AQA GCSE Chemistry Chemistry Paper 1 (Higher), November 2021: Question 3

10 marks · Standard Demand difficulty · Extended Answer

Calculate the relative atomic mass and identify an unknown metal, determine percentage atom economy for tin extraction, and evaluate three methods for extracting tungsten.

Practise this question

Question

Question 3 contains four parts about metal extraction. Part 03.1 gives the reaction 3 H2 + RO3 -> R + 3 H2O and states that the sum of relative formula masses of the reactants is 150, asking for the Ar of R. Part 03.2 asks to identify element R. Part 03.3 gives the reaction SnO2 + C -> Sn + CO2 and asks to calculate the percentage atom economy for extracting tin. Part 03.4 provides Table 2 detailing three extraction methods for tungsten oxide (Method 1 using Carbon with low cost producing tungsten solid, carbon dioxide gas, and tungsten carbide solid; Method 2 using Hydrogen with high cost producing tungsten solid and water vapour; Method 3 using Iron with low cost producing tungsten solid and iron oxide solid) and asks candidates to evaluate the three methods.
Question text

03 This question is about the extraction of metals.

Element R is extracted from its oxide by reduction with hydrogen.

The equation for the reaction is:

3H2 + RO3 → R + 3H2O

03.1 The sum of the relative formula masses (Mr) of the reactants (3H2 + RO3) is 150

Calculate the relative atomic mass (Ar) of R.

Relative atomic masses (Ar): H = 1 O = 16

[2 marks]

Relative atomic mass (Ar) of R =

03.2 Identify element R.

You should use:

• your answer to question 03.1

• the periodic table.

[1 mark]

Identity of R =

03.3 Carbon is used to extract tin (Sn) from tin oxide (SnO2).

The equation for the reaction is:

SnO2 + C → Sn + CO2

Calculate the percentage atom economy for extracting tin in this reaction.

Relative atomic masses (Ar): C = 12 O = 16 Sn = 119

[3 marks]

Percentage atom economy =10 %

03.4 Tungsten (W) is a metal.

Tungsten is extracted from tungsten oxide (WO3).

All other solid products from the extraction method must be separated from the

tungsten.

Table 2 shows information about three possible methods to extract tungsten from

tungsten oxide.

Table 2

Method Reactant Relative cost of reactant Products

Tungsten solid

1 Carbon Low Carbon dioxide gas

Tungsten carbide solid

Tungsten solid

2 Hydrogen High

Water vapour

Tungsten solid

3 Iron Low

Iron oxide solid

Evaluate the three possible methods for extracting tungsten from tungsten oxide.

[4 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 3 showing marking criteria for 03.1 (calculating Ar = 96, 2 marks), 03.2 (identifying R as molybdenum/Mo, 1 mark), 03.3 (calculating total Mr = 163, % atom economy = (119/163)*100 = 73%, 3 marks), and 03.4 (levels of response table for 4 marks with indicative content evaluating cost of reactants, separation of solid products, and loss of tungsten). Total 10 marks.

Question 3

AO /

Question Answers Extra information Mark

Spec. Ref.

03.1 (3 × Mr H2O = 3 × (2 + 16) =) 54 1 AO2

4.3.1.1

(Ar R = 150 – 54 =) 96 ignore units 1 4.3.1.2

alternative approach:

(Mr RO3 = 150 – 6 =) 144 (1)

(Ar R = 144 – (3 × 16) =) 96 (1) ignore units

AO /

Spec. Ref.

03.2 (R =) molybdenum / Mo allow ecf from question 03.1 1 AO3

4.1.1.1

AO /

Spec. Ref.

03.3 (total Mr of reactants) = 163 1 AO2

4.3.1.2

119 allow correct use of an 1 4.3.3.2

(% atom economy =) 163 (×100)

incorrectly calculated value of

total Mr

= 73 (%) allow 73.00613– (%)HEMISTRYcorrectly – 1 –

rounded to at least 2 significant

figures

Question 3 continued

AO/

Question Answers Mark

Spec. Ref

Level 2: Some logically linked reasons are given. There may also AO3

03.4 3-4

be a simple judgement. 4.4.1.3

Level 1: Relevant points are made. They are not logically linked. 1–2

No relevant content 0

Indicative content

• carbon and iron are the cheapest reactants

• hydrogen is the most expensive reactant

• separating solid products is expensive

• separating solid products is time consuming

• in method 1, tungsten needs to be separated from tungsten

carbide

• in method 1, some tungsten is lost as tungsten carbide

• in method 1, the carbon dioxide produced will escape

• in method 2, the water vapour produced will escape

• in method 2, no separation of solids is needed

• in method 3, tungsten needs to be separated from iron oxide

Total 10

How to answer it

Extracting Metals & Chemical Calculations

📌 What this question tests

This question assesses key quantitative chemistry and industrial application skills from AQA Chemistry:

  • Deducing relative atomic mass (Aᵣ): Using conservation of mass, balanced equations, and formula mass arithmetic.
  • Periodic Table identification: Matching calculated atomic masses to chemical symbols.
  • Percentage atom economy: Calculating reactant formula mass and applying the atom economy formula.
  • Industrial method evaluation: Comparing competing metal extraction processes by weighing reactant costs against purification difficulties and product yield.
Question 03.1

Calculating Relative Atomic Mass (Aᵣ) of Unknown Element R

Reaction: 3 H₂ + RO₃ → R + 3 H₂O  |  Sum of reactant masses = 150  |  2 Marks

📐 Step-by-Step Calculation

Method A (Using Reactants):

  1. Find mass of 3 H₂:
    3 × (2 × 1) = 6
  2. Subtract hydrogen from total reactant mass to find Mᵣ of RO₃:
    Mᵣ(RO₃) = 150 - 6 = 144 (1 mark)
  3. Subtract the 3 oxygen atoms (3 × 16 = 48) to find Aᵣ of R:
    Aᵣ(R) = 144 - 48 = 96 (1 mark)

Method B (Using Products & Conservation of Mass):

  1. Total mass of products must also equal 150 .
  2. Mass of 3 H₂O = 3 × (2 + 16) = 54 (1 mark).
  3. Mass of element R = 150 - 54 = 96 (1 mark).

✅ Correct Answer

Relative atomic mass (Aᵣ) of R = 96

(Aᵣ values have no units, so ignore units if written).

❌ Common Errors

  • Forgetting the balancing number: Using H₂ = 2 instead of 3 H₂ = 6 .
  • Miscounting oxygen atoms: Subtracting only one oxygen atom (16) from 144 rather than three (48).
Mark Breakdown:
• [1 mark] for determining Mᵣ(RO₃) = 144 OR total mass of water = 54.
• [1 mark] for calculating final atomic mass = 96.
Question 03.2

Identifying Element R from the Periodic Table

Periodic Table deduction  |  1 Mark

✅ Correct Answer

Molybdenum (or Mo)

🧠 Exam Technique: Error Carried Forward (ecf)

If you made a mathematical mistake in 03.1 and got an incorrect number, you can still gain this 1 mark by correctly naming whatever element has an Aᵣ matching your incorrect number! Always write an element name, never leave it blank.

Mark Breakdown: [1 mark] for Molybdenum / Mo (or correct ecf element from 03.1).
Question 03.3

Calculating Percentage Atom Economy

Reaction: SnO₂ + C → Sn + CO₂  |  Aᵣ: C = 12, O = 16, Sn = 119  |  3 Marks

📐 Step-by-Step Calculation

  1. Calculate total Mᵣ of all reactants:
    Reactants are 1 mole of SnO₂ and 1 mole of C.
    Mᵣ(SnO₂) = 119 + (2 × 16) = 119 + 32 = 151
    Total Reactant Mᵣ = 151 + 12 = 163 (1 mark)
  2. Apply the percentage atom economy formula:
    % Atom Economy = (Mᵣ of Desired Product / Total Mᵣ of All Reactants) × 100
    Desired product is Tin (Sn) = 119.
    % Atom Economy = (119 / 163) × 100 (1 mark)
  3. Calculate final value:
    = 73.006...% → 73% (or 73.0%) (1 mark)

💡 Key Knowledge: Atom Economy

  • Atom economy measures the proportion of starting materials that end up as useful products.
  • Always sum all reactants shown in the balanced equation (including balancing coefficients, if any).
  • Alternatively, the sum of all product Mᵣ values equals the total reactant Mᵣ due to conservation of mass.

❌ Common Errors

  • Confusing atom economy with percentage yield: Atom economy is theoretical based on the equation; percentage yield is experimental.
  • Leaving out carbon: Only using 151 (SnO₂) as the denominator rather than 163 (SnO₂ + C).
Mark Breakdown:
• [1 mark] for calculating total Mᵣ of reactants = 163.
• [1 mark] for setting up calculation correctly: (119 / 163) × 100.
• [1 mark] for final answer 73% (accept 73.006% rounded correctly to at least 2 sig figs).
Question 03.4

Evaluating Extraction Methods for Tungsten (W)

Extended Evaluation  |  4 Marks (Level 2: 3–4 marks, Level 1: 1–2 marks)

Method Reactant Relative Cost Products Formed
1 Carbon Low Tungsten solid, Carbon dioxide gas, Tungsten carbide solid
2 Hydrogen High Tungsten solid, Water vapour
3 Iron Low Tungsten solid, Iron oxide solid

✅ Model Evaluation Response (4 Marks)

A top-level response must compare reactants and separation of products across the methods and give a reasoned judgement:

  • Reactant cost: Carbon and iron are cheap/low-cost reactants, whereas hydrogen is expensive.
  • Separation of solid products: In Method 2, tungsten is the only solid product because water vapour is a gas and escapes easily. No costly separation is needed.
  • Method 1 drawbacks: Method 1 produces an unwanted solid byproduct (tungsten carbide). Separating solids is difficult, expensive, and time-consuming. Furthermore, some tungsten is wasted in the carbide, lowering yield.
  • Method 3 drawbacks: Method 3 produces solid iron oxide mixed with solid tungsten, which requires an extra separation step.
  • Judgement: Even though hydrogen is expensive, Method 2 is the best method because it yields pure solid tungsten without needing an expensive, difficult separation stage.

🧠 Exam Technique: Level 2 Criteria (3–4 Marks)

To reach Level 2 (3–4 marks), examiners require:

  • Points from both themes: cost of reactants AND separation of solid products.
  • Logical links: e.g., linking "water vapour escapes" to "no need to separate solid products".
  • A clear, justified conclusion/judgement stating which method is best and why.

❌ Common Errors

  • Merely copying the table without explaining consequences (e.g. saying "Method 2 makes water vapour" without noting that gases escape freely).
  • Ignoring the prompt clue: "All other solid products... must be separated from tungsten."
Mark Breakdown:
• Level 2 (3–4 marks): Logically linked comparisons addressing reactant cost and product separation with a reasoned conclusion.
• Level 1 (1–2 marks): Isolated points made (e.g., stating hydrogen is dearer or carbon is cheaper) without clear links or evaluation.

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), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.