AQA GCSE Combined Science: Trilogy Chemistry Paper 1 (Higher), 2021: Question 6

12 marks · Standard Demand difficulty · Short Answer

Calculate masses and moles in reactions of metals with oxygen, explain the effect of using a crucible lid, and determine the formula and balanced equation for an iron oxide from mole data.

Practise this question

Question

The question page shows Question 06 about metal oxides formed when metals are heated in air. At the top is a labelled diagram of apparatus: a crucible containing magnesium with a lid resting slightly open, supported on a tripod above heat, and a separate balance; labels point to lid, crucible, magnesium, tripod, heat and balance. The page contains four parts: 06.1 asks students to calculate the moles of oxygen gas reacting when 0.12 g magnesium produces 0.20 g magnesium oxide, with relative atomic mass O = 16; 06.2 asks why the mass of magnesium oxide would be different without a lid on the crucible; 06.3 asks for the mass of copper oxide from 0.50 g copper given that 63.5 g copper produces 79.5 g copper oxide, to 3 significant figures; 06.4 gives 0.015 moles iron reacting with 0.010 moles oxygen gas and asks for the formula of the iron oxide and the balanced symbol equation.
Question text

06 Metal oxides are produced when metals are heated in air.

A student investigated the change in mass when 0.12 g of magnesium was

heated in air.

Figure 5 shows the apparatus.

Figure 5

The student measured the mass of magnesium oxide produced.

06.1 0.12 g of magnesium reacted to produce 0.20 g of magnesium oxide.

Calculate the number of moles of oxygen gas (O2) that reacted.

Relative atomic mass (Ar): O = 16

[3 marks]

Moles of oxygen gas =19

06.2 The student repeated the experiment without a lid on the crucible.

Suggest why the mass of magnesium oxide produced would be different without a lid

on the crucible.

[2 marks]

06.3 Copper reacts with oxygen to produce copper oxide.

63.5 g of copper produces 79.5 g of copper oxide.

Calculate the mass of copper oxide produced when 0.50 g of copper

reacts with oxygen.

Give your answer to 3 significant figures.

[3 marks]

20 Mass (3 significant figures) = g

06.4 Iron reacts with oxygen to produce an oxide of iron.

0.015 moles of iron reacts with 0.010 moles of oxygen gas (O2).

Determine:

• the formula of the iron oxide produced

• the balanced symbol equation for the reaction.

[4 marks]

Formula of iron oxide =

Balanced symbol equation

Mark scheme

Show the mark scheme The mark scheme is a table listing answers, extra information, marks and AO/specification references for parts 06.1 to 06.4, totalling 12 marks. For 06.1 it gives mass of oxygen as 0.20 minus 0.12 equals 0.08 g, then moles of oxygen as 0.08 divided by 32 equals 0.0025, allowing 1 mark for 0.005 if obtained from dividing by 16; for 06.2 it states the magnesium oxide mass would be less because product escaped, allowing 'magnesium oxide escaped'. For 06.3 it gives 79.5 divided by 63.5 times 0.5 equals 0.62598 g, rounded to 0.626 g, with allowance for correct 3 significant figure rounding from a full-working error; for 06.4 it gives the ratio 3:2 for Fe to O2 molecules or 3:4 for Fe to O atoms, the formula Fe3O4, and the balanced equation 3Fe + 2O2 → Fe3O4, with partial credit notes for an incorrectly determined formula used consistently.

AO /

Question Answers Extra information Mark

Spec. Ref.

06.1 (mass of oxygen = 0.20 – 0.12) AO2

= 0.08 (g) 1 5.1.1.1

5.3.1.1

0.08 5.3.1.3

(moles of oxygen) = 1

32 5.3.2.1

allow 1 mark for 0.005 5.4.1.1

0.08

= 0.0025 if derived from 1

06.2 (without a lid the) mass of 1 AO3

magnesium oxide was less 5.4.1.1

(because) products escaped allow magnesium oxide escaped 1

06.3 (mass of copper oxide =) AO2

79.5 5.1.1.1

× 0.5 1 5.3.1.1

63.5 5.3.1.3

= 0.62598 (g) 5.3.2.1

5.3.2.2

= 0.626 (g) 5.3.2.3

allow an answer correctly 1

5.4.1.1

rounded to 3 significant figures

from an incorrect calculation

– OMBINED SCIENCE: TRIwhich usesall thevalues in theLOGY– – JUNE 2021

question

06.4 3:2 ratio Fe : O2 (molecules) 1 AO2

or 5.1.1.1

3:4 ratio Fe : O (atoms) 5.3.1.1

5.3.1.3

5.3.2.1

(formula) Fe3O4 allow 1 mark for Fe3O2 1 5.3.2.2

from 3:2 ratio Fe : O (atoms) 5.3.2.3

(MP2 but not MP1) 5.4.1.1

3 Fe + 2O2 Fe3O4 allow multiples 2

allow correct use of incorrectly

determined formula

allow 1 mark for Fe, O2 and

Fe3O4

or

14 allow 1 mark for Fe, O2 and

incorrectly determined formula

Total 12

How to answer it

Metals Reacting with Oxygen: Mass, Moles and Formulae

What this question tests

You need to use mass changes to find the mass of oxygen gained, convert between mass and moles, explain why a lid matters in practical work, use a ratio to scale up a mass calculation, and work out an unknown formula and balanced equation from mole numbers.

Question 06.1–06.4

Magnesium, copper and iron oxides

Using mass changes, ratios and equations

💡 Key knowledge

  • When a metal reacts with oxygen, the mass increases because oxygen is added.
  • moles = mass ÷ Mr
  • For oxygen gas, O₂, Mr = 32
  • To find a formula, compare mole ratios of elements in the compound.
  • To balance an equation, change coefficients only, not formulas.

🧠 Exam technique

  • Show every step clearly for calculation marks.
  • Write units at each stage: g, mol, and final answer.
  • Round only at the end, unless the question tells you otherwise.
  • In 3-mark calculations, marks are often for: method, calculation, and final answer.

❌ Common errors

  • Using 16 instead of 32 for O₂.
  • Finding the mass of magnesium oxide instead of the mass of oxygen gained.
  • Forgetting that without a lid, some product can escape.
  • Writing a formula from masses instead of mole ratios.
  • Changing the subscripts in the formula when balancing an equation.
Part 06.1

Calculate the number of moles of oxygen gas reacted

📐 Calculations: step-by-step

  1. Find the mass of oxygen gained
    Mass of oxygen = 0.20 g − 0.12 g = 0.08 g
  2. Use moles = mass ÷ Mr
    Mr of O₂ = 16 × 2 = 32
    Moles of O₂ = 0.08 ÷ 32 = 0.0025 mol

✅ Correct answer

0.0025 mol

Mark breakdown: 1 mark for mass of oxygen, 1 mark for using Mr = 32, 1 mark for final answer.

🧠 What the examiner wanted

Top answers clearly showed the mass gain first, then converted to moles using the correct formula. The best responses included 0.08 ÷ 32 and a final answer in mol.

❌ Common traps

  • Using 0.20 ÷ 32 instead of the mass gained.
  • Using 16 instead of 32 for O₂.
  • Giving the answer as 0.08 or 0.0025 g instead of mol.
Part 06.2

Why would the mass be different without a lid?

✅ Correct answer

  • Without a lid the mass of magnesium oxide would be less.
  • Because product escaped from the crucible.

💡 Key knowledge

A lid helps stop the solid product being lost when the magnesium reacts and the contents may spit or move out of the crucible.

🧠 How marks were awarded

This is a 2-mark explanation: one mark for saying the mass would be less, and one mark for explaining that product escaped. “Magnesium oxide escaped” was accepted.

❌ Common errors

  • Saying the mass would be greater.
  • Only saying “because oxygen gets in” — this does not explain the lower measured product mass.
  • Not linking the lid to preventing loss of solid product.
Part 06.3

Calculate the mass of copper oxide produced

📐 Calculations: step-by-step

  1. Use the mass ratio from the question
    63.5 g copper → 79.5 g copper oxide
  2. Scale up from 0.50 g copper
    Mass of copper oxide = (79.5 ÷ 63.5) × 0.50
  3. Calculate
    = 0.625984... g
  4. Round to 3 significant figures
    = 0.626 g

✅ Correct answer

0.626 g

Answer must be to 3 significant figures.

🧠 Exam technique

  • Keep the ratio the same: if copper is divided by 63.5, copper oxide must also be scaled the same way.
  • Show the fraction clearly: 79.5 ÷ 63.5 × 0.50
  • Use the values in the question exactly, then round at the end.

❌ Common calculation traps

  • Subtracting 63.5 from 79.5 instead of using a ratio.
  • Forgetting to multiply by 0.50.
  • Rounding too early and losing accuracy.
  • Not giving the answer to 3 significant figures.
Part 06.4

Determine the formula and balanced equation for iron oxide

📐 Working out the formula

  1. Find the mole ratio of Fe to O₂
    0.015 mol Fe reacts with 0.010 mol O₂
    Ratio Fe : O₂ = 0.015 : 0.010
  2. Simplify the ratio
    Divide both by 0.005 → 3 : 2
  3. Convert to atoms in the formula
    3 Fe atoms react with 2 O₂ molecules = 4 oxygen atoms
    So the oxide formula is Fe₃O₄

✅ Correct answers

Formula of iron oxide = Fe₃O₄

Balanced symbol equation: 3Fe + 2O₂ → Fe₃O₄

Marks: 1 for ratio/formula, 2 for correct balanced equation, with method credit if the formula is correct.

💡 Key knowledge

  • Oxygen gas is diatomic, so it is written as O₂.
  • In Fe₃O₄, there are 3 iron atoms and 4 oxygen atoms.
  • The balanced equation must conserve atoms on both sides.

🧠 What distinguished top-level responses

  • They used the mole ratio correctly: 3:2 for Fe : O₂.
  • They linked the ratio to the formula, not just the equation.
  • They wrote the equation with correct coefficients: 3Fe + 2O₂ → Fe₃O₄

❌ Common errors

  • Writing FeO or Fe₂O₃ without using the ratio given.
  • Using the numbers 0.015 and 0.010 directly in the formula.
  • Writing 3Fe + O₂ → Fe₃O₄ without balancing oxygen.
  • Changing the formula when balancing instead of changing coefficients.
Quick Revision

How to score full marks on questions like this

💡 Remember these facts

  • Mass gained in a reaction with oxygen = mass of oxygen added.
  • O₂ has Mr 32.
  • Mass ratio questions often need a direct scaling calculation.
  • Formula questions need mole ratios first, then the balanced equation.

🧠 Best exam habit

Always write the working out clearly. Even if the final answer is wrong, showing the correct method can still earn marks.

❌ Final warning

Check units, keep oxygen as O₂, and round only at the end. These are the most common places marks are lost.

Topics

Chemistry · C3: Quantitative Chemistry · C4: Chemical Changes

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Chemistry Paper 1 (Higher), 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.