Ch 2 · Gas/liquid chromatography · AS & A Level Chemistry 9701
Topic 2 of 4 in Analytical techniques — AS & A Level Chemistry 9701 — 48 lessons.
Introduction
Gas/liquid chromatography (GLC) separates volatile mixtures by carrying them as a gas through a column coated with a liquid stationary phase, giving both a way to identify components by their retention time and to measure how much of each is present.
Key ideas — Key terms in GLC
The mobile phase is an unreactive carrier gas, typically nitrogen or helium, which carries the vaporised sample through a long, narrow column. The stationary phase is a high-boiling liquid coated as a thin film onto an inert solid support packed inside, or lining, the column. The retention time is the time taken, from injection of the sample, for a particular component to travel through the column and reach the detector, recorded as the position of its peak on the chromatogram.
🧑🏫 Teacher's explanation
Imagine a busy airport terminal where people are walking down a long hallway (the column) toward an exit (the detector). The people are your sample components. Now, imagine there are several different types of moving walkways or sticky floor patches along that hallway—these are the stationary phase. Some people walk through very quickly because they don't interact with the floor, while others move slowly because they keep stopping to look at things or get stuck to the floor.
📖 Definitions
🔍 Worked example — Easy
A mixture contains two components, A and B. Component A has a retention time of 2.5 min and component B has a retention time of 5.0 min. Calculate the percentage composition of component A in the mixture if the area under peak A is 120 units and the area under peak B is 30 units.
🔍 Worked example — Standard
A chromatogram shows three peaks with areas of 4.5 cm^2, 3.0 cm^2, and 2.5 cm^2. Determine the percentage of the third component.
🔍 Worked example — Exam-style
Explain why a compound with a boiling point of 50^°C has a shorter retention time than a compound with a boiling point of 120^°C in GLC.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the percentage composition of the second component. [2] (b) Explain why the component with the lower boiling point typically appears as the first peak. [2] [4 marks]
Key ideas — Interpreting percentage composition
A GLC chromatogram plots detector signal against retention time, giving one peak for each component present in the mixture, and the area under each peak is proportional to the amount (moles or mass, depending on the detector) of that component. The percentage composition of the mixture is found by expressing each peak's area as a percentage of the total area of all peaks: \%\ component = area of that peak/sum of all peak areas × 100
🧑🏫 Teacher's explanation
Imagine you are watching a race where runners are moving through a crowd. The 'mobile phase' is the crowd pushing the runners forward, and the 'stationary phase' is the crowd itself, which some runners might get stuck in or slow down in. The 'retention time' is simply how long it takes each runner to cross the finish line.
📖 Definitions
🔍 Worked example — Easy
A gas chromatogram shows two peaks. The area of Peak A is 2.5 cm^2 and the area of Peak B is 7.5 cm^2. Calculate the percentage composition of component A.
🔍 Worked example — Standard
A mixture contains three components with peak areas of 12 mm^2, 18 mm^2, and 20 mm^2. Calculate the percentage composition of the third component.
🔍 Worked example — Exam-style
A sample contains two compounds. Compound X has a peak area of 4.0 units and Compound Y has a peak area of 1.0 unit. If the total mass of the original sample was 5.0 g, calculate the mass of Compound X in grams.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the percentage composition of component A in the mixture. [2] (b) If component A has a boiling point of 50 ° C and component B has a boiling point of 85 ° C, explain which component will have a longer retention time. [2] [4 marks]
Key ideas — Retention time and the stationary phase
A component's retention time depends on how strongly it interacts with the liquid stationary phase relative to how readily it remains in the gaseous mobile phase: a component that dissolves more strongly in, or forms stronger intermolecular forces with, the stationary phase spends more time out of the gas phase and is retained on the column longer, giving a longer retention time, while a component that interacts weakly with the stationary phase remains mostly in the gas phase and passes through quickly, giving a short retention time.
🧑🏫 Teacher's explanation
Imagine a busy airport terminal where people are moving from the entrance to the gates. The 'mobile phase' is the moving walkway that carries everyone forward. The 'stationary phase' is like a series of duty-free shops along the walkway. Some people are in a rush and walk straight through without stopping (short retention time), while others are very interested in the shops and spend a lot of time browsing (long retention time).
📖 Definitions
🔍 Worked example — Easy
A mixture contains two components, A and B. Component A has a retention time of 2.5 minutes, and Component B has a retention time of 5.0 minutes. Calculate the ratio of the retention times of B to A.
🔍 Worked example — Standard
A chromatogram shows three peaks. The area under the peaks for components X, Y, and Z are 15.0 mm^2, 30.0 mm^2, and 5.0 mm^2 respectively. Calculate the percentage composition by volume of component Y in the mixture.
🔍 Worked example — Exam-style
In a gas chromatography analysis, Component 1 has a boiling point of 50 ° C and Component 2 has a boiling point of 120 ° C. Assuming the stationary phase is a non-polar liquid, which component will have a shorter retention time and why?
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the percentage by volume of component B in the mixture. [2] (b) If component A has a lower boiling point than component C, explain why A appears earlier on the chromatogram. [2] [4 marks]
Key ideas — Boiling point and retention time
A compound's boiling point reflects the strength of the intermolecular forces between its own molecules, which correlates closely with how strongly it interacts with the stationary phase: a compound with a lower boiling point is more volatile, remains in the gas phase for a greater proportion of the time, and is therefore only weakly retained by the stationary phase, giving it a shorter retention time. A higher-boiling compound condenses into, or interacts more strongly with, the stationary phase and takes longer to travel through the column, giving a longer retention time; this is why, for a series of similar compounds, retention time generally increases with boiling point.
🧑🏫 Teacher's explanation
Imagine a crowded shopping mall where people are walking along a long corridor (the mobile phase) while some people stop at various shop windows (the stationary phase) to look at displays. If you are in a huge rush and don't care about the shops, you will reach the exit very quickly. If you stop to look at every single shop window, it will take you much longer to reach the exit.
📖 Definitions
🔍 Worked example — Easy
A mixture contains two components, A and B. Component A has a retention time of 2.5 min and Component B has a retention time of 5.0 min. If the total area under the chromatogram peaks is 100 units and the area for Component A is 30 units, calculate the percentage composition of Component A.
🔍 Worked example — Standard
A chromatogram shows three peaks with retention times of 1.2 min, 2.8 min, and 4.5 min. If these peaks represent a homologous series of alkanes, which alkane is likely to have the highest boiling point?
🔍 Worked example — Exam-style
A sample is analyzed via gas chromatography. The peaks for two components have areas of 15.0 mm^2 and 35.0 mm^2. Calculate the mass ratio of the two components if the molar masses are M_r(comp 1) = 60 and M_r(comp 2) = 120.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Arrange the compounds in order of increasing boiling point. [1] (b) Explain your answer to part (a) in terms of intermolecular forces. [2] [3 marks]
Worked example
Question. A mixture of three esters is analysed by GLC. The chromatogram shows three peaks with retention times of 1.2 min, 2.5 min and 4.0 min, and peak areas of 15.0, 45.0 and 40.0 (arbitrary units) respectively. (a) Calculate the percentage of the mixture that is due to the ester with the shortest retention time. [2] (b) The ester giving the peak at 1.2 min has the lowest boiling point of the three. Explain why it has the shortest retention time. [2]
🔍 Worked example — Easy
A chromatogram shows two peaks with areas of 2.5 cm^2 and 7.5 cm^2. Calculate the percentage composition of the component corresponding to the first peak.
🔍 Worked example — Standard
In a gas chromatogram, Component A has a retention time of 5.0 min and Component B has a retention time of 12.0 min. If Component A has a boiling point of 50circC and Component B has a boiling point of 110circC, explain the difference in retention times.
🔍 Worked example — Exam-style
A mixture contains three compounds with peak areas of 12.0, 38.0, and 50.0 units. Calculate the percentage of the mixture that is NOT the component with the largest peak area.
⚠️ Common errors
Common mistakes
Practice
Answers