Compare column and gas chromatography, interpret retention times and explain how GC–MS and calibration strengthen chemical analysis.
Column chromatography collects separated fractions
A column is packed with a suitable stationary phase such as silica. Dissolve the mixture in a small amount of compatible solvent and apply it as a narrow band. Pass a liquid mobile phase through the column and collect the emerging liquid in successive fractions.
Components with a stronger preference for the mobile phase relative to the stationary phase travel through faster. More strongly retained components emerge later. TLC can help check which collected fractions contain the desired component. Poor packing, channels or an overloaded band can reduce separation.
The time taken to emerge is affected by the stationary phase, solvent composition, column dimensions and flow. Match conditions when comparing retention behaviour. High-performance liquid chromatography, HPLC, is an extension using controlled high-pressure liquid flow and specialised columns; it is not a gas technique.
Diagram placeholder
Diagram placeholder — a column separating two bands
Labels to include:
- Solvent reservoir
- Evenly packed silica stationary phase
- Narrow starting sample band
- Two separated component bands
- Support plug or frit and outlet
- Successive labelled collecting vessels
Draw the more strongly retained band higher up the column at the same time. Show the faster band emerging first and being collected as a fraction. The diagram should not suggest that the column chemically transforms the sample.
Gas chromatography uses a gaseous mobile phase
GC separates suitable volatile, thermally stable components. The sample is vaporised and carried through a column by an inert gas such as helium or nitrogen. The stationary phase can be a solid or a high-boiling liquid film supported on a solid or column surface, depending on the system.
In gas–liquid chromatography, components partition between the moving gas and the stationary liquid. Both volatility and interaction with the stationary phase influence retention. A high boiling point often increases retention in comparable conditions, but boiling point alone cannot predict the full order when stationary-phase interactions differ.
The column is held in a controlled-temperature oven and carrier gas is supplied under controlled pressure or flow. A detector produces a signal as components leave the column. The sample needs to be vapour-phase without unacceptable decomposition; not every organic substance is directly suitable for GC.
Diagram placeholder
Diagram placeholder — GC apparatus and chromatogram
Labels to include:
- Inert carrier gas supply and flow control
- Heated sample inlet
- Column inside a temperature-controlled oven
- Detector at the column outlet
- Signal plotted vertically against time horizontally
- Retention time measured from injection to each peak maximum
Connect the detector output to a chromatogram with several peaks. Label one possible solvent peak separately from analyte peaks. The apparatus schematic should show gas flow through the column rather than a liquid solvent rising up a plate.
Use retention times under matched conditions
Retention time is the elapsed time between injection and detection of a component, conventionally using its peak maximum. Compare an unknown with standards using the same column, stationary phase, temperature programme and carrier-gas flow.
Increasing oven temperature generally shortens retention for a given analyte under otherwise comparable conditions because it favours the gas phase. Increasing carrier-gas flow usually shortens transit times. A longer column generally increases retention at the same linear flow velocity, but pressure and flow conditions must be stated for a meaningful comparison.
Two compounds can have similar or identical retention times. A matching time supports an identification; it is not unique proof. Temperature changes can alter selectivity and peak resolution as well as timing, so “hotter is always better” is not a sensible optimisation rule.
Peak count and area need interpretation
Each resolved analyte peak suggests a detected component. A solvent peak, reference substance or contaminant should not be counted as an unknown sample component. Co-elution and undetected substances mean peak count can underestimate the number of components.
For a given compound and calibrated detector within its linear working range, peak area can be related to amount or concentration. Different compounds can have different response factors, so 40% of the total raw peak area does not automatically mean 40% by mass. Use calibration and the units stated in the question.
Original example: a 2.00 mg cm−3 standard of compound X gives peak area 3000 units. An unknown gives 2400 units under the same injection and detector conditions. Assuming a proportional calibration through the origin, concentration = 2.00 × 2400/3000 = 1.60 mg cm−3. If the analysed solution was a tenfold dilution, the original concentration was 16.0 mg cm−3.
If the calibration has a non-zero intercept, use its equation or graph instead of this simple ratio. This quantitative example is an application of calibration, not a universal law about every detector or mixture.
GC separates; mass spectrometry helps identify
Coupling GC to a mass spectrometer allows the mass spectrum of material emerging at a particular time to be recorded. The molecular ion, where observable, and fragmentation pattern provide much more information than retention time alone. A computer can compare spectra with a reference library.
A good library match can support a positive identification, especially alongside retention data and an authentic standard. Co-elution, weak molecular ions, similar spectra or incomplete library coverage can still limit confidence. The instrument does not make ambiguous data infallible.
GC–MS is used in contexts such as forensic, environmental and food analysis. Explain the relevant analytical benefit: separating a mixture before measuring spectra makes it easier to associate spectral evidence with its components. Do not say that the chromatographic column itself measures molecular mass.
Choose the technique for the question
For identification, combine independent evidence wherever the question permits. For obtaining a product, consider whether the technique collects material in the needed quantity. Analytical detection and preparative purification are related tasks but not identical.
| Method | Mobile / stationary phases | Useful output | Key limitation |
|---|---|---|---|
| TLC | Liquid / solid coating | Spots and Rf; comparisons with standards | Co-elution and detection limitations; conditions must match |
| Column chromatography | Liquid / packed stationary phase | Separated fractions for collection or analysis | Incomplete separation and losses during collection |
| GC | Gas / solid or supported liquid | Retention times and calibrated peak areas | Requires suitable volatility and thermal stability |
| GC–MS | GC followed by mass spectrometry | Retention data plus molecular/fragment information | Co-elution and spectral ambiguity can remain |
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.
Q1. In a silica column with a given liquid mobile phase, which component normally emerges first?Show answer
The component that is less strongly retained relative to its affinity for the moving solvent. It spends a greater proportion of time travelling with the mobile phase.
Q2. Why should retention times be compared under the same GC conditions?Show answer
Temperature, column/stationary phase and carrier-gas flow affect retention. A difference caused by changed conditions must not be mistaken for a different identity.
Q3. Why might a three-peak chromatogram come from a sample containing more than three components?Show answer
Some components may co-elute into the same peak, fail to be detected or be unsuitable for the method. Check whether any observed peak is solvent or a reference as well.
Q4. A standard at 5.00 mg cm−3 gives area 4000. The same analyte in an unknown gives area 2800. With equal injection volumes and proportional response, find its concentration.Show answer
5.00 × 2800/4000 = 3.50 mg cm−3. This assumes the same analyte, matched conditions and a valid proportional calibration.
Q5. What information does GC–MS add beyond ordinary retention-time comparison?Show answer
It provides a mass spectrum with molecular-ion and fragmentation information for the eluting material, which can be interpreted or compared with reference spectra. This strengthens identification but still requires assessment of overlap and match quality.
Sources
Sources and examiner guidance (reviewed 2 October 2026)
- AQA 7405 organic chemistry specification — 3.3.16 coverage and required skills.
- Chemrevise: Chromatography — Coverage checklist; explanations, data exercises and quick checks on this page are original Finesse material.
- AQA June 2023 Paper 3 mark scheme — Q03.4–03.7 (p21) and examiner report pp4–5: TLC cover, two-phase separation, locating agent and Rf.
- AQA June 2023 Paper 3 examiner report — Read alongside the question-specific marking guidance; not a universal wording checklist.
- AQA practical handbook: Required Practical 12 — RP12 pp156–161, particularly the medicines TLC method pp158–159. Explanations and numerical examples here are independently written.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.
