CHEMICAL UNITS OF WEIGHT. FORMULAE 79 



+ 35.46 = 36.468) is correct, since 36.468 g. is the amount filling 

 the cube. Again, the formula for oxygen gas itself must represent 

 32 g. (= 2 X 16), the weight of 22.4 1., and is therefore 2 . Half 

 as much as this may enter into a compound, H 2 0, ZnO, etc., but, 

 logically, the formula for free oxygen must record the double 

 weight required to fill the cube when the gas is present alone, in the 

 free condition (see pp. 86, 105). Similarly the formula for hydrogen 

 gas is H 2 (2 X 1.008 = 2.016, the weight of 22.4 1.). The formula 

 of every volatile substance must thus be written so as to show the 

 weight of the chemical unit quantity (the molecular weight, p. 84). 

 When the substance is not easily volatilized, this unit cannot be 

 measured, and the simplest formula is employed. 



Information Contained in Each Formula. A formula 

 thus contains, in condensed form, several items of information. 

 It shows: 



1. The elements making up the substance, 



2. The proportion by weight of those elements, 



3. The total unit weight (molecular weight) of the substance. 



Given the formula, we can read these facts in it. 



Thus, if we are given the formula of carbon dioxide, C0 2 , we 

 consult the table of reacting or atomic weights (inside rear cover), 

 and learn that C = 12 parts by weight of carbon and 2 = 2 X 16 

 parts by weight of oxygen. The proportion by weight of the ele- 

 ments in this compound is, therefore, 12 of carbon to 32 of oxygen. 

 The total weight (molecular weight) is 12 + 32 = 44, and this 

 must be the weight filling 22.4 1. (at and 760 mm.). 



Why was 22.4 Liters Chosen as the Unit Volume? It is 



clearly advantageous to employ a unit volume of such dimen- 

 sions that no element shall receive a unit weight smaller than 

 unity. At first sight, indeed, it would seem self-evident that we 

 ought to choose, for our own convenience, a unit volume based on 



