IB Chemistry SLTopic 8 — Collecting and Processing DataInternal assessmentPractical skill~14 min read
Processing Data
Processing is where a page of readings becomes a single answer. It is also where a perfectly good experiment gets thrown away, usually by copying nine digits off a calculator and forgetting to say how sure you are about any of them.
📚 What you need to know
Processed data goes in its own table, separate from the raw one.
Show one full worked calculation for every type of calculation you do.
Average only concordant results. The rough titre and justified anomalies are left out.
Adding or subtracting → add the absolute uncertainties.
Multiplying or dividing → add the percentage uncertainties.
The final answer carries the significant figures of your least precise measurement.
Quote the result as value ± uncertainty, rounded so the two match.
Two tables, not one
Your raw table is evidence and must not be edited. So processing produces a second table beside it, and the reader can look from one to the other and follow exactly what you did. Merging them saves half a page and costs you the ability to prove anything.
The processed table is where units get invented. A rate in s−1 did not come off any instrument — you made it, so you have to label it.
Spreadsheets are fine, and nobody is asking you to do fifteen calculations by hand. But a column of numbers a computer produced proves nothing about whether you understand the chemistry. Write out one full calculation, longhand, for each type. That single worked example is what the marks are actually for.
Averaging, and what to leave out
A mean is only meaningful if the values going into it were measuring the same thing under the same conditions. Two results belong in a mean; two results and a mistake do not.
In titrations there is a firm convention. Results are concordant when they agree to within 0.10 cm3, and only concordant titres are averaged. Two things are always excluded:
The rough titre. It was a pilot, run fast to find roughly where the endpoint lives. It was never meant to be accurate.
Any anomaly you can justify. Not any result you dislike — one you can point to a reason for.
Watch the trap here. If your three “proper” titres are 21.55, 21.60 and 22.40 cm3, you do not average all three and you do not simply bin the odd one. You run more trials until you have three that agree. Concordancy is something you achieve in the lab, not something you arrange afterwards.
Carrying the uncertainty through
Every measurement arrived with an uncertainty attached, and those uncertainties do not evaporate when you press the equals key. They travel through the calculation into your final answer. There are only two rules to learn, and which one you use depends entirely on what the calculator is doing.
The two rules
adding or subtracting → add the absolute uncertainties
multiplying or dividing → add the percentage uncertainties
The left-hand box is why a burette titre carries ±0.10 cm3 and not ±0.05. You read the scale twice, so you were uncertain twice.
To turn an absolute uncertainty into a percentage, divide it by the measurement and multiply by 100. To go back the other way at the end, take that percentage of your final answer. Students lose marks by leaving the answer as “±3.18%” — convert it back into real units, because that is the form the reader can actually judge.
Significant figures and the final quote
Your calculator does not know how good your thermometer was. Rounding is not tidying up; it is the last honest statement you make about the quality of your data.
The rule is that your answer carries the number of significant figures of the least precise measurement that went into it. If one value in the chain was 3 s.f., the answer is 3 s.f., however many digits everything else had.
Uncertainties are quoted to one significant figure, and then the value is rounded to sit at the same decimal place. Do it in that order and it always works.
WORKED EXAMPLE
A student titrating diluted lemon drink against 0.0500 mol dm−3 NaOH records burette readings giving a rough titre of 22.10 cm3 and then 21.55, 21.60 and 21.50 cm3. Find the mean titre to use in the calculation.
Which values qualifyThe rough titre is a pilot run and is always excluded. The other three span 21.50 to 21.60, a spread of 0.10 cm³, so they are concordant.The mean(21.55 + 21.60 + 21.50) ÷ 3 = 21.55 cm³mean titre = 21.55 cm³ (±0.10)Why the decimal places stayThe readings had two, so the mean keeps two. Rounding to 21.6 here would throw away precision you actually earned at the burette.
WORKED EXAMPLE
25.00 cm3 of the diluted drink needed 21.55 cm3 of 0.0500 mol dm−3 NaOH. Citric acid reacts with NaOH in a 1:3 ratio, and the drink had been diluted five-fold before titrating. Find the concentration of citric acid in the original drink.
Step 1: moles of NaOHn = c × V = 0.0500 × (21.55 ÷ 1000) = 1.0775 × 10−3 molStep 2: use the ratioCitric acid has three acidic hydrogens, so one mole of it neutralises three moles of NaOH.n(citric acid) = 1.0775 × 10−3 ÷ 3 = 3.592 × 10−4 molStep 3: concentration in the diluted samplec = n ÷ V = 3.592 × 10−4 ÷ 0.02500 = 0.01437 mol dm−3Step 4: undo the dilution0.01437 × 5 = 0.07183 mol dm−3Step 5: significant figuresThe least precise value used was 0.0500 mol dm−3, which has 3 s.f., so the answer does too.0.0718 mol dm−3Forgetting the 1:3 ratio is the classic error here, and it makes the answer three times too big.
WORKED EXAMPLE
25.00 cm3 of 1.00 mol dm−3 HCl (pipette ±0.03 cm3) was mixed with 25.00 cm3 of 1.00 mol dm−3 NaOH in a polystyrene cup. The temperature rose by 6.8 °C, measured with a thermometer reading to ±0.1 °C. Take the density as 1.00 g cm−3 and c as 4.18 J g−1 K−1. Find the enthalpy of neutralisation with its uncertainty.
Step 1: heat releasedq = mcΔT = 50.00 × 4.18 × 6.8 = 1421.2 J = 1.4212 kJThe mass is the whole 50.00 cm³ of mixture, not just the acid.Step 2: moles reactingn = 1.00 × 0.02500 = 0.02500 molStep 3: enthalpy changeΔH = −1.4212 ÷ 0.02500 = −56.848 kJ mol−1Step 4: percentage uncertaintiesMass: (0.06 ÷ 50.00) × 100 = 0.12%. Temperature rise, two readings so ±0.2: (0.2 ÷ 6.8) × 100 = 2.94%. Moles: (0.03 ÷ 25.00) × 100 = 0.12%.total = 0.12 + 2.94 + 0.12 = 3.18%Step 5: back to real units3.18% of 56.848 = 1.81 → ±2 (1 s.f.)ΔH = −57 ± 2 kJ mol−1The insight worth writing downThe temperature rise supplies 2.94 of that 3.18% — over 90% of the total. A more expensive pipette would change nothing. A finer thermometer, or a bigger temperature rise from more concentrated solutions, would change everything.
💡 Exam tip
Give the processed data its own table, with derived units in the headers.
Show one full calculation per type, even if a spreadsheet did the rest.
Say which results you averaged and why the others were left out.
Pick the rule by the operation: absolutes for ±, percentages for × and ÷.
Round the uncertainty first, to 1 s.f., then match the value to it.
Name the largest percentage uncertainty — it hands you your evaluation for free.
⚠️ Common mix-up
Averaging everything, rough titre and anomaly included.
Adding percentage uncertainties in a subtraction, or absolutes in a multiplication.
Forgetting a titre involves two readings, so it carries ±0.10, not ±0.05.
Copying the full calculator display into the final answer.
Leaving the uncertainty as a percentage instead of converting it back to units.
Losing the units on a quantity you invented yourself, like a rate.
Up next: Interpreting Results — you have one number and an uncertainty. The next page is about plotting it, saying what the pattern is, and then saying why the chemistry produced that pattern.
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