Chromatography: GC and HPLC (SACE Stage 2 Chemistry)

Syllabus dot point

“Explain the principles of chromatography, including gas chromatography (GC) and high-performance liquid chromatography (HPLC), and interpret chromatograms.”

9 min readUpdated 2 Jun 2026

Quick answer

Chromatography separates a mixture by repeatedly partitioning its components between a moving mobile phase and a fixed stationary phase. Components that interact more strongly with the stationary phase move slowly and have longer retention times. In gas chromatography (GC) the mobile phase is an inert gas, suited to volatile compounds; in HPLC it is a high-pressure liquid, suited to non-volatile or heat-sensitive ones. Peak position identifies a component (by retention time) and peak area quantifies it (against standards).

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What this dot point is asking
Lead worked calculation
The principle: partitioning
Gas chromatography (GC)
High-performance liquid chromatography (HPLC)
Interpreting a chromatogram
Why it matters for monitoring
Exam-style questions

What this dot point is asking

SACE expects you to explain partitioning and retention, distinguish GC from HPLC and their uses, and interpret a chromatogram, including calculating a concentration from peak areas.

Lead worked calculation

Quantifying a pesticide by GC peak area

A GC calibration for a pesticide gives peak areas of $4.0 \times 10^{3}$ at $0.20\ \text{ppm}$ , $8.0 \times 10^{3}$ at $0.40\ \text{ppm}$ and $1.2 \times 10^{4}$ at $0.60\ \text{ppm}$ . A river sample gives a peak area of $9.0 \times 10^{3}$ for the same pesticide. Find its concentration.

Find the calibration gradient

The points are linear through the origin: $k = \dfrac{4.0 \times 10^{3}}{0.20} = 2.0 \times 10^{4}\ \text{area per ppm}$ .

Read off the sample

c = \frac{\text{area}}{k} = \frac{9.0 \times 10^{3}}{2.0 \times 10^{4}} = 0.45\ \text{ppm}

The pesticide is present at $0.45\ \text{ppm}$ . Peak area, not height, is used because area is proportional to the amount of substance reaching the detector.

The principle: partitioning

Mobile and stationary phases

The stationary phase is fixed in place (a coating on the inside of the GC column, or packed particles in an HPLC column). The mobile phase flows through, carrying the sample. Each component constantly partitions between the two phases according to its relative attraction for each. A component held more strongly by the stationary phase spends more time stuck and travels more slowly; one held more strongly by the mobile phase moves quickly. This difference in partitioning is what separates the mixture.

The output is a chromatogram: a plot of detector signal against time, with one peak per separated component.

Retention time ( $t_R$ ): the time from injection to the peak maximum. It is characteristic of a compound under fixed conditions, so matching $t_R$ to a known standard identifies a component.
Peak area: proportional to the amount of that component, so it is used (with a calibration curve) to find concentration.

Gas chromatography (GC)

In GC the mobile phase is an inert carrier gas (helium or nitrogen) and the column sits in a temperature-controlled oven. The sample is vaporised on injection, so GC suits volatile, thermally stable compounds: petrochemicals, solvents, volatile pollutants, alcohol in blood. Components separate by a combination of their volatility and their attraction to the stationary phase, then reach a detector that produces the chromatogram.

High-performance liquid chromatography (HPLC)

In HPLC the mobile phase is a liquid solvent pumped through a packed column at high pressure. Because no vaporisation is needed, HPLC handles non-volatile, large, or heat-sensitive molecules that would decompose in GC: sugars, proteins, pharmaceuticals, caffeine, vitamins. Separation depends on the components' relative attraction to the liquid mobile phase and the solid stationary phase.

Choosing GC or HPLC

Use GC for small, volatile, heat-stable molecules; use HPLC for large, non-volatile or heat-sensitive ones. The deciding factor is whether the analyte can be vaporised intact. Both rely on the same partitioning principle and both quantify by peak area against a calibration curve.

Interpreting a chromatogram

To read a chromatogram: count the peaks (number of components), use each retention time to identify a component by comparison with standards run under identical conditions, and use each peak area with a calibration curve to find its concentration. Conditions (temperature, flow rate, column) must be identical for retention times to be comparable.

Using peak height and ignoring dilution

A common error is reading peak height instead of area for quantification; area accounts for peak broadening and is what is proportional to amount. Another is forgetting that retention times only match between runs done under identical conditions. As with AAS, if the sample was diluted before injection, multiply the calibration result by the dilution factor to recover the original concentration.

Why it matters for monitoring

Chromatography separates complex environmental mixtures so that individual pollutants, such as pesticides, hydrocarbons and drug residues, can be both identified (retention time) and quantified (peak area) at trace levels. GC and HPLC complement AAS, which targets metals, giving analysts the tools to profile almost any contaminant in air, water or soil.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2021-style4 marks

In an HPLC analysis of caffeine, standards gave peak areas:

5.0\ \text{ppm} \rightarrow 1.20 \times 10^{4}

;

10.0\ \text{ppm} \rightarrow 2.40 \times 10^{4}

;

15.0\ \text{ppm} \rightarrow 3.60 \times 10^{4}

(arbitrary units). An energy-drink sample, diluted

50

-fold, gave a caffeine peak area of

2.04 \times 10^{4}

. Calculate the caffeine concentration in the undiluted drink in

\text{ppm}

Show worked answer →

Step 1: the standards are linear through the origin; gradient $= \dfrac{1.20 \times 10^{4}}{5.0} = 2.40 \times 10^{3}\ \text{area per ppm}$ . (2 marks)

Step 2: concentration of the diluted sample $= \dfrac{\text{area}}{\text{gradient}} = \dfrac{2.04 \times 10^{4}}{2.40 \times 10^{3}} = 8.5\ \text{ppm}$ . (1 mark)

Step 3: correct for the $50$ -fold dilution: $c = 8.5 \times 50 = 425\ \text{ppm}$ . (1 mark)

SACE 2019-style3 marks

A chromatogram of a two-component mixture shows peaks at retention times

2.4\ \text{min}

and

5.1\ \text{min}

. Explain what retention time depends on, and why the component eluting at

5.1\ \text{min}

took longer to pass through the column.

Show worked answer →

Retention time is the time a component takes to travel from injection to the detector. It depends on how strongly the component is attracted to (interacts with) the stationary phase relative to the mobile phase. (1 mark)

A component that interacts more strongly with the stationary phase spends more time bound to it and less time moving with the mobile phase, so it moves more slowly through the column. (1 mark)

The component eluting at $5.1\ \text{min}$ therefore has a greater affinity for the stationary phase (or weaker affinity for the mobile phase) than the one eluting at $2.4\ \text{min}$ , so it is retained longer. (1 mark)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board