APPLICATION OF THE MOLECULAR HYPOTHESIS 85 



For example, 190 c.c. of a gas at and 760 mm. weigh 1.23 g 

 If x be the weight of 22.4 liters (= 22,400 c.c.), 



190 : 1.23 :: 22,400 : x (= 145 g.). 



Again, 210 c.c. of a vapor at 100 and 743 mm. weigh 1.12g. 

 This volume at and 760 mm. would become: 



210 x |f x|| = i50.3,c. 



and 



150.3 : 1.12 :: 22,400 : x (= 167 g.). 



Molecular Weight and Density. Density is the term used 

 in physics for the weight of 1 cubic centimeter of a substance. 

 Thus, the density of water at 4 C. is 1, because at 4^ C., 1 c.c. 6T 

 water weighs 1 g. The density of ammonia gas (wt. in g. of 1 c.c. 

 at and 760 mm.) is 0.000759. 



Since the molecular weight is the weight of 22.4 1., or 22,400 c.c., 

 the molecular weight of a gas is obtained by multiplying the 

 density (if known) by 22,400. Thus, the molecular weight of 

 ammonia is 0.000759 X 22,400 = 17.0. 



Densities are often given on the scale, density-of-air = 1. 

 Now 22.4 liters of air weigh 28.95 g. If a gas has a density twice 

 that of air, 22.4 liters of this gas would weigh 28.95 X 2 g. For 

 example, the density of carbon dioxide (air = 1) is 1.52. The 

 molecular weight is therefore 28.95 X 1.52, or 44.0 (see p. 79). 



Atomic Weights. Since a compound substance can be 

 formed by union of the elementary substances, and decomposed 

 to give these substances, its molecule may be assumed to contain 

 those constituent elements as distinct parts of the mass. Those 

 elementary parts of a molecule are called atoms. When two ele- 

 mentary substances combine, the process involves the union of 

 the two kinds of atoms to form compound molecules. Now, in 

 the table (p. 76), we recorded the weights of the constituent ele- 

 ments which together made up the molecular weights of the com- 



