6.9.2Organic Chemistry and Analysis
Gas Chromatography, GC
5 min read 5 parts Lesson 29 of 33 Virtual lab
Part 1 of 5How GC Works
A sample is vaporised and carried through a long column in a moving gas stream.
Mobile phase
- An unreactive carrier gas such as helium or nitrogen carries the vaporised sample through the system.
Stationary phase
- A liquid or solid coated on the inside of a long, thin column held in an oven, which is heated to keep the sample as a gas.
Separation
- Components with a greater attraction to the stationary phase spend longer in the column and emerge later.
Detector
- A detector at the end of the column produces a signal as each component leaves, giving a chromatogram.
Part 2 of 5Reading a Chromatogram
Reading a Chromatogram
Retention time
- The time from injection to the peak maximum;
- it is used to identify a compound by comparison with standards run under the same conditions.
Peak area
- The area under a peak is proportional to the amount of that component in the sample.
Calibration
- Running samples of known concentration allows peak areas to be converted into actual concentrations.
Number of peaks
- Each separate peak shows a different component, so a single peak suggests a pure substance.
Limitation
- Only compounds that can be vaporised without decomposing can be analysed by gas chromatography.
| Peak | Retention time / min | Area (arbitrary units) |
|---|---|---|
| 1 | 1.8 | 20 |
| 2 | 3.4 | 50 |
| 3 | 5.1 | 30 |
Part 3 of 5Using GC Results
1
Inject a small volume of the sample, which is vaporised in a heated port and swept onto the column.
2
The components separate in the oven as they interact differently with the stationary phase.
3
The detector records a peak for each component as it emerges from the column.
4
Compare retention times with known compounds to name the components in the mixture.
5
Calculate the proportion of each component from its peak area, using calibration if accurate quantities are needed.
Part 4 of 5Worked Example
Question
- A GC trace shows three peaks with areas 20, 50 and 30 units.
- Assuming the detector responds equally to each component, calculate the percentage of the middle peak in the mixture.
- Total area = 20 + 50 + 30 = 100 units.
- Fraction of the middle peak = 50 ÷ 100 = 0.50.
- Percentage = 0.50 × 100 = 50%.
Answer The middle peak is 50% of the mixture.
Part 5 of 5Exam Checklist & Tips
6.9.2 – Key Points for the Exam
- Retention time The time a component takes to pass through the column.
- Carrier gas The unreactive gas that moves the sample along the column.
- Chromatogram The detector output plotted against time.
- Peak area A measure of the quantity of each component.
Common Mistakes
- Saying the stationary phase is the gas; the gas is the mobile phase.
- Using peak height rather than area to compare amounts.
- Assuming that equal retention times prove two compounds are the same under different conditions.
- Applying GC to substances that decompose or cannot be vaporised on heating.