AQA A-level Chemistry (7405) · Amount of Substance
Mini-Lesson
Amount of Substance
This mini-lesson covers AQA 3.1.2 Amount of substance: the mole and the Avogadro constant, the ideal gas equation pV = nRT, empirical and molecular formulae, percentage yield and atom economy, and titration calculations — the maths engine of the whole A-level.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Everything here is A-level standard — the maths is done properly, not skipped. Press Start when you're ready.
Amount of substance · the mole
Moles, mass and concentration
The mole is the amount of substance that contains 6.022 × 10²³ particles — that number is the Avogadro constant, L = 6.022 × 10²³ mol⁻¹. (Historically it was defined as the number of atoms in 12 g of carbon-12.)
n = m / Mn = moles · m = mass in g · M = molar mass in g mol⁻¹
n = c × Vc = concentration in mol dm⁻³ · V = volume in dm³ (÷1000 to convert cm³ → dm³)
Unit discipline is where marks are lost. Concentration is per dm³, so a 25.0 cm³ aliquot is 0.0250 dm³. Mass in kg must become g. Get the units right before you touch a calculator.
Calculate
Your turn
1Calculate the mass of 0.250 mol of calcium carbonate, CaCO₃ (Mr = 100.1).
g
Hint: m = n × M = 0.250 × 100.1.
Amount of substance · gases
The ideal gas equation
For a gas we cannot weigh out moles conveniently, so we use pressure, volume and temperature:
pV = nRTp in pascals (Pa) · V in cubic metres (m³) · T in kelvin (K) · R = 8.31 J K⁻¹ mol⁻¹
The three conversions that cost marks:
kPa → Pa: × 1000 (100 kPa = 1.00 × 10⁵ Pa)
cm³ → m³: × 10⁻⁶ · dm³ → m³: × 10⁻³
°C → K: + 273
An ideal gas assumes the particles have negligible volume, have no intermolecular forces, and undergo perfectly elastic collisions. Real gases deviate most at high pressure and low temperature, where those assumptions break down.
2Calculate the volume, in dm³, occupied by 0.500 mol of an ideal gas at 100 kPa and 300 K. (R = 8.31 J K⁻¹ mol⁻¹)
dm³
Hint: V = nRT/p = (0.500 × 8.31 × 300) ÷ 100 000 m³, then × 1000 to get dm³.
Quick check
Quick check
?A student measures the volume of a gas in cm³ and its pressure in kPa, then substitutes straight into pV = nRT with R = 8.31. What will happen?
Amount of substance · formulae
Empirical and molecular formulae
The empirical formula is the simplest whole-number ratio of atoms. The molecular formula is the actual number of each atom in a molecule — always a whole-number multiple of the empirical formula.
Worked example — 2.40 g C, 0.40 g H, 3.20 g O
Moles: C = 2.40 ÷ 12.0 = 0.200 · H = 0.40 ÷ 1.0 = 0.40 · O = 3.20 ÷ 16.0 = 0.200
Divide by the smallest (0.200): C = 1, H = 2, O = 1 → empirical formula CH₂O (Mr = 30.0)
If the true Mr is 180.0, then 180.0 ÷ 30.0 = 6, so the molecular formula is C₆H₁₂O₆.
Percentages work exactly the same way — treat the % as if it were a mass in grams (out of 100 g), because the ratio is what matters, not the total.
Sort it
Sort each compound by its relative molecular mass
Tap a formula, work out its M_r, then tap the right bin. Use C = 12.0, H = 1.0, N = 14.0, O = 16.0.
🟩 M_r ≈ 44.0
🟪 M_r ≈ 46.0
🟦 M_r ≈ 28.0
Amount of substance · yield & efficiency
Percentage yield and atom economy
Two different measures of how good a reaction is — do not confuse them.
% yield = (actual moles of product ÷ theoretical moles) × 100measures how much of the possible product you actually got — losses, side reactions, reversibility
% atom economy = (Mr of desired product ÷ Σ Mr of ALL reactants) × 100use the AQA form — mass of desired product ÷ total mass of reactants. Include the balancing numbers. (Because mass is conserved, the sum of the Mr of all the products gives the same answer.)
A reaction can have a 100% yield but a terrible atom economy (lots of waste by-product), or a 100% atom economy but a poor yield (an addition reaction that does not go to completion). Addition reactions always have 100% atom economy because there is only one product.
Why industry cares: a high atom economy means less waste to treat, cheaper raw materials and a smaller environmental impact — it is designed in at the equation stage, not fixed later.
Calculate
Your turn
324.0 g of magnesium (Ar = 24.3) is burned in excess oxygen: 2Mg + O₂ → 2MgO. Calculate the theoretical mass of MgO (Mr = 40.3) formed.
g
Hint: n(Mg) = 24.0 ÷ 24.3 = 0.988 mol. The ratio Mg : MgO is 1 : 1, so m = 0.988 × 40.3.
Calculate
Your turn
4In that experiment only 34.0 g of MgO is collected. Using your theoretical mass of 39.8 g, calculate the percentage yield.
%
Hint: % yield = (34.0 ÷ 39.8) × 100.
Amount of substance · titrations
Titration calculations
A titration finds an unknown concentration precisely. A pipette delivers a fixed aliquot (e.g. 25.0 cm³); a burette delivers a variable, measured volume (readings to ±0.05 cm³, so a titre has an uncertainty of ±0.10 cm³).
Do a rough titration first, then repeat carefully.
Use only concordant titres — within 0.10 cm³ of each other — and take their mean. Do not average in the rough.
Rinse the burette with the solution it will hold; rinse the conical flask with distilled water only.
n = c × Vthen use the balanced equation ratio to cross to the other reagent
Method, every time: (1) moles of the reagent you know everything about; (2) ratio from the balanced equation; (3) moles of the unknown; (4) divide by its volume in dm³.
Calculate
Your turn
5A 25.0 cm³ portion of NaOH(aq) is exactly neutralised by 22.50 cm³ of 0.100 mol dm⁻³ HCl. NaOH + HCl → NaCl + H₂O. Calculate the concentration of the NaOH.
mol dm⁻³
Hint: n(HCl) = 0.02250 × 0.100 = 2.25 × 10⁻³ mol. Ratio is 1 : 1, so n(NaOH) = 2.25 × 10⁻³ mol; c = n ÷ 0.0250.
Match it
Match the quantity to its definition
Tap a quantity on the left, then its definition on the right.
Quantity
Definition
Quick check
Quick check
?Which reaction necessarily has an atom economy of 100%?
Quick check
Quick check
?A 25.0 cm³ aliquot of NaOH(aq) is pipetted into a conical flask and titrated with HCl from the burette. The student had rinsed the conical flask with NaOH(aq) instead of distilled water. What is the effect on the titre?
Quick check
Quick check
?A gas deviates most from ideal behaviour under which conditions?
Amount of substance · exam traps
Uncertainty, significant figures and the classic errors
Burette uncertainty. A burette is read to ±0.05 cm³, and a titre needs two readings, so the uncertainty in the titre is ±0.10 cm³.
This is why a larger titre reduces the percentage uncertainty — the same absolute error over a bigger number.
Significant figures. Give your answer to the fewest number of significant figures used in the data. 3 s.f. data → a 3 s.f. answer.
Do not round mid-calculation. Carry the full value through and round only at the end.
Empirical from percentages: treat each % as a mass in grams. It works because you only need the ratio.
Gas-syringe and mass-loss experiments have their own uncertainties. Always quote the uncertainty of the instrument, and remember that a mass measured by difference (two weighings) doubles the absolute uncertainty.
Quick check
Quick check
?A titre of 12.50 cm³ and a titre of 25.00 cm³ both have an uncertainty of ±0.10 cm³. Which has the smaller percentage uncertainty, and why?
Calculate
Your turn
6A compound contains 2.40 g C, 0.40 g H and 3.20 g O, giving the empirical formula CH₂O (Mr = 30.0). Its true Mr is 180.0. By what number must the empirical formula be multiplied to get the molecular formula?
Hint: 180.0 ÷ 30.0.
Recap
The big ideas to know
Moles: n = m / M · n = c × V (V in dm³) · n = pV / RT
Ideal gas: pV = nRT with p in Pa, V in m³, T in K, R = 8.31 J K⁻¹ mol⁻¹
Empirical formula: simplest whole-number ratio: divide masses (or %) by A_r, then by the smallest