A hydrate of Na₂SO₃ has 50% water by mass. Formula of the compound is

Assume that the formula of Na₂SO₃ containing 50% of water by mass is Na₂SO₃.xH₂O. 

As the compound contains, 50% of Na₂SO₃ and 50% of water, thereby the ratio of weight of Na₂SO₃ and H₂O in the compound is 1:1.

$\frac{\mathrm{Weight} \mathrm{of} {\mathrm{Na}}_2{\mathrm{SO}}_3}{\mathrm{Weight} \mathrm{of} {\mathrm{H}}_2\mathrm{O}}=\frac{1}{1}$

Formula of the compound is determined from the mole ratio of Na₂SO₃ and water. Their mole ratio is written as:

$\begin{array}{ccc}\frac{n_{{\mathrm{Na}}_2{\mathrm{SO}}_3}}{n_{{\mathrm{H}}_2\mathrm{O}}}& =& \frac{{\displaystyle \frac{\mathrm{Weight} \mathrm{of} {\mathrm{Na}}_2{\mathrm{SO}}_3}{\mathrm{Molecular} \mathrm{weight} \mathrm{of} {\mathrm{Na}}_2{\mathrm{SO}}_3}}}{{\displaystyle \frac{\mathrm{Weight} \mathrm{of} {\mathrm{H}}_2\mathrm{O}}{\mathrm{Molecular} \mathrm{weight} \mathrm{of} {\mathrm{H}}_2\mathrm{O}}}}\\ & =& \frac{\mathrm{Weight} \mathrm{of} {\mathrm{Na}}_2{\mathrm{SO}}_3}{\mathrm{Weight} \mathrm{of} {\mathrm{H}}_2\mathrm{O}}\times \frac{\mathrm{Molecular} \mathrm{weight} \mathrm{of} {\mathrm{H}}_2\mathrm{O}}{\mathrm{Molecular} \mathrm{weight} \mathrm{of} {\mathrm{Na}}_2{\mathrm{SO}}_3} ....\left(1\right)\end{array}$

Molecular weight of Na₂SO₃ is calculated as:

$\begin{array}{ccc}\mathrm{Molecular} \mathrm{weight} \mathrm{of} {\mathrm{Na}}_2{\mathrm{SO}}_3& =& 2(\mathrm{atomic} \mathrm{weight} \mathrm{of} \mathrm{Na})+\mathrm{atomic} \mathrm{weight} \mathrm{of} \mathrm{S}+3(\mathrm{atomic} \mathrm{weight} \mathrm{of} \mathrm{O})\\ & =& 2\left(23\right)+32+3\left(16\right)\\ & =& 126\end{array}$

Molecular weight of H₂O is calculated as:

$\begin{array}{ccc}\mathrm{Molecular} \mathrm{weight} \mathrm{of} {\mathrm{H}}_2\mathrm{O}& =& 2(\mathrm{atomic} \mathrm{weight} \mathrm{of} \mathrm{H})+\mathrm{atomic} \mathrm{weight} \mathrm{of} \mathrm{O}\\ & =& 2\left(1\right)+16\\ & =& 18\end{array}$

Substitute the values in equation (1).

$\begin{array}{ccc}\frac{n_{{\mathrm{Na}}_2{\mathrm{SO}}_3}}{n_{{\mathrm{H}}_2\mathrm{O}}}& =& \frac{1}{1}\times \frac{18}{126}\\ & =& \frac{1}{7}\end{array}$

Above ratio states that one fraction of Na₂SO₃ has 7 fractions of H₂O. Therefore, the formula of compound becomes Na₂SO₃.7H₂O.