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

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

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.

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.

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

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.

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

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.

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