1. The Uncertainty of a Reading: Rules Used on This Site
Every reading has an uncertainty. This site uses one set of rules on every Unit 5 and required practical page.
2. Digital reading (balance, digital thermometer, stopwatch): the uncertainty is ± the last digit.
3. A value found by difference counts two readings, so double the uncertainty. This applies to a burette titre, a mass found by weighing by difference and a temperature change ΔT.
Use these typical values unless a question gives you a different value. If a question or the apparatus gives a value, always use that value.
| Apparatus | Uncertainty per reading | Readings for one result |
|---|---|---|
| Burette (0.1 cm³ divisions) | ±0.05 cm³ | 2 (initial and final), so ±0.10 cm³ for a titre |
| Pipette (25 cm³) | ±0.06 cm³, or the value printed on the pipette | 1 |
| Balance (3 decimal places) | ±0.001 g | 2 when weighing by difference |
| Balance (2 decimal places) | ±0.01 g | 2 when weighing by difference |
| Thermometer (1 °C divisions) | ±0.5 °C | 2 for a temperature change ΔT, so ±1.0 °C |
2. Random vs Systematic Errors
Experimental error represents the unavoidable difference between a measured value and the true physical value. In OxfordAQA Unit 5, candidates must classify errors accurately and evaluate their impact on calculated results.
Random Errors
Definition: Unpredictable fluctuations that cause individual measurements to fall randomly above or below the true value.
Typical Causes:
- Human reaction time when starting/stopping a stopwatch (RP8).
- Parallax error when viewing a burette meniscus from slightly different angles.
- Temperature fluctuations in the laboratory during an experiment.
Remedy: Repeat the experiment multiple times and calculate a mean of concordant results (e.g. within 0.10 cm³ in titrations). Repeating averages out random scatter.
Systematic Errors
Definition: Errors that cause measurements to be consistently too high or consistently too low by the same margin in every trial.
Typical Causes:
- Calorimeter heat loss to surroundings without insulation (RP2).
- Balance zero error (tare not set to 0.00 g).
- Gas dissolving in water trough instead of collecting in syringe.
- Incompletely dried precipitate containing residual solvent.
Remedy: Repeating trials does NOT remove systematic error. The apparatus must be recalibrated or the experimental technique improved (e.g. using cooling curve extrapolation or polystyrene lids).
3. Volumetric Glassware Percentage Uncertainty Formulas
Every piece of volumetric equipment carries an inherent apparatus tolerance (uncertainty) marked on the glass. The general percentage uncertainty equation is:
| Apparatus | Tolerance | Readings Taken | Delivered Volume | Exact Percentage Uncertainty Formula | Typical % Uncertainty |
|---|---|---|---|---|---|
| Volumetric Pipette (25.0 cm³) | ±0.06 cm³ | 1 reading | 25.00 cm³ | [ (0.06 × 1) / 25.00 ] × 100 | 0.24% |
| Volumetric Flask (250.0 cm³) | ±0.20 cm³ | 1 reading | 250.0 cm³ | [ (0.20 × 1) / 250.0 ] × 100 | 0.08% |
| Burette Titration (Titre = 20.00 cm³) | ±0.05 cm³ | 2 readings (initial & final) | 20.00 cm³ | [ (0.05 × 2) / 20.00 ] × 100 = [ 0.10 / 20.00 ] × 100 | 0.50% |
| Small Burette Titre (Titre = 5.00 cm³) | ±0.05 cm³ | 2 readings (initial & final) | 5.00 cm³ | [ (0.05 × 2) / 5.00 ] × 100 = [ 0.10 / 5.00 ] × 100 | 2.00% (4x higher!) |
A burette delivers volume by difference: Titre = Final Reading - Initial Reading. Even if the initial reading is set to 0.00 cm³, that 0.00 reading still has an uncertainty of ±0.05 cm³. Because both the initial and final levels must be judged by the human eye against the calibration marks, the absolute uncertainty doubles: 0.05 × 2 = 0.10 cm³. Candidates who write 0.05 / Titre lose full marks on Unit 5.
4. Balance Uncertainty & Weighing by Difference
In standard solution preparation (RP1) and gravimetric analysis, solid compounds are weighed on a digital analytical balance. A typical school balance measures to two decimal places (±0.01 g). An analytical balance measures to three decimal places (±0.001 g).
Mass Transferred = Mass of boat with solid - Mass of boat after emptying
Because two separate mass readings are recorded on the balance:
Percentage Uncertainty = [ (Balance Uncertainty × 2) / Mass Transferred ] × 100
A candidate prepares a standard volumetric solution of anhydrous sodium carbonate (Na₂CO₃) using a balance with an apparatus uncertainty of ±0.01 g:
- Mass of weighing boat + solid = 14.85 g
- Mass of weighing boat after transfer = 12.35 g
- Mass of Na₂CO₃ transferred = 14.85 - 12.35 = 2.50 g
Calculate the percentage uncertainty in the mass of sodium carbonate transferred:
Percentage Uncertainty = [ 0.02 / 2.50 ] × 100 = 0.80%
Answer: 0.80%
5. Thermometer Uncertainty & Temperature Rises (ΔT)
In calorimetry experiments (RP2), the temperature change ΔT is calculated as ΔT = T_final - T_initial. Because temperature change requires reading the thermometer twice, the apparatus uncertainty doubles:
Consider a standard laboratory thermometer calibrated in 1 °C divisions, with a reading tolerance of ±0.5 °C:
- For a large temperature change (ΔT = 25.0 °C):
% Uncertainty = [ (0.5 × 2) / 25.0 ] × 100 = [ 1.0 / 25.0 ] × 100 = 4.0% - For a very small temperature change (ΔT = 2.5 °C):
% Uncertainty = [ (0.5 × 2) / 2.5 ] × 100 = [ 1.0 / 2.5 ] × 100 = 40.0% (unacceptable error!)
6. Error Propagation Rules in Chemical Calculations
When multiple experimental quantities are combined in mathematical formulas (such as calculating enthalpy change from q = mcΔT and n = cV), individual percentage uncertainties propagate through the calculation:
| Mathematical Operation | General Formula | Uncertainty Rule |
|---|---|---|
| Multiplication or Division | Z = A × B or Z = A / B | Add the percentage uncertainties: % Uncertainty in Z = (% Uncertainty in A) + (% Uncertainty in B) |
| Power / Exponent | Z = A^n | Multiply percentage uncertainty by the power: % Uncertainty in Z = n × (% Uncertainty in A) |
| Addition or Subtraction | Z = A + B or Z = A - B | Add the absolute uncertainties: Absolute Uncertainty in Z = (Absolute in A) + (Absolute in B) |
7. Strategies to Reduce Experimental Uncertainty
A classic Unit 5 question presents an experimental scenario and asks candidates to suggest two practical modifications to reduce the overall percentage uncertainty.
| Experimental Technique | Problem Causing High Uncertainty | Modification to Reduce Percentage Uncertainty | Explanation & Justification |
|---|---|---|---|
| Titration (RP1) | Titre volume is too small (e.g. 6.20 cm³, % uncertainty = 1.61%). | 1. Decrease the concentration of the titrant in the burette. 2. Increase the volume/concentration of the sample in the conical flask. |
Both changes force a larger titre volume (e.g. ~25.0 cm³). Because delivered volume appears in the denominator, increasing the titre reduces percentage uncertainty to ~0.40%. |
| Weighing Solid (RP1, RP10) | Mass weighed is very small (e.g. 0.15 g on 2-decimal balance). | 1. Use a balance with 3 or 4 decimal places (±0.001 g). 2. Weigh a larger mass of solid and dissolve in a larger volumetric flask. |
Using a more sensitive balance reduces the numerator; weighing a larger sample increases the denominator. Both reduce % uncertainty. |
| Calorimetry (RP2) | Temperature rise is too small (e.g. ΔT = 1.8 °C). | 1. Increase the concentration of the reacting solutions. 2. Increase the mass of solid reactant added. |
Releasing more heat energy into the same volume of water produces a substantially larger temperature change, decreasing percentage uncertainty. |
8. Worked Error Analysis Problem
Problem: In an experiment to determine the enthalpy of neutralisation (RP2), a student mixes 25.0 cm³ of 1.00 mol dm⁻³ HCl with 25.0 cm³ of 1.00 mol dm⁻³ NaOH in a polystyrene cup. The student records the following measurements:
- Volume of HCl (pipette tolerance ±0.06 cm³) = 25.00 cm³
- Volume of NaOH (pipette tolerance ±0.06 cm³) = 25.00 cm³
- Initial temperature = 19.5 °C (thermometer ±0.5 °C)
- Maximum temperature = 26.0 °C (thermometer ±0.5 °C)
Calculate the percentage uncertainty in each measurement and evaluate the overall combined apparatus percentage uncertainty in the calculated enthalpy change.
Step 1: Calculate percentage uncertainty in HCl volume
% Uncertainty (HCl) = (0.06 / 25.00) × 100 = 0.24%
Step 2: Calculate percentage uncertainty in NaOH volume
% Uncertainty (NaOH) = (0.06 / 25.00) × 100 = 0.24%
Step 3: Calculate percentage uncertainty in temperature rise (ΔT)
ΔT = 26.0 - 19.5 = 6.5 °C
% Uncertainty (ΔT) = [ (0.5 × 2) / 6.5 ] × 100 = (1.0 / 6.5) × 100 = 15.38%
Step 4: Percentage uncertainty in the mass of solution (m)
Total volume = 25.0 + 25.0 = 50.0 cm³, with absolute uncertainty = 0.06 + 0.06 = 0.12 cm³
% Uncertainty (m) = (0.12 / 50.0) × 100 = 0.24%
Step 5: Percentage uncertainty in the amount of acid (n)
n = 25.00 cm³ × 1.00 mol dm⁻³, so n comes from the pipetted volume of HCl
% Uncertainty (n) = (0.06 / 25.00) × 100 = 0.24%
Step 6: Combine percentage uncertainties for ΔH = mcΔT / n
Total % Uncertainty = % (m) + % (ΔT) + % (n)
Total % Uncertainty = 0.24% + 15.38% + 0.24% = 15.86% ≈ 15.9%
Conclusion: The thermometer reading is by far the largest source of apparatus uncertainty (15.4% out of 15.9% total).
9. Practice Exam Questions
Question 1: A student records a mean titre of 18.40 cm³ using a burette with an uncertainty of ±0.05 cm³ per reading. What is the percentage uncertainty in this titre?
Show Answer & Explanation
Correct Answer: B
A burette titre requires two readings (initial and final), so the absolute uncertainty is 0.05 × 2 = 0.10 cm³. Percentage uncertainty = (0.10 / 18.40) × 100 = 0.543% ~ 0.54%.
Question 2: Which procedural modification would most effectively reduce the percentage uncertainty in a titration where the mean titre is only 4.50 cm³?
Show Answer & Explanation
Correct Answer: B
Diluting the titrant quadruples the required volume of titrant to reach neutralisation (from 4.50 cm³ to 18.00 cm³), reducing percentage uncertainty from 2.22% down to 0.56%.