Further experiments on dissociation will be found in Part III. (Chap. II.). Meanwhile an experiment may be performed to illustrate the general statement that the amount of dissociation of a dissociable compound increases as temperature increases (up to a certain limit), pressure being kept constant.

Exp. 3. Make 3 determinations of the specific gravity of the vapour obtained by heating amylic bromide by V. Meyer's method, (1) using a bath of methyl salicylate which boils at 222o, (2) using a bath of bromonaphthalene which boils at 280°, and (3) using a bath of sulphur which boils at 450°. (These boiling points are approximately correct, pressure being 760 mm.)

The formula of amylic bromide is C.H1Br; when the compound is heated considerably above its boiling point (129) it is dissociated into amylene (C,H) and hydrobromic acid (HBr); dissociation begins at about 160°, and at about 360° the vapour consists wholly of a mixture in equal volumes of CH10 and HBr; at intermediate temperatures the vapour consists of CH1Br, mixed with C,H10 and HBr. The specific gravity of gaseous C,H,,Br, referred to air as unity, is 5.22; the specific gravity of a mixture of equal volumes of CH and HBr gases is 2.61. If no dissociation occurs at a specified temperature the observed specific gravity will be 5-22; if complete dissociation occurs the observed specific gravity will be 261; and if partial dissociation occurs the observed specific gravity will be a number between 5.22 and 2·61.

The vapour obtained by heating amylic bromide to about 220° has a specific gravity of about 4.2; at 280° the specific gravity is about 3·1; and at 450° it is 2.6; pressure in each case being normal.

Reference to "ELEMENTARY CHEMISTRY." Chap. XII. pars. 233 to 236; also Chap. XVI. pars. 333 to 337.

CHAPTER V.

REACTING WEIGHTS OF COMPOUNDS DETERMINED BY CHEMICAL METHODS.

THE relative weight of the smallest mass of a compound which takes part in chemical interactions is determined by quantitatively studying some of the typical reactions of the compound.

Exp. 1. Determine the reacting weight of acetic acid (a compound of H, C, and O); assuming the acid to be monobasic, and assuming the atomic weight of silver to be 108.

Silver acetate is decomposed by heating in air; silver remains, and the carbon hydrogen and oxygen are burnt away. Prepare pure silver acetate by adding a solution of recrystallised sodium acetate to a concentrated solution of silver nitrate, collecting the crystals which form, and crystallising once or twice from hot water. Dry the crystals of silver acetate by pressing between filter paper and then heating to 100°. Weigh out about 5 gram of the dry salt into a weighed porcelain crucible with a lid, then heat gently to full redness over a Bunsen-lamp, and then for a few minutes over the blowpipe (keeping the lid on the crucible), cool, and weigh the silver in the crucible.

Calculate the percentage of silver, and deduct this from 100; the difference gives the carbon, hydrogen, and oxygen, combined with the silver to form 100 parts of silver acetate. The atomic weight of silver is 108; calculate the mass of carbon, hydrogen, and oxygen, combined with 108 parts by weight of silver to form silver acetate. As acetic acid is monobasic, the acid would be obtained from the silver salt

by replacing 108 of silver by 1 of hydrogen: make the necessary calculation, and thus find the reacting weight of acetic acid.

Exp. 2. Assuming the composition of a reacting weight of oxalic acid to be represented by the formula nHCO, (where n = a whole number), find the probable value of n.

Prepare a considerable quantity of a fairly concentrated solution of potash; about 150 grams potash in 300 c.c. water. Dissolve in water about 60 grams of oxalic acid; divide the solution into three equal parts; exactly neutralise one part by the potash solution, and to another part add half as much of the same potash solution; evaporate both solutions to the crystallising point; collect the crystals which form, and recrystallise them twice from hot water. Dry each set of crystals by pressure between paper, and then at 100°; weigh out equal quantities of the two solids, add sulphuric acid to each, warm, and run in a standardised solution of potassium permanganate until a faint pink colour is permanently produced.

The fact that a certain mass of one of the solids requires a quantity of permanganate to completely oxidise it very different from that required by an equal mass of the other solid, shews that the two sets of crystals are different salts. (It is assumed that oxalates are wholly oxidised by permanganate in presence of sulphuric acid.) Hence at least two potassium salts can be obtained by interactions between potash and oxalic acid hence the composition of a reacting weight of oxalic acid is expressed by a formula not less than H,C,O4

To the third quantity of oxalic acid (equal to those already used) add twice as much potash as you know is required for neutralisation; evaporate; collect the crystals which form; recrystallise them three or four times from hot water; dry; weigh out a quantity equal to that taken of the former crystals, and titrate with permanganate as before. The result shews that these crystals are the same salt as was produced by exactly neutralising oxalic acid with potash.

So far as these experiments go, oxalic acid appears to form two, and not more than two, potassium salts; hence you conclude that the acid is dibasic; and hence that the probable value of n in the formula nHCO, is two, that is, the composition of the reacting weight of oxalic acid is probably expressed by the formula H.CO.

Exp. 3. Find the reacting weight of barium oxide, by examining the reaction which occurs between this base and an acid the reacting weight of which is known.

It will be necessary to determine the simplest formula to be given to barium oxide, assuming the atomic weights of barium and oxygen to be 137 and 16 respectively, and knowing that barium and oxygen are combined in barium oxide in the ratio 137 16.

:

Weigh out accurately about 5 gram of barium oxide; add a measured quantity of dilute sulphuric acid the strength of which is known, adding so much that the liquid above the solid barium sulphate which forms is strongly acid to litmus; dilute; warm for some time; then allow the pp. of barium sulphate to settle; add a drop or two of litmus solution, and determine the amount of sulphuric acid which remains unchanged by means of a standardised alkali solution.

Calculate the mass of sulphuric acid which has been used to convert the barium oxide into sulphate; then calculate the mass of barium oxide which has interacted with each reacting weight (H2SO4 = 98) of sulphuric acid.

The result is that 98 parts by weight of sulphuric acid react with 153 parts of barium oxide; but 98 is the reacting weight of sulphuric acid; therefore the reacting weight of barium oxide is probably not greater than 153. The formula BaO (Ba = 137, O=16) expresses the composition of 153 parts by weight of barium oxide; therefore the formula BaO probably expresses the composition of the reacting weight of barium oxide: the experiment with sulphuric acid shews that the reacting weight is probably not greater than 153; but it cannot be less than this if the atomic weights of barium and oxygen are 137 and 16, respectively.

Supplement the result obtained by performing a similar experiment, using standardised hydrochloric acid instead of sulphuric acid. In this case the barium chloride formed remains in solution; after adding a decided excess of hydrochloric acid, determine the acid which remains by means of a standardised ammonia solution. Assuming the reacting weight of hydrochloric acid to be 36.5 (HCl), the results of the experiment shew that one reacting weight of this acid interacts with 15376.5 parts by weight of barium oxide. But the reacting weight of barium oxide cannot be less than 153; therefore we conclude that this oxide and hydrochloric acid interact to form barium chloride in the ratio, expressed in reacting

weights, of 12. This result confirms that obtained by using sulphuric acid.

Exp. 4. Find the reacting weight of aniline; assuming that the double compound which this base forms with hydrochloric acid and platinic chloride has the composition 2XHCl. PtCl ̧, where X = one reacting weight of aniline.

It is first necessary to purify the 'pure' aniline of commerce as follows. Equal masses (say 25 grams) of aniline and glacial acetic acid are mixed in a retort which is tilted upwards and attached to a long glass tube which serves as a condenser. The contents of the retort are heated to boiling for 12 hours, after which time the formation of acetanilide may be presumed to be complete. While the retort is still hot it is connected with a condenser, and heating is continued until the liquid which passes over begins to solidify. This will happen at about 280°-290o. The condenser is now removed and the receiver changed. The impure acetanilide which collects in the receiver is purified by recrystallising once or twice from a large quantity of boiling water. The acetanilide is reconverted into aniline by heating it with potash in a flask fitted with an inverted condenser, and the aniline is blown over with steam. The aniline is separated from the water as completely as possible, dehydrated by means of solid potash, and finally distilled; the portion boiling at 181°-182° is pure aniline. To this pure aniline is added a slight excess of hydrochloric acid, and the solution is evaporated to dryness at 100°; the residue is dissolved in a small quantity of water, and the liquid is filtered if necessary. To the filtrate a fairly strong solution of platinic chloride is added; but care must be taken that the platinum sult is not in excess. The double chloride of aniline and platinum which is precipitated is strained off from the mother liquor, and a small quantity is dried at 100° in a weighed crucible. The crucible plus dried salt is weighed; and the salt is then decomposed by heating it at first gently and finally strongly over a Bunsen-flame. The residual metallic platinum is weighed. Let w' weight of double chloride

=

and let X represent the reacting weight of aniline. Since the platinum double salts of nitrogenous bases are constituted on the same type as the platinum double salts of ammonia, it follows that 2XHCl. PtCl, represents the composition of the