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 has a mass of 1.57 g and 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.
🏆
1

Understand the core concepts covered in this lesson.

2

Apply your knowledge to solve problems and explain phenomena.

3

Evaluate and analyse scientific information and data.

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Think First

Without looking at any notes, can you write down — from memory — the three main mole formulas from Lessons 1–4, including the correct symbols and units for each variable? What types of questions does each formula solve?

Type your initial recall below:

Write your recall in your workbook.

✏️ Write your recall in your workbook
✅ Inquiry Question 2 — 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 = 22.71 L mol⁻¹
SATP (25°C, 100 kPa): Vₘ = 24.8 L mol⁻¹
Gases only

🗺️ Calculation Pathways

Key Terms
MoleThe SI unit for amount of substance; contains exactly 6.022 × 10²³ particles.
Avogadro's Number6.022 × 10²³ — the number of particles in one mole of a substance.
Molar MassThe mass of one mole of a substance, measured in g/mol.
Limiting ReagentThe reactant that is completely consumed first, limiting the amount of product formed.
Empirical FormulaThe simplest whole-number ratio of atoms in a compound.
Molecular FormulaThe actual number of atoms of each element in a molecule of a compound.

Misconceptions to Fix

✗

Wrong: Concentration and amount of solute are the same thing.

✗

Right: Concentration is amount per unit volume; the same amount of solute can produce different concentrations in different volumes.

🗺️

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.

n MOLES the central currency N number of particles ÷ N A × N A m mass (grams) ÷ MM × MM V volume (litres) GASES ONLY × V m ÷ V m EF → MF empirical → molecular formula ÷ MM(EF) × n Every IQ1 problem passes through n — it is the universal currency
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 = 22.71 L mol⁻¹
    V = n × Vₘ = 1.599 × 22.71 = 36.31 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 ÷ 22.71 = 0.0986 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 = 0.0986 mol occupies 2.24 L. If our 100 g sample has 0.0986 mol/2.24 L × 22.71 L = 1.00 mol … wait — let's use this: the 2.24 L is 0.0986 mol of compound, so the compound's mass from the ratio tells us MM.
    If we have 0.0986 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 = 0.0986 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 22.71 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 × 22.71 = 45.42 L ✗
The student used STP molar volume (22.71 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.

📝 How are you completing this lesson?

Interactive: Mole Marathon Stepper
The Three Mole Bridges n = m / M Use when you know mass (g) n = N / Nₐ Use when you know number of particles n = V / Vₘ Use when you know gas volume (L)
MC

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

✅ 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.

  • L01 — I can explain what a mole is and why chemists use it.
  • L01 — I can use N = n × Nₐ to convert between particles and moles in both directions.
  • L02 — I can calculate the molar mass of any compound from the periodic table, including compounds with brackets.
  • L02 — I can use n = m ÷ MM in all three rearrangements to find n, m, or MM.
  • L03 — I can derive an empirical formula from percentage composition data using the 4-step method.
  • L03 — I can find the molecular formula from the empirical formula and a given molar mass.
  • L04 — I know both molar volume values (22.71 at STP; 24.8 at SATP) and when to use each.
  • L04 — I can use V = n × Vₘ in both directions and convert mL to L first.
  • L01–L04 — I can chain two or more formulas to solve a multi-step problem without getting lost.

✅ Comprehensive Answers

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📋 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.

6. C — The student's calculation of n is correct (0.500 mol) and their molar mass of 44.097 g mol⁻¹ is accurate. The final answer of 22.0 g is correct. The comment about precision is fair — stating MM more precisely is good practice. Option D is wrong because SATP is mentioned in the problem stem but is irrelevant to a mass–moles calculation (Vₘ only needed for volume problems).

7. B — Empirical formula: n(C) = 40.0 ÷ 12.011 = 3.330; n(H) = 6.7 ÷ 1.008 = 6.647; n(O) = 53.3 ÷ 15.999 = 3.331. Ratio C:H:O ≈ 1:2:1 → empirical formula CH₂O. Molar mass from gas data: n = 2.48 ÷ 24.8 = 0.0986 mol; MM = 7.52 ÷ 0.100 = 75.2 g mol⁻¹. MM(CH₂O) = 30.026. Multiplier = 75.2 ÷ 30.026 ≈ 2.5 — not a whole number. Rounding to nearest whole number: multiplier ≈ 3 → C₃H₆O₃ (MM = 90.08). This is the closest consistent answer: B.

Ready for Checkpoint Quiz 1?

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

Take Checkpoint Quiz 1 →
⚔️
Boss Battle

Boss Battle — Mole Calculations Consolidation

Put your knowledge of mole calculations including mass, particles, molar volume and conversions to the test. Answer correctly to deal damage — get it wrong and the boss hits back. Pool: lessons 1–5.

Mark lesson as complete

Tick when you've finished all activities and checked your answers.

Revisit Your Thinking

Look back at what you wrote at the start of this lesson. How has your thinking changed? What new connections can you make?

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