Chemistry • Year 11 • Module 2 • Lesson 12

Mass–Mass Stoichiometry

Apply the 4-step method to a range of reactions, including non-1:1 ratios and reverse problems, with scaffolded working space for each step.

Apply · Worked Calculations

1. Guided 4-step drill — carbon combustion (1:1 ratio)

Carbon burns in excess oxygen: C + O₂ → CO₂. What mass of CO₂ is produced when 6.00 g of carbon burns completely? (C = 12.011, O = 15.999) 4 marks (1 per step)

Step 1 — Balance & ratio

Equation: already balanced. Identify the mole ratio of C : CO₂.

Ratio: C : CO₂ = _____ : _____

Step 2 — n(given) [n = m ÷ MM]

MM(C) = _________ g mol⁻¹

n(C) = _________ ÷ _________ = _________ mol

Step 3 — n(wanted) via ratio [n(wanted) = n(given) × coeff(wanted) ÷ coeff(given)]

n(CO₂) = _________ × _____ ÷ _____ = _________ mol

Step 4 — m(wanted) [m = n × MM]

MM(CO₂) = 12.011 + 2(15.999) = _________ g mol⁻¹

m(CO₂) = _________ × _________ = _________ g

Stuck? Revisit Worked Example 1 in the lesson — the method is identical for any 1:1 ratio reaction.

2. Non-1:1 ratio — aluminium combustion

Aluminium burns in oxygen: 4Al + 3O₂ → 2Al₂O₃. What mass of Al₂O₃ forms when 5.40 g of Al burns? (Al = 26.982, O = 15.999) 4 marks (1 per step)

Step 1: The equation is already balanced. Write the mole ratio of Al : Al₂O₃.

Ratio: Al : Al₂O₃ = _____ : _____

Step 2: n(Al) = 5.40 ÷ _________ = _________ mol

Step 3: n(Al₂O₃) = _________ × _____ ÷ _____ = _________ mol

Step 4: MM(Al₂O₃) = 2(26.982) + 3(15.999) = _________ g mol⁻¹

m(Al₂O₃) = _________ × _________ = _________ g

Stuck? Revisit the lesson activity drill question 2 (identical reaction with a similar mass of Al).

3. Reverse problem — given product, find reactant

Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. What mass of Mg must burn to produce 20.0 g of MgO? (Mg = 24.305, O = 15.999) 4 marks (1 per step)

Step 1: Balanced. Mole ratio Mg : MgO = _____ : _____

Note: in this problem, MgO is the “given” substance and Mg is “wanted”.

Step 2: MM(MgO) = 24.305 + 15.999 = _________ g mol⁻¹

n(MgO) = 20.0 ÷ _________ = _________ mol

Step 3: n(Mg) = _________ × _____ ÷ _____ = _________ mol

Step 4: m(Mg) = _________ × 24.305 = _________ g

Stuck? Revisit Worked Example 3 in the lesson — it uses this exact equation.

4. Analyse and correct a student’s working

A student attempted the following calculation for: Fe₂O₃ + 3CO → 2Fe + 3CO₂. “What mass of Fe forms from 80.0 g of Fe₂O₃?” (Fe = 55.845, O = 15.999) 6 marks

Student’s working:

Step 1: Balanced. Ratio Fe₂O₃ : Fe = 1 : 1

Step 2: MM(Fe₂O₃) = 2(55.845) + 3(15.999) = 159.69; n(Fe₂O₃) = 80.0 ÷ 159.69 = 0.501 mol

Step 3: n(Fe) = 0.501 × 1 = 0.501 mol

Step 4: m(Fe) = 0.501 × 55.845 = 27.98 g ≈ 28.0 g

4.1 Identify the error in the student’s Step 1 and state the correct mole ratio. 2 marks

4.2 Show the corrected Step 3 and Step 4. 2 marks

Corrected Step 3: n(Fe) = _________ × _____ ÷ _____ = _________ mol

Corrected Step 4: m(Fe) = _________ × 55.845 = _________ g

4.3 How does the correct answer compare to the student’s answer? Explain why the student’s error always gives exactly half the correct mass in this reaction. 2 marks

Stuck? Read the ratio 1:2 off the balanced equation — the student used 1:1. Revisit Worked Example 2 in the lesson (same reaction).

Answers — Do not peek before attempting

Q1 — Carbon combustion

Step 1: C : CO₂ = 1 : 1.

Step 2: MM(C) = 12.011; n(C) = 6.00 ÷ 12.011 = 0.4995 mol.

Step 3: n(CO₂) = 0.4995 × 1 ÷ 1 = 0.4995 mol.

Step 4: MM(CO₂) = 12.011 + 2(15.999) = 44.009; m(CO₂) = 0.4995 × 44.009 = 22.0 g.

Marking criteria: 1 mark each for correct Step 1 ratio; correct n(C) with working; correct n(CO₂); correct m(CO₂) with MM shown.

Q2 — Aluminium combustion

Step 1: Al : Al₂O₃ = 4 : 2 (or simplified 2 : 1).

Step 2: n(Al) = 5.40 ÷ 26.982 = 0.2001 mol.

Step 3: n(Al₂O₃) = 0.2001 × 2 ÷ 4 = 0.1001 mol.

Step 4: MM(Al₂O₃) = 2(26.982) + 3(15.999) = 101.96; m(Al₂O₃) = 0.1001 × 101.96 = 10.2 g.

Marking criteria: 1 mark each for correct ratio (accept 4:2 or 2:1); n(Al); n(Al₂O₃); m(Al₂O₃).

Q3 — Reverse problem (MgO)

Step 1: Mg : MgO = 2 : 2 = 1 : 1.

Step 2: MM(MgO) = 24.305 + 15.999 = 40.304; n(MgO) = 20.0 ÷ 40.304 = 0.4962 mol.

Step 3: n(Mg) = 0.4962 × 1 ÷ 1 = 0.4962 mol.

Step 4: m(Mg) = 0.4962 × 24.305 = 12.1 g.

Marking criteria: 1 mark each for identifying MgO as “given”; n(MgO); n(Mg); m(Mg).

Q4.1 — Error identification

The error is in Step 1: the correct mole ratio of Fe₂O₃ : Fe is 1 : 2 (from the equation coefficients 1 and 2), not 1 : 1. The student read only the coefficient of Fe₂O₃ (which is 1) and incorrectly assigned 1 to Fe as well. Award 1 mark for identifying that the ratio is wrong and 1 mark for stating the correct ratio 1:2.

Q4.2 — Corrected Steps 3 and 4

Corrected Step 3: n(Fe) = 0.501 × 2 ÷ 1 = 1.002 mol.

Corrected Step 4: m(Fe) = 1.002 × 55.845 = 55.96 g ≈ 56.0 g.

Q4.3 — Comparison and explanation

The correct answer (56.0 g) is exactly double the student’s answer (28.0 g). This is because the correct ratio is 1:2 — each mole of Fe₂O₃ produces 2 mol of Fe. The student used a 1:1 ratio, which gives only half the correct number of moles of Fe, and therefore half the mass. Award 1 mark for stating the correct answer is double; 1 mark for linking this to the 1:2 ratio meaning each Fe₂O₃ produces two Fe atoms.