Chemistry • Year 11 • Module 2 • Lesson 19
Module 2 Synthesis & Exam Practice
Lock in every formula, key term and formula-web connection from Module 2 before tackling multi-step problems in later worksheets.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: mole, molar mass, Avogadro’s number, empirical formula, molar volume, concentration, stoichiometry, limiting reagent, percentage yield, percentage purity, titration, primary standard. 12 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The SI unit for amount of substance; equal to 6.022 × 1023 entities. | |
| 1.2 | The mass in grams of one mole of a substance; numerically equal to relative formula mass. Units: g mol−1. | |
| 1.3 | The number of particles in one mole: 6.022 × 1023 mol−1. | |
| 1.4 | The simplest whole-number ratio of atoms of each element in a compound. | |
| 1.5 | The volume occupied by one mole of any ideal gas at a specified temperature and pressure; 22.71 L mol−1 at STP, 24.8 L mol−1 at RTP. | |
| 1.6 | The amount of solute (in mol) dissolved per litre of solution; units mol L−1. | |
| 1.7 | The use of mole ratios from a balanced equation to relate amounts of reactants and products. | |
| 1.8 | The reactant that is completely consumed first, controlling the maximum amount of product formed. | |
| 1.9 | (actual yield ÷ theoretical yield) × 100%; always ≤ 100% due to practical losses. | |
| 1.10 | (mass of pure substance ÷ mass of sample) × 100%; used to correct for impurities before calculating moles. | |
| 1.11 | A volumetric analysis technique in which a solution of known concentration is reacted with a solution of unknown concentration until the equivalence point is reached. | |
| 1.12 | A highly pure, stable compound of known formula used to prepare a standard solution of precisely known concentration (e.g. anhydrous Na2CO3). |
2. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 9 marks (1 per blank)
Word bank:
Avogadro · coefficients · concentration · equivalence · limiting · molar mass · moles · purity · stoichiometry
In Module 2, the central quantity is ___________, which connects mass, volume, particles and concentration through four formulas. The mass formula uses ___________ (MM) as the conversion factor. For gases, the molar volume (Vm) is based on ___________’s principle that equal volumes of gases at the same conditions contain equal numbers of molecules. The ___________ formula c = n ÷ V converts between moles and solution volume. Once moles are known, ___________ uses the ___________ of the balanced equation to find moles of any other reactant or product. If a sample is impure, the percentage ___________ must be applied before calculating moles. In a multi-reactant scenario the ___________ reagent controls the theoretical yield. In titration, the endpoint signals the ___________ point where moles of acid equal moles of base (in a 1:1 reaction).
3. True or false — with correction
Circle T or F. If false, write the corrected version on the line below. 12 marks (1 T/F + 1 correction each)
3.1 The mole is a measure of mass. T / F
3.2 One mole of any ideal gas occupies 24.8 L at STP. T / F
3.3 To dilute a solution, the equation c1V1 = c2V2 relies on the fact that the number of moles of solute is conserved during dilution. T / F
3.4 When a reactant is impure, the percentage purity should be applied after the stoichiometry step. T / F
3.5 The percentage yield can exceed 100% if the reaction is highly exothermic. T / F
3.6 In a titration back-calculation, the moles of titrant at the equivalence point equal the moles of analyte for a 1:1 reaction. T / F
4. Function recall
Answer each question in 1–2 sentences using precise Module 2 terminology. 8 marks (2 each)
4.1 What is the function of a primary standard in preparing a standard solution, and why must it be highly pure?
4.2 State the four “mole formulas” and identify which physical quantity each connects to moles.
4.3 How do you identify the limiting reagent when given amounts of two reactants and a balanced equation?
4.4 Why is the volume in the concentration formula c = n ÷ V always expressed in litres, not millilitres?
5. Build the formula web
Draw labelled arrows between the seven terms below to show how they connect via the mole concept. Each arrow must carry a formula or linking phrase (e.g. “n = m ÷ MM”, “via mole ratio”). Aim for at least 7 labelled arrows. 7 marks (1 per valid labelled arrow)
Supplied terms: moles (n) · mass (m) · particles (N) · gas volume (V) · concentration (c) · n(other substance) · empirical formula.
Q1 — Term–definition match
1.1 mole • 1.2 molar mass • 1.3 Avogadro’s number • 1.4 empirical formula • 1.5 molar volume • 1.6 concentration • 1.7 stoichiometry • 1.8 limiting reagent • 1.9 percentage yield • 1.10 percentage purity • 1.11 titration • 1.12 primary standard.
Q2 — Cloze paragraph
In order: moles / molar mass / Avogadro / concentration / stoichiometry / coefficients / purity / limiting / equivalence.
Q3 — True / false
3.1 False. The mole is a measure of amount of substance. One mole contains 6.022 × 1023 particles; mass depends on which substance (n × MM = m).
3.2 False. 24.8 L mol−1 is the molar volume at RTP (25 °C, 100 kPa). At STP (0 °C, 100 kPa) the molar volume is 22.71 L mol−1.
3.3 True.
3.4 False. Percentage purity must be applied before stoichiometry: m(pure) = m(sample) × (% purity ÷ 100). Applying it after gives the wrong number of moles.
3.5 False. Percentage yield can never exceed 100%. It equals (actual yield ÷ theoretical yield) × 100%, so it can only equal or be less than the theoretical maximum.
3.6 True.
Q4.1 — Primary standard
A primary standard is used to prepare a standard solution of precisely known concentration. It must be highly pure so that its molar mass can be used directly to calculate the exact number of moles weighed out; any impurity would mean the calculated concentration is too high, making all downstream titration results inaccurate.
Q4.2 — Four mole formulas
(1) n = m ÷ MM — connects moles to mass. (2) n = N ÷ NA — connects moles to number of particles. (3) n = V ÷ Vm — connects moles to gas volume. (4) n = c × V — connects moles to solution concentration and volume.
Q4.3 — Identifying the limiting reagent
Divide the moles of each reactant by its stoichiometric coefficient from the balanced equation. The reactant with the smallest resulting value is the limiting reagent, because it runs out first relative to its stoichiometric demand.
Q4.4 — Concentration formula requires litres
Concentration is defined as moles per litre (mol L−1). The formula c = n ÷ V requires V in litres so the units cancel correctly: mol ÷ L = mol L−1. Using millilitres without converting gives an answer 1000 times too large.
Q5 — Formula web (sample arrows)
- moles (n) — n = m ÷ MM → mass (m)
- moles (n) — n = N ÷ NA → particles (N)
- moles (n) — n = V ÷ Vm → gas volume (V)
- moles (n) — n = c × V ← concentration (c)
- moles (n) — via mole ratio (coeff B ÷ coeff A) → n(other substance)
- mass (m) — ÷ MM → simplest ratio → empirical formula
- n(other substance) — × MM → mass (m)
Award 1 mark per labelled arrow with a correct formula or phrase. Accept any logically correct direction and phrasing.