**Solve the following problem:**

Find the percentage composition of constituent green vitriol crystals (FeSO_{4}.7H_{2}O). 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 FeSO_{4}.7H_{2}O = 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 7H_{2}O 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 FeSO
_{4}.7H_{2}O 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**