Quantitative Chemistry: Exam Practice

This lesson brings together all the key concepts from the Quantitative Chemistry topic in the AQA GCSE Chemistry specification (4.3). It provides a comprehensive review of the essential knowledge, common question types, and exam strategies you need to succeed. Use this lesson to consolidate your understanding and prepare effectively for your exams.

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Topic Overview

Quantitative chemistry covers the following areas in the AQA specification:

Specification Section	Topics Covered
4.3.1	Conservation of mass, relative formula mass, the mole [H], amounts in equations
4.3.2	Reacting masses, limiting reactants [H], percentage yield [H], atom economy [H]
4.3.3	Concentration of solutions, moles and gas volumes [H], titration calculations [H]

Items marked [H] are Higher Tier only.

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Key Formulae Summary

All the essential formulae you need for quantitative chemistry:

Formula	What It Calculates
Mr = sum of Ar values	Relative formula mass
moles = mass / Mr [H]	Number of moles from mass
mass = moles x Mr [H]	Mass from number of moles
Number of particles = moles x 6.02 x 10^23 [H]	Number of particles from moles
concentration (g/dm3) = mass / volume	Concentration in grams per dm3
concentration (mol/dm3) = moles / volume [H]	Concentration in moles per dm3
moles = concentration x volume [H]	Moles of solute in solution
volume (dm3) = moles x 24 [H]	Volume of gas at RTP
percentage yield = (actual / theoretical) x 100 [H]	Percentage yield
atom economy = (Mr desired / total Mr all products) x 100 [H]	Atom economy

Exam Tip: You are not given a formula sheet in the AQA Chemistry exam. You must memorise all the key formulae. Write them on flash cards and test yourself regularly. Being fluent with formulae saves time and reduces errors in the exam.

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Common Question Types

graph TD
    A["Quantitative Chemistry Questions"] --> B["Mr Calculations"]
    A --> C["Reacting Masses"]
    A --> D["Mole Calculations H"]
    A --> E["Concentration"]
    A --> F["Gas Volumes H"]
    A --> G["Percentage Yield H"]
    A --> H["Atom Economy H"]
    A --> I["Titrations H"]
    A --> J["Conservation of Mass"]

    style A fill:#2c3e50,color:#fff
    style B fill:#3498db,color:#fff
    style C fill:#3498db,color:#fff
    style D fill:#e67e22,color:#fff
    style E fill:#3498db,color:#fff
    style F fill:#e67e22,color:#fff
    style G fill:#e67e22,color:#fff
    style H fill:#e67e22,color:#fff
    style I fill:#e67e22,color:#fff
    style J fill:#3498db,color:#fff

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Practice Question 1: Relative Formula Mass

Calculate the relative formula mass (Mr) of magnesium nitrate, Mg(NO3)2.

Model Answer:

• Mg(NO3)2 contains: 1 Mg, 2 N, 6 O
• Mr = (1 x 24) + (2 x 14) + (6 x 16)
• Mr = 24 + 28 + 96
• Mr = 148

Key points: Show each element separately. Multiply atoms inside brackets by the subscript outside.

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Practice Question 2: Reacting Masses

Iron oxide reacts with carbon monoxide: Fe2O3 + 3CO --> 2Fe + 3CO2. Calculate the mass of iron produced from 80 g of iron oxide (Fe2O3).

Model Answer:

Step 1: Mr of Fe2O3 = (2 x 56) + (3 x 16) = 112 + 48 = 160

Step 2: moles of Fe2O3 = 80 / 160 = 0.5 mol

Step 3: From the equation, Fe2O3 : Fe = 1 : 2 So moles of Fe = 0.5 x 2 = 1.0 mol

Step 4: mass of Fe = 1.0 x 56 = 56 g

Exam Tip: In reacting mass questions, always start by writing the balanced equation and identifying the mole ratio. Then follow the steps: Mr --> moles of known --> mole ratio --> moles of unknown --> mass of unknown. Show every step for full marks.

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Practice Question 3: Concentration

5.85 g of sodium chloride is dissolved in 250 cm3 of water. Calculate the concentration in g/dm3.

Model Answer:

Step 1: Convert volume: 250 / 1000 = 0.25 dm3

Step 2: concentration = mass / volume = 5.85 / 0.25 = 23.4 g/dm3

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Practice Question 4: Moles and Gas Volumes [H]

What volume of hydrogen gas (in cm3) is produced when 1.3 g of zinc reacts with excess hydrochloric acid at RTP?

Balanced equation: Zn + 2HCl --> ZnCl2 + H2

Model Answer:

Step 1: Mr of Zn = 65. moles of Zn = 1.3 / 65 = 0.02 mol

Step 2: From the equation, Zn : H2 = 1 : 1. So moles of H2 = 0.02 mol

Step 3: Volume = 0.02 x 24,000 = 480 cm3

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Practice Question 5: Percentage Yield [H]

A student reacted 10 g of calcium carbonate with excess hydrochloric acid. They obtained 4.0 g of carbon dioxide. Calculate the percentage yield.

CaCO3 + 2HCl --> CaCl2 + H2O + CO2

Model Answer:

Step 1: Calculate theoretical yield of CO2.

• moles of CaCO3 = 10 / 100 = 0.1 mol
• From equation, CaCO3 : CO2 = 1 : 1, so moles of CO2 = 0.1 mol
• Theoretical mass of CO2 = 0.1 x 44 = 4.4 g

Step 2: percentage yield = (4.0 / 4.4) x 100 = 90.9%

Exam Tip: For percentage yield, always calculate the theoretical yield first (from the balanced equation), then apply the percentage yield formula. Never skip the theoretical yield step.

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Practice Question 6: Atom Economy [H]

Calculate the atom economy for the production of iron in the thermite reaction: Fe2O3 + 2Al --> 2Fe + Al2O3

Model Answer:

• Mr of desired product (2Fe) = 2 x 56 = 112
• Total Mr of all products = 112 + (2 x 27 + 3 x 16) = 112 + 102 = 214
• Atom economy = (112 / 214) x 100 = 52.3%

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Practice Question 7: Titration [H]

25.0 cm3 of potassium hydroxide (KOH) was neutralised by 22.5 cm3 of 0.10 mol/dm3 hydrochloric acid. Calculate the concentration of KOH in mol/dm3.

KOH + HCl --> KCl + H2O

Model Answer:

Step 1: moles of HCl = 0.10 x (22.5 / 1000) = 0.00225 mol

Step 2: Ratio KOH : HCl = 1 : 1, so moles of KOH = 0.00225 mol

Step 3: concentration of KOH = 0.00225 / (25.0 / 1000) = 0.00225 / 0.025 = 0.09 mol/dm3

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Practice Question 8: Limiting Reactant [H]

2.4 g of magnesium is added to 50 cm3 of 1.0 mol/dm3 sulfuric acid. Identify the limiting reactant and calculate the mass of magnesium sulfate produced.

Mg + H2SO4 --> MgSO4 + H2

Model Answer:

Step 1: moles of Mg = 2.4 / 24 = 0.1 mol moles of H2SO4 = 1.0 x (50 / 1000) = 0.05 mol

Step 2: From the equation, Mg : H2SO4 = 1 : 1 0.1 mol Mg needs 0.1 mol H2SO4, but only 0.05 mol is available.

H2SO4 is the limiting reactant.

Step 3: moles of MgSO4 = 0.05 mol (based on limiting reactant) Mr of MgSO4 = 24 + 32 + (4 x 16) = 120 mass of MgSO4 = 0.05 x 120 = 6.0 g

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Exam Strategy Checklist

Follow this checklist for every quantitative chemistry calculation:

Step	Action
1	Read the question twice — identify what you are given and what you need to find
2	Write the balanced equation (if not given, balance it yourself)
3	Identify the mole ratio from the equation
4	Calculate Mr for all relevant substances
5	Show all working — one step at a time
6	Convert units if needed (cm3 to dm3, g to kg)
7	State your answer with correct units
8	Check your answer — is it reasonable?

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Common Mistakes to Avoid