EQUIVALENTS, CHEMICAL 303 



density of hydrogen is 0-0692, half of which is 0-0346. Then 0-0346 x 81 - 2-8026. 

 Experiment gave 2'73. 



To obtain the density of any vapour or gas having a condensation to two volumes. 

 Multiply the density of hydrogen by the atomic weight of the gas or vapour. 

 EXAMPLE : Find the density of chlorine gas. The atomic weight of chlorine being 

 35-5, and the density of hydrogen 0-0692, we have by the rule, 0-0692 x 35'5 = 2-4566. 

 The density by experiment is 2'44. 



To obtain the density of any vapour or gas having a condensation to one volume. 

 Multiply twice the density of hydrogen by the atomic weight of the gas or vapour. 

 EXAMPLE : Find the density of the vapour of oxygen. The atomic weight of oxygen 

 being 8, and twice the density of hydrogen being 0-1384, we have 0-1384 x 8= 1-1072. 

 Experiment has yielded 1-1056. 



The above methods of calculating the densities of vapours and gases are those always 

 employed by the writer of this article, and will be found incomparably shorter and 

 more convenient than any other. 



It is perfectly plain that, by a simple inversion of the above rules, ife is equally easy 

 from the known density of a gas or vapour to calculate its atomic weight. Never- 

 theless, for the sake of those who are unaccustomed to calculations of this kind, we 

 append the following rules : 



To calculate the atomic weight of any gas or vapour having a condensation to four 

 volumes. Divide the density of the gas by half the density of hydrogen. EXAMPLE : 

 Find the atomic weight of hydrobromic acid gas, the density of which is 2-8026 ; 



^5-81-000. 



0-0346 



To calculate the atomic weight of any gas or vapour having a condensation to two 

 volumes. Divide the density of the gas by the density of hydrogen. 



To calculate the atomic weight of any gas or vapour having a condensation to one 

 volume. Divide the density of the gas by twice the density of hydrogen. 



It is plain, then, that if we are in possession of the atomic weight and vapour- volume 

 of any substance, it is easy to determine the density of its vapour or gas. Also, that 

 having the density of the vapour and the vapour-volume, it is easy to calculate the 

 atomic weight. If we consider for an instant what is meant by the term ' density of a 

 vapour or gas,' it will appear equally easy to find, from the density of the gas, the 

 weight of 100 cubic inches at the standard temperature and pressure. By the density 

 of a gas is meant the number expressing how much it is heavier or lighter, bulk for 

 bulk, than air. If, therefore, we multiply the density of a gas by the weight of 100 

 cubic inches of air, at the standard temperature and pressure ( = 30'00 grains), we 

 immediately find the number required. EXAMPLE : The density of hydrogen is 

 0-0692 and 0'0692 x 30 = 2-0760, or the weight of 100 cubic inches of hydrogen, at a 

 temperature of 60 Fahr., and 30 inches of the barometer. 



From what has been said, it is evident that no difficulty exists in determining the 

 equivalents of bodies which can be obtained in a gaseous state. Where the equivalent 

 of a fixed body is to be ascertained, or where it is desired to proceed in a different 

 manner, the method employed must depend upon the nature of the substance. We 

 shall consider three of the most simple and general cases, namely, an acid, an alkali, 

 and a neutral body. 



1. Mode of determining the equivalent of an acid. For this purpose it is necessary 

 to analyse a salt, the constitution of which is known. If the base or metallic oxide in 

 the salt is one of which the atomic weight is well established, it is very easy to deter- 

 mine the combining proportion of the acid. We say, as the percentage of oxide is to 

 the percentage of acid, so is the atomic weight of the oxide to the atomic weight of the 

 acid. EXAMPLE : Butyrato of silver has the following composition : 



Oxide of silver . . ..... . V / .v . 69-487 



Butyric acid 40-613 



100-000 

 We therefore say : 



69-487 : 40-513 :: 116 : 79-000 



Percentage of oxide Percentage of acid. Equivalent of oxide Equivalent of the 

 of silver. of silver. acid. 



It must be remembered that the atomic weight so obtained is that of the anhydrous 

 acid, so that one equivalent of water must be added to find the atomic weight of the 

 acid in its ordinary condition. If the equivalent desired be that of a hydrogen acid, 

 the method of proceeding must be slightly modified, but the details need not be given, 

 as they are self-evident. 



