Amount of substance
A-level Chemistry (7405) · exam-style practice, examiner-report intelligence and the tools that drill it.
The topic on one screen
- One mole = 6.022 × 1023 particles (the Avogadro constant). Moles link every quantity in this topic: m = n × Mr.
- Solutions: n = c × V, with V in dm3 — the cm3 → dm3 conversion is the single most common dropped mark.
- Gases: pV = nRT with p in Pa, V in m3, T in K. Convert units before you substitute, not after.
- Empirical formula = simplest whole-number ratio. Divide each % by Ar, then divide through by the smallest.
- Titrations: moles of known → ratio from the balanced equation → moles of unknown → concentration.
- Percentage uncertainty = (uncertainty ÷ measured value) × 100, summed per piece of apparatus — not per total volume.
Where students actually lose marks
Mass-for-a-standard-solution calculations were answered well except for one thing: forgetting to convert 250 cm3 to 0.250 dm3 before using n = c × V.
June 2023 Paper 1 examiner report (Q06.1)
Describing how to make a standard solution loses marks in predictable places: making the solution up in the beaker instead of transferring to a volumetric flask, skipping the rinse-and-transfer of washings, and forgetting to invert the flask to mix.
June 2023 Paper 1 examiner report (Q06.2)
Only 30% of students got full marks on the titration calculation. The killers: using the wrong Mr, ignoring the 2:1 ratio in the equation, and dividing by the wrong volume at the final step.
June 2023 Paper 1 examiner report (Q06.5)
Percentage uncertainty answers were generally poor — most students added both uncertainties and divided by the combined volume. Each apparatus gets its own % first; then you add the percentages.
June 2023 Paper 1 examiner report (Q06.6)
Try it — exam-style
Calculate the mass, in g, of anhydrous sodium carbonate (Mr = 106.0) needed to prepare 250 cm3 of 0.150 mol dm−3 sodium carbonate solution.
The solid is dissolved in a beaker containing about 100 cm3 of distilled water. Describe how this solution is used to make exactly 250 cm3 of standard solution.
0.663 g of anhydrous sodium carbonate (Mr = 106.0) is dissolved and made up to 250 cm3. A 25.0 cm3 portion reacts with exactly 23.30 cm3 of hydrochloric acid: Na2CO3 + 2HCl → 2NaCl + H2O + CO2. Calculate the concentration of the hydrochloric acid.
The uncertainty in a 25.0 cm3 pipette reading is ±0.05 cm3. The total uncertainty in a 21.40 cm3 burette titre is ±0.10 cm3. Calculate the total percentage error.
A compound contains 54.53% carbon and 9.15% hydrogen by mass. The rest is oxygen. Calculate its empirical formula.
Calculate the volume, in dm3, occupied by 0.0350 mol of an ideal gas at 100 kPa and 298 K. (R = 8.31 J K−1 mol−1)
Questions are written in the style of past AQA papers (source shown on each) — never copied from them.
Get the printable question packfree accountDrill it properly
Stuck on amount of substance?
Titration calculations are where I see the most dropped marks in lessons — I teach the exact working layout examiners reward, and your first lesson is free.