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AS · Practice questions · The mole and the Avogadro constant

Count it by mass.

Six original Cambridge-style questions on the mole and the Avogadro constant: conversions in both directions, counting atoms and ions, and working back to an unknown element.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
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Go through the mole every time.

01
Recall
[1 marks]

Define the mole.

  • The amount of substance that contains 6.02 × 1023 particles (the Avogadro constant), the same number as there are atoms in exactly 12 g of carbon-12. ✓
02
Calculation
[3 marks]

For 4.40 g of carbon dioxide, CO2, calculate (a) the amount in moles, (b) the number of molecules and (c) the total number of atoms. (Ar: C 12.0, O 16.0)

  • (a) Mr = 44.0, so n = 4.40 ÷ 44.0 = 0.100 mol. ✓
  • (b) 0.100 × 6.02 × 1023 = 6.02 × 1022 molecules. ✓
  • (c) Three atoms per molecule: 3 × 6.02 × 1022 = 1.81 × 1023 atoms. ✓
03
Calculation
[2 marks]

Calculate the mass of 3.01 × 1022 atoms of magnesium. (Ar: Mg 24.3)

  • n = 3.01 × 1022 ÷ 6.02 × 1023 = 0.0500 mol. ✓
  • Mass = 0.0500 × 24.3 = 1.22 g. ✓
04
Analysis
[3 marks]

0.0200 mol of an oxide with the formula X2O3 has a mass of 3.20 g. Calculate the relative formula mass of the oxide and identify element X. (Ar: O 16.0)

  • M = m ÷ n = 3.20 ÷ 0.0200 = 160 g mol−1, so the relative formula mass is 160. ✓
  • 2Ar(X) + 3 × 16.0 = 160, so Ar(X) = (160 − 48.0) ÷ 2 = 56.0. ✓
  • X is iron (Ar 55.8); the oxide is Fe2O3. ✓
05
Calculation
[3 marks]

Calculate the total number of ions in 2.00 g of calcium chloride, CaCl2. (Ar: Cl 35.5, Ca 40.1)

  • Mr = 40.1 + 2 × 35.5 = 111.1, so n = 2.00 ÷ 111.1 = 0.0180 mol. ✓
  • Each formula unit gives one Ca2+ and two Cl−: 3 × 0.0180 = 0.0540 mol of ions. ✓
  • 0.0540 × 6.02 × 1023 = 3.25 × 1022 ions. ✓
06
Calculation
[2 marks]

Calculate the mass, in grams, of a single molecule of water. (Ar: H 1.0, O 16.0)

  • One mole of water, 6.02 × 1023 molecules, has a mass of 18.0 g. ✓
  • Mass of one molecule = 18.0 ÷ 6.02 × 1023 = 2.99 × 10−23 g. ✓

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