← AQA GCSE Chemistry/Topic 3

FROM PARTICLES TO NUMBERS.

Quantitative Chemistry

Make sense of masses, moles and the calculations behind chemical reactions.

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Quick reference: key definitions
Key Definitions

Topic Core Terms

Relative formula mass (Mr)
The sum of the relative atomic masses of all atoms in a formula.
Mole
The amount of substance containing 6.02 × 10²³ particles (Avogadro's constant).
Concentration
The amount of solute dissolved in a given volume of solution (g/dm³ or mol/dm³).
Limiting reactant
The reactant that is completely used up first and determines the amount of product formed.
Percentage yield
The actual yield as a percentage of the theoretical (maximum) yield.
Atom economy
The percentage of reactant atoms that form the desired (useful) product.
CHAPTER 01 · 3 sectionsMass and measurementConservation, uncertainty and formula mass

Conservation of Mass

The law of conservation of mass states that no atoms are lost or made during a chemical reaction. The total mass of the products equals the total mass of reactants.

This is because the same atoms are present before and after the reaction: they have just been rearranged. This is why we must balance chemical equations.

Conservation of mass: a level balance with carbon and oxygen on one pan and carbon dioxide on the other Carbon reacts with oxygen to make carbon dioxide. The left pan holds one carbon atom and one oxygen molecule, 12 g plus 32 g. The right pan holds one carbon dioxide molecule, 44 g. Both sides contain one carbon atom and two oxygen atoms, so both sides have a mass of 44 g and the balance stays level. C + O₂ → CO₂ REACTANTS PRODUCTS C + O O O C O C + O₂ CO₂ 1 C atom · 2 O atoms 1 C atom · 2 O atoms 12 g + 32 g = 44 g 44 g =

The same atoms exist before and after: mass is always conserved

Key Concept

Key Fact

Atoms are neither created nor destroyed in a chemical reaction: only rearranged into different substances.

Conservation of Mass and Relative Formula Mass (Mᵣ)

Because mass is conserved, in any balanced chemical equation, the sum of the Mᵣ of all reactants equals the sum of the Mᵣ of all products (taking stoichiometry numbers into account).

Example: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

  • Reactants Mᵣ sum: Mᵣ(CH₄) + 2 × Mᵣ(O₂) = 16 + (2 × 32) = 80
  • Products Mᵣ sum: Mᵣ(CO₂) + 2 × Mᵣ(H₂O) = 44 + (2 × 18) = 80
Worked Example

Deducing Balancing Coefficients from Reacting Masses [HT Only]

8.1 g of zinc (Aᵣ = 65) reacts with 2.0 g of oxygen gas (Mᵣ = 32) to form 10.1 g of zinc oxide (Mᵣ = 81). Deduce the balanced equation.

Step 1 (Moles): Calculate moles of each substance:

Moles of Zn = 8.1 / 65 = 0.125 mol

Moles of O₂ = 2.0 / 32 = 0.0625 mol

Moles of ZnO = 10.1 / 81 = 0.125 mol

Step 2 (Simplest Whole Number Ratio): Divide all moles by the smallest mole value (0.0625):

Zn: 0.125 / 0.0625 = 2 | O₂: 0.0625 / 0.0625 = 1 | ZnO: 0.125 / 0.0625 = 2

Balanced Equation: 2Zn(s) + O₂(g) → 2ZnO(s)

Apparent Changes in Mass

Sometimes, the mass of a reaction vessel appears to change:

  • If a gas escapes (e.g., CO₂ from a thermal decomposition), mass appears to decrease.
  • If a gas is gained from the air (e.g., oxygen during oxidation of metals), mass appears to increase.

In reality, if the system were sealed, the total mass would remain unchanged.

Chemical Measurements & Uncertainty

Whenever a measurement is made, there is always some uncertainty about the result obtained. No measurement is perfectly accurate; all instruments have limitations and all humans introduce small errors when reading scales.

Improving the Quality of Measurements

To increase confidence in results, scientists:

  • Repeat measurements and calculate a mean (average).
  • Identify and exclude anomalous results (outliers): values that don't fit the pattern. These are usually caused by human error or equipment malfunction.
  • Use the range of the measurements about the mean as a measure of uncertainty.
Key Concept

Key Fact

Uncertainty = range ÷ 2. The range is the difference between the highest and lowest values in a set of repeat measurements (after removing anomalies).

Worked Example

A student records three titre values: 24.50, 24.60, and 24.40 cm³. Calculate the uncertainty.

Step 1: Mean = (24.50 + 24.60 + 24.40) ÷ 3 = 24.50 cm³

Step 2: Range = 24.60 - 24.40 = 0.20 cm³

Step 3: Uncertainty = 0.20 ÷ 2 = ± 0.10 cm³

Result: 24.50 ± 0.10 cm³

Common Mistake

Examiner Trap

If a question asks you to "estimate the uncertainty," calculate half the range. A smaller uncertainty means the results are more precise (close together). Don't confuse precision with accuracy: accurate means close to the true value.

Relative Formula Mass (Mr)

The relative formula mass of a compound is the sum of the relative atomic masses of all the atoms in its formula. Ar values are found on the periodic table.

Worked Example

Calculate the Mr of calcium carbonate (CaCO₃)

Ca = 40, C = 12, O = 16

Mr = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100

Worked Example

Calculate the Mr of magnesium hydroxide (Mg(OH)₂)

Mg = 24, O = 16, H = 1

Mr = 24 + 2 × (16 + 1) = 24 + 34 = 58

Common Mistake

Examiner Trap

Watch out for brackets! In Mg(OH)₂, the subscript 2 applies to both the O and the H inside the brackets, so there are 2 oxygen atoms and 2 hydrogen atoms.

Worked Example

Calculate the Mr of aluminium sulfate, Al₂(SO₄)₃

Al = 27, S = 32, O = 16

There are: 2 Al, 3 S, 12 O (3 × 4 = 12)

Mr = (2 × 27) + (3 × 32) + (12 × 16) = 54 + 96 + 192 = 342

Exam Tip

Exam Technique

For formulae like Al₂(SO₄)₃, multiply everything inside the brackets by the subscript outside. Here the 3 applies to one S and four O atoms, giving 3 S and 12 O.

CHAPTER 02 · 3 sectionsMoles and reacting amountsCount particles and use chemical equations

Moles & Avogadro's Constant

A mole is simply a number: 6.02 × 10²³ (Avogadro's constant). One mole of any substance contains exactly this number of particles (atoms, molecules, or ions).

moles = mass ÷ Mr

The Moles Triangle

Moles formula triangle: mass = moles × Mr mass (g) moles (mol) × Mr (no units)
mass = mol × Mr
mol = mass ÷ Mr
Mr = mass ÷ mol
Worked Example

How many moles in 11 g of CO₂?

Mr of CO₂ = 12 + (2 × 16) = 44

Moles = 11 ÷ 44 = 0.25 mol

Worked Example

What mass is 3 moles of water (H₂O)?

Mr of H₂O = (2 × 1) + 16 = 18

Mass = mol × Mr = 3 × 18 = 54 g

Exam Tip

Exam Technique

This triangle is your best friend for calculations. Cover what you want to find, and the remaining two values show you what to do. If they are side by side, multiply. If one is on top, divide.

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Check your mole calculations instantly with our Moles Calculator - converts between mass, moles, and molar mass.

Reacting Masses

Use a balanced equation plus the moles formula to predict masses used or produced in reactions. Follow these three steps every time:

Step-by-step process flowchart for reacting mass calculations 1 2 3 g → mol 2 : 2 e.g. 2Mg : 2MgO mol → g Find moles of the known substance moles = mass ÷ Mr Use the mole ratio from the balanced equation Convert to mass (or gas volume) mass = moles × Mr
Worked Example

What mass of magnesium oxide is produced from 6 g of magnesium?

Equation: 2Mg(s) + O₂(g) → 2MgO(s)

Step 1: Moles of Mg = 6 ÷ 24 = 0.25 mol

Step 2: Ratio: 2Mg : 2MgO = 1:1, so moles of MgO = 0.25 mol

Step 3: Mass of MgO = 0.25 × 40 = 10 g

Worked Example

What mass of iron is produced when 80 g of iron(III) oxide is reduced?

Equation: Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g)

Step 1: Mr of Fe₂O₃ = (2 × 56) + (3 × 16) = 160.
Moles = 80 ÷ 160 = 0.5 mol

Step 2: Ratio: 1 Fe₂O₃ : 2 Fe, so moles of Fe = 0.5 × 2 = 1.0 mol

Step 3: Mass of Fe = 1.0 × 56 = 56 g

Exam Tip

Exam Technique

Always check the mole ratio carefully - it is NOT always 1:1! Read the balanced equation coefficients to find the correct ratio.

Limiting Reactants

In many reactions, one reactant is used up before the others. This reactant is called the limiting reactant: it limits the amount of product that can be formed.

The other reactant(s) are said to be in excess.

Diagram showing limiting reactants representation where one of the red particles is crossed out,…

Limiting

Used up first

Diagram showing excess reactants representation where three green particles remain, indicating there are…

Excess

Left over

Key Concept

Key Fact

The amount of product formed is always determined by the limiting reactant, not the one in excess.

Worked Example

4.8 g of Mg reacts with 14.6 g of HCl. Which is the limiting reactant?

Equation: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Step 1: Moles of Mg = 4.8 ÷ 24 = 0.2 mol

Step 2: Moles of HCl = 14.6 ÷ 36.5 = 0.4 mol

Step 3: The ratio is 1 Mg : 2 HCl. So 0.2 mol Mg needs 0.2 × 2 = 0.4 mol HCl.

Result: We have exactly 0.4 mol HCl - both reactants run out at the same time. Neither is in excess.

If we only had 0.3 mol HCl, then HCl would be limiting (not enough to react with all the Mg).

Exam Tip

Exam Technique

To identify the limiting reactant: (1) calculate moles of each reactant, (2) use the balanced equation ratios to compare, (3) the one that runs out first is the limiting reactant.

CHAPTER 03 · 3 sectionsSolutions and efficiencyConcentration, yield and atom economy

Concentration

Concentration tells you how much solute is dissolved in a given volume of solution.

Comparison of dilute and concentrated solutions Dilute low concentration few solute particles Concentrated high concentration many solute particles solute particle (same volume of solution in each beaker)
Dilute vs concentrated solutions at the same total volume.
concentration (g/dm³) = mass of solute (g) ÷ volume (dm³)

The Concentration Triangle

Concentration formula triangle: mass = concentration × volume mass (g) concentration (g/dm³) × volume (dm³)
mass = conc × vol
conc = mass ÷ vol
vol = mass ÷ conc
Key Concept

Key Fact

Remember: 1 dm³ = 1000 cm³. To convert cm³ to dm³, divide by 1000.

Worked Example

2.5 g of NaOH dissolved in 500 cm³. Find concentration in g/dm³.

Volume = 500 ÷ 1000 = 0.5 dm³

Concentration = 2.5 ÷ 0.5 = 5 g/dm³

Concentration in mol/dm³

concentration (mol/dm³) = moles of solute ÷ volume (dm³)
Worked Example

4 g of NaOH is dissolved in water and made up to 250 cm³ of solution. Find the concentration in mol/dm³.

Step 1: Mr of NaOH = 23 + 16 + 1 = 40

Step 2: Moles = 4 ÷ 40 = 0.1 mol

Step 3: Volume = 250 ÷ 1000 = 0.25 dm³

Step 4: Concentration = 0.1 ÷ 0.25 = 0.4 mol/dm³

Exam Tip

Exam Technique

If a question gives you mass in grams and asks for mol/dm³, you must first convert mass to moles using mol = mass ÷ Mr, then divide by the volume in dm³. Always convert cm³ to dm³ first!

🧪

Need to work through a titration calculation? Use our Titration Calculator for step-by-step working with automatic unit conversion.

Percentage Yield Chemistry Only

The percentage yield compares the actual amount of product obtained to the theoretical maximum (calculated from stoichiometry).

% yield = (actual yield ÷ theoretical yield) × 100

Three main reasons why yields are always less than 100%:

  • The reaction is reversible and does not go to completion.
  • Some product is lost during transfer (e.g., filtration, evaporation).
  • Side reactions produce unwanted by-products.
Worked Example

A student calculates they should make 10 g of copper sulfate. They actually collect 7.5 g. What is the percentage yield?

% yield = (7.5 ÷ 10) × 100 = 75%

Atom Economy Chemistry Only

Atom economy measures the proportion of reactant atoms that become useful product. It evaluates the efficiency of the reaction pathway itself.

atom economy = [(stoichiometric coefficient × Mr of desired product) ÷ total Mr of all reactants from balanced equation] × 100
Addition vs Substitution Reactions Addition reaction Substitution reaction + desired product 100% all atoms end up in one product + + desired waste less than 100% a waste by-product also forms atom economy = (Mᵣ of desired product ÷ total Mᵣ of all reactants) × 100
High atom economy means fewer waste products and maximum efficiency.

High atom economy is desirable. It means less waste, lower costs, and a more sustainable process.

Yield vs Atom Economy

Feature % Yield Atom Economy
Measures Lab performance Pathway efficiency
Needs experiment? Yes No (equation only)
Type of waste Spills, losses By-products
Ideal value 100% 100%
Worked Example

Calculate the atom economy for producing hydrogen from the reaction:

Zn + H₂SO₄ → ZnSO₄ + H₂

Step 1: Mr of desired product (H₂) = 2

Step 2: Mr of all products = ZnSO₄ (161) + H₂ (2) = 163

Step 3: Atom economy = (2 ÷ 163) × 100 = 1.2%

This is very low - most of the atoms end up in the by-product (ZnSO₄), not in the desired product (H₂).

Exam Tip

Exam Technique

You need both high percentage yield and high atom economy for a truly efficient, sustainable reaction. AQA loves comparing these two metrics.

CHAPTER 04 · 2 sectionsTitrations and gasesFurther calculations for separate chemistry

Titrations Chemistry Only

A titration is a technique used to find the concentration of an unknown acid or alkali by reacting it with one of known concentration.

Labelled titration apparatus Burette contains acid of known concentration Tap (stopcock) controls the flow of acid Conical flask contains alkali + indicator (pink phenolphthalein shown) White tile shows the colour change clearly Clamp stand
The burette delivers a measured volume of acid into the alkali below.

The Method

  1. Use a pipette to measure a fixed volume of alkali into a conical flask.
  2. Add a few drops of indicator (e.g., phenolphthalein or methyl orange).
  3. Fill a burette with acid of known concentration.
  4. Add acid gradually, swirling, until the indicator permanently changes colour: the end point.
  5. Record the titre (volume of acid added). Repeat until concordant results are achieved (within 0.10 cm³).
Exam Tip

Exam Technique

When describing the titration method, always mention: using a pipette for the fixed volume, a burette for the variable volume, and an indicator for the end point. Repeat to get concordant results.

Titration Calculations

Worked Example

25 cm³ of NaOH (unknown concentration) is neutralised by 20 cm³ of 0.5 mol/dm³ HCl. Find the concentration of NaOH.

Equation: NaOH + HCl → NaCl + H₂O

Step 1: Moles of HCl = conc × vol = 0.5 × (20 ÷ 1000)
= 0.5 × 0.02 = 0.01 mol

Step 2: Ratio NaOH : HCl = 1:1, so moles of NaOH = 0.01 mol

Step 3: Volume of NaOH = 25 ÷ 1000 = 0.025 dm³

Step 4: Concentration of NaOH = 0.01 ÷ 0.025 = 0.4 mol/dm³

Exam Tip

Exam Technique

Titration calculations always follow the same pattern: find moles of the known substance → use the mole ratio → calculate the unknown concentration. Be careful to convert cm³ to dm³!

Molar Volume of Gases Chemistry Only Higher Tier

At room temperature and pressure (RTP: 20°C, 1 atm), one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³).

volume (dm³) = moles × 24
moles = volume (dm³) ÷ 24

Gas Volume Triangle

Gas volume triangle: volume = moles × 24 (at RTP) volume (dm³) moles (mol) × 24 (dm³/mol at RTP)
vol = mol × 24
mol = vol ÷ 24
(at RTP)
Worked Example

What volume does 0.5 mol of oxygen gas occupy at RTP?

Volume = 0.5 × 24 = 12 dm³

Worked Example

What volume of CO₂ is produced when 5 g of CaCO₃ is decomposed? (at RTP)

Equation: CaCO₃ → CaO + CO₂

Step 1: Mr of CaCO₃ = 40 + 12 + (3 × 16) = 100

Step 2: Moles of CaCO₃ = 5 ÷ 100 = 0.05 mol

Step 3: Ratio 1:1, so moles of CO₂ = 0.05 mol

Step 4: Volume = 0.05 × 24 = 1.2 dm³ (or 1200 cm³)

Exam Tip

Exam Technique

The molar volume (24 dm³) only applies at RTP. If a question states different conditions, you may need the ideal gas equation instead (not required at GCSE). Questions often combine mass→moles→volume in multi-step problems.

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