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 IQ2 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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You're here
Warm up and think first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
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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The four formulas, and the words for them
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Inquiry Question 2, 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
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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 IQ2 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
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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.
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n is the hub
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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 IQ2 calculations: N ↔ n (× or ÷ NA); m ↔ n (× or ÷ MM); V ↔ n (× or ÷ Vm, 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?
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The six most costly errors in IQ2
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The six most costly errors in IQ2
core concept
We just saw that n is the central hub connecting every IQ2 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 Vm = 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.
nNₐMMVₘ
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.
Model answer: Each of m, V (gas) and N can be converted to n using a single relationship (n = m ÷ MM, n = V ÷ Vₘ, n = N ÷ Nₐ). Because every quantity passes through n, you can step from any one of them to any other by first finding moles, making n the central pivot of every multi-step mole calculation.
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Three multi-step worked examples
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)
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.
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The three traps that cost marks
Top 3 traps · most-lost-marks
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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.
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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.
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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.
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Quick-fire drill, then revisit your thinking
Quick-fire multi-step practice · 5 reps +2 XP per reveal
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Calculate the number of O atoms in 8.00 g of sulfur dioxide (SO₂). (S = 32.06, O = 15.999)
MM(SO₂) = 32.06 + 2(15.999) = 64.058 g mol⁻¹ · n = 8.00 ÷ 64.058 = 0.1249 mol · N(SO₂) = 0.1249 × 6.022 × 10²³ = 7.52 × 10²² molecules · each has 2 O → N(O) = 1.50 × 10²³ O atoms
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A 12.4 L sample of propane (C₃H₈) gas is collected at SATP. Calculate its mass. (C = 12.011, H = 1.008)
SATP → Vₘ = 24.8 L mol⁻¹ · n = 12.4 ÷ 24.8 = 0.500 mol · MM(C₃H₈) = 3(12.011) + 8(1.008) = 44.097 g mol⁻¹ · m = 0.500 × 44.097 = 22.05 g
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A 1.984 L sample of a hydrocarbon at SATP has mass 4.49 g. Combustion shows 85.63% C and 14.37% H. Find the molecular formula.
n(C) = 85.63÷12.011 = 7.130; n(H) = 14.37÷1.008 = 14.256; H:C = 2.00 → EF = CH₂. n = 1.984÷24.8 = 0.0800 mol; MM = 4.49÷0.0800 = 56.1 g mol⁻¹. MM(CH₂) = 14.027; multiplier = 56.1÷14.027 = 4.00 → Molecular formula = C₄H₈
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How many molecules of CO₂ are in 88 g of CO₂ at any conditions? (C = 12.011, O = 15.999)
MM(CO₂) = 44.009 g mol⁻¹ · n = 88 ÷ 44.009 = 2.00 mol · N = n × Nₐ = 2.00 × 6.022 × 10²³ = 1.204 × 10²⁴ molecules
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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)
n(CO₂) = 22.0 ÷ 44.009 = 0.500 mol · same n for SO₃ · MM(SO₃) = 32.06 + 3(15.999) = 80.057 g mol⁻¹ · m = 0.500 × 80.057 = 40.0 g of SO₃
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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?
Use the Mole Marathon Stepper. A compound is 37.5% C, 12.6% H and 49.9% O by mass. What is its empirical formula?
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Practice questions
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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.
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Short answer · multi-step consolidation
ApplyLower-order3 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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ApplyMid-order4 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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AnalyseHigher-order6 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 4.49 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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EvaluateHigher-order4 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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CreateHigher-order5 marks
Q5. Design a multi-step calculation problem that requires a student to use all four IQ2 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
Particles to moles. n = 1.204 × 10²⁴ ÷ 6.022 × 10²³ = 2.00 mol.
Empirical from molecular formula. HCF of 4 and 10 is 2: C₄H₁₀ ÷ 2 = C₂H₅.
Same moles, different gas. MM(CO₂) = 44.009; n = 22.0 ÷ 44.009 = 0.500 mol; MM(SO₃) = 80.057; m = 0.500 × 80.057 = 40.0 g.
Largest gas volume. 3.0 mol He gives V = 74.4 L, the largest of the four.
Molecular formula from percentage composition. 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].
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.
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Retrieve, then finish
Check what actually stuck
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Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
