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

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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
$Percentage of water = \frac{Mass of water in compound}{Molecular mass of {FeSO}_{4} . 7 H_{2} O} \times 100 = \frac{126}{278} \times 100 = 45.3 %$

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