Question

A compound on analysis gave the following percentage
composition: Na=14.31% S = 9.97%, H = 6.22%, O = 69.5%, calcualte the
molecular formula of the compound on the assumption that all the
hydrogen in the compound is present in combination with oxygen as water
of crystallisation. Molecular mass of the compound is 322 [Na = 23, S = 32,
H = 1, 0 = 16].

Answer 1

mole of Na = 14.31/23 = 0.622
mole of S = 9.97/32 = 0.311
mole of H = 6.22/1 = 6.22
mole of O = 69.5/16 = 4.34
relative number of atom of each atom :
for Na = 0.622/0.311 = 2
for S = 0.311/0.311 = 1
for H = 6.22/0.311 = 20
for O = 4.34/0.311 = 14
empirical formula = Na2SH20O14
empirical formula mass = 322
n*empirical formula mass = molecular formula mass
n*322 = 322
n = 1
it is given that all H is present as a H2O 
hence H20O10 = 10H2O
molecular formula = Na2SO4 .10H2O 

Answer 2

Na+S+O=14.31+9.97+69.5=93.78

rest part is H

100-93.78=6.22•/•

Na=14.31/23 = 0.622 = 0.622/0.311= 2

S =9.97/32 = 0.311 = 0.311/0.311= 1

O =69.5/16 =4.34 = 4.34/0.311= 14

H =6.22/1 = 6.22 = 6.22/0.311= 20

Empirical formula :-Na2SH20O14

Emphirical mass:-23*2 +32 +20+16*14=322

(Na2SH20O14) *n = 322

322*n=322

n=1

Molecular formula :-Na2SH20O14

The compound consist water of crystallization so H20O14 CAN BE SOLVED AS

Na2SO4. 10H2O