Question

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.

Answer 1

1. Convert mass percentages to moles

Assume 100 g of the compound:

• Mass of Fe = 69.9 g
• Mass of O = 30.1 g

Moles of Fe: \[ n_{\text{Fe}} = \frac{69.9}{55.85} \approx 1.25 \] Moles of O: \[ n_{\text{O}} = \frac{30.1}{16.00} \approx 1.88 \]

2. Find simplest whole-number ratio

Divide each by the smaller value (≈ 1.25):

• Fe: \(\dfrac{1.25}{1.25} \approx 1\)
• O: \(\dfrac{1.88}{1.25} \approx 1.5\)

So the ratio Fe : O ≈ 1 : 1.5. Multiply both by 2 to clear the fraction:

• Fe : O = 2 : 3

3. Empirical / molecular formula

The simplest whole-number ratio is Fe : O = 2 : 3, so the formula of the oxide is: \[ \boxed{\ce{Fe2O3}} \]