Work out what each instrument reading contributes to the answer, combine uncertainties appropriately and separate random scatter from systematic error.
A reading is not always the final measurement
A titre is the difference between two burette readings. A mass delivered by difference comes from two balance readings. A temperature rise comes from two temperatures. Under the usual maximum-uncertainty model, add the absolute uncertainties of the two readings when calculating the uncertainty of the difference.
Do not automatically double every uncertainty. A pipette tolerance normally describes one delivered volume, not two independent readings to subtract. Use the uncertainty stated for the actual equipment or question and identify whether it applies per reading or per delivered measurement.
Absolute and percentage uncertainty
Absolute uncertainty has the same unit as the measured quantity. Relative uncertainty compares that interval with the measured value; multiply by 100 to express it as a percentage. This explains why the same absolute uncertainty matters more for a small measurement.
A burette with ±0.05 cm³ per reading gives a maximum titre uncertainty of ±0.10 cm³. For a 24.60 cm³ titre, the percentage uncertainty is 0.10/24.60 × 100 = 0.407%, approximately 0.41%. For a 5.00 cm³ titre with the same instrument, it is 2.0%. A larger sensible titre reduces the percentage contribution, provided the reaction and apparatus remain suitable.
Combine repeated deliveries on the correct total
Suppose a 20.00 cm³ pipette has a stated tolerance of ±0.03 cm³ and is used three times. The total is 60.00 cm³ with a maximum uncertainty of ±0.09 cm³; 0.09/60.00 × 100 = 0.15%. Compare this with one 60.00 cm³ delivered measurement only if its stated tolerance is known.
Use the total delivered volume in the denominator. Dividing ±0.09 cm³ by 20.00 cm³ would combine the uncertainty of three transfers with the volume of only one.
How uncertainties enter a calculated result
For addition or subtraction, add absolute uncertainties in a simple maximum estimate. For multiplication or division, add percentage uncertainties. If a measured quantity is raised to a power, multiply its percentage uncertainty by the magnitude of that power. These are the common assessment approximations for small uncertainties, not a full statistical treatment of every experiment.
For concentration c = n/V, if the amount contributes 0.30% uncertainty and the volume contributes 0.20%, the combined maximum estimate is 0.50%. State your assumptions. If a value is given as exact for the exercise, do not invent a new uncertainty for it.
Choose a measurement with a useful signal
A temperature rise from 20.2 °C to 26.6 °C is 6.4 °C. If each reading is uncertain by ±0.1 °C, the rise has maximum uncertainty ±0.2 °C, giving 3.125%, approximately 3.1%. A rise of only 1.0 °C with the same instrument would have 20% uncertainty.
This is why a larger measurable temperature change can improve a calorimetry result. It must not be achieved by introducing a new unaccounted heat loss or a change of reaction. Insulation, appropriate quantities and a suitable sensor address different limitations; identify which one matters.
Random and systematic effects need different remedies
Random effects vary from trial to trial and contribute to scatter. Repeat readings and a justified mean can reduce their influence on an estimate. Systematic effects bias results in a consistent direction; examples include an offset sensor and persistent loss before measurement begins. Calibration or method changes target these causes.
An endpoint overshoot is not automatically random or systematic without knowing how the procedure is repeated. Consistent overshooting biases titres high, whereas inconsistent judgment can add scatter. Name the behaviour and its effect rather than memorising a label.
Retain information until the final answer
Keep extra digits during a calculation and round once at the end to a precision justified by the inputs and the question. Decimal places and significant figures are different: 0.00450 has three significant figures, while its leading zeros only locate the decimal point. Exact stoichiometric coefficients do not usually limit the precision.
A displayed calculator result such as 0.002473829 mol is not evidence that the amount is known to ten significant figures. Give a sensible final value and its unit, retain unrounded values internally, and show intermediate expressions so a reader can follow your method.
Worked example: a concentration with an uncertainty estimate
An original measurement gives 1.260 g of a pure monoprotic acid of molar mass 126.0 g mol⁻¹, dissolved to 250.0 cm³. The acid amount is 1.260/126.0 = 0.01000 mol. The solution concentration is 0.01000/0.2500 = 0.04000 mol dm⁻³. For this illustration, take molar mass and purity as exact.
If the delivered mass has uncertainty ±0.002 g and the final volume ±0.15 cm³, their percentage contributions are 0.1587% and 0.0600%. The maximum combined estimate is 0.2187%, approximately 0.22%. The corresponding absolute concentration uncertainty is 0.04000 × 0.002187 ≈ 0.00009 mol dm⁻³. The solution preparation also needs complete transfer and mixing; this estimate does not account for an unnoticed spill.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.
Q1. A 0.486 g mass is found by difference using two readings, each ±0.001 g. Find its maximum percentage uncertainty.Show answer
Absolute uncertainty = 0.001 + 0.001 = 0.002 g. Percentage = 0.002/0.486 × 100 = 0.4115%, about 0.41%. Using ±0.001 g would count only one reading.
Q2. A 10.00 cm³ pipette with tolerance ±0.02 cm³ is used four times. Find the total volume and maximum percentage uncertainty.Show answer
Total volume = 40.00 cm³. Maximum uncertainty = 4 × 0.02 = 0.08 cm³. Percentage = 0.08/40.00 × 100 = 0.20%.
Q3. For q = mcΔT, the stated percentage uncertainties in m and ΔT are 0.4% and 2.5%. Treat c as exact. Estimate the maximum percentage uncertainty in q.Show answer
For this multiplication, add the contributions: 0.4 + 2.5 = 2.9%. This estimate describes the stated measurements and does not include unmeasured energy loss.
Q4. Why can repeating a titration twenty times fail to correct a calculated concentration that is consistently too high?Show answer
A systematic cause can persist, such as a procedure that consistently overshoots or an incorrectly standardised reagent. Repetition assesses scatter but does not remove bias. Identify which measured quantity and calculation produce the high concentration, then correct the cause.
Q5. Convert 35.0 cm³ to dm³ and m³ without losing its stated significant figures.Show answer
35.0 cm³ = 0.0350 dm³ = 3.50 × 10⁻⁵ m³. Each form retains three significant figures. The factor from cm³ to m³ is 10⁻⁶, not 10⁻³.
Sources
Sources and examiner guidance (reviewed 9 October 2026)
- Pearson Edexcel 9CH0 specification, Issue 3 — Appendix 5a, pp77–78; mathematical skills, pp85–89; command words, pp91–92. AS students use the common skills, with advanced examples signposted.
- Chemrevise: Edexcel practical guide — Secondary practical coverage check; pp1–8 for measurement and volumetric techniques. Methods and conclusions here are original Finesse teaching.
- Pearson June 2023 9CH0/03 mark scheme — Q7(a)(iii), PDF p27: comparing percentage uncertainty for a combined delivered volume.
- Pearson June 2023 9CH0/03 examiner report — Introduction, p3; Q7(a)(ii–iii), pp61–62: calculation presentation, rounding and choosing the correct total-volume denominator.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.
