AQA · GCSE · Combined Science

    Course version: 8464 · Version from Sept 2018

    The masses of reactants and products can be calculated from balanced…

    Full specification wording

    Amounts of substances in equations (HT only): The masses of reactants and products can be calculated from balanced symbol equations.

    A balanced symbol equation fixes the ratio in which particles react, so it also fixes the ratio of amounts in moles.

    Checked against the AQA 8464 specification (opens in a new tab) on .

    • Point 1 of 6
    • 4 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    A balanced symbol equation fixes the ratio in which particles react, so it also fixes the ratio of amounts in moles. To calculate masses, first balance the equation, then read the coefficients as the number of moles of each substance. Convert any given mass to moles using n = m ÷ Mᵣ, apply the mole ratio from the coefficients, and convert back to mass using m = n × Mᵣ. For example, in 2Mg + O₂ → 2MgO, burning 4.8 g of Mg (Mᵣ = 24) gives n(Mg) = 4.8 ÷ 24 = 0.20 mol; the ratio Mg : MgO is 2 : 2, so n(MgO) = 0.20 mol and m(MgO) = 0.20 × 40 = 8.0 g. The same method works for any reactant or product once the equation is balanced.

    What earns the marks

    • A correct balanced symbol equation is required before any mass calculation; coefficients give the reacting mole ratio.
    • Converting a given mass to moles uses n = m ÷ Mᵣ, with Mᵣ values taken from the periodic table and summed correctly for compounds.
    • The mole ratio from the balanced equation is applied to the substance whose mass is required, not assumed to be 1 : 1.
    • The final answer is converted from moles back to mass using m = n × Mᵣ and given with a suitable unit such as g or kg.
    • Where a reactant is in excess, the limiting reactant determines the amount of product formed.
    • Answers should be checked for sensible magnitude and, where appropriate, given to a reasonable number of significant figures.

    Common mistakes and how to put them right

    • Using an unbalanced equation.

      Put it right: Balance the equation first and use the final coefficients.

    • Assuming the mole ratio is always 1.

      Put it right: Read the coefficients directly from the balanced equation.

    • Multiplying the given mass by Mᵣ instead of dividing.

      Put it right: Use n = m ÷ Mᵣ to find moles, then m = n × Mᵣ to return to mass.

    • Forgetting to multiply by the coefficient when finding the mass of a product.

      Put it right: Include the coefficient in the mole ratio before converting to mass.

    Exam tips

    • Write the balanced equation and label the relevant mole ratio before doing any arithmetic; this makes method marks easier to award.
    • Show each step: mass → moles → mole ratio → moles → mass, with units at every stage.
    • Check that the relative formula mass you use matches the formula in the equation, but remember that coefficients do not affect the Mᵣ calculation to avoid double-counting.
    • If the question gives two reactant masses, identify the limiting reactant before calculating product mass.

    Try the lesson

    The start of the Amounts of substances in equations (HT only): The masses of reactants and products can be calculated from balanced symbol equations. lesson, made from checked teaching notes for this specification point.

    Have a go first

    Calcium carbonate decomposes as shown: CaCO₃ → CaO + CO₂. What mass of carbon dioxide is produced when 25 g of calcium carbonate decomposes completely? (Mᵣ: CaCO₃ = 100, CO₂ = 44)

    Listen

    A balanced symbol equation fixes the ratio in which particles react, so it also fixes the ratio of amounts in moles. To calculate masses, first balance the equation, then read the coefficients as the number of moles of each substance. Convert any given mass to moles using n = m ÷ Mᵣ, apply the mole ratio from the coefficients, and convert back to mass using m = n × Mᵣ. For example, in 2Mg + O₂ → 2MgO, burning 4.8 g of Mg (Mᵣ = 24) gives n(Mg) = 4.8 ÷ 24 = 0.20 mol; the ratio Mg : MgO is 2 : 2, so n(MgO) = 0.20 mol and m(MgO) = 0.20 × 40 = 8.0 g. The same method works for any reactant or product once the equation is balanced. A strong answer earns marks for these points. A correct balanced symbol equation is required before any mass calculation; coefficients give the reacting mole ratio. Converting a given mass to moles uses n = m ÷ Mᵣ, with Mᵣ values taken from the periodic table and summed correctly for compounds. The mole ratio from the balanced equation is applied to the substance whose mass is required, not assumed to be 1 : 1. The final answer is converted from moles back to mass using m = n × Mᵣ and given with a suitable unit such as g or kg. A common slip: Using an unbalanced equation: this gives the wrong mole ratio. Instead, balance the equation first and use the final coefficients. Exam tip: Write the balanced equation and label the relevant mole ratio before doing any arithmetic; this makes method marks easier to award.

    See it

    A flowchart running from top to bottom with six steps. First, balance the symbol equation. Second, read the coefficients as the mole ratio. Third, convert the given mass to moles using n = m ÷ Mᵣ. Fourth, apply the mole ratio from the coefficients. Fifth, convert moles back to mass using m = n × Mᵣ. Sixth, give the answer with a unit such as g or kg.
    It shows the sequence of steps for calculating the mass of a reactant or product from a balanced symbol equation.

    Put it in order

    Put these steps in the order they happen.

    1. 1. Convert given mass to moles: n = m ÷ Mᵣ
    2. 2. Read coefficients as the mole ratio
    3. 3. Balance the symbol equation
    4. 4. Give the answer with a unit such as g or kg
    5. 5. Convert moles back to mass: m = n × Mᵣ
    6. 6. Apply the mole ratio from the coefficients

    Key words

    Say what each word means before you turn the card.

    Balanced symbol equation

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    Mole ratio

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    Limiting reactant

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    Relative formula mass

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    See every definition
    Balanced symbol equation
    A correct equation that must be written before any mass calculation, whose coefficients give the reacting mole ratio.
    Mole ratio
    The ratio of amounts in moles, read from the coefficients of a balanced equation and applied to the substance whose mass is required.
    Limiting reactant
    Where a reactant is in excess, this reactant determines the amount of product formed.
    Relative formula mass
    A value taken from the periodic table and summed correctly for compounds, used when converting a given mass to moles using n = m ÷ Mᵣ.

    Label the diagram

    Some boxes have lost their names. Pick the right name for each number.

    The diagram again, with 3 boxes numbered instead of named. A flowchart running from top to bottom with six steps. First, balance the symbol equation. Second, read the coefficients as the mole ratio. Third, convert the given mass to moles using n = m ÷ Mᵣ. Fourth, apply the mole ratio from the coefficients. Fifth, convert moles back to mass using m = n × Mᵣ. Sixth, give the answer with a unit such as g or kg.
    • 1Box 1

    • 2Box 2

    • 3Box 3

    0 of 3 labelled

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