IB Chemistry SL Topic 7 — Mathematics Paper 1 & 2 Core skill ~13 min read

Working with Uncertainties

An uncertainty is not an admission that you did something wrong. It is a statement of how sharply your equipment could ever have answered the question — and comparing it against how far off you actually were is the most useful thing you can do with a set of results.

📚 What you need to know

Uncertainty is not error

The words get swapped constantly and they describe different things. An uncertainty is a property of the measurement: a burette marked every 0.10 cm3 simply cannot resolve better than about ±0.05 cm3, no matter how careful you are. An error is something that pushed the reading away from the truth — heat lost to the room, an uncalibrated probe, a reading taken from above.

The practical consequence: you can reduce an uncertainty by choosing better apparatus, but you reduce an error by changing the method.

Where the number comes from

THREE RULES, AND THE INSTRUMENT PICKS ONEANALOGUE SCALEDIGITAL DISPLAYREPEATED DATA± half a division± the last digit shown± half the range2.43burette: ±0.05 cm³balance: ±0.01 ghalf of max − minlook at the instrument first, then decide which of the three rules appliesa difference needs two readings, so its uncertainty is doubleda titre, a temperature rise and a mass change are all differences
That last line covers most of the practical work you will do. A titre, a ΔT and a change in mass are all found by reading an instrument twice, so each carries twice the single-reading uncertainty.

Absolute, fractional, percentage

Three ways of saying the same thing. Take a burette reading of 19.60 cm3 on a scale divided every 0.10 cm3.

The same uncertainty, three ways absolute = 0.10 ÷ 2 = ±0.05 cm3
fractional = 0.05 ÷ 19.60 = 0.0026
percentage = 0.0026 × 100 = 0.26%

The percentage form is the one that lets you compare. An uncertainty of ±0.05 cm3 is trivial on a 25 cm3 titre and serious on a 2 cm3 one, and only the percentage tells you which situation you are in.

This is exactly why percentage uncertainty falls when you measure more. The ±0.05 cm3 stays the same whatever you do, so making the titre bigger — by using a more dilute titrant, say — puts a bigger number underneath it. The other route is finer equipment, which shrinks the top instead.

Combining uncertainties

WHICH UNCERTAINTY DO YOU ADD?the operation decides, so check it before you combine anything+ or −× or ÷poweradd the ABSOLUTE uncertaintiestwo burette readings at ±0.05 give a titre of ±0.10add the PERCENTAGE uncertaintiesn = cV, so the two percentages are added togethermultiply the percentage by the powersquaring a value doubles its percentage uncertaintynever add an absolute to a percentage; convert first, then combineand convert the final percentage back to an absolute for the answer
The last step matters. A result quoted as 2.45 × 10–3 ± 0.91% is not wrong, but the expected form is an absolute uncertainty in the same units as the result.

Uncertainty against error: the useful comparison

Here is the move that turns a set of numbers into an evaluation. Work out the total percentage uncertainty of your result, then work out the percentage error against the literature value, and compare them.

DOES YOUR UNCERTAINTY EXPLAIN YOUR ERROR?ERROR WITHIN UNCERTAINTYERROR BEYOND UNCERTAINTY% error is smaller than thetotal % uncertainty% error is larger than thetotal % uncertaintyrandom errors alone explain ita systematic error is presentthe result agrees within its uncertaintysomething is biasing every readingthe band is the total uncertainty and the red mark is the percentage errorcomparing the two tells you which kind of error you are dealing withwhich is exactly what an evaluation section is asking you to do
If the error is far larger than the uncertainty, no amount of careful reading would have saved the result. Something in the method is biasing it, and that is what the evaluation should name.
Uncertainty bars on a graph make the same point visually. Draw each point with a bar of ± its absolute uncertainty; if a straight line can be drawn passing through every bar, the data supports a linear relationship. If it cannot, either the relationship is not linear or something has been underestimated.

The coefficient of determination

Spreadsheets will offer you R2 alongside a trend line. It measures how well the line fits the points: 0 means no predictive value at all, 1 means the line passes through every point exactly, and anything between describes the quality of the fit.

Treat it carefully. A high R2 says the line describes those points well; it does not confirm that a straight line was the right model in the first place, and it says nothing about whether a systematic error shifted them all together.

WORKED EXAMPLE

A titration delivers 24.50 ± 0.10 cm3 of a solution of concentration 0.1000 ± 0.0005 mol dm–3. Calculate the amount in moles and its absolute uncertainty.

Step 1 — the two percentage uncertainties volume: 0.10 ÷ 24.50 × 100 = 0.41% concentration: 0.0005 ÷ 0.1000 × 100 = 0.50% Step 2 — the operation is a multiplication n = cV, so add the percentages. 0.41 + 0.50 = 0.91% Step 3 — the value n = 0.1000 × 0.02450 = 2.450 × 10⁻³ mol Step 4 — convert back to absolute 0.91% of 2.450 × 10⁻³ = 0.022 × 10⁻³ n = (2.450 ± 0.022) × 10⁻³ mol The titre uncertainty is ±0.10 not ±0.05, because two burette readings were taken.
WORKED EXAMPLE

In a calorimetry experiment a thermometer reading to ±0.5 °C gives an initial temperature of 21.5 °C and a final temperature of 34.0 °C. Calculate the temperature rise, its absolute uncertainty and its percentage uncertainty.

Step 1 — the rise ∆T = 34.0 − 21.5 = 12.5 °C Step 2 — a subtraction, so add the absolutes 0.5 + 0.5 = ±1.0 °C Step 3 — convert to a percentage 1.0 ÷ 12.5 × 100 = 8.0% ∆T = 12.5 ± 1.0 °C, or ±8.0% 8% is large, and it comes straight from the thermometer. A probe reading to ±0.1 °C would cut it to 1.6% without changing anything else.
WORKED EXAMPLE

Using the temperature rise above, a student obtains an enthalpy of combustion of –520 kJ mol–1 where the literature value is –726 kJ mol–1. The total percentage uncertainty in the experiment is about 10%. Comment on the result.

Step 1 — the percentage error (726 − 520) ÷ 726 × 100 = 28.4% Step 2 — compare it with the uncertainty 28.4% is much larger than 10% the uncertainty cannot account for the discrepancy Step 3 — so what is left? A systematic error must be present. The result is far less exothermic than it should be, which is consistent with heat being lost to the surroundings and to the apparatus rather than reaching the water. Notice the logic: the comparison is what licenses you to say “systematic”. Without it, “there must have been heat loss” is only a guess.

💡 Exam tip

⚠️ Common mix-up

Up next: Graphing Skills — uncertainty bars, gradients and intercepts all live on a graph, and a graph drawn properly answers questions that a table of numbers cannot.

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