B.1 Handling data
- Syllabus
- 2017
- Topic
- —
- Level
- A2
Significant figures communicate how precisely a value is supported, not how many digits a calculator can display. Keep extra digits through intermediate working, then round the final result once, using the raw measurements and any explicit instruction to decide the justified precision.
| Situation | Reporting guide |
|---|---|
| multiplication or division | match the measured input with the fewest significant figures |
| addition or subtraction | match the least precise decimal place |
| exact count or stoichiometric coefficient | does not limit significant figures |
| stated answer precision | follow the instruction after completing the calculation |
For 80.0−15.846=64.154kg, the subtraction is limited by 80.0kg to the tenths place, so 64.2kg is appropriate. Writing 64.154kg claims precision that the first measurement did not provide.
Precision must also fit the quantity. A calculated number of protons, neutrons, atoms or molecules represents a count, so a physically interpreted answer may need to be a whole number even if earlier data contain several significant figures.
Do not round every intermediate value to the final precision: accumulated rounding can change the answer. Decimal places and significant figures are different, and trailing zeros after a decimal may be essential evidence of precision.
\bar{x}=\frac{\sum x}{n}\qquad \text{weighted mean}=\frac{\sum(x_iw_i)}{\sum w_i}
| Data situation | Mean to use |
|---|---|
| repeated comparable measurements | arithmetic mean of the accepted values |
| isotopes with different abundances | mass weighted by abundance |
| mixture components with stated fractions | property weighted by component fraction |
| titration containing an outlier | mean of concordant accurate titres only |
For isotopes of masses 35 and 37 with abundances 75% and 25%, Ar=(35imes75+37imes25)/100=35.5. The result lies between the isotope masses and closer to the more abundant isotope, providing a useful check.
Calculate each titre as final minus initial burette reading. Exclude the rough value, then identify concordant accurate titres: in this Edexcel chemistry context, accepted titres agree within 0.20cm3. For 24.30, 23.80 and 24.20cm3, use 24.30 and 24.20 only, giving a mean of 24.25cm3.
Do not average every recorded value automatically. Excluding a value requires a stated concordance or outlier rule, while a weighted mean must divide by the total weight rather than merely by the number of entries.
\text{percentage uncertainty}=\frac{\text{absolute uncertainty}}{|\text{measured change or value}|}\times100%
| Derived value | Simple uncertainty treatment |
|---|---|
| one direct reading | use the stated uncertainty for that reading |
| difference of two readings | add their absolute uncertainties |
| mass by difference | include both balance readings |
| titre or temperature change | include both initial and final readings |
A titre is final burette reading minus initial reading. If each reading is ±0.05cm3, the titre uncertainty is ±0.10cm3. For an 18.95cm3 titre, the percentage uncertainty is (0.10/18.95)imes100=0.53%.
If a reported mass of 9.53g came from two balance readings, each with ±0.01g uncertainty, the mass by difference is 9.53±0.02g, giving the possible range 9.51 to 9.55g.
With the same apparatus, reduce percentage uncertainty by measuring a larger change while keeping the chemistry valid—for example, a larger temperature rise makes a fixed thermometer uncertainty a smaller fraction of the result.
Subtract readings to obtain the measured change, but add their absolute uncertainties. Repeating can reveal scatter, yet it does not halve the stated uncertainty of each instrument reading.