GCSE · GCSE Chemistry
GCSE Chemistry mole calculations
What is mole calculations?
Mole calculations convert between mass, moles and particles in chemistry. The three core equations: n = m / Mr (moles from mass + relative formula mass), n = C × V (concentration × volume in dm³ for solutions), and n = V / 24 (gases at room temperature and pressure). GCSE Higher tier asks for mole ratios from balanced equations — the equation tells you how many moles of each reactant produce how many moles of each product.
Worked example
What mass of magnesium oxide is produced when 4.8 g of magnesium burns in excess oxygen? (Ar Mg = 24, O = 16; balanced: 2Mg + O₂ → 2MgO)
- Find moles of Mg: n = m/Mr = 4.8 / 24 = 0.20 mol.
- Use the mole ratio: 2 Mg → 2 MgO, so 0.20 mol Mg → 0.20 mol MgO.
- Find Mr of MgO: 24 + 16 = 40.
- Find mass of MgO: m = n × Mr = 0.20 × 40 = 8.0 g.
The three core mole equations
Memorise these for GCSE Higher Chemistry:
- n = m / Mr (moles from mass; m in grams; Mr is relative formula mass from the periodic table).
- n = C × V (moles from concentration × volume; C in mol/dm³; V in dm³ — convert from cm³ by ÷ 1000).
- n = V / 24 (moles of gas at room temperature 25°C and pressure 1 atm; V in dm³).
Balancing equations first
Mole ratios come from a correctly-balanced equation. If the equation isn't balanced, your mole ratios are wrong and the whole calculation fails. Always check the equation balances atoms on both sides before applying ratios.
Limiting reactant
When the question gives you amounts of TWO reactants, one will run out first (the limiting reactant). Calculate moles of each, then use the smaller mole quantity divided by the equation coefficient to identify which limits. Higher-tier GCSE Maths-style algebra.
Where it appears in the GCSE exam
AQA 8462, Edexcel 1CH0 and OCR J248 examine mole calculations on Higher tier across atomic structure, quantitative chemistry, electrolysis, energy changes and rates of reaction. Around 20-25% of Higher-tier marks are calculation-based.
Common mistakes
- Forgetting to balance the equation before applying mole ratios.
- Mixing up grams and kg, or dm³ and cm³ (divide by 1000).
- Rounding intermediate values — keep extra decimal places until the final answer.
- Confusing Ar (atomic mass) and Mr (formula mass) — for compounds, Mr is the sum.
- Using the molar volume of 24 dm³/mol outside room temperature (it changes with temperature/pressure).
Frequently asked
- What's the difference between Ar and Mr?
- Ar is the relative atomic mass of one element (e.g. Ar Mg = 24). Mr is the relative formula mass of a compound — sum of the Ar values of all atoms in the formula (e.g. Mr H₂O = 2×1 + 16 = 18).
- Why does 1 mole of gas have a volume of 24 dm³?
- At room temperature (25°C) and atmospheric pressure, one mole of any ideal gas occupies 24.0 dm³ (more precisely 24.46 at 25°C). The figure 24 is the value used on UK GCSE papers. At 0°C and 1 atm, the volume is 22.4 dm³.
- Do I need to memorise the periodic table?
- No — a periodic table is provided in every GCSE Chemistry exam. You do need to be confident reading atomic numbers, mass numbers, group/period information and the relative atomic masses for the most common elements.
GCSE Chemistry glossary terms
- Mole calculationsMole calculations convert between mass, moles and particles in chemistry. Core equations: n = m / Mr (moles from mass + relative formula mass), n = C × V (for solutions, with C in mol/dm³ and V in dm³), n = V / 24 (for gases at room temperature and pressure). Higher-tier GCSE Chemistry asks for mole ratios from balanced equations — read the equation as 'per mole of A, you need x moles of B'.
- Avogadro's numberAvogadro's number is the number of particles in one mole of any substance: 6.022 × 10²³ mol⁻¹ (named after the chemist Amedeo Avogadro). One mole of any element or compound contains exactly this many atoms, molecules or formula units. GCSE Chemistry uses it for mole-particle calculations; A-Level extends to lattice energy and molar enthalpy calculations.
- Percentage yieldPercentage yield compares the actual mass of product obtained from a reaction with the maximum theoretical mass calculated from stoichiometry: (actual mass / theoretical mass) × 100%. Real-world yields are usually below 100% due to incomplete reactions, side products, or losses during purification. A high percentage yield is a key indicator of an efficient industrial process.
- Atom economyAtom economy measures the proportion of reactant atoms that end up in the desired product: (Mr of desired product / sum of Mr of all products) × 100%. Unlike percentage yield (which accounts for reaction efficiency), atom economy is a theoretical maximum based on stoichiometry. High atom-economy reactions are greener — fewer waste atoms. Addition reactions typically have 100% atom economy; substitution reactions produce by-products and have lower atom economy.
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