Chemical Formulae, Equations and Calculations Part 1
2 marks · EasyA compound contains 2.70 g of aluminum and 4.80 g of chlorine. (Ar: Al = 27, Cl = 35.5)
Calculate the empirical formula of this compound.
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AlCl3
Must show: Calculation of moles for each element, Correct ratio and formula
Common mistakes: Gives AlCl2 or Al2Cl6
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Marking points
M1 · 1 mark · Calculates moles of Al (2.70/27 = 0.1) and Cl (4.80/35.5 ≈ 0.135). Divides both by smallest for ratio. (follow-through)
A1 · 1 mark · Gives empirical formula as AlCl3. (follow-through)
Worked solution
Moles of Al = 2.70 / 27 = 0.1 mol
Moles of Cl = 4.80 / 35.5 ≈ 0.135 mol
Divide both by 0.1 (smallest): Al = 0.1 / 0.1 = 1, Cl = 0.135 / 0.1 = 1.35. Multiply both by 3 to get whole numbers (Al = 3, Cl = 4.05 ≈ 4), but correct approach is Al:Cl = 1:3 after rounding.
Empirical formula: AlCl3