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Module 2 · L19 of 20 ~45 min ⚡ +50 XP in Learn · +25 to complete

Module 2 Synthesis & Exam Practice

Module 2 mastery means solving a 5-step problem without losing the thread. This lesson chains all four inquiry questions together in exam-style problems, the same kind you'll face in your assessment tasks.

Today's hook, The mole (n) is the universal bridge. Every formula in Module 2 either converts something into moles or back out of moles. Master that bridge and the rest follows.

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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.

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

Recall, your gut answer first

+5 XP warm-up

A chemist reacts a 92% pure sample of Na₂CO₃ with HCl to produce CO₂ gas. If she mixes up the order of operations, applying the stoichiometry first, then correcting for purity afterwards, what would happen to her final answer? And how many distinct "steps" do you think are involved in going from a concentration of HCl to a volume of gas produced?

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The master formula web, and the words for it

Master formula web · n is the universal bridge

core formula

Master mole formula web migrated to tree layout

What you'll master

Know

Key facts

The four mole formulas (n = m/MM, N = n×Nₐ, V = n×Vₘ, c = n/V)
Purity comes BEFORE stoichiometry; yield comes AFTER
STP molar volume = 22.71 L/mol (NESA); RTP = 24.8 L/mol
The mole (n) is the universal bridge for every calculation

Understand

Concepts

Why the same calculation pathway works across all four IQs
How to track 5–7 step problems without losing the thread
When to apply purity vs % yield in chained problems

Can do

Skills

Chain concentration → stoichiometry → mass or gas volume in one problem
Apply purity before stoichiometry to find theoretical yield
Solve standard solution → titration back-calc → purity problems

Key terms

Mole concept

1 mol = 6.022 × 10²³ particles; the bridge between atomic/molecular scale and measurable laboratory quantities.

Four mole formulas

n = m/M (mass); n = V/Vₘ (gas); n = c × V (solution); n = N/Nₐ (particles); each used in different contexts.

Stoichiometry

Using the balanced equation and mole ratios to calculate masses, volumes, and concentrations of reactants and products.

Limiting reagent

The reactant consumed first; controls theoretical yield; identified by comparing mole-to-coefficient ratios.

Percentage yield

Actual yield / theoretical yield × 100%; always ≤ 100% due to practical losses.

Volumetric analysis

Determination of unknown concentration using titration against a standard solution at the equivalence point.

How the four inquiry questions connect

core concept · +3 XP at end

Module 2 is built around four inquiry questions. Each one has its own toolkit, but they share a common centre: the mole. Every formula and every calculation eventually passes through n.

1IQ1

Chemical Reactions & Stoichiometry
Mole ratios
Limiting reagent
% yield & % purity
Gas & solution stoich

2IQ2

The Mole Concept
n = m ÷ MM
N = n × Nₐ
V = n × molar vol
Empirical formula

3IQ3

Concentration & Molarity
c = n ÷ V
c₁V₁ = c₂V₂
Primary standards
Titration & gravimetric

4IQ4

Gas Laws
Boyle, Charles, Gay-Lussac
Avogadro's law
Combined gas law
Ideal gas law PV = nRT

The universal currency: the mole (n) is the central intermediate. Every formula either converts into n (from m, V, c, or N) or out of n (back to m, V, c, or N). The gas laws (IQ4) then let you find that gas volume under any temperature and pressure, not just standard conditions. Master that pivot and chained problems become bookkeeping rather than puzzles.

Module 2 has four inquiry questions: IQ1 (stoichiometry + yield/purity), IQ2 (mole formulas), IQ3 (concentration + analysis), IQ4 (gas laws: Boyle, Charles, Gay-Lussac, Avogadro, PV = nRT). The mole (n) is the universal bridge, every calculation passes through n. Core formulas: n = m ÷ MM; N = n × N_A; V = n × V_m; n = c × V; PV = nRT. In chained problems: convert known quantity → n → apply ratio → n(target) → convert back.

Pause, copy the highlighted summary into your book before moving on.

Did you get this? True or false: every Module 2 calculation passes through the moles (n) of at least one substance as an intermediate step.

Purity comes first, yield comes last

Purity comes first · yield comes last

core concept

We just saw the complete map of Module 2 formulas, with n as the central hub. That raises a question: when purity and yield are both involved in the same problem, does the order you apply them matter? This card answers it → purity must come before stoichiometry; yield comes after.

In chained problems, the order in which you apply purity and yield decides whether your answer is right or off by a fixed factor. The rule is simple:

Purity → BEFORE stoichiometry

m(pure) = m(sample) × (% purity ÷ 100). You can't calculate moles of reactant until you've worked out how much of the sample is actually the reactant.

% yield → AFTER stoichiometry

% yield = (actual ÷ theoretical) × 100. You can only compare to the theoretical yield once you've actually calculated it from the stoichiometry of the (pure) reactant.

So a complete chained problem with an impure reactant looks like: m(sample) → ×purity → m(pure) → n → ratio → n(product) → m(theoretical) → compare with m(actual) → % yield.

Purity is applied BEFORE stoichiometry: m(pure) = m(sample) × (% purity ÷ 100), use m(pure) in the moles step. % yield is applied AFTER calculating theoretical yield: % yield = (actual ÷ theoretical) × 100. Applying them in the wrong order overestimates or underestimates theoretical yield by a constant factor. Also: convert mL → L before c = n ÷ V; use V_m = 24.8 L mol⁻¹ at SATP (NESA default).

Pause, copy the highlighted rules into your book before moving on.

Quick check: A student applies purity AFTER calculating the theoretical yield from the impure sample mass. What happens to their final answer for theoretical yield?

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

purity % yield moles theoretical

Apply ___ before converting to ___. Apply ___ at the end, after you have calculated the ___ yield.

Interactive · Module 2 Mastery Marathon

try it

Chain all Module 2 skills in exam-style problems.

Three chained exam-style problems

Worked examples · chained exam-style problems

Chain Problem 1 · concentration → stoichiometry → mass (IQ2 + IQ3, 5 steps) +5 XP on full reveal

A student mixes 35.0 mL of 0.400 mol/L H₂SO₄ with excess Zn metal. The reaction produces ZnSO₄ and H₂ gas. Zn + H₂SO₄ → ZnSO₄ + H₂. (a) Calculate n(H₂SO₄). (b) Find n(H₂). (c) Calculate V(H₂) at RTP. (d) Calculate the mass of Zn consumed.

$V = 35.0\ \mathrm{mL} = 0.0350\ \mathrm{L}$; $n(\mathrm{H_2SO_4}) = 0.400 \times 0.0350 =$ 0.01400 mol

(a) c × V; convert mL → L first

Ratio $\mathrm{H_2SO_4}:\mathrm{H_2} = 1:1$ → $n(\mathrm{H_2}) =$ 0.01400 mol

(b) Apply mole ratio

$V(\mathrm{H_2}) = 0.01400 \times 24.8 =$ 0.347 L at RTP

Ratio $\mathrm{Zn}:\mathrm{H_2SO_4} = 1:1$ → $n(\mathrm{Zn}) = 0.01400\ \mathrm{mol}$; $\mathrm{MM(Zn)} = 65.38$; $m = 0.01400 \times 65.38 =$ 0.915 g

(d) Apply ratio, convert back to mass

Chain Problem 2 · purity → stoichiometry → % yield (IQ3 full integration, 6 steps) +5 XP on full reveal

An iron ore sample is 76.0% Fe₂O₃ by mass. 500 g of the ore is reduced in a blast furnace: Fe₂O₃ + 3CO → 2Fe + 3CO₂. (a) Calculate the mass of pure Fe₂O₃. (b) Calculate the theoretical yield of Fe. (Fe = 55.845, O = 15.999). (c) Actual yield = 245 g. Calculate % yield. (d) Give one reason actual yield is less than theoretical.

$m(\mathrm{Fe_2O_3}) = 500 \times 0.760 =$ 380 g

(a) Apply purity BEFORE stoichiometry

$\mathrm{MM(Fe_2O_3)} = 2(55.845) + 3(15.999) = 159.69$; $n(\mathrm{Fe_2O_3}) = 380 \div 159.69 = 2.379\ \mathrm{mol}$

(b) Convert mass → moles

Ratio $\mathrm{Fe_2O_3}:\mathrm{Fe} = 1:2$; $n(\mathrm{Fe}) = 4.758\ \mathrm{mol}$; theoretical $m(\mathrm{Fe}) = 4.758 \times 55.845 =$ 265.8 g

(b cont.) Apply ratio, convert back to mass

% yield $= (245 \div 265.8) \times 100 =$ 92.2%

(d) Any valid: incomplete reduction; product lost during collection; side reactions; impurities reacting with CO.

Real-world losses below 100% yield

Chain Problem 3 · standard solution → titration back-calc → purity (advanced, 7 steps) +5 XP on full reveal

A chemist wants to determine the purity of a commercial NaOH sample. She dissolves 2.40 g of the sample in water and makes it up to 250 mL. 25.0 mL aliquots are titrated against 0.0980 mol/L HCl. Average concordant titre = 23.6 mL. HCl + NaOH → NaCl + H₂O. (Na = 22.990, O = 15.999, H = 1.008). (a) Find c(NaOH). (b) Find m(pure NaOH) in 250 mL. (c) Calculate % purity.

$n(\mathrm{HCl}) = 0.0980 \times 0.0236 = 2.313 \times 10^{-3}\ \mathrm{mol}$; ratio 1:1 → $n(\mathrm{NaOH})$ in aliquot $= 2.313 \times 10^{-3}$

(a) n(standard) → n(unknown)

$c(\mathrm{NaOH}) = 2.313 \times 10^{-3} \div 0.0250 =$ 0.09250 mol/L

(a cont.) c = n ÷ V(flask)

$n(\mathrm{NaOH})$ in 250 mL $= 0.09250 \times 0.250 = 0.02313\ \mathrm{mol}$

(b) Scale up to full prepared volume

$\mathrm{MM(NaOH)} = 22.990 + 15.999 + 1.008 = 39.997$; $m(\mathrm{NaOH}) = 0.02313 \times 39.997 =$ 0.9250 g

(b cont.) n → m for pure NaOH

% purity $= (0.9250 \div 2.40) \times 100 =$ 38.5%

Match each Module 2 quantity to the formula or relationship that defines it.

Moles from mass
Moles from gas volume at STP (0 °C, 100 kPa)
Moles from solution concentration
Percentage yield

n = c × V (V in litres)
n = m ÷ MM
(actual ÷ theoretical) × 100
n = V ÷ 22.71 (V in litres)

Sort the calculation chain

Sort the steps given an impure reactant mass and asked for the actual mass of product (factoring in % purity and % yield), put the chain in order. Click two steps to swap them.

n(reactant) = m(pure reactant) ÷ MM(reactant)
Write the balanced equation and note the mole ratio
Apply % yield: actual m(product) = theoretical m(product) × (% yield ÷ 100)
Apply % purity FIRST: m(pure reactant) = m(sample) × (% purity ÷ 100)
Use mole ratio → n(product), then m(product) = n × MM (theoretical yield)

The three traps that cost marks

Common errors · the 3 traps that cost marks

Applying purity AFTER stoichiometry

If you use the impure sample mass to calculate moles, your n is too large, and so is your theoretical yield. The whole chain is shifted upward.

Fix: Always m(pure) = m(sample) × (purity ÷ 100) FIRST, then convert to moles.

Forgetting mL → L in concentration calculations

n = c × V requires V in litres. Using 35 instead of 0.035 gives an answer 1000× too large. This happens every year in HSC exams.

Fix: Write "V = ___ mL = ___ L" before any calculation. Force the conversion to be visible.

Skipping the balanced equation

The mole ratio lives in the coefficients. Without writing the equation, you can't extract it, and 1:1 isn't always right (H₂SO₄ + 2NaOH, Fe₂O₃ → 2Fe, Na₂CO₃ + 2HCl are all non-1:1).

Fix: Write the balanced equation at the top of every chained problem. Circle the coefficients you'll use.

Quick-fire drill, then revisit your thinking

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

An ore sample is 65.0% FeS₂ by mass. Find the mass of pure FeS₂ in 200.0 g of ore.

50.0 mL of 0.300 mol/L Ca(OH)₂ is mixed with 50.0 mL of 0.250 mol/L HCl. Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O. Identify the limiting reagent.

1.06 g of impure Na₂CO₃ (92.0% pure, MM = 105.99) reacts with excess HCl: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. Calculate V(CO₂) at STP (22.71 L/mol).

The actual mass of Fe collected is 148 g, when the theoretical yield was 178.3 g. Calculate % yield.

0.0250 mol of NaOH reacts with 0.0250 mol of HCl (1:1). Theoretical m(NaCl) = 1.46 g; actual m(NaCl) = 1.38 g. Calculate % yield.

Revisit your thinking

At the start of this lesson, you thought about what happens when purity is applied in the wrong order, and how many steps are in a chained stoichiometry problem.

Purity must be applied before stoichiometry, you need the mass of pure reactant before you can calculate moles. A full chained problem typically involves 5–7 steps: convert concentration to moles, apply purity if needed, use the mole ratio, convert moles to the required quantity (mass, volume, or concentration). Keeping track of what you know and what you need at each step is the key to solving these problems without losing the thread.

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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)?

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Practice questions

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 · exam-style

ApplyLower-order4 marks

Q1. A student dissolves 1.06 g of impure Na₂CO₃ (92.0% pure, MM = 105.99) in water and reacts it with excess HCl: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (a) Find the mass of pure Na₂CO₃ in the sample. (b) Calculate the theoretical volume of CO₂ at STP.

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AnalyseMid-order5 marks

Q2. 30.0 mL of 0.300 mol/L Ca(OH)₂ is mixed with 50.0 mL of 0.250 mol/L HCl: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O. (a) Identify the limiting reagent. (b) Calculate the mass of CaCl₂ formed. (Ca = 40.078, Cl = 35.453)

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AnalyseMid-order5 marks

Q3. An ore sample (85.0% Fe₂O₃ by mass) is reacted with excess CO in a blast furnace: Fe₂O₃ + 3CO → 2Fe + 3CO₂. The actual mass of Fe collected from 300.0 g of the ore is 148 g. (a) Calculate the theoretical yield of Fe. (b) Calculate the percentage yield. (Fe = 55.845, O = 15.999)

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EvaluateHigher-order5 marks

Q4. A chemist calculates that 0.0250 mol of NaOH should react with 0.0250 mol of HCl (1:1 ratio). After mixing 25.0 mL of 1.00 mol/L NaOH with 25.0 mL of 1.00 mol/L HCl, she evaporates the solution and finds 1.38 g of NaCl instead of the theoretical 1.46 g. (a) Calculate the percentage yield of NaCl. (b) The chemist suggests two possible explanations: (i) some NaCl was lost when evaporating the solution, or (ii) the HCl solution was slightly less concentrated than 1.00 mol/L. For each explanation, identify whether it represents a random or systematic error and predict the effect on all future trials. (Na = 22.990, Cl = 35.453)

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📖 Comprehensive answers (click to reveal)

Multiple choice, drill bank

MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.

Short answer model answers

Q1 (4 marks):

(a) m(pure Na₂CO₃) = 1.06 × 0.920 = 0.9752 g.

(b) n(Na₂CO₃) = 0.9752 ÷ 105.99 = 9.201×10⁻³ mol; ratio 1:1; n(CO₂) = 9.201×10⁻³. V(CO₂) = 9.201×10⁻³ × 22.71 = 0.209 L at STP.

Q2 (5 marks):

(a) n(Ca(OH)₂) = 0.300 × 0.0300 = 9.00×10⁻³; ÷ 1 = 9.00×10⁻³. n(HCl) = 0.250 × 0.0500 = 1.25×10⁻²; ÷ 2 = 6.25×10⁻³. HCl has smaller n÷coeff (6.25×10⁻³ < 9.00×10⁻³) → HCl is the limiting reagent.

(b) Ratio HCl:CaCl₂ = 2:1; n(CaCl₂) = 1.25×10⁻² ÷ 2 = 6.25×10⁻³ mol. MM(CaCl₂) = 110.98; m = 6.25×10⁻³ × 110.98 = 0.694 g.

Q3 (5 marks):

(a) m(Fe₂O₃) = 300.0 × 0.850 = 255.0 g. MM(Fe₂O₃) = 2(55.845) + 3(15.999) = 159.69; n(Fe₂O₃) = 255.0 ÷ 159.69 = 1.597 mol. Ratio Fe₂O₃:Fe = 1:2; n(Fe) = 3.193 mol; theoretical m(Fe) = 3.193 × 55.845 = 178.3 g.

(b) % yield = (148 ÷ 178.3) × 100 = 83.0%.

Q4 (5 marks):

(a) MM(NaCl) = 22.990 + 35.453 = 58.443; theoretical m = 0.0250 × 58.443 = 1.461 g. % yield = (1.38 ÷ 1.461) × 100 = 94.5%.

(b) Explanation (i), NaCl lost during evaporation: this is a random error if the amount lost varies between trials. It could also be systematic if the student consistently loses the same amount (e.g., always stops heating too early). It would generally give a yield below 100% in every trial, with the magnitude varying randomly.

Explanation (ii), HCl less concentrated than 1.00 mol/L: this is a systematic error. Every trial uses the same solution; every trial would have the same deficit in HCl moles. The theoretical yield calculated from 0.0250 mol would be wrong in the same way every time, the % yield would be consistently below the true value in a predictable, reproducible way.

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