AQA GCSE Chemistry 8462: Quantitative chemistry – Revision Notes

These are condensed recall notes for section 4.3 Quantitative chemistry (4.3.1.1 to 4.3.5) of the AQA GCSE Chemistry (8462) specification, for teaching from September 2016 and exams from June 2018 onwards. The topic is assessed on Paper 1, which is set at Foundation and Higher Tier. Points the specification marks (HT only) are labelled Higher tier only. Sections 4.3.3 to 4.3.5 are chemistry-only content, which is part of this course.

For full explanations, read the Quantitative chemistry study guide first. Then test yourself with the Quantitative chemistry practice questions. The course hub is AQA GCSE Chemistry, and the printable checklist lists every specification point. The next topic, Chemical changes, uses these calculations in titrations.

Relative atomic masses used below: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, S = 32, Ca = 40, Fe = 56. In the exam you are given a periodic table.

Formula sheet

Quantity	Formula	Tier
Relative formula mass	Mr = sum of Ar of every atom in the formula	Both
Percentage by mass of an element	(Ar × number of atoms of that element) ÷ Mr × 100	Both
Uncertainty	± (range ÷ 2), quoted about the mean	Both
Amount in moles	moles = mass (g) ÷ Mr	Higher tier only
Number of particles	moles × 6.02 × 10²³	Higher tier only
Concentration (mass)	concentration (g/dm³) = mass of solute (g) ÷ volume (dm³)	Both
Percentage yield	mass of product actually made ÷ maximum theoretical mass of product × 100	Both
Atom economy	Mr of desired product from equation ÷ sum of Mr of all reactants from equation × 100	Both
Concentration (moles)	concentration (mol/dm³) = moles ÷ volume (dm³)	Higher tier only
Gas volume at RTP	volume (dm³) = moles × 24	Higher tier only

Volume conversion: 1 dm³ = 1000 cm³, so divide cm³ by 1000 to get dm³. RTP means 20 °C and 1 atmosphere.

4.3.1.1 Conservation of mass and balancing

• Law of conservation of mass: no atoms are lost or made in a chemical reaction, so the mass of the products equals the mass of the reactants.
• A balanced symbol equation has the same number of atoms of each element on both sides.
• Multiplier in normal script (the 2 in 2H₂O) multiplies the whole formula. Subscript (the ₂ in H₂O) belongs to the formula. Balance by changing multipliers only. Never change a subscript.

Example: in 3Ca(OH)₂ there are 3 Ca, 6 O and 6 H atoms.

4.3.1.2 Relative formula mass

• Mr is the sum of the relative atomic masses of the atoms in the numbers shown in the formula. No units.
• In a balanced equation, the total Mr of the reactants (in the quantities shown) equals the total Mr of the products.

Worked reminders

Mr of Ca(OH)2 = 40 + 2 × (16 + 1) = 74

% N in NH4NO3:  Mr = 14 + 4 + 14 + 48 = 80
                % N = (2 × 14) ÷ 80 × 100 = 35%

4.3.1.3 Mass changes when a gas is involved

In a non-enclosed system, the mass on the balance can change. Mass is still conserved; a gas has simply entered or left.

Reaction	What the balance shows	Particle explanation
Metal heated in air, e.g. 2Mg + O₂ → 2MgO	Mass increases	Oxygen atoms from the air join the metal atoms in the solid oxide
Metal carbonate decomposes, e.g. CaCO₃ → CaO + CO₂	Mass decreases	Carbon dioxide molecules escape into the air; the oxide is the only solid left

Always use the equation: point to the gas (g) and say where its particles went or came from.

4.3.1.4 Chemical measurements

• Every measurement has some uncertainty.
• Use the range about the mean: uncertainty = ± half the range.

Readings: 31.2, 30.8, 31.6 cm³
Mean = 31.2 cm³   Range = 31.6 − 30.8 = 0.8 cm³
Result = 31.2 ± 0.4 cm³

4.3.2.1 Moles (Higher tier only)

• The unit is the mole, symbol mol.
• The mass of one mole of a substance in grams equals its Mr (numerically).
• One mole of any substance contains the same number of stated particles: the Avogadro constant, 6.02 × 10²³ per mole.
• Moles can count atoms, molecules, ions, electrons or formulae. One mole of C contains as many atoms as one mole of CO₂ contains molecules.

12.6 g of HNO3 (Mr 63):  moles = 12.6 ÷ 63 = 0.200 mol
molecules = 0.200 × 6.02 × 10^23 = 1.20 × 10^23

4.3.2.2 Reacting masses – method in steps (Higher tier only)

1. Write the balanced equation.
2. Moles of the known substance = mass ÷ Mr.
3. Use the ratio of the balancing numbers to get moles of the unknown.
4. Mass of the unknown = moles × Mr.

CaCO3 → CaO + CO2.  Mass of CaO from 25 g of CaCO3?
moles CaCO3 = 25 ÷ 100 = 0.25 mol  →  ratio 1 : 1  →  0.25 mol CaO
mass CaO = 0.25 × 56 = 14 g

4.3.2.3 Balancing from masses – method in steps (Higher tier only)

1. Convert each mass to moles (mass ÷ Mr).
2. Divide every value by the smallest.
3. Scale to whole numbers. These are the balancing numbers.

5.6 g Fe + 2.4 g O2 → 8.0 g Fe2O3
moles: Fe 0.10, O2 0.075, Fe2O3 0.050
÷ 0.050:  2 : 1.5 : 1   × 2:  4 : 3 : 2   →   4Fe + 3O2 → 2Fe2O3

4.3.2.4 Limiting reactants (Higher tier only)

• An excess of one reactant is often used so that all of the other is used.
• The reactant used up completely is the limiting reactant. It fixes the amount of product.
• Test: find moles of each, then use the ratio to see which runs out first.

2Mg + O2 → 2MgO with 4.8 g Mg and 4.8 g O2
Mg = 0.20 mol; O2 = 0.15 mol; 0.20 mol Mg needs only 0.10 mol O2
Mg is limiting; MgO = 0.20 mol × 40 = 8.0 g

4.3.2.5 Concentration of solutions

• Concentration in g/dm³ = mass of solute ÷ volume of solution in dm³.
• 15 g dissolved to make 250 cm³ → 15 ÷ 0.250 = 60 g/dm³. Mass in 40 cm³ of it: 60 × 0.040 = 2.4 g.
• Higher tier only: more solute in the same volume gives a higher concentration; the same mass in a larger volume gives a lower concentration.

4.3.3 Yield and atom economy

Percentage yield. You rarely get the calculated mass because:

• the reaction may not go to completion because it is reversible
• some product is lost when it is separated from the reaction mixture
• some reactants react in ways different from the expected reaction.

Example: 14 g of CaO expected, 11.9 g obtained → 11.9 ÷ 14 × 100 = 85%. Calculating the theoretical mass from a reactant mass and the equation is Higher tier only.

Atom economy (atom utilisation) measures how much of the starting materials ends up as useful products. High atom economy matters for sustainable development and for economic reasons.

Fermentation: C6H12O6 → 2C2H5OH + 2CO2, ethanol wanted
atom economy = (2 × 46) ÷ 180 × 100 = 51.1%

Higher tier only: choose a reaction pathway by weighing atom economy, yield, rate, equilibrium position and whether the by-products are useful.

4.3.4 Concentrations in mol/dm³ (Higher tier only)

• moles = concentration (mol/dm³) × volume (dm³). 0.050 mol in 250 cm³ → 0.050 ÷ 0.250 = 0.20 mol/dm³.
• Convert to g/dm³ by multiplying by Mr.
• If you know both volumes and one concentration of two solutions that react completely, you can find the other concentration.

20.0 cm³ of 0.100 mol/dm³ H2SO4 neutralises 25.0 cm³ of NaOH
H2SO4 + 2NaOH → Na2SO4 + 2H2O
moles H2SO4 = 0.100 × 0.0200 = 0.00200 mol  →  NaOH = 0.00400 mol
[NaOH] = 0.00400 ÷ 0.0250 = 0.160 mol/dm³  = 0.160 × 40 = 6.40 g/dm³

4.3.5 Volumes of gases (Higher tier only)

• Equal amounts in moles of gases occupy the same volume at the same temperature and pressure.
• One mole of any gas occupies 24 dm³ at RTP.
• 3.2 g of O₂ = 0.10 mol → 0.10 × 24 = 2.4 dm³.
• The mole ratio for gases in an equation is also their volume ratio. CH₄ + 2O₂ → CO₂ + 2H₂O: 40 cm³ of methane needs 80 cm³ of oxygen and makes 40 cm³ of carbon dioxide (water is a liquid at RTP).

Must-know distinctions

• Mr vs mass: Mr has no units; mass of one mole is in grams.
• Multiplier vs subscript: only multipliers change when balancing.
• Percentage yield vs atom economy: yield compares what you got with the maximum possible; atom economy comes from the equation alone and ignores practical losses.
• g/dm³ vs mol/dm³: multiply mol/dm³ by Mr to get g/dm³.
• Limiting vs excess: the limiting reactant runs out; some of the excess reactant is left over.
• cm³ vs dm³: divide by 1000 before any concentration or gas calculation.

Quick self-test

1. Calculate the Mr of Al₂(SO₄)₃. (Al = 27)
2. Calculate the percentage by mass of oxygen in CaCO₃.
3. Magnesium is heated in an open crucible. Why does the mass go up?
4. Four readings are 12.4, 12.8, 12.6 and 12.2 s. Give the mean and its uncertainty.
5. (Higher tier only) How many moles are in 9.0 g of water?
6. (Higher tier only) How many molecules are in 0.50 mol of CO₂?
7. (Higher tier only) 2Mg + O₂ → 2MgO. What mass of MgO forms from 6.0 g of Mg?
8. 8.0 g of solute is dissolved to make 400 cm³ of solution. Give the concentration in g/dm³.
9. The theoretical mass is 22.0 g and 16.5 g is collected. Calculate the percentage yield.
10. Calculate the atom economy for making CaO by CaCO₃ → CaO + CO₂.
11. (Higher tier only) How many moles of NaOH are in 25.0 cm³ of 0.40 mol/dm³ solution?
12. (Higher tier only) What volume does 7.0 g of N₂ occupy at RTP?

Answers

1. 2 × 27 + 3 × (32 + 4 × 16) = 342.
2. 48 ÷ 100 × 100 = 48%.
3. Oxygen atoms from the air combine with the magnesium, so the solid oxide has more mass than the metal.
4. Mean = 12.5 s; range 0.6 s, so ± 0.3 s.
5. 9.0 ÷ 18 = 0.50 mol.
6. 0.50 × 6.02 × 10²³ = 3.01 × 10²³.
7. 6.0 ÷ 24 = 0.25 mol Mg → 0.25 mol MgO → 0.25 × 40 = 10 g.
8. 8.0 ÷ 0.400 = 20 g/dm³.
9. 16.5 ÷ 22.0 × 100 = 75%.
10. 56 ÷ 100 × 100 = 56%.
11. 0.40 × 0.0250 = 0.010 mol.
12. 7.0 ÷ 28 = 0.25 mol → 0.25 × 24 = 6.0 dm³.

Where marks are usually lost

• Changing a subscript (H₂O to H₂O₂) to balance an equation instead of a multiplier.
• Forgetting the brackets when working out Mr, e.g. counting one OH in Ca(OH)₂ instead of two.
• Saying mass is “lost” or “destroyed” when a carbonate is heated, rather than naming the gas that escapes.
• Quoting the full range as the uncertainty instead of half of it.
• Leaving volumes in cm³ in a g/dm³ or mol/dm³ calculation.
• Skipping the mole ratio and going straight from mass of one substance to mass of another.
• Picking the limiting reactant as the one with the smaller mass rather than comparing moles against the ratio.
• Dividing theoretical by actual mass in percentage yield, giving a value over 100%.
• In atom economy, leaving out the balancing number of the desired product or of a reactant.
• Multiplying moles by 24 cm³ instead of 24 dm³ (or 24 000 cm³) for a gas volume.

Official syllabus

AQA GCSE Chemistry (8462) specification, Version 1.1 (October 2019), for teaching from September 2016 and exams from June 2018 onwards (AQA), section 4.3 Quantitative chemistry. Check your recall with the free 10-minute diagnostics.