Chromatography is the separating half of analysis; mass spectrometry and infrared and NMR are the identifying half. Every technique here runs on one idea, and it is an idea you already have — polarity and the forces between molecules decide how strongly something sticks to a surface and how readily it dissolves. You have already used it once, to separate the amino acids from a hydrolysed protein.
Two phases, one balance
Every chromatography experiment has the same two parts. A stationary phase stays put; a mobile phase moves through or over it, carrying the mixture along. Each component of the mixture is pulled two ways at once — dissolved in the moving phase it travels, stuck to the stationary phase it does not — and it spends its time switching between the two.
Separation depends on the balance between solubility in the moving phase and retention by the stationary phase.
That one sentence answers a surprising number of questions, and it is worth learning in exactly those words. A component that is more soluble in the mobile phase, or less strongly retained by the stationary phase, spends more of its time moving — so it travels further, or comes off sooner. One held tightly by the stationary phase lags behind. Give only half of that balance and you get half the marks.
AQA asks you about three types. They differ only in what the two phases are and in how you measure the result.
| Type | Stationary phase | Mobile phase | Result measured as |
|---|---|---|---|
| Thin-layer, TLC | a solid coating on a plate | a solvent moving up the plate | Rf value |
| Column, CC | a solid packed into a column | a solvent moving down the column | time (or volume of solvent) to come off |
| Gas, GC | a solid, or a solid coated with a liquid, in a long column | an unreactive carrier gas, under pressure at high temperature | retention time |
- Name both halves of the balance. “It is more soluble in the solvent” on its own is half an answer; so is “it is held less strongly by the plate”.
- The mobile phase in GC is a gas and it is unreactive — nitrogen or helium, not air.
- In TLC the solvent moves up; in column chromatography it moves down. The stationary phase is a solid in both.
- Chromatography separates. On its own it identifies nothing — you identify by comparing with a standard run under the same conditions.
Exam question
State in general terms what determines the distance travelled by a spot in TLC.
Show answer
The (balance between) solubility in the moving phase and retention by the stationary phase. 1 mark
The alternative wording AQA allows is the relative affinity for the stationary (solid) phase and the mobile (liquid / solvent) phase. One phase on its own does not score.
Source: AQA A-Level Chemistry past papers.
Thin-layer chromatography & Required Practical 12
Two contexts AQA suggest for this are worth knowing, because both turn up in questions: separating the analgesics in a painkiller tablet — aspirin, paracetamol, ibuprofen and caffeine, where a combination tablet gives one spot per ingredient — and identifying transition-metal ions in solution, where the spots are coloured and need no locating agent at all.
In TLC the stationary phase is a thin layer of a solid — usually silica or alumina, both of which carry polar bonds — spread on a plate. A spot of the mixture goes on a start line near the bottom, the plate stands in a shallow layer of solvent, and the solvent creeps up the plate by capillary action, carrying the components with it.
- Draw the start line in pencil. Ink would dissolve in the solvent and run up the plate with the sample.
- Keep the solvent level below the start line. If the spot is submerged it dissolves off into the solvent instead of travelling up the plate.
- Put a lid on the beaker. It keeps the atmosphere inside saturated with solvent vapour, which stops the solvent evaporating off the plate before it reaches the top.
- Handle the plate by the edges — grease from your fingers is itself a substance the solvent can carry.
- Mark the solvent front in pencil the moment the plate comes out. It moves no further, but it evaporates, and once it has you cannot measure it.
- Colourless spots are located with a developing agent — ninhydrin for amino acids, iodine vapour, or a UV lamp.
The full practical, with the other eleven, is on the required practicals page.
The Rf value
A spot’s position is reported as its Rf value — how far it went as a fraction of how far the solvent went. Both distances are measured from the start line, and the spot is measured to its centre.
Rf = distance moved by the spot ÷ distance moved by the solvent front
Because it is a ratio of two lengths, Rf has no units, and it can never be greater than 1 — a spot cannot outrun the solvent carrying it. Under the same conditions — same stationary phase, same solvent, same temperature — a substance always gives the same Rf, so you identify a spot by running a standard alongside it and comparing.
On a developed plate the start line is at 0.0 cm, the centre of the spot is 5.4 cm above it and the solvent front is 8.7 cm above it. Calculate the Rf value and state what else you would need in order to name the compound.
Step 1 — both distances from the start line: spot 5.4 cm, solvent front 8.7 cm.
Rf = 5.4 ÷ 8.7 = 0.62
Step 2 — say what Rf alone cannot do: 0.62 is only a number until you have something to compare it with. You need the Rf of a known standard run on the same plate, in the same solvent — or a table of Rf values measured under those conditions.
Quote Rf to two significant figures and never with a unit. If your answer comes out above 1, you have divided the wrong way round.
A mixture is run alongside four standards on one plate, in the same solvent. The mixture gives two spots, at Rf = 0.29 and Rf = 0.71. Identify them.
| Standard | Rf in this solvent |
|---|---|
| P | 0.15 |
| Q | 0.29 |
| R | 0.52 |
| S | 0.71 |
Step 1 — match each Rf: 0.29 matches standard Q; 0.71 matches standard S.
Step 2 — state the condition that makes the match valid: the standards were run on the same plate in the same solvent. Rf is only reproducible under identical conditions, so a value from a textbook run in a different solvent would prove nothing.
Step 3 — say what has not been shown: matching Rf is consistent with Q and S, but two different compounds can happen to share an Rf. That is exactly why a second solvent, or a second technique, is used to confirm.
- Both distances are measured from the start line — not from the bottom edge of the plate. That single slip produces most of the wrong answers in a multiple-choice Rf question.
- Measure to the centre of the spot, not its leading edge.
- Quote Rf with no unit. A value above 1 means you divided the wrong way round.
- An Rf only identifies something against a standard run under the same conditions. Change the solvent and every value changes.
- Two substances can share an Rf. A match is consistent with an identification, not proof of one — which is the reason for running a second solvent.
Exam questions
A concentrated solution of pure 1,4-dinitrobenzene was spotted on a TLC plate coated with a solid that contains polar bonds. Hexane was used as the solvent, in a beaker with a lid. The start line, drawn in pencil, the final position of the spot and the final solvent front are shown on the chromatogram.
Use the chromatogram to deduce the Rf value of 1,4-dinitrobenzene in this experiment.
To obtain the chromatogram, the TLC plate was held by the edges and placed in the solvent in the beaker in the fume cupboard. The lid was then replaced on the beaker. Give one other practical requirement when placing the plate in the beaker.
Show answer
The solvent depth must be below the start line. 1 mark
Safety points are ignored here — the question is about what makes the chromatogram work, and the fume cupboard and the lid have already been given to you.
Source: AQA A-Level Chemistry past papers.
Column chromatography
Column chromatography uses the same two phases as TLC, turned on end and scaled up. A glass column is packed with the solid stationary phase; the mixture goes on at the top; solvent — the mobile phase — runs down through the packing. Components separate into bands as they travel, and each one washes out of the bottom at its own time. The one that is least strongly retained comes off first.
The practical advantage over TLC is that you can put a beaker under the tap and collect each component as it comes off, instead of just looking at where it stopped. Instead of an Rf you quote the time, or the volume of solvent, needed to wash a component through.
A mixture of hexane, hexan-1-ol and hexanal is run down a silica column with a non-polar solvent. In what order do they come off?
Step 1 — rank the polarities: hexane is an alkane, so non-polar. Hexanal has a polar C=O. Hexan-1-ol has a polar –OH and can hydrogen bond, so it is the most polar.
Step 2 — remember which phase is polar: the silica is polar, the solvent is not. Polar compounds are the ones that stick.
Step 3 — least retained comes off first: hexane, then hexanal, then hexan-1-ol.
The reasoning is identical for a TLC plate; only the words change — on a column you talk about what comes off first, on a plate about what travels furthest.
Exam question
A sample of cyclohexene has been contaminated with cyclohexanol. The cyclohexene can be separated from the cyclohexanol by column chromatography. Silica gel is used as the stationary phase and hexane as the mobile phase. Explain why cyclohexene has a shorter retention time than cyclohexanol.
Show answer
Cyclohexene is less polar than cyclohexanol. 1 mark
So cyclohexene has a greater affinity for the mobile phase / hexane — or, equivalently, cyclohexanol has a greater affinity for the stationary phase / silica. 1 mark
“Cyclohexanol is held in the stationary phase for longer” is allowed, as is “cyclohexene is more soluble in the mobile phase”, and so is a reference to hydrogen bonds between cyclohexanol and the silica — which is the underlying reason: the –OH can hydrogen bond to the polar surface and the C=C cannot.
Source: AQA A-Level Chemistry past papers.
Gas chromatography & GC–MS
In gas chromatography the mobile phase is an unreactive carrier gas — nitrogen or helium — pushed under pressure through a long, thin column held in an oven at high temperature. The stationary phase inside that column is a solid, or a solid coated with a liquid. The sample is injected, vaporises instantly in the heat, and is swept along by the gas.
Each component takes its own time to pass through and reach the detector at the far end — its retention time — decided by the same balance as always: the more time it spends dissolved in the carrier gas rather than held by the stationary phase, the sooner it arrives.
- Retention time — where a peak sits along the axis — identifies the component, but only by comparison with a standard run under identical conditions.
- The area under a peak — its integration — is proportional to how much of that component is present. The peak with the largest area is the most abundant.
- Area, not height. Peaks broaden the longer a component is held on the column, so a late peak can be short and wide yet still enclose more area than an early tall, narrow one. Where the peaks are equally sharp the tallest is also the largest, which is why the tallest peak is usually the right answer — but write area.
- Conditions must match exactly. Change the temperature, the carrier gas or the column and every retention time changes with it.
A fuel sample is analysed by GC. Three peaks appear, at retention times 1.8, 3.1 and 4.6 minutes, and the peak at 3.1 minutes is by far the largest. Standards run on the same instrument give retention times of 1.8 minutes for hexane, 3.1 minutes for heptane and 4.6 minutes for octane. What can you conclude?
Step 1 — identify from the retention times: each peak matches a standard, so the sample contains hexane, heptane and octane. This only holds because the standards were run on the same instrument under the same conditions.
Step 2 — quantify from the peak areas: the peak at 3.1 minutes encloses by far the largest area, and area is proportional to amount, so the fuel is mostly heptane.
Step 3 — note the elution order makes sense: retention time rises from hexane to octane, in order of chain length — the larger the molecule, the stronger its van der Waals attraction to the stationary phase and the longer it is held.
Two different questions, two different features of the trace: where a peak is identifies it, the area under it says how much.
GC–MS — and the one thing it cannot do
Feed what comes off the GC column straight into a mass spectrometer and you have GC–MS. The two instruments do different halves of the job: the mass spectrometer identifies each component from its mass spectrum, but only because the GC separated them first, so they arrive one at a time.
Isomers have the same molecular formula, so the same Mr, so the same m/z for the molecular ion. A mass spectrum cannot separate them. Gas chromatography can — isomers differ in polarity and shape, so they have different retention times. This is precisely why the two techniques are used together.
- The carrier gas must be described as unreactive. It is there to carry, not to react.
- Retention time identifies only by comparison. On its own a retention time of 3.1 minutes names nothing.
- Do not say a mass spectrometer can identify isomers. It cannot — same molecular formula, same Mr, same m/z. Separating them is the GC’s job.
- When you are asked why two components separate, the answer is still the balance between the two phases, not their boiling points or their masses.
- Quantify by peak area, not peak height. “The tallest peak” only happens to be right when the peaks are equally sharp; the area is what is proportional to the amount every time.
Exam questions
The dipeptide cysteine–aspartic acid (cys-asp), J, and the dipeptide aspartic acid–cysteine (asp-cys), K, are isomers. A mixture of the two is analysed by gas chromatography followed by mass spectrometry (GC–MS). Explain why J and K can be separated by gas chromatography, and why mass spectrometry using electrospray ionisation does not enable you to identify them.
Show answer
Gas chromatography. They have different retention times — the dipeptides appear at different times. 1 mark
Because they have a different balance between solubility in the moving phase / carrier gas and retention by the stationary phase / column — a different relative affinity for the two phases. 1 mark
Mass spectrometry. They give the same m/z values. 1 mark
Because both dipeptides have the same molecular formula / Mr. 1 mark
Two marks for the separation, two for the limitation. Note the structure of the answer: each claim is followed by its reason, and the reason is the phase balance in one case and the molecular formula in the other.
The chromatogram below is for a sample containing four isomers with the molecular formula C6H12O2. The carrier gas is nitrogen and the stationary phase is polar. Which of the four isomers in this sample is the most abundant?
Source: AQA A-Level Chemistry past papers.
Why one component runs further than another
Everything so far has been descriptive. The questions that separate grades ask you to predict — which spot will be higher, what happens if the solvent is changed — and that always comes down to matching polarities.
Silica and alumina are polar. A polar compound is attracted to them — it can hydrogen bond or form dipole–dipole attractions to the surface — so it is strongly retained and moves a short distance: a low Rf. A non-polar compound has little to hold it there, stays dissolved in the solvent, and travels further: a high Rf.
Change the solvent to a more polar one and it competes for the polar solutes. They spend more time dissolved, so every Rf rises — and the polar components rise the most.
1,2-dinitrobenzene has its two –NO2 groups adjacent; in 1,4-dinitrobenzene they are opposite. On a silica plate run in hexane, which has the higher Rf?
Step 1 — compare the polarities: in the 1,4- isomer the two polar C–NO2 bonds point in opposite directions, so their dipoles cancel and the molecule is non-polar overall. In the 1,2- isomer they are adjacent, do not cancel, and leave a resultant dipole — so 1,2- is the more polar.
Step 2 — apply the rule: the plate is polar and hexane is not. The less polar 1,4- isomer is held less strongly by the plate and is more soluble in the hexane, so it travels further.
Step 3 — answer in the direction asked: 1,4-dinitrobenzene has the higher Rf.
This is the same argument as cyclohexene versus cyclohexanol on a column — polarity first, then which phase that makes it prefer.
- State the polarities before you explain the Rf. In AQA’s mark scheme the explanation mark is dependent on getting the polarity comparison right first.
- Say which phase, not just “it is attracted more”: attracted to the polar stationary phase / plate, or more soluble in the mobile phase / hexane.
- Do not describe a solute as “bonded to the mobile phase” — it is dissolved in it, or attracted to it.
- “Yes” or “no” on its own never scores on a justify question, and a wrong conclusion cancels the explanation.
Exam questions
A second TLC experiment was carried out using 1,2-dinitrobenzene and 1,4-dinitrobenzene. An identical plate was used under the same conditions with the same solvent (hexane). In this experiment, the Rf value of 1,4-dinitrobenzene was found to be greater than that of 1,2-dinitrobenzene. Deduce the relative polarities of the two isomers and explain why 1,4-dinitrobenzene has the greater Rf value.
Show answer
1,2- is more polar, or 1,4- is less polar. 1 mark
1,4- (being the less polar or non-polar one) is less attracted to the polar plate / stationary phase — or is more attracted to, or more soluble in, the non-polar solvent / mobile phase / hexane. 1 mark
The second mark is dependent on the first: get the polarities the wrong way round and the explanation cannot score. The equivalent argument written about 1,2- instead is accepted. “1,2- is polar, 1,4- is non-polar” also scores the first mark.
A third TLC experiment was carried out using 1,2-dinitrobenzene. An identical plate was used under the same conditions, but the solvent contained a mixture of hexane and ethyl ethanoate. A student stated that the Rf value of 1,2-dinitrobenzene in this third experiment would be greater than in the previous one. Is the student correct? Justify your answer.
Show answer
Yes — but there is no mark for “yes” on its own.
The solvent is (more) polar — ethyl ethanoate is polar. 1 mark
So the polar isomer is more attracted to, more soluble in, or has a stronger affinity for the solvent than before — so it travels further and Rf rises. 1 mark
Answering “no” is a contradictory error and scores zero for the whole question. Saying the solute is “bonded to the mobile phase” is penalised in the second mark.
Source: AQA A-Level Chemistry past papers. This topic is examined mostly inside longer organic and analysis questions, so the questions above are the chromatography parts of those.
Run the plate yourself
Everything on this page comes together in one experiment. Pick a solvent, run the plate, and watch where three components of different polarity end up — then reveal the Rf measurements and change the solvent to see the whole pattern shift.
- The principle, in AQA’s words: separation depends on the balance between solubility in the moving phase and retention by the stationary phase. Give both halves.
- TLC — solid coating on a plate, solvent moves up. Pencil start line, solvent below it, lid on the beaker, plate held by the edges, solvent front marked as soon as the plate is out. This is Required practical 12.
- Rf = distance moved by the spot ÷ distance moved by the solvent front, both from the start line, spot measured to its centre. No units, always less than 1, and only meaningful against a standard run under the same conditions.
- Column chromatography — solid packed in a column, solvent moves down; the least strongly retained component comes off first, and you can collect each one.
- Gas chromatography — unreactive carrier gas under pressure at high temperature, stationary phase a solid or a solid coated with a liquid. Retention time identifies (against a standard); peak area quantifies.
- GC–MS — the GC separates, the mass spectrometer identifies. It cannot distinguish isomers: same molecular formula → same Mr → same m/z. GC can, because isomers have different retention times.
- Predicting: silica is polar, so polar compounds are retained and give a low Rf; non-polar compounds run further. A more polar solvent raises every Rf, and raises the polar components most. State the polarities first — the explanation mark depends on it.