Question

30. A compound on analysis gave the following % composition:
Na-43.4%, C=11.4% and 0-45.3%.
Determine the empirical and molecular formulae. ( molecular mass of the compound = 106)​

Answer 1

Answer:

Dividing their percentages by their atomic weights.

Sodium=2343.4=1.88

Carbon=1211.3=0.941

Oxygen=1645.3=2.83

Dividing all these numbers with simple numbers

Sodium=0.941.88=2

Carbon=0.940.94=1

Oxygen=0.942.83=3

Their ratio 2:1:3 

Empirical formula = Na2CO3