Question 3 of 36intermediate🔧 ApplyNumerical3 marks

Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Correct Answer

The empirical formula of the oxide of iron is Fe2O3\text{Fe}_2\text{O}_3.

Exercise: EXERCISES | Q: 1.3 | (Page 25)
For More Understanding

Explanation

To determine the empirical formula, we first convert the given mass percentages into grams by assuming a 100 g sample of the compound. Next, we convert these masses into the number of moles by dividing them by their respective atomic masses. Finally, we determine the simplest whole number ratio of the moles of each element to write the empirical formula.

Solution Steps

  1. Step 1: Conversion of mass per cent to grams Assume a 100 g sample of the iron oxide. This sample will contain 69.9 g of iron and 30.1 g of dioxygen.

  2. Step 2: Convert into number of moles of each element Divide the masses by their respective atomic masses (Fe = 55.85 g/mol, O = 16.00 g/mol). Moles of iron = 69.9 g55.85 g mol−1=1.25\frac{69.9 \text{ g}}{55.85 \text{ g mol}^{-1}} = 1.25 mol Moles of oxygen = 30.1 g16.00 g mol−1=1.88\frac{30.1 \text{ g}}{16.00 \text{ g mol}^{-1}} = 1.88 mol

  3. Step 3: Find the simplest whole number ratio Divide the number of moles by the smallest value (1.25): Ratio of Fe = 1.251.25=1\frac{1.25}{1.25} = 1 Ratio of O = 1.881.25=1.5\frac{1.88}{1.25} = 1.5 Since 1.5 is not a whole number, multiply both ratios by 2 to get whole numbers: Fe = 1×2=21 \times 2 = 2 O = 1.5×2=31.5 \times 2 = 3

  4. Step 4: Write the empirical formula The simplest whole number ratio of iron to oxygen is 2:3. Thus, the empirical formula of the oxide is Fe2O3\text{Fe}_2\text{O}_3.

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