MAKING OF FORMULAE AND EQUATIONS 99 



Making Formulas. In the formulae, these proportions are 

 to be replaced by multiples of the atomic weights by whole num- 

 bers. We therefore divide the quantity of each element by the 

 corresponding atomic weight. This operation gives us the factors 

 by which the atomic weights are to be multiplied. The atomic 

 weights we find in the table, where the values determined by the 

 most expert chemists are collected. 



For example. In the case of tin oxide the proportion of tin to 



100 

 oxygen is The atomic weights are 119 and 16, respectively. 



2() .Oa 



100 



100 -T- 119 = 0.84, and 26.89 -T- 16 = 1.68. The proportion ^-^ 



^o . oy 



119 X 0.84 

 now becomes 16 x 1 gg 



Now this proportion like all chemical proportions must be 

 expressed in multiples of the atomic weight by whole numbers. 

 Hence, we next find the greatest common measure of the two 

 factors. It is 0.84. Indeed, in this simple instance, we can see 

 that the ratio of the factors is 1 : 2. Dividing above and below 



. 119 X 1 

 by 0.84, we get Ig -_ 



Now, the symbols stand for the atomic weights. Substi- 

 tuting these symbols, the proportion becomes ^ - The 



U X ^ 



formula is therefore SnO 2 . 

 Applying the same process to the lead oxide, we get 



JLOO 207.1 X 0.483 207.1 X 1 Pb X 1 p , o 

 7.72 16 X 0.483 16 X 1 O X 1 ' 



Treating the other data in the same manner, we find : ferric oxide 

 Fe2O 3 , zinc sulphide ZnS and mercuric oxide HgO. 



If the composition of the substance is given in percentages, 

 the same process is used. Thus, the case of sodium sulphate 

 works out as follows: 



