Chemistry • Year 11 • Module 3 • Lesson 5
Mole Calculations — Consolidation
Apply all four calculation pathways to real data, multi-step scenarios, and a worked-solution critique. Show full working at every step.
1. Interpret calculation data — identifying errors and selecting pathways
The table below shows five student mole calculations. Each row gives the given data and the student’s stated answer. Some answers are correct; others contain an error. 10 marks
| # | Given data | Student’s answer | Correct? (Y/N) — if N, state the error |
|---|---|---|---|
| 1.1 | 3.01 × 1023 molecules of H2O; find n. | n = 0.500 mol | |
| 1.2 | 44.8 g of N2 (MM = 28.014 g mol−1); find n. | n = 0.625 mol | |
| 1.3 | 2.00 mol of CO2 at SATP; find V. | V = 45.42 L | |
| 1.4 | Empirical formula CH2O; MM = 180.16 g mol−1; find MF. | MF = C6H12O6 | |
| 1.5 | 0.250 mol of Fe (MM = 55.845 g mol−1); find mass. | m = 13.96 g |
1.6 For any incorrect answer you identified above, show the correct working and final answer. 5 marks (max 1 per corrected row)
2. Interpret graph — moles vs mass for three substances
The graph below plots n (moles) against m (mass in grams) for three substances: iron (Fe), water (H2O), and glucose (C6H12O6). Each line starts at the origin. 8 marks
Figure 1. Amount of substance (mol) vs mass (g) for Fe, H2O and C6H12O6. Lines are n = m ÷ MM plotted from the origin.
2.1 Explain why the three lines have different gradients, and identify which line has the steepest gradient. Link your answer to the formula n = m ÷ MM. 3 marks
2.2 Use the graph to estimate the number of moles in 90 g of iron (Line A). Then verify your estimate by calculation. 3 marks
2.3 A student says: “Because glucose has a higher molar mass, the same mass of glucose contains more moles than the same mass of iron.” Assess whether this is correct and explain your reasoning using the graph. 2 marks
3. Compare the four IQ1 calculation pathways
Complete the table below. For each pathway, state the formula used, the quantities given and found, the unit of Vm if applicable, and one common error to avoid. 12 marks (1 per cell)
| Pathway | Formula (with rearrangement used) | Quantity given → quantity found | One common error to avoid |
|---|---|---|---|
| L01: Particles ↔ Moles | |||
| L02: Mass ↔ Moles | |||
| L03: EF → MF | |||
| L04: Gas volume ↔ Moles |
4. Predict and justify — a multi-step Pilbara scenario
Iron ore mined at the Pilbara, Western Australia, is mostly haematite (Fe2O3, MM = 159.69 g mol−1). A mining company says one tonne (1.00 × 106 g) of haematite contains enough iron atoms to make several steel beams.
4.1 Calculate the number of moles of Fe2O3 in one tonne of haematite. Show full working. 2 marks
4.2 Each formula unit of Fe2O3 contains 2 iron atoms. Calculate the total number of iron atoms in one tonne of haematite. 3 marks
4.3 Predict whether one tonne of aluminium oxide (Al2O3, MM = 101.96 g mol−1) contains more or fewer moles than one tonne of haematite. Justify without a full calculation. 2 marks
Q1 — Data-table — correct / incorrect
1.1 Correct. n = N ÷ NA = 3.01 × 1023 ÷ 6.022 × 1023 = 0.500 mol. ✓
1.2 Incorrect. n = 44.8 ÷ 28.014 = 1.599 mol. Student likely divided by a single N atomic mass (14.007) instead of diatomic N2 (28.014).
1.3 Incorrect. Student used STP molar volume (22.71) instead of SATP (24.8). Correct: V = 2.00 × 24.8 = 49.6 L.
1.4 Correct. MM(CH2O) = 30.026 g mol−1. n = 180.16 ÷ 30.026 = 6.00. MF = C6H12O6. ✓
1.5 Correct. m = 0.250 × 55.845 = 13.96 g. ✓
Marking criteria: 1 mark per row for correct Y/N identification (5 marks). Q1.6: 1 mark per correct working and corrected answer for each incorrect row identified (max 2 marks for rows 1.2 and 1.3).
Q2.1 — Graph gradients (3 marks)
The gradient of each line equals 1/MM (from n = m × (1/MM)). A substance with a smaller MM has a steeper gradient because each gram provides more moles [1]. Line B (H2O, MM = 18.015) has the steepest gradient [1] because it has the smallest molar mass of the three. Line C (glucose, MM = 180.16) has the shallowest gradient [1].
Q2.2 — Reading graph and verifying (3 marks)
From graph: at m = 90 g on Line A, read n ≈ 1.6 mol [1 mark for reasonable estimate, 1.5–1.7 mol accepted]. Calculation: n = 90 ÷ 55.845 = 1.612 mol [1 mark for correct formula]; consistent with graph estimate [1 mark for correct comparison].
Q2.3 — Student claim assessment (2 marks)
Incorrect [1]. A higher molar mass means fewer moles per gram, not more. Because glucose (MM = 180.16) has a much larger molar mass than iron (MM = 55.845), the same mass of glucose gives fewer moles. This is visible on the graph: Line C (glucose) has a shallower gradient than Line A (iron), meaning for any given mass, Line C gives a lower n value [1].
Q3 — Four pathway comparison table
L01: N = n × NA (rearranges to n = N ÷ NA) | Number of particles ↔ moles | Common error: treating N and n as the same quantity; forgetting to divide/multiply by NA.
L02: n = m ÷ MM (rearranges to m = n × MM or MM = m ÷ n) | Mass (g) ↔ moles | Common error: not expanding subscript brackets when calculating MM (e.g. Al2(SO4)3).
L03: n(multiplier) = MM(compound) ÷ MM(EF); MF = EF × n | % composition → EF → MF | Common error: forgetting to find the missing-element percentage (e.g. oxygen when given only C and H %).
L04: n = V ÷ Vm (rearranges to V = n × Vm) | Gas volume (L) ↔ moles | Common error: using STP value (22.71) when conditions are SATP (24.8), or forgetting to convert mL to L.
Marking: 1 mark per correctly completed cell (3 cells per row × 4 rows = 12). Accept equivalent phrasing.
Q4.1 — Moles of Fe2O3 (2 marks)
n = m ÷ MM = 1.00 × 106 ÷ 159.69 = 6261 mol [1 mark formula + substitution; 1 mark correct answer with unit]. Accept 6.26 × 103 mol.
Q4.2 — Number of iron atoms (3 marks)
N(formula units) = n × NA = 6261 × 6.022 × 1023 = 3.770 × 1027 formula units [1]. Each unit contains 2 Fe atoms [1]. N(Fe) = 2 × 3.770 × 1027 = 7.54 × 1027 Fe atoms [1]. Full precision maintained until final answer.
Q4.3 — Comparison prediction (2 marks)
One tonne of Al2O3 contains more moles than one tonne of Fe2O3 [1]. Because Al2O3 has a smaller molar mass (101.96 vs 159.69 g mol−1), dividing the same mass by a smaller MM gives a larger n (n = m ÷ MM) [1]. Accept any reasoning that correctly links smaller MM to more moles per gram.