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Module 2 · L5 of 20 · ★ Consolidation ~35 min ⚡ +50 XP in Learn · +25 to complete

Mole Calculations — Consolidation

No new content today. Instead: a challenge. A 2.24 L sample of gas at STP, mass 1.57 g, contains only carbon and hydrogen in a 3:1 mass ratio. Find the molecular formula. This problem requires L01–L04 working together.

Today's hook — Every IQ1 problem passes through n. Master the connections between N, m, V and EF/MF and no multi-step problem can stop you.

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Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Recall — your gut answer first

+5 XP warm-up

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?

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Inquiry Question 1 — complete formula reference (L01–L04)

all four formulas

Formula

Variables

Conditions / notes

L01

N = n × Nₐ

N = particles · n = mol · Nₐ = 6.022 × 10²³ mol⁻¹

Rearranges to n = N ÷ Nₐ · atoms, molecules, ions, formula units

L02

n = m ÷ MM

n = mol · m = g · MM = g mol⁻¹

m = n × MM · MM = m ÷ n · MM from periodic table

L03

MF = EF × n

n = MM(MF) ÷ MM(EF) · whole number

4-step: % → g → mol → ratio → simplify

L04

V = n × Vₘ

V = L · n = mol · Vₘ = L mol⁻¹

STP: Vₘ = 22.71 L mol⁻¹ · SATP: Vₘ = 24.8 L mol⁻¹ · gases only

What you'll master

Know

Key facts

The four mole-calculation pathways (N↔n, m↔n, EF/MF, V↔n)
Correct formula and units for each pathway
n is the universal currency for every IQ1 problem

Understand

Concepts

How to select the right pathway for a multi-step problem
How to chain two or more formulas together
Why intermediate values should not be rounded

Can do

Skills

Solve multi-step mole problems that require two or more conversions
Move fluently between particles, mass, volume and formula
Self-assess readiness for Checkpoint Quiz 1

Key terms

Calculation pathway

A sequence of formula rearrangements used to convert a given quantity into a required quantity via moles.

Multi-step problem

A problem requiring two or more conversions; always pass through moles (n) as the central unit.

Intermediate value

A calculated result (e.g. moles) that is not the final answer but is needed for the next step.

Significant figures

The number of meaningful digits in a measurement; do not round intermediate values — only round the final answer.

Unit analysis

Checking that units cancel correctly at each step to confirm you used the right formula and rearrangement.

Formula selection

Identifying which of N=nNₐ, n=m/M, n=V/Vₘ, or MF=n×EF applies based on the quantities given and required.

Connecting the formulas — n is the hub

core concept · +3 XP at end

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.

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?"

The mole is the universal hub of all IQ1 calculations: N ↔ n (× or ÷ N_A); m ↔ n (× or ÷ MM); V ↔ n (× or ÷ V_m, gases only); EF ↔ MF (× or ÷ multiplier). Every multi-step problem passes through n — never skip the middle step.

Pause — copy the highlighted hub diagram into your book before moving on.

Did you get this? True or false: in a multi-step problem that converts mass to volume of a gas, you must pass through moles (n) in the middle.

Quick check: You're given the mass of a gas and asked for the number of molecules. Which two formulas do you need in order?

The six most costly errors in IQ1

core concept

We just saw that n is the central hub connecting every IQ1 formula. That raises a question: even knowing the pathways, what mistakes do students make most often in the HSC? This card answers it → six specific, examinable errors to recognise and avoid.

These are the errors that cost students marks in the HSC exam — ranked by how often they appear.

1. Using 22.71 when conditions are SATP (25 °C)

The student used STP molar volume (22.71 L mol⁻¹) but the question specifies SATP. At higher T, gas expands — the correct volume is larger. Fix: Use V = 2.0 × 24.8 = 49.6 L. Underline the temperature given in the question before picking a Vₘ value. Default for NSW HSC: 24.8 unless explicitly told 0 °C or STP.

2. Not expanding brackets when calculating MM

For Al₂(SO₄)₃, the 3 outside the bracket multiplies both S and O inside. There are 3 sulfate groups → 3 S and 12 O. Fix: Expand to 2 Al + 3 S + 12 O. MM = 2(26.98) + 3(32.06) + 12(16.00) = 342.14 g mol⁻¹.

3. Rounding too early in a multi-step calculation

Rounding n = 15 ÷ 18.02 to 0.83 mol then 0.8 mol gives V = 19.84 L. Carrying full precision (0.8327) gives V = 20.65 L. Fix: Keep all decimals in your calculator throughout. Only round the final answer to an appropriate sig fig count.

4. Forgetting to find the missing % element

A compound labelled "40% C and 6.7% H" almost certainly contains oxygen too: % O = 100 − 40 − 6.7 = 53.3%. Ignoring it produces a completely wrong empirical formula. Fix: Always add the given percentages first.

5. Confusing N (particles) with n (moles)

"How many molecules in 2.0 mol of H₂O?" requires N = n × Nₐ = 2.0 × 6.022 × 10²³ = 1.204 × 10²⁴, not 2.0. Fix: Ask "moles or particles?" and label your answer accordingly.

6. Not multiplying every subscript when scaling EF → MF

Empirical CH₂ × multiplier 4 → C₄H₈, not CH₈. Every atom (including implicit subscript 1) is multiplied. Fix: Write C₁H₂ explicitly before multiplying.

Key rules to avoid exam errors: default V_m = 24.8 L mol⁻¹ (SATP) unless told otherwise; always expand brackets in MM calculations; never round intermediate values — only the final answer; if % composition doesn't sum to 100%, calculate the remainder as oxygen.

Add the highlighted rules to your notes before the check below.

Fill the blanks: drag each token into the matching blank.

n Nₐ MM Vₘ

Moles (___) is the central currency. From mass, divide by ___. From volume of gas, divide by ___. From particles, divide by ___.

Did you get this? True or false: in a 3-step multi-step problem, rounding to 3 sig figs after step 1 is fine because it makes the numbers easier to read.

Lock-in task: In one or two sentences, explain why n (moles) acts as the "hub" linking mass, volume of gas, and number of particles in mole calculations.

Multi-step worked examples · reveal as you go

Worked example 1 · particles → moles → mass +5 XP on full reveal

A sample contains 1.806 × 10²⁴ molecules of ammonia (NH₃). Calculate the mass of this sample. (N = 14.007, H = 1.008)

n = N ÷ Nₐ = 1.806 × 10²⁴ ÷ 6.022 × 10²³ = 3.00 mol

Convert particles → moles · L01

MM(NH₃) = 14.007 + 3(1.008) = 17.031 g mol⁻¹

Calculate molar mass · L02

m = n × MM = 3.00 × 17.031 = 51.09 g ✓

Convert moles → mass · L02

Worked example 2 · mass → moles → volume of gas +5 XP on full reveal

What volume does 44.8 g of nitrogen gas (N₂) occupy at STP? (N = 14.007)

MM(N₂) = 2 × 14.007 = 28.014 g mol⁻¹

Calculate molar mass · L02

n = m ÷ MM = 44.8 ÷ 28.014 = 1.599 mol

Convert mass → moles · L02

STP → Vₘ = 22.71 L mol⁻¹

Pick the right Vₘ · L04

V = n × Vₘ = 1.599 × 22.71 = 36.3 L at STP ✓

Convert moles → volume · L04

Worked example 3 · opening challenge (uses L01+L02+L03+L04) +5 XP on full reveal

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.

n = V ÷ Vₘ = 2.24 ÷ 22.71 = 0.0986 mol

Find moles of gas from V at STP · L04

Treat C:H mass ratio 3:1 as 75 g C and 25 g H per 100 g sample

Set up empirical-formula calculation · L03

n(C) = 75 ÷ 12.011 = 6.245 mol; n(H) = 25 ÷ 1.008 = 24.80 mol; ratio H/C = 3.97 ≈ 4 → EF = CH₄

Derive empirical formula · L03

MM(CH₄) = 12.011 + 4(1.008) = 16.043 g mol⁻¹

MM of empirical unit · L02

Multiplier = MM(compound) ÷ MM(CH₄) = 16.04 ÷ 16.04 = 1 → Molecular formula: CH₄ (methane) ✓

Scale EF → MF · L03

Sort the steps+7 XP

Question: 44.8 g of N₂ at STP — find its volume. Put the multi-step solution in order.

Select Vₘ for STP conditions: Vₘ = 22.71 L mol⁻¹.
Calculate the molar mass: MM(N₂) = 2 × 14.007 = 28.014 g mol⁻¹.
Apply V = n × Vₘ = 1.599 × 22.71 = 36.3 L at STP.
Convert mass to moles using n = m ÷ MM = 44.8 ÷ 28.014 = 1.599 mol.

Top 3 traps · most-lost-marks

Using the wrong Vₘ for the conditions

Using 22.71 L mol⁻¹ at SATP (or vice versa) gives the wrong answer even with perfect method. NSW HSC default is 24.8 L mol⁻¹ unless STP / 0 °C is stated.

Fix: Underline conditions before substituting. Confirm Vₘ matches the stated temperature.

Rounding intermediate values

Rounding to 3 sig figs at each step compounds error. In a 3-step problem this can shift the final answer by 5 % or more.

Fix: Carry full precision through every step; round only the final value to the question's sig-fig demand.

Mixing up N and n

Capital N is the raw count (e.g. 6.0 × 10²³). Lowercase n is moles (e.g. 1.0 mol). Asking "how many molecules" needs N; asking "how many moles" needs n.

Fix: Re-read the question and circle moles or particles before answering.

Work mode · how are you completing this lesson?

Quick-fire multi-step practice · 5 reps +2 XP per reveal

Calculate the number of O atoms in 8.00 g of sulfur dioxide (SO₂). (S = 32.06, O = 15.999)

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

A 1.984 L sample of a hydrocarbon at SATP has mass 4.29 g. Combustion shows 85.63% C and 14.37% H. Find the molecular formula.

How many molecules of CO₂ are in 88 g of CO₂ at any conditions? (C = 12.011, O = 15.999)

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

Revisit your thinking

Look back at what you wrote at the start of this lesson. How has your thinking changed? Can you now move fluently between particles, mass, volume and formula via moles?

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Interactive Tool — Stoichiometry Calculator Open fullscreen ↗

Use the Stoichiometry Calculator. How many moles are in 44 g of CO₂ (molar mass = 44 g/mol)?

Multiple choice

+2 XP per correct · +5 bonus if perfect

Pick your answer, then rate your confidence — that tells the system what to drill next.

Short answer · multi-step consolidation

ApplyBand 33 marks

Q1. Calculate the number of moles in 1.204 × 10²⁴ molecules of water (H₂O). Then state the mass of water this corresponds to. (H = 1.008, O = 15.999)

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ApplyBand 44 marks

Q2. A 12.4 L sample of propane gas (C₃H₈) is collected at SATP. Calculate (a) the moles of propane, (b) the mass of propane, and (c) the total number of H atoms present. (C = 12.011, H = 1.008)

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AnalyseBand 56 marks

Q3. An unknown hydrocarbon contains 85.63% C and 14.37% H by mass. A 1.984 L sample of the gas at SATP has a mass of 4.29 g. (a) Determine the empirical formula. (b) Use the gas data to determine the molar mass of the compound. (c) Determine the molecular formula. (C = 12.011, H = 1.008)

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EvaluateBand 54 marks

Q4. A student claims: "1 mol of helium at SATP and 1 mol of methane (CH₄) at SATP have the same mass because they have the same volume." Evaluate this claim and correct any errors. (He = 4.003, C = 12.011, H = 1.008)

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CreateBand 65 marks

Q5. Design a multi-step calculation problem that requires a student to use all four IQ1 formulas (N = n × Nₐ; n = m ÷ MM; V = n × Vₘ; MF = EF × n). Write the problem clearly, then solve it showing the full pathway. State the conditions you choose and the molar volume value used.

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📖 Comprehensive answers (click to reveal)

Multiple choice — drill bank

1. C — n = 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; MM(SO₃) = 80.057; m = 0.500 × 80.057 = 40.0 g.

4. A — 3.0 mol He gives V = 74.4 L, the largest of the four.

5. C — Empirical from %: 2:4 = CO₂; multiplier = 132.07 ÷ 44.009 = 3.00 → C₃O₆.

Short answer model answers

Q1 (3 marks): n = 1.204 × 10²⁴ ÷ 6.022 × 10²³ = 2.00 mol [1]. MM(H₂O) = 2(1.008) + 15.999 = 18.015 g mol⁻¹ [1]. m = 2.00 × 18.015 = 36.03 g [1].

Q2 (4 marks): (a) SATP → Vₘ = 24.8 L mol⁻¹; n = 12.4 ÷ 24.8 = 0.500 mol [1]. (b) MM(C₃H₈) = 3(12.011) + 8(1.008) = 44.097 g mol⁻¹; m = 0.500 × 44.097 = 22.05 g [1]. (c) N(C₃H₈) = 0.500 × 6.022 × 10²³ = 3.011 × 10²³; each has 8 H atoms → N(H) = 2.409 × 10²⁴ H atoms [2].

Q3 (6 marks): (a) Assume 100 g: n(C) = 85.63 ÷ 12.011 = 7.130; n(H) = 14.37 ÷ 1.008 = 14.256; H:C = 2.00 → EF = CH₂ [2]. (b) SATP: n = 1.984 ÷ 24.8 = 0.0800 mol; MM = 4.29 ÷ 0.0800 = 53.6 g mol⁻¹ [2]. (c) MM(CH₂) = 14.027; multiplier = 53.6 ÷ 14.027 = 3.82 ≈ 4 → Molecular formula = C₄H₈ [2].

Q4 (4 marks): Same volume claim is correct — 1 mol of any ideal gas at SATP occupies 24.8 L [1]. Same mass claim is incorrect: mass = n × MM, and MMs differ. m(He) = 1 × 4.003 = 4.003 g; m(CH₄) = 1 × 16.043 = 16.043 g [2]. Same number of molecules, different masses because individual molecule masses differ [1].

Q5 (5 marks): Mark using a rubric: 1 mark for a clear problem that requires all four formulas; 1 mark for correctly using each formula (4 marks total) with correct conditions and units throughout.

Boss battle · mole calculations consolidation

earn bronze · silver · gold

Five timed multi-step problems drawn from L01–L05. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).

⚔ Enter the arena

Science Jump · consolidation pool

arcade practice

Climb platforms, hit checkpoints, and answer multi-step questions drawn from the L01–L05 pool.

Mark lesson as complete

Tick when you've finished the practice and review.

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