OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 4
1 mark · Medium difficulty · Multiple Choice
Identify which reaction to extract iron has the highest atom economy.
Practise this questionQuestion
Question text
4 Which reaction to extract iron (Fe) has the highest atom economy?
A 2Al + Fe2O3 2Fe + Al2O3
B FeO + CO Fe + CO2
C Fe2O3 + 3CO 2Fe + 3CO2
D 2Fe2O3 + 3C 4Fe + 3CO2
Your answer [1]
Mark scheme
Show the mark scheme
4 D 1
How to answer it
Extracting Iron: Calculating & Comparing Atom Economy
- Definition of Atom Economy: Understanding that atom economy measures the proportion of reactant mass converted into desired products.
- Stoichiometric Calculations: Using balanced chemical equations and relative atomic/formula masses ( Mᵣ ) correctly.
- Efficiency & Sustainability: Comparing multiple synthetic routes to identify the greenest reaction pathway in terms of waste generation.
Question 4 (Multiple Choice)
Identifying the Extraction Route with the Highest Atom Economy
✅ Correct Answer
Option D: 2Fe₂O₃ + 3C → 4Fe + 3CO₂
💡 Key Knowledge
The standard formula for atom economy is:
Atom Economy (%) = [ (Sum of Mᵣ of desired products) / (Sum of Mᵣ of all products) ] × 100
- Desired product: Only iron, Fe ( Aᵣ = 55.8 ).
- Total mass: Due to conservation of mass, sum of Mᵣ of all products = sum of Mᵣ of all reactants.
- Stoichiometric coefficients (big numbers in front) must be included in both numerator and denominator.
🧠 Fast MCQ Technique
In a multiple-choice question, doing 4 full percentage calculations wastes precious time. Use the "Waste per Mole of Desired Product" shortcut:
- Since the desired product is always Fe , the reaction that produces the least waste mass per mole of Fe has the highest atom economy.
- A: 102.0 g Al₂O₃ / 2 mol Fe = 51.0 g waste/mol Fe
- B: 44.0 g CO₂ / 1 mol Fe = 44.0 g waste/mol Fe
- C: 132.0 g CO₂ / 2 mol Fe = 66.0 g waste/mol Fe
- D: 132.0 g CO₂ / 4 mol Fe = 33.0 g waste/mol Fe
Option D yields the lowest waste per mole of iron, so it must have the highest atom economy!
📐 Step-by-Step Calculation for All Options
Using standard OCR Data Sheet values: Fe = 55.8 , Al = 27.0 , O = 16.0 , C = 12.0
| Option | Equation | Mass of Fe (Desired) | Total Mass of Products | Atom Economy (%) |
|---|---|---|---|---|
| A | 2Al + Fe₂O₃ → 2Fe + Al₂O₃ | 2 × 55.8 = 111.6 | 111.6 + 102.0 = 213.6 | (111.6 / 213.6) × 100 = 52.2% |
| B | FeO + CO → Fe + CO₂ | 1 × 55.8 = 55.8 | 55.8 + 44.0 = 99.8 | (55.8 / 99.8) × 100 = 55.9% |
| C | Fe₂O₃ + 3CO → 2Fe + 3CO₂ | 2 × 55.8 = 111.6 | 111.6 + (3 × 44.0) = 243.6 | (111.6 / 243.6) × 100 = 45.8% |
| D | 2Fe₂O₃ + 3C → 4Fe + 3CO₂ | 4 × 55.8 = 223.2 | 223.2 + (3 × 44.0) = 355.2 | (223.2 / 355.2) × 100 = 62.8% |
❌ Common Misconceptions & Traps
- Confusing Atom Economy with Percentage Yield: Percentage yield is an experimental value based on actual mass collected vs theoretical mass. Atom economy is purely theoretical based on the balanced stoichiometric equation.
- Forgetting Stoichiometry: A frequent mistake is dividing Aᵣ(Fe) by Mᵣ(CO₂) without multiplying by their balancing numbers (e.g. ignoring the 4 in front of Fe or the 3 in front of CO₂ in D).
- Assuming "Fewer Reactants = Higher Atom Economy": Many students pick B on intuition alone because it has simple 1:1 ratios. However, CO₂ makes up a larger proportion of the product mass in B (44.1%) than in D (37.2%).
Topics
Module 2: Foundations in chemistry · 2.1 Atoms and reactions
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.