AQA A-Level Chemistry Paper 3, 2024: Question 34

Gas chromatography: identify the most abundant isomer

AO3 Mark scheme Q34 1 mark

What this question tests

Reading a gas chromatogram: recognising that peak area (or height) represents relative/percentage abundance.
Linking retention time to intermolecular interactions with the stationary phase (here: polar).
Using chemical structure features (e.g. –OH, ethers) to make a justified identification.

Given: carrier gas = nitrogen (inert mobile phase), stationary phase = polar. Four isomers all have formula C₆H₁₂O₂.

Part (a): Which isomer is the most abundant? (1 mark)

Mark scheme award: B (1 AO3).

✅ Correct answer (what to write)

(This gains the 1 mark. No explanation is required for a multiple-choice-style selection, but you must select the correct option.)

💡 Key knowledge

Most abundant component corresponds to the largest peak on the chromatogram (in practice, peak area; many exam diagrams use peak height as a visual proxy).
Retention time increases when a molecule interacts more strongly with the stationary phase.
With a polar stationary phase, more polar molecules (especially those able to H-bond) typically have longer retention times.

🧠 Exam technique (how to secure the mark fast)

Step 1: Find the tallest/largest peak → that is the highest percentage abundance.
Step 2: Match that peak to the correct option using the information given (polar stationary phase can help if the question expects reasoning).
When options are structures, quickly scan for functional groups: –OH (H-bond donor) is usually “more polar” than just ethers ( –O– ).

❌ Common errors (examiner-style)

Choosing the latest peak (largest retention time) instead of the largest peak. Retention time ≠ abundance.
Mixing up axes: the y-axis is percentage abundance; the x-axis is retention time.
Assuming “most polar = most abundant”. Polarity affects separation, not how much of each is in the sample.

📐 Calculations?

None required. This is a data interpretation question: read the chromatogram and select the isomer corresponding to the largest peak.

🧠 Examiner insight: what distinguished full-mark responses

Top responses did not overthink the chemistry: they correctly interpreted the chromatogram and chose the option linked to the dominant peak.
Students who lost marks typically confused retention time with abundance, or picked based on “polar stationary phase” alone.
Where students added extra notes, the best ones stated: “largest peak = most abundant”. Incorrect added reasoning can create doubt—stick to the marking point.

💡 Quick reminder: how polarity relates to retention time (useful context)

Polar stationary phase retains more polar molecules longer (greater attractions such as dipole–dipole and H-bonding).
A molecule with an –OH group can hydrogen bond strongly with a polar phase, often increasing retention time compared with molecules that only have ether oxygens.
But: the question asked most abundant, not “most strongly retained”. Abundance comes from the peak size.

Checklist to self-mark (1 mark total)

✅ Mark breakdown (from mark scheme)

1 mark: Select B.

❌ Last-minute traps

If two peaks look close in height, remember that in real GC you compare area, but in exam sketches the largest visible peak is intended.
Don’t assume the first peak is “the most” because it elutes first. Early elution usually means weaker attraction, not higher concentration.

Mini revision: GC in one paragraph

In gas chromatography, compounds partition between a gas mobile phase (carrier gas, here nitrogen) and a stationary phase (here polar). Compounds that interact more strongly with the stationary phase spend longer adsorbed and so have a larger retention time. The detector response is plotted as peaks; the peak area (or height in simplified diagrams) is proportional to the amount of that compound, so the largest peak corresponds to the most abundant component.