ORGANIC CHEMISTRY. [Three hours allowed for this Paper.] 1. What is the action of heat upon formiate of am monium? 2. Describe the processes for transforming benzol into aniline. 3. An organic acid has been found to contain in ICO The analysis of the silver salt of this acid has given the following result: 1052 parts of the silver salt left upon ignition 0.4637 parts of metallic silver. What is the formula of the acid? Give details of calculation. Solution. Dividing by each equivalent we obtain the following quotients, expressing the proportionate number of equiva lents in the compound: Eqs. of C combined with 1 eq. of O or H = 2'33 Multiplying these numbers by 3, to clear of fractional The empirical formula is therefore C, H, O.. Again, 1052 parts of the silver salt yield 4637 parts of silver; the quantity of the salt necessary to yield 108 parts or 1 eq. of Ag may be easily found by the following proportion : Ag. Ag. Silver Salt. Silver Salt. •4637 : 108 :: 1'052 : 245 This 245 is then the equivalent of the silver salt, and deducting the equivalent of AgO, 108 + 8 = 116, we +8 obtain 245-116 129 for the equivalent of the dry acid or anhydride. == The real acid will contain in addition HO, and its equivalent will therefore be 129 + 9 = 138. = Now the equivalent of the empirical formula, C, H, O., given above, is 42 + 3 + 24 69, which is just half the real equivalent 138. So that it is evident this formula C, H, O, must be doubled so as to express one equivalent of the acid. It then becomes C, H, O or C, H, O, HO. SECTION II. 1. Describe the processes of transforming ferrocyanide of potassium respectively into urea and cyanuric acid. 2. State the preparation, the composition, the properties, and the principal transformations of oxalic acid. 12 18 12 3. A certain variety of malt has been found to contain 10 per cent. of starch (C12 H10 Ojo) and 20 per cent. of sugar (C12 H1s O12). Describe the changes which starch and sugar undergo during their transformation into acetic acid (C, H, O.), and state the quantity of malt required for the production of 100 pounds of acetic acid, supposing the whole amount of starch and sugar to be converted. Equation (1) expressing the change of sugar into alcohol; (1.) C,H,Os = 2(C,H,O,) + 400. Equations (2, 3) expressing the change of starch into sugar and alcohol;" 10 12 (2.) C12H110 + 2HO = C12H1O19 + Equation (4) showing the conversion of alcohol into acetic acid; (4.) CH ̧O2 + Ọn = CHO, + 2HO. It will be seen from these equations, that I eq. sugar produces a eqs. alcohol; and as I eq. alcohol yields 1 eq. acetic acid, therefore from I eq, sugar 2 eqs. of acetic acid are formed. Similarly, I eq, starch produces a eqs. acetic acid. Therefore 180 lbs. sugar yield 120 lbs. acetic acid; and 162 lbs. starch yield the same quantity. The quantity of starch in the malt is 10 per cent., and of sugar 20 per cent. Then byproportion, Total quantity of acetic acid from 100 lbs. malt = 7}} +13 2014 lbs. The quantity of malt necessary for the production of 100 lbs. acetic acid can then easily be found by proportion; 1. Express in formulæ the transformation of oxalic ether into oxamat of ethyl and oxamide. 2. Describe the method of preparing the chlorides of the acid-radicals, and their deportment under the influence of water, alcohol, and ammonia. 3. In determining the vapour-density of an organic com pound, according to Dumas's process, the following observations were recorded: 100 cubic centimetres of air at o° C. and bar. 0.760 metres weigh o12932. What is the vapour-density of the compound referred to hydrogen as unity? Give details of calculation. Solution. 1st. To find the weight of the air contained in the globes, viz., 292 c. c. at 12° C. and 760 m. Co-efficient of expansion of air for 1°='00367. 1. c. c. of air at o° becomes + (00367 × 12)= 1'044 c. c. at 12°. ... 1044 c. c. at 12o, reduced to o° become 1 c. c. and 292 c. c. at 12°, reduced to o° become Correcting for barometric pressure; as the volumes of gases are inversely as the pressure, 279.694 c. c. at a pressure of 762 m. become at 760 m. Now 100. c. c. air at 0° and 760 m. weigh 12932 grm. .. 280 43 c. c. air at o° and 760 m. weigh •3626 grm. i. e., the weight of air in globe 2nd. To find the weight of substance in the state of vapour. = globe and air = 386 grm. 7486 grm. 762 m. As the globe and vapour .. Weight of vapour 386+ 3626 = Hence 292 c. c. of vapour at 152° C. and weigh 7486 grm. 3rd. To reduce this volume to o° and 760 m. I. c. c. at 0° becomes 1 + (00367 × 152)=1°55784 at 152°, i. e., 155784 at 152°, when cooled to o°, become I C. C. |