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.
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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
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
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 ).
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".
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
- Step 1: Calculate total M r of reactants
Total reactant M r = M r(CdO) + A r(C) = 128 + 12 = 140 [1 mark] - Step 2: Identify desired product M r
The desired product is cadmium ( Cd ) → A r = 112 - 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.
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!
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
- Step 1: Calculate total mass of reactants (left side)
Mass of WO₃ + 3 × Mass of H₂
= 232 + (3 × 2) = 232 + 6 = 238 [1 mark] - Step 2: Calculate total mass of products (right side)
Mass of W + 3 × Mass of H₂O
= 184 + (3 × 18) = 184 + 54 = 238 [1 mark] - 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.
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.
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.