Topic 4 of 8
Measurements and recording
Keep raw readings, temperature changes and observations interpretable.
A-Level 9476 (2026-2027)
Record measurements that another person can interpret
Use the instrument scale, a clear table and observations that preserve the sequence.
Choose apparatus for the needed measurement: a volumetric pipette delivers a fixed accurate volume, a burette delivers a variable accurately measured volume, a measuring cylinder is suitable for less exact volume work, and a balance measures mass. Read the scale at eye level to avoid parallax; use the bottom of the meniscus for a clear aqueous solution unless the instructions specify otherwise.
| Trial | Initial burette / cm3 | Final burette / cm3 | Titre / cm3 |
|---|---|---|---|
| Rough | 0.00 | 24.80 | 24.80 |
| 1 | 1.10 | 25.50 | 24.40 |
| 2 | 0.20 | 24.65 | 24.45 |
| 3 | 0.30 | 24.70 | 24.40 |
Use the reading convention appropriate to the burette and the practical instructions; a 0.1 cm3-graduated burette is commonly estimated to 0.05 cm3. Keep the same decimal places in comparable raw readings. Agreement between titres is checked using the criterion in the task, not by deleting a value solely to make the mean look tidy.
Worked example
Select a justified mean titre
Use the precise titrations in the table; their maximum separation is 0.05 cm3.
- Trials 1, 2 and 3 form a close group. The rough trial was for locating the end-point.
- Mean = (24.40 + 24.45 + 24.40) / 3 = 24.4166... cm3.
- Record a suitable rounded mean of 24.42 cm3; retain the unrounded value through subsequent calculations where useful.
24.42 cm3, with the selected trials clearly identified.
Separate raw observations from interpretation. "A white precipitate forms after three drops and dissolves in excess" preserves information that "positive test" loses. Record colour, physical state, changes on standing or warming, and any confirmed gas test. Put units in table headings; do not attach a unit to a pure ratio.
An instrument resolution is not automatically the total experimental uncertainty. If each burette reading has a stated uncertainty of ±0.05 cm3, a conservative worst-case uncertainty in a difference is ±0.10 cm3. For a 24.40 cm3 titre that is about 0.41%. This estimate does not include endpoint judgement or chemical bias; use the uncertainty convention supplied in the task.
Check your understandingThe initial reading is 2.35 cm3 and the final reading is 26.70 cm3. What volume was delivered?Think it through, then reveal the answer
Measure temperature changes and changes in mass
Distinguish a reading from the chemical quantity it represents.
For a temperature-change experiment, measure a stable starting temperature, combine the stated quantities, stir consistently and follow temperature with time. Use a suitable thermometer or probe, keeping it immersed appropriately without resting on the vessel. A lid and insulation reduce exchange with the surroundings. Do not treat a single late reading as the maximum change if the mixture has already cooled.
A temperature rise indicates energy transferred to the measured surroundings in an exothermic process; a fall indicates an endothermic process under comparable conditions. To compare reactions fairly, account for the amounts reacting and the mass being warmed, rather than ranking enthalpy changes by temperature rise alone.
Worked example
Convert a temperature rise to an enthalpy estimate
50.0 g of solution warms by 6.00 K when 0.0200 mol reacts. Assume c = 4.18 J g-1 K-1, negligible vessel heat capacity and negligible heat exchange.
- q(solution) = mcΔT = 50.0 × 4.18 × 6.00 = 1254 J.
- q(reaction) = -1254 J because the solution receives the energy.
- ΔH = -1.254 / 0.0200 = -62.7 kJ mol-1. Heat loss would make the observed rise smaller and the estimate less negative.
Estimated ΔH = -62.7 kJ mol-1, subject to the stated assumptions.
In a thermometric titration, add measured portions of titrant, mix consistently and record temperature against the cumulative added volume. For an exothermic neutralisation, temperature initially rises as reaction releases energy. Once the limiting reagent is used up, further titrant produces no further neutralisation heat; dilution and heat exchange can lower the temperature. Estimate equivalence by extending the appropriate before- and after-reaction trends to their intersection.
Estimate the endpoint between measured additions
Six illustrative measurements lie on two local trends. The rising trend through volumes 8, 12 and 16 cubic centimetres meets the falling trend through 24, 28 and 32 at 20 cubic centimetres and 24 degrees Celsius. No reading was taken exactly at that intersection.
Worked example
Read the intersection, not just the highest measured point
Near the endpoint, the fitted trends in the graph are T = 22.0 + 0.100V and T = 24.5 - 0.0250V, with T in °C and V in cm3. Estimate the reacting volume.
- At the intersection, the temperatures agree: 22.0 + 0.100V = 24.5 - 0.0250V.
- 0.125V = 2.50, so V = 20.0 cm3 and T = 24.0 °C.
- The highest measured point is at 24 cm3, where T = 23.9 °C. Choosing that addition as the endpoint would miss the intersection between measurements.
- Use the inferred volume with the balanced reacting ratio if calculating concentration. Inspect whether straight trends are justified: changing solution mass, heat loss, dilution or a slow response can cause curvature. Smaller volume increments near the endpoint and consistent mixing and timing improve the evidence.
Estimated equivalence volume: 20.0 cm3. The construction is justified by the two local trends and their assumptions.
In volatilisation gravimetry, heat a sample, cool appropriately and weigh, then repeat to a constant mass. Constant mass supports completion under those conditions; it does not prove that the remaining solid has the assumed formula. Loss of water by a hydrate is useful only if the salt does not also decompose or reabsorb moisture before weighing.
Worked example
Use mass loss to find water of crystallisation
A crucible is 25.42 g. With a hydrated salt it is 27.92 g; after heating to constant mass it is 27.02 g. The anhydrous salt has molar mass 160 g mol-1; assume only water is lost.
- Hydrate mass = 2.50 g; anhydrous salt mass = 1.60 g; water lost = 0.90 g.
- n(salt) = 1.60 / 160 = 0.0100 mol. n(water) = 0.90 / 18.0 = 0.0500 mol.
- The water:salt ratio is 5:1. Incomplete dehydration would underestimate water lost and overestimate residue, giving a ratio that is too small.
The formula is salt·5H2O under the assumptions. Losing solid by spitting would bias the inferred water content upward.