Question

Determine the percentage of water of crystallisation, iron, sulphur and oxygen in pure ferrous sulphate $(FeS{O}_4\cdot 7H_{2}O)$.

Answer 1

The formula mass of ferrous sulphate
$At.massofFe+At.,massofS+At.massofoxygen+7\times Mol.massof{H}_{2}O$
$56.0+32.0+4\times 16.0+7\times 18.0=278.0$
So % of water of crystallisation $=\frac{126}{278}\times 100=45.32$
% of iron $=\frac{56}{278}\times 100=20.14$
% of sulphur $=\frac{32}{278}\times 100=11.51$
% of oxygen $=\frac{64}{278}\times 100=23.02$
(Oxygen present in water molecules is not taken into account.)