Solution

Element mass % Atomic mass(g) m%/At. mass Integral ratio
Na 14.31 23 0.622 2
S 9.97 32 0.311 1
H 6.22 1 6.22 20
O 69.5 16 4.34 14
Empirical formula is therefore $$Na_2SH_{20}O_{14}$$
empirical mass=$$2\times 23+32+20 \times 1+ 14 \times 16=322$$g
Given:
Molecular mass = 322 g
$$\therefore$$ molecular formula =empirical formula=$$Na_2SH_{20}O_{14}$$
Given that all the hydrogen in the compound is reset in combination with oxygen as water of crystallisation
If water of crystallization are $$nH_2O$$ then, $$2n=20$$
or $$n=10$$
Crystallized water is $$10H_2O$$ remaining atoms are $$Na_2SO_{4}$$
therefore, the molecule is $$Na_2SO_4.10H_2O$$