AQA GCSE Chemistry Chemistry Paper 1 (Higher), June 2025: Question 3

11 marks · Standard Demand difficulty · Short Answer

Answer questions on metal extraction, redox, percentage atom economy, differences between transition and alkali metals, conservation of mass, and the importance of high atom economy.

Practise this question

Question

Question 3 covers metal extraction and chemical calculations across six parts: 03.1 asks to identify the substance reduced and give a reason in terms of oxygen for the reaction SnO2 + C -> Sn + CO2 (2 marks); 03.2 asks why carbon can be used to extract tin from tin oxide (1 mark); 03.3 asks to calculate the percentage atom economy for producing cadmium from CdO + C -> Cd + CO given relative atomic and formula masses (3 marks); 03.4 asks for two differences in properties between tungsten (transition metal) and potassium (Group 1 metal) (2 marks); 03.5 asks to show that WO3 + 3 H2 -> W + 3 H2O obeys the law of conservation of mass using provided relative formula masses (2 marks); and 03.6 asks why it is important to use reactions with a high atom economy in industry (1 mark).
Question text

03 This question is about metals and metal oxides.

Tin (Sn) is extracted from tin oxide using carbon.

The equation for the reaction is:

SnO2 + C ⟶ Sn + CO2

03.1 Which substance is reduced in this reaction?

Give one reason for your answer.

Answer in terms of oxygen.

[2 marks]

Substance reduced

Reason

03.2 Why can carbon be used to extract tin from tin oxide?

[1 mark]

03.3 Cadmium (Cd) can be extracted by the reaction of cadmium oxide with carbon.

The equation for the reaction is:

CdO + C ⟶ Cd + CO

Calculate the percentage atom economy for the production of cadmium in

this reaction.

Relative atomic masses (Ar): C = 12 Cd = 112

Relative formula mass (Mr): CdO = 128

[3 marks]

Percentage atom economy = %

03.4 Tungsten is a transition metal.

Potassium is a Group 1 metal.

Give two differences between the properties of tungsten and

the properties of potassium.

[2 marks]

Tungsten oxide reacts with hydrogen to give tungsten (W) and water.

03.5 The equation for the reaction is:

WO3 + 3 H2 ⟶ W + 3 H2O

The law of conservation of mass states:

*10* ‘The mass of the products equals the mass of the reactants during a

chemical reaction.’

Show that the equation obeys the law of conservation of mass.

Relative atomic mass (Ar): W = 184

Relative formula masses (Mr): H2 = 2 WO3 = 232 H2O = 18

[2 marks]

03.6 The reaction producing tungsten has a high atom economy.

Why is it important to use reactions with a high atom economy in industry?

[1 mark]

Mark scheme

Show the mark scheme Mark scheme for Question 3: 03.1 awards 1 mark for SnO2 / tin oxide and 1 mark for losing oxygen; 03.2 awards 1 mark for carbon being more reactive than tin; 03.3 awards 1 mark for total Mr of reactants = 140, 1 mark for (112/140) x 100, and 1 mark for 80%; 03.4 awards 2 marks for any two property comparisons between tungsten and potassium (higher melting/boiling point, denser, harder, stronger, less reactive, forms ions with different charges, forms coloured compounds, can be a catalyst); 03.5 awards 1 mark for sum of reactant masses (232 + 3x2 = 238) and 1 mark for product masses (184 + 3x18 = 238); 03.6 awards 1 mark for sustainable development or economic reasons (minimise waste, conserve resources).

Question 3

AO /

Question Answers Extra information Mark

Spec. Ref.

03.1 (substance reduced)

SnO2 allow tin oxide 1 AO2

4.4.1.1

4.4.1.3

(reason)

(SnO2) loses oxygen allow (tin oxide) loses oxygen 1

MP2 is dependent upon MP1

being awarded

AO /

Spec. Ref.

03.2 carbon is more reactive than tin allow tin is less reactive than 1 AO3

carbon 4.4.1.2

4.4.1.3

AO /

Spec. Ref.

03.3 (total Mr = 128 + 12 =) 140 allow (total Mr = 112 + 12 + 16 1 AO2

=) 140 4.3.3.2

(% atom economy =)

112 allow correct use of an 1

× 100 incorrectly determined total M

140 r

= 80 (%) 1

AO /

Spec. Ref.

03.4 allow the converse for

potassium

allow the transition metal for

tungsten

allow the Group 1 metal for

potassium

ignore references to atomic

structure

any two from: 2 AO1

12 4.1.2.5

tungsten 4.1.3.1

• has a higher melting / boiling 4.1.3.2

point

• is denser

• is harder allow is less malleable / ductile

• is stronger

• is less reactive allow specific reactions showing

difference in reactivity

• has ions with different

charges

• forms coloured compounds

• can be a catalyst

AO /

Spec. Ref.

03.5 (sum of relative formula masses allow (sum of relative formula 1 AO2

on left hand side =) masses on left hand side =) 4.3.1.1

232 + (3 × 2) = 238 232 + 6 = 238 4.3.1.2

(is equal to

sum of relative formula masses allow (is equal to 1

on right hand side =) sum of relative formula masses

184 + (3 × 18) = 238 on right hand side =)

184 + 54 = 238

if no other mark awarded,

allow 1 mark for

232 + (3 × 2) = 184 + (3 × 18)

or

232 + 6 = 184 + 54 13

or

allow 1 mark for both sides =

AO /

Spec. Ref.

03.6 for sustainable development allow to minimise use of limited 1 AO1

resources 4.3.3.2

allow to minimise use of energy

allow to minimise waste

or

for economic reasons

ignore references to yield

Total Question 3 11

How to answer it

Metals, Reactivity Series & Quantitative Calculations

Topic Summary

What this question tests:

  • Redox in terms of oxygen: Identifying oxidation and reduction based on oxygen gain or loss.
  • Metal extraction & reactivity: Explaining why carbon can displace certain metals from their oxides.
  • Quantitative chemistry: Calculating percentage atom economy using relative formula masses ( M r).
  • Periodic Table trends: Contrasting transition metals with Group 1 alkali metals.
  • Conservation of mass: Demonstrating through calculation that reactant mass equals product mass in a balanced equation.
  • Industrial sustainability: Explaining why high atom economy processes are preferred in manufacturing.

Question 03.1

Reduction in Terms of Oxygen [2 Marks]

Equation: SnO₂ + C → Sn + CO₂

✅ Correct Answer

Substance reduced: SnO₂ (or tin oxide) [1 mark]

Reason: It loses oxygen [1 mark]

💡 Key Knowledge

  • Oxidation = gain of oxygen.
  • Reduction = loss of oxygen.
  • Carbon ( C ) gains oxygen to form CO₂ (oxidised).
  • Tin oxide ( SnO₂ ) loses oxygen to form tin metal ( Sn ) (reduced).

🧠 Exam Technique

Always name the reactant that starts with oxygen as the substance reduced (e.g. tin oxide or SnO₂). Note that Mark 2 depends on having Mark 1 correct.

❌ Common Errors

Writing just "tin" or "Sn" as the substance reduced. Tin is the product; the substance that actually loses oxygen during the reaction is tin oxide ( SnO₂ ).

Mark scheme: 1 mark for SnO₂ / tin oxide; 1 mark for (SnO₂) loses oxygen. MP2 is dependent on MP1.

Question 03.2

Extraction of Metals Using Carbon [1 Mark]

✅ Correct Answer

Carbon is more reactive than tin
(or tin is less reactive than carbon).

💡 Key Knowledge

Metals less reactive than carbon can be extracted from their oxides by reduction with carbon. A more reactive element displaces a less reactive element from its compound.

🧠 Exam Technique

When asked why a non-metal/metal can extract another, the answer is almost always a direct comparative statement about their relative reactivity.

❌ Common Errors

Stating only that "carbon is reactive" without comparing it to tin. It must be a comparative statement: "more reactive than tin".

Mark scheme: 1 mark for carbon is more reactive than tin (or converse).

Question 03.3

Percentage Atom Economy Calculation [3 Marks]

Equation: CdO + C → Cd + CO
Given: A r values: C = 12, Cd = 112; M r: CdO = 128

📐 Step-by-Step Calculation

Percentage Atom Economy = (Total M r of desired product ÷ Total M r of all reactants) × 100
  1. Step 1: Calculate total M r of reactants
    Total reactant M r = M r(CdO) + A r(C) = 128 + 12 = 140 [1 mark]
  2. Step 2: Identify desired product M r
    The desired product is cadmium ( Cd ) → A r = 112
  3. Step 3: Calculate percentage atom economy
    (112 ÷ 140) × 100 = 80% [2 marks: 1 for correct fraction/substitution, 1 for final answer]

✅ Final Answer

80%

❌ Common Errors

  • Forgetting to include carbon in the total reactant mass (using 128 instead of 140).
  • Dividing by product mass instead of reactant mass (though here both sum to 140, forming good habits is essential).
  • Confusing atom economy with percentage yield. Atom economy is theoretical and based entirely on the balanced equation.
Mark scheme: 1 mark for total M r = 140; 1 mark for (112 ÷ 140) × 100; 1 mark for 80(%). Allow ECF for arithmetic slips.

Question 03.4

Properties of Transition Metals vs Group 1 [2 Marks]

Compare tungsten (transition metal) with potassium (Group 1 metal).

✅ Correct Differences (any two)

Tungsten (compared to potassium):

  • Has a higher melting / boiling point
  • Is denser / has higher density
  • Is harder / stronger (potassium is soft and can be cut with a knife)
  • Is less reactive (potassium reacts violently with water)
  • Can form ions with different charges (variable oxidation states)
  • Forms coloured compounds
  • Can act as a catalyst

🧠 Exam Technique

Always state which metal has which property! You can phrase it in terms of tungsten (e.g., "tungsten has a higher melting point") or potassium (e.g., "potassium is less dense"). Avoid vague words like "different strength" without qualifying which is higher.

💡 Key Knowledge

Transition elements are typical strong, dense engineering metals with high melting points. Group 1 alkali metals are uniquely soft, have low densities (lithium, sodium, and potassium float on water), and have low melting points.

❌ Common Errors

Giving electronic structure differences (e.g. "potassium has 1 outer electron"). The question asks for differences in properties, not atomic structure!

Mark scheme: Any 2 valid points comparing properties [1 mark each]. Ignore atomic structure references.

Question 03.5

Proving the Law of Conservation of Mass [2 Marks]

Equation: WO₃ + 3 H₂ → W + 3 H₂O
Given: A r: W = 184; M r: H₂ = 2, WO₃ = 232, H₂O = 18

📐 Step-by-Step Calculation

  1. Step 1: Calculate total mass of reactants (left side)
    Mass of WO₃ + 3 × Mass of H₂
    = 232 + (3 × 2) = 232 + 6 = 238 [1 mark]
  2. Step 2: Calculate total mass of products (right side)
    Mass of W + 3 × Mass of H₂O
    = 184 + (3 × 18) = 184 + 54 = 238 [1 mark]
  3. Conclusion:
    Since reactant mass (238) = product mass (238), the law of conservation of mass is obeyed.

🧠 Exam Technique

Remember the big balancing numbers! The ' 3 ' in front of H₂ and H₂O means you must multiply their formula masses by 3. Show both left and right totals clearly.

❌ Common Errors

Forgetting balancing coefficients: calculating 232 + 2 = 234 and 184 + 18 = 202 , which leads to unequal totals and loss of marks.

Mark scheme: 1 mark for sum of reactants = 238; 1 mark for sum of products = 238. (Allow 1 mark if complete equation substitution shown without totals).

Question 03.6

Importance of High Atom Economy in Industry [1 Mark]

✅ Correct Answer (any one)

  • For sustainable development
  • For economic reasons / to reduce costs
  • To minimise waste produced
  • To minimise use of limited resources / conserve raw materials
  • To minimise energy use in processing waste

❌ Common Errors

Confusing atom economy with percentage yield. Do NOT write "gives a higher yield" or "produces more product" – atom economy is about the proportion of reactant atoms that become useful products, not reaction efficiency.

Mark scheme: 1 mark for sustainable development OR economic reasons OR minimise waste/resource use. Ignore references to yield.

Topics

Chemistry · C1: Atomic Structure and the Periodic Table · C3: Quantitative Chemistry · C4: Chemical Changes

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.