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Module 2 · L20 of 20 ~15 min ⚡ +15 XP in Learn · +25 to complete

Module 2 Review

You'll be able to solve multi-step problems that chain several inquiry questions.

The complete reference for Module 2, Introduction to Quantitative Chemistry. Every formula, every key term, every exam question type, and worked examples of each. Use this as your study reference before the module quiz and the HSC.

Today's hook, The complete reference for Module 2, Introduction to Quantitative Chemistry. Every formula, every key term, every exam question type, and worked examples of each. Use this as your study reference before the module quiz and the HSC.
0/5QUESTS
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Warm up and recall the four formulas

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
01
Recall, your gut answer first
+5 XP warm-up

Without looking at any notes, can you list the four key formulas that connect moles (n) to mass, particles, gas volume, and concentration? And which single concept acts as the "universal bridge" that links every calculation in this module?

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What you should be able to do, and the words for it

03
What you'll master
Know

Key facts

  • The four Module 2 inquiry questions and what each covers
  • Core formulas: n = m ÷ MM, N = n × Nₐ, V = n × Vₘ, n = c × V, PV = nRT
  • STP = 22.71 L/mol; SATP = 24.8 L/mol; concordant titres within 0.10 mL
Understand

Concepts

  • How the mole links mass, particles, gas volume and concentration
  • When to apply purity, percentage yield, limiting reagent and the gas laws
  • Why a true yield is ≤ 100% and an apparent value above 100% signals error
Can do

Skills

  • Solve multi-step problems that chain several inquiry questions
  • Choose the correct molar volume and convert temperatures to kelvin
  • Check answers for physical plausibility (yield ≤ 100%, concordant titres)
04
Key terms
Mole (mol)
SI unit of amount of substance; 1 mol = 6.022 × 10²³ elementary entities (Avogadro's constant).
Molar mass (M)
Mass per mole of a substance in g mol⁻¹; numerically equal to the relative atomic or molecular mass.
Concentration (c)
Amount of solute per unit volume; c = n/V in mol L⁻¹; ppm and ppb used for trace concentrations.
Stoichiometric ratio
The ratio of moles of substances from the balanced equation; used to convert between moles of different species.
Limiting reagent
Determines the maximum theoretical yield; identified by the smallest mole-to-coefficient ratio.
Gravimetric and volumetric analysis
Two quantitative techniques: gravimetric uses mass of precipitate; volumetric uses volume of titrant at the equivalence point.
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The three traps that cost marks

Common errors · the 3 traps that cost marks

1

Misconception to fix

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

2

Misconception to fix

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

3

Applying purity and yield at the wrong stage

Percentage purity must be applied to the impure sample mass before converting to moles, while percentage yield is applied only at the very end (to the theoretical product). Students often calculate moles from the full impure mass, or multiply by yield too early, which makes every downstream value wrong.

Fix: Follow the order: impure mass → pure mass (× % purity) → moles → mole ratio → theoretical product → × % yield. Purity first, yield last.

4

Two chain problems, then a quick-fire drill

Worked examples · two chain problems +5 XP on full reveal

Chain Problem 1 · limiting reagent → gas volume → percentage yield (IQ1 + IQ3, 5 steps) +5 XP on full reveal

5.00 g of zinc is added to 50.0 mL of 1.20 mol/L hydrochloric acid. Zn + 2HCl → ZnCl₂ + H₂. (a) Calculate n(Zn) and n(HCl). (b) Identify the limiting reagent. (c) Calculate the theoretical volume of H₂ at SATP. (d) A student collects 0.380 L. Calculate the percentage yield.

1
$n(\mathrm{Zn}) = 5.00 \div 65.38 =$ 0.0765 mol
(a) n = m ÷ MM. MM(Zn) = 65.38
2
$n(\mathrm{HCl}) = 1.20 \times 0.0500 =$ 0.0600 mol
(a) n = c × V. Convert 50.0 mL to 0.0500 L first
3
$2 \times 0.0765 = 0.153 \gt 0.0600$
(b) Zn would need 0.153 mol of HCl to react fully, but only 0.0600 mol is present, so HCl is the limiting reagent. Compare what each reagent WOULD need against what is actually there — never assume the smaller number of moles is limiting, the ratio decides
4
$\mathrm{HCl}:\mathrm{H_2} = 2:1$ → $n(\mathrm{H_2}) = 0.0600 \div 2 = 0.0300\ \mathrm{mol}$; $V = 0.0300 \times 24.8 =$ 0.744 L
(c) Work from the LIMITING reagent only, then V = n × Vₘ at SATP
5
$(0.380 \div 0.744) \times 100 =$ 51.1%
(d) Percentage yield is actual ÷ theoretical × 100. The theoretical value is the one from step 4, not one worked from the excess reagent
Step 1 of 5
Chain Problem 2 · concordant titres → stoichiometry → concentration (IQ3, 5 steps) +5 XP on full reveal

A NaOH solution is standardised against 25.00 mL aliquots of 0.0500 mol/L H₂SO₄. Four titres are recorded: 24.15, 23.80, 23.85 and 23.82 mL. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. (a) Select the concordant titres and calculate the mean. (b) Calculate n(H₂SO₄) in one aliquot. (c) Calculate n(NaOH) delivered. (d) Calculate c(NaOH).

1
$23.85 - 23.80 = 0.05 \lt 0.10$
(a) Concordant means within 0.10 mL of each other, and these three span 0.05 mL. The 24.15 mL titre is the rough one: discard it and do not average it in
2
$\bar{V} = (23.80 + 23.85 + 23.82) \div 3 =$ 23.82 mL
(a) Mean of the concordant titres only — averaging all four would drag the answer high
3
$n(\mathrm{H_2SO_4}) = 0.0500 \times 0.02500 =$ 1.250 × 10⁻³ mol
(b) n = c × V for the aliquot in the flask, the known solution
4
$\mathrm{H_2SO_4}:\mathrm{NaOH} = 1:2$ → $n(\mathrm{NaOH}) = 2 \times 1.250 \times 10^{-3} =$ 2.500 × 10⁻³ mol
(c) Diprotic acid, so two moles of base per mole of acid. This is the single most common place marks are lost
5
$c(\mathrm{NaOH}) = (2.500 \times 10^{-3}) \div 0.02382 =$ 0.105 mol/L
(d) c = n ÷ V, using the mean titre in litres as the volume of base delivered
Step 1 of 5

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

1

Calculate the mass of 0.25 mol of calcium carbonate, CaCO₃. (M = 100.09)

2

A solution contains 0.40 mol of NaCl dissolved in 250 mL. Calculate its concentration in mol L⁻¹.

3

A 3.0 L sample of gas at 100 kPa is compressed at constant temperature until the pressure is 150 kPa. Find the new volume.

4

How many molecules are present in 8.0 g of methane, CH₄? (M = 16.0)

5

Complete combustion CH₄ + 2O₂ → CO₂ + 2H₂O: what volume of CO₂ (25 °C, 100 kPa) forms from 0.50 mol of CH₄? (Vₘ = 24.79 L mol⁻¹)

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5

Revisit your thinking, then check the calculator

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Revisit your thinking

At the start of this lesson, you tried to recall the four key formulas and identify the concept that bridges all of Module 2.

The mole (n) is the universal bridge, every formula in Module 2 converts something into moles or from moles: n = m ÷ MM (mass), N = n × Nₐ (particles), V = n × molar volume (gas), and n = c × V (solution). Every calculation, whether IQ1, IQ2, IQ3 or IQ4 (the gas laws, where n appears in PV = nRT), passes through n as the central intermediate step. If you can automatically and accurately perform each of these four conversions, the rest of every stoichiometry or concentration problem follows.

Reflect: how did your initial thinking compare to what you've learned?

Write a reflection in your workbook.

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Interactive Tool, Module 2 Blitz Open fullscreen ↗
Use the Module 2 Blitz. Magnesium burns in oxygen: 2Mg + O2 → 2MgO. If 0.40 mol of Mg reacts completely, how many moles of MgO form?
✓

Practice questions

01
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.

02
Short answer
✓

Retrieve, then finish

Check what actually stuck
01
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.

Start the module quiz →

Mark lesson as complete

Tick when you've finished the practice and review.

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