Analysing: Trends and Relationships¶
Syllabus mapping
Working Scientifically outcome — Analysing (part 1 of 2). Covers identifying trends and relationships in processed data, comparing a result against a known/accepted value, and recognising the difference between interpolation and extrapolation. See Evaluating Data and Conclusions for the second half of Analysing — reliability, validity, limitations, and drawing a conclusion.
Outcome codes: CH-11WS-05 (Year 11) and CH-12WS-05 (Year 12) — Analysing data and information. Same outcome wording in both years, with a separate code for each. Wording on this page is paraphrased for teaching use — check the official syllabus before treating any line as verbatim NESA text.
You need to know: how to describe what a graph is actually telling you in precise language, and how to compare an experimental result against an accepted value properly — not just "close" or "not close."
Core¶
Describing a trend precisely¶
"The graph goes up" isn't an analysis — it doesn't say how it goes up, which is usually the entire point of the investigation. Once a relationship is linearised (see Representing Data), describe it in terms of proportionality:
- Directly proportional — a straight line through the origin; doubling \(x\) doubles \(y\). (Volume of a gas vs kelvin temperature; mass of product vs mass of limiting reagent.)
- Linear, but not directly proportional — a straight line with a non-zero y-intercept; a change in \(x\) still produces a predictable change in \(y\), but \(y\) isn't simply a multiple of \(x\). (Volume of a gas vs temperature in °C.)
- Inversely proportional — \(y\) decreases as \(x\) increases, in a \(y \propto 1/x\) pattern (a curve when plotted directly; a straight line through the origin when plotted against \(1/x\)). (Pressure vs volume of a gas.)
A trend statement should name the pattern and point at the evidence for it: "Pressure is inversely proportional to volume, shown by the straight line of best fit through the origin on the graph of \(P\) against \(\dfrac{1}{V}\)" is analysis; "the graph is a straight line" alone is just description.
Interpolation vs extrapolation¶
- Interpolating — reading a value from within the range of data actually collected. Generally safe, since you have evidence the relationship holds there.
- Extrapolating — reading a value from beyond the tested range. Riskier: you're assuming the same relationship continues past the last data point, which isn't guaranteed. Many chemistry relationships genuinely break down outside a certain range — a gas stops behaving ideally at very low temperatures (it turns into a liquid), and water can't be used as a solvent above 100 °C at normal pressure.
Worked example
A calibration curve of absorbance against concentration is a straight line for standard solutions from 0.010 to 0.050 mol L⁻¹. Reading the concentration of an unknown with an absorbance in the middle of that range is interpolation — safe. Reading one with an absorbance twice the highest standard is extrapolation, and risky: at high concentrations the line often starts to curve. The fix is to dilute the unknown until it falls within the calibrated range.
Comparing a result to an accepted value¶
Where a quantity has a known accepted value (e.g. \(\Delta H\) for a neutralisation, the molar volume of a gas, the formula of magnesium oxide), a result is compared using percentage error, not percentage uncertainty (see Precision and Uncertainty — these are two different calculations that get mixed up constantly):
Worked example
A calorimetry experiment gives \(\Delta H_{\text{neutralisation}} = -52 \pm 4\ \text{kJ mol}^{-1}\) (see Energy Changes). The accepted value is \(-57\ \text{kJ mol}^{-1}\).
This alone doesn't say whether the result is "good" — that depends on comparing it against the uncertainty. The experimental range is \(-48\) to \(-56\ \text{kJ mol}^{-1}\), which does not include \(-57\). So the result does not agree with the accepted value within experimental uncertainty — a sign of a systematic error. In calorimetry, the obvious suspect is heat lost to the surroundings, which always makes \(\Delta H\) come out too small.
Percentage error vs percentage difference¶
Percentage error assumes there's a known accepted value to compare against. When comparing two experimental results against each other instead — such as two groups' values, or the same reaction measured in two different calorimeters — neither one is "correct," so the comparison is a percentage difference, using the mean of the two as the reference point:
Advanced
When "agreement within uncertainty" isn't enough. Two results can agree within their stated uncertainties and still both be systematically wrong in the same direction — uncertainty ranges only capture random error, not systematic error (see Precision and Uncertainty). Two groups who both use uninsulated beakers will get similar \(\Delta H\) values that are both too small. A result that agrees with another within uncertainty is good evidence the random error was reasonably estimated; it says nothing about whether a systematic error was present, which is why a full evaluation (see Evaluating Data and Conclusions) has to consider both separately.
The boxes below are optional, tertiary-level statistical tools — beyond this syllabus, but exactly the kind of thing the strongest students reach for in a depth study. Each stands alone and builds on the mean/standard deviation content from Precision and Uncertainty.
Extension — Standard error of the mean: how much to trust your average
Beyond the syllabus for this course — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
Standard deviation, \(\sigma\), describes how spread out individual readings are. It does not, by itself, say how confident you should be in the mean of those readings. The standard error of the mean (SEM) answers that:
SEM shrinks as \(n\) grows, even though \(\sigma\) doesn't — averaging more repeats narrows your confidence in the mean, without making any individual reading more precise.
Worked example
Five calorimetry results for \(\Delta H_{\text{neutralisation}}\): −53, −55, −52, −56, −54 kJ mol⁻¹.
Mean \(\bar{x} = -54.0\). Standard deviation (calculator, \(\sigma_{n-1}\) key): \(\sigma \approx 1.58\).
Report as \(\Delta H = -54.0 \pm 0.7\ \text{kJ mol}^{-1}\).
Extension — The one-sample t-test: is your result really different from the accepted value?
Beyond the syllabus for this course — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
The one-sample t-test checks rigorously whether your mean is genuinely different from an accepted value, by calculating how many standard errors apart they are:
where \(\mu\) is the accepted value. Compare \(|t|\) with a critical value from a t-table (degrees of freedom \(df = n - 1\), usually at 95% confidence). If \(|t|\) is larger, the difference is statistically significant — unlikely to be random chance alone.
Worked example
Using the five results above: \(\bar{x} = -54.0\), \(SE = 0.71\), accepted value \(\mu = -57.0\).
With \(df = 4\), the critical value at 95% confidence is \(2.776\). Since \(4.2 > 2.776\), the difference is statistically significant — the results are consistently too small, which is strong evidence of a systematic error (heat loss), not just random scatter.
Extension — The two-sample t-test: comparing two experimental results
Beyond the syllabus for this course — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
A simplified two-sample t-test compares two means, each with its own spread and number of repeats:
(This simplified approach uses \(df \approx n_1 + n_2 - 2\). A full statistics course uses Welch's t-test, which adjusts the degrees of freedom when the two spreads differ.)
Worked example
The same neutralisation in two calorimeters: a polystyrene cup gives \(\bar{x}_1 = -54.0\), \(s_1 = 1.6\), \(n_1 = 5\); a glass beaker gives \(\bar{x}_2 = -49.0\), \(s_2 = 2.0\), \(n_2 = 5\).
With \(df \approx 8\), the critical value at 95% confidence is \(2.306\). Since \(|-4.4| > 2.306\), the two calorimeters give significantly different results — evidence that the glass beaker loses noticeably more heat.
Extension — Chi-squared goodness-of-fit
Beyond the syllabus for this course — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
Comparing a whole set of data points against a predicted line (for example, a full calibration curve, or gas law data against \(PV = nRT\)) gets more rigorous treatment at university via chi-squared goodness-of-fit testing — a single statistic that weighs how far every point sits from the prediction, accounting for each point's uncertainty. Worth researching if a depth study collects enough data points (10+) that a single percentage-error comparison feels like it's throwing away information.
Video/visual resources¶
- 🎥 Khan Academy — TODO: source a percentage error / comparing results explainer that distinguishes percentage error (vs an accepted value) from percentage difference (between two experimental results).
Check yourself¶
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An experiment at 25 °C and 100 kPa gives the molar volume of a gas as \(23.9 \pm 0.6\ \text{L mol}^{-1}\). The accepted value at those conditions is \(24.79\ \text{L mol}^{-1}\). Calculate the percentage error, and state whether the result agrees with the accepted value within uncertainty.
Answer
\(\%\text{ error} = \dfrac{|23.9 - 24.79|}{24.79} \times 100\% \approx 3.6\%\). The experimental range is \(23.3\) to \(24.5\ \text{L mol}^{-1}\), which does not include \(24.79\) — so the result does not agree within uncertainty, despite the small percentage error. (A likely cause: some gas dissolved in the water it was collected over.)
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A solubility curve for potassium nitrate was measured between 20 °C and 60 °C. Is using it to find the solubility at 45 °C interpolation or extrapolation? What about at 90 °C?
Answer
45 °C is within the measured range, so it's interpolation — reasonably safe. 90 °C is outside the range, so it's extrapolation — less reliable, because there's no evidence the curve keeps the same shape that far.
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Two groups measure \(\Delta H\) for the same reaction and get \(-54\) and \(-49\ \text{kJ mol}^{-1}\). Calculate the percentage difference.
Answer
Mean \(= \dfrac{-54 + (-49)}{2} = -51.5\). \(\%\text{ difference} = \dfrac{|-54 - (-49)|}{51.5} \times 100\% \approx 9.7\%\).
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(Stretch — uses the Extension boxes above.) Five repeated readings give a mean of \(24.6\) with standard deviation \(\sigma = 0.5\). Calculate the standard error of the mean, and explain in one sentence why it's smaller than \(\sigma\).
Answer
\(SE = \dfrac{0.5}{\sqrt{5}} \approx 0.22\). It's smaller than \(\sigma\) because SEM measures confidence in the mean of several readings, not the spread of individual readings — averaging narrows the uncertainty in the mean.