1. IC5.4.4 Empirical formulae from experimental data © Oxford University Press Empirical formulae from experimental data

2. IC5.4.4 Empirical formulae from experimental data © Oxford University Press The empirical formula of a compound can be calculated from experimental data. For example, magnesium is oxidised to magnesium oxide when it is heated in air.

3. IC5.4.4 Empirical formulae from experimental data © Oxford University Press Mass (g) Aempty crucible23.70 Bcrucible + magnesium24.42 Ccrucible + magnesium oxide24.90 Here are some example results. From these results: mass of magnesium is (24.42 – 23.70) = 0.72 g mass of oxygen is (24.90 – 24.42) = 0.48 g The empirical formula can be worked out using the following steps.

4. IC5.4.4 Empirical formulae from experimental data © Oxford University Press Write each symbolMgO Write each mass in g0.720.48 Write each A r 0.4816 Find the number of moles0.72 ÷ 24 = 0.030.48 ÷ 16 = 0.03 Divide by the smallest number 0.03 ÷ 0.03 = 1 Check for whole numbers, write the formula Mg0

5. IC5.4.4 Empirical formulae from experimental data © Oxford University Press Use these A r values to help you answer the following question: O = 16, S = 32 1.An oxide of sulfur contains 50% sulfur and 50% oxygen by mass. Work out its empirical formula.

6. IC5.4.4 Empirical formulae from experimental data © Oxford University Press 1. SymbolSO 2. Mass in gSO 2 0.50 3. A r 3216 4. Number of moles0.50 ÷ 32 = 0.0150.50 ÷ 16 = 0.030 5. Divide by smallest number 0.015 ÷ 0.015 = 10.030 ÷ 0.015 = 2 6. Check for whole numbers, write the formula SO 2 For 100 g of the compound:

7. IC5.4.4 Empirical formulae from experimental data © Oxford University Press A r values: H = 1, C = 12, O = 16 2.A 2.3 g sample of compound X contains 1.2 g carbon, 0.3 g hydrogen, and 0.8 g oxygen. Work out its empirical formula.

8. IC5.4.4 Empirical formulae from experimental data © Oxford University Press 1. SymbolCHO 2. Mass in g1.20.30.8 3. A r 12116 4. Number of moles 1.2 ÷ 12 = 0.1 0.8 ÷ 16 = 0.05 5. Divide by smallest number 0.1 ÷ 0.05 = 20.3 ÷ 0.05 = 60.05 ÷ 0.05 = 1 6. Check for whole numbers, write the formula C2H6OC2H6O For 100 g of the compound: