AQA GCSE Chemistry · free printable

Every GCSE Chemistry calculation in one printable guide

All the AQA GCSE Chemistry and Combined Science: Trilogy calculations, from relative formula mass to Rf values, each with its equation, units and Higher and Separate flags, with worked examples from our notes.

25 calculations
14 Higher tier only
8 Separate Chemistry only

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How to read the guide

Calculations marked Combined + Separate are for everyone. Higher tier only calculations are not on Foundation papers, and Separate Chemistry only ones are not in Combined Science. Worked examples come from Examva’s notes for that topic.

Combined + SeparateHigher tier onlySeparate Chemistry only

Atomic structure

Revise atomic structure notes →

Combined + Separate

Protons, neutrons and electrons in an atom

number of neutrons = mass number − atomic number

mass number (protons + neutrons): no unit
atomic number (protons): no unit

Worked example

Sodium is written as ²³₁₁Na. Work out its numbers of protons, neutrons and electrons.

Atomic number (bottom) = 11, so there are 11 protons.

Neutrons = mass number − atomic number = 23 − 11 = 12.

A sodium atom is neutral, so it has 11 electrons.

In a neutral atom, electrons = protons.

Spec ref 8462 4.1.1.5

Combined + Separate

Electrons in an ion

electrons in an ion = protons − charge on the ion

protons (atomic number): no unit
charge on the ion: e.g. +2 or −2

Worked example

Group 2 metals lose 2 electrons, so they form +2 ions (e.g. Mg²⁺, Ca²⁺).

Group 6 non-metals gain 2 electrons, so they form −2 ions (e.g. O²⁻, the oxide ion - not 'oxygen ion').

Mg²⁺ (atomic number 12) has 12 − 2 = 10 electrons; O²⁻ (atomic number 8) has 8 + 2 = 10 electrons.

Spec ref 8462 4.1.1.5

Combined + Separate

Relative atomic mass from isotopes

Ar = sum of (isotope mass × % abundance) ÷ 100

relative atomic mass, Ar: no unit
abundance: per cent, %

Worked example

A sample has 20% boron-10 and 80% boron-11: Ar = (20 × 10 + 80 × 11) ÷ 100 = 10.8, between the two isotope masses.

Spec ref 8462 4.1.1.6

Quantitative chemistry

Revise quantitative chemistry notes →

Combined + Separate

Relative formula mass

Mr = sum of the Ar of every atom in the formula

relative formula mass, Mr: no unit
relative atomic mass, Ar: no unit

Worked example

For Ca(OH)₂ the Mr is 40 + 2 × (16 + 1) = 40 + 34 = 74 using Ca = 40, O = 16, H = 1.

Relative formula mass has NO units - never add g or kg.

Spec ref 8462 4.3.1.2

Combined + Separate

Percentage by mass of an element

% by mass of an element = (Ar × number of atoms of that element ÷ Mr) × 100

percentage by mass: per cent, %
Ar and Mr: no unit

Worked example

% by mass of iron in Fe₂O₃

Mr = (2 × 56) + (3 × 16) = 160; iron = 2 × 56 = 112; 112 ÷ 160 × 100 = 70%.

Count both Fe atoms (not 35%).

Spec ref 8462 4.3.1.2

Combined + Separate

Mean, range and uncertainty

mean = total of the repeats ÷ number of repeats; range = highest − lowest

mean, range and uncertainty: same unit as the readings

Worked example

Mean = (24.8 + 25.0 + 25.2) ÷ 3 = 75.0 ÷ 3 = 25.0 cm³.

Range = highest − lowest = 25.2 − 24.8 = 0.4 cm³.

Estimate uncertainty as half the range: 0.4 ÷ 2 = 0.2 cm³.

Spec ref 8462 4.3.1.4

Higher tier only

Moles from mass

moles = mass (g) ÷ Mr

amount: moles, mol
mass: grams, g
relative formula mass, Mr: no unit

Worked example

For 80 g of calcium (Ar Ca = 40): moles = mass ÷ Mr = 80 ÷ 40 = 2 mol.

Rearrange it to find mass from moles: mass (g) = moles × Mr.

Spec ref 8462 4.3.2.1

Higher tier only

Number of particles (Avogadro constant)

number of particles = moles × 6.02 × 10²³

number of particles: no unit
amount: moles, mol

Worked example

Two moles of a substance therefore contain 2 × 6.02 × 10²³ particles.

2 × 6.02 × 10²³ = 1.204 × 10²⁴ particles.

That number of particles is the Avogadro constant, 6.02 × 10²³ per mole.

Spec ref 8462 4.3.2.1

Higher tier only

Reacting masses from a balanced equation

moles = mass ÷ Mr, then use the equation's ratio, then mass = moles × Mr

mass: grams, g
amount: moles, mol
relative formula mass, Mr: no unit

Worked example

CaCO₃ → CaO + CO₂. For 50 g CaCO₃ (Mr 100): moles = 50 ÷ 100 = 0.5. Ratio 1 : 1, so 0.5 mol CaO. Mr CaO = 56, mass = 0.5 × 56 = 28 g.

Spec ref 8462 4.3.2.2

Higher tier only

Balancing an equation from masses

moles of each substance = mass ÷ Mr; divide by the smallest; scale to the simplest whole-number ratio; then balance any other formulae from the atoms

mass: grams, g
amount: moles, mol
relative formula mass, Mr: no unit

Worked example

Moles of Mg = 4.8 ÷ 24 = 0.2 mol.

Moles of O₂ = 3.2 ÷ 32 = 0.1 mol.

Divide by the smallest (0.1): Mg = 0.2 ÷ 0.1 = 2; O₂ = 0.1 ÷ 0.1 = 1.

Ratio Mg : O₂ = 2 : 1; balancing the atoms then gives 2Mg + O₂ → 2MgO.

If a ratio comes out as 1 : 1.5, scale both up (× 2) to reach whole numbers such as 2 : 3 before writing the equation.

Spec ref 8462 4.3.2.3

Higher tier only

Limiting reactant and product mass

moles of each reactant = mass ÷ Mr; use the ratio to find which runs out; the product amount comes from the limiting reactant

mass: grams, g
amount: moles, mol
relative formula mass, Mr: no unit

Worked example

For Mg + 2HCl → MgCl₂ + H₂, 0.10 mol Mg reacts with 0.30 mol HCl. Find the limiting reactant and the mass of H₂ made. (Ar: H = 1.)

The ratio is Mg : HCl = 1 : 2, so 0.10 mol Mg needs 2 × 0.10 = 0.20 mol HCl.

0.30 mol HCl is available, which is more than 0.20 mol, so HCl is in excess.

Ratio Mg : H₂ = 1 : 1, so 0.10 mol Mg makes 0.10 mol H₂.

Mass of H₂ = moles × Mr = 0.10 × 2 = 0.20 g.

Always compare moles, not masses, because it is the mole ratio in the equation that decides which reactant runs out.

Spec ref 8462 4.3.2.4

Combined + Separate

Concentration in g/dm³

concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)

concentration: grams per cubic decimetre, g/dm³
mass of solute: grams, g
volume of solution: cubic decimetres, dm³ (cm³ ÷ 1000)

Worked example

Keep 20 g of solute but make the volume 0.5 dm³: concentration = 20 ÷ 0.5 = 40 g/dm³.

To convert a volume from cm³ into dm³, divide by 1000: e.g. 250 cm³ = 0.25 dm³, and 500 cm³ = 0.5 dm³.

Spec ref 8462 4.3.2.5

Separate Chemistry only

Percentage yield

% yield = (mass of product actually made ÷ maximum theoretical mass) × 100

percentage yield: per cent, %
mass: grams, g

Worked example

10.0 g of product is possible but only 9.0 g is collected because some sticks to the filter paper: yield = (9.0 ÷ 10.0) × 100 = 90%.

Spec ref 8462 4.3.3.1

Separate Chemistry onlyHigher tier only

Theoretical mass of product (for % yield)

moles of reactant = mass ÷ Mr; use the mole ratio; maximum theoretical mass of product = moles × Mr

mass: grams, g
amount: moles, mol
relative formula mass, Mr: no unit

Worked example

Example: heating 25.0 g of CaCO₃ (Mr 100) gives 0.250 mol, so CaCO₃ → CaO + CO₂ makes 0.250 mol of CaO (Mr 56) - a theoretical mass of 14.0 g.

Only then can you use % yield = actual ÷ theoretical × 100 when just the reactant mass is given.

Spec ref 8462 4.3.3.1

Separate Chemistry only

Atom economy

% atom economy = (Mr of the desired product ÷ sum of the Mr of all the reactants) × 100, each Mr multiplied by its coefficient in the balanced equation

atom economy: per cent, %
relative formula mass, Mr: no unit

Worked example

Zn + H₂SO₄ → ZnSO₄ + H₂.

Mr of desired product H₂ = 2 × 1 = 2.

Sum of Mr of all reactants = 65 + 98 = 163, then % = (2 ÷ 163) × 100.

Atom economy = 1.23% (3 s.f.) - most reactant atoms end up in the ZnSO₄ by-product.

Use the relative formula masses in the amounts shown by the balanced equation - multiply each Mr by its coefficient.

Spec ref 8462 4.3.3.2

Separate Chemistry onlyHigher tier only

Concentration in mol/dm³

concentration (mol/dm³) = amount of solute (mol) ÷ volume of solution (dm³)

concentration: moles per cubic decimetre, mol/dm³
amount of solute: moles, mol
volume of solution: cubic decimetres, dm³ (cm³ ÷ 1000)

Worked example

A NaOH solution has a concentration of 0.5 mol/dm³. Find the moles of NaOH in 2 dm³, then give the concentration in g/dm³. (Ar: Na = 23, O = 16, H = 1.)

Moles = concentration × volume = 0.5 × 2 = 1 mol of NaOH.

Mr of NaOH = 23 + 16 + 1 = 40.

Concentration in g/dm³ = mol/dm³ × Mr = 0.5 × 40 = 20 g/dm³.

Rearrange it to find moles = concentration × volume, or volume = moles ÷ concentration.

Spec ref 8462 4.3.4

Separate Chemistry onlyHigher tier only

Converting mol/dm³ to g/dm³

concentration (g/dm³) = concentration (mol/dm³) × Mr

concentration: g/dm³ and mol/dm³
relative formula mass, Mr: no unit

Worked example

A solution is 0.1 mol/dm³ of NaCl (Mr = 58.5). Concentration in g/dm³ = 0.1 × 58.5 = 5.85 g/dm³, found by multiplying the molar concentration by Mr.

Convert g/dm³ back to mol/dm³ by dividing the mass concentration by the solute's Mr.

Spec ref 8462 4.3.4

Separate Chemistry onlyHigher tier only

Volume of a gas at room temperature and pressure

volume of gas at rtp (dm³) = moles × 24

volume of gas: cubic decimetres, dm³
amount: moles, mol

Worked example

Example: what volume does 8 g of methane, CH₄ (Mr = 16), occupy at rtp?

moles = 8 ÷ 16 = 0.5 mol, so volume = 0.5 × 24 = 12 dm³ at rtp.

1 dm³ = 1000 cm³, so 24 dm³ is the same as 24 000 cm³ at rtp.

Spec ref 8462 4.3.5

Separate Chemistry onlyHigher tier only

Gas volumes from a balanced equation

volume ratio of gases = mole ratio in the balanced equation (same temperature and pressure)

gas volumes: same unit for each gas, e.g. dm³

Worked example

In 2H₂ + O₂ → 2H₂O, 48 dm³ of H₂ needs 24 dm³ of O₂ at rtp, because the 2:1 mole ratio is also the 2:1 volume ratio for gases.

Spec ref 8462 4.3.5

Chemical changes

Revise chemical changes notes →

Higher tier only

pH and hydrogen ion concentration

each 1 unit decrease in pH = 10 times the H⁺ concentration

pH: no unit (0–14 scale)
H⁺ concentration: factor of 10 per pH unit

Worked example

So a solution of pH 2 has ten times the H⁺ concentration of one at pH 3.

From pH 4 to pH 1 is 3 units, so the H⁺ concentration is 10 × 10 × 10 = 1000 times greater.

Spec ref 8462 4.4.2.6

Separate Chemistry onlyHigher tier only

Titration calculations

moles of known solution = concentration × volume (dm³); use the equation's ratio; unknown concentration = moles ÷ volume (dm³)

concentration: mol/dm³ (or g/dm³ after × Mr)
amount: moles, mol
volume of solution: cubic decimetres, dm³ (cm³ ÷ 1000)

Worked example

25.0 cm³ of sodium hydroxide is exactly neutralised by 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid.

HCl + NaOH → NaCl + H₂O shows the acid and alkali react in a 1 : 1 ratio.

Moles of HCl = concentration × volume in dm³ = 0.100 × 20.0/1000 = 0.00200 mol.

1 : 1 ratio → moles of NaOH = 0.00200 mol.

Concentration of NaOH = moles ÷ volume in dm³ = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³.

Always convert cm³ to dm³ (÷1000) before using concentration = moles ÷ volume.

Spec ref 8462 4.4.2.5

Energy changes

Revise energy changes notes →

Higher tier only

Energy change from bond energies

overall energy change = energy needed to break bonds − energy released making bonds

energy: kilojoules per mole, kJ/mol

Worked example

Breaking bonds takes 500 kJ/mol; forming bonds releases 650 kJ/mol. Overall change = 500 − 650 = −150 kJ/mol: exothermic.

Break = takes IN (endo). Make = gives OUT (exo).

Spec ref 8462 4.5.1.3

Rates of reaction

Revise rates of reaction notes →

Combined + Separate

Mean rate of reaction

mean rate of reaction = quantity of reactant used (or product formed) ÷ time taken

rate: g/s or cm³/s (mol/s at Higher tier)
quantity: grams, g, or cubic centimetres, cm³ (moles, mol, at Higher tier)
time: seconds, s

Worked example

24 cm³ gas collected in 40 s gives a mean rate of 24 ÷ 40 = 0.60 cm³/s.

Mean rate = quantity used or formed ÷ time. Watch the units: g/s or cm³/s.

Spec ref 8462 4.6.1.1

Higher tier only

Rate from the gradient of a tangent

rate at one moment = |gradient of the tangent| = |change in quantity ÷ change in time|, both changes read along the tangent

rate: vertical-axis unit ÷ horizontal-axis unit, e.g. cm³/s

On a reactant-remaining curve the gradient is negative - the rate is the magnitude of that negative gradient.

Spec ref 8462 4.6.1.1

Chemical analysis

Revise chemical analysis notes →

Combined + Separate

Rf value (chromatography)

Rf = distance moved by the substance ÷ distance moved by the solvent

Rf value: no unit (ratio, never more than 1)
distances: same unit for both, e.g. mm or cm

Worked example

A spot travels 54 mm and the solvent front 90 mm: Rf = 54 ÷ 90 = 0.60. Measure from the same baseline; compare under the same conditions.

Measure from the baseline to the centre of each spot, and from the baseline to the solvent front.

Spec ref 8462 4.8.1.3

Based on the AQA GCSE Chemistry (8462) and Combined Science: Trilogy (8464) specifications as used in Examva’s notes. Examva is independent and not affiliated with or endorsed by AQA.

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