Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10 ppm of iron in 100 kg of wheat is _______. (Round off to the nearest integer)
Atomic weight : Fe = 55.85; S = 32.00; O = 16.00

Solution

Chemistry ma’am · 3 steps

Find the grams of iron first, then scale up by the molar mass of the whole hydrated salt.

1. Iron needed. 10 ppm by mass means 10 g of iron in every 10⁶ g of wheat. The wheat is 100 kg = 10⁵ g, so iron = (10 × 10⁵)/10⁶ = 1 g.
2. Molar mass of salt. The salt is FeSO₄.7H₂O. Taking H as 1, its molar mass is 55.85 + 32 + 64 + 7 × 18 = 277.85 g mol⁻¹. Each mole of it holds one mole of Fe, that is 55.85 g of iron.

   mₛₐₗₜ = 1 g × 277.85/55.85 = 4.975 g

3. So. About 4.98 g of ferrous sulphate heptahydrate is needed, which rounds to 5.

The common slip: 3. Leaving out the 7 waters and using FeSO₄ (151.85) gives 2.72 g, which rounds to 3, not 5.

The idea to keep. Turn ppm into grams of the element first, then scale by the compound's molar mass over the element's mass per mole. Water of crystallisation counts in what you weigh.

Answer from the NTA final answer key. Solution by PrepWiser.