Year 11 ChemistryModule 2📚 Final ReviewLesson 20 of 20
Module 2 Review
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.
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Choose how you work — type your answers below or write in your book.
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Choose how you work — type your answers below or write in your book.
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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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?
Type your initial thoughts below:
Record your ideas in your workbook.
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📐 Complete Formula Reference
IQ2 — Mole Concept
N = n × Nₐparticles = mol × 6.022×10²³
n = m ÷ MMmol = g ÷ (g/mol)
V = n × 22.71volume at STP — 0°C, 100 kPa (NESA)
V = n × 24.8volume at RTP (litres)
% comp = (m(element) ÷ MM) × 100percentage by mass
IQ3 — Concentration & Analysis
c = n ÷ Vmol/L; V must be in litres
n = c × Valways convert mL → L
c₁V₁ = c₂V₂dilution formula
c(unknown) = n ÷ V(flask)titration back-calc
n(standard) → ratio → n(unknown)via balanced equation
The SI unit for amount of substance; one mole contains 6.022×10²³ particles (Avogadro's number).
Molar Mass (MM)
The mass of one mole of a substance in g/mol; numerically equal to the sum of atomic masses × subscripts.
Empirical Formula
The simplest whole-number ratio of atoms in a compound. Derived from % composition via the 4-step method.
Molar Volume
Volume occupied by 1 mol of gas: 22.71 L/mol at STP (0°C, 100 kPa — NESA standard); 24.8 L/mol at RTP (25°C, 100 kPa). Note: 22.4 L/mol is the older value for 0°C and 1 atm.
Concentration (c)
Amount of solute per unit volume of solution, measured in mol/L. Calculated by c = n ÷ V (V in litres).
Primary Standard
A pure, stable substance of known composition used to prepare a standard solution of precisely known concentration.
Gravimetric Analysis
Quantitative technique using mass of a precipitate formed in a precipitation reaction to determine concentration of an analyte.
Titration
Volumetric technique where a solution of known concentration (standard) is added to an unknown until stoichiometric equivalence is reached (equivalence point/endpoint).
Concordant Titres
Titration results that agree within 0.10 mL of each other. The rough titre is discarded; concordant results are averaged.
Limiting Reagent
The reactant that is completely consumed first, stopping the reaction. Identified by comparing n ÷ coefficient — the smaller value is the limiting reagent.
Theoretical Yield
The maximum mass of product that could form if the limiting reagent reacted completely with 100% efficiency.
Percentage Yield
The ratio of actual yield to theoretical yield, expressed as a percentage. Always ≤ 100% in a real experiment.
Percentage Purity
The mass of pure substance as a percentage of the total sample mass. Applied before stoichiometry to find the mass of pure reactant available.
Equivalence Point
The point in a titration at which the moles of titrant added are stoichiometrically equivalent to the moles of analyte in the flask — i.e. exactly enough has been added for complete reaction.
📋 Common HSC Exam Question Types
Mass–Mass Stoichiometry🔥 Very common
Method: (1) Balance equation → (2) n(given) = m ÷ MM → (3) n(wanted) = n × ratio → (4) m = n × MM. Most common error: skipping the mole ratio and assuming 1:1.
Limiting Reagent🔥 Very common
Method: Convert both reactants to moles. Divide each by its coefficient. The smaller value is the limiting reagent. Calculate all products from the LR only. Common error: picking LR by smaller mass, not smaller n÷coefficient.
Titration Back-Calculation🔥 Very common
Method: n(standard) = c × V(titre) → apply ratio → n(unknown) → c(unknown) = n ÷ V(flask). Discard rough titre; average concordant. Common error: using titre volume for c(unknown) instead of flask volume.
% Yield + % Purity Combined🔥 Very common
Method: m(pure) = m(sample) × (purity ÷ 100) → n(pure) → theoretical yield → % yield = (actual ÷ theoretical) × 100. Purity is applied BEFORE stoichiometry; yield is calculated AFTER. Common error: confusing which to apply when.
Solution Stoichiometry🔥 Very common
Method: n = c × V (in litres) → ratio → n(product) → m = n × MM or c = n ÷ V(total). Common error: not converting mL to L; using incorrect volume for product concentration.
Gas Stoichiometry📊 Common
Method: Add step 0 (n = V ÷ molar volume) if gas is given, or step 4 (V = n × molar volume) if gas is the answer. Use 22.71 L/mol for STP (NESA); 24.8 L/mol for RTP. Common error: using wrong molar volume for the stated conditions.
Dilution Calculations📊 Common
Method: c₁V₁ = c₂V₂. V₁ is the volume of concentrated solution; V₂ is the total final volume. Common error: using the volume of water added instead of the total final volume for V₂.
Empirical & Molecular Formula📊 Common
Method: % → assume 100 g → m → n = m÷MM → divide by smallest → whole number ratio = empirical formula. Then: molecular formula = empirical × (MM(compound) ÷ MM(empirical)).
Interactive: Module 2 Blitz
Revisit — Think First
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, or IQ3 — 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.
✏️ Write a reflection in your workbook
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
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Wrong: Concentration and amount of solute are the same thing.
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Right: Concentration is amount per unit volume; the same amount of solute can produce different concentrations in different volumes.
Key Point
Pay close attention to reaction conditions and reagents. These determine the products formed.
MC
Multiple Choice
5 random questions from a replayable lesson bank — feedback shown immediately
📝 HSC-Style Practice Questions
HSC-Style Q1 — Stoichiometry + % Purity
ApplyBand 3
5 MARKS
A sample of iron pyrite (FeS₂) ore contains 65.0% FeS₂ by mass. The ore undergoes roasting: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂. (a) Calculate the mass of SO₂ produced from 200.0 g of the ore sample. (b) Calculate the volume of SO₂ at RTP. (Fe = 55.845, S = 32.06, O = 15.999)
A student dissolves 0.265 g of anhydrous Na₂CO₃ (MM = 105.99) in water, making it up to 250 mL. She pipettes 25.0 mL aliquots and titrates against HCl solution. Titres recorded: Rough = 20.5 mL, T1 = 22.1 mL, T2 = 22.2 mL, T3 = 22.0 mL. Equation: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (a) Calculate the average concordant titre. (b) Find [HCl].
(a) Discard rough (20.5). T1, T2, T3 are within 0.2 mL — concordant.
50.0 mL of 0.300 mol/L Pb(NO₃)₂ is mixed with 50.0 mL of 0.400 mol/L KI. Pb(NO₃)₂ + 2KI → PbI₂↓ + 2KNO₃. (a) Show which reagent is in excess. (b) Calculate the mass of PbI₂ precipitate formed. (c) Calculate the concentration of excess reagent remaining in solution. (Pb = 207.2, I = 126.90)
(a) n(Pb(NO₃)₂) = 0.300 × 0.0500 = 0.01500 mol; ÷ 1 = 0.01500n(KI) = 0.400 × 0.0500 = 0.02000 mol; ÷ 2 = 0.01000
KI has smaller quotient (0.01000 < 0.01500) → KI is the limiting reagent; Pb(NO₃)₂ is in excess(b) KI:PbI₂ = 2:1; n(PbI₂) = 0.02000 ÷ 2 = 0.01000 molMM(PbI₂) = 207.2 + 2(126.90) = 461.0; m = 0.01000 × 461.0 = 4.61 g(c) n(Pb(NO₃)₂ consumed) = n(KI) × (1÷2) = 0.02000 × 0.500 = 0.01000 moln(Pb(NO₃)₂ excess) = 0.01500 − 0.01000 = 0.00500 molV(total) = 100.0 mL = 0.100 L; c(excess) = 0.00500 ÷ 0.100 = 0.0500 mol/LMark allocation: 1 — LR identified with comparison; 1 — n(PbI₂); 1 — m(PbI₂); 1 — n(Pb(NO₃)₂ excess); 1 — c(excess) using total volume
HSC-Style Q4 — Evaluate Experimental Design
EvaluateBand 5
5 MARKS
A student attempts to determine the concentration of a sulfuric acid solution using a sodium hydroxide standard. She prepares her NaOH solution by dissolving solid NaOH in water without weighing it on an analytical balance — instead she measures approximately 4 g on a top-pan balance. She then uses this solution as her standard in a titration against the H₂SO₄. (a) Identify ONE property of a primary standard that her NaOH solution fails to meet. (b) Explain how this failure would affect the calculated concentration of H₂SO₄, and in which direction. (c) Suggest an improvement to her procedure that would give a valid result.
(a) A primary standard must have an accurately known concentration (i.e., it must be weighed precisely and have a known, stable molar mass). NaOH fails this because it is hygroscopic — it absorbs water from the atmosphere — and because it was only roughly weighed. The true mass of NaOH is not accurately known, so the concentration of the standard is uncertain.
(b) If the NaOH solution is more dilute than assumed (e.g., 4 g weighed imprecisely on a top-pan balance, or NaOH has absorbed water so its effective mass is less), then more volume of NaOH will be required to reach the endpoint. This makes the titre larger. Using the inflated titre to back-calculate c(H₂SO₄) will give a value that is too high — an overestimate of the true concentration.
(c) Replace NaOH with a primary standard such as anhydrous sodium carbonate (Na₂CO₃) or potassium hydrogen phthalate (KHP). Weigh an accurately known mass on an analytical balance (±0.0001 g) and dissolve in a volumetric flask. This gives a standard solution with a precisely known concentration to use in the titration against H₂SO₄.
Mark allocation: 1 — correct property identified (known concentration / not hygroscopic); 1 — direction of effect on titre; 1 — reasoning linking titre to c(H₂SO₄); 1 — valid improvement named; 1 — explanation of how improvement fixes the problem
HSC-Style Q5 — Design a Procedure
CreateBand 6
6 MARKS
Design a complete experimental procedure to determine the percentage purity of a commercial sample of citric acid (C₆H₈O₇, MM = 192.12). You have access to: anhydrous Na₂CO₃ (primary standard, MM = 105.99), universal indicator, standard laboratory glassware, and an analytical balance. The relevant equation is: Na₂CO₃ + C₆H₈O₇ → Na₂C₆H₆O₇ + H₂O + CO₂. Your procedure should enable you to calculate the mass of pure citric acid in the sample. (Na = 22.990, C = 12.011, H = 1.008, O = 15.999)
Step 1 — Prepare standard Na₂CO₃ solution: Weigh approximately 2.65 g Na₂CO₃ accurately on analytical balance (±0.0001 g). Dissolve in ~150 mL distilled water. Transfer quantitatively to a 250 mL volumetric flask. Make up to the mark. Calculate c(Na₂CO₃) = n ÷ 0.250.
Step 2 — Prepare citric acid solution: Weigh accurately ~1–2 g of the commercial citric acid sample on the analytical balance. Record mass precisely. Dissolve in distilled water and transfer to a clean conical flask.
Step 3 — Fill burette with Na₂CO₃ standard: Rinse burette with Na₂CO₃ solution, then fill. Check for air bubbles; record initial reading.
Step 4 — Titration: Pipette 25.0 mL of citric acid solution into a conical flask. Add 2–3 drops of indicator. Perform a rough titration, then at least two concordant titrations (within 0.10 mL). Record concordant titre average.
Step 5 — Calculation: n(Na₂CO₃) = c × V(titre) → ratio 1:1 → n(citric acid in 25 mL) → scale up to total volume → m(pure citric acid) = n × 192.12 → % purity = (m(pure) ÷ m(sample)) × 100.
Mark allocation: 1 — standard solution prepared with analytical balance + volumetric flask; 1 — burette rinsed, air bubble check; 1 — rough titre + concordant procedure described; 1 — correct titre averaged (rough excluded); 1 — correct stoichiometric pathway to m(pure); 1 — % purity formula correctly applied