An oxide of iron contains 69.9% iron, its empirical formula, is: (Given : Molar mass of Fe and O are 56 and 16 g mol-1 respectively.)
- FeO
- Fe2O3
- Fe3O4
- FeO3
Answer
(2) Fe2O3
Solution
Mass % of O = 100 − 69.9 = 30.1%
Taking 100 g of the oxide:
Moles of Fe = 69.956 = 1.248
Moles of O = 30.116 = 1.881
Simplest ratio: Fe : O = 1.2481.248 : 1.8811.248 = 1 : 1.507 ≈ 1 : 1.5
⇒ Fe : O = 2 : 3
∴ Empirical formula = Fe2O3
Answer: (2)