
 oroDvlic " - 30 
 
 Boiling 
 1. point. 
 
 150 F. 
 173 
 
 205 
 233 
 270 
 305 
 
 butylic " ... 
 
 37 
 
 amylic . . . 
 
 44 
 
 caprylic " ... 
 
 51 
 
 
 
 5. Aldehydes of the acetic series (alcohol dehydrogena- 
 tum) : Alcohols minus two atoms of H. Under the influence 
 of oxidizing agents alcohols give up two atoms of H, forming 
 aldehydes, viz., CH 3 OH + O = CH 2 O + H 2 O.
 
 NOTES ON CHEMICAL LECTURES. 27 
 
 1. CH 2 O, mcthylic or formic aldehyde. 
 
 2. C 2 H A ethylic or acetic 
 
 3. C 3 H 6 O, propylic 
 
 4. C 4 HA butylic 
 
 5. C 3 H 10 O, valeric 
 
 Etc. 
 
 G. Fat acids, acetic series : Aldehydes + one atom of 
 oxygen, monatomic. 
 
 1. CH 2 O 2 , formic acid, found in red ants. 
 
 2. C 2 H 4 O,, acetic " " " vinegar. 
 
 3. C 3 H 6 O,, propylic " " " oils. 
 
 4. C 4 H 8 O,, butyric " " " rancid butter. 
 
 5. C 5 H 10 O 2 , valeric " " " valerian. 
 
 6. QH 12 O 2 , caproic " " " rancid butter and sweat. 
 
 And continuing regularly to 
 
 20. C M H W O 2 , butic acid, found in butter. 
 30. C 30 H W A, melissic " " " beeswax. 
 
 7. Ethers, of monohydric alcohols ; Oxides of the alcohol 
 radicals, after the type of H 2 O, in which both atoms of H in 
 the H 2 O are replaced by alcohol radicals. 
 
 1. (CH 3 ) 2 O, methylic ether. 
 
 2. (C 2 H 5 )A ethylic 
 
 3. (C 3 H-) 2 0, propylic " 
 
 4. (C 4 H 9 ) 2 O, butylic 
 
 5. (C 5 H n ) 2 O, amylic 
 
 Every alcohol has its corresponding aldehyde, acid, and 
 ether. 
 
 The paraffins are saturated hydrocarbon compounds. 
 
 By replacing 3 atoms of H in a paraffin with 3 hydroxyl 
 (OH) groups a triatomic (trihydric) alcohol is formed. 
 Glycerine (C^H-^OH).,) is a triatomic alcohol. 
 
 Replacing 3 atoms of H in the paraffin C 3 H.<, propane, with 
 
 OH 
 3 hydroxyl groups, a triatomic alcohol, C 3 H-OH = C 3 HAs 
 
 OH 
 glycerine (propenyl alcohol), is formed.
 
 28 A'OTES ON CHEMICAL LECTURES. 
 
 Glycerine occurs in most animal and vegetable fats in com- 
 bination with the acids of the acetic and oleic series, as 
 glycerides. Suet contains stearin, C 3 H 5 (C 18 H 35 O 2 )3, a glyceride 
 of stearic acid. 
 
 By the action of superheated steam on the stearin in fats 
 glycerine and stearic acid are set free. 
 
 QH^H^s + 3H 2 = C 3 H 5 (OH) 3 + SC^O, 
 
 Stearin. Water. Glycerine. Stearic acid. 
 
 All the monatomic alcohols of the ethylic series (the ordi- 
 nary alcohols) excepting the first two members of the series 
 have numerous isomeric modifications. They are distinguished 
 especially by their behavior on oxidation. Kolbe gave to 
 methylic alcohol, CH 3 OH, the name carbinol, whilst all the 
 succeeding alcohols of the series he termed carbinols, regard- 
 ing them as derivatives of the first term CH 3 OH, methylic 
 alcohol, and formed by the replacement of hydrogen by monad 
 alcohol radicals. 
 
 1 . Primary alcohols ; Compounds in which one atom of H 
 of the CH 3 of carbinol (CH 3 OH) is replaced by an alcohol 
 radical. 
 
 a. CH 2 CH 3 OH = C 2 H 5 OH, methyl carbinol (ethylic alco- 
 hol). 
 
 b. CH 2 C 2 H r> OH = C 3 H-OH, ethyl carbinol (propylic alco- 
 hol). 
 
 Primary alcohols on oxidation yield 
 
 1. An aldehyde. 
 
 2. A fat acid. 
 
 3. An ethereal salt. 
 
 1. C 2 H 5 OH + O = C 2 H 4 O + H 2 O 
 
 Ethylic alcohol. Acetic aldehyde. 
 
 2. C 2 H 4 + O = C 2 H A 
 
 Acetic aldehyde. Acetic acid. 
 
 3. C 2 H 4 O 2 + C 2 H 5 OH = C 2 H 5 C 2 H 3 O 2 + H 2 O * 
 
 Acetic acid. Ethylic alcohol. Ethylic acetate (acetic ether) 
 
 2. Secondary alcohols : Compounds in which two atoms 
 of H of the CH 3 of carbinol are replaced by alcohol radicals.
 
 NOTES ON CHEMICAL LECTURES. 29 
 
 a. CHCH 3 CH 3 OH = C 3 H 7 OH, dimethyl carbinol (isomeric 
 with propylic alcohol). 
 
 b. CHCH,C,H,OH = C 4 H 9 OH, ethyl methyl carbinol (iso- 
 meric with butylic alcohol). 
 
 Secondary alcohols on oxidation yield 
 
 1. No aldehyde. 
 
 2. A ketone. 
 
 3. An acid containing a less number of carbon atoms than 
 the alcohol oxidized, i.e., an acid of the fat series. 
 
 Ketone : An aldehyde in which one atom of hydrogen is 
 replaced by an alcohol radical^ ^ Oj-0 ~ H^ -+C? /V, 
 
 C 2 H 4 O = C 2 H 3 CH 3 O 
 
 Acetic aldehyde. Acetic ketone. 
 
 3. Tertiary alcohols : Compounds in which three atoms of 
 H in the CH 3 of carbinol are replaced by alcohol radicals. 
 
 a. CCH,CH 3 CH 3 OH = QH 9 OH, trimethyl carbinol (iso- 
 
 CH 3 
 
 I 
 
 meric with butylic alcohol), CH 3 C OH = C 4 H 9 OH, tri- 
 
 CH 3 
 
 methyl carbinol (isomeric with butylic alcohol). 
 
 b. CCH 3 CH 3 C 2 H 5 OH = C 5 H U OH, ethyl dimethyl carbinol 
 (isomeric with amylic alcohol). 
 
 Tertiary alcohols on oxidation yield 
 
 1. No aldehyde. 
 
 2. No ketone. 
 
 3. One or more acids of the acetic series. 
 
 Four primary alcohols ; 
 Three secondary alcohols ; 
 One tertiary alcohol ; 
 
 are possible, having the same empirical and molecular formula. 
 Six of these are well known, the other two have not yet been 
 discovered.
 
 30 NOTES ON CHEMICAL LECTURES. 
 
 DECOMPOSITION OF ORGANIC SUBSTANCES. 
 
 All organic substances are naturally prone to undergo de- 
 composition. If the organic substance contain nitrogen, the 
 tendency towards decomposition is increased. 
 
 DECOMPOSING AGENTS, 
 i. Oxygen. 
 
 1. Direct combustion. 
 
 2. Slow combustion as in the eremacausis (slow oxidation) 
 of oak wood. 
 
 QsH^On -f O 2 = 2H 2 O -f C 18 H 18 O U 
 
 Oak wood. 
 
 then 
 
 Ci$H, 8 O u = CO 2 + Cj 7 H 18 O 9 
 
 /. e., as soon as two atoms of oxygen have taken away four 
 atoms of hydrogen, one atom of carbon unites with two atoms 
 of oxygen, and so on until finally nothing remains of the oak 
 wood but carbon. This is sometimes used as an illustration 
 of the formation of coal. 
 
 3. When nitrogen is present in the compound, fermenta- 
 tion or putrefaction may take place. Ammoniacal gas, NH 3 , 
 may be given off. 
 
 Chlorine, bromine, and iodine may produce decomposition. 
 New compounds are formed. 
 
 C 6 H 6 + C1 2 = C 6 H 5 C1 + HC1 
 
 Continuing the addition of chlorine, all of the hydrogen in 
 C 6 H 6 may be replaced, leaving C 6 C1 6 , hexachlorbenzol (per- 
 chlorbenzol). 
 
 2. Heat. 
 
 1. Some compounds when heated to a certain temperature 
 volatilize, and when cooled sublime unchanged ; examples : 
 strychnine, morphine, benzoic acid. 
 
 2. Other compounds when heated directly in the air undergo 
 decomposition (burn). If inorganic matter be absent, no resi- 
 due remains.
 
 NOTES ON CHEMICAL LECTURES. 31 
 
 3. When organic compounds are heated in a closed vessel 
 they undergo destructive distillation, as in the destructive dis- 
 tillation of coal in the manufacture of illuminating gas. Pyro- 
 acids may be formed as in the production of pyroligneous acid 
 in the destructive distillation of wood. 
 
 3. Acids. 
 
 1. Nitric acid. 
 
 a. If the organic compound be basic, it may combine with 
 it and form a salt, as 
 
 C^H^NAHNOg, strychnine nitrate. 
 
 b. It may effect the oxidation of the organic compound, as 
 
 Q.H^On -f 9O 2 = 6H 2 C 2 O 4 -f 5H 2 O 
 
 Cane sugar. Oxalic acid. 
 
 c. It may form substitution compounds, as 
 
 C 6 H 8 + HN0 3 H 2 + C 6 H 5 N0 2 
 
 Benzol. Nitrobenzol (oil of mirbane). 
 
 C 3 HA + 3HNO 3 C 3 H 3 (NO 2 ) 3 O 3 + 3H 2 O 
 
 Glycerine. Trinitroglycenue. 
 
 C, ; H 5 OH + 3HNO 3 = C 6 H 2 (NO 2 ) 3 OH + 3H 2 O 
 
 Phenol (carbolic acid). Trinitrophenol (picric acid). 
 
 2. Sulphuric acid. 
 
 a. If the organic compound be basic, it may combine with 
 it and form a salt, as 
 
 (C 17 H 19 NO 3 ) 2 H 2 SO 4 -j~ 5H 2 O, morphine sulphate. 
 (C^H^NA^risO, + 8H 2 O, quinine 
 
 b. It may decompose the organic compound, as 
 CH 2 2 + H 2 S0 4 = CO -I- H 2 + H 2 S0 4 
 
 Formic acid. Carbon monoxide. 
 
 c. It may abstract the elements of water from the organic 
 compound, as 
 
 H 2 C 2 O 4 + H 2 SO 4 = CO 2 + CO + H 2 O + H 2 SO 4 
 
 Oxalic acid. 
 
 d. It may introduce the elements of water into the organic 
 compound (assimilation of the elements of water), as 
 
 C 6 H 10 5 -f H 2 S0 4 + H 2 = C 6 H 12 6 + H 2 SO 4 
 
 Starch. Glucose.
 
 32 NOTES ON CHEMICAL LECTURES. 
 
 The property possessed by H 2 SO 4 of converting starch into 
 glucose is made use of in determining starch quantitatively, 
 /. e., converting the starch into glucose and determining the 
 quantity of the latter, and calculating the amount of starch 
 from the amount of glucose obtained. 
 
 C 6 H 5 OH, phenol (carbolic acid), treated with H 2 SO 4 forms 
 C, ; H 5 HSO 4 , phenol-sulphuric acid. 
 
 C 6 H 5 OH + H 2 SO 4 = C 6 H 5 HSO 4 + H 2 O 
 
 This acid (C 6 H-HSO 4 ) is formed when carbolic acid is in- 
 gested. 
 
 The salts of phenol-sulphuric acid are not poisonous. Anti- 
 dotes for carbolic acid, sodium sulphate, magnesium sulphate, 
 or any soluble non-poisonous sulphate. 
 
 4. Alkalies. 
 
 a. If the organic compound be an acid, alkalies will 'com- 
 bine with it and form salts, as 
 
 H 2 C 2 O 4 + 2NaOH == Na 2 C 2 O 4 -f 2H 2 O 
 
 Oxlic acid. Sodium oxalate. 
 
 b. Alkalies may cause combination of the elements in a 
 compound. 
 
 KOH -f CO = KCHO 2 
 
 Potassium formate. 
 
 c. If the compound acted upon by the alkali contain nitro- 
 gen directly combined, the nascent hydrogen evolved in the 
 decomposition combines with the nitrogen to form NH 3 
 ammonia. 
 
 The alkaline substance usually employed is soda-lime, com- 
 posed of 
 
 One part NaOH (caustic soda). 
 Two parts CaO (caustic lime). 
 
 PUTREFACTION AND FERMENTATION. 
 
 Putrefaction ; Chemical decomposition of nitrogenous or- 
 ganic compounds, under certain conditions, by bacteria, with 
 the evolution of more or less disagreeable odors. 
 
 Professor Sir Henry Roscoe, F. R. S., says, in connection 
 with the causation of the symptoms in infectious diseases, that

 
 NOTES ON CHEMICAL LECTURES. 33 
 
 " the symptoms of infectious diseases are no more due to the 
 microbes which constitute the infection than alcoholic intoxi- 
 cation is produced by the yeast-cell, but these symptoms are 
 due to the presence of definite chemical compounds, the result of 
 the life of these microscopic organisms." 
 
 Fermentation : Decomposition of certain non-nitrogenous 
 organic compounds in the presence of certain nitrogenized 
 substances known as fermentation fungi. 
 
 Ferment : A nitrogenous body capable of inducing fermen- 
 tation in a non-nitrogenous body. The yeast-cell is an 
 example of a ferment. 
 
 Ordinary yeast is composed principally of two varieties of 
 cells, 
 
 Torula cerevisiae (large round cells). 
 Penicilium glaucum (small oval cells). 
 
 Fermentescible body : A non-nitrogenous body capable 
 of undergoing fermentation. Glucose is an example of a 
 fermentescible bod^^ ^ f^^ , (J^u^ tuw 
 
 Amygdalin, aJferrnentesciDle body, is'broken $p by emu/sin, 
 a ferment (both being constituents of bitter almonds, peach- 
 kernels, etc.), into 
 
 + 2H 2 : : C 7 H 6 + HCN + 2C 6 H 12 O 6 
 
 Amygdalin. Oil of bitter almonds. Hydrocyanic acid. Glucose. 
 
 Certain conditions are necessary in a fermentation, 
 
 viz., the presence of a ferment, a fermentescible body, and 
 moisture, a certain temperature, 20 to 40 C. (70 to 100 F.). 
 Air must be present, at least at the beginning of the fermenta- 
 tion. The presence of a small quantity of salts of the alkaline 
 earths facilitates the progress of fermentation. 
 
 Fermentation may be prevented by metallic salts, as HgCl 2 , 
 CuSO 4 , etc., a temperature above 100 F. and below 70 F. 
 
 There are five varieties of fermentation, their distinctive 
 names being derived from the principal product furnished : 
 
 1 . Alcoholic, or vinous, in which alcohol is produced. 
 
 2. Acetous, " " acetic acid 
 
 3. Lactic, " " lactic acid 
 
 4. Butyric, " " butyric acid " 
 
 5. Viscous, " " a gummy matter "
 
 34 NO TES ON CHEMICAL LECTURES. 
 
 1. Alcoholic fermentation is fermentation characterized by 
 the formation of alcohol. It results from the action of yeast 
 on a solution of glucose. The active agent or ferment is the 
 torula cerevisice of the yeast. Only 95 per cent, of the glucose 
 is fermented. The remaining five per cent, is converted into 
 aldehydes, fat acids and perhaps other alcohols. 
 
 1 molecule. 2 molecules. 2 molecules. 
 
 180. 2 X 46 = 92. 2 X 44 = 88. 
 
 C 6 H 12 O 6 : : 2C 2 H 5 OH + 2CO 2 
 
 Glucose. Ethyl alcohol. Carbon dioxide. 
 
 Advantage is taken of this action to determine the quantity 
 of glucose in urine. 
 
 When the alcohol in the solution reaches 20 per cent., fer- 
 mentation ceases. A solution containing 25 per cent, of glucose 
 will not undergo fermentation. 
 
 2. Acetous fermentation is fermentation characterized by 
 the formation of acetic acid. It is an advanced stage of the 
 alcoholic fermentation. The active agent or ferment is the 
 mycodermce aceti. It appears to act as a carrier of oxygen. 
 
 C 2 H 5 OH + 2 = C 2 H 4 2 + H 2 
 
 Ethyl alcohol. Acetic acid. 
 
 3. Lactic acid fermentation is fermentation- characterized 
 by the formation of lactic acid. It results from the action of 
 putrefying cheese or milk on glucose or milk sugar. The 
 active agent or ferment is the penicilium glaucum. 
 
 CuN^Ou + H 2 = 4C 3 H 6 3 
 
 Milk sugar. Lactic acid. 
 
 C 6 H 12 O 6 = 2C 3 H 6 O 3 
 
 Glucose Lactic acid. 
 
 Milk sugar and lactose C^H^On + H 2 O are synonymous terms 
 for the same compound. Galactose, C 6 H 12 O 6 , is derived from 
 milk sugar. It results from boiling milk sugar with dilute 
 sulphuric acid. 
 
 4. Butyric acid fermentation is fermentation characterized 
 by the formation of butyric acid. It is an advanced stage of 
 the lactic acid fermentation. The ferment is the penicilium 
 glaucum, the same as in lactic acid fermentation. 
 
 2C 3 H 6 3 = C 4 H 8 2 + 2C0 2 + 2H 2 
 
 Lactic acid. Butyric acid.
 
 (J
 
 NOTES ON CHEMICAL LECTURES. 35 
 
 5. Viscous fermentation is a fermentation characterized by 
 the formation of gummy matters. It occurs in the fermenta- 
 tion of the juice of the sugar-beet, and also in sweet white 
 wines, the liquid becoming " ropy." It may be arrested by 
 the addition of a little alum or calcium sulphite. It does not 
 occur in red wines because of the presence of astringent sub- 
 stances. The particular ferment causing this fermentation is 
 unknown. 
 
 ORGANIC ANALYSIS. 
 
 PROXIMATE AND ULTIMATE. 
 
 Proximate analysis : The separation and determination 
 of the organic compounds contained in an organic body, as 
 the separation of morphine, etc., from opium. 
 
 Ultimate analysis : The detection and determination of 
 the ultimate elements entering into the composition of an 
 organic compound, as the quantity of C, H, O, and N in 
 morphine. 
 
 That the compound is organic may be shown by heating it 
 on platinum foil ; if it chars it is organic. Some compounds 
 volatilize before the temperature of the charring-point is 
 reached, and others undergo decomposition without charring. 
 Such compounds must be heated in a sealed glass tube or 
 with cupric oxide. If CO 2 or H 2 O are given off when the 
 compound is heated with cupric oxide it is organic. 
 
 Organic compounds may be composed of C and H or C and 
 N ; C, H, and O ; C, H, N, and O ; C, H, N, O, and S ; C, H, 
 N, O, S, and P. 
 
 Compounds artificially prepared may contain Cl, Br, I, As, 
 Sb, etc. 
 
 QUALITATIVE ANALYSIS. 
 
 Presence of C and H : Shown by the compound charring 
 when heated alone, or producing CO 2 and H 2 O when heated 
 with cupric oxide. 
 
 Presence of nitrogen : a. Many compounds containing 
 nitrogen, when burned, evolve an odor similar to that of burnt 
 feathers.
 
 36 NOTES ON CHEMICAL LECTURES. 
 
 b. Many compounds containing nitrogen, when heated 
 with an alkali or with soda-lime, give rise to the formation 
 of ammoniacal gas (NH 3 ), which may be detected by its 
 odor, and, when in solution, by its forming a precipitate of 
 (NH 4 Cl) 2 PtCl 4 on the addition of platinic chloride. 
 
 c. To detect nitrogen in other compounds, they are heated 
 with a small piece of metallic potassium or sodium, thereby 
 forming cyanogen, which combines with the potassium or 
 sodium forming a cyanide, the residue is dissolved in water 
 and the solution tested for a cyanide with FeSO 4 -f Fe 2 Cl 6 + 
 HC1, = formation of prussian blue. 
 
 Presence of sulphur : a. If the compound be a solid, it 
 is heated with a mixture of solid potassium hydroxide (KOH) 
 (twelve parts) and solid potassium nitrate (KNO 3 ) (six parts) : 
 the sulphur is oxidized to sulphuric acid (which combines 
 with the potassium to form K 2 SO 4 ). The fused mass is dis- 
 solved in water and tested with barium chloride for a sulphate. 
 
 b. If the compound be a liquid, it is boiled with nitric acid 
 alone or with potassium chlorate : the sulphur is oxidized to 
 sulphuric acid. The liquid is tested with barium chloride for 
 a sulphate. 
 
 If the compound be a volatile liquid, it is heated in a 
 sealed glass tube with about twenty to thirty times its 
 volume of nitric acid and the liquid is diluted with water and 
 tested for a sulphate. 
 
 c. To determine if the sulphur in the compound is directly 
 (unoxidized) or indirectly (oxidized) combined with the carbon, 
 the compound is heated with a solution of potassium hydrox- 
 ide (KOH). If the sulphur is directly combined, a sulphide 
 (K 2 S) will be formed. The solution is tested for a sulphide 
 with lead acetate, Pb(C 2 H 3 O 2 ) 2 , black lead sulphide (PbS) will 
 be formed, or tested for a sulphide with sodium nitroprusside 
 (Na 2 NOFe(CN) 5 ), with which an intense purple-red color is 
 produced. If the sulphur is indirectly combined, the solution 
 will not respond to the tests for a sulphide, but in some cases 
 may respond to the tests for a sulphate. 
 
 Presence of phosphorus : a. The substance is fused with 
 the mixture before stated of KOH and KNO 3 , (a) or is boiled 
 with nitric acid, as in the case of sulphur, and the aqueous
 
 ^*S <r tfay a^ 
 
 iMu^.._, 
 
 a. 
 
 a
 
 NOTES ON CHEMICAL LECTURES. 37 
 
 solution tested with ammonium chloride, ammonium hydrox- 
 ide, and magnesium sulphate (magnesia mixture) for a phos- 
 phate (formation of a crystalline precipitate of MgNH 4 PO 4 ). 
 
 b. If the organic compound be a volatile liquid, it is heated 
 in a sealed glass tube with about twenty to thirty times its 
 volume of nitric acid and the liquid is diluted with water and 
 tested for a phosphate. 
 
 Presence of inorganic matter ; a. The substance is 
 heated on platinum foil, thereby burning off the organic 
 matter, and leaving the inorganic matter as a fixed residue. 
 
 Presence of chlorine, iodine, or bromine : a. The 
 organic compound is mixed with caustic lime (CaO), and 
 heated in a combustion-tube. The mixture is suspended in 
 water, slightly acidulated with nitric acid, filtered, and the 
 filtrate tested with argentic nitrate for a chloride, iodide or 
 bromide. 
 
 QUANTITATIVE ORGANIC ANALYSIS. 
 
 Ultimate or elementary analysis : 
 Carbon is determined as CO 2 . 
 Nitrogen " " " NH 3 or as N. 
 
 23'{ 32 
 
 Sulphur " " " SO 3 , (BaSO 4 = S). 
 
 222 62 
 
 Phosphorus" " " P 2 O 5 , (Mg 2 P 2 O 7 = P 2 ). 
 
 Hydrogen " " " H 2 O. 
 
 Oxygen is determined by difference, /. e., after the per- 
 centages of the elements in the compound have been deter- 
 mined the percentages are added together, and if the result 
 does not foot up 100, the difference between the footing and 
 100 is ascribed to oxygen. 
 
 Conditions to be observed in ultimate analysis: 1. 
 The compound to be analyzed must be pure and dry. A crys- 
 talline substance maybe purified by repeated recrystallizations. 
 The compound may be dried by allowing it to remain some 
 time over sulphuric acid or calcium chloride in a desiccator, or 
 it may be dried in a hot-water or air oven. When two weigh- 
 ings agree with each other, /. e., the compound ceases to lose 
 weight, it is considered dry. 
 
 2. The compound must be completely burned.
 
 38 NOTES ON CHEMICAL LECTURES. 
 
 3. The products of the combustion must be accurately col- 
 lected and weighed or measured. 
 
 Requisites for an elementary analysis : 1. A combus- 
 tion-tube of difficultly fusible Bohemian glass, about eighteen 
 inches in length and drawn out at one end in a bayonet-like 
 form and sealed at the drawn out end. 
 
 2. A combustion-furnace. 
 
 3. An aspirator. 
 
 4. A gas-holder containing air or oxygen. 
 
 5. Cupric oxide (CuO) or fused granular lead chromate 
 (PbCrO 4 ) to furnish oxygen during the progress of the com- 
 bustion. 
 
 6. A U-shaped tube containing calcium chloride to absorb 
 the H 2 O formed in the combustion. 
 
 7. Geissler's or Liebig's bulbs containing a strong solution 
 of potassium hydroxide to absorb the CO 2 formed in the com- 
 bustion. 
 
 Method. Determination of carbon and hydrogen : A 
 Bohemian glass combustion-tube is about half filled with 
 freshly heated, perfectly dry, granular cupric oxide (or lead 
 chromate). The accurately weighed organic compound to be 
 analyzed is placed in the tube on top of the cupric oxide, some 
 fine cupric oxide added, and the compound thoroughly mixed 
 with the cupric oxide by means of a copper wire terminating 
 in a spiral. The tube is then filled to within nearly an inch of 
 the end with more granular cupric oxide, a plug of loose 
 asbestos inserted, the tube placed on a combustion-furnace, 
 and a previously weighed tube, containing calcium chloride in 
 small pieces to absorb the H 2 O produced in the combustion, 
 is attached to it by means of a closely-fitting perforated 
 rubber stopper. Previously weighed Geissler's or Liebig's 
 bulbs, containing a solution of KOH (specific gravity 1.27) to 
 absorb the CO 2 formed in the combustion, having a tube 
 attached containing small pieces of KOH (previously weighed 
 with the Geissler's bulbs) to absorb the last traces of CO 2 , and 
 also to hold any moisture that might be carried over from the 
 KOH solution in the bulbs by the current of gas, are attached 
 to the calcium chloride tube. 
 
 The combustion-tube is heated first at the end to which the 
 calcium chloride tube is attached ; when the cupric oxide in
 
 NO TES ON CHEMICAL LECTURES. 39 
 
 this part of the tube is of a dull red heat the heating is com- 
 menced at the other end of the tube, and continued until the 
 cupric oxide in that part is also of a dull red heat. The heat 
 is then gradually extended to the middle of the tube until 
 finally the whole tube is heated. The heating must be gradual, 
 so that the combustion is not too rapid, or some of the products 
 may pass through the absorption-bulbs unabsorbed. When 
 the combustion is completed the liquid in the bulb of the 
 Geissler's bulbs nearest the calcium chloride tube will ascend. 
 An aspirator is attached to the Geissler's bulbs, and a rubber 
 tube leading from a drying apparatus is attached to the bayonet- 
 like end of the combustion-tube, the end of the glass tube 
 broken off while in the rubber tube, and the aspirator started. 
 Oxygen or air, free from CO 2 or H 2 O, is slowly drawn from the 
 gas-holder through the combustion-tube to burn any of the 
 organic compound which may have escaped decomposition, 
 and also to convey any CO 2 or vapor of H 2 O remaining in the 
 combustion-tube into the Geissler's bulbs and calcium chloride 
 tube. The drying apparatus through which the oxygen or air 
 is caused to pass before entering the combustion-tube is com- 
 posed of a series of three cylinders, the first containing a solu- 
 tion of KOH to absorb CO 2 ; the second and third containing 
 respectively sulphuric acid and pieces of calcium chloride to 
 absorb H 2 O. After drawing oxygen or air through the tube a 
 few minutes the flames are extinguished, the Geissler's bulbs 
 and the calcium chloride tube are detached (the openings stop- 
 pered) and allowed to cool. When cool they are unstoppered 
 and weighed. The increase in weight of the Geissler's bulbs 
 containing the KOH will indicate the quantity by weight of 
 CO 2 which resulted from the combustion of the carbon in 
 the organic substance employed. 
 
 The increase in weight of the tube containing calcium 
 chloride will indicate the quantity by weight of H 2 O which 
 resulted from the combustion of the hydrogen in the organic 
 substance employed. 
 
 The quantity of carbon is calculated from the weight of CO 2 
 obtained, as follows : 
 
 44 12 
 
 CO 2 : C :: wt. of CO 2 obtained : X
 
 40 NOTES ON CHEMICAL LECTURES. 
 
 The quantity of hydrogen is calculated from the weight of 
 H 2 O obtained, as follows : 
 
 18 2 
 
 H 2 O : H 2 :: wt. of H 2 O obtained : X 
 
 If the compound for analysis be a liquid, it is placed in a 
 small weighed glass bulb having a drawn-out tube. This is 
 accomplished by rarefying the air in the bulb by heating and 
 holding the drawn-out end in the liquid. On cooling, the 
 liquid will ascend and occupy the space of the expelled air. 
 The tube is then sealed over a Bunsen flame and the bulb 
 weighed. The increase of weight is the weight of the com- 
 pound contained in the bulb. The bulb containing the liquid 
 is dropped into the combustion-tube containing the cupric 
 oxide, and the method as before described employed. 
 
 If the compound to be analyzed contain nitrogen in addi- 
 tion to the carbon and hydrogen or oxygen, oxides of nitrogen 
 may be formed during the combustion, and these oxides being 
 absorbed with the CO 2 by the KOH solution in the Geissler's 
 bulbs, would lead to inaccurate results by apparently increasing 
 the quantity of CO 2 . In this case, metallic copper (in the 
 form of turnings or gauze) is heated in a current of hydrogen 
 and then placed in the combustion-tube, at the end at which 
 the calcium chloride tube is attached, in order to decompose 
 the oxides of nitrogen into free oxygen and nitrogen. 
 
 If sulphur be present in the compound, SO 3 (sulphuric 
 anhydride) may be formed in the combustion and be absorbed 
 with the CO 2 by the KOH in the Geissler's bulbs. In this 
 case, PbO 2 (lead dioxide) should be placed in the combustion- 
 tube containing the cupric oxide, at the end at which the 
 calcium chloride tube is attached. The SO 3 will combine 
 with, and be held by, the lead as PbSO 4 . Lead dioxide need 
 not be placed in the tube if lead chromate instead of cupric 
 oxide is used in the combustion. 
 
 Compounds containing chlorine, iodine, or bromine when 
 burned with cupric oxide form cupric chloride,, iodide, or 
 bromide, which volatilize at a high temperature and condense 
 in the calcium chloride tube, thus causing a fictitious increase 
 in the weight of the H 2 O. Compounds containing these
 
 NOTES ON CHEMICAL LECTURES. 41 
 
 elements must be burned with lead chromate (PbCrO 4 ) ; non- 
 volatile plumbic chloride, iodide, or bromide will be formed 
 and retained in the combustion-tube. 
 
 Compounds difficult of combustion should be burned with 
 lead chromate or with pure oxygen. The latter is employed 
 especially for coals and coke. The substance to be burned 
 is placed in a porcelain boat, and, instead of glass, an iron 
 combustion-tube is often employed. 
 
 DETERMINATION OF NITROGEN METHODS : 
 
 1. Dumas' method: Depends upon the decomposition of 
 the organic substance with the evolution of all of the nitrogen, 
 which is collected and measured in a eudiometer. 
 
 The nitrogenous compound is burned in a combustion-tube, 
 sealed at one end, with cupric oxide and copper-wire gauze as 
 in the determination of carbon and hydrogen, except that a 
 layer of about three inches of acid sodium carbonate (NaHCO 3 ) 
 or of magnesite (magnesium carbonate, MgCO 3 ) is placed at 
 the closed end of the tube. When filled the tube contains 
 the following substances in the given order: 
 
 1. NaHCO 3 or MgCO 3 (to the depth of about three inches). 
 
 2. Cupric oxide (nearly to the middle of the tube). 
 
 3. Mixture of organic substance and cupric oxide. 
 
 4. Cupric oxide. 
 
 5. Copper gauze to decompose oxides of nitrogen. 
 
 A delivery-tube is attached to the open end of the combus- 
 tion-tube (the exit of the delivery-tube being brought below 
 the surface of mercury contained in a trough), and a glass 
 tube, containing mercury and a solution of potassium or 
 sodium hydroxide, is inverted over the exit of the delivery- 
 tube. The acid sodium carbonate, or the magnesium carbonate 
 is heated first. Decomposition takes place and CO 2 is evolved. 
 
 2NaHCO 3 = Na 2 CO 3 -f H 2 O + CO 2 
 MgC0 3 = MgO + C0 2 
 
 The heating of the carbonate is continued until all the 
 air is driven out of the combustion-tube by CO 2 . This is
 
 42 NO TES ON CHE MIC A L L E C 7 URES. 
 
 accomplished when the bubbles of evolved gas are com- 
 pletely absorbed by the KOH solution. 
 
 When all the air is driven out, the heating of the carbonate 
 is discontinued, and the KOH tube is removed. A eudiometer, 
 containing mercury and a layer (about three or four inches in 
 thickness) of solution of KOH, is now inverted over the exit 
 of the delivery-tube. Heat is applied to the end of the tube 
 (No. 5) containing the copper gauze, and also to the part con- 
 taining cupric oxide (No. 2) next to the carbonate. The heat 
 is gradually extended to the middle of the tube, until the entire 
 tube, except (No. 1) carbonate part, is heated. The heating 
 is discontinued when gas-bubbles cease to come over into the 
 eudiometer. At this time the acid sodium carbonate is again 
 heated, more CO 2 is evolved which drives out the remaining 
 nitrogen. The heating is continued until the gas-bubbles 
 entering the eudiometer are completely absorbed by the KOH. 
 
 The CO 2 and H 2 O produced in the combustion are absorbed 
 by the solution of KOH. The nitrogen is not absorbed, and 
 collects in the upper part of the eudiometer. 
 
 The eudiometer containing the nitrogen is transferred to a 
 vessel of water, the atmospheric pressure is equalized, and the 
 volume of nitrogen read off and corrected for temperature 
 and pressure. 
 
 14 grm. nitrogen = 11160 c.c. 
 11160 : 14 :: 1 c.c. : 0.001256 grm. 
 1 c.c. of nitrogen at C. and 760 mm. weighs 0.001256 grm. 
 
 2. Will and Varrentrapp's method : Depends upon the 
 formation of NH 3 (ammoniacal gas) when an organic sub- 
 stance containing nitrogen is heated with soda-lime. 
 
 This method is not applicable for the determination of nitro- 
 gen in compounds in which the nitrogen is present as a nitro 
 group. 
 
 A combustion-tube is nearly half-filled with dry soda-lime. 
 The weighed organic substance is added and thoroughly 
 mixed, by means of a wire stirrer, with the soda-lime. More 
 soda-lime is added until the tube is within about an inch of 
 being filled, and a plug of loose asbestos is placed in the end.
 
 NOTES ON CHEMICAL LECTURES. 43 
 
 The tube now contains the following substances in the fol- 
 lowing order : 
 
 1. Soda-lime. 
 
 2. Mixture of organic substance and soda-lime. 
 
 3. Soda-lime. 
 
 4. Plug of loose asbestos. 
 
 a. Gravimetric determination of nitrogen. 
 
 The combustion-tube is connected with Will's bulbs, con- 
 taining dilute hydrochloric acid, and is heated in a combustion- 
 furnace. The heating should be commenced simultaneously at 
 each end of the tube, and gradually extended to the middle. 
 
 The NH 3 is absorbed by the HC1 in the Will's bulbs, forming 
 NH 4 C1. 
 
 The contents of the Will's bulbs are emptied into a dish ; 
 excess of platinum chloride, which precipitates the ammonia, 
 is added, and the whole evaporated to dryness on a water-bath. 
 The residue is collected on a weighed filter by means of a 
 mixture of alcohol and ether, and thoroughly washed with a 
 mixture of the same liquid. 
 
 When the precipitate, (NH 4 Cl) 2 PtCl 4 , is dry it is weighed 
 while on the filter. Every 100 parts of (NH 4 Cl) 2 PtCl 4 con- 
 tains 6.318 parts of nitrogen. 
 
 (NH 4 CI) 2 PtC] 4 . N 2 . 
 
 443.3 : 28 :: 100 : 6.318 
 or the quantity of nitrogen may be calculated as follows : 
 
 (NH 4 Cn 2 P'Cl 4 . N 2 . 
 
 443.3 : 28 : weight of precipitate : X 
 
 Instead of weighing the precipitate as (NH 4 Cl) 2 PtCl 4 ,*it may 
 be incinerated and weighed as metallic platinum ; then 
 
 Every 100 parts of platinum are equivalent to 14.410 parts 
 of nitrogen. 
 
 Pt. N 2 . 
 
 194.3 : 28 :: 100 : 14.410 
 or the quantity of nitrogen may be calculated as follows : 
 
 Pt. N 2 . 
 
 194.3 : 28 :: weight of metallic platinum : X 
 
 6
 
 44 NOTES ON CHEMICAL LECTURES. 
 
 b. Volumetric determination of nitrogen. 
 
 A measured volume of a normal solution of oxalic acid 
 (H 2 C 2 O 4 + 2H 2 O) may be used in the Will's bulbs, instead of 
 dilute HC1, to absorb the NH 3 , and the quantity of nitrogen 
 calculated from the number of cubic centimetres of oxalic 
 acid solution, neutralized by the NH 3 , as determined by titer- 
 ing with a normal solution of NaOH. 
 
 H 2 C 2 O 4 -f 2H.,O, a dibasic acid. 
 
 126 -63 
 ~2~ 
 
 H 2 C 2 O 4 + 2H 2 O + 2NH 3 = (NH 4 ) 2 C 2 O 4 + 2H 2 O 
 
 Oxalic acid. NH 3 . 
 
 126 grm. = 34 grm. 
 63 " =-17 " 
 
 A normal solution of oxalic acid is prepared by dissolving 
 63 grm. of pure oxalic acid in 1000 c.c. water ; then 
 
 Oxalic acid. NH 3 . N. 
 
 1000 c.c. = 63.0 grm. == 17.0 grm. == 14.0 grm. 
 1 c.c. = 0.063 " = 0.017 " = 0.014 " 
 
 10 c.c. of the oxalic acid solution are placed in the Will's 
 bulbs. The combustion of the organic compound is performed 
 and the NH 3 evolved is absorbed by the oxalic acid solution, 
 
 When the combustion is completed the oxalic acid solution 
 in the Will's bulbs is emptied into a beaker, the bulbs rinsed 
 with water and the wash water also placed in the beaker, a 
 few drops of litmus added, and the oxalic acid solution titered 
 with a normal solution of NaOH. 
 
 1000 c.c. of normal sol. of NaOH contains 40.0 grm. NaOH. 
 1 c.c. " " " " " 0.040 " 
 
 0.040 NaOH = 0.063 oxalic acid. 
 
 Consequently 1 c.c. normal solution of NaOH is equal to 
 (will neutralize) 1 c.c. normal oxalic acid solution, and 10 c.c. 
 normal NaOH solution are equal to 10 c.c. normal oxalic acid 
 solution. 
 
 On titering the 10 c.c. oxalic acid solution from the Will's 
 bulbs, it will now be found that less than 10 c.c. of normal 
 
 *
 
 NOTES ON CHEMICAL LECTURES. 45 
 
 NaOH solution will be required, showing that some of the 
 oxalic acid has been neutralized by the NH 3 . 
 
 Suppose only 6 c.c. of normal NaOH solution were required 
 to neutralize the 10 c.c. oxalic acid solution from the Will's 
 bulbs, then the difference between 6 and 10 = 4 indicates the 
 number of cubic centimetres of oxalic acid solution neutralized 
 by the NH 3 which resulted from the combustion of the organic 
 compound. 
 
 1 c.c. of oxalic acid solution is neutralized by 0.017 NH 3 , 
 equivalent to 0.014 N, and as 4 c.c. were neutralized then 
 4 X 0.014 = 0.056 grm. nitrogen present in the weight of 
 organic compound analyzed. 
 
 3. Kjeldahl's method: Depends upon the conversion of 
 the nitrogen in a compound into ammonia, by boiling it with 
 sulphuric acid. The ammonia combines with the sulphuric 
 acid to form ammonium sulphate. The ammonium sulphate, 
 (NH 4 ) 2 SO 4 , is decomposed into NH 3 and Na 2 SO 4 by boiling it 
 with sodium hydroxide. The evolved ammoniacal gas (NH 3 ) 
 is collected in dilute hydrochloric acid and precipitated with 
 platinum chloride as (NH 4 Cl) 2 PtCl 4 . 
 
 The precipitate is collected on a filter, washed with a mix- 
 ture of alcohol and ether, and treated exactly as in Will and 
 Varrentrapp's method. 
 
 The NH 3 may also be collected in a measured volume of 
 normal oxalic acid solution and titered with a normal solution 
 of sodium hydroxide, as in Will and Varrentrapp's method. 
 
 DETERMINATION OF SULPHUR. 
 
 Sulphur in an organic compound is determined quantita- 
 tively by fusing the compound with about 12 parts of potas- 
 sium hydroxide (KOH) and 6 parts of potassium nitrate 
 (KNO 3 ), or by boiling the compound with nitric acid, as in 
 the qualitative detection of sulphur. The sulphur is oxidized 
 to sulphuric acid, the latter is then precipitated with barium 
 chloride (BaCl 2 ) as barium sulphate (BaSO 4 ), and the quantity 
 of sulphur is calculated from the amount of BaSO 4 obtained. 
 233 parts of BaSO 4 (molec. wt.) = 32 parts of S. 
 
 BaSO 4 . S. 
 
 233 : 32 : : wt. of precipitate (BaSO 4 ) : x
 
 46 NOTES ON CHEMICAL LECTURES. 
 
 DETERMINATION OF PHOSPHORUS. 
 
 Phosphorus is determined quantitatively by fusing the 
 compound with 1 2 parts of KOH and 6 parts of KNO 3 , or by 
 boiling it with HNO 3 , as in the determination of sulphur. 
 The phosphorus is oxidized to phosphoric acid, which is 
 precipitated by the magnesia mixture (NH 4 C1 -j- NH 4 HO -f- 
 MgSO 4 ), and weighed as magnesium pyrophosphate, Mg 2 P 2 O 7 . 
 
 222 parts of Mg 2 P 2 O 7 (molec. wt.) = 62 parts of P. 
 
 MgP 2 7 . P. 
 
 222 : 62 :: wt. of magnesium pyrophosphate : x 
 
 DETERMINATION OF CHLORINE, BROMINE, AND IODINE. 
 
 Chlorine, bromine, and iodine are determined quantita- 
 tively by heating the compound in which the element is con- 
 tained, in a combustion-tube with caustic lime, as in the 
 qualitative detection of these elements, dissolving in water, 
 filtering, neutralizing with HNO 3 and precipitating with argentic 
 nitrate (AgNO 3 ), and weighing as argentic chloride (AgCl), 
 argentic bromide (AgBr), and argentic iodide (Agl). 
 
 CALCULATION OF RESULTS AND DEDUCTION OF 
 FORMULAS. 
 
 Rule : Determine the percentages of all the elements in the 
 compound. Divide the percentage of each element by its 
 atomic weight. Then divide the quotients obtained by the 
 smallest quotient, and, if necessary, multiply these final 
 quotients by the least number that will make all of them 
 whole or nearly whole numbers. The formula thus obtained 
 is the empirical formula, and at the same time may be the 
 molecular formula. The molecular formula is determined by 
 observing the vapor density of the compound, etc. 
 
 Example : A combustion of 0.340 grm. of oil of turpen- 
 tine, a compound containing only carbon and hydrogen, 
 furnished 
 
 CO 2 1.100 grm. 
 
 H 2 O . 0.360 "
 
 NOTES ON CHEMICAL LECTURES. 47 
 
 CO,. C. Wt. ofCO 2 . C. 
 
 then, a. 44 : 12 :: 1.100 : 0.300 grm. 
 
 HoO. H.,. Wt.ofH 2 O. H. 
 
 18 : 2 :: 0.360 : 0.040 grm. 
 Hence 0.340 grm. of the oil contained 
 
 0.300 carbon. 
 0.040 hydrogen. 
 
 Oil. C. 
 
 b. 0340 : 0.300 :: 100 : 88.235 per cent, carbon. 
 
 Oil. H. 
 
 0.340 : 0.040 :: 100 : 11.765 per cent, hydrogen. 
 The percentage composition is, 
 
 Carbon 88.235 
 
 Hydrogen 11 765 
 
 100.000 
 
 c. 88.235 H- 12 = 7.353 -*- 7.353 =1x5 = 5 atoms C. 
 11.765 -H 1 = 11.765 -H 7.353 = 1.599 X 5 = 7.995 atoms H, 
 or, approximately, 8 atoms of H. 
 
 Hence the empirical formula of oil of turpentine is C 5 H S . 
 (The molecular formula is twice C 3 H 8 , or C 10 H 16 .) 
 Or, if 0.340 grm. oil of turpentine furnished 
 
 CO 2 *. . . 1.100 grm. 
 
 H 2 O 0.360 " 
 
 Oil. CO 2 . COo. 
 
 then, a. 0.340 : 1.100 :: 100 : 323.53 grm. 
 
 Oil. Hj-O. H 2 O. 
 
 0.340 : 0.360 :: 100 : 105.88 grm. 
 
 If 100 grm. of the oil yields 323.53 grm. CO 2 and 105.88 
 grm. H 2 O, then to obtain the percentage proportions of carbon 
 and hydrogen, 
 
 co.,. c. 
 b. 44 : 12 :: 323.53 : 88.235 per cent, carbon. 
 
 H 2 O. H 2 . 
 
 18 : 2 :: 105.88 : 11.765 percent, hydrogen.
 
 48 NOTES ON CHEMICAL LECTURES. 
 
 And from these percentages of carbon and hydrogen we find, 
 
 ^.88.235-^-12= 7.353 -f- 7,353 = 1 X 5 = 5 atoms C. 
 11.765-- 1 = 11. 765 -T- 7.353 = 1.599 X 5 = 7.995 atoms H, 
 
 or, approximately, 8 atoms of H. 
 
 Hence the empirical formula for the compound is C 5 H S . 
 
 Oxalic acid is composed of, 
 
 Carbon 26.66 per cent. 
 
 Hydrogen 2.22 
 
 Oxygen 71.12 
 
 100.00 
 Then, 
 
 C 26.66 ~l'2 = 2.22 -*- 2.22 = 1 atom carbon. 
 H 2.22 -f- 1 = 2.22 ~- 2.22 = 1 " hydrogen. 
 O 71.12 -4- 16 = 4.44 -4- 2 22 = 2 atoms oxygen. 
 
 Hence the empirical formula of oxalic acid is HCO 2 . (The 
 molecular formula is twice HCO 2 , or H 2 C 2 O 4 .) 
 
 Composition of cane sugar according to Liebig, 
 
 Carbon 42.30 = 1.001 
 
 Hydrogen ....... 6.45 = 2.009 
 
 Oxygen 51.25 = 1 OOP 
 
 100.00 
 
 Formula for cane sugar according to different chemists, 
 
 Berzelius . C^H^On 
 
 Doebereiner C 6 H 12 O 6 
 
 Dumas C 5 H 10 O 5 
 
 Prout C 8 H 16 O 8 
 
 THE URINE. 
 
 The urine is the secretion of the kidneys, in which effete 
 nitrogenized products are thrown out from the system. 
 
 The color of normal urine is yellow. The intensity of color 
 is proportionate to the specific gravity, except diabetic urine, 
 in which the specific gravity may be high and the color pale 
 yellow.
 
 NOTES ON CHEMICAL LECTURES. 49 
 
 The reaction of normal urine just voided is usually acid. 
 The acid reaction is not due to the presence of free acid but to 
 acid salts, principally to acid sodium phospJiate (NaH 2 PO 4 ) 
 arising from the sodium phosphate (Na 2 HPO 4 ) of the blood 
 coming in contact with the uric acid of the urine. 
 
 2Na 2 HPO 4 + H 2 C 5 H 2 N 4 O 3 = 2NaH 2 PO 4 + Na 2 C 5 H 2 N 4 O 3 
 
 Neutral sodium phosphate. Uric acid. Acid sodium phosphate. Neutral sodium urate. 
 
 Urine having an alkaline reaction may be voided, especially 
 after a very hearty meal, and also after the ingestion of alkalies 
 and substances, as the acetates, tartrates, and citrates (lemon- 
 juice) of the alkalies, which in the organism are converted 
 into and voided as alkaline carbonates. 
 
 On the ingestion of acids, urine having a strong acid reac- 
 tion may be voided. 
 
 Normal urine, on standing some time, may become alkaline 
 in reaction, due to the presence of ammonium carbonate 
 resulting from the action of certain micro-organisms (micro- 
 coccus ureae) on the urea. 
 
 CO(NH 2 ) 2 + 2H 2 = (NH 4 ) 2 C0 3 
 
 Urea. Ammon. carbonate. 
 
 Normal urine may have an amphoteric reaction, /. e., 
 changing blue litmus-paper to red, and red litmus to blue, 
 due to the simultaneous presence in the urine of acid and 
 neutral phosphates. 
 
 In collecting urine for a quantitative determination of its 
 constituents, the total quantity voided in 24 hours should be 
 collected and, that the liquid collected shall be homogeneous 
 in constitution, the separate portions collected during the 24 
 hours must finally be placed in one bottle and thoroughly 
 mixed. 
 
 In collecting the urine passed in 24 hours the collection 
 should be begun at a certain time. For example, at 8 a. m. of 
 one day the bladder should be emptied and this portion of 
 urine discarded. The urine voided from this time until 8 
 a. m. the next day (at which time the bladder should be finally 
 emptied), should be collected in a bottle, including that voided 
 in the final emptying of the bladder. During the period of 
 collection the urine, voided should be kept in a cold place. 

 
 50 NOTES ON CHEMICAL LECTURES. 
 
 Acidity of the urine : The acidity of the average quantity 
 of urine voided in a day will require from 1.0 to 1.5 grm. 
 NaOH to neutralize it. 
 
 The average quantity of sodium acid phosphate (NaH 2 PO 4 ) 
 daily eliminated in the urine is equivalent to about 2.3 grm. 
 of phosphoric anhydride (P 2 O 5 ), and to convert 2.3 grm. P 2 O 5 
 as NaH 2 PO 4 into Na 2 HPO 4 will require 1.3 grm. NaOH. 
 
 The acidity of the urine is determined by means of a deci- 
 normal solution of sodium hydroxide (NaOH). 
 
 A normal solution of NaOH contains 40.0 grm. of NaOH 
 in 1000 c.c. distilled water. A deci-normal solution contains 
 T V of 40.0, or 4.0 grm. of NaOH in 1000 c.c. of distilled 
 water. 
 
 NaOH. 
 
 1000 c.c. == 4.0 grm. 
 1 c.c. = 0.004 " 
 
 Method : To 100 c.c. of urine, two or three drops of rosolic 
 acid, or of phenol-phthalein solution are added, a deci-normal 
 solution of NaOH is run in from a burette until a change 
 takes place in the color of the indicator (rosolic acid, or phenol- 
 phthalein), the number of c.c. NaOH solution required to 
 produce the change in color is read off, and this number is 
 multiplied by the value of 1 c.c. of the deci-normal solution 
 expressed in NaOH units, namely, 0.004. 
 
 Example : 24 hours urine = 1200 c.c. 
 
 100 c.c. of the urine required 30. c.c. NaOH solution. 
 
 30 X 0.004 = 0.120 grm. NaOH required for 100 c.c. urine. 
 
 12 X 0.120 = 1.440 grm. NaOH required to neutralize the 
 
 acidity of 1200 c.c. urine (quantity of urine voided in 24 
 
 hours). 
 
 Specific gravity of urine : The specific gravity of normal 
 urine varies from 1015 to 1025. The average is about 4017, 
 water taken as 1000. 
 
 The specific gravity may be as low as 1003 or even lower, 
 and as high as 1030 or higher. 
 
 The specific gravity may be determined by 
 
 a. Urinometer. 
 
 b. Specific gravity bottle (picnometer).

 
 NOTES ON CHEMICAL LECTURES. 51 
 
 The specific gravity is affected by temperature. It should 
 be taken with the liquid at a temperature of 15 C. or 
 60 F. 
 
 The specific gravity is lowered one unit (the specific gravity 
 of water expressed as 1000) for every increase of 3 C. tem- 
 perature. 
 
 Determination of quantity of solid matter in urine ; 
 1. A ready method is to evaporate 10 c.c. urine, in a weighed 
 dish, to dryness on a water-bath. The residue is further dried 
 in a drying-oven at 100 C.,and placed in a dessicator. After 
 a time it is weighed, and the drying operation and weighing 
 are repeated until a constant weight is obtained. The final 
 weight, minus the weight of the dish, is the amount of solid 
 matter in 10 c.c. of the urine. 
 
 This method is not very accurate, because during the evapo- 
 ration some of the urea is decomposed with the evolution of 
 ammoniacal gas (NH 3 ), thus 
 
 a. CO(Nfi -f H 2 O = 2NH 3 + CO 2 
 
 b. NH 3 4 CO 2 + H 2 O = NH 4 HCO 3 
 
 c. NaH 2 PO 4 + NH 3 "= NaNH 4 HPO 4 
 
 d. NaNH 4 HPO 4 at 100 C. = NaH 2 PO 4 + NH 3 
 
 2. An accurate method is to evaporate to dryness 2 c.c. 
 urine in a weighed porcelain boat in Neubauer's special drying 
 apparatus, collecting the NH 3 which results from the decom- 
 position of some of the urea and finally weighing the boat 
 with the dry residue. The NH 3 evolved from the decom- 
 position of the urea in the process of drying is collected in a 
 normal solution of sulphuric acid and the quantity of the latter 
 neutralized by the NH 3 is determined by means of a normal 
 NaOH solution, and from the quantity neutralized the equiv- 
 alent of urea is deduced and the weight of urea found to hr\v 
 been decomposed in the operation of drying is added to the 
 weight of the dry residue in the porcelain boat. 
 
 2 molecules ammonia. 1 molecule urea. 
 
 2NH 3 CO(NH 2 ) 2 
 Or, 
 
 34 parts of ammonia = 60 parts of urea.
 
 52 NOTES ON CHEMICAL LECTURES. 
 
 Example : 2 c.c. urine employed. 
 
 Weight of boat plus residue .... 0.640 grm. 
 
 minus " ... . 0.600 " 
 Weight of dry residue .... 0.040 " 
 
 The NH 3 evolved was collected in 10 c.c. normal solution 
 of sulphuric acid. After collecting the NH 3 in the 10 c.c. 
 of normal solution of sulphuric acid, the latter was titered 
 with a normal solution of sodium hydroxide. 8 c.c. of the 
 NaOH solution were required to neutralize the 10 c.c. of 
 sulphuric acid solution. Previous to the collection of the 
 NH 3 in the 10 c.c. normal solution of sulphuric acid, 10 c.c. 
 of normal solution of sodium hydroxide would have been 
 required to neutralize it. Now, however, some of the sulphuric 
 acid has been neutralized by the NH 3 and consequently a less 
 volume of sodium hydroxide solution will be required to 
 neutralize the 10 c.c. of normal sulphuric acid solution. As 
 8 c.c. of normal sodium hydroxide solution were required it 
 indicates that 2 c.c. of the normal sulphuric acid solution were 
 neutralized by the ammonia evolved. 
 
 Thus, 
 
 1 c.c. of normal sulphuric acid solution = 0.017 grm. NH 3 
 
 2 c.c. = 2 X 0.017 = 0.034 grm. NH 3 
 
 As 34 parts of NH 3 equal 60 parts of urea, therefore, 
 
 NH 3 . Urea NH S . 
 
 34 : 60 :: 0.034 : x = 0.060 grm. urea. 
 
 Weight of residue in porcelain boat 0.040 grm. 
 
 Weight of urea as determined by amount of NH 3 
 
 evolved 0.060 " 
 
 Weight of solid matter in 2 c.c. of the urine . 0.100 " 
 
 c.c. grm. c.c. 
 
 2 : 0.100 :: 100 : x = 5.0 grm. solid matter. 
 
 3. The solid matter may be approximately determined by 
 multiplying the last two figures of the specific gravity by 
 Haeser and Neubauer's factor, 2.33. Thus if the specific 
 gravity is 1017, 
 
 17 X 2.33 = 39.61 grm. solid matter in 1000 c.c. of the urine.
 
 iV\ 
 
 NOTES ON CHEMICAL LECTURES. 
 
 53 
 
 Or more closely, (if the specific gravity be expressed with five 
 figures), by multiplying the last three figures by the factor 
 0.233. Thus if the specific gravity is 1020.3, 
 
 203 X 0.233 = 47.299 grm. solid matter in 1000 c.c. of the urine. 
 
 AVERAGE COMPOSITION OF HUMAN URINE. 
 
 Water 
 
 950.00 
 28.00 ^ 
 0.60 j 
 0.35 
 0.651 
 8.00 ) 
 8.00 
 2.00 
 1.25 
 0.25 
 0.30 
 
 O.GOy 
 
 Voided per day. 
 Grains. Grammes. 
 
 , 520.80 35.00 
 
 (Organic 11 16 075 
 ' matter. v> ' J 
 
 37.60 6.51 0.44 
 | 12.09 0.81 
 148.80 10.00 
 148.80 10.00 
 37.20 2.50 
 
 'maTtef 23.45 1.56 
 
 12.40 4.65 0.31 
 5.58 0.37 
 
 11.16 0.75 
 
 Urea . 
 
 Uric acid 
 
 Hippuric acid 
 
 Creatinine 
 
 Kxtractives 
 
 Sodium chloride . . . 
 Phosphoric acid .... 
 Sulphuric acid .... 
 Lime (CaO) . . . . 
 
 Magnesia (MgO) . . . 
 Potash (K 2 O) and soda 
 (Na 2 O) 
 
 Water 
 
 1000.00 930.20 
 
 950.00 
 37.60 699.36 
 12.40 230,34 
 
 62.49 
 
 47.00 
 15.49 
 
 Organic matter .... 
 Inorganic matter . . . 
 
 Average quantity of urine voided per day, about 40 fluid- 
 ounces, or 1200 c.c. 
 
 ANALYSIS OF THE URINE. 
 
 The analysis of urine may have for its object 
 
 1. The qualitative detection of the normal and abnormal 
 constituents. 
 
 2. The quantitative determination of the normal and ab- 
 normal constituents. 
 
 The percentages ot solids in urine are expressed as per- 
 centages by volume, /. e., 100 c.c. by volume contain a given 
 weight of a solid.
 
 54 NOTES ON CHEMICAL^LECTURES. 
 
 DETERMINATION OF SODIUM CHLORIDE (NaCI) IN 
 
 URINE. 
 
 The chlorine in urine is in combination as a chloride, chiefly 
 as sodium chloride (NaCI). 
 
 It may be detected qualitatively by strongly acidulating the 
 urine with nitric acid and adding a solution of argentic nitrate 
 (AgNO 3 ). If NaCI be present in rather large quantity, a curdy 
 precipitate of argentic chloride (AgCl) will be formed ; if the 
 quantity present be small, only a milkiness will be produced. 
 If the urine contain albumen, it must be removed before 
 making this test. 
 
 QUANTITATIVE,, DETERMINATION OF SODIUM CHLORIDE. 
 
 1. Gravimetric method ; Principle : when argentic nitrate 
 (AgNO 3 ) is added to sodium chloride (NaCI) all of the chlorine 
 is precipitated as argentic chloride (AgCl). 
 
 143.5. 
 
 AgNO 3 + NaCI = AgCl -j- NaNO 3 
 
 The quantity of sodium chloride, or of chlorine, is calculated 
 from the quantity of AgCl obtained. 
 
 Every 143.5 parts of AgCl = 58.5 NaCI or 35.5 Cl. 
 Hence 
 
 AgCl. NaCI. NaCI. 
 
 143.5 : 58.5 :: weight of precipitate of AgCl : X 
 or, 
 
 AgCl. Cl. Cl. 
 
 143.5 I 35.5 :: weight of precipitate of AgCl : X 
 
 Method : 10 c.c. of urine, to which is added about 2 grm. 
 potassium nitrate (KNO 3 ) to destroy the organic matter, are 
 evaporated to dryness in a small porcelain evaporating dish on 
 a water-bath (absolute dryness cannot be obtained) or over a 
 flame, care being taken to 'avoid loss from spurting. (The 
 contents of the dish must not be stirred.) The dish con- 
 taining the residue is held by means of a crucible tongs over 
 a flame and heated, very gently at first, then with strong heat, 
 until all the organic matter is destroyed as indicated by the 
 disappearance of the charred material. The dish containing the 
 residue while cooling is held at the outer edge by a crucible
 
 NOTES ON CHEMICAL LECTURES. 55 
 
 tongs and given a rotary motion until the fused material has 
 distributed itself as a thin layer over the dish and finally 
 solidified. By this procedure the cracking of the dish will be 
 prevented. Care must be taken that in this procedure none ot 
 the fused material comes in contact with the crucible tongs or 
 is projected from the dish. When the dish is thoroughly 
 cooled the residue is treated with 5 c.c. of water and eight or 
 ten drops of strong nitric acid. The nitric acid is added to 
 neutralize the KOH and K 2 CO 3 , which are produced in the 
 decomposition of the KNO 3 , with the organic matter. The 
 liquid in the dish is warmed to hasten solution. This first por- 
 tion of 5 c.c. is poured into a beaker (keeping back any enamel 
 that may have separated from the dish), and washing the dish 
 with portions of 5 c.c. of water is repeated (without further 
 addition of nitric acid), and heating, if necessary, until the dish 
 has been thoroughly washed. Each portion of 5 c.c. is poured 
 into the beaker until 30 c.c. have been employed and collected 
 in the beaker. If the liquid is not acid in reaction, it must be 
 acidulated with nitric acid. The solution is warmed and excess 
 of AgNO 3 solution is added. The liquid is stirred with a glass 
 rod until the precipitate separates in curd-like masses. The 
 precipitate is collected on a filter, and washed with distilled 
 water until the filtrate shows no trace of the presence of silver. 
 This is determined by collecting in a test tube a portion of the 
 filtrate as it comes from the funnel and adding hydrochloric 
 acid. The precipitate is dried while still on the filter, as much 
 as is possible, is detached from the filter, and placed in a 
 weighed crucible. The folded filter is held by means of a 
 platinum wire over the crucible, ignited, and the ash allowed 
 to fall into the crucible. The ash is moistened with two or 
 three drops of strong nitric acid to dissolve the metallic silver 
 which may have resulted from the reduction of AgCl in burn- 
 ing the filter, and the crucible is warmed very gently. When 
 cool, the ash is moistened with two or three drops of hydro- 
 chloric acid to reprecipitate, as chloride, the silver dissolved 
 by the nitric acid. The excess of acid is expelled by gently 
 heating the crucible over a very small flame. The crucible is 
 allowed to cool and is then weighed. The weight of the 
 empty crucible and the filter ash is deducted from the weight
 
 56 NOTES ON CHEMICAL LECTURES. 
 
 of the crucible, precipitate and ash. The difference in weights 
 is the weight of the argentic chloride. 
 Example : 
 
 Weight of crucible + prec. + filter ash ... 10.623 
 " " " " " "... 10.420 
 
 " " prec. -f- filter ash 0.203 
 
 " " filter ash 0.003 
 
 " " prec. of AgCl 0.200 
 
 Calculation : 
 
 AgCl. NaCl. 
 
 143.5 : 58.5 :: weight of precipitate of AgCl : X = quantity 
 of sodium chloride in the volume (10 c.c.) of urine employed. 
 Multiply by 10 = percentage of NaCl. 
 
 Or, 
 
 AgCl. Cl. 
 
 143.5 : 35.5 :: weight of precipitate of AgCl : X = quantity of 
 chlorine in the 10 c.c. urine employed. Multiply by 10 = per- 
 centage of chlorine. 
 
 If 0.200 grm. AgCl obtained, then 
 
 AgCl. NaCl. Wt. of prec. NaCl. NaCl. 
 
 143.5 : 58.5 :: 0.200 : 0.08153 X 10 = 0.8153 per cent, in a 
 
 volume of 100 c.c. urine. 
 In 100 grm. of urine. 
 
 Suppose specific gravity of urine was 1020. 
 
 NaCl. 
 
 Then, 1020 : 1000 :: 0.815 : 0.799 per cent, in 100 grm. of 
 the urine. 
 
 2. Mohr's volumetric method for the determination of 
 sodium chloride ; Depends upon the precipitation of the 
 chlorine of the sodium chloride by means of a standard solu- 
 tion of argentic nitrate and calculating the quantity of sodium 
 chloride present from the quantity of standard solution 
 required for complete precipitation. 
 
 The standard solution of argentic nitrate is prepared of 
 convenient strength, /. e., so that 1 c.c. of the solution shall 
 equal 0.010 of NaCl. 1 c.c. of the solution is also equal to 
 0.006068 chlorine,/, e., 
 
 NaCl. Cl. NaCl. Cl. 
 
 58.5 : 35.5 :: 0.010 : 0.006068
 
 NOTES ON CHEMICAL LECTURES. 57 
 
 Preparation of the standard solution of AgNO 3 : 
 
 58.5. 170. 
 
 NaCl + AgNO 3 = AgCl + NaNO 3 
 
 58.5. At. wt.of Ag, 108. 
 
 NaCl + AgNO 3 = AgCl + NaNO 3 
 
 170 grm. of AgNO 3 are equal to 58.5 grm. of NaCl. 
 
 170 grm. of AgNO 3 (equal to 58.5 grm. NaCl) contain 108 
 grm. of metallic silver, consequently 108 grm. of silver are 
 also equal to 58.5 grm. of NaCl. 
 
 The quantity of AgNO 3 necessary to prepare 1000 c.c. of 
 standard solution so that 1 c.c. of it shall equal 0.010 NaCl 
 or 0.006068 Cl is determined by 
 
 NaCl. AgNO 3 . Grm. NaCl. AgNO 3 . 
 
 58.5 : 170 :: 10 : 29.059 grm. 
 
 Or of metallic silver to. be converted into nitrate, by treating 
 with nitric acid, evaporating excess of acid, and dissolving the 
 residue in water, and diluting to 1000 c.c. 
 
 NaCl. Ag. Grm. NaCl. Metallic silver. 
 
 58.5 : 108 :: 10 : 18.461 grm. 
 
 Preparation of standard solution of argentic nitrate : 
 29.059 grammes pure, dry AgNO 3 are dissolved in distilled 
 water and the solution diluted with distilled water to a volume 
 of 1000 c.c. 
 
 Or, 18.461 grammes pure metallic silver are dissolved in 
 nitric acid, the excess of acid is expelled by evaporation on a 
 water-bath and the residue (which is AgNO 3 ) is dissolved in 
 distilled water, and the solution diluted with water to 1000 c.c. 
 Then, 
 
 AgNO 3 . NaCl. 
 
 1000 c.c. = 29.059 grm. = 10.0 
 10 c.c. = 0.29059 " ; 0.100 
 1 c.c. = 0.029059 " 0.010 
 
 Chlorine. 
 
 1 c.c. = : 0.029059 " 0.006068 
 
 Sometimes it is necessary to prove the accuracy of the solu- 
 tion by standardizing it with a standard solution of sodium 
 chloride.
 
 58 NOTES ON CHEMICAL LECTURES. 
 
 Thus : 
 
 1.0'grm. Nad is dissolved in distilled water and the solution 
 diluted to 100 c.c. 
 
 NaCl. 
 
 100 c.c. = 1.0 grm. 
 10 c.c. = 0.100 " 
 1 c.c. = 0.010 " 
 
 Standardizing the argentic nitrate solution : To 10 c.c. 
 of the NaCl solution two drops of potassium chromate (K 2 CrO 4 ) 
 solution (the indicator), are added, and the AgNO 3 solution run 
 into the solution from a burette until the lemon-yellow color 
 of the liquid is changed to a very slight orange-red. The 
 red color is due to the formation of red argentic chromate 
 (Ag 2 CrO 4 ), and indicates that all of the NaCl has been decom- 
 posed and a slight excess of the AgNO 3 solution has acted on 
 the potassium chromate present. 
 
 The 10 c.c. of NaCl solution, which contains 0.100 grm. of 
 NaCl should require exactly 10 c.c. of the AgNO 3 solution to 
 completely precipitate the chlorine and act upon the indicator 
 
 a. The standard solution may be too strong, i. e., re- 
 quiring the addition of less than 10 c.c. of the AgNO 3 to the 
 10 c.c. of solution containing 0.100 grm. NaCl. It may be 
 corrected by determining and adding the volume of water 
 necessary to dilute it to the proper strength. 
 
 Suppose 8 c.c. instead of 10 c.c. of the AgNO 3 solution have 
 been required for the 10 c.c. of solution containing 0.100 grm. 
 NaCl. Then to every 8 c.c. of standard solution remaining 
 a volume of distilled water equal to the difference between 8 
 and 10 c.c., or 2 c.c. is added. 
 
 1000 c.c. solution prepared. 
 
 8 c.c. used. 
 8)992 c.c. on hand. 
 
 124 X 2 = 248 c.c. water to be added to the 
 992 c.c. standard solution. 
 
 10 c.c. will now equal 0.100 grm. of NaCl. 
 1 c.c. " 0.010 " " " 
 
 b. The standard solution may be too weak, i. e., requiring 
 the addition of more than 10 c.c. of the AgNO 3 solution to the
 

 
 NOTES ON CHEMICAL LECTURES. 59 
 
 10 c.c. of solution containing 0.100 grm. NaCl. The AgNO 3 
 solution may be corrected by adding more crystallized AgNO 3 
 to the solution, titering with 10 c.c. of standard NaCl solution 
 and then correcting after the manner of the correction of a 
 standard solution that is too strong (see a before mentioned). 
 
 If the AgNO 3 solution is thus standardized and the residue 
 from 10 c.c. urine is dissolved in .30 c.c. water, then, in actual 
 analysis, because of increased dilution, 0.2 c.c. must be deduc- 
 ted (0.1 for every 10 c.c. above a fixed 10 c.c.) from the num- 
 ber of c.c. AgNO 3 solution required. 
 
 c. The AgNO 3 solution may be standardized so that in 
 actual analysis no correction (deduction of 0.2 c.c.) will be 
 necessary. 
 
 10 c.c. standard solution of NaCl (containing 0.100 grm. 
 NaCl) are diluted with 20 c.c. water, so that the volume (30 c.c.) 
 shall equal the volume of the final solution in the actual work 
 with the urine. Two drops of K 2 CrO 4 solution are added and 
 the AgNO 3 solution from a burette is run into the NaCl solu- 
 tion until the lemon-yellow color of the liquid is changed to a 
 slight orange-red. 
 
 If more or less than 10 c.c. of standard AgNO 3 solution 
 should be required to give the reaction with the indicator, the 
 AgNO 3 solution should be corrected, after the manner of the 
 correction of standard solutions that are too weak or too strong 
 as before described, so that exactly 10 c.c. shall be required for 
 the 10 c.c. of standard solution of NaCl diluted with 20 c.c. of 
 water. 
 
 Method: 10 c.c. of urine, to which is added about 2 grm. 
 potassium nitrate (KNO 3 ) to destroy the organic matter, are 
 evaporated to dryness in a small porcelain evaporating dish 
 on a water-bath (absolute dryness cannot be obtained) or over 
 a flame, care being taken to avoid loss from spurting. (The 
 contents of the dish must not be stirred.) The dish con- 
 taining the residue is held by means of a crucible tongs over 
 a flame and heated, very gently at first, then with strong heat 
 until all of the organic matter is destroyed as indicated by the 
 disappearance of the charred material. The dish containing 
 the residue, while cooling, is held at the outer edge by a cruci- 
 ble tongs and given a rotary motion until the fused material
 
 60 NOTES ON CHEMICAL LECTURES. 
 
 has distributed itself as a thin layer over the dish and finally 
 solidified. By this procedure the cracking of the dish will be 
 prevented. Care must be taken that in this procedure none of 
 the fused material comes in contact with the crucible tongs or 
 is projected from the dish. When the dish is thoroughly 
 cooled the residue is treated with 5 c.c. of water and eight or 
 ten drops of strong nitric acid. The nitric acid is added to 
 neutralize KOH and K 2 CO 3 which are produced in the decom- 
 position of the KNO 3 with the organic matter. The liquid in 
 the dish is warmed to hasten solution. This first portion of 
 5 c.c. is poured into a beaker (disregarding any enamel which 
 may have separated from the dish), and washing the dish with 
 portions of 5 c.c. of water is repeated (without further addition 
 of nitric acid), heating, if necessary, until the dish has been 
 thoroughly washed. Each portion of 5 c.c. is poured into the 
 beaker until 30 c.c. have been employed and collected in the 
 beaker. If the liquid is not acid in reaction it must be veiy 
 slightly acidulated with nitric acid. (Up to this point, ex- 
 cept disregarding the enamel, the method is practically similar 
 to the one employed in the gravimetric determination of 
 sodium chloride). 
 
 The liquid is neutralized by the addition of an excess of 
 precipitated calcium carbonate (CaCO 3 ). The point of neu- 
 tralization is determined by the cessation of effervescence and 
 the remaining of a portion of undecomposed calcium carbonate 
 at the bottom of the beaker. (Neutralization of the solution 
 is necessary because the nitric acid present would dissolve the 
 orange-red argentic chromate produced, and thus prevent a 
 visible reaction on the indicator). 
 
 Two drops of solution of potassium chromate (the indicator) 
 are added and the standard solution of AgNO 3 is run from the 
 burette into the 30 c.c. of liquid containing the sodium chloride 
 until the lemon-yellow color of the liquid is changed to a 
 slight orange-red. 
 
 The number of cubic centimetres of standard AgNO 3 solution 
 required is read off and this number is multiplied by the value 
 of 1 c.c. of the standard AgNO 3 solution expressed in terms of 
 NaCl, namely, 0.010. The result obtained is multiplied by 
 10 in order to obtain the percentage of NaCl in the urine.
 
 NOTES ON CHEMICAL LECTURES. 61 
 
 If the standard AgNO 3 solution has been standardized with 
 10 c.c. of standard NaCl solution diluted with 20 c.c. of water 
 making a volume of 30 c.c. no correction is necessary. 
 
 Example : Suppose 8.3 c.c. AgNO 3 solution have been 
 required. Then 8.3 X 0.010 = 0.083 grm. NaCl in the 10 c.c. 
 urine employed. 0.083 X 10 = 0.83 per cent. NaCl. 
 
 3. Liebig's method for the determination of sodium chloride : Depends 
 upon the formation of a soluble salt of mercury (mercuric chloride, HgCl 2 ) on the 
 addition of mercuric nitrate, Hg(NO 3 ) 2 , to a solution of sodium chloride. 
 
 117. 
 2NaCl - Hg(NO 3 )., = HgCI 2 + 2NaNO 3 
 
 Liebig's method for the determination of NaCl in the urine is not accurate and 
 is rarely used. 
 
 The mercuric nitrate solution is prepaied of such strength that 1 c.c. of it shall 
 equal 0.010 grm. NaCl. 
 
 117 grm. NaCl require 200 grm. of mercury (once the atomic weight of Hg 
 expressed in grammes). 
 
 Hence 
 
 2NaCl. Hg. NaCl. Hg. 
 
 117 : 200 :: 10 : 17.094 grm. 
 
 Or, 117 grm. NaCl require 216 grm. mercuric oxide (HgO) (once the molecular 
 weight of HgO expressed in grammes). 
 
 2NaCl. HgO. NaCl. HgO. 
 
 117 : 216 :: 10 : 18.461 grm. 
 
 17.094 grm. of metallic mercury, or 18.461 grm. of mercuric oxide, are equal to 
 10 grm. NaCl. 
 
 Preparation of standard solution of mercuric nitrate : 17.094 grm. of 
 mercury, or 18.461 grm. of mercuric oxide, are dissolved in an excess cf nitric 
 acid to which is added a small quantity of water. The excess of acid is expelled 
 by evaporation and the residue is dissolved in distilled water and slowly diluted 
 to 1000 c.c. If on dilution with water a canary-yellow precipitate of basic 
 mercuric nitrate should separate, it is allowed to subside, the supernatant liquid is 
 poured off and the precipitate is dissolved in a few drops of strong nitric acid, and 
 the solution returned to the previously poured off supernatant liquid. 
 
 Hg(N0 3 ),, sol. NaCl. 
 1000 c.c. == 10 grm. 
 10 c.c. = 100 
 1 c.c. = 010 " 
 
 Chlrrine. 
 
 1 c.c. = 0.006068 " 
 
 The mercuric nitrate solution may be standardized with a solution of NaCl con- 
 taining 1.0 grm. NaCl in 100 c.c. water, using a pinch of urea as the indicator.
 
 62 NOTES ON CHEMICAL LECTURES. 
 
 The phosphates, sulphates^and carbonates in the urine inttrfere with the appli- 
 cation of the method. They are removed by the baryta mixture, composed of 
 
 2 volumes of a cold saturated solution of barium hydroxide (Ba(OH) 2 ). 
 1 volume of a cold saturated solution of barium nitrate (Ba(NO 3 ) 2 ). 
 
 The first action of the mercuric solution will be on the NaCl in the urine to 
 form soluble HgCl 2 ; as soon as all the NaCl present has been decomposed, any 
 excess of mercuric nitrate solution added will act upon the urea to form a white 
 insoluble compound of mercuric oxide and urea, (HgO) 2 CO(NH 2 ) 2 , 
 
 2Hg(N0 3 ) 2 + CO(NH 2 ) 2 + 2H 2 = (HgO) 2 CO(NH 2 ) 2 + 4HNO 3 
 
 and thus the urea acts as the indicator. 
 
 Method : 40 c.c. of urine are mixed with 20 c.c. of baryta mixture and filtered 
 through a filter that has not been previously moistened with water. The filtrate 
 is neutralized with a drop or two of nitric acid. 15 c.c. of the filtrate (representing 
 10 c.c. of urine and 5 c.c. of baryta mixture) are transferred by means of a pipette 
 to a beaker. The standard mercuric nitrate solution is run from the burette into 
 the liquid until a permanent milky; turbidity is produced. (Action on the urea.) 
 
 The number of cubic centimetres of standard mercuric nitrate solution required 
 is multiplied by 0.010 and the result multiplied by 10 furnishes the percentage of 
 NaCl in the urine. 
 
 UREA (Carbamide). 
 
 CO(NH 2 ) 2 . Molec. wt., 60. 
 
 CON 2 H 4 , urea. Generally considered to be CO(NH 2 ) 2 , 
 carbamide. Gamgee considers that it is not carbamide, be- 
 cause, when urea is heated with potassium permanganate 
 (K 2 Mn 2 O 8 ) and potassium hydroxide, all of its nitrogen is 
 evolved as free nitrogen, whereas salts of ammonium (NH 4 ) 
 and amides yield their nitrogen as N 2 O 5 : 
 
 Compounds isomeric with urea : 
 
 NH 4 CNO, ammonium cyanate. 
 CON 2 H 4 , isuretin. 
 
 Isuretin, CON 2 H 4 , is not urea, because it is 
 
 NH 
 NH 2 OH + HCN = CH NH, hydroxyl-methenyldiamide. 
 
 Hydroxylamide. Hydrocyanic acid. OFT 
 
 Urea was first recognized in the urine, and obtained in an 
 impure state, by Rouelle, in 1773. It was obtained in a purer 
 condition, by Fourcroy and Vauquelin, in 1799. It was pre- 
 pared artificially, by Woehler, in J 828, it being the first organic 
 compound produced artificially.

 
 NOTES ON CHEMICAL LECTURES. 63 
 
 Urea is found in the urine of all mammals. The urine of 
 birds and reptiles contains it in small quantity. Thirty per 
 cent, of the solid matter of the vitreous humor of the eye is 
 urea. It is contained in almost all the animal fluids, blood, 
 lymph, chyle, etc., and in the liver and spleen. It is the 
 principal solid constituent of the urine. Ninety per cent, of the 
 nitrogen eliminated by the urine is in the form of urea. 
 
 Urea may be prepared artificially by fusing and oxidizing 
 potassium ferrocyanide (K 4 Fe(CN) 6 ) with manganese dioxide 
 (MnO 2 ) or with minium (Pb 3 O 4 = 2PbO + PbO 2 ), with the 
 production of potassium cyanate (KCNO), which, when warmed 
 with a solution of ammonium sulphate ((NH 4 ) 2 SO 4 ), forms 
 ammonium cyanate (NH 4 CNO), and by continued warming 
 changes into urea (CO(NH 2 ) 2 ). The solution is evaporated to 
 dryness on a water-bath, and the residue extracted with 
 strong alcohol. The alcohol dissolves only the urea, which 
 may be obtained in crystalline form by allowing the alcohol to 
 evaporate at ordinary temperature. 
 
 1. K 4 Fe(CN) 6 + O 9 = 4KCNO + 2CO 2 + FeO + N 2 
 
 2. 2KCNO + (NH 4 ) 2 SO 4 = 2NH 4 CNO + K 2 SO 4 
 
 3. NH 4 CNO = CO(NH 2 ) 2 
 
 It may also be prepared by warming plumbic cyanate 
 (Pb(CNO) 2 ) with a solution of ammonium sulphate, 
 
 Pb(CNO) 2 + (NH 4 ) 2 SO 4 = PbSO 4 + 2NH 4 CNO 
 on further warming, the NH 4 CNO is converted into urea. 
 
 UREA MAY BE OBTAINED FROM THE URINE BY THE 
 FOLLOWING METHODS. 
 
 1. As urea : Baryta mixture is added to the urine, the liquid 
 is filtered and the filtrate is evaporated to dryness on a water- 
 bath. The residue is treated with strong alcohol and the solu- 
 tion filtered and evaporated to dryness. The residue is dissolved 
 in water and the solution decolorized by being passed through 
 animal charcoal. The decolorized liquid is evaporated to dry- 
 ness and the residue treated with strong alcohol and the liquid 
 filtered. The filtrate is allowed to evaporate at ordinary tem- 
 perature, and needle-like crystals of urea will separate
 
 64 NOTES ON CHEMICAL LECTURES. 
 
 2. As urea nitrate : (soluble in 8 parts of water). 250 c.c., 
 or more, of urine are evaporated on a water-bath to about 
 one-sixth its original volume. When cold, nitric acid, of about 
 1.25 specific gravity, is added to the liquid and the mixture 
 kept cold. Crystals of CO(NH 2 ) 2 HNO 3 will separate. The 
 mass of crystals is collected on a moistened piece of muslin 
 and the excess of liquid squeezed out. The mass is scraped 
 off the muslin and dissolved in water. Barium carbonate 
 (BaCO 3 ) is added to the solution to separate the HNO 3 from 
 the CO(NH 2 ) 2 HNO 3 
 
 2CO(NH 2 ) 2 HNO 3 + BaCO 3 = Ba(NO 3 ) 2 + H 2 O + CO 2 + 
 2CO(NH 2 ) 2 
 
 The solution is decolorized by being passed through animal 
 charcoal. The decolorized liquid is evaporated to dryness on 
 a water-bath, the residue is treated with strong alcohol and the 
 liquid is filtered, and the filtrate allowed to evaporate at ordi- 
 dinary temperature. Crystals of urea will separate. 
 
 3. As urea oxalate : (soluble in 25 parts of water). The 
 method is similar to that with nitric acid except that oxalic 
 acid (H 2 C 2 O 4 ) (strong solution or in powder) is used instead of 
 nitric acid, and calcium carbonate (CaCO 3 ) is used instead of 
 barium carbonate. 
 
 (CO(NH 2 ) 2 ) 2 H 2 C 2 O 4 + CaCO 3 = CaC 2 O 4 + H 2 O + CO 2 
 
 Urea oxalate. Calcium oxalate. 
 
 + 2CO(NH 2 ) 2 
 
 Properties of urea : Urea is a white, odorless compound, 
 crystallizing in four-sided prisms. It has a cooling, bitter-like 
 taste, somewhat resembling potassium nitrate. It is very 
 soluble in water, 1 in 1 ; soluble in alcohol, 1 in 5 ; insoluble 
 in ether. Melts at a temperature of 130 C. 
 
 Urea in solution has no action on blue or red litmus-paper, 
 is neutral, yet it will combine with acids, bases, and salts. 
 
 a. CO(NH 2 ) 2 HNO 3 
 
 b. (HgO) 2 CO(NH 2 ) 2 
 
 c. CO(NH 2 ) 2 NaCl 
 
 It unites with acids without displacing the hydrogen in the 
 acid.
 
 NOTES ON CHEMICAL LECTURES. 65 
 
 a. Urea in solution is decomposed by sodium hypochlorite 
 with the evolution of CO 2 and N. 
 
 CO(NH 2 ) 2 + SNaCIO = 3NaCl + 2H 2 O + CO 2 + N 2 
 
 b. Urea in solution is decomposed by nitrous acid (HNO 2 ) 
 with the evolution of CO 9 and N. 
 
 CO(NH,) 2 + 2HNO 2 = 3H 2 O -f CO 2 + 2N 
 
 c. Urea heated with water in a glass tube sealed at both 
 ends is converted into ammonium carbonate. 
 
 CO(NH 2 ) 2 + 2H 2 O = (NH 4 ) 2 CO 3 
 
 A like conversion occurs when urea in solution is exposed 
 for a time to the air, due to the action of micrococcus ureae. 
 
 a. Urea in crystals heated to a temperature of 150 170 
 C. fuses and yields ammonia (NH 3 ) and a compound called 
 biuret (C 2 H 5 N 3 O 2 ). 
 
 2CO(NH 2 ) 2 = C 2 H 5 N 3 O 2 + NH 3 
 
 Biuret in solution produces a violet-red color on the addi- 
 tion of a few drops of very dilute cupric sulphate solution, 
 and afterwards a solution of potassium or sodium hydroxide. 
 (Biuret reaction.) 
 
 Peptones and albumoses respond to the same test. 
 
 b. Urea heated to a higher temperature (over 170 C.) 
 yields ammonia NH 3 , and cyanuric acid (II 3 C 3 N 3 O 3 ). 
 
 3CO(NH 2 ) 2 = 3NH, + H 3 C 3 N 3 O 3 
 
 c. Urea heated to a still higher temperature yields am- 
 monia, and melanuric acid, H 4 C 3 N 4 O 2 . 
 
 4CO(NH 2 ) 2 = 4NH 3 -f CO 2 -f- H 4 C 3 N 4 O 2 
 QUANTITATIVE DETERMINATION OF UREA IN URINE. 
 
 1. Davy's method, introduced in 1854, depends upon the decomposition of 
 the urea in the urine by means of sodium hypochlorite with the evolution of carbon 
 dioxide and nitrogen. The CO., is ahsor'ied by the excess of alkali (NaOH) in 
 the NaCIO solution, the nitrogen remains unabsorbed, and is collected, and meas- 
 ured in a tube graduated in cubic inches. 
 
 Davy's method is inaccurate, is of historical interest only and is not used.
 
 66 NOTES ON CHEMICAL LECTURES. 
 
 The volume of nitrogen is corrected for temperature (60 F.) and barometric 
 pressure (30 inches), and the quantity of urea calculated from the volume of 
 nitrogen obtained. 
 
 60. 2s. 
 
 CO(NH 2 ) 2 -f SNaCIO = 3NaCl + 2H 2 O + CO 2 + N 2 
 
 1 molecule (60) of urea contains two atoms (2 X 14 = 28) of nitrogen. 
 28 grains N = 93.33 cubic inches. 
 
 60 grains urea contain 28 grains N = 93.33 cubic inches nitrogen, at temperature 
 of 60 F. and barometric pressure of 30 inches. 
 
 Grains urea. Cubic inches. Grain urea. Nitrogen. 
 
 60 : 93.33 :: 1 : 1.55 cu. in. 
 
 Consequently 1.55 cubic inch of nitrogen is evolved from 1 grain of urea, or 1 
 cubic inch of nitrogen is evolved from 0.645 grain of urea. 
 Cu. in. N. Grain urea. Cu. in N. Urea. 
 
 1.55 : 1 :: 1 : 0.645 grain. 
 
 Method : A tube graduated in cubic inches is filled with mercury to one third 
 of its capacity. 100 grains of urine are added and the remainder of the tube is 
 rapidly filled with sodium hypochlorite solution. The thumb is quickly placed 
 over the opening of ihe tube and the latter is inverted in a trough containing 
 mercury. Decomposition of the urea occurs and the evolved nitrogen collects in 
 the upper part of the tube. 
 
 After the lapse of about half an hour the opening of the tube is closed with the 
 thumb and the tube transferred to a vessel containing water. The atmospheric 
 pressure is equalized and the number of cubic inches of nitrogen evolved is read 
 off. The number of cubic inches of nitrogen evolved is divided by 1.55 or multi- 
 plied by 645, and the result will be the percentage of urea in the urine. (100 
 grains of urine having been employed.) 
 
 Suppose 4.65 cubic inches nitrogen evolved. 
 
 4.65 -5- 1.55 -= 3 per cent. urea. 
 Or, 4.65 X 0-645 2.99 per cent. (3 per cent.) urea. 
 
 Objections to this method : According to Fenton, NaCIO in presence of 
 caustic alkalies causes the evolution of only one-half of the nitrogen of urea, the 
 remainder being retained as cyanate, thus : 
 
 2CO(NH 2 ) 2 + SNaCIO + 2NaOH = 3 NaCl + 2NaCNO + 5H 2 O + N 2 
 
 2. Fowler's modification of Davy's method for the determination of urea : 
 Depends upon the decomposition of urea in solution by sodium hypochlorite 
 (NaCIO), thereby causing a reduction in the density of the solution. Fowler 
 found that a loss of one degree in specific gravity in a mixture of one volume of 
 urine and seven volumes of hypochlorite solution represented the presence (decom- 
 position) of 0.77 per cent, of urea. 
 
 Seven volumes of hypochlorite solution of 1035 specific gravity will decompose 
 the urea in one volume of an average urine. 
 
 The method is inaccurate and is rarely used. 
 
 Method : The specific gravities of the urine and of the hypochlorite solution 
 are determined separately. To one volume of the urine seven volumes of hypo- 
 chlorite solution are added, say 10 c.c. urine and 70 c.c. hypochlorite solution.
 
 NOTES ON CHEMICAL LECTURES. 67 
 
 (The specific gravity of this mixture is determined by calculation.) After allowing 
 the mixture to stand umil the decomposition is completed ^ about half an hour) the 
 specific gravity of the mixture is determined. This latter specific gravity is deducted 
 from the specific gravity (obtained by calculation) of the mixture. The loss of 
 specific gravity multiplied by 0.77 furnishes the percentage of urea in the urine. 
 Example : 
 
 7 volumes of NaCIO sol., specific gravity . . . 1036 X 7 = 7252 
 1 volume of urine, " " ... 1025 X 1 = 1025 
 
 8 volumes 8)8277 
 
 Specific gravity of mixture before decomposition . . 1034.6 
 " " " after " . . 1030.0 
 
 4.6 
 
 Hence 4.6 X 0.77 --= 3.524 per cent. urea. 
 
 3. Hypobromite method for the quantitative determination 
 of urea : Depends upon the decomposition of the urea in the 
 urine by means of sodium hypobromite (NaBrO) with the 
 production of sodium bromide (NaBr), water, carbon dioxide 
 (CO 2 ), and the evolution of nitrogen. 
 
 This is the most accurate method for the quantitative deter- 
 mination of urea. 
 
 The CO 2 is absorbed by the excess of alkali (NaOH) in the 
 NaBrO solution, and the nitrogen remains unabsorbed, and is 
 collected, and measured in a tube graduated in cubic centi- 
 metres. 
 
 The volume of nitrogen is corrected for temperature (0 C.) 
 and barometric pressure (760 mm.), and the quantity of urea 
 is calculated from the volume of nitrogen obtained. 
 
 60. 28. 
 
 CO(NH 2 ) 2 + 3NaBrO == 3NaBr + 2H 2 O + CO 2 + N 2 
 
 1 molecule urea. 1 molecule (2 atoms). 
 60. 28. 
 
 CO(NH 2 ) 2 = N 2 
 
 00 grammes of urea contain 28 grar^mes nitrogen. 
 28 grammes nitrogen = 22.32 litres or 22320 c.c. 
 
 60 grammes urea evolve 22.32 22320 c.c. nitrogen. 
 
 Grms. urea. Grm. urea. Litres. Litre. 
 
 60 : 1 :: 22.32 : 0.372 or 372 cubic centimetres. 
 
 1 gramme of urea will evolve 372 c.c. of nitrogen (at C. 
 and 760 mm.). 2. 
 
 c.c. N Grm. urea, c c. N. Grm. urea. 
 
 372 1 1 : 0.002688
 
 68 A'OTES ON CHEMICAL LECTURES. 
 
 1 cubic centimetre of nitrogen, measured at C. temper- 
 ature and 760 mm. pressure, is evolved from 0.002688 grm. 
 urea (2.688 milligrammes urea). 
 
 1.0 gramme urea evolves 372.0 c.c. nitrogen. 
 0.001 " 0.372 " 
 
 0.010 " " 3.72 " 
 
 0.020 " " " 7.44 " "' 
 0.030 " 11.16 " 
 
 1 c.c. of a 1 per cent, solution of urea (containing 0.010 
 urea) will evolve 3.72 c.c. nitrogen. 
 
 1 c.c. of a 2 per cent, solution of urea (containing 0.020 
 urea) will evolve 7.44 c.c. nitrogen. 
 
 The sodium hypobromite solution is prepared by dissolv- 
 ing 100 grammes sodium hydroxide (NaOH) in 250 c.c. water, 
 cooling the solution, an^ adding 25 c.c (75 grammes) of 
 bromine. f 
 
 The reagent (NaBriB^lution) should be freshly prepared. 
 
 160. 80. 
 
 2Br + 2NaOlt'pj 
 
 2Br. 2NaOH.J 
 
 160 80 
 
 Hence the 75 grammes OT 
 
 NaBr + HO 
 
 rm. Br. NaOH. 
 
 75 : 37.5 grm. 
 v. " 
 
 bromine will combine with only 
 37.5 grammes of the 100 grammes of NaOH employed in the 
 preparation of the solution, leaving the difference between 
 37.5 and 100, or 62.5 grammes, free NaOH in the solution, to 
 absorb the CO 2 evolved in the application of the method. 
 
 There are many forms of apparatus employed in the deter- 
 mination of urea by sodium hypobromite, 
 
 Russell & West's. 
 Huefner's. 
 Greene's. 
 Marshall's. 
 Etc. 
 
 Method, with the use of Marshall's apparatus : The side 
 opening of the bulbed tube is closed with the thumb and the 
 tube is filled with hypobromite solution (which may previously 
 be diluted with an equal volume of distilled water). The upper

 
 .VOTES ON CHEMICAL LECTURES. 09 
 
 opening of the tube is closed with a rubber stopper and the 
 tube is inclined so as to allow any air-bubbles which may be 
 just below the rubber stopper to escape through the side 
 opening. The tube is inverted and the end closed with the 
 rubber stopper is fixed in the saucer-shaped vessel. 
 
 1 c.c. of the urine is slowly run from the graduated pipette 
 through the side opening of the tube into the hypobromite 
 solution, and the pipette quickly withdrawn. 
 
 1 c.c. of urine, in the great majority of cases, suffices; if, 
 however, the urine contain very little urea, more than 1 c.c. 
 may be employed. 
 
 When the decomposition is completed (about twenty min- 
 utes) and all the gas-bubbles have collected in the upper part 
 of the tube (may be facilitated by gently tapping the tube 
 with the finger), the atmospheric pressure is equalized by 
 attaching the funnel-tube to the side opening of the hypo- 
 bromite tube, and pouring hypobromite solution into it until 
 the surfaces of the liquid in both tubes are equal in height. 
 The number of cubic centimetres of nitrogen is read off, the 
 temperature of the air and the barometric pressure are ob- 
 served and the amount of urea calculated from the volume of 
 nitrogen evolved after having been corrected for temperature 
 and pressure. 
 
 Example : Suppose 1 c.c. urine evoived 12.5 c.c. nitrogen, 
 the temperature being 20 C. and barometric pressure 750 mm. 
 
 A. Then, correcting simply for temperature and pres- 
 sure : $ v . 
 
 20. 0. 20. 
 
 a. 293 : 273 :: J2.5 : 11.64 c.c. at C. temp. 
 and, p y y < 
 
 b. 760 : 750 :: 11.64 : 11.48 c.c. at C.temp. and 760 mm. 
 Hence 
 
 11.48 c.c. N X O.OQ2688 = 0.03085 urea (in 1 c.c. urine). 
 0.03085 X 100 = 3.085 per cent, urea (in 100 c.c. urine). 
 
 B. Correcting for temperature, pressure and tension of 
 aqueous vapor : 
 
 20. 0. 20. 
 
 a. 293 : 273 :: i2.5 : 11.64 c.c. at C. temp.
 
 70 NOTES ON CHEMICAL LECTURES. 
 
 b. 750.0 mm. observed barometric pressure, 
 
 1 7.4 correc. in mm. for tension of aqueous vapor at 20 C. 
 732.6 mm. corrected barometric pressure. 
 
 760 : 732.6 :: 11.64 : 11.23 c.c. at C. temp. 760 mm. 
 pressure and corrected for tension of aqueous vapor. 
 Hence 
 
 11.23 c.c. N X 0.002688 = 0.03018 urea (in 1 c.c. urine). 
 0.03018 X 100 = 3.018 per cent, urea (in 100 c.c. urine). 
 
 C. Correcting for temperature, barometric pressure, and 
 tension of aqueous vapor using the following formula : 
 
 V' = 
 
 v X (B-T) 
 
 760 X (1 + 0.003665 1) 
 in which 
 
 V is the corrected volume of nitrogen in c.c. 
 
 v " observed 
 
 B " barometric pressure in mm. 
 
 T " tension of aqueous vapor for temp. t. 
 
 t " observed temperature. 
 
 0.003665 is the coefficient of expansion of gases for each 
 degree Centigrade. 
 
 Hence, for the above example, 
 
 12.5 X (750 17.4) _ 12.5 X 732.6 9157.50 
 
 760 X 1.073 760 X 1.073 815.48 
 
 ^ n 93 ~ 
 
 ~~ 
 
 11.23 c.c. N X 0.002688 = 0.03018 (in 1 c.c. urine). 
 0.03018 X 100 = 3.018 per cent, urea (in 100 c.c. urine). 
 
 D. To correct for the expansion of mercury at 20 C. in the barometer- tube, 
 
 0.000171 X t 
 thus, 0.000171 X 20 = 0.0034 
 
 0.0034 X 750 = 2.56 mm. 
 so, 
 
 Barom. = 750 2 56 = 747.44. 
 
 750:747.44 :: 11.23 : 11.19 c.c. N. 
 
 11.19 X 0.002688 = 0.03007 grm. urea. 
 
 0.03007 X 100 = 3 007 per cent, urea (in 100 c.c. urine).
 
 NOTES ON CHEMICAL LECTURES. 71 
 
 Table of Tension of Aqueous Vapor Expressed in Millimetres for 
 Certain Temperatures Centigrade. 
 
 Temp. Temp. 
 
 Tension rp Tension 
 
 in mm. in mm. 
 
 10 C 9.139 23 C 20.857 
 
 11C 9.767 24 C 22.152 
 
 12 C 10.432 25 C 23.517 
 
 13 C 11.137 26 C 24.955 
 
 14 C 11.883 27 C 26.470 
 
 15 C 12.673 28 C 28.065 
 
 16 C 13.510 29 C 29.743 
 
 17 C 14,395 30 C 31.509 
 
 18 C. ....... 15.330 31 C 33.366 
 
 19 C 16.318 3i C 35.318 
 
 20 C 17.363 33 C 37.368 
 
 21 C 18.465 34 C 39.522 
 
 22 C 19.629 35 C 41.784 
 
 4. Liebig's method for the quantitative determination of 
 urea : Depends upon the production of an insoluble compound 
 of mercuric oxide and urea (HgO) 2 CO(NH 2 ) 2 on the addition 
 of mercuric nitrate Hg(NO 3 )., to a solution of urea. 
 
 q?,VlOj-l,vla*jl. - H-j tf i-h 2. +1 Co *T. Oj 
 
 2Hg(N0 3 ) 2 +CO(NH 2 ) 2 ^2H 2 = (HgO) 2 CO(NH 2 ) 2 +4HN0 3 
 
 In the practical application of the method the Hg(NO 3 ) 2 
 acts first on the NaCl of the urine to form soluble HgCL. 
 When sufficient Hg(NO 3 ) 2 has been added to combine with all 
 the NaCl present it then acts on the urea. 
 
 H g(NO 3 ) 2 + 2NaCl = HgCl._, + 2NaNO 
 
 The precipitate formed after the NaCl is satisfied is com- 
 posed of 2 molecules of HgO in combination with 1 molecule 
 of CO(NH 2 ) 2 . 
 
 216. 
 
 Twice the molecular weight of HgO is 432 
 
 Once " " " CO(NH 2 ) 2 " 60 
 
 In the formation of the compound (HgO) 2 CO(NH 2 ) 2 , 432 
 parts by weight of HgO enter into combination with 60 parts 
 by weight of CO(NH 2 ) 2 .
 
 72 NOTES ON CHEMICAL LECTURES. 
 
 Preparation of the standard Hg(NO 3 ) 2 solution; The 
 quantity of HgO necessary to prepare 1000 c.c. of standard 
 Hg(NO 3 ) 2 solution so that 1 c.c. of it shall equal 0.010 urea, 
 is determined by 
 
 Urea. HgO. Urea. HgO. 
 
 60 : 432 :: 10 : 72.0 grm. = 66.66 grm. metallic Hg. 
 
 In the preparation of the solution 5.2 grm. HgO must be 
 added to the 72.0 grm. to act on the indicator. 
 
 72.0 + 5.2 77.2 grm. HgO = 71.48 grm. metallic Hg. 
 Hence 
 
 1000 c.c. containing 77.2 grm. HgO = 10.0 grm. urea. 
 1 c.c. " 0.0772 " " = 0.010 " 
 
 This same solution may be used for the quantitative deter- 
 mination of sodium chloride. 
 
 HgO. 2NaCl. HgO. NaCl. 
 
 216 : 117 :: 77.2 : 41.817 grm. 
 
 NaCl. 
 
 1000 c.c. containing 77.2 grm. HgO = 41.817 grm. 
 1 c.c. " 0.0772" " = 0.041817 " 
 
 The insoluble HgO, or the metallic Hg, is converted into 
 soluble Hg(NO 3 ) 2 by dissolving in nitric acid. 
 
 77.2 grm. HgO or 71.48 grm. metallic Hg are dissolved in 
 an excess of nitric acid to which is added a small quantity of 
 water. The excess of acid is expelled by evaporation on a 
 sand-bath and the residue is dissolved in distilled water, and 
 slowly diluted to a volume of 900 c.c. If on dilution with 
 water a canary-yellow precipitate of basic mercuric nitrate 
 should separate, it is allowed to subside, the supernatant liquid 
 is poured off and reserved. The canary-yellow precipitate is 
 dissolved in a few drops of strong nitric acid and the solution 
 returned to the previously poured off supernatant liquid. 
 Should there be a recurrence of the precipitation, the precipi- 
 tate must be allowed to subside, the supernatant liquid poured 
 off, etc., as above described. 
 
 The mercuric nitrate solution must be standardized with a 
 standard solution of urea of 2 per cent, strength.
 

 
 NOTES ON CHEMICAL LECTURES. 73 
 
 Preparation of the standard urea solution : 2.0 grm. dry 
 urea are dissolved in distilled water and the solution is diluted 
 to 100 c.c. 
 
 Urea. 
 
 100 c.c. = 2.0 grm. 
 
 10 c.c. = 0.200 " 
 
 1 c.c. = 0.020 " 
 
 The indicator is a strong solution of sodium carbonate. 
 With mercuric nitrate it produces yellowish-brown basic 
 mercuric oxycarbonate, HgCO 3 (HgO) 3 . 
 
 4Hg(NO 3 ) 2 + 4Na 2 CO 3 = HgCO 3 (HgO) 3 + 8NaNO 3 + 3CO 2 
 
 Standardizing the mercuric nitrate solution : To stand- 
 ardize the mercuric nitrate solution, 10 c.c. of the standard 
 urea solution (== 0.200 urea) are placed in a beaker, and the 
 mercuric nitrate solution is run into it from a burette, stirring 
 after each addition, until a drop of the liquid in the beaker 
 produces a slight yellow color when brought, by means of a 
 glass rod, in contact with sodium carbonate solution on a 
 porcelain tablet. 
 
 If 1 c.c. of the mercuric nitrate solution is equal to 0.010 
 urea, then 20 c.c. should be required to combine with the urea 
 (0.200) present in the 10 c.c. urea solution and act on the 
 indicator. 
 
 1000 c.c. contains 5.2 grm. HgO to act on the indicator. 
 1 c.c. " 0.0052 " " 
 
 20 c.c. HgO solution having been required for the 10 c.c. 
 urea solution, and as each c.c. HgO solution contains 0.0052 
 HgO to act on the indicator, 20 X 0.0052 or 0.104 HgO must 
 have been added with the 20 c.c. HgO solution to the 10 c.c. 
 urea solution. Hence, in the combined volumes, 20 c.c. -f- 10 
 c.c. = 30 c.c., there were present 0.104 HgO, and in each c.c. 
 0.104 -r- 30 = 0.003466 mgrm. HgO to act on the indicator. 
 
 If a larger or smaller quantity than 20 c.c. should be 
 required, the mercuric nitrate solution must be corrected after 
 the manner given for solutions too strong or too weak in the 
 correction of the argentic nitrate solution for the determination 
 of sodium chloride.
 
 74 NOTES ON CHEMICAL LECTURES. 
 
 The phosphates, sulphates, and carbonates in the urine inter- 
 fere with" the application of the method. They are removed 
 by the baryta mixture, composed of 
 
 2 volumes of a cold saturated solution of barium hydroxide 
 (Ba(OH) 2 ). 
 
 1 volume of a cold saturated solution of barium nitrate 
 (Ba(N0 3 ) 2 ). 
 
 Method : 40 c.c. urine are mixed with 20 c.c. baryta mix- 
 ture, and filtered through a filter that has not been previously 
 moistened with water. The filtrate is neutralized with a drop 
 or more of nitric acid. 15 c.c. of the filtrate (representing 
 10 c.c. urine and 5 c.c. baryta mixture) are transferred by 
 means of a pipette to a beaker, and the mercuric nitrate solu- 
 tion is run into it from a burette, stirring after each addition, 
 until a permanent milky turbidity is produced (action on the 
 urea). The quantity of mercuric nitrate solution added up to 
 this point was required by the sodium chloride. 
 
 The number of c.c. mercuric nitrate solution added up to 
 this point is noted and the number noted multiplied by 0.0418, 
 will represent the quantity of sodium chloride in 10 c.c. of the 
 urine, and this, multiplied by }0, gives the percentage. 
 
 The addition of the mercuric nitrate solution is continued, 
 stirring after each addition, until a drop of the liquid in the 
 beaker produces a slight yellow color when brought, by means 
 of a glass rod, in contact with the indicator (sodium carbonate) 
 on a porcelain tablet. 
 
 The total number of c.c. of mercuric nitrate solution required 
 for the sodium chloride and urea is read off. From this number 
 the quantity required for the sodium chloride is deducted and 
 the remainder will be the quantity required for the urea. 
 
 If more than 30 c.c. of mercuric nitrate solution have been 
 required for the urea alone, then for every 4 c.c. required over 
 30 c.c. add 0.1 c c. to the number of c.c. of mercuric nitrate 
 solution required for the urea alone. 
 
 If less than 30 c.c. of mercuric nitrate solution have been 
 required for the urea alone, then for every 4 c.c. required less 
 than 30 c.c. deduct 0.1 c.c. from the number of c.c. of mercuric 
 nitrate solution required.
 
 NOTES ON CHEMICAL LECTURES. 75 
 
 The corrected number of c.c. mercuric nitrate solution is 
 multiplied by 0.010, and the result will represent the quantity 
 of urea in 10 c.c. of the urine, and this result multiplied by 10, 
 gives the percentage of urea in the urine. 
 
 Example : Suppose a total quantity of 36 c.c. mercuric 
 nitrate solution had been required, 
 
 36 c.c. 
 
 _2 c.c. for NaCl. 
 
 34 c.c. for urea. 
 
 30 
 
 4 (once 4 over 30). 
 Hence 
 
 34.0 c.c. + 0.1 = 34.1 c.c. 
 341 X 0.010 = 0.341 grm. 
 0.341 X 10 = 3.41 per cent. urea. 
 
 These corrections are necessary to make the results corre- 
 spond to the end reaction obtained in standardizing the mer- 
 curic nitrate solution, in which there were just two volumes 
 of the mercuric nitrate solution employed for one volume of 
 the urea solution. 
 
 CORRECTIONS FOR VARYING QUANTITIES OF UREA. 
 
 a. If the undiluted urine under examination contains over 
 3 per cent, of urea, then on mixing it with the baryta fluid in 
 the proportion of 10 c.c. of the urine with 5 c.c. of the baryta 
 fluid, the resulting mixture will contain over 2 per cent, of urea. 
 
 Under these circumstances 15 c.c., or say one volume, of 
 the urine mixture will require over 30 c.c., or two volumes, of 
 the mercuric nitrate solution for complete precipitation of the 
 urea. 
 
 Consequently, the excess of the mercuric oxide originally 
 present in the mercuric nitrate solution for the purpose of 
 acting upon the indicator (sodium carbonate) will be under a 
 less degree of dilution than was present when the mercuric 
 nitrate solution was standardized, and hence a portion of this 
 excess will be consumed in the precipitation of the urea, there 
 still being left in excess a quantity of mercuric oxide equal to 
 10
 
 76 NOTES ON CHEMICAL LECTURES. 
 
 that originally present when the mercuric nitrate was stand- 
 ardized ; namely, 3.466 milligrammes in each c.c. of the end 
 mixture. 
 
 The quantity of the excess of mercuric oxide thus consumed 
 by urea can be accurately compensated for by adding to the 
 number of cubic centimetres of mercuric nitrate solution em- 
 ployed 0.1 c.c. for every 4 c.c. of the mercuric nitrate solution 
 required above 30 c.c. 
 
 Thus if 34 c.c. of the mercuric nitrate solution are required, 
 it should be read 34.1 c.c., which would indicate that the un- 
 diluted urine contained 3.41 per cent, of urea. 
 
 b. Should the undiluted urine contain less than 3 per cent, 
 of urea, then on mixing 10 c.c. of the urine with 5 c.c. of the 
 baryta fluid, the resulting mixture will contain less than 2 per 
 cent, of urea, and, consequently, 15 c.c. of the mixture will 
 require less than 30 c.c. of standard mercuric nitrate solution 
 for the complete precipitation of the urea. 
 
 Under these conditions the original excess of mercuric oxide, 
 in the mercuric nitrate solution for the purpose of acting upon 
 the indicator, will be under a greater degree of dilution than 
 existed when the mercuric nitrate solution was standardized, 
 and, therefore, a portion of the mercuric oxide intended for the 
 precipitation of the urea, will be consumed by acting on the 
 indicator. Hence, when less than 30 c.c. of the mercuric 
 nitrate solution are required for a mixture of 10 c.c. of urine 
 and 5 c.c. of baryta fluid, deduct from the reading 0.1 c.c. for 
 every 4 c.c. of mercuric nitrate solution required less than 
 30 c.c. Thus, if 26 c.c. of mercuric nitrate solution were 
 required, it should be read 25.9 c.c., which would indicate that 
 the undiluted urine contained 2.59 per cent, of urea. 
 
 In other words, these corrections may be stated as follou's : 
 For each c.c. of mercuric nitrate solution required above two 
 volumes of the mercuric nitrate solution for one volume of 
 the urine mixture, add 0.025 c.c. to the reading ; and for each 
 c.c. of the mercuric nitrate solution required less than two 
 volumes of mercuric nitrate solution for one volume of the 
 urine mixture, deduct 0.025 c.c. from the reading. 
 
 Examples : A. 10 c.c. of 2 per cent, urea solution, plus 
 20 c.c. of the HgO solution. Thus, in each c.c. of the HgO
 
 NOTES ON CHEMICAL LECTURES. 77 
 
 solution there are 5.2 milligrammes of HgO in excess to act 
 on the indicator. In 20 c.c. there are 5.2 X 20 = 104 milli- 
 grammes excess of HgO in the 30 c.c. of end mixture. There- 
 fore in each c.c. of the 30 c.c. of mixture there are 104 -=- 30 or 
 3.46(3 milligrammes of HgO in excess to act on the indicator. 
 
 B. 10 c.c. of urine containing 3 per cent, of urea, plus 5 c.c. 
 of baryta niixf2ire, plus 30 c.c. of HgO solution. In each c.c. 
 of HgO solution there are 5.2 milligrammes excess of HgO. 
 In 30 c.c. there are 30 X 5.2 milligrammes = 156 milligrammes 
 excess in 45 c.c. of end mixture. Therefore in 1 c.c. there are 
 156 -*- 45 = 3.466 milligrammes. 
 
 C. 10 c.c. of urine containing 4 per cent, of urea, plus 5 c.c* 
 of baryta mixture, plus 40 c.c. of HgO solution. In 40 c.c. of 
 HgO solution there are 40 X 5.2 = i08 milligrammes excess 
 of HgO in 55 c.c. of end mixture. Therefore in 1 c.c. there 
 are 208 -4- 55 = 3.781. But adding 5 c.c. of H,O to the mix- 
 ture, there are 208 milligrammes of HgO excess in 60 c.c. 
 mixture, and, therefore, in 1 c.c. there are 208 -=- 60 = 3.466. 
 
 In the case of a urine containing 4 per cent, of urea, 10 c.c. 
 of the urine would contain 400 milligrammes of urea and, 
 with the dilution of 5 c.c. of baryta mixture, should require 
 40 c.c. of standard mercuric nitrate solution. As a matter of 
 fact the action on the indicator will occur when 39.76 c.c. of 
 mercuric nitrate solution have been added, for after that quan- 
 tity has been added, each c.c. of the end mixture will contain 
 3.460 milligrammes of HgO. Each c.c. of the mercury solu- 
 tion contains 72 mgrms. of HgO * which is designed to com- 
 bine with 10 mgrms. urea, therefore, 400 mgrms. of urea 
 would require 40 X 0.072 = 2.880 grms. HgO. In the 39.76 
 c.c. mercury solution actually required there are 39.76 X 
 0.0772 HgO = 3.069 grms. HgO. From this quantity of 
 HgO, which is designed for both urea and indicator, must be 
 deducted the quantity of HgO required for the 400 mgrms. of 
 urea, namely, 2.880 grms. Thus, 3.069 2.880 = 0.189472 
 mgrms., which is the quantity of HgO distributed in the end 
 mixture to act upon the indicator, consequently in each c.c. of 
 the end mixture (composed of 10 c.c. urine -f 5 c.c. baryta 
 mixture -j- 39.76 c.c. mercury solution 54.76 c.c.) there are 
 
 * Othp than the 5.2 mgrms. HgO for the indicator.
 
 78 NOTES ON CHEMICAL LECTURES. 
 
 0.189472 -r- 54.76 = 0.00346 mgrms. HgO to act upon the 
 indicator. 39.76 c.c. mercury solution^ employed 30 c.c. 
 mercury solution the basis for which no correction is neces- 
 sary, = 9.76 -=- 4 = 2.44 the number of tenths which must be 
 added to the number 39.76 to make it correspond to the 
 correct number of cubic centimetres, namely 40. Therefore, 
 in the correction, each 4 c.c. of mercury solution used above 
 30 c.c. are equivalent to one-tenth of a cubic centimetre of the 
 standard mercuric nitrate solution. 
 
 D. 10 c.c. of urine containing 2 per cent, of urea, plus 5 c.c. 
 of baryta mixture, plus 20 c.c. of HgO solution. In 20 c.c. of 
 HgO solution there are 5.2 X 20 or 104 milligrammes of HgO 
 excess in 35 c.c. of end mixture. Therefore in each c.c. there 
 are 104 -=- 35 = 2.971. But 10 c.c. of a 2 per cent, urea solu- 
 tion plus 5 c.c. of baryta mixture will require 20.23 c.c. of 
 HgO. In 20 c.c. of HgO solution there are 20 X 5.2 = 104 
 milligrammes in excess. Each c.c. of the standard solution of 
 HgO contains 77.2 milligrammes of HgO, therefore 0.23 c.c. 
 will contain 77.2 X 0.23 = 17.75 in excess, plus 104 = 121.75 
 -4- 35.23 = 3.466 mgrms. 
 
 PHOSPHORIC ACID. 
 
 Phosphoric acid occurs in the urine partly in combination 
 with sodium, potassium, and ammonium (alkaline phosphates), 
 and partly with calcium and magnesium (earthy phosphates). 
 
 It also occurs in small quantity in the urine as glycerine 
 phosphoric acid (C 3 H 9 PO 6 ). 
 
 /OH 
 
 PO OH = Glycerine phosphoric acid. 
 
 \OC 3 H 5 (OH) 2 
 
 About four-fifths of the phosphoric acid in the urine is in 
 combination as alkaline phosphates, and one-fifth as earthy 
 phosphates. 
 
 Phosphoric acid in the urine is determined in terms of P 2 O 5 
 (phosphoric anhydride). 
 
 The volumetric method for the quantitative deter- 
 mination of phosphoric acid in the urine depends upon the
 
 f 
 
 : 
 
 -f-
 
 NOTES ON CHEMICAL LECTURES. 79 
 
 principle that P 2 O 5 in the presence of free acetic acid and an 
 acetate of an alkali, on the application of heat, is precipitated 
 by uranium acetate, or nitrate, as insoluble (UrO 3 ) 2 P 2 O 5 . 
 
 It is always necessary to standardize the uranium solution 
 with a standard solution of P 2 O 3 . 
 
 The standard solution of P 2 O 5 is prepared so that it shall 
 contain 0.2 per cent. P 2 O 5 . 
 
 1. Preparation of standard solution of P 2 O 5 : Di-sodic 
 hydrogen phosphate (Na 2 HPO 4 + 12H 2 O) is selected for this 
 purpose. 
 
 Two molecules of Na 2 HPO 4 -j- 12H 2 O contain one molecule 
 of P 2 5 . 
 
 Na 2 HPO 4 + 12H 2 Q = 358 X 2 = 716 
 PA = 142 
 
 Hence, to determine the quantity of Na 2 HPO 4 -f- 12H 2 O 
 necessary to prepare 1000 c.c. solution, so that it shall contain 
 2 grm. P 2 O 5 (0.2 per cent.), 
 
 P 2 C . 2NaHPO 4 + 12H 2 O. P 2 O 5 . Na 2 HPO 4 + 12H 2 O. 
 
 142 : 716 :: 2 : 10.085 grm. 
 or to prepare only 250 c.c. of the standard Na 2 HPO 4 + 12H 2 O 
 solution, 
 
 NajjHP0 4 + 12H,O. 
 
 142 : 716 :: 0.5 : 2.521 grm. 
 
 10.085 grm. pure non-effloresced di-sodic hydrogen phos- 
 phate are dissolved in water and the solution diluted to 1000 
 c.c., or to prepare only 250 c.c. of the solution, 2.521 grm. of 
 the phosphate are dissolved in water and diluted to 250 c.c. 
 
 P 2 5 . 
 
 1000 c.c. = 2.000 grm. 
 100 c.c. = 0.200 " 
 50 c.c. = 0.100 " 
 1 c.c. = 0.002 " 
 
 2. Preparation of uranium acetate, or nitrate solution ; 
 The precipitate formed when uranium acetate, or nitrate, is 
 added to a solution containing P 2 O 5 is (UrO 3 ) 2 P 2 O 5 , /. e., two 
 molecules of UrO 3 in combination with one molecule of P 2 O 5 . 
 
 288. 
 
 Twice the molecular weight of UrO 3 is 576 
 
 142. 
 
 Once " " P 2 O 5 " 142
 
 80 NOTES ON CHEMICAL LECTURES. 
 
 The quantity of uranium oxide (UrO 3 ) required to prepare 
 1000 c.c. of standard solution so that 1 c.c. shall" equal 0.010 
 P 2 O 5 is determined : 
 
 P 2 O 5 . 2UrO 3 . PoO 5 . UrO 3 . 
 
 142 : 576 :: 10 : 40.56 grm. = to 10 grm. P 2 O 5 
 The quantity of uranium acetate (UrO 3 (C 2 H 3 O 2 ) 2 + 2H 2 O) 
 or of the nitrate (UrO 3 N 2 O 5 + 6H 2 O) equivalent to 40.56 grm. 
 UrO 3 is determined : 
 
 Uran. acetate. UrOs. Uran. acetate. 
 
 288 : 442 :: 40.56 : 62.24 grm. 
 
 UrOs- Uran. nitrate. UrOg. Uran. nitrate 
 
 288 : 504 :: 40.56 : 70.98 grm. 
 
 The quantity of uranium acetate or of nitrate necessary 
 to prepare 1000 c.c. of solution may also be calculated by one 
 proportion : 
 
 P 2 O 5 . 2 Uran. acetate. P 2 O 5 . Uran. acetate. fl / W / "7 &4* 
 
 142 : 884 :: 10 : 62.24 grm.^^ * 
 
 PoO 5 . 2 Uran. nitrate. PoOs- Uran. nitrate. 
 
 142 : 1008 :: 10 : 70.98 grm. 
 
 Because of uranium acetate and nitrate being contaminated 
 with oxides of uranium the respective quantities, as indicated 
 by the above proportions, are dissolved in 900 c.c. water 
 instead of 1000 c.c. 
 
 Ammonium salts in the urine interfere with uranium nitrate, 
 therefore uranium acetate is preferred in the preparation of the 
 standard solution. 
 
 62.24 grm. uranium acetate or 70.98 grm. uranium nitrate 
 are boiled with about 850 c.c. water, the liquid allowed to cool, 
 the insoluble oxides of uranium are removed by filtration and 
 the solution diluted to 900 c.c. 
 
 3. Preparation of the solution containing an acetate of 
 an alkali and free acetic acid : 50 grm. sodium acetate are 
 dissolved in 450 c.c. water, and acetic acid is added until the 
 volume reaches 500 c.c. 
 
 4. The indicator is a solution of potassium ferrocyanide of 
 about ten per cent, strength. With a soluble uranium salt it 
 produces a chocolate color (due to formation of uranium ferro- 
 cyanide).
 
 
 
 
 V
 
 NOTES ON CHEMICAL LECTURES. 81 
 
 Standardizing the uranium acetate solution : To stand- 
 ardize the uranium acetate solution, 50 c.c. of the standard 
 phosphoric acid solution (containing 0.100 grm. P 2 O 5 ) are 
 placed in a beaker, 5 c.c. of the acetate of an alkali solution 
 are added and the liquid is heated to the simmering-point. 
 The uranium acetate solution (about one-half c.c. at a time) is 
 run into it from a burette, stirring after each addition, until a 
 drop of the liquid in the beaker produces a slight chocolate 
 color when brought, by means of a glass rod, in contact with 
 the potassium ferrocyanide solution on a porcelain tablet. 
 
 The number of cubic centimetres of uranium acetate solu- 
 tion required to effect this result is noted. 
 
 Example : Suppose 8 c.c. of uranium solution were re- 
 quired. The solution is too strong and must be diluted. For 
 every 8 c.c. of the uranium solution (of the 900 c.c.) remaining, 
 a volume of water equal to the difference between 8 and 10, 
 or 2 c.c. must be added. 
 
 900 c.c. original volume of solution. 
 8 c.c. volume used. 
 
 8)892 c.c. volume remaining. 
 
 111.5 X 2 = 223 c.c. water to be added to the 892 c.c. 
 uranium solution. 
 
 P 2 5 . 
 
 10 c.c. will now equal 0.100 grm. 
 1 c.c. " " " 0.010 " 
 
 Method : 50 c.c. urine are placed in a beaker, 5 c.c. acetate 
 of an alkali solution are added and the liquid is heated to the 
 simmering-point. The uranium acetate solution (about one- 
 half c.c. or less, at a time) is slowly run from a burette into the 
 liquid in the beaker, stirring after each addition, until a drop 
 of the liquid in the beaker produces a slight chocolate color 
 when brought, by means of a glass rod, in contact with potas- 
 sium ferrocyanide solution on a porcelain tablet. 
 
 The first titration (adding about 0.5 c.c. at a time) usually 
 gives only approximate results, unless performed with great 
 care. A second titration should be made, and the uranium 
 solution slowly run in until within 0.5 c.c. of the quantity 
 required in the first titration. The uranium solution is now 
 added, 0.1 c.c. at a time, and continued, testing with the indicator
 
 82 NOTES ON CHEMICAL LECTURES. 
 
 after each addition, until a slight chocolate color is obtained in 
 the potassium ferrocyanide solution on the porcelain tablet. 
 
 The number of cubic centimetres of uranium acetate solution 
 required is read off the burette and this number is multiplied 
 -by 0.010, and the result will represent the quantity of P 2 O 5 in 
 50 c.c. urine. This result multiplied by 2 gives the percentage 
 of P 2 O 5 in the urine. 
 
 Example : Suppose 8.6 c.c. uranium solution were required. 
 Then 
 
 8.6 X 0.010 = 0.086 X 2 = 0.172 per cent. P 2 O 5 (total). 
 
 QUANTITATIVE DETERMINATION OF THE EARTHY 
 PHOSPHATES IN URINE. 
 
 To 200 c.c. urine excess of ammonium hydroxide (NH 4 OH) 
 (about 10 c.c. ordinary strength NH 4 OH), is added and the 
 liquid allowed to stand twelve hours. The precipitated earthy 
 phosphates (phosphates of calcium and magnesium) are col- 
 lected on a filter and washed several times with small quanti- 
 ties of water containing a few drops of NH 4 OH. A beaker 
 with a 50 c.c. mark on it is now placed under the funnel. The 
 filter, while in the funnel, is pierced with a glass rod, and the 
 precipitate treated drop by drop with about 3 or 4 c.c. acetic 
 acid. This is for the purpose of dissolving the earthy phos- 
 phates. The remainder of the precipitate on the filter is 
 washed with water into the beaker below until 50 c.c. liquid 
 are collected in the beaker. Any insoluble matter in the liquid 
 in the beaker is mostly mucous from the urine. No attention 
 need be paid to this. 
 
 5 c.c. of acetate of alkali solution are added and the liquid 
 is heated to the simmering-point. 
 
 The uranium acetate solution (less than one-half c.c. at a 
 time) is slowly run from a burette into the liquid in the beaker, 
 stirring after each addition, until a drop of the liquid in the 
 beaker when brought, by means of a glass rod, in contact with 
 potassium ferrocyanide solution on a porcelain tablet, produces 
 a slight chocolate color. 
 
 The number of cubic centimetres of uranium acetate solution 
 required is read off the burette and this number is multiplied
 
 NOTES ON CHEMICAL LECTURES. 83 
 
 by 0.010, the result is divided by 2 (because 200 c.c. of urine 
 had been originally employed). The final result will represent 
 the percentage of P 2 O 5 in combination with the alkaline earths 
 in the urine. 
 
 Example : Suppose 5.4 c.c. uranium acetate solution were 
 required. 
 
 5.4 X 0.010 = 0.054 -5- 2 = 0.027 per cent, of P 2 O 5 
 
 The percentage of earthy phosphates is deducted from the 
 percentage of total phosphates (alkaline + earthy), and the 
 remainder is the percentage of P 2 O 5 in combination with the 
 alkalies. 
 
 Suppose the percentage of total P 2 O 3 = 0.172 per cent. 
 
 earthy " =0.027 
 
 alkaline " =0.145 
 Phosphoric acid often occurs in urinary sediments as triple 
 
 phosphate (magnesium ammonium phosphate, MgNH 4 PO 4 ). 
 Phosphoric acid may occur in combination in the form of 
 
 urinary calculi. Phosphatic calculi do not occur as frequently 
 
 as uric acid calculi. 
 
 PHOSPHATIC CALCULI MAY OCCUR IN THREE 
 FORMS. 
 
 1. Composed of calcium phosphate : Three varieties, 
 acid, neutral, and basic. 
 
 a. CaH 4 (P0 4 ) 2 
 
 b. CaHPO 4 
 ft Ca 3 (P0 4 ) 2 
 
 All are soluble in nitric acid ; when their solution is neutral- 
 ized with NH 4 OH they are reprecipitated. 
 Heat does not affect them. 
 
 2. Composed of magnesium ammonium phosphate 
 (MgNH 4 PO 4 ), (triple phosphate) the most common form. 
 
 Soluble in hydrochloric acid. When the hydrochloric acid 
 solution is neutralized with NH 4 OH, crystalline triple phos- 
 phate appears. When heated, ammoniacal gas is evolved. 
 
 3. Fusible phosphates, usually composed of i and 2 
 (above). Contain more or less organic matter. 
 
 11
 
 84 NOTES ON CHEMICAL LECTURES. 
 
 URIC ACID, (Lithic acid). C.H 4 N 4 O 3 . 
 
 Molecular weight, 168. 
 
 A dibasic acid, /. e., contains two replaceable atoms of 
 hydrogen, H 2 C 5 H 2 N 4 O 3 - 
 
 Found in urine chiefly as sodium acid urate, NaHC-H 2 N 4 O 3 . 
 
 Present from 0.2 part to 1.0 part in 1000 parts urine. 
 
 Average percentage in urine 0.05 per cent. 
 
 Recognized as a distinct compound by Scheele in 177<). 
 Studied more fully by Liebig in 1838. Was prepared syn- 
 thetically in 1882. 
 
 Formerly called lithic acid. 
 
 It is contained in the solid excrement of birds and serpents, 
 from which source it is most readily obtained by treating the 
 excrement with a 5 per cent, solution of NaOH, filtering, and 
 adding HC1 to the filtrate. The uric acid is precipitated in an 
 amorphous state or, on standing, may separate in crystalline 
 form. 
 
 It may be deposited in the joints, as in the uric acid diathesis 
 (gout). 
 
 To obtain pure and colorless crystals the acid obtained from 
 about 200 c.c. of urine is dissolved in about 75 c.c. water to 
 which about 1 c.c. of a 10 per cent, solution of NaOH has 
 been added. The solution is slightly acidulated with HC1, 
 and allowed to stand twenty-four hours in a cool place. Uric 
 acid will separate. If the crystals are not colorless, they are 
 collected on a filter, redissolved in water with the aid of NaOH, 
 and HC1 added as before. The operation is repeated, if neces- 
 sary, until colorless crystals are obtained. 
 
 Uric acid crystallizes in many forms, most often in wedge- 
 shaped crystals. 
 
 It is soluble in 15,000 parts of cold and 2000 parts of hot 
 water. Insoluble in alcohol and ether. Soluble in an alkaline 
 solution and in sulphuric acid. Insoluble in hydrochloric 
 acid. 
 
 When burned, an odor similar to burnt feathers is pro- 
 duced. 
 
 It is dibasic ; forms neutral and acid salts.
 
 NOTES ON CHEMICAL LECTURES. 85 
 
 Neutral salts : a. K 2 C 5 H 2 N 4 O 3 , potassium urate, soluble in 
 44 parts of cold water. 
 
 ~& Na 2 C 5 H 2 N 4 O 3 , sodium urate, less soluble than the corre- 
 sponding neutral potassium salt. 
 
 c. The neutral ammonium salt is unknown. 
 
 Acid salts ; a. KHC 5 H 2 N 4 O 3 ,, acid potassium urate, soluble 
 in SpOparts of water. 
 
 b~. NaTlC 5 H 2 N 4 O 3 , acid sodium urate, less soluble than the 
 corresponding potassium salt. 
 
 c. NH 4 HC 5 H 2 N 4 O 3 , acid ammonium urate, soluble in 1800, 
 parts of water. 
 
 Uric acid treated with cold strong nitric acid (specific 
 gravity 1.41) is oxidized, with the formation of alloxan 
 (C 4 H 2 N 2 O 4 ). Effervescence occurs, and urea is produced at 
 the same time. 
 
 C 5 H 4 N 4 3 + H 2 + O = C 4 H 2 N 2 4 +/6o(NH 2 ) 
 
 Uric acid. Alloxan. \ 
 
 s 
 
 Uric acid treated with hot dilute nitric acid is oxidized, with 
 the formation of alloxantin (C 8 H 4 N 4 O 7 ). 
 
 QUALITATIVE TESTS FOR URIC ACID. 
 
 1. Murexide test: The solid uric acid, or its solution 
 evaporated to dryness, is placed in a small porcelain dish and 
 the uric acid covered with strong nitric acid. The mixture is 
 evaporated to dryness on a water-bath, the dish allowed to 
 cool and the residue is moistened with a drop of very dilute 
 ammonium hydroxide. A beautiful red color will appear, 
 due to the formation of murexide (ammonium purpurate), 
 C 8 H 8 N 6 6 = (NH 4 C 8 H 4 N 5 6 ). 
 
 2. SchifFs test : The uric acid is dissolved in a solution of 
 sodium carbonate, and a drop of the solution is brought in 
 contact with filter-paper which has been previously saturated 
 with a solution of argentic nitrate. Spots of reduced silver of 
 a yellowish-brown or deep-black color, depending upon the 
 quantity of uric acid in the solution, will appear on the paper. 
 
 Hippuric acid (C 9 H 9 NO 3 ) is contained in very small quantity 
 in human urine.
 
 86 NOTES ON CHEMICAL LECTURES. 
 
 Benzole acid (C 7 H 6 O 2 ) when taken into the human organism 
 is converted into hippuric acid through the agency of glycocol 
 (C 2 H 5 NO 2 ), a substance formed in the liver, thus : 
 
 C 7 H 6 O 2 + C 2 H 5 NO 2 = C 9 H 9 NO 3 + H 2 O 
 
 Benzoic acid. Glycocol. Hippuric acid. 
 
 Quinic acid (C 7 H 12 O 6 ), one of the acids contained in cinchona 
 bark, and also in coffee-beans, is eliminated as hippuric acid. 
 Hippuric acid appears in the urine after eating cranberries. 
 
 QUANTITATIVE DETERMINATION OF URIC ACID BY MEANS 
 
 OF HCI. 
 
 1. Heintz's method: If albumen be present it must be 
 removed by coagulation by heat and filtration. To 200 c.c. 
 urine (if not clear, filter,) 5 or 10 c.c. HCI are added and the 
 liquid is allowed to stand about twenty-four hours. Uric acid 
 separates in crystalline form. Some of the crystals float on 
 the surface of the liquid. The crystals separated in this 
 manner are brownish-red in color owing to urine coloring 
 matter which they have taken up in the process of crystalliza- 
 tion. 
 
 The crystals are collected on a washed, or on an equipoised, 
 or a weighed washed filter. The filtrate is used to wash the 
 crystals out of the beaker. The crystals are washed with 
 small portions of water (5 c.c.) at a time until the last portions 
 of the filtrate coming from the funnel are free from hydro- 
 chloric acid (test with AgNO 3 solution). 
 
 The two equipoised filters are separated, the crystals allowed 
 to dry on the filter, and the two filters are weighed. 
 
 The weight obtained is the quantity of uric acid in 200 c.c. 
 urine. This weight divided by 2 furnishes the percentage of 
 uric acid in the urine. 
 
 If over 30 c.c. of water are required in the washing of the 
 crystals, then for every cubic centimetre of water employed 
 over 30 c.c. 0.000045 grm. must be added to the weight of 
 uric acid obtained. (1 c.c. water applied in this manner dis- 
 solves 0.000045 grm. uric acid). 
 
 No correction is necessary when only 30 c.c. are required 
 in the washing.
 
 NOTES ON CHEMICAL LECTURES. 87 
 
 The uric acid in crystallizing takes up coloring-matter from 
 the urine, but the amount of uric acid dissolved by the 30 c.c. 
 water used in washing is just sufficient to compensate for the 
 increase of weight due to the coloring-matter taken up. 
 
 2. x Salkowski-Ludwig method : Depends upon the prin- 
 ciple that when an ammoniacal solution of argentic nitrate is 
 added to x -^ solution of uric acid, to which an ammoniacal 
 mixture of magnesium sulphate and ammonium chloride has 
 been previously added, the uric acid is precipitated as a 
 magnesio-silver salt. This is collected on a filter, washed, 
 and decomposed by means of sodium or potassium sulphide, 
 whereupon the uric acid passes into solution as sodium or 
 potassium urate. On the addition of an excess of hydrochloric 
 acid to this solution the urate is broken up and uric acid 
 separates. The uric acid is collected on a previously weighed 
 or an equipoised filter and weighed. 
 
 3. Haycraft's method : Depends upon the principle that 
 when uric acid is precipitated by an ammoniacal solution of 
 argentic nitrate, in the presence of an ammoniacal mixture of 
 magnesium sulphate and ammonium chloride, the precipitate 
 is considered to contain one atom of silver to each molecule 
 of uric acid. The precipitate is collected on a filter and dis- 
 solved in nitric acid of about 1.12 to 1.18 specific gravity, in 
 which solution the uric acid is determined indirectly by deter- 
 mining the amount of silver by means of a one-fiftieth normal 
 solution of potassium sulphocyanate. 
 
 Other methods for the quantitative determination of uric acid 
 are those of Fokker, Camerer, and Czapek. 
 
 CREATININE. C 4 H 7 N 3 O. 
 
 Molecular weight, 113. 
 
 First recognized in 1844. Isolated by Liebig in 1847. 
 
 Occurs in the urine of man (0.06 per cent.), horse, cow, 
 sheep, and dog. 
 
 A substance supposed to be creatinine was found in muscle, 
 but was afterwards shown to be creatine (C 4 H 9 N 3 O 2 ). 
 
 Creatinine crystallizes in colorless prisms. Soluble in 12 
 parts of water and in 100 parts of alcohol.
 
 88 NOTES ON CHEMICAL LECTURES. 
 
 Its solution is strongly alkaline and will change red litmus 
 to blue, and will change turmeric paper brown. It has a 
 caustic taste, somewhat like dilute ammonium hydroxide. It 
 
 L&-4f4Ls\s*s^* 
 
 is the strongest of all bases of animal origin. 
 
 It unites with acidsyXvithout displacing the hydrogen in the 
 acid, to form salts, as 
 
 C 4 H-N 3 O HC1 (creatinine hydrochloride). 
 It combines with one molecule of water to form creatine : 
 C 4 H 7 N 3 O + H 2 O = C 4 H 9 N 3 O 2 
 
 Creatinine. Creatine. 
 
 It combines with zinc chloride to form creatinine zinc 
 chloride, (C 4 H 7 N 3 O) 2 ZnCl 2 , which contains 62.432 per cent, of 
 creatinine. 
 
 QUALITATIVE TESTS FOR CREATININE. 
 
 1 . Picric acid test : A solution of creatinine treated with 
 a few drops of sodium hydroxide and a little picric acid and 
 then warmed is colored claret-red. 
 
 2. Weyl's test ; A dilute creatinine solution treated with a 
 few drops of very dilute sodium nitroprusside solution, and 
 then with a dilute solution of sodium hydroxide, added drop 
 by drop, becomes ruby-red in color, changing in a few moments 
 to an intense straw color, which in turn becomes green when 
 warmed with acetic acid. 
 
 Creatinine reduces Fehling's solution changing the blue 
 liquid to yellow. The cuprous oxide does not separate, but 
 remains in solution. 
 
 QUANTITATIVE DETERMINATION OF CREATININE IN URINE. 
 
 300 c.c. urine are treated with milk of lime until slightly 
 alkaline, and a solution of calcium chloride is added as long 
 as a precipitate forms (may require 5 to 8 c.c. of 5 per cent. 
 CaCl 2 solution). The liquid is filtered and the filtrate is 
 evaporated to syrup-like consistence (to about 20 c.c.) on a 
 water-bath and, while warm, about 50 c.c. alcohol (95 per cent.) 
 are added. 
 
 The liquid is stirred with a glass rod until a precipitate is 
 formed. (The precipitate may form only after long stirring).
 
 NOTES ON CHEMICAL LETCURES. 89 
 
 The precipitate is collected on a filter and the dish and the 
 filter are washed with about 10 c.c. alcohol (in 2 or 3 portions), 
 collecting the wash-alcohol with the filtrate. (The precipitate 
 may be thrown away.) The beaker containing the filtrate is 
 covered and allowed to stand twenty-four hours in a cool 
 place. 
 
 A precipitate separating after the liquid has stood about 
 twenty-four hours, is collected on a small filter and washed 
 with about 10 c.c. of alcohol (in 2 or 3 portions), and the 
 washings collected with the filtrate. 
 
 The filtrate is concentrated by evaporation to about 60 c.c., 
 and, when cold, 1 c.c. of acid free zinc chloride solution of 
 about 1.2 specific gravity is added. 
 
 The liquid is stirred with a glass rod until a precipitate 
 (cloudiness) begins to appear. (The precipitate may appear 
 only after long stirring.) The beaker is then covered and 
 allowed to stand two or three days in a cool place to permit 
 crystallization to occur. Creatinine zinc chloride separates in 
 crystalline tufts or rosettes. 
 
 The precipitate of creatinine zinc chloride is collected on an 
 equipoised filter (using the filtrate, if necessary, to transfer the 
 precipitate from the beaker). 
 
 The precipitate on the filter is washed with alcohol (2 or 
 3 c.c. at a time) until the filtrate is colorless and free from 
 chlorine (test last portions of the filtrate coming from the 
 funnel with AgNO 3 ). The precipitate is dried on the filter at 
 a temperature of 100 C. and, when dry, weighed. 
 
 Every 100 parts (C 4 H 7 N 3 O) 2 ZnCl 2 contain 62.432 parts 
 creatinine. 
 
 Then, 
 
 100 : 62.432 :: weight of precipitate : x = quantity of 
 creatinine in the 300 c.c. urine employed. 
 
 To obtain the percentage of creatinine divide the result by 
 3, because 300 c.c. of urine were originally employed. 
 
 The creatinine may be obtained from the creatinine zinc 
 chloride by dissolving the latter in hot water and boiling it 
 from one-quarter to one-half hour with freshly precipitated, 
 well-washed plumbic hydroxide (prepared by precipitating
 
 90 NOTES ON CHEMICAL LECTURES. 
 
 plumbic acetate with ammonium hydroxide) or with basic 
 plumbic carbonate (prepared by precipitating plumbic acetate 
 with sodium carbonate). When cool, the liquid is filtered and 
 the filtrate decolorized by warming it with animal charcoal. 
 The filtrate from the animal charcoal is evaporated to dryness 
 on a water-bath. The residue, in addition to creatinine, con- 
 tains creatine, which has resulted from the breaking up of 
 some of the creatinine in the boiling with plumbic hydroxide. 
 To separate them the residue is treated with cold concentrated 
 alcohol, which dissolves the creatinine and leaves the creatine 
 undissolved. The liquid is filtered and the filtrate is allowed 
 to evaporate at ordinary temperature. Creatinine will separate 
 from the liquid in crystalline form. 
 
 SUGARS. 
 
 1. Glucose group : C 6 H 12 O 6 . 
 
 Dextrose (polarizes light to the right). 
 Lactose. 
 Grape sugar. 
 
 Laevulose (polarizes light to the left). 
 Fruit sugar. 
 Mannitose. 
 Inosite. 
 Sorbine. 
 Etc. 
 
 Glucoses, by the action of a ferment, are converted into 
 alcohol and carbon dioxide. 
 
 C 6 H 12 O 6 = 2C 2 H 5 OH + 2CO 2 
 
 Diastase, a ferment, converts starch (C 6 H 10 O 5 ) into dextrose 
 (C 6 H 12 O 6 ). 
 
 Starch warmed with dilute sulphuric acid or dilute hydro- 
 chloric acid is converted into dextrose (glucose). 
 
 3C 6 H 10 5 + 3H 2 + (H 2 S0 4 ) X = 3C 6 H 12 O 6 + (H 2 SO 4 ) X 
 
 Dextrine may be formed at the same time. 
 3C 6 H 10 5 + H 2 +(H 2 S0 4 ) X = C 6 H 12 G + 2C 6 H 10 O 5 + (H 2 SO 4 ) X 
 
 Dextrine. 
 
 Dextrine is a gummy substance.
 
 NOTES ON CHEMICAL LECTURES. 91 
 
 2. Saccharose group: C^H^On. 
 
 Saccharose (cane sugar). 
 Milk sugar. 
 Maltose. 
 Etc. 
 
 Cane sugar, C l2 H. a O llt warmed with dilute sulphuric acid is 
 converted into glucose and laevulose, the mixture of the two 
 latter is termed fruit sugar. 
 
 C M H a O u + H 2 + (H 2 S0 4 ) X = C 6 H 12 O ti + C 6 H 12 O 6 
 
 Glucose. Laevulose. 
 
 3. Amylose group : C 6 H 10 O;;. 
 
 Starch. 
 Dextrine. 
 Cellulose. 
 Arabinose. 
 Etc. 
 
 According to some investigators, the formula for starch, 
 instead of C 6 H 10 O 5 , should be 3C 6 H 10 O 5 = C^H^O^. 
 
 Other investigators suggest as the most probable formula 
 
 4C 6 H 10 O 5 = C^H^OjjQ. 
 
 Starch is not determined directly, but is converted into 
 dextrose (glucose) by being warmed with dilute sulphuric acid 
 or dilute hydrochloric acid. The quantity of starch is calcu- 
 lated from the quantity of dextrose (glucose) produced. 
 
 ABNORMAL CONSTITUENTS OF URINE. 
 
 The substances occurring in the urine only pathologically, 
 or only very seldom in soluble form, which need be considered 
 are as follows : 
 
 Albumin ; globulin ; hemialbumose ; peptone ; mucin ; glu- 
 cose ; milk sugar ; inosite ; dextrin ; bile acids ; bile coloring- 
 matters ; blood coloring-matter ; urorubrohaematin and uroru- 
 brofuscin ; melanin ; leucin ; t^rosin ; allantoin ; fat, lecithin, 
 and cholesterin ; acetone and alcohol ; Baumstark's substance, 
 C 3 H 8 N 2 O 8 ; urocaninic acid ; hydrogen sulphide ; diacetic acid ; 
 homogentisic acid C 8 H 8 O 4 . 
 
 12
 
 92 NOTES ON CHEMICAL LECTURES. 
 
 The abnormal constituents of the urine occurring most 
 frequently, and of the greatest importance to the physician, 
 are glucose and albumin. 
 
 DIABETIC URINE. 
 
 Urine containing glucose is diabetic urine. 
 
 Synonyms of glucose : dextrose, grape sugar, diabetic 
 sugar. 
 
 Glucose is an abnormal constituent of urine. 
 
 It is not contained in normal urine, although Pavy claims 
 that it is present in minute quantity in all urine. 
 
 Diabetic urine is generally paler in color than normal urine 
 and has a sweet taste. 
 
 The quantity of urine voided in diabetes mellitus is usually 
 greater than in health, and the urinous odor of normal urine 
 is generally absent. 
 
 Its specific gravity, as a rule, is higher than that of normal 
 urine (1030-1050, and even 1060). 
 
 Glucose may be present, varying from a trace up to 1012 
 per cent., and even 14 per cent. 
 
 When diabetic urine is exposed to the air for a time, at not 
 too low temperature, the surface often becomes covered with 
 a scum or mould (fungus), due to the multiplication of cells 
 of the torula cerevisiae, originating from cells derived from 
 the air. 
 
 A scum may also form on urine not containing glucose, but 
 the scum in such case is not due to the torula cerevisiae. 
 
 QUALITATIVE TESTS FOR GLUCOSE IN URINE. 
 
 1. Moore's test : To the urine contained in a test-tube 
 about one-fourth its volume of sodium or potassium hydroxide 
 solution (about 10 per cent, strength) is added and the liquid 
 is boiled. If glucose be present the liquid will become brown 
 or black in color, depending upon the quantity of glucose. 
 
 Delicacy of the test, 0.3 per cent, glucose. 
 
 The test is unreliable, because normal constituents of the 
 urine, particularly mucin, may give a similar coloration.
 
 NOTES ON CHEMICAL LECTURES. 93 
 
 2. Johnson's picric acid test ; Picric acid (carbazotic acid, 
 trinitrophenol), QH^NO^OH, is a derivative of carbolic acid 
 (phenol), C 6 H 5 OH. " 
 
 A few drops of a saturated aqueous solution of picric acid 
 and sufficient sodium hydroxide (NaOH), or KOH solution, 
 to render the urine alkaline are added to the urine and the 
 liquid is warmed. 
 
 If glucose be present the liquid will become claret-rgjdtjn color, 
 owing to the production of picramic acid (C 6 H 2 (NO 2 ) 2 NH 2 OH), 
 (resulting from the replacement of one of the NO 2 radicals in 
 picric acid by the NH 2 radical,) which enters into combination 
 with the alkali present to form a salt, a picraminate of the 
 alkali. 
 
 Delicacy of the test, 0.01 per cent, glucose. 
 
 This test is unreliable, because creatinine, a normal con- 
 stituent of the urine, responds in a similar manner, /. e., gives 
 a claret-red color. 
 
 3. Boettcher's test (bismuth test) : To the urine about 
 one^fourth its volume of sodium or potassium hydroxide solu- 
 tion is added. A portion of bismuth oxynitrate (subnitrate) 
 about the size of a mustard-seed is then added and the liquid 
 is boiled. 
 
 If glucose be present the bismuth oxynitrate will be reduced 
 to metallic bismuth, coloring the liquid first gray and then 
 black, or the reduced metallic bismuth may settle in the bottom 
 of the tube. 
 
 Delicacy of the test, 0.4 per cent, glucose. 
 
 Albumin must be removed before employing this test. 
 
 Albumin interferes with this test by producing a similar 
 change of color, due to the formation of black bismuth sul- 
 
 o 
 
 phide (Bi 2 S 3 ). The albumin undergoes decomposition when 
 heated with the caustic alkali, forming sodium sulphide (Na 2 S) 
 (the sulphur being derived from the albumin), which acts on 
 the bismuth compound to form bismuth sulphide. 
 
 4. The fermentation test ; Depends upon the breaking up 
 of glucose into alcohol and carbon dioxide by the action of a 
 ferment, /. e., yeast. 
 
 C G H 12 O 6 = 2C 2 H 5 OH + 2CO 2
 
 94 NOTES ON CHEMICAL LECTURES. 
 
 A test-tube is filled with the urine, and a drop or two of 
 brewer's yeast or a piece of compressed yeast about the size 
 of a pea is added. The tube is inverted over some of the 
 same urine contained in a dish, and stood aside for six or eight 
 hours in a place where the temperature is between 70 and 
 100 F. 
 
 If sugar be present it will undergo fermentation with the 
 evolution of carbon dioxide, which will collect in the upper 
 part of the inverted tube. 
 
 To prove that the gas in the tube is carbon dioxide, a piece 
 of sodium hydroxide is inserted in the tube, the opening of 
 the tube is closed with the thumb, and the tube agitated. The 
 carbon dioxide will be absorbed by the sodium hydroxide, 
 forming sodium carbonate. 
 
 As yeast itself may give off gas, a control experiment may 
 be performed by testing the yeast in the same manner, but 
 employing water in the test-tube in the place of urine. 
 
 Delicacy of the test, 0.4 per cent, glucose. 
 
 The quantity of carbon dioxide evolved from 0.4 per cent, of 
 glucose is just sufficient, at ordinary temperature, to saturate 
 the water in which it is contained and, therefore, will not appear 
 as gas in the upper part of the tube, hence the delicacy of the 
 test, under ordinary conditions, is 0.4 per cent, of glucose. 
 
 5. Trommer's test : Depends upon the reduction of cupric 
 oxide (CuO) in alkaline solution by glucose to red cuprous 
 oxide (Cu 2 O) or yellow cuprous hydroxide (Cu 2 (OH) 2 ). 
 ^xA-lbumin must be removed from the urine before making 
 Trommer's or Fehling's test. 
 
 To about 5 c.c. of the urine about one-fourth its volume of 
 sodium or potassium hydroxide solution is added. Then, dr0p 
 by drop, a solution of cupric sulphate (about 10 per cent, solu- 
 tion) is added and the liquid is agitated until the bluish-white 
 precipitate of cupric hydroxide (Cu(OH) 2 ) which first appears 
 ceases to be dissolved and the liquid presents a slightly turbid 
 or opaque appearance. If on the addition of cupric sulphate 
 the bluish-white precipitate of cupric hydroxide should dis- 
 solve on agitating the liquid and impart a purplish color to 
 the liquid instead of a pure blue, it may be taken as a fair 
 indication of the presence of glucose. The liquid is heated and 

 
 NOTES ON CHEMICAL LECTURES. 95 
 
 if glucose be present the cupric oxide will be reduced to red 
 or brownish-red cuprous oxide or yellow cuprous hydroxide. 
 
 Delicacy of the test, 0.01 per cent, of glucose in the urine. 
 
 The cupric sulphate solution must never be added to urine 
 containing sodium hydroxide while hot, or black cupric oxide 
 will be produced. 
 
 Uric acid has the property of reducing cupric oxide in alka- 
 line solution to cuprous oxide. 
 
 Creatinine has the property of reducing cupric oxide to 
 cuprous oxide and redissolving the latter. 
 
 If on the addition of one drop of cupric sulphate solution 
 to the urine rendered alkaline, and agitating the liquid, the 
 bluish-white precipitate of cupric hydroxide is dissolved, the 
 presence of glucose may be inferred. 
 
 If the liquid is heated to the boiling-point, and glucose be 
 present, cuprous oxide should appear. The test, however, 
 may fail to give visible cuprous oxide, or cuprous hydroxide 
 even though glucose be present. 
 
 In such a case several drops (3 to 6) of cupric sulphate 
 solution are added to a fresh mixture of urine and sodium 
 hydroxide, and the liquid agitated. If the bluish-white pre- 
 cipitate is dissolved the addition of the cupric sulphate is 
 continued until the liquid presents a slightly turbid or opaque 
 appearance. On heating the liquid, if glucose be present, a 
 precipitate of cuprous oxide or hydroxide should appear. 
 
 If the bluish-white precipitate of cupric hydroxide produced 
 on adding a few drops of cupric sulphate solution does not 
 wholly dissolve, and on heating the liquid cuprous oxide or 
 hydroxide fails to appear, it may be concluded that glucose is 
 not present. Its presence, however, is to be suspected when 
 the specific gravity of the urine is high and the cupric hydrox- 
 ide is dissolved. 
 
 If the cupric hydroxide is dissolved, and heating the liquid 
 fails to give cuprous oxide or hydroxide, glucose may still be 
 present, especially if the urine possesses a high specific gravity. 
 
 A. If more than 5 to 10 drops cupric sulphate solution are 
 required to produce the turbidity, a fresh portion of the urine 
 is diluted with \ or 9 volumes of water and Trommer's test 
 (adding sodium hydroxide and 5 to 10 drops of cupric
 
 96 NOTES ON CHEMICAL LECTURES. 
 
 sulphate) is applied. If the cupric hydroxide wholly dissolves 
 and, on heating, unsatisfactory results are obtained, a portion 
 of the fresh urine is more largely diluted with water, and the 
 test again applied. 
 
 ' B. If the results obtained by the foregoing methods are not 
 satisfactory, as may be the case even when glucose is present 
 in large quantity, a portion of Fehling's solution is diluted 
 with about 4 volumes of water, heated to the boiling-point, 
 and several drops of the suspected urine added to the hot 
 diluted Fehling's solution. This test may fail even when 
 glucose is present, but the results are generally satisfactory. 
 
 C. If the results are still unsatisfactory, the urine is passed 
 through animal charcoal and the filtrate tested for glucose. 
 
 D. Bruecke's lead process for the removal of interfering 
 substances from the urine. 
 
 If the* results are unsatisfactory, after having followed the 
 foregoing directions, the urine is examined by the lead process. 
 
 Lead process : 50 c.c. of the urine are treated with 60 c.c. 
 of about 10 per cent, solution of commercial plumbic acetate, 
 which precipitates the sulphates, phosphates, carbonates, 
 coloring-matter, and some of the uric acid and creatinine. 
 The liquid is filtered and the precipitate is washed once or 
 twice with water. Excess of ammonium hydroxide, which 
 precipitates the glucose in combination with lead as plumbic 
 saccharate (PbO) 3 (C 6 H 12 O 6 ) 2 , is added to the filtrate. 
 
 The precipitate is collected on a filter and washed until free 
 from ammonia. The filter paper is pierced with a glass rod 
 and the precipitate is washed through the aperture into a 
 beaker placed under the funnel. A stream of hydrogen sul- 
 phide (H 2 S) is passed through the mixture until all of the lead 
 is precipitated as black plumbic sulphide (PbS). 
 
 The black plumbic sulphide is filtered off and the precipitate 
 washed once or twice with water. The filtrate, with wash- 
 water, is evaporated to a volume of about 25 c.c., or until free 
 from hydrogen sulphide. 
 
 a. Trommer's test is applied to 5 or 6 c.c. of the liquid. 
 
 b. Fehling's test is applied to another portion. 
 
 After subjecting normal urine to the lead process and testing 
 the final solution, a slight reduction of the cupric oxide may
 

 
 NOTES ON CHEMICAL LECTURES. 97 
 
 occur, due to the presence of a small quantity of uric acid 
 which may have escaped removal in the process. This 
 probably is the reason for the statement that glucose is a 
 normal constituent of urine. The final solution may be 
 allowed to stand twenty-four or forty-eight hours and the uric 
 acid will separate in crystals, which may be filtered off and 
 the tests for glucose applied to the filtrate, or the final solu- 
 tion is evaporated to dryness on a water-bath, the glucose 
 is dissolved from the residue with alcohol, the liquid is 
 filtered, the alcohol solution evaporated to dryness on a 
 water-bath, the residue dissolved in about 25 c.c. of water and 
 Trommer's and Fehling's tests applied to portions of the 
 solution. 
 
 An inconstant loss of about 50 per cent, of the glucose 
 occurs in the course of the lead process. 
 
 6. Fehling's test : Depends upon the reduction of cupric 
 oxide (CuO) in alkaline solution by glucose to red cuprous 
 oxide (Cu 2 O) or yellow cuprous hydroxide (Cu 2 (OH) 2 ). 
 
 Fehling's solution must always be tested before being used, 
 by diluting it with about four volumes of water and heating to 
 the boiling-point. If it has undergone decomposition, a reduc- 
 tion of the cupric oxide with the separation of cuprous oxide 
 or hydroxide will occur on heating the diluted solution. 
 
 Fehling's solution which has undergone decomposition is 
 unfit for use. 
 
 Application of Fehling's test : About 1 c.c. of Fehling's 
 solution is diluted with about 4 c.c. of water and heated to the 
 boiling-point, and the urine, 2 or 3 drops at a time, is added 
 to the liquid which is heated to the boiling-point after each 
 addition of urine. 
 
 If glucose be present reduction will occur, and a precipitate 
 of red cuprous oxide or yellow cuprous hydroxide will be 
 formed. 
 
 Delicacy of the test, 0.001 per cent, glucose or one part of 
 glucose in 60,000 parts of dilution. 
 
 Often in applying Trommer's test and also Fehling's test to 
 the urine, a brownish-red flocculent precipitate, due to phos- 
 phates, is produced. This is often mistaken for cuprous oxide 
 and, in consequence, glucose believed to be present in the
 
 98 NOTES ON CHEMICAL LECTURES. 
 
 specimen of urine. In differentiating between the precipitate 
 caused by phosphates and the precipitate of cuprous oxide 
 caused by the presence of glucose it may be observed, that 
 the precipitate of phosphates is flocculent, is unevenly distrib- 
 uted throughout the liquid, settles to the bottom slowly and 
 does not form a compact layer at the bottom of the tube. The 
 precipitate of cuprous oxide is not flocculent, is evenly dis- 
 tributed throughout the liquid, settles to the bottom rapidly 
 and forms a compact layer at the bottom of the tube. 
 
 A considerable number of substances occur under normal 
 or pathological conditions in the urine which possess the 
 .property of reducing cupric oxide in alkaline solution (Trom- 
 mer's and Fehling's tests), such as uric acid, creatinine, 
 creatine, allantoin, mucin, milk sugar, pyro-catechin, hydro- 
 chinon, glycuronic acid, bile coloring-matters, and homo- 
 gentisic acid. 
 
 Homogentisic acid, C 8 H 8 O 4 , may be detected in urine by 
 placing the urine in a test-tube, rendering it alkaline by the 
 addition of sodium hydroxide, closing the tube with the thumb 
 and shaking the tube so that the liquid is brought intimately 
 in contact with the air in the tube. In the presence of homo- 
 gentisic acid oxidation products will be formed and the liquid 
 will become brown or black in color depending upon the 
 quantity of homogentisic acid present. (Pyrocatechin and 
 hydrochinon respond in a similar manner to the test.) 
 
 On the ingestion of certain compounds, such as benzoic 
 acid, salicylic acid, balsam copaiba, oxalic acid, oil of turpen- 
 tine, glycerine, chloral and chloroform, substances appear in 
 the urine which possess the property of reducing cupric oxide 
 
 in alkaline solution. 
 y 
 
 7. Test with phenylhydrazine hydrochloride (C 6 H 8 N.,HC1 or 
 
 C 6 H 5 NHNH 2 HC1) (Fischer's test, 1883): Depends upon the formation of a 
 crystalline compound, phenylglukosazon, when phenylhydrazine hydrochloride 
 is brought in contact with glucose. The compound usually separates in rosettes 
 composed of yellow needle-shaped crystals, which melt at a temperature of 204 
 to 205 C. 
 
 Twice as much phenylhydrazine hydrochloride as will cover the end of a pen- 
 knife-blade is placed in a test-tube and also into the same tube three times as much 
 sodium acetate as will cover the end of a penknife-blade is placed. The test-tube 
 is filled to about one-third its capacity with water, the liquid slightly warmed, and 
 a volume of the urine equal in volume to the liquid in the tube is added.
 
 NOTES ON CHEMICAL LECTURES. 99 
 
 The tube containing the mixture is placed during fifteen or twenty minutes in 
 boiling hot water, and then in a vessel containing cold water. 
 
 If the urine contain a considerable quantity of glucose a yellow crystalline pre- 
 cipitate (C 6 H 10 O 4 (N 2 HC 6 H.). 2 phenylglukosazon) will almost immediately appear. 
 
 Sometimes the precipitate may appear amorphous macroscopically, and in such 
 cases should be examined microscopically. 
 
 If the urine contain a small quantity of glucose, the liquid in the test-tube 
 should be emptied into a conical glass and the sediment examined microscopically 
 for yellow, needle-shaped crystals. The occurrence of rather large yellow plates 
 or brown globules does not indicate the presence of glucose. 
 
 Albumin does not interfere with this test, but if it be present in large quantity 
 it is better to remove the greater part of it by boiling and filtering. 
 
 This test is unreliable as Fischer himself has found that other substances in the 
 urine will yield yellow condensation products with phenylhydrazine. 
 
 3. Molisch's tests : 
 
 a. Test with alpha-naphthol (C 10 H 7 OH). 
 Alpha-naphthol is a derivative of naphthalene. 
 
 H H 
 
 I i 
 
 C C 
 
 // \/ ^ 
 H C C C H 
 
 | || | = naphthalene. 
 H C C C H 
 
 ^ /\ // 
 C C 
 
 I I 
 
 H H 
 
 H OH 
 
 I I 
 
 C C 
 
 ^ \/ ^ 
 
 H C C C H 
 
 | | = alpha-naphthol. 
 
 H C C C H 
 
 ^ /\ // 
 C C 
 
 H H 
 
 The urine is diluted with water (about 100 of water to 1 of urine), and to 1 or 
 2 c.c. of the dilute urine 2 drops of a 15 or 20 per cent, alcoholic alpha-naphthol 
 solution are added. (The liquid may become turbid owing to the separation of 
 some of the alpha-naphthol.) A quantity of concentrated sulphuric acid equal to 
 
 13
 
 ICO NOTES ON CHEMICAL LECTURES. 
 
 the volume of liquid in the test-tube is added, and if glucose be present a deep 
 violet color, transitory in nature, will be produced. On diluting the liquid with 
 water a bluish-violet precipitate will be formed. 
 
 Delicacy of the test, 0.00001 per cent, of glucose (1 part glucose in 10,000,000 
 parts of water). 
 
 The test is not very reliable when applied to the urine. 
 
 b. Test with thymol (C, H U O = C 6 H 3 CH 3 C 3 H.OH, methylpropylphenol). 
 
 The test is performed in a manner similar to the alpha-naphthol test. 
 
 The urine is considerably diluted with water (1-100), and to 1 or 2 c.c. of the 
 liquid 2 drops of a 15 or 20 per cent, alcoholic thymol solution are added. A 
 quantity of concentrated sulphuric acid equal in volume to the liquid in the test- 
 tube is added and, if glucose be present, a deep cinnabar-red color will be pro- 
 duced, quickly changing to ruby-red, and then to carmine. On diluting with 
 water the carmine color remains. 
 
 Delicacy of the test, the same as the alpha-naphthol test. 
 
 The test is not very reliable when applied to the urine. 
 
 Both of Molisch's tests fail to give a reaction with urea, uric acid, hippuric acid, 
 creatinine, allantoin, pyro-catechin, and indican. 
 
 With cane sugar, fruit sugar, and maltose both these tests give reactions similar 
 to those with glucose. 
 
 QUANTITATIVE DETERMINATION OF GLUCOSE IN URINE. 
 
 1. Fermentation method by loss in weight : Depends upon the fermenta- 
 tion of glucose in the urine, thereby causing a loss in the weight of the urine, 
 due to the formation of alcohol and the escape of carbon dioxide. 
 
 180. 92. 88. 
 
 C 6 H 12 O 6 = 2C 2 H 5 OH + 2CO 2 
 
 Theoretically, in fermentation, 180 parts glucose evolve 92 parts alcohol and 88 
 parts carbon dioxide. Hence 1 part by weight of carbon dioxide lost is equivalent 
 to the fermentation of 2.045 parts of glucose. 
 
 CO. CeHi,0 6 . CO.,. C 6 H ]2 O 6 . 
 
 88 : 180 :: 1 : 2.045 
 
 Method : 50 c.c. of urine are placed in a small flask, and a small portion of 
 yeast is added. The flask is closed with a perforated cork to which a small calcium 
 chloride tube is attached (to collect moisture). 
 
 The entire apparatus is weighed and stood aside in a warm place that the glucose 
 may ferment. 
 
 When the fermentation is completed, the cork is taken out of the bottle, and, by 
 means of a glass tube, the carbon dioxide which may occupy the air-space in the 
 flask, is sucked out. The cork is replaced and the entire apparatus is weighed. 
 Each gramme lost is equivalent to 2.045 grm. glucose. To obtain the percentage 
 of glucose, 50 c.c. of urine having been used, the result must be multiplied by 2. 
 
 Water dissolves its own volume of carbon dioxide, therefore the 50 c.c. of liquid 
 holds in solution 50 c.c. of carbon dioxide, which is weighed with the apparatus, 
 and thus causes an error by adding excess of weight equal to the weight of the 
 carbon dioxide retained by the liquid.
 
 o :
 
 NOTES ON CHEMICAL LECTURES. 101 
 
 To correct this error, the weight of 50 c.c. of carbon dioxide must be added to 
 the loss of weight before multiplying by the factor 2.045. 
 
 1 c.c. CO 2 weighs 0.001971 grm. 
 50 X 0.001971 = 0985 grm. 
 
 Example : Weight of apparatus before fermentation, 67.6 grm. 
 " after " 65.6 " 
 
 2.0 " loss in weight. 
 -f weight of 50 c.c. CO 2 = 0985 
 2.0985 
 
 2.0985 X 2.045 = 4.29 grm. glucose in 50 c.c. urine ; then to obtain the percentage 
 (quantity in 100 c.c. urine), 
 
 2 X 4.29 = 8.58 per cent, glucose. 
 
 Instead of the foregoing the correction may be made by adding the number 0.4 
 to the apparent percentage. 0.4 grm. glucose will evolve 0.1955 grm. CO 2 , or 
 sufficient CO 2 to saturate 100 c.c. water. 
 
 Glucose. Carbon dioxide. Glucose. Carbon dioxide. 
 
 1. 180 : 88 : 0.4 : 0.1955 grm. 
 
 2. 
 
 CO 2 . CO* CO 2 . CO 2 . 
 
 22320 c.c. : 44 grm. :; 100 c.c. : 0.1971 grm. 
 
 The 50 c.c. liquid absorbed 50 c.c. CO 2 , or 50 X 0.001971 grm. = 0.0985 grm. 
 CO 2 , and as the calculation is made on the basis of 100 c.c. of liquid, 2 X 0.0985 
 = 0.1971 grm. CO 2 in 100 c.c. liquid, equivalent to 0.4 grm. glucose. 
 
 The first of the foregoing proportions shows that 0.1955 grm. CO 2 is evolved 
 from 0.4 grm. glucose, and as 0.1971 grm. excess of weight of CO 2 is retained by 
 100 c.c. of liquid, the loss is compensated by adding the fixed number 0.4 to the 
 apparent percentage. 
 
 Example : Weight of apparatus before fermentation, 67.6 grm. 
 " " after " 65.6 " 
 
 2.0 " loss in weight. 
 Then 
 
 2 grm. X 2.045 = 4.090 grm. X 2 = 8.18 
 
 Of- 
 
 8.58 per cent, glucose. 
 
 To obtain percentage in 100 parts by weight of urine. 
 Example : 
 
 Spec. grav. of urine. Spec. grav. of water. 
 
 1033 : 1000 :: 8.58 : X 
 
 2. Roberts' differential density method for the deter- 
 mination of glucose in urine : Depends upon the loss in 
 specific gravity of the urine, due to the fermentation of glucose 
 with the formation of alcohol, and evolution of carbon dioxide.
 
 102 NOTES ON CHEMICAL LECTURES. 
 
 Each degree in specific gravity lost is equivalent to 1 grain 
 of glucose in 437.5 grains (one imperial fluid ounce) of urine. 
 Or referred to percentage by volume : 
 
 437.5 : 1 :: 100 : 0.23 per cent. 
 
 Therefore one degree of specific 'gravity lost is equivalent 
 to 0.23 per cent, of glucose. 
 
 Method : To 60 or 70 c.c. of urine a small quantity of yeast 
 is added and the liquid stood aside in a moderately warm 
 place to ferment. As a control test another portion of 60 or 
 70 c.c. of the same urine is taken and stood aside without the 
 addition of yeast. 
 
 When the fermentation is completed the specific gravities of 
 the fermented and unfermented urines are taken separately. 
 The specific gravity of the fermented urine is deducted from 
 the specific gravity of the unfermented urine, and the result is 
 multiplied by the factor 0.23. The result of the multiplication 
 will be the percentage of glucose. 
 
 This method is the one most easily performed by the physi- 
 cian and affords fairly accurate results. 
 
 3. Fehling's method : Depends upon the reduction of 
 cupric oxide in alkaline solution by glucose to cuprous 
 oxide. 
 
 Fehling's solution is prepared so that 1 c.c. of it shall equal 
 0.005 grm. glucose, /. e., 0.005 grm. glucose will be required 
 to reduce the cupric oxide in 1 c.c. of the solution to cuprous 
 oxide. 
 
 Preparation of the solution : 5 molecules of crystallized 
 cupric sulphate are reduced to cuprous oxide by 1 molecule 
 of glucose. 
 
 Then 
 
 CeHjoOg. 5CuSO 4 + 5H 2 O. Glucose. Cupric sulphate. 
 
 180 : 1247.5 :: 5 grm. : 34.652 grm. 
 
 Therefore 34 : .652 grm. cupric sulphate will be reduced by 5 
 grm. glucose. 
 
 The cupric sulphate crystals, before being weighed, should 
 be deprived of water, held mechanically, by being crushed and 
 dried between bibulous paper.
 
 ^dr2V a 
 
 k J 
 
 *k V ^-^f (Cj(*J ft i <-*> ^t^ 
 
 &,> -~^JU jfM*Ji^ - ^ y'<0 
 
 / /. J 

 
 NOTES ON CHEMICAL LECTURES. 103 
 
 One molecule of glucose is equivalent to 5 molecules of 
 cupric sulphate, CuSO 4 -f~ 5H 2 O. 
 
 One molecule of glucose is equivalent to 5 molecules of 
 cupric oxide, CuO. 
 
 A. 34.652 grm. pure crystallized cupric sulphate are dissolved 
 in about 200 c.c. water. 
 
 B. About J 73 grm. sodic potassium tartrate (KNaC 4 H 4 O 6 ) 
 (Rochelle salt) are dissolved in about 480 c.c. of sodium 
 hydroxide solution of 1.14 specific gravity. The cupric sul- 
 phate solution is slowly added to the Rochelle salt solution } 
 at the same time the solution is constantly stirred. 
 
 The bluish-white precipitate of cupric hydroxide which 
 appears will be completely dissolved by the liquid. 
 
 The object of the Rochelle salt is to hold the cupric 
 hydroxide in solution. 
 
 The blue liquid is diluted with water to 1000 c.c., then 
 
 1000 c.c. = 5.0 grm. glucose. 
 10 c.c. = 0.050 " 
 1 c.c. = 0.005 " 
 
 Fehling's solution is prone to undergo spontaneous de- 
 composition. It may, however, be preserved so as to avoid 
 its undergoing decomposition by keeping the cupric sulphate 
 and Rochelle salt solutions separately and mixing equal 
 volumes of the two solutions when needed, viz. : 
 
 34.652 grm. cupric sulphate are dissolved in water and 
 diluted to 500 c.c. The Rochelle salt solution is also diluted 
 with water to 500 c.c. The two solutions must be kept in 
 separate bottles closed with rubber stoppers. 
 
 To employ the solutions : 1 volume of the cupric sulphate 
 solution is mixed with an equal volume of Rochelle salt 
 solution. 
 
 Example : 
 
 5 c.c. cupric sulphate solution. 
 _5 c.c. Rochelle salt 
 10 c.c. Fehling's " 
 
 The accuracy of the solution may be determined by titering 
 it with a standard solution of glucose, prepared by dissolving 
 0.5 grm. glucose in 100 c.c. water.
 
 104 NOTES ON CHEMICAL LECTURES. 
 
 The cupric oxide in 10 c.c. Fehling's solution should be 
 exactly reduced by 10 c.c. of the glucose solution. 
 
 In using Fehling's solution the glucose solution must be 
 added to the Fehling's solution, and not the Fehling's to the 
 urine. 
 
 If the diabetic urine have a specific gravity of about 1035, 
 it should be diluted with 4 volumes of water, /. e., 10 c.c. of 
 urine + 40 c.c. of water. 
 
 If the diabetic urine have a specific gravity of about 1040, 
 it should be diluted with 9 volumes of water, i. e., 10 c.c. of 
 urine -f- 90 c.c. of water. 
 
 Albumin, if present in the urine, must previously be re- 
 moved by coagulation and filtration as it interferes with 
 Fehling's test. 
 
 Method : 10 c.c. of Fehling's solution (= 0.050 grm. 
 glucose) are placed in a beaker or dish and diluted with about 
 40 c.c. of water and the liquid is heated to the boiling-point. 
 If necessary, the urine should be diluted with 4 or 9 volumes of 
 water then placed in a burette. From 0.5 to 1.0 c.c. at a time 
 of the liquid in the burette should be run into the hot diluted 
 Fehling's solution. Heat is applied to the Fehling's solution 
 after each addition of urine to keep it at about the boiling- 
 point. The addition of the urine to the hot Fehling's solu- 
 tion is continued until the reduction of the cupric oxide is 
 completed. This is recognized by the complete disappearance 
 of the blue color of the liquid. 
 
 As the point of complete reduction is approached, the pre- 
 cipitate of cuprous oxide will subside more rapidly, thereby 
 allowing the easy observance of the disappearance of the blue 
 color of the liquid. 
 
 The first titration usually gives only approximate results 
 unless performed with the greatest care. A second titration 
 should be made, in which the urine (from 0.5 to 1.0 c.c. at a 
 time) is run into the hot Fehling's solution until the quantity 
 added is about 1 c.c. less than that employed in the first titration. 
 The addition of the urine, in small portions, is continued until 
 the blue color of the liquid is completely discharged. 
 
 Example : Suppose the urine employed had been diluted 
 in the proportion of 1 volume urine to 9 volumes water, and
 
 J IW (,0*L 
 4 O V 
 
 U 
 
 i
 
 NOTES ON CHEMICAL LECTURES. 105 
 
 8 c.c. of this diluted urine had been required to reduce the 
 cupric oxide in 10 c.c. Fehling's solution. 
 
 One-tenth of the 8 c.c. of liquid was urine, and must have 
 contained 0.050 grm. glucose (the quantity of glucose required 
 to reduce 10 c.c. Fehling's solution). 
 
 Urine. Glucose. Urine. Glucose. 
 
 Then 0.8 c.c. : 0.050 grm. :: 100 c.c. : 6.25 per cent. 
 
 The process should be performed as rapidly as is consistent 
 with accuracy. 
 
 If the urine mixture employed contain more than 0.5 per 
 cent, glucose, it should be still further diluted with a measured 
 volume of water. 
 
 If albumin be present, it must be removed before making 
 the determination. 
 
 CLINICAL APPLICATION OF FEHLING'S METHOD. 
 
 i c.c. Fehling's solution = 0.005 g rm - glucose. 
 
 a. 1 c.c. of Fehling's solution is placed in a test-tube and 
 diluted with 4 c.c. water (or 5 c.c. of diluted Fehling's solution, 
 consisting of 1 c.c. Fehling's solution and 4 c.c. water). 
 
 The liquid is heated to the boiling-point and 1 c.c. of undi- 
 luted urine is added and the liquid is again heated. 
 
 If the blue color is entirely discharged, 0.5 per cent, or 
 more of glucose is present. 
 
 If the blue color is not entirely discharged, less than 0.5 per 
 cent, of glucose is present. 
 
 b. To 1 c.c. Fehling's solution -j- 4 c.c. water heated to the 
 boiling-point 0.1 c.c. of the same urine is added and the liquid 
 heated. 
 
 If the blue color is entirely discharged, 5.0 per cent, or more 
 glucose is present. 
 
 If the blue color is not entirely discharged, less than 5.0 per 
 cent, glucose is present. 
 
 If the color is not discharged, add another 0.1 c.c. urine and 
 again heat the liquid. 
 
 If the blue color is entirely discharged, 2.5 per cent, or more 
 glucose is present.
 
 106 NOTES ON CHEMICAL LECTURES. 
 
 If the blue color is not entirely discharged, less than 2.5 per 
 cent, glucose is present. 
 
 The addition of the urine 0.1 c.c. at a time, keeping account 
 of the quantity added, and heating the liquid is continued until 
 the blue color is discharged. 
 
 c. 1 c.c. of Fehling's solution is placed in a test-tube and 
 diluted with 4 c.c. of water and the liquid heated to the boiling- 
 point. The urine is added from a pipette, a drop at a time, 
 (the number of drops added is noted?) and the liquid is heated 
 after the addition of each drop. This addition of a drop at a 
 time of the urine and heating the liquid after the addition ot 
 each drop is repeated until finally the blue color of the liquid 
 has disappeared. As two drops of urine are approximately 
 equal to 0.1 c.c., therefore, the number of drops employed 
 divided by 2 will approximately furnish the number of tenths 
 of a cubic centimetre employed. In calculating results the 
 following rule should be used. 
 
 Rule : Divide the number 5 by the number of tenths of 
 urine employed converted into whole numbers. The result 
 will be the approximate percentage of glucose. 
 
 To obtain more accurate results, the urine should be diluted 
 and titered in the usual manner with 10 c.c. Fehling's solution 
 -j- 40 c.c. water. 
 
 Percentage amount of glucose present in urine, as in- 
 dicated by the quantity of the urine required to exactly 
 decolorize i c.c. of Fehling's standard solution diluted 
 with 4 c.c. of water. 
 
 OF UNDILUTED URINE. 
 
 0.1 (5 -=- 1) 5.0 
 
 0.12 . 4.2 
 
 0.14 3.5 
 
 0.16 . . 3.1 
 
 0.18 2.7 
 
 0.2 2.5 
 
 0.25 2.0 
 
 0.3 1.66 
 
 0.35 . 1.4 
 
 c.c. urine. 
 
 0.4 .... 
 
 Glucose, 
 
 per cent. 
 
 . . . . 1.25 
 
 0.45 .... 
 
 .... 1 10 
 
 05 .... 
 
 . . . . 10 
 
 06 .... 
 
 . ... 083 
 
 0.7 .... 
 
 .... 0.71 
 
 0.8 .... 
 
 . . . . 0.62 
 
 09 .... 
 
 . . . . 0-55 
 
 1.0 
 
 0.5
 
 NOTES ON CHEMICAL LECTURES. 107 
 
 OF DILUTED URINE (i to 10). 
 
 c.c. urine. 
 
 0.4 (50 ~ 4) 
 0.5 .... 
 
 Glucose, 
 per cent. 
 
 . . . . 12.5 
 . . . .100 
 
 0.6 .... 
 
 . . . . 833 
 
 0.7 .... 
 
 . . . . 7.14 
 
 0.8 .... 
 
 . . . 625 
 
 0.9 .... 
 
 . ... 555 
 
 1.0 .... 
 
 . ... 50 
 
 1.2 ... 
 
 . ... 4.2 
 
 1.4 .... 
 
 . ... 35 
 
 1.6 . . . . 
 
 . ... 31 
 
 1.8 .... 
 
 . ... 27 
 
 2.0 .... 
 
 . ... 2.5 
 
 2.25 . 
 
 2.2 
 
 
 Glucose, 
 
 
 per cent. 
 
 2.50 . . . . 
 
 .... 2.0 
 
 2.75 . . . . 
 
 .... 1.8 
 
 3.00 . . . . 
 
 .... 1.6 
 
 3.5 . . . . 
 
 . .1.4 
 
 4.0 . . . . 
 
 .... 1.25 
 
 4.5 . . . . 
 
 .... 1.1 
 
 5.0 . . . . 
 
 .... 1.0 
 
 6.0 . . . . 
 
 .... 0.83 
 
 7.0 ' . . . . 
 
 .... 0.7 
 
 8.0 . . . . 
 
 .... 0.6 
 
 9.0 . . . . 
 
 .... 0.55 
 
 10.0 
 
 . 0.5 
 
 4. Determining glucose quantitatively by means of the 
 saccharimeter : If light which has undergone double refrac- 
 tion, as in passing through a crystal of Iceland spar, is exam- 
 ined with an analyzer, it is found that both the ordinary and 
 extraordinary rays are completely polarized at right angles to 
 each other. Advantage is taken of this in the construction of 
 the saccharimeter or polariscope. 
 
 When 20.51 grm. anhydrous glucose are dissolved in 
 water and diluted to 100 c.c., and an observation-tube 200 
 mm. in length filled with the solution is placed in Laurent's 
 saccharimeter, the ray of light will be deflected to the 
 right 100 markings or divisions as indicated on the vernier- 
 scale. 
 
 Hence 
 
 100 divisions on the scale = 20.51 grm. glucose. 
 1 division " = 0.2051 " 
 
 Method ; 100 c.c. urine are mixed with 10 c.c. basic 
 acetate of lead (Pb 3 O 2 (C 2 H 3 O 2 )2) solution, and filtered through 
 a filter which has not been previously moistened with water. 
 A 200 mm. (standard) observation-tube is filled with the 
 filtrate. 
 
 14
 
 108 NOTES ON CHEMICAL LECTURES. 
 
 The field of vision in the saccharimeter must be homoge- 
 neous in color, and the zero divisions on the vernier must 
 correspond before the tube, is placed in the saccharimeter. 
 
 The observation-tube containing the urine is placed in the 
 saccharimeter, and the effect on the field of vision is noted. 
 If the urine contain glucose, one-half of the field of vision 
 will be darker than the other half. The large thumb-screw is 
 then rotated until the field of vision again becomes homoge- 
 neous in color. 
 
 The reading on the vernier is noted. As the urine was 
 diluted one-tenth by the lead solution, the reading, therefore, 
 is one-tenth too low. This is corrected by adding one-tenth of 
 the reading to the reading. The corrected reading is then 
 multiplied by the value of each division on the vernier, 
 namely, 0.2051 grm. 
 
 Example : 
 
 Suppose the reading was 2.3.1 
 
 ytg- of the reading added to the reading . 2.31 
 
 25.41 
 25.41 X 0.2051 = 5.211 per cent, glucose. 
 
 If the amount of glucose in the urine be very small, a 400 
 mm. observation-tube may be used. In such an event the 
 corrected reading must be divided by 2 before multiplying by 
 
 0.2051. 
 
 RELATION OF SACCHAROSES AND AMYLOSES TO GLUCOSE^ 
 
 Mol - w V?.(f 
 Glucose, C 12 H 24 O 12 (2C 6 H 12 O 6 ), = 360^= 100 glucose. 
 
 Saccharoses, C 12 H 22 O n , = 342^ 95 
 
 Amyloses, C 12 H 20 O 10 (2C 6 H 10 O 5 ), = 324^, 90 
 
 SYNOPSIS OF ALBUMINS. 
 
 1. Native albumins, soluble in water. 
 
 a. Serum albumin, not precipitated by ether. 
 /;. Egg albumin, precipitated by ether. 
 c. Peptones. 

 

 
 NOTES ON CHEMICAL LECTURES. 109 
 
 2. Globulins, insoluble in water, but soluble in 1 per cent, 
 solution of sodium chloride. 
 
 a. Fibrinogen. 
 
 b. Fibrinoplastin. 
 
 c. Myosin. 
 
 d. Vitellin. 
 
 e. Crystallin. 
 
 3. Derived "albumins or albuminates, insoluble in water 
 orjjdilute sodium chloride solution, but soluble in acids or 
 gastric juice. 
 
 a. Acid albumin (syntonin). 
 
 b. Alkali albumin. 
 
 c. Casein. 
 
 d. Fibrin. 
 
 c. Coagulated albumin. 
 '/. Amyloid substance. 
 
 ALBUAilE IN URINE. 
 
 The albumin in urine in Bright's disease is serum albumin. 
 
 Serum globulin in small quantity may also be present. 
 
 The principal forms of albumin present in urine are serum 
 albumin, globulin, hemialbumose and nucleo-albumin (mucin). 
 
 The quantity of albumen varies from a slight trace to 3 per 
 cent. 
 
 Albuminous urine is usually pale in color and has a low 
 specific gravity (1006 to 1014). 
 
 QUALITATIVE TESTS FOR ALBUMIN. 
 
 1. By heating the urine to the boiling-point (boiling- 
 test). 
 
 a. If a flocculent precipitate appear, it is due either to earthy 
 phosphates or coagulated albumin. 
 
 The warm urine in which precipitation has occurred is 
 treated with one or two drops of nitric acid. If the precipitate 
 be dissolved by the acid, it is composed of phosphates ; if 
 unclissolved, it is albumin.
 
 110 .VOTES ON CHEMICAL LECTURES. 
 
 b. If no precipitate appear on boiling, the warm urine is 
 treated with one or two drops of nitric acid, and if albumin 
 be present it will be precipitated. If the quantity of nitric 
 acid be either too large or too small the precipitation of the 
 albumin may not occur. 
 
 Serum albumin and globulin respond to this test. Peptone 
 does not. 
 
 If the precipitate separates only after cooling, it is albumose. 
 
 2^ Heller's method. 
 
 If not clear, the urine must be filtered before making the 
 test. 
 
 A test-tube, containing strong nitric acid to the depth of at 
 least an inch, or better, a conical glass, containing strong nitric 
 acid, is inclined and the urine is slowly poured down the inner 
 side of the vessel so that it shall form a layer above the nitric 
 acid. If albumin be present, a milky zone will be produced 
 at the point of contact of the two liquids. 
 
 Delicacy of the test, 0.00*25 per cent, albumin. 
 
 If the urine contain excess of urates a zone, due to the sepa- 
 ration of uric acid, may be formed. This zone is brownish-red 
 in color (the albumin zone is white), and forms, not at the 
 point of contact of the urine and acid, but some distance 
 above the point of contact of the urine and acid. The 
 under part of the zone is not so sharply defined as the 
 albumin zone. If in doubt, dilute the urine nearly one-half 
 with water and repeat the test. In the diluted urine uric acid 
 will not interfere. 
 
 In the presence of albumin and excess of urates two zones 
 will be produced, the uric acid zone above the albumin. 
 
 On the administration of balsams such as copaiba a sub- 
 stance (abietic acid) appears in the urine which, with nitric 
 acid, forms a zone similar to that formed by albumin. This 
 substance also interferes with the first test, by separating as a 
 flocculent precipitate on boiling the urine. The precipitate is 
 soluble in alcohol, albumin is not soluble in alcohol. 
 
 If indican be present, the urine will become violet in color 
 at the point of contact with the nitric acid. In presence of 
 biliary matter it will become green, and then change to red- 
 brown.
 
 NOTES ON CHEMICAL LECTURES. Ill 
 
 J5. Picric acid test (C 6 H 2 (NOj 3 OH) (carbazotic acid) Gal- 
 
 lipe'.s test). \^JKA llU^ljGVAM' 
 
 If the urine is turbid, it must be filtered. 
 
 a. Urine to the depth of an inch is placed in a test-tube and 
 a saturated aqueous solution of picric acid is slowly poured 
 down the side of the tube so that it collects as a distinct layer 
 above the urine. If albumin be present, a white zone will be 
 produced at the point of contact of the two liquids. 
 
 b. Urine is added, drop by drop, to a saturated solution of 
 picric acid, and, if albumin be present, a sharply defined tur- 
 bidity will be produced at the point of contact of the drop of 
 urine with the acid. 
 
 jl. Potassium ferrocyanide test : If the urine is turbid, it 
 must be filtered. ttWA ^MJU&X . 
 
 The urine is strongly acidulated with acetic acid (to 1 vol- 
 ume of urine add about \ volume of acid) and, without heating, 
 3 or 4 drops of potassium ferrocyanide solution are added. 
 
 If albumin be present, a turbidity or a flocculent precipitate, 
 depending upon the quantity of albumin, will be produced. 
 
 Delicacy of the test, 0.0025 per cent, albumin. 
 
 Serum albumin, globulin, and albumose respond to the test, 
 but peptone does not. Nucleo-albumin responds to the test. 
 Xucleo-albumin may be removed by precipitating with plumbic 
 acetate, filtering, removing the lead from the filtrate by means 
 of hydrogen sulphide and applying the ferrocyanide test to 
 the filtrate. 
 
 5. Trichloracetic acid test, HCClgCCX (Raabe). 5 c.c. of 
 urine filtered through a filter, which has been previously 
 thoroughly washed with water to remove vegetable albumin, 
 are placed in a narrow test-tube and a piece, about the size 
 of a large pea, of trichloracetic acid is placed in the urine. In 
 the presence of albumin a white zone or cloud will be pro- 
 duced at the point where the resulting solution of acid comes 
 in contact with the urine. 
 
 Urine containing an excess of uric acid may produce a 
 diffuse turbidity which disappears on heating the liquid. This 
 turbidity, due to excess of uric acid, may not occur if the 
 urine be previously diluted with water. Nucleo-albumin 
 (mucin) responds to the test and as nearly every normal urine 
 
 TPT 
 ^W^T
 
 112 
 
 NOTES ON CHEMICAL LECTURES. 
 
 contains more or less nucleo-albumin the test is not of nuicli 
 I'ahie unless the nucleo-albumin be removed by precipitation 
 with plumbic acetate, etc., previous to applying the trichlor- 
 acetic acid test. 
 
 Jk_ Biuret test for albumin : The urine is rendered alkaline 
 with sodium or potassium hydroxide, and a few drops of dilute 
 cupric sulphate solution are added. If albumin be present, 
 the precipitate of cupric hydroxide will dissolve on shaking 
 and a violet-red color will be imparted to the liquid. 
 Delicacy of test, 0.01 per cent. 
 
 Serum albumin, globulin, albumose, and peptone respond 
 to this test. 
 
 To distinguish between serum albumin, globulin, hemi- 
 albumose and nucleo-albumin (mucin). 
 
 1. The cold urine is treated with an excess of acetic acid. 
 
 Serum albumin. Globulin. Hemialbumose. 
 
 Nucleo-albumin. 
 (Afucin.) 
 
 Not Not Not 
 
 precipitated, precipitated, precipitated. 
 
 2. The urine is filtered and solution of potassium ferro- 
 cyanide is added to the cold filtrate. 
 
 Serum albtimin. Globulin. Hemialbumose. 
 
 Precipitated. Precipitated. 
 
 Precipitated. 
 
 The liquid is heated to the boiling temperature. 
 
 Serum albumin. 
 
 Globulin. Hemialbumose. 
 
 Precipitate 
 
 remains 
 undissolved. 
 
 Precipitate 
 remains 
 undissolved. 
 
 Precipitate 
 dissolved.
 
 NOTES ON CHEMICAL LECTURES. 113 
 
 To separate serum albumin from globulin, a slight excess 
 of ammonium hydroxide is added to the urine to precipitate 
 the phosphates of the alkaline earths. After the liquid has 
 stood several hours the phosphates are filtered off and the 
 filtrate is treated with its own volume of a saturated solution 
 of ammonium sulphate which will precipitate the globulin 
 leaving the serum albumin in solution. 
 
 QUANTITATIVE DETERMINATION OF ALBUMIN IN URINE. 
 
 1 . By weighing ; If the urine is turbid, it must be filtered. 
 50 c.c. or 100 c.c. urine, depending upon the quantity of 
 
 albumin present, are warmed to about blood-heat on a water- 
 bath, and acetic acid added, drop by drop, until the albumin 
 separates in flocculent masses. (The acetic acid keeps the 
 earthy phosphates in solution.) The liquid is heated to the 
 boiling-point, and the coagulated albumin is collected on an 
 equipoised or a weighed washed filter and washed with water 
 (portions of 5 c.c. at a time), until a last portion of filtrate 
 coming from the funnel fails to respond to the test for a 
 chloride with argentic nitrate, or a drop of the filtrate fails to 
 leave a residue when evaporated on platinum foil. The coag- 
 ulated albumin is washed while on the filter with alcohol, 
 the two (equipoised) filters separated and dried at a temper- 
 ature not above 100 C., and weighed. 
 
 The weight of albumin obtained is the quantity in the 
 volume of urine employed. 
 
 2. Esbach's method : Depends upon the precipitation of 
 albumin by picric acid. 
 
 The quantity of albumin is determined by measuring the 
 height of the precipitate in a specially graduated tube, Esbach's 
 albuminimeter. 
 
 The picric acid solution is prepared by dissolving 10 grm. 
 picric acid and 20 grm. citric acid in 900 c.c. hot water, and, 
 after cooling, diluting the solution to 1000 c.c. 
 
 The albuminimeter-tube is filled to the mark U with urine, 
 and upon this the picric acid solution is poured until the level 
 of the liquid reaches the mark R. The opening of the tube is 
 closed with the thumb, and the liquids are mixed by inverting 
 the tube several times.
 
 114 NOTES ON CHEMICAL LECTURES. 
 
 The tube is closed with a rubber stopper and stood aside 
 for twenty-four hours. The depth of the sediment is ascer- 
 tained by observing where the top of the sediment comes in 
 contact with a line on the scale. 
 
 The figures on the scale represent the number of grammes 
 of albumin in 1000 c.c. urine. 
 
 * 
 
 The results are not absolutely accurate because of the pre- 
 cipitation of creatinine by picric acid, but sufficiently accurate 
 for clinical purposes. - 
 
 QUALITATIVE TESTS FOR BILIARY COLORING MATTER 
 IN URINE. 
 
 1. Urine containing biliary coloring matter when violently 
 shaken in a test-tube yields a yellow foam ; normal urine 
 yields a white foam. 
 
 2. The urine is filtered through a white filter paper. The 
 filter paper is taken from the funnel and unfolded and spread 
 on a large porcelain dish or plate. Slightly yellow, strong- 
 nitric acid is dropped on the filter. If unaltered biliary color- 
 ing matter is present, concentric rings of color will be pro- 
 duced where the drop of nitric acid comes in contact with the 
 filter. The colors from the centre to the periphery are red, 
 violet, blue, and green. Albumin in the urine interferes with 
 this test. It may be removed by boiling and filtering. Small 
 quantities of biliary coloring matter may be precipitated with 
 the albumin but may be recovered by extracting the coag- 
 ulated albumin with chloroform, evaporating the solution to 
 dryness in a porcelain dish and applying the nitric acid test 
 directly to the residue. 
 
 3. The urine is treated with a few drops of sodium hydrox- 
 ide, to render it slightly alkaline, and calcium chloride is added 
 in excess. The precipitate is collected on a. filter. (The pre- 
 cipitate from icteric urine is yellow ; from normal urine, white.) 
 When all of the liquid has passed through, the filter is un- 
 folded and spread on a large porcelain dish or plate and slightly 
 yellow, strong nitric acid is dropped on the filter. If unaltered 
 biliary coloring matter is present, concentric rings of color will 
 be produced where the drops of nitric acid come in contact
 

 
 ^ 
 
 i(*^J&l~CX^4 
 (^ fa 
 
 

 
 NOTES ON CHEMICAL LECTURES. 115 
 
 with the filter. The colors from centre to periphery are red, 
 violet, blue, and 
 
 Acetone, CjHgO. Defection in urine : About 600 c.c. of 
 urine are acidulated with acetic acid (about 2 c.c. of acid 
 to every 100 c.c. of urine), and distilled until about 50 c.c. 
 of distillate are collected in the receiver. To this distillate a 
 few drops of a dilute solution of sodium hydroxide are added 
 and then a slight excess of concentrated solution of iodo- 
 potassium iodide (iodine dissolved in potassium iodide solu- 
 tion) is added. In the presence of acetone, the solution, after 
 the lapse of a little time, becomes slightly yellow in color and, 
 after standing some time longer, a yellow, crystalline pre- 
 cipitate of iodoform, which may be recognized by its peculiar 
 odor, appears. 
 
 Diacetic acid, C 4 H O 3 . Detection in urine : Urine con- 
 taining an appreciable quantity of diacetic acid when treated \ .- 
 with a solution of ferric chloride becomes claret red in color. 
 The substances which appear in the urine after the ingestion 
 of antipyrin, kairin, and thallin also produce a claret-red color 
 with ferric chloride. 
 
 TOXICOLOGY. 
 
 A poison is any substance which, when taken into the 
 body and either being absorbed or by its direct chemical action 
 upon the parts with which in contact, or when applied externally 
 and entering the circulation, is capable of producing deleterious 
 effects. 
 
 Poisons vary greatly in regard to their toxic action. 
 
 Thus, aconitinc, one of the most active poisons, has proved 
 fatal to an adult in the dose of -fa grain. The activity of 
 aconitinc is about seven times greater than that of strychnine, 
 and about forty times greater than that of arsenic. The activity 
 of poisons may be expressed by their fatal toxic action per 
 kilogramme of body weight as follows : 
 
 ti. Aconitine, 0.056 mgrm. per kilo., or about 1 to 18,000,000 
 parts of body weight. 
 
 15
 
 116 NOTES ON CHEMICAL LECTURES. 
 
 b. Strychnine, 0.400 mgrm. per kilo., or about 1 to 2,500,000 
 
 parts of body weight. 
 
 c. Arsenic, 2.24 mgrm. per kilo., or about 1 to 500,000 parts 
 
 of body weight. 
 
 CAUSES WHICH MODIFY THE EFFECTS OF POISONS. 
 
 a. Idiosyncrasy, or a peculiarity of constitution, may 
 variously modify the effects of poisons. 
 
 b. Habit may render certain poisons harmless in doses 
 which to most persons would prove rapidly fatal. 
 
 c. Disease : In certain diseased conditions of the system 
 there is a diminished susceptibility to the action of certain 
 poisons, whilst in others there is an increased susceptibility, 
 even to the action of the same substance. 
 
 d. Condition of the stomach : The presence of another 
 substance or poison ; sleep. 
 
 CLASSIFICATION OF POISONS. 
 
 a. Irritant poisons, as a class, produce irritation and 
 inflammation of the stomach and bowels, attended or followed 
 by intense pain in these parts, tenderness of the abdomen, and 
 violent vomiting and purging, the matters evacuated being 
 often tinged with blood. 
 
 The irritant poisons may be divided into three sections, 
 namely, mineral, vegetable, and animal. 
 
 b. Narcotic or cerebral poisons are such as act prin- 
 cipally on the brain and spinal marrow, more especially on the 
 former. 
 
 c. Narcotico-irritants partake, as indicated by their name, 
 of the action of both the preceding classes. 
 
 SOURCES OF EVIDENCE OF POISONING. 
 
 1. Evidence from symptoms. 
 
 a. The symptoms occur suddenly, and soon after the taking 
 of some solid or liquid. 
 
 b. The symptoms rapidly run their course.
 
 . uro 
 
 77; " -III a^ 
 
 ^ 
 
 -*.
 
 1 
 

 
 NOTES ON CHEMICAL LECTURES. 117 
 
 2. Evidence from post-mortem appearances. 
 
 a. The irritant poisons, as a class, usually produce irritation 
 and inflammation of one or more portions of the alimentary 
 canal, the effects being sometimes confined to the stomach, 
 while at other times they extend to a greater or less degree 
 throughout the entire canal. 
 
 b. Narcotic poisons, in some instances, produce more or less 
 distension of the veins of the brain, but in others they leave 
 no marked morbid appearances, and in none are the appear- 
 ances peculiar. ^ 
 
 c. Narcotico-irritants partake, in the nature of their effects, 
 of both the preceding classes. 
 
 APPEARANCES COMMON TO POISONING AND DISEASE. 
 
 a. Redness of the stomach and intestines as the effect of 
 poisoning cannot in itself be distinguished from that arising 
 from natural disease. 
 
 b. Softening of the stomach is another appearance which 
 may give rise to embarrassment. When due to the action of 
 poison, it is usually accompanied by other appearances which 
 readily distinguish it from the effects of ordinary disease or 
 post-mortem changes. 
 
 c. Ulceration and perforation of the stomach are not un- 
 frequently produced by corrosive poisons, but they, especially 
 the latter, are rarely met with as the result of the action of 
 the simple irritants. As the effect of natural disease or post- 
 mortem action they are not uncommon. 
 
 d. Points to be observed in post-mortem examinations : 
 All investigations of this kind should be made in the presence 
 of the proper law officer ; and it is well for the examiner to 
 have the assistance and corroboration of another physician. 
 All appearances observed, whether abnormal or otherwise, 
 should be fully written down at the time of their observance. 
 
 All the organs and blood removed for the purpose of examina- 
 tion should be collected in separate, new, clean glass vessels, 
 great care being taken tJiat none of the reserved substances at any 
 time be brought in contact with any substance that afterwards 
 might give rise to suspicion. Before passing out of the sight
 
 118 NOTES ON CHEMICAL LECTURES. 
 
 of the examiner, the bottles should be securely sealed and fully 
 labeled. They should then be retained in /iis sole possession 
 until delivered to the proper legal official. 
 
 3. Evidence from chemical analysis. 
 
 a. In most charges of poisoning the final issue depends 
 upon the results of the chemical analysis. In fact, in many 
 instances in which the evidence from symptoms, post-mortem 
 appearances, and moral circumstances is very equivocal or in 
 part wanting, a chemical examination may at once determine 
 the true cause of death.. It must be remembered, however, a 
 person may die from the effects of poison and not a trace of 
 its presence be discoverable in any part of the body ; while, on 
 the other hand, the mere discovery of a poison in the food or 
 drink taken, or in the body after death, is not in itself positive 
 proof that it occasioned death. 
 
 b. Substances requiring analysis : The substances that 
 may directly become the subject of a chemical analysis in a 
 case of suspected poisoning are : the pure poison in its solid 
 or liquid state ; suspected articles of food or medicine ; matters 
 ejected from the body by vomiting or purging ; the urine ; 
 suspected solids found in the stomach or intestines after death ; 
 the contents of the stomach or bowels ; any of the soft organs 
 of the body, as the liver, spleen, etc., and the blood. 
 
 There are some poisons for which no definite chemical test 
 is known. 
 
 Some poisons are detected by a combination of tests, and 
 others by a single test or tests. 
 
 4. Limit of tests. 
 
 1 part of arsenic may be detected in 5,000,000 parts water. 
 Strychnine, -^fam of a grain. 
 Hydrocyanic acid, ronVrU" ^ a g ram - 
 
 5. Limit of recovery. 
 
 6. Failure to detect a poison. 
 
 Numerous instances of poisoning are reported in which 
 persons died from the effects of poison and none was dis- 
 covered by chemical analysis in the body after death. This 
 result has most frequently been observed in poisoning with 
 organic substances, but it has happened when mineral poisons,
 
 la^^^x/v-T. ^
 
 NOTES ON CHEMICAL LECTURES. 119 
 
 and even those which are most easily detected by chemical 
 tests, had been taken in large quantity. 
 
 A failure of this kind may be due to any of the following 
 circumstances: 1. The poison may have been one of the 
 organic poisons, which cannot at present be recognized by 
 chemical tests. 2. The quantity present in the part examined 
 may have been so minute as under the circumstances not to 
 admit of recovery, or at least in a state sufficiently pure to 
 permit its true nature to be established: 3. The poison may 
 have been removed from the stomach and intestines by vomit- 
 ing and purging or by absorption. 4. The absorbed poison 
 may have been carried out of the system with the excretions. 
 5. If volatile, like hydrocyanic acid and some few other poisons, 
 it may have been dissipated in the form of vapor. 6. It may 
 have undergone a chemical change in the living body, or, 
 especially if of organic origin, have decomposed in the dead 
 body if far advanced in putrefaction. 
 
 7. Caution regarding the purity of reagents. 
 
 8. Preservation of chemical results and material. 
 
 9. Duties and rights of experts. 
 
 An expert can not be compelled to make a post-mortem 
 examination or chemical analysis. 
 
 Having made the examination, the knowledge acquired is 
 common property, J=d( ~i~~-~--r- 4**. X>J(^{ A^" t^-^-C--/., 
 
 The expert should be cautious in expressing opinions before 
 the case is called for trial. 
 
 ARSENICUM. 
 
 Atomic weight, 75. 
 
 In its pure state arsenicum has a steel-gray color, a bright 
 metallic lustre, and has a crystalline structure. In dry air it 
 remains unchanged, but in the presence of moisture it slowly 
 absorbs oxygen and assumes a dark-gray appearance. It 
 volatilizes at a temperature of 110 C. (230 F.). 
 
 Metallic arsenic, when taken into the stomach, is capable of 
 acting as a powerful poison, but perhaps only in so far as the 
 metal becomes oxidized and converted into arsenious acid.
 
 120 NOTES ON CHEMICAL LECTURES. 
 
 Compounds of arsenicum and oxygen : 
 
 As 2 O 3 , arsenious oxide, molecular weight, 198. 
 As 2 O 5 , arsenic " " " 230. 
 
 Both form acids with the elements of water : 
 
 As 2 O 3 + 3H 2 O = 2H 3 AsO 3 , arsenious acid. 
 As 2 O 5 -f 3H 2 O = 2H 3 AsO 4 , arsenic acid. 
 
 Arsenious oxide (As 2 O 3 ) is readily obtained by volatilizing 
 metallic arsenicum in a free supply of air. 
 
 It is found in commerce either as a white or dull white, 
 opaque powder, or in the form of large, hard masses. 
 
 Symptoms of poisoning with As 2 O 3 : These are subject 
 to great variation. Sooner or later after a large dose of the 
 poison has been swallowed there is usually a sense of heat 
 and constriction in the throat, with thirst, nausea, and burn- 
 ing pain in the stomach. The pain becomes excruciating, 
 and is attended with violent vomiting and retching ; the 
 matters vomited present various appearances, being some- 
 times streaked with blood, and at other times of a bilious 
 character; the pain in the stomach is increased by pressure. 
 As the case progresses the pain extends throughout the ab- 
 domen, and there is generally severe purging and tenesmus ; 
 the matters passed from the bowels not unfrequently contain 
 blood. The thirst usually becomes very intense ; in some 
 instances there is great difficulty in swallowing. The features 
 are collapsed and expressive of great anxiety ; the pulse is 
 quick, small, and irregular ; the eyes red ; the tongue dry and 
 furred ; the skin cold and clammy, but sometimes hot ; the 
 respiration difficult ; and sometimes there are violent cramps 
 of the legs and arms. The urine is frequently diminished in 
 quantity and its passage attended with great pain. Stupor, 
 delirium, paralysis, and convulsions have also been observed. 
 In many cases death takes place calmly, and the intellectual 
 faculties remain clear to the last. 
 
 Fatal quantity : 2 grains taken in divided doses during a 
 period of five days have proved fatal. On the other hand, 
 2 ounces have been ingested and recovery occurred in six 
 hours.
 
 NOTES ON CHEMICAL LECTURES. 121 
 
 When fatal : Usually in twelve to twenty-four hours, 
 although in three cases death occurred in two hours, and in 
 one case in twenty minutes. 
 
 Antidote : Hydrated ferric oxide (ferric hydroxide) is the 
 most important chemical antidote. 
 
 The antidotal action of this substance is due to its forming 
 an insoluble compound, ferric arsenate, with the arsenious 
 oxide. 
 
 Thus, 
 
 2Fe 2 O 3 3H 2 O -f 2H 3 AsO 3 = Fe 3 2AsO 4 -f FeO + 9H 2 O - 
 
 The antidote should be given in its moist state and administered 
 in large excess. The antidote has no action on arsenious oxide 
 in its solid state, but only when in solution. 
 
 Hydrated ferric oxide (ferric hydroxide) may readily be 
 prepared by treating ordinary tincture of ferric chloride with 
 slight excess of ammonia, collecting the precipitate on a 
 muslin strainer, and washing it with water until it no longer 
 emits the odor of ammonia. A tablespoonful or more of the 
 moist magma, mixed with a little water, may be given at a 
 dose. The antidote should always be freshly prepared. 
 
 The ordinary magnesia of the shops may be used to add to 
 the tincture of ferric chloride instead of ammonia, and the 
 mixture containing excess of magnesia and hydrated ferric 
 oxide may be administered at once without filtering. 
 
 Antiseptic properties of arsenic : The preservative power 
 of arsenic when brought in direct contact with animal 
 textures is well known : and the poison seems to exert a 
 similar action when carried by means of the circulation to 
 the different tissues of the body. The bodies, therefore, of 
 those who have died from the effects of this poison are not 
 unfrequently found in a good state of preservation, even long 
 periods after death. 
 
 Solid arsenious oxide. Tests. 
 
 a. It is volatile at a temperature of 138 C. (280 F.). 
 
 b. When heated on charcoal in the reducing flame it is 
 dissipated in the form of white fumes, and emits a garlic-like 
 odor.
 
 122 NOTES ON CHEMICAL LECTURES. 
 
 | c. When heated in a reduction-tube, arsenious oxide vola- 
 tilizes without fusing, and recondenses in the cooler portion of 
 I the tube in the form of minute, octahedral crystals. 
 
 d. When heated in a reduction-tube containing a small 
 
 I piece of ignited charcoal, the arsenious oxide is volatilized 
 
 \ and deoxidized in its passage over the ignited charcoal, and 
 
 1 deposits in the cooler portion of the tube as a sublimate of 
 
 Imetallic arsenic. 
 
 A similar reaction occurs when arsenious oxide is heated in 
 la tube with a perfectly dry mixture of powdered charcoal and 
 podium carbonate. 
 
 If the closed end of the tube be removed (by breaking it 
 off) and the metallic sublimate then heated, it is readily vola- 
 tilized and oxidized into arsenious oxide, which condenses in 
 octahedral crystals. 
 
 The metallic sublimate is soluble in a solution of either 
 sodium or calcium hypochlorite. 
 
 - c. When arsenious oxide is heated in a reduction-tube with 
 a perfectly dry mixture of about equal parts sodium carbonate 
 and potassium cyanide, reduction will take place and a sub- 
 limate of metallic arsenic will form in the cooler portion of 
 ^he tube. 
 
 f. Potassium ferrocyanide may be employed as a reducing 
 agent instead of potassium cyanide. The arsenious oxide is 
 mixed with about 6 or 8 times its volume of dry potassium 
 ferrocyanide, and heated in a tube as in the preceding test. 
 Solutions of arsenious oxide. Tests. 
 
 a. Ammonio-silver nitrate test : Ammpnio-silver nitrate 
 throws down from aqueous solutions of arsenious acid a bright- 
 yellow precipitate of tribasic silver arsenite (Ag 3 AsO 3 ), the 
 reaction being, perhaps, 
 
 2H 3 AsO 3 + 6AgNH 3 NO 3 = 2Ag 3 AsO 3 + 6NH 4 NO 3 
 
 The precipitate is readily soluble, forming a colorless solu- 
 tion, in ammonium hydroxide and in nitric and acetic acids, 
 sparingly soluble in ammonium nitrate, and insoluble in the 
 fixed caustic alkalies. 
 
 b. Ammonio-copper sulphate test : Ammonio-copper 
 sulphate produces in solutions of arsenious acid a green,
 
 7. 

 
 NOTES ON CHEMICAL LECTURES. 123 
 
 amorphous precipitate of copper arsenite (CuHAsO 3 ), known 
 also as Scheele's green, the reaction being, perhaps, 
 
 2H 3 AsO 3 + 2CuH 2 O 2 (NH 3 ) 2 , (NH 4 ) 2 SO 4 = 2CuHAsO 3 + 
 2(NH 4 ) 2 S0 4 + 4NH 4 OH 
 
 The precipitate is nearly insoluble even in large excess of 
 the precipitant, but readily soluble in ammonia and in free 
 acids. From very dilute solutions of the poison the precipitate 
 does not appear of its characteristic color until the mixture 
 has stood some time. The same precipitate is thrown down, 
 from solutions of neutral arsenites by copper sulphate alone. 
 
 c. Hydrogen sulphide test : Hydrogen sulphide throws 
 down from solutions of arsenious acid, previously acidulated 
 with hydrochloric acid, a bright-yellow, amorphous precipitate 
 of arsenious sulphide (As 2 S 3 ), the reaction being 
 
 2H 3 AsO 3 + 3H 2 S j== As 2 S 3 + 6H 2 O 
 
 d. Reinsch's test ; When a solution of free arsenious acid 
 or an arsenite is strongly acidulated with hydrochloric acid, 
 and the mixture boiled with a slip of bright metallic copper, 
 the latter decomposes the arsenical compound and receives a 
 coating of metallic arsenic, or an alloy of copper and arsenic 
 (As 2 Cu 5 ). The arsenical nature of the deposit may be shown 
 in the following manner : the coated copper, after having been 
 carefully washed with pure water and dried in a water-oven, 
 is heated by means of a spirit lamp or the small flame of a 
 Bunsen burner, in a narrow and perfectly dry and clean reduc- 
 tion-tube, when the arsenic volatilizes, and, becoming oxidized, 
 yields a sublimate of octahedral crystals of arsenious oxide. 
 This sublimate usually forms within from a quarter to half an 
 inch above the .point at which the heat is applied. When the 
 sublimate is not exceedingly minute, it presents a well-defined 
 ring of sparkling crystals to the naked eye. 
 
 c. Marsh's test : When metallic zinc is treated with diluted 
 sulphuric acid, the hydrogen of the latter is displaced by the 
 metal with the formation of zinc sulphate, the hydrogen dis- 
 placed passing off in its free state : 
 
 Zn + H 2 SO 4 = ZnSO 4 + H 2 
 
 16
 
 124 NOTES ON CHEMICAL LECTURES. 
 
 If, however, arsenious acid or arsenic acid, or any of the 
 soluble compounds of the metal, be present, the nascent 
 hydrogen decomposes the arsenical compound, and, uniting 
 with the metal, forms arsenuretted hydrogen gas (AsH 3 ), 
 which is evolved in its free state. 
 
 The reaction in the case of arsenious acid is as follows : 
 
 3Zn + 3H,SO 4 -f- H 3 AsO 3 = 3ZnSO 4 + 4H 2 O + AsH 3 
 
 Metallic zinc is placed in the flask of the apparatus and 
 a quantity of a cooled mixture of 1 volume of pure concen- 
 trated sulphuric acid and 4 volumes of distilled water, suffi- 
 cient to cover the zinc, is poured in through a funnel tube and 
 the decomposition of the acid allowed to proceed. If the zinc 
 should act very slowly upon the acid, as is frequently the case 
 with the pure metal, the action may be hastened by the addi- 
 tion of a few drops of platinic chloride. 
 
 Should the zinc or sulphuric acid be contaminated with 
 arsenic this will give rise to arsenuretted hydrogen. There- 
 fore, before applying the test to a suspected solution, the purity 
 of the materials must be fully established. For this purpose, 
 after the apparatus has become completely filled with hydrogen 
 and while the gas is still being, evolved, the outer uncontracted 
 portion of the reduction-tube is heated to redness for about 
 fifteen minutes or longer. If this fails to produce a metallic 
 deposit or stain in the contracted part of the tube, in advance 
 of the part heated, the material may be considered free from 
 arsenic. The purity of the materials having been thus estab- 
 lished, it may be necessary to wash and renew the zinc, dry 
 the tubes, and add a fresh portion of the diluted acid. 
 
 The apparatus being adjusted and completely filled with 
 evolved hydrogen, the jet of gas, as it issues from the drawn- 
 out end of the reduction-tube, is ignited, care being taken not 
 to apply a light until the whole of the atmospheric air is 
 expelled from the apparatus, as otherwise an explosion might 
 occur. A small quantity of the arsenical solution is then 
 introduced into the funnel-tube and washed into the flask by 
 the subsequent addition of a few drops of the diluted sulphuric 
 acid. The decomposition of the arsenical compound, with 
 the evolution of arsenuretted hydrogen, will commence
 
 
 j2y r " v 
 
 LC^V^- 
 
 <*l 

 
 NOTES ON CHEMICAL LECTURES. 125 
 
 immediately. The presence of the arsenuretted gas may be 
 established by three different methods, namely, 
 
 a. By the properties of the ignited jet. 
 
 b. By decomposing it by heat applied to the reduction-tube. 
 
 c. By its action upon a solution of argentic nitrate. 
 
 a. The ignited jet ; As soon as the arsenical solution is 
 introduced into the flask the evolution of gas increases. The 
 flame of the jet will now increase in size, acquire a bluish tint, 
 and, unless only a minute quantity of arsenic is present, evolve 
 white fumes of arsenious oxide ; so, also, the flame sometimes 
 emits a peculiar garlic- like odor. If the white fumes be re- 
 ceived upon a cold surface they condense to a white powder, 
 which sometimes contains octahedral crystals. This is not a 
 delicate method for detecting the presence of arsenic, and 
 should never be employed to the exclusion of the following 
 modification, viz. : 
 
 If the flame be allowed to strike against a cold body, as a 
 piece of white porcelain, it yields a brown to black deposit of 
 metallic arsenic. 
 
 Fallacy ; Solutions of antimony, under these same condi- 
 tions undergo decomposition with -the production of anti- 
 monuretted hydrogen, which, like arsenuretted hydrogen, 
 burns with the evolution of white fumes, and yields metallic 
 deposits upon cold surfaces applied to the flame. The spots 
 produced by arsenic are readily soluble in a solution of either 
 sodium or calcium hypochlorite, whereas those from antimony 
 are insoluble, or dissolve only after prolonged digestion in the 
 hypochlorite solution. 
 
 b. Decomposition of the gas by heat : When arsenuretted 
 hydrogen, as evolved in Marsh's test, comes in contact with 
 the red-hot portion of the reduction-tube, it is decomposed 
 with the production of a deposit of metallic arsenic in the 
 contracted part of the tube, in advance of the flame. 
 
 Fallacy : Antimonuretted hydrogen also is decomposed 
 under the above conditions with the deposition of metallic 
 antimony. Antimony, however, is almost wholly deposited 
 before reaching the part of the reduction-tube to which the 
 flame is applied ; when it yields deposits on both sides of the
 
 126 NOTES ON CHEMICAL LECTURES. 
 
 flame, the outer one is quite near the flame. On the other 
 hand, arsenic deposits about one-half to three-quarters of an 
 inch in advance, or on the outer side of the flame, and never 
 before reaching the part of the tube to which the heat is 
 directly applied. 
 
 c. Decomposition by argentic nitrate ; If the reduction- 
 tube of the apparatus be substituted by a tube bent at a right 
 angle, and the arsenuretted hydrogen conducted into a solu- 
 tion of argentic nitrate, both the gas and the silver salt undergo 
 decomposition with the production of arsenious acid, which 
 remains in solution, and the separation of metallic silver, which 
 falls as a black precipitate. The reaction is 
 
 AsH 3 + 6AgNO 3 + 3H 2 O = 6Ag + H 3 AsO 3 + 6HNO 3 
 
 The presence of arsenious acid in the solution may be shown 
 by the usual tests for that substance. 
 
 SEPARATION OF ARSENIC FROM ORGANIC TISSUES. 
 
 / Disintegrate the organic matter by means of hydrochloric 
 cid and potassium chlorate, under the action of heat. 
 
 The following proportions of tissue, acid, and the chlorate 
 yield very satisfactory results : 
 
 Treat 300 grm., or 10 ounces, of the solid tissue, as of the 
 iver, cut into very small pieces and placed in a clean porcelain 
 dish, with a mixture of 60 c.c., or 2 fluid ounces, of strong 
 lydrochloric acid and 240 c.c., or 8 fluid ounces, of water. 
 
 Heat the mixture on a sand-bath, and when at about the 
 Doiling temperature, add about 1 grm., or 15 grains, of pow- 
 dered potassium chlorate, and repeat the addition at intervals 
 of a few minutes, with frequent stirring, until about 6 or 7 
 rm., or 100 grains, have been added and the mass has become 
 homogeneous and of a light-yellow color. During this process 
 water should occasionally be added to replace that lost by 
 evaporation. 
 
 The disintegrating action of this mixture is chiefly due to 
 the free chlorine and chlorine peroxide evolved by the mutual 
 iecomposition of the chlorate and a portion of the hydro- 
 chloric acid. 
 
 12HC1 = 4KC1 + 6H 2 O + 3C1O 2 -f C1 9
 
 NOTES ON CHEMICAL LECTURES. 127 
 
 Moderately heat the disintegrated mass until the odor of 
 chlorine has entirely disappeared, and then allow to cool. 
 Transfer the cooled mixture to a moistened linen strainer, 
 and, when the liquid has all passed, wash the solids with a 
 little warm water, the washings being collected separately. 
 Concentrate the washings on a water-bath to a small volume, 
 allow to cool, then add the washings to the first strained 
 liquid, and filter the mixed liquid through paper. Any arsenic 
 present will now exist as arsenic acid. 
 
 From the filtrate thus obtained free chlorine must be com- 
 pletely expelled, otherwise in the subsequent treatment of 
 the solution with sulphurous acid the latter will be acted upon 
 by the free chlorine resulting in the formation of hydrochloric 
 and sulphuric acids and, consequently, the sulphurous acid 
 would have no action upon the arsenic acid until all the free 
 chlorine was satisfied. 
 
 SO 2 -f C1 2 -f 2H 2 O = 2HC1 -f H 2 SO 4 
 
 To the liquid free from uncombined chlorine add a solution 
 of sulphurous acid until it smells strongly of the gas (sul- 
 phurous anhydride). Any arsenic acid present will be reduced 
 to arsenious acid, in which form the metal is more rapidly and 
 more completely precipitated by hydrogen sulphide than when 
 it exists in the form of arsenic acid. The reducing action of 
 the gas may be represented, 
 
 H 3 AsO 4 + SO 2 -f H 2 O = H 2 SO 4 -f H 3 AsO 3 
 
 Concentrate the liquid on a water-bath to a volume twice that 
 of the hydrochloric acid employed in preparing the mixture. 
 In this concentrating of the liquid the sulphurous acid must 
 be completely expelled, otherwise in the subsequent addition 
 of hydrogen sulphide decomposition of both compounds 
 occurs with the formation of pentathionic acid and separation 
 of sulphur. 
 
 5SO 2 H- 5H 2 S = 4H 2 O + H 2 S 5 O 6 -f S 5 
 
 Allow the concentrated liquid to cool, and then filter. 
 
 Pass a slow stream of washed hydrogen sulphide through 
 the filtrate for several hours ; then gently warm it, and allow 
 to stand twelve to twenty-four hours. Any arsenic present
 
 12S NOTES ON CHEMICAL LECTURES. 
 
 will be precipitated as arsenious sulphide (As,S 3 ), together with 
 more or less organic matter and free sulphur. Should the 
 liquid contain mercury, antimony, copper, or lead, these metals 
 would also be precipitated as sulphides by the hydrogen sul- 
 phide. Liquids prepared as the above may yield with hydrogen 
 sulphide a brownish or yellowish precipitate of organic matter 
 and free sulphur, even in the absence of any metal. 
 
 Collect the precipitate on a small filter and wash, at first 
 with water containing a little hydrogen sulphide, until the 
 washings no longer contain chlorine (test with argentic 
 nitrate). 
 
 Dissolve the moist precipitate on the filter with dilute am- 
 monium hydroxide (1 to 10), which will dissolve any arsenious 
 sulphide present, with more or less of the organic matter and 
 free sulphur. The sulphides of mercury, antimony, copper, 
 and lead which might be present would remain undissolved, 
 except perhaps a slight trace of the antimony sulphide. 
 
 Collect the ammoniacal liquid, usually of a dark -brown color, 
 in a small porcelain dish and evaporate to dryness on a water- 
 bath, treat the residue with a small quantity of strong nitric 
 acid, and again evaporate the liquid to dryness ; repeat the 
 operation with nitric acid, if necessary, until the moist residue 
 has a yellow color. 
 
 Moisten the residue with a few drops of concentrated solu- 
 tion of sodium hydroxide and evaporate to dryness. Treat 
 the residue with several drops of concentrated sulphuric acid, 
 and heat the mass on a sand-bath until it becomes about dry ; 
 again treat the residue with sulphuric acid and heat in the 
 same manner until fumes of the acid are no longer evolved. 
 
 Pulverize, if necessary, the carbonaceous residue, and boil 
 it with a small quantity of water containing a drop or two of 
 sulphuric acid, cool the liquid, filter off, and wash the insoluble 
 carbonaceous residue. If in the carbonization the whole of 
 the free sulphuric acid was expelled, the resulting solution 
 (filtrate) will be colorless and entirely free from organic matter. 
 Add 1 or 2 c.c. of solution of SO 2 to reduce As 2 O 5 to As 2 O 3 , 
 and then heat until all the SO 2 has been expelled. Concen- 
 trate the solution to a small and definite volume (say 20 c.c.), 
 and divide it into two equal portions. Make the qualitative
 
 O - 
 vf Qs is fj C^A^VVO ^UtL^( O*- ^JUU( 
 
 V 
 
 /3
 
 NOTES ON CHEMICAL LECTURES. 129 
 
 tests with one portion and the quantitative determination with 
 the other portion. 
 
 With the first portion 
 
 a. Apply Reinsch's test. 
 
 b. Marsh's " 
 
 With the second portion of 10 c.c. make the quantitative 
 determination. 
 
 QUANTITATIVE DETERMINATION. 
 
 Acidulate the solution with a few drops of hydrochloric acid, 
 warm to about blood-heat, and pass a stream of hydrogen sul- 
 phide through the liquid. The arsenic will be precipitated as 
 arsenious sulphide. Collect the precipitate on an equipoised 
 or on a weighed, washed filter, wash at first with water con- 
 taining a little hydrogen sulphide, then with pure water until 
 the washings are free from chlorine. Dry the precipitate on 
 the filter at a temperature of about 100 C., and weigh. 
 
 The quantity of As 2 O 3 corresponding to the weight of As 2 S 3 
 obtained is determined by 
 
 As 2 S :! . As. 2 O s . 
 
 246 : 198 :: weight of precipitate : X = one-half, 
 the amount of arsenic as As 2 O 3 recovered from the quantity of 
 organic tissue employed. 
 
 TESTS FOR SOME OF THE MORE COMMON ORGANIC 
 POISONS. 
 
 Opium,. 
 
 The presence of opium may be inferred by showing the 
 presence of meconic acid, an organic acid peculiar to the drug. 
 For this purpose, the aqueous extract of the substance is 
 treated with lead acetate, which will precipitate the acid as 
 meconate of lead. This is collected on a filter and washed, 
 the filtrate being reserved for the examination for morphine. 
 The contents of the filter are diffused in a small quantity of 
 water, and the lead precipitated by H 3 S. The filtrate from this 
 precipitate is concentrated to a small volume and treated with 
 a drop of ferric chloride solution, when, if meconic acid is 
 present, a deep blood-red coloration will be produced. Limit of 
 reaction, y^-jj^nj- solution.
 
 130 NOTES ON CHEMICAL LECTURES. 
 
 I Morphine.* <a ^-^ ev:< Lj" 
 
 tf. Sulpho-molybdic acid (Froehde) : The reagent is pre- 
 pared by heating 1 part of molybdic acid with 100 parts of 
 strong H 2 SO 4 C. P., until complete solution has taken place. 
 On the addition of a drop of the reagent to morphine or any 
 :>f its salts in the solid state, on a porcelain test tablet, a purple 
 ir crimson color appears immediately, passing through various 
 shades and finally to blue, which appears first at the margin of 
 he mixture. Limit, YoVoTo g ram - 
 
 b. Neutral ferric chloride added to morphine or any of its 
 ialts in the solid state, produces a deep blue color, which is 
 
 discharged by acids, alkalies and heat. Limit, ^o JJ-QO" g ra i n - 
 
 c. lodic acid in strong solution, added to morphine or any 
 of its salts in the solid state, the acid is decomposed with the 
 liberation of iodine, forming a brown or reddish-brown pre- 
 cipitate. Limit, yoVg- grain. 
 
 The presence of free iodine may be shown by agitating the 
 mixture with either, carbon disulphide or chloroform, which 
 will dissolve he iodine and assume a purple color. 
 
 Strychnine. 
 
 a. " Color-test; " Strychnine or any of its salts dissolved 
 in a drop of strong, chemically pure H 2 SO 4 , on a porcelain . 
 
 test tablet, and a minute crystal of potassium dichromatejBrawri v 
 
 through the solution by means of a glass rod, immediately 
 produces a cJiaracteristic succession of colors beginning with 
 blue, passing into purple, violet and greenish-yellow. Limit, 
 
 b. The caustic alkalies precipitate the free alkaloid from 
 solutions of salts of strychnine in the form of a white po\vder, 
 which soon assumes the crystalline form. Limit, ^^ 00 grain. 
 
 To prove the true nature of the precipitate apply the color 
 test. 
 
 c. Potassium dichromate produces a bright yellow pre- 
 cipitate of strychnine chromate, which quickly becomes crys- 
 talline. Limit, 10000 g ram - 
 
 This precipitate treated with a drop of strong, chemically 
 pure H,SO 4 alone gives the color reaction.
 
 
 
 
 
 ' 
 
 T^^ > 
 
 jl U-*-f- W 
 
 */
 
 Errata. Page 24, first line, read acetic instead of sulphuric. 
 
 INDEX. 
 
 PAGE PAGE 
 
 Acetic acid 27 Reinsch's test for 123 
 
 Acetone 115 separation of, from organic tissues . . 126 
 
 tests for 115 tests for 121 
 
 Albumin 108 Azotized compounds 6 
 
 biuret test for iia Baryta mixture, composition of 74 
 
 boiling test for 109 Beckmann's method 18 
 
 Esbach's method for quantitative Benzoic acid 86 
 
 determination of 113 Benzol, empirical formula of ai 
 
 Heller's test for no graphic formula of 21 
 
 native 108 molecular formula oi 21 
 
 picric acid test for in ring 21 
 
 potassium ferrocyanide test for . . . in Bile, tests for 114 
 
 qualitative tests for 109 Biuret 65 
 
 quantitative determination oi .... 113 test 112 
 
 trichloracetic acid test for in Bismuth test 93 
 
 Albumins, derived 109 Black precipitate la 
 
 Albuminates 109 Boettcher's test 93 
 
 Alcohol, amylic 26 Bromine, qualitative test for 37 
 
 butylic 26 quantitative determination of .... 46 
 
 definition of 26 Bruecke's lead process 96 
 
 ethylic 26 Butane 26 
 
 methylic 26 Butene 25 
 
 propenyl 27 Butyl 26 
 
 propylic 20 _ . , , . , 
 
 radicals, definition of . .26 Calculi, phosphatic, forms of 83 
 
 tables oi .... .26 Cane sugar . . 9 x 
 
 synthetical production of 8 _ composition of 91 
 
 Alcohols, monohydric .26 Carbamic acid 14 
 
 primary ... . 8 Carbamide 14 
 
 secondary 28 Carbinol .... - 28 
 
 table of 36 Carbohydrates, definition of ....... 6 
 
 tertiary. '. '. 29 Carbolic acid a 3 
 
 triatomic . 27 antidote for a 3 
 
 Aldehyde, butylic a 7 Carbon, qualitative test for 35 
 
 ethylic 27 quantitative determination of .... 38 
 
 propylic ! 1 a 7 Cellulose ...... 13 
 
 methylic 27 Chlorine, qualitative test for 37 
 
 valeric ' 27 quantitative determination of .... 46 
 
 Aldehydes, definition of '. '. '. 26 Composition of compounds . 35 
 
 table of 27 Compound radical, definition of 5 
 
 Alkarsin ' 10 Correction for barometric pressure .... 69 
 
 Alloxan ' 85 Correction for temperature 69 
 
 Alloxantin .' .' .' .' .' .' . ." .' . .' .' .' .' .' 85 Creatinine 87 
 
 Alpha-naphthol . . . 99 properties of .......... 88 
 
 Amido-benzol 21 quantitative determination of .... 88 
 
 mercuric chioride '. 12 l ? sts f r f 
 
 mercurous chloride .12 zinc chloride ........ 88 
 
 Ammonium carbamate 14 Cyanogen, synthetical production of ... 7 
 
 Amygdalin 33 Davy's method for determining urea . . 65 
 
 Analysis, elementary 37 Decomposing agents, acids . . 31 
 
 method of 38 alkalies . . oa 
 
 requisites for 38 heat | 
 
 organic 35 oxyge n '. |o 
 
 proximate 35 Derived albumins . 100 
 
 qualitative . . 35 Dextrine . . oi 
 
 ultimate -.--. 35 Dextrose ... .90 
 
 conditions observed in 37 Diacetic acid its 
 
 Aniline ... 21 test for .'.'.'.'. 115 
 
 Antipynn, formula of 14 Diastase no 
 
 Aqueous vapor, correction for 70 Dimethylamide '. '13 
 
 table of tension of 71 Dimethylarsin . '. 10 
 
 Arabinose 91 Dumas' method for determining nitrogen 41 
 
 Arsemcum 19 
 
 antidote for ai Earthy phosphates 78 
 
 fatal quantity of 20 quantitative determination of .... 8a 
 
 Marsh's test for 23 Empirical formula, definition of 15 
 
 quantitative determination of .... 29 method of determining 46 
 
 17 (131)
 
 132 
 
 INDEX. 
 
 PAGE PAGE 
 
 Eremacausis 30 Isomeric compounds 24 
 
 Ethane 26 divisions of 24 
 
 Ethene 25 examples of 24 
 
 Ether, amylic 37 Isomerism, definition of 24 
 
 definition of 27 Isuretine 62 
 
 ethylic 27 
 
 methylic 27 Johnson's test . 93 
 
 propylic 27 
 
 Ethers, definition of 27 , , . 
 
 Ethers table of 27 Kakodyl . . to 
 
 Ethyl 26 compounds of 10 
 
 urethan 14 Kakodyhc acid 10 
 
 properties of 10 
 
 Fat acids, definition of 27 Kekulg's benzol ring .... 
 
 table of 27 Ketone, definition of 29 
 
 Fehling's solution, clinical use of' .' .' .' .'105 Kjeldahl's method for determining nitre- 
 
 preparation of 102 8 en 45 
 
 table for clinical method to6 
 
 Fehling's test 97 Lactose 34 
 
 method of applying 97 Laevulose 90 
 
 Ferment, definition of 33 Lead process 96 
 
 Fermentation, acetous 34 Liebig's method for the determination of 
 
 alcoholic 34 sodium chloride 61 
 
 butyric 34 practical application of 62 
 
 conditions necessary for ... . . . 53 Liebig's method for the determination of 
 
 definition of 33 urea 71 
 
 lactic 34 corrections for 74 
 
 test for glucose 93 practical application of 74 
 
 varieties of 33 Lithic acid 84 
 
 vinous 34 
 
 viscous 35 Maltose 91 
 
 Fermentescible body ... 33 Mannitose 90 
 
 Formic acid 27 Marshall's apparatus, method of using . . 68 
 
 synthetical production of 8 Marsh's test 123 
 
 Formulas, deduction of 46 Meta compounds 22 
 
 Fowler's modification of Davy's method Metameric compounds, definition of ... 24 
 
 for determining urea 66 Methane 26 
 
 Fruit sugar 91 Methene 25 
 
 Methyl 25 
 
 Globulins 109 Methylamide 13 
 
 test for 112 Methyl-ethylamide 13 
 
 Glucose 9 Methyl-ethyl-propylamide 13 
 
 alpha-naphthol test for 99 Methyl urethan .14 
 
 bismuth test for 93 Micrococcus ureae 40 
 
 Boettcher s test for 93 Milk sugar 
 
 determination of, by saccharimeter . . 107 Mohr's method for determining sodium 
 
 fermentation test for 93 chloride ... 56 
 
 Fehling's test for 97 practical application of 59 
 
 Johnson s test for 93 Molecular formula, definition of 15 
 
 Moore s test for 9* method of determining 15 
 
 phenylhydrazmehydrochloride test for 98 of non-vaporizable substances . ... 17 
 
 picric acid test for 93 Molecular weight, method of determining 15 
 
 qualitative tests for . ...... 92 Molisch's tests 99 
 
 quantitative determination of, by clin- Moore's test 92 
 
 ical method 105 Morphine, tests for .' .' 130 
 
 Fehlmg's solution 102 Mucin 91 
 
 fermentation ......... 93 Murexide test 85 
 
 Robert g differential density me- Mycodermae aceti 
 
 thod TOI 
 
 thymol test for 100 Naphthalene . . 99 
 
 Trommer s test for 94 Native albumins 108 
 
 Glycendes 08 Nitrogen, Dumas' method for determining 41 
 
 Glycerine 27 gravimetric determination of .... 43 
 
 Grape sugar -..- 9 Kjeldahl's method for determining . . 45 
 
 Graphic formula, definition of 20 qualitative tests for 35 
 
 Haycraft's method 87 quantitative determination of .... 41 
 
 Heller's method no volumetric determination of ..... 44 
 
 Hemialbumose . . 112 W and . Varrentrapp s method for 
 
 Hippuric acid 86 . determining 4 z 
 
 Homogentisic Acid, test for 98 Nitrogenous compounds 6 
 
 Homologous series, definition of . . . . . 25 . , , ,. . . ' ,. 
 
 table of an 25 defines, definition of 25 
 
 Hydrocarbons, definition of _ . table of 25 
 
 Hydrogen, qualitative detection of . ... 35 Opium, tests for. 129 
 
 quantitative determination of . .38 Organic body, definition of. 7 
 
 Hypobromite method for determining urea 67 chemistry, definitions of 5 
 
 compounds, saturated n 
 
 Inosite 90 matter, tests for 35 
 
 Iodine, qualitative test for 37 substances, decomposition of .... 30 
 
 quantitative determination of . . .46 Ortho compounds 22
 
 INDEX. 
 
 133 
 
 PAGE 
 
 Para compounds 23 
 
 Paraffins, definition of 26 
 
 table of 26 
 
 Pentane 26 
 
 Pentene 25 
 
 Penicilium glaucum 34 
 
 Phenol 23 
 
 Phenol-sulphuric acid 23 
 
 Phenylhydrazine hydrochloride 98 
 
 Phosphoric acid 78 
 
 indicator in 80 
 
 practical method for determining . . Si 
 principles of volumetric determination 
 
 of 78 
 
 volumetric determination of 81 
 
 Phosphorus, qualitative tests for 36 
 
 quantitative determination of .... 46 
 
 Picric acid test for glucose ....... 93 
 
 15 
 
 '7 
 16 
 
 Poison, definition o 
 
 Poisoning, appearances common to 
 
 sources of evidence in 
 
 symptoms of arsenical .... 
 Poisons, causes which modify . . . 
 
 classification of 
 
 limit of tests for 
 
 toxic action of 
 
 Polymeric compounds, definition of . . 24 
 
 Potass-amide 13 
 
 Potassium ferrocyanide test for albumin . in 
 
 Pressure, correction for barometric ... 69 
 
 Propane 26 
 
 Propene 25 
 
 Propyl 26 
 
 Propylic acid 27 
 
 Proximate principles, definition of . ... 7 
 
 analysis, definition of 35 
 
 Putrefaction, definition of 32 
 
 Radical, definition of 8 
 
 electrical condition of 8 
 
 Radicals, equivalence of 8 
 
 types of 9 
 
 Rational formula, definition of 20 
 
 Raoult's method for determining molecular 
 
 formula 18 
 
 Reinsch's test 123 
 
 Results, calculation of 46 
 
 Roberts' differential density method for 
 
 determining glucose 101 
 
 Saccharimeter 107 
 
 Saccharose 91 
 
 Salkowski-Ludwig method 87 
 
 Schiff 's test for uric acid 85 
 
 Soda-lime, composition of 32 
 
 Sodium chloride, Liebig's method for 
 
 determining 61 
 
 gravimetric determination of .... 54 
 
 Mohr's method for determining ... 56 
 
 Sodium hypobromite solution 68 
 
 Sorbine 90 
 
 Standard solution of mercuric nitrate for 
 
 determining sodium chloride . . . 61 
 of mercuric nitrate for determining urea 72 
 
 of phosphoric acid 79 
 
 of silver nitrate 57 
 
 corrections for 58 
 
 of uranium acetate 79 
 
 of uranium nitrate 79 
 
 Starch 91 
 
 Stearin 28 
 
 Strychnine, tests for 130 
 
 Substitution, definition of II 
 
 examples of 12 
 
 Sugars 90 
 
 FAGK 
 
 Sulphur, qualitative analysis of 36 
 
 quantitative determination of .... 45 
 
 Synopsis of albumins 108 
 
 Synthetical production of alcohol .... 8 
 
 benzaldehyde 8 
 
 cyanogen 7 
 
 formic acid 8 
 
 Synthetical production of urea 7 
 
 Temperature, correction for 69 
 
 Thymol test 100 
 
 Toxicology 115 
 
 Trimethylamide 13 
 
 Trinitrocellulose 13 
 
 Trinitroglycerine 13 
 
 Trommer's test 94 
 
 Type, definition of n 
 
 Uranium acetate solution, standardization 
 
 of 81 
 
 Urea 6a 
 
 artificial preparation of 63 
 
 Davy's method for determining ... 65 
 Fowler's modification of Davy's meth- 
 od for determining 66 
 
 hypobromite method for determining . 67 
 
 Liebig's method for determining . . 71 
 
 methods of obtaining from urine ... 63 
 
 nitrate of 64 
 
 oxalate of .' . 64 
 
 preparation of standard solution of . . 73 
 
 properties of 64 
 
 qualitative tests for 65 
 
 quantitative determination of .... 65 
 
 special history of 62 
 
 synthetical production of ...... 7 
 
 Urethan 14 
 
 Urethans 14 
 
 Uric acid 84 
 
 properties of 84 
 
 qualitative tests for 85 
 
 quantitative determination of .... 86 
 
 salts of 85 
 
 Urine 48 
 
 abnormal constituents of 91 
 
 accurate method for determining quan- 
 tity of solid matter in 51 
 
 acidity of 50 
 
 amphoteric reaction of 49 
 
 analysis of 33 
 
 approximate method for determining 
 
 quantity of solid matter in 52 
 
 average quantity voided 53 
 
 collection of 49 
 
 determination of quantity of solid 
 
 matter s 1 
 
 diabetic 92 
 
 method of determining acidity of . . 50 
 
 methods of obtaining urea from ... 63 
 
 properties of diabetic . . 92 
 
 reaction of . . . . 49 
 
 specific gravity of 59 
 
 table of average composition of ... 53 
 
 Valeric acid 27 
 
 Victor Meyer's method 15 
 
 Vital force 5 
 
 Weyl's test for creatinine 88 
 
 White precipitate 12 
 
 Will and Varrentrapp's method for deter- 
 mining nitrogen 42
 
 SL 
 

 
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