Question

A substance, on analysis, gave the following percent composition : 

Na = 43.4 %, C = 11.3 % and O = 45.3 %. 

Calculate the empirical formula. 

(At. mass Na = 23 u, C = 12 u, O = 16 u).

Answer 1

Given : 

Atomic mass of Na = 23 u, 

C = 12 u, and 

O = 16 u 

Percentage of Na, C and O = 43.4%, 11.3% and 45.3% respectively. 

To find :

The empirical formula of the compound 

Calculation :

Hence,

The ratio of number of moles of Na:C:O is

Hence, 

Empirical formula is Na₂CO₃.

∴ Empirical formula of the compound = Na₂CO₃