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Question
Solve the following problem:
Find the percentage composition of constituent green vitriol crystals (FeSO4.7H2O). Also, find out the mass of iron and the water of crystallisation in 4.54 kg of the crystals. (At. mass : Fe = 56; S = 32; O = 16)
Solution
Given: i. Atomic mass: Fe = 56; S = 32; O = 16
ii. Mass of crystal = 4.54 kg
To find: i. Mass percentage of Fe, S, H, and O
ii. Mass of iron and water of crystallisation in 4.54 kg of crystal
Formula: Percentage (by weight) = `"Mass of the element in 1 mole of compound"/"Molar mass of the compound"xx100`
i. Molar mass of FeSO4.7H2O = 1 × (56) + 1 × (32) + 14 × (1) + 11 × (16)
= 56 + 32 + 14 + 176
= 278 g mol−1
Percentage of Fe = `56/278xx100` = 20.14%
Percentage of S = `32/278xx100` = 11.51%
Percentage of H = `14/278xx100` = 5.04%
Percentage of O = `176/278xx100` = 63.31%
ii. 278 kg green vitriol = 56 kg iron
∴ 4.54 kg green vitriol = x
∴ x = `(56xx4.54)/278` = 0.915 kg
Mass of 7H2O in 278 kg green vitriol = 7 × 18 = 126 kg
∴ 4.54 kg green vitriol = y
∴ y = `(126xx4.54)/278` = 2.058 kg
- Mass percentage of Fe, S, H and O in FeSO4.7H2O are 20.14%, 11.51%, 5.04% and 63.31% respectively.
- Mass of iron in 4.54 kg green vitriol = 0.915 kg
Mass of water of crystallisation in 4.54 kg green vitriol = 2.058 kg
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