SACE Board SACE 2021-style practice: In an HPLC analysis of caffeine, standards gave peak areas: 5.0\ ppm → 1.20 × 10^4; 10.0\ ppm → 2.40 × 10^4; 15.0\ ppm → 3.60 × 10^4 (arbitrary units). An energy-drink sample, diluted 50-fold, gave a caffeine peak area of 2.04 × 10^4. Calculate the caffeine concentration in the undiluted drink in ppm.

Step 1: the standards are linear through the origin; gradient = 1.20 × 10^45.0 = 2.40 × 10^3\ area per ppm. (2 marks) Step 2: concentration of the diluted sample = areagradient = 2.04 × 10^42.40 × 10^3 = 8.5\ ppm. (1 mark) Step 3: correct for the 50-fold dilution: c = 8.5 × 50 = 425\ ppm. (1 mark)

§-Syllabus dot point

SAChemistrySyllabus dot point

How does chromatography separate and identify the components of a mixture?

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

Mobile and stationary phases, partitioning, retention time, and the difference between GC and HPLC, with worked SACE-style chromatogram interpretation and calibration calculations of component concentration.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

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

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

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.

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.

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 practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20214 marksIn 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 20193 marksA 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)