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Eduqas GCSE Chemistry · Topic 3 — Chemical formulae, equations and amount of substance
Mini-Lesson

Chemical formulae, equations & amount of substance

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.

mass (grams) moles (amount) ÷ Mₕ the mole bridges the two

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.

Chemical formulae

Building a formula from ions

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).

Ca²⁺ 2+ charge + Cl⁻ Cl⁻ CaCl₂
One Ca²⁺ needs two Cl⁻ to balance, giving the formula CaCl₂. The little 2 is a subscript on Cl.

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).

Quick check

Pick the right formula

?Aluminium forms Al³⁺ ions and oxygen forms O²⁻ ions. What is the formula of aluminium oxide?
Symbol equations

Writing & balancing equations

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.

2Mg(s) + O₂(g) → 2MgO(s)2 Mg + 2 O on the left = 2 Mg + 2 O on the right ✓

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.

How to balance, step by step

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.

Sort it

Is it balanced?

Tap the equation that is correctly balanced in each pair.

Conservation of mass

Atoms are never lost

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.

reactants 48 g products 48 g = the pans balance — total mass is unchanged
Mass is conserved: nothing is added or lost, the atoms are simply rearranged.

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.

Quick check

Where did the mass go?

?A lump of marble chips (calcium carbonate) is dropped into acid in an open flask on a balance. The reading falls. Why?
Relative formula mass

Relative formula mass, Mₕ

The relative formula mass (Mₕ) of a compound is found by adding up the relative atomic masses (Aₕ) of every atom in its formula.

Mₕ = sum of all the Aₕ valuesjust add up an Aₕ for every atom in the formula
Ca 40 + C 12 + 3 × O 3×16 = 48 = 100
CaCO₃: 40 + 12 + (3 × 16) = 100.

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.

Calculate

Your turn — relative formula mass

1Calculate the relative formula mass (Mₕ) of sulfuric acid, H₂SO₄. (Aₕ: H = 1, S = 32, O = 16)
Hint: (2 × 1) + 32 + (4 × 16). Remember Mₕ has no units.
The mole

The mole & Avogadro's constant

Chemists count particles in moles. One mole is simply a fixed number of particles — Avogadro's constant:

1 mole = 6.02 × 10²³ particlesAvogadro's constant — atoms, molecules, ions… any particle

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:

moles = mass ÷ Mₕamount (mol) = mass (g) ÷ relative formula mass
mass moles Mₕ ×
Cover the quantity you want: mass = moles × Mₕ; moles = mass ÷ Mₕ; Mₕ = mass ÷ moles.
Worked example

How many moles are in 36 g of water (H₂O, Mₕ = 18)?

moles = 36 ÷ 18 = 2 mol

Calculate

Your turn — moles from mass

2How many moles are there in 80 g of sodium hydroxide, NaOH? (Mₕ of NaOH = 40)
mol
Hint: moles = mass ÷ Mₕ = 80 ÷ 40.
Reacting masses

Reacting masses from equations

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.

2Mg + O₂ → 2MgO 2 mol Mg 48 g 2 mol MgO 80 g ratio 2 : 2
2 mol Mg (Aₕ 24 → 48 g) makes 2 mol MgO (Mₕ 40 → 80 g).
Worked example

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

Calculate

Your turn — reacting masses

3Hydrogen burns: 2H₂ + O₂ → 2H₂O. What mass of water forms from 4 g of hydrogen gas? (Mₕ: H₂ = 2, H₂O = 18)
g
Hint: moles H₂ = 4 ÷ 2 = 2. Ratio 2H₂ : 2H₂O is 2 : 2, so 2 mol H₂O. Mass = 2 × 18.
Higher only

Balancing from masses / moles

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.

Worked example

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).

Higher only

The limiting reactant

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).

limiting all used up in excess some left over product amount set by the limiting one

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.

Quick check

Which one limits?

?Higher 3 mol of hydrogen reacts with 2 mol of nitrogen: N₂ + 3H₂ → 2NH₃. Which is the limiting reactant?
Concentration of solutions

Concentration: g/dm³ and mol/dm³

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:

concentration (g/dm³) = mass (g) ÷ volume (dm³)and: concentration (mol/dm³) = moles ÷ volume (dm³)
1 dm³ of solution solute (green dots) more solute = higher conc.
Pack more solute into the same volume and the concentration rises.

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.

Worked example

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)

Calculate

Your turn — concentration in g/dm³

415 g of salt is dissolved to make 0.25 dm³ of solution. Calculate the concentration in g/dm³.
g/dm³
Hint: concentration = mass ÷ volume = 15 ÷ 0.25.
Calculate

Your turn — concentration in mol/dm³

50.5 mol of HCl is dissolved to make 0.25 dm³ of solution. Calculate the concentration in mol/dm³.
mol/dm³
Hint: concentration = moles ÷ volume = 0.5 ÷ 0.25.
Higher only

Percentage yield

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:

% yield = (actual ÷ theoretical) × 100actual mass of product ÷ maximum possible mass × 100
Worked example

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.

Calculate

Your turn — percentage yield

6Higher A reaction has a theoretical yield of 50 g but produces 40 g of product. Calculate the percentage yield.
%
Hint: (actual ÷ theoretical) × 100 = (40 ÷ 50) × 100.
Higher only

Atom economy

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.

% atom economy = (Mₕ of useful product ÷ total Mₕ of all products) × 100using the balanced equation's mole ratios
Worked example

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.

Calculate

Your turn — atom economy

7Higher In 2H₂ + O₂ → 2H₂O the only product is water, Mₕ 18 (×2 = 36). What is the atom economy for making water?
%
Hint: water is the only product, so useful product Mₕ ÷ total product Mₕ = 36 ÷ 36, then × 100.
Higher only

Empirical & molecular formulae from data

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.

Worked example

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.

Higher only

Titration calculations

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.

acid (known) in burette alkali (unknown) + indicator end-point: indicator changes colour
Add acid until the indicator just changes colour — that volume is the titre.
Worked example

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

Calculate

Your turn — titration

8Higher 25 cm³ of HCl is neutralised by 20 cm³ of 0.50 mol/dm³ NaOH (ratio 1 : 1). Calculate the concentration of the HCl in mol/dm³.
mol/dm³
Hint: moles NaOH = 0.50 × (20 ÷ 1000) = 0.01. That equals moles HCl. conc HCl = 0.01 ÷ (25 ÷ 1000).
Higher only

The molar volume of a gas

At room temperature and pressure (rtp), one mole of any gas occupies the same volume — the molar volume:

volume of gas (dm³) = moles × 241 mole of any gas occupies 24 dm³ at rtp
Worked example

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.

Calculate

Your turn — gas volume

9Higher Calculate the volume occupied by 2 mol of oxygen gas at room temperature and pressure (rtp).
dm³
Hint: volume = moles × 24 = 2 × 24.
Match-up

Match each tool to its job

Tap a quantity on the left, then its correct formula on the right.

Recap

The equations to know

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.

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