This mini-lesson walks you through the whole of Eduqas Topic 3: writing formulae from ions, building and balancing symbol equations, conservation of mass, the mole, and every quantitative calculation the exam can throw at you.
Work through each screen, answer the questions as you go (most are calculations) and collect ⭐ stars. Higher-tier-only ideas are flagged with a purple Higher badge. Press Start when you're ready.
A chemical formula shows the ratio of atoms in a substance. For an ionic compound the positive and negative charges must cancel out — so you balance the ion charges (valencies).
Common ions to know: Na⁺, K⁺, Ca²⁺, Mg²⁺, Al³⁺, H⁺ · Cl⁻, O²⁻, OH⁻, NO₃⁻, CO₃²⁻, SO₄²⁻. For a "swap-and-drop" shortcut: the charge of one ion becomes the subscript of the other (then cancel down).
A balanced symbol equation shows what reacts and what forms, with state symbols — (s) solid, (l) liquid, (g) gas, (aq) aqueous. You balance it by adding big numbers in front of formulae until every element has equal atoms on both sides.
Golden rule: you can only change the big numbers in front — never the small subscripts inside a formula. Changing a subscript would change the substance itself.
Magnesium burns in oxygen: Mg + O₂ → MgO.
O₂ brings 2 oxygens, so we need 2 MgO on the right. That needs 2 Mg on the left.
Final: 2Mg + O₂ → 2MgO — balanced.
Tap the equation that is correctly balanced in each pair.
In a chemical reaction atoms are only rearranged — none are created or destroyed. So the total mass of products = total mass of reactants. This is why a balanced equation must have the same atoms on each side.
Watch out — "the mass changed!": if a reaction looks like it gains or loses mass, a gas has escaped or joined in. Burning magnesium gains mass because O₂ from the air is added; a fizzing acid–carbonate reaction loses mass because CO₂ gas escapes. Seal the container and the mass stays the same.
The relative formula mass (Mₕ) of a compound is found by adding up the relative atomic masses (Aₕ) of every atom in its formula.
Watch out: Mₕ has no units — it is a relative mass compared to carbon-12, so it is just a number. Multiply a subscript (like the 3 in CO₃) through the Aₕ before you add.
Chemists count particles in moles. One mole is simply a fixed number of particles — Avogadro's constant:
Conveniently, the mass of one mole of a substance in grams is numerically equal to its Mₕ. So 1 mol of water (Mₕ = 18) weighs 18 g. That gives the key equation:
How many moles are in 36 g of water (H₂O, Mₕ = 18)?
moles = 36 ÷ 18 = 2 mol
The big numbers in a balanced equation give the mole ratio of the substances. Convert a known mass to moles, use the ratio, then convert back to a mass.
What mass of MgO forms when 48 g of Mg burns? (Aₕ Mg = 24, Mₕ MgO = 40)
moles Mg = 48 ÷ 24 = 2 mol → ratio 2 : 2 → 2 mol MgO
mass MgO = 2 × 40 = 80 g
On the Higher tier you can be given the masses that react and asked to deduce the balancing numbers. Convert each mass to moles, then write the moles as the simplest whole-number ratio — that is the equation ratio.
2 g of hydrogen reacts exactly with 16 g of oxygen to form water.
moles H₂ = 2 ÷ 2 = 1 mol · moles O₂ = 16 ÷ 32 = 0.5 mol
ratio H₂ : O₂ = 1 : 0.5 = 2 : 1 → 2H₂ + O₂ → 2H₂O
Tip: always simplify the mole ratio to the smallest whole numbers — multiply through to clear any decimals (here ×2).
When two reactants are mixed, one usually runs out first — the limiting reactant. It is completely used up, and it controls how much product can form. The other reactant is in excess (some is left over).
Watch out: the limiting reactant is the one that gives the fewest moles of product — not always the one with the smaller mass. Always work in moles, applying the equation ratio, to decide which runs out.
The concentration of a solution tells you how much solute is dissolved in each dm³ (1 dm³ = 1 litre = 1000 cm³). You can measure it two ways:
Watch out: g/dm³ and mol/dm³ are different units — convert between them with Mₕ: divide a g/dm³ value by Mₕ to get mol/dm³ (and ×Mₕ to go back). Also convert cm³ → dm³ by dividing by 1000.
20 g of NaOH is dissolved to make 0.5 dm³ of solution.
concentration = 20 ÷ 0.5 = 40 g/dm³ (= 40 ÷ 40 = 1 mol/dm³, since Mₕ NaOH = 40)
You rarely get every gram a reaction promises — some product is lost, reactions don't fully finish, or side-reactions occur. Percentage yield compares the actual amount you got with the theoretical maximum:
A reaction should make 8 g of product, but only 6 g is collected.
% yield = (6 ÷ 8) × 100 = 75%
Why never 100%? Product is lost on filtering/transfer, the reaction may be reversible or incomplete, and reactants may react in unexpected (side) reactions.
Atom economy measures how much of the reactant mass ends up as the useful product (rather than waste). A high atom economy means a more sustainable, less wasteful process.
CaCO₃ → CaO + CO₂. We want the CaO. (Mₕ: CaCO₃ = 100, CaO = 56, CO₂ = 44)
total product mass = 56 + 44 = 100
atom economy = (56 ÷ 100) × 100 = 56%
Don't confuse them: percentage yield is about how much you actually made; atom economy is about how much of the reactant mass becomes useful product in theory. A reaction can have 100% yield but poor atom economy.
The empirical formula is the simplest whole-number ratio of atoms in a compound. From masses or percentages: divide each element's mass by its Aₕ (to get moles), then divide by the smallest to get the ratio.
A compound is 80% carbon, 20% hydrogen by mass. (Aₕ: C = 12, H = 1)
C: 80 ÷ 12 = 6.67 · H: 20 ÷ 1 = 20
divide by smallest (6.67): C = 1, H = 3 → empirical formula CH₃
Empirical → molecular: the molecular formula is a whole-number multiple of the empirical one. If the empirical Mₕ is 15 (CH₃) and the true Mₕ is 30, the multiplier is 30 ÷ 15 = 2, so the molecular formula is C₂H₆.
A titration finds an unknown concentration by reacting an acid and alkali exactly to the end-point. From the known solution: find its moles, use the equation ratio to find the unknown's moles, then divide by its volume.
25 cm³ of NaOH is exactly neutralised by 20 cm³ of 0.10 mol/dm³ HCl. NaOH + HCl → NaCl + H₂O (ratio 1 : 1).
moles HCl = 0.10 × (20 ÷ 1000) = 0.002 mol → moles NaOH = 0.002 mol
conc NaOH = 0.002 ÷ (25 ÷ 1000) = 0.002 ÷ 0.025 = 0.08 mol/dm³
At room temperature and pressure (rtp), one mole of any gas occupies the same volume — the molar volume:
What volume does 0.5 mol of carbon dioxide occupy at rtp?
volume = 0.5 × 24 = 12 dm³
Handy reverse: moles = volume ÷ 24. So 48 dm³ of any gas at rtp is 48 ÷ 24 = 2 mol — whatever the gas, because equal volumes contain equal numbers of molecules.
Tap a quantity on the left, then its correct formula on the right.
Relative formula mass: Mₕ = sum of all Aₕ values (no units)
The mole: moles = mass ÷ Mₕ · 1 mol = 6.02 × 10²³ particles
Concentration: g/dm³ = mass ÷ volume · mol/dm³ = moles ÷ volume
Percentage yield (H): (actual ÷ theoretical) × 100
Atom economy (H): (Mₕ useful product ÷ total Mₕ) × 100
Gas volume (H): volume (dm³) = moles × 24 at rtp
You've covered all of Eduqas Topic 3 — formulae & balancing, conservation of mass, the mole, reacting masses, concentration, plus the Higher-tier (H) calculations: limiting reactants, yield, atom economy, empirical formulae, titrations and gas volumes. Press Finish to see your score.
You've worked through Chemical formulae, equations & amount of substance for Eduqas GCSE Chemistry. 🎉
Your stars: 0 / 0
Next: test yourself in the Evaluate stage Confidence Quiz, then lock it in with Verify.