Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively.

Let the molecular formula of the iron oxide be FexOy.

Mass percentage of Fe = 69.9%
Mass percentage of O = 30.1%

Assuming 100g of the oxide, we have:
Mass of Fe = 69.9 g
Mass of O = 30.1 g

Number of moles of Fe = mass / atomic mass = 69.9 g / 55.85 g/mol ≈ 1.25 mol
Number of moles of O = mass / atomic mass = 30.1 g / 16 g/mol ≈ 1.88 mol

Now, we find the simplest mole ratio by dividing both values by the smaller value (1.25 mol):
Fe: 1.25 mol / 1.25 mol = 1
O: 1.88 mol / 1.25 mol ≈ 1.5

Since we need whole numbers, we multiply both ratios by 2:
Fe: 1 * 2 = 2
O: 1.5 * 2 = 3

Therefore, the empirical formula of the iron oxide is Fe2O3.