AQA GCSE Chemistry Chemistry Paper 1 (Foundation), November 2021: Question 10

📐 Step-by-Step Calculation

Given: Sum of reactant formula masses = 150; A_r(H) = 1 , A_r(O) = 16 .

Method A (Using Reactants):

1. Find mass of 3H₂: 3 × (2 × 1) = 6
2. Find formula mass of RO₃: 150 − 6 = 144 [1 mark]
3. Subtract oxygen mass (3 × 16 = 48) to find R:
   A_r(R) = 144 − 48 = 96 [1 mark]

Method B (Law of Conservation of Mass):

Mass of reactants = Mass of products = 150

1. Find mass of 3H₂O: 3 × [ (2 × 1) + 16 ] = 3 × 18 = 54 [1 mark]
2. Subtract water mass from total product mass:
   A_r(R) = 150 − 54 = 96 [1 mark]

❌ Common Errors & Pitfalls

• Forgetting the stoichiometric '3': Calculating H₂ as 2 instead of 3 × 2 = 6 .
• Miscounting oxygen atoms: Subtracting 16 instead of 3 × 16 = 48 from the formula mass of RO₃.
• Putting units: Relative atomic mass is dimensionless. Do not add grams ( g ). (Though mark schemes usually ignore units, keep it clean!)

✅ Correct Answer

Molybdenum (or chemical symbol Mo)

🧠 Exam Technique

• Look up atomic mass 96 (not atomic number) on the periodic table.
• Element number 42 has a relative atomic mass of 96.
• Error Carried Forward (ECF): If you calculated an incorrect value in 10.1, you still gain this mark if you name the element matching your calculated number!

📐 Step-by-Step Calculation

1. Step 1: Calculate total M_r of all reactants
   M_r(SnO₂) = 119 + (2 × 16) = 151
   M_r(C) = 12
   Total reactant M_r = 151 + 12 = 163 [1 mark]
2. Step 2: Identify desired product and write expression
   Desired product is Tin (Sn), which has A_r = 119 .
   % Atom Economy = (119 / 163) × 100 [1 mark]
3. Step 3: Calculate and round the final answer
   119 / 163 × 100 = 73.006...%
   Final Answer = 73% (or 73.0%) [1 mark]

💡 Key Formula

Percentage Atom Economy:

• Remember: Total M_r of reactants = Total M_r of products.
• You could also find the denominator by adding products: Sn (119) + CO₂ (44) = 163 .

✅ Model Answer Structure

1. Evaluate Method 1 (Carbon):

• Advantage: Carbon is cheap / low cost. Carbon dioxide escapes naturally as a gas.
• Disadvantage: Produces solid tungsten carbide, meaning some tungsten is lost and extra energy/cost is required to separate the two solids.

2. Evaluate Method 2 (Hydrogen):

• Advantage: The only byproduct is water vapour which escapes as a gas, meaning no separation of solids is needed and high-purity tungsten is produced.
• Disadvantage: Hydrogen reactant is expensive and poses safety risks (flammable/explosive).

3. Evaluate Method 3 (Iron):

• Advantage: Iron is a cheap / low cost reactant.
• Disadvantage: Produces solid iron oxide alongside solid tungsten; separating two solid products is difficult, expensive, and time-consuming.

Conclusion / Judgement:

Method 2 is the best method to produce pure tungsten because water vapour simply escapes, eliminating the expensive and difficult process of separating solid mixtures, despite hydrogen's higher initial cost.

🧠 Level of Response Marking Guide

This is a 4-mark level-of-response question:

• Level 2 (3–4 marks): Gives logically linked points comparing both cost and separation difficulties across multiple methods, supported by a clear, reasoned judgement.
• Level 1 (1–2 marks): Mentions isolated relevant points (e.g. stating simply "hydrogen is expensive" or "iron oxide is a solid") without comparing the methods or discussing why solid separation matters.

💡 Top Examiner Insight

The prompt gave a critical clue: "All other solid products must be separated from the tungsten." Top-scoring students always linked this instruction to the table, identifying that gases ( CO₂ , water vapour) easily escape on their own, while solid byproducts ( tungsten carbide , iron oxide ) demand costly physical/chemical separation.