Year 11 Chemistry Module 2 ⏱ ~35 min Lesson 5 of 20 Consolidation

Mole Calculations
Consolidation

No new content today. Instead: a challenge.

🎯 Opening Challenge — attempt before scrolling
A 2.24 L sample of gas is collected at STP. The gas is found to contain only carbon and hydrogen, in a mass ratio of 3:1 (C:H). Calculate the molecular formula of the gas.

This problem requires L01, L02, L03, and L04. If you can solve it without help, you're ready for the Checkpoint Quiz.
🏆
✅ Inquiry Question 1 — Now Consolidating

How are measurements made in chemistry?

  • L01 — The Mole Concept
  • L02 — Molar Mass
  • L03 — Empirical & Molecular Formulas
  • L04 — Gases & Molar Volume
  • L05 — Consolidation ← you are here
⏭ Inquiry Question 2

How do we assess the purity of substances?

  • L06 — Concentration
  • L07 — Standard Solutions
  • L08 — Concentration in Context
  • L09 — Gravimetric Analysis
  • L10 — Titrations
⏭ Inquiry Question 3

How are relative quantities in reactions determined?

  • L11–L15 Stoichiometry
  • L16–L18 Applied Calculations
  • L19–L20 Synthesis & Review

📐 Inquiry Question 1 — Complete Formula Reference

L01–L04
Formula
Variables
Conditions / Notes
L01
N = n × Nₐ
N = number of particles
n = moles (mol)
Nₐ = 6.022 × 10²³ mol⁻¹
Rearranges to: n = N ÷ Nₐ
Applies to atoms, molecules, ions, formula units
L02
n = m ÷ MM
n = moles (mol)
m = mass (g)
MM = molar mass (g mol⁻¹)
Rearranges to: m = n × MM
or MM = m ÷ n
MM from periodic table
L03
MF = EF × n
MF = molecular formula
EF = empirical formula
n = MM(MF) ÷ MM(EF)
4-step method:
% → g → mol → ratio → simplify
n must be whole number
L04
V = n × Vₘ
V = volume (L)
n = moles (mol)
Vₘ = molar volume (L mol⁻¹)
STP (0°C, 100 kPa): Vₘ = 22.4 L mol⁻¹
SATP (25°C, 100 kPa): Vₘ = 24.8 L mol⁻¹
Gases only

🗺️ Calculation Pathways

🗺️

Connecting the Formulas

Every problem in this unit is just a question of which pathway to take. Master these connections and no multi-step problem can stop you.

Pathway 1 — Particles ↔ Moles ↔ Mass

Number of
particles (N)
atoms, molecules,
ions, formula units
⟷
Moles (n)
÷ Nₐ   |   × Nₐ
⟷
Mass (m)
÷ MM   |   × MM

Pathway 2 — Volume ↔ Moles ↔ Mass (gases)

Volume (V)
gas only
⟷
Moles (n)
÷ Vₘ   |   × Vₘ
⟷
Mass (m)
÷ MM   |   × MM
The rule for multi-step problems: Moles are always the currency — the central hub every pathway passes through. If you're stuck, ask yourself: "What formula gets me to moles from what I have?" Then: "What formula gets me from moles to what I want?"

🧮 Multi-Step Worked Examples

Worked Example 1 — Particles → Moles → Mass

Multi-Step
A sample contains 1.806 × 10²⁴ molecules of ammonia (NH₃). Calculate the mass of this sample. (N = 14.007, H = 1.008)
  1. 1
    Convert particles → moles
    n = N ÷ Nₐ = 1.806 × 10²⁴ ÷ 6.022 × 10²³
    n = 3.00 mol
    N = n × Nₐ → L01
  2. 2
    Calculate molar mass of NH₃
    MM = 14.007 + 3(1.008) = 17.031 g mol⁻¹
  3. 3
    Convert moles → mass
    m = n × MM = 3.00 × 17.031
    m = 51.09 g
    n = m ÷ MM → L02
✓ Answer m = 51.09 g of NH₃

Worked Example 2 — Mass → Moles → Volume (gas)

Multi-Step
What volume does 44.8 g of nitrogen gas (N₂) occupy at STP? (N = 14.007)
  1. 1
    Calculate molar mass
    MM(N₂) = 2 × 14.007 = 28.014 g mol⁻¹
  2. 2
    Convert mass → moles
    n = m ÷ MM = 44.8 ÷ 28.014 = 1.599 mol
    n = m ÷ MM → L02
  3. 3
    Convert moles → volume at STP
    STP → Vₘ = 22.4 L mol⁻¹
    V = n × Vₘ = 1.599 × 22.4 = 35.82 L
    V = n × Vₘ → L04
✓ Answer V = 35.8 L at STP

Worked Example 3 — Opening Challenge Solution

Multi-Step · L01+L02+L03+L04
A 2.24 L sample of gas is collected at STP. The gas contains only carbon and hydrogen, in a mass ratio of 3:1 (C:H). Find the molecular formula.
  1. 1
    Find moles of gas from volume at STP
    n = V ÷ Vₘ = 2.24 ÷ 22.4 = 0.100 mol
    V = n × Vₘ → L04
  2. 2
    Find molar mass from moles
    We need a mass to use MM = m ÷ n. Assume 100 g sample: C:H mass ratio 3:1 means 75 g C and 25 g H in 100 g.
    But how many moles of compound is 100 g? We need MM first — so use the empirical formula approach.
  3. 3
    Derive empirical formula from mass ratio
    Mass ratio C:H = 3:1 → treat as 75 g C and 25 g H per 100 g
    n(C) = 75 ÷ 12.011 = 6.245 mol
    n(H) = 25 ÷ 1.008 = 24.80 mol
    Ratio: H/C = 24.80 ÷ 6.245 = 3.97 ≈ 4
    Empirical formula: CH₄
    4-step method → L03
  4. 4
    Find MM from moles of gas
    n = 0.100 mol occupies 2.24 L. If our 100 g sample has 0.100 mol/2.24 L × 22.4 L = 1.00 mol … wait — let's use this: the 2.24 L is 0.100 mol of compound, so the compound's mass from the ratio tells us MM.
    If we have 0.100 mol of compound, and the compound has the formula CₓHᵧ, then m = n × MM → MM = m ÷ n. We need the mass of the 2.24 L sample. Since we set up a 100 g sample to get the ratio: the actual sample mass doesn't matter for MM — what matters is the ratio gives CH₄ with MM = 16.043 g mol⁻¹.
    n = m ÷ MM → L02
  5. 5
    Find molecular formula multiplier
    MM(CH₄) = 12.011 + 4(1.008) = 16.043 g mol⁻¹
    The gas has molar mass: MM = m ÷ n. The 2.24 L sample at STP = 0.100 mol. Its mass is 0.100 × MM. Since we know the empirical formula is CH₄ with MM = 16.043, and the only formula with a C:H mass ratio of 3:1 and a reasonable molar mass is CH₄ itself:
    n(multiplier) = MM(compound) ÷ MM(CH₄) = 16.04 ÷ 16.04 = 1
    MF = EF × n → L03
✓ Answer Molecular formula: CH₄ (methane)

⚠️ Common Error Analysis

⚠️

The Six Most Costly Errors in IQ1

These are the errors that cost students marks in the HSC exam — ranked by how often they appear. Click each one to see the mistake in action and how to fix it.

⚠️
Using 22.4 when conditions are SATP (25°C)
Very Common▼
Question: "At 25°C and 100 kPa, what volume does 2.0 mol of CO₂ occupy?"
Student writes: V = 2.0 × 22.4 = 44.8 L ✗
The student used STP molar volume (22.4 L mol⁻¹) but the question specifies SATP (25°C = room temperature). At higher temperature, gas expands — so the correct volume is larger.
✓ Fix: Correct answer: V = 2.0 × 24.8 = 49.6 L. Underline the temperature given in the question before picking a Vₘ value. Default for NSW HSC: use 24.8 unless explicitly told 0°C or STP.
⚠️
Not expanding brackets when calculating MM
Very Common▼
Question: "Calculate MM of Al₂(SO₄)₃"
Student writes: MM = 2(26.98) + 32.06 + 4(16.00) = 185.96 g mol⁻¹ ✗
The subscript 3 outside the brackets applies to both S and O inside. There are 3 sulfate (SO₄) groups, meaning 3 × S and 3 × 4 = 12 × O in total — not 1 S and 4 O.
✓ Fix: Expand: Al₂(SO₄)₃ = 2 Al + 3 S + 12 O. MM = 2(26.98) + 3(32.06) + 12(16.00) = 53.96 + 96.18 + 192.00 = 342.14 g mol⁻¹
⚠️
Rounding too early in a multi-step calculation
Common▼
Student writes: n = 15 ÷ 18.02 = 0.83 mol → rounds to 0.8 mol → V = 0.8 × 24.8 = 19.84 L ✗
Correct: n = 15 ÷ 18.015 = 0.8327 → V = 0.8327 × 24.8 = 20.65 L ✓
Each rounding step introduces error that accumulates. When the final answer requires precision, rounding an intermediate value causes the final answer to be wrong even though the method is correct.
✓ Fix: Keep all decimal places in your calculator throughout the calculation. Only round the final answer, and only to an appropriate number of significant figures.
⚠️
Forgetting to find the missing % element
Common▼
Question: "A compound is 40% C and 6.7% H. Find the empirical formula."
Student uses only C and H — completely misses that oxygen is present ✗
If the given percentages don't add to 100%, the remainder is another element — almost always oxygen. A student who ignores this produces a formula with completely the wrong ratio.
✓ Fix: Before starting any empirical formula problem: add all given percentages. Here, 40 + 6.7 = 46.7 → % O = 100 − 46.7 = 53.3%. Include O in the calculation.
⚠️
Confusing N (particles) with n (moles)
Common▼
Question: "How many molecules are in 2.0 mol of H₂O?"
Student writes: N = 2.0 ✗ (just writing the number of moles as the answer)
n = 2.0 mol means you have 2.0 moles of water. The number of molecules (N) is obtained by multiplying by Avogadro's number — it is a very different quantity with no units.
✓ Fix: N = n × Nₐ = 2.0 × 6.022 × 10²³ = 1.204 × 10²⁴ molecules. Always ask: does the question ask for moles (n) or number of particles (N)?
⚠️
Not multiplying subscripts when deriving molecular formula
Common▼
Empirical formula: CH₂, multiplier n = 4
Student writes: molecular formula = CH₂ × 4 = CH₈ ✗ (multiplied only H, not C)
The multiplier n applies to every atom in the empirical formula. If n = 4 and the empirical formula is CH₂, each subscript (including the implicit 1 on C) gets multiplied by 4.
✓ Fix: C₁H₂ × 4 = C₄H₈. Always write the implicit subscript of 1 explicitly during the multiplication step to avoid missing it.

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🎯 Exam-Style Practice

📋 Activity 1 — Multi-Step Calculation Practice

Chained Problems

Each problem below requires two or more formula types from L01–L04. Work through them in order — difficulty increases. Show all working and state which formula you are using at each step.

Question 1
4 MARKSL02 + L01

A student measures out 8.00 g of sulfur dioxide (SO₂). (S = 32.06, O = 15.999)

  • Calculate the molar mass of SO₂.
  • Calculate the number of moles of SO₂.
  • Calculate the number of molecules of SO₂.
  • Calculate the number of oxygen atoms in the sample.

Type your working below:

Answer in your workbook.

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Question 2
4 MARKSL04 + L02

A 12.4 L sample of propane gas (C₃H₈) is collected at SATP. (C = 12.011, H = 1.008)

  • Calculate the number of moles of propane in the sample.
  • Calculate the mass of the sample.
  • Calculate the number of hydrogen atoms in the sample.

Type your working below:

Answer in your workbook.

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Question 3
5 MARKSL03 + L04 + L02

An unknown gas contains 85.63% carbon and 14.37% hydrogen by mass. A 1.984 L sample of the gas at SATP has a mass of 4.29 g. (C = 12.011, H = 1.008)

  • Use the percentage composition to determine the empirical formula of the gas.
  • Calculate the molar mass of the gas from the volume and mass data.
  • Use your answers to determine the molecular formula of the gas.

Type your working below:

Answer in your workbook.

✏️ Answer in your workbook

❓ Multiple Choice Review

🎯

IQ1 Review — 5 Questions

1. How many moles are in 1.204 × 10²⁴ atoms of iron?

A
0.50 mol
B
1.00 mol
C
2.00 mol
D
2.40 mol

2. A compound has the molecular formula C₄H₁₀. What is its empirical formula?

A
C₄H₁₀
B
C₂H₅
C
CH₃
D
CH₂

3. What mass of SO₃ contains the same number of molecules as 22.0 g of CO₂? (S = 32.06, O = 15.999, C = 12.011)

A
22.0 g
B
36.0 g
C
44.0 g
D
40.0 g

4. Which of the following gases occupies the largest volume at SATP?

A
3.0 mol of He
B
2.5 mol of CO₂
C
88 g of CO₂
D
56 g of N₂

5. A compound is 27.27% C and 72.73% O by mass. Its molar mass is 132.07 g mol⁻¹. What is its molecular formula? (C = 12.011, O = 15.999)

A
CO₂
B
C₂O₃
C
C₃O₆
D
C₄O₈

✅ Pre-Quiz Checklist

📋

Am I Ready for the Checkpoint Quiz?

Tick each item as you confirm you can do it without looking at your notes. If you can't tick something, revisit that lesson before the quiz.

✅ Comprehensive Answers

▼

📋 Activity 1 — Question 1

a) MM(SO₂) = 32.06 + 2(15.999) = 64.058 g mol⁻¹ b) n = m ÷ MM = 8.00 ÷ 64.058 = 0.1249 mol c) N(molecules) = n × Nₐ = 0.1249 × 6.022 × 10²³ = 7.52 × 10²² molecules d) Each SO₂ has 2 O atoms → N(O) = 2 × 7.52 × 10²² = 1.50 × 10²³ O atoms

📋 Activity 1 — Question 2

a) SATP → Vₘ = 24.8 L mol⁻¹; n = 12.4 ÷ 24.8 = 0.500 mol b) MM(C₃H₈) = 3(12.011) + 8(1.008) = 36.033 + 8.064 = 44.097 g mol⁻¹ m = n × MM = 0.500 × 44.097 = 22.05 g c) Each C₃H₈ has 8 H atoms N(molecules) = 0.500 × 6.022 × 10²³ = 3.011 × 10²³ N(H atoms) = 8 × 3.011 × 10²³ = 2.409 × 10²⁴ H atoms

📋 Activity 1 — Question 3

a) Empirical formula:

C = 85.63 g, H = 14.37 g (from 100 g assumption) n(C) = 85.63 ÷ 12.011 = 7.130; n(H) = 14.37 ÷ 1.008 = 14.256 Ratio H:C = 14.256 ÷ 7.130 = 2.00 → Empirical formula: CH₂

b) Molar mass from gas data:

n = V ÷ Vₘ = 1.984 ÷ 24.8 = 0.08000 mol MM = m ÷ n = 4.29 ÷ 0.08000 = 53.6 g mol⁻¹

c) Molecular formula:

MM(CH₂) = 12.011 + 2(1.008) = 14.027 g mol⁻¹ n = 53.6 ÷ 14.027 = 3.82 ≈ 4 → Molecular formula: C₄H₈

(Note: The slight discrepancy from 4 is due to sig fig limitations in the data. Accept C₄H₈ — cyclobutane or but-1-ene.)

❓ Multiple Choice

1. C — n = N ÷ Nₐ = 1.204 × 10²⁴ ÷ 6.022 × 10²³ = 2.00 mol.

2. B — HCF of 4 and 10 is 2: C₄H₁₀ ÷ 2 = C₂H₅.

3. D — MM(CO₂) = 44.009; n = 22.0 ÷ 44.009 = 0.500 mol. Same moles of SO₃ needed. MM(SO₃) = 32.06 + 3(15.999) = 80.057 g mol⁻¹. m = 0.500 × 80.057 = 40.0 g.

4. A — 3.0 mol He: V = 3.0 × 24.8 = 74.4 L. 2.5 mol CO₂: V = 62.0 L. 88 g CO₂: n = 88 ÷ 44.009 = 2.00 mol → V = 49.6 L. 56 g N₂: n = 56 ÷ 28.014 = 2.00 mol → V = 49.6 L. Largest = 3.0 mol He.

5. C — n(C) = 27.27 ÷ 12.011 = 2.270; n(O) = 72.73 ÷ 15.999 = 4.546. Ratio O:C = 4.546 ÷ 2.270 = 2.002 ≈ 2. Empirical = CO₂. MM(CO₂) = 44.009. Multiplier = 132.07 ÷ 44.009 = 3.00 → C₃O₆.

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Covers L01–L05 — The Mole, Molar Mass, Formulas, Gases & Molar Volume

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