Question

2) Calculate the percent composition for each element in the following two compounds: a) SO: b) NaH PO

Answer 1

a) SO3:
Molar mass of SO3 = 32.07 + (16.0*3) = 80.07 g/mol

In 80.0632 g of SO3 there is 32.07 g of S. Hence % composition of S = (32.07/80.07)*100 = 40.05%

There is no other element in the compound hence % composition of O = 100-40.05 = 59.95%

b) NaH2PO3:

Molar mass = 22.99 + (1.01*2) + 31.00 + (16.0*3) = 104.01 g/mol

In 104.01 g of sample we will have 22.99 g of Na. So, % composition of Na = (22.99/104.01)*100 = 22.10%

In 104.01 g of sample we will have 31.00 g of P. So, % composition of P = (31/104.01)*100 = 29.81%

In 104.01 g of sample we will have 2.02 g of H. So, % composition of H = (2.02/104.01)*100 = 1.94%  

Now only remaining element is oxygen hence % composition of O = 100-(22.1+29.81+1.94) = 46.15%  

please note that I have used litrature values of atomic masses in this calculations. And our refernces for this may be different and so the atomic mass values may differ by some decimal point. Eventually, it may lead to slightly different values of % composition values.