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AQA GCSE Chemistry

Moles, masses and conservation — AQA GCSE Chemistry revision

Free revision notes, key terms, common exam traps and 5 practice questions with answers. About 4 minutes to read.

Core equations

  • Moles = mass ÷ relative formula mass (Mr).
  • Relative formula mass = add up the relative atomic masses in the formula.
  • Mass is conserved: total mass of reactants = total mass of products in a closed system.

Yield and concentration

  • Percentage yield = (actual yield ÷ theoretical yield) × 100.
  • Concentration (g/dm³) = mass ÷ volume in dm³. 1000 cm³ = 1 dm³.
  • A reaction may look like mass changed if a gas escapes or is taken in from the air.

Key terms

Mole
An amount of substance containing 6.02 × 10²³ particles.
Limiting reactant
The reactant used up first, which stops the reaction.
Percentage yield
How much product you actually got compared with the maximum possible.
Molar mass
The mass of one mole of a substance in grams, numerically equal to its relative formula mass.
Avogadro's constant
6.02 × 10²³ — the number of particles in one mole of a substance.
Concentration
The amount of solute dissolved in a given volume of solution, in g/dm³ or mol/dm³.

Common exam traps

  • Forgetting to convert cm³ to dm³ (divide by 1000).
  • Using mass instead of moles when balancing ratios.
  • Rounding too early in multi-step calculations.

Practice questions with answers

  1. 1. Calculate the number of moles in 88 g of CO₂ (Mr = 44).

    Answer: 2 mol

    moles = mass ÷ Mr = 88 ÷ 44.

  2. 2. A sealed flask reaction shows no mass change. Why?

    • • Atoms are destroyed
    • • Atoms are rearranged, not created or destroyed
    • • Gas escaped
    • • Energy has mass

    Answer: Atoms are rearranged, not created or destroyed

    Conservation of mass in a closed system.

  3. 3. 25 g of product was made when 40 g was possible. Calculate percentage yield.

    Answer: 62.5%

    (25 ÷ 40) × 100.

  4. 4. Convert 250 cm³ to dm³.

    Answer: 0.25 dm³

    Divide by 1000.

  5. 5. Calculate the relative formula mass of CaCO3 (Ca=40, C=12, O=16).

    Answer: 100

    40 + 12 + (16 × 3) = 40 + 12 + 48 = 100.

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