Amounts of Substances in Equations and Using Moles to Balance Equations [HT only]

⬢ Higher Tier Only

SPEC 4.3.2.2–3 • QUANTITATIVE CHEMISTRY • PAPER 1

Mole Ratios in Equations

The core of quantitative chemistry. Given a balanced equation and the mass of one substance, calculate the mass of any other. Three-step method works every time. This lesson is the most-tested calculation type on Higher Tier Paper 1.

The Mole Ratio Rule

The coefficients in a balanced equation aren't just for balancing — they tell you the mole ratio in which substances react and are produced.

Mole Ratio

The ratio of moles of reactants and products in a balanced equation, given by the coefficients (the big numbers in front of formulae).

Mg + 2HCl → MgCl₂ + H₂

Reading the Equation in Moles

The spec's example: 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of magnesium chloride and 1 mole of hydrogen gas.

Mole ratio Mg : HCl : MgCl₂ : H₂ = 1 : 2 : 1 : 1

⚠️ Critical: the ratio applies to MOLES, not MASSES. 1 mole of Mg has mass 24 g, but 2 moles of HCl have mass 2 × 36.5 = 73 g — totally different gram amounts. The moles are in 1:2 ratio. Always convert mass to moles before applying the ratio.

The Three-Step Method

Every mass-to-mass calculation in Module 3 follows the same flow. Memorise these three steps.

Why this works

You can't compare grams of different substances directly because each has a different M_r. Moles let you compare amounts because the equation tells you the mole ratio. The first and last steps are just unit conversions; the middle step is where the chemistry happens.

Worked Example 1 — Magnesium + Oxygen

What mass of magnesium oxide is produced when 6.0 g of magnesium burns completely in oxygen? [A_r: Mg = 24, O = 16]

2Mg + O₂ → 2MgO

1

Moles of known (Mg).

moles of Mg = mass ÷ A_r = 6.0 ÷ 24 = 0.25 mol

2

Apply the mole ratio. From the equation: 2Mg : 2MgO simplifies to 1 : 1.

moles of MgO = 0.25 mol

3

Mass of unknown (MgO). M_r(MgO) = 24 + 16 = 40

mass of MgO = moles × M_r = 0.25 × 40 = 10 g

mass of MgO = 10 g

Reading the Mole Ratio

Most exam equations don't give you a 1:1 ratio. You must read the coefficients carefully.

Balanced EquationSubstances of InterestMole Ratio

2Mg + O₂ → 2MgOMg : MgO2 : 2 = 1 : 1

2Mg + O₂ → 2MgOMg : O₂2 : 1

Mg + 2HCl → MgCl₂ + H₂HCl : H₂2 : 1

2Fe₂O₃ + 3C → 4Fe + 3CO₂Fe₂O₃ : Fe2 : 4 = 1 : 2

2Fe₂O₃ + 3C → 4Fe + 3CO₂Fe₂O₃ : CO₂2 : 3

Method

1. Find the coefficient of substance A. 2. Find the coefficient of substance B. 3. Write the ratio A : B. 4. Simplify if both share a common factor (2:2 → 1:1; 2:4 → 1:2).

If a substance has no visible coefficient, it's 1. So Mg + 2HCl... → 1 mole Mg.

Worked Example 2 — Iron(III) Oxide + Carbon

Calculate the mass of CO₂ produced when 16.0 g of Fe₂O₃ reacts completely with excess carbon. [M_r(Fe₂O₃) = 160; M_r(CO₂) = 44]

2Fe₂O₃ + 3C → 4Fe + 3CO₂

1

Moles of known (Fe₂O₃).

moles of Fe₂O₃ = 16.0 ÷ 160 = 0.10 mol

2

Apply the mole ratio. From the equation: Fe₂O₃ : CO₂ = 2 : 3.

moles of CO₂ = 0.10 × (3 ÷ 2) = 0.15 mol

3

Mass of unknown (CO₂).

mass of CO₂ = 0.15 × 44 = 6.6 g

mass of CO₂ = 6.6 g

The ratio multiplier

For ratio A : B = 2 : 3, multiply moles of A by (3 ÷ 2) to get moles of B. The ratio is read as a fraction: "moles of B per mole of A". This works in either direction.

Examiner Alert — Use the Question's Values

Two of the most common Higher Tier errors come from ignoring values given in the question and computing them yourself (often wrongly).

⚠️ EXAMINER ALERT (P1 Jun24 Q07.6 — Fe₂O₃ question): "Despite being given the M_r of Fe₂O₃ as 160, many students used 320 instead." Doubling Fe₂O₃ in the equation does NOT mean doubling its M_r. The equation coefficient and the M_r are separate things.

⚠️ EXAMINER ALERT (P1 Jun22 Q07.4 — Mg + SiO₂): students used A_r(Mg) = 48 instead of 24. Wrong A_r doubles every subsequent step. Magnesium has A_r = 24 — read it from the periodic table on your exam paper.

Two Rules

If a value is given in the question, USE IT. Even if your own calculation gives a different number — the mark scheme is built around the given value.

If a value isn't given, read A_r from the periodic table printed on your exam paper. Don't rely on memory under exam pressure.

Examiner Alert — Don't Guess Mole Ratios

The mole ratio comes from the balanced equation. There is no shortcut, no intuition, no guessing.

⚠️ EXAMINER ALERT (P1 Jun23 Q09.2): the examiner report stated that incorrect answers "suggest a large degree of guessing rather than using the equation to work out the mole ratio." Students wrote 1:1 because it looked simple, or 2:1 without checking the coefficients.

The Discipline

Write out the balanced equation FIRST.

Underline or circle the coefficients.

Pick out the two substances you care about.

Write the ratio explicitly: e.g. "Fe₂O₃ : Fe = 2 : 4 = 1 : 2".

THEN apply it.

Showing the ratio is a mark

In a 4-mark mass-to-mass calculation, ONE of the marks is awarded for explicitly stating the correct mole ratio. Even if your final answer is wrong, writing "ratio = 2:3" earns the method mark. Hide the ratio in your head and you lose it.

Using Moles to Balance Equations

The reverse problem (spec 4.3.2.3): given masses of reactants and products, work out the balanced equation. This appears on roughly half of HT papers.

The Method

Convert each mass to moles (moles = mass ÷ M_r).

Write down the moles ratio of all substances.

Divide all moles by the smallest value.

If the ratios aren't whole numbers, multiply through by the smallest factor that makes them whole (×2, ×3, etc.).

The whole-number ratio gives the coefficients in the balanced equation.

Why it works

The mole ratio in a balanced equation IS the coefficient ratio. Working backwards from masses gives you the moles, which gives you the coefficients. It's the same three-step method run in reverse.

Worked Example 3 — Balancing from Masses

In a reaction, 0.48 g of magnesium combined with 0.32 g of oxygen to produce 0.80 g of magnesium oxide. Determine the balanced equation. [A_r: Mg = 24, O = 16]

1

Convert each mass to moles.

moles Mg = 0.48 ÷ 24 = 0.020 mol

moles O₂ = 0.32 ÷ 32 = 0.010 mol (M_r O₂ = 32)

moles MgO = 0.80 ÷ 40 = 0.020 mol (M_r MgO = 40)

2

Find the simplest ratio. Divide all by the smallest (0.010):

Mg : O₂ : MgO = 2 : 1 : 2

3

Write the balanced equation.

2Mg + O₂ → 2MgO ✓

Sanity Check

Mass conservation: 0.48 + 0.32 = 0.80 g ✓. Atoms balance: LHS has 2 Mg + 2 O; RHS has 2 Mg + 2 O ✓.

Critical Pitfall — Ratio to Moles, Not Masses

The single most common mistake on HT mass-to-mass questions: applying the mole ratio directly to grams. This always gives a wrong answer.

WRONG — ratio applied to mass

"6.0 g of Mg, ratio 2:2 = 1:1, so 6.0 g of MgO"

This treats grams as if they're directly comparable across different substances. They're not — Mg has A_r=24, MgO has M_r=40, so 1 mole of each weighs different amounts.

Wrong answer: 6.0 g

RIGHT — ratio applied to moles

"6.0 g Mg → 0.25 mol → 0.25 mol MgO → 10 g MgO"

Convert mass to moles first; apply the ratio in moles; convert moles back to mass. The mole is the universal unit that makes the ratio meaningful.

Correct answer: 10 g

⚠️ Test yourself: if a question gives you grams and asks for grams, your working must visit "mol" twice — once for the known, once for the unknown. If you can't see "mol" written in your working, you've shortcut the calculation.

Practice Question 1

Hydrogen burns in oxygen to produce water. The balanced equation is:

2H₂ + O₂ → 2H₂O

Calculate the mass of water produced when 4.0 g of hydrogen reacts completely.

A_r: H = 1, O = 16. 4 marks

1

Moles of known (H₂). M_r(H₂) = 2.

moles H₂ = 4.0 ÷ 2 = 2.0 mol1 mark

2

Mole ratio H₂ : H₂O = 2 : 2 = 1 : 1.

moles H₂O = 2.0 mol1 mark

3

Mass of unknown. M_r(H₂O) = 18.

mass H₂O = 2.0 × 18 = 36 g2 marks (1 method, 1 answer)

Practice Question 2

2Mg + SiO₂ → 2MgO + Si

1.2 kg

5 marks

(Modelled on AQA P1 Jun22 Q07.4)

0

Convert kg to g. 1.2 kg = 1200 g

1

M_r(SiO₂) = 28 + 2(16) = 601 mark

2

Moles of SiO₂.

moles SiO₂ = 1200 ÷ 60 = 20 mol1 mark

3

Mole ratio Mg : SiO₂ = 2 : 1.

moles Mg = 20 × 2 = 40 mol1 mark

4

Mass of Mg.

mass Mg = 40 × 24 = 960 g2 marks (1 method, 1 answer)

(960 g, or equivalently 0.96 kg.)

Practice Question 3

A reaction produces sodium chloride from its elements. 4.6 g of sodium reacted with 7.1 g of chlorine (Cl₂) to form 11.7 g of sodium chloride.

Use this data to determine the balanced symbol equation.

A_r: Na = 23, Cl = 35.5. 3 marks

1

Convert each mass to moles. A_r(Na) = 23, M_r(Cl₂) = 71, M_r(NaCl) = 58.5.

moles Na = 4.6 ÷ 23 = 0.20 mol

moles Cl₂ = 7.1 ÷ 71 = 0.10 mol

moles NaCl = 11.7 ÷ 58.5 = 0.20 mol1 mark

2

Simplify the ratio. Divide all by the smallest (0.10):

Na : Cl₂ : NaCl = 2 : 1 : 21 mark

3

Write the balanced equation.

2Na + Cl₂ → 2NaCl1 mark

Common Errors — Mark-Scheme Verbatim

Five recurring error patterns examiners flag every year on Higher Tier mass-to-mass questions.

Students often writeCorrect versionWhy it matters

Apply mole ratio directly to MASSESApply ratio to moles only ; convert mass ↔ moles at each endDifferent substances have different M_r — grams aren't comparable

Used 320 for M_r(Fe₂O₃) when 160 was givenUse the M_r given in the questionP1 Jun24 Q07.6 — coefficient and M_r are separate

Used A_r(Mg) = 48 instead of 24Read A_r from the periodic table on the exam paperP1 Jun22 Q07.4 — wrong A_r doubles every step

Guessed mole ratio (1:1, 2:1) without checkingRead coefficients from the balanced equationP1 Jun23 Q09.2 — the equation is the source of the ratio

Numbers scattered with no labelsLabel every step: "moles Mg = ", "moles MgO = ", "mass = "Cross-cutting theme — partial credit needs visible method

Summary

Key Concepts

Mole ratio: the ratio of moles of substances in a balanced equation, given by the coefficients.

The Three-Step Method: mass of known → ÷M_r → moles of known → ×ratio → moles of unknown → ×M_r → mass of unknown.

Reverse problem (4.3.2.3): given masses → convert to moles → simplify ratio to whole numbers → that's the coefficient ratio.

The Three Steps

moles known = mass ÷ M_r

moles unknown = moles known × (ratio)

mass unknown = moles unknown × M_r

Use Given Values

If the question gives M_r, USE IT. Read A_r from the periodic table on the paper. Don't recalculate or guess.

Don't Guess Ratios

Mole ratios come ONLY from the coefficients in the balanced equation. Write the equation first; circle the coefficients; write the ratio explicitly.

Label Everything

"moles Mg = ", "moles MgO = ", "mass = ". Each labelled step earns its method mark — even if your final number slips.

Exam Tip

The 4-mark mass-to-mass question is the highest-frequency calculation type on HT Paper 1. Master the three-step method and the four common errors and you've banked 4 marks every paper. The next lesson (3.05) extends this to limiting reactants — when one reactant runs out before the others.

Ready for 3.05 [partial HT]

Next: Limiting Reactants and Concentration of Solutions — using the three-step method to identify which reactant runs out first, and calculating concentration in mol/dm³.