Question

# Calculate the percentage of water in ferrous sulphate crystals [Fe = 56, S = 32, O = 16, H = 1].

Open in App
Solution

## Formula of ferrous sulphate crystal is FeSO4.7H2O. Mass of water in one mole of the crystal = 7{2(1)+16} g = 7(18) g = 126 g Molecular mass of ferrous sulphate crystal = {56+32+16(4)+126} g = 278 g $\mathrm{Percentage}\mathrm{of}\mathrm{water}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{compound}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{FeSO}}_{4}.7{\mathrm{H}}_{2}\mathrm{O}}×100=\frac{126}{278}×100=45.3%$

Suggest Corrections
4
Explore more