All-in-One Learning Compendium

Year: 11, Subject: Chemistry, Topic: Investigating solubility Part-2 (Chromatography)

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1. Previous Knowledge

1.1 Mixtures and pure substances

A mixture contains two or more substances that are physically mixed but not joined by chemical bonds. A pure substance contains only one kind of particle. Ink, air, sea water and orange juice are mixtures. Distilled water and pure table salt are pure substances.

Because the substances in a mixture keep their own properties, we can separate them using physical methods such as filtration, evaporation, distillation and chromatography. No new substance is made when we separate a mixture.

1.2 Solutions and concentration

A solution forms when a solute dissolves in a solvent. In sugar water, sugar is the solute and water is the solvent. When water is the solvent we call the solution aqueous.

Concentration tells us how much solute is in a certain volume of solution. In this topic we use concentration (mg L⁻¹) = mass of solute (mg) ÷ volume of solution (L). Remember 1 g = 1000 mg and 1 L = 1000 mL. For example, 25 mg of caffeine in 1.0 L of solution is 25 mg L⁻¹.

1.3 Polar and non-polar molecules

Atoms pull on the electrons they share. When one atom pulls harder, the bond is uneven and has a slightly negative end (δ−) and a slightly positive end (δ+). A molecule with an uneven spread of charge is polar. Water, H₂O, and ethanol are polar. A molecule with an even spread of charge is non-polar. Hexane, C₆H₁₄, and oils are non-polar.

1.4 Forces between molecules

Intermolecular forces are the attractions between molecules. They are weaker than the bonds inside molecules. There are three kinds you need:

  • Dispersion forces: weak attractions present in all molecules. They get stronger as molecules get larger.
  • Dipole–dipole forces: attractions between the δ+ end of one polar molecule and the δ− end of another.
  • Hydrogen bonds: the strongest of the three. They form when hydrogen is bonded to nitrogen, oxygen or fluorine and is attracted to a lone pair on a nearby N, O or F atom.

Rough strength order: hydrogen bonds > dipole–dipole forces > dispersion forces (for molecules of similar size).

1.5 “Like dissolves like”

A substance dissolves best in a solvent with similar intermolecular forces. This is the like dissolves like rule. Polar solutes dissolve in polar solvents. Non-polar solutes dissolve in non-polar solvents.

Examples: sugar and ethanol dissolve in water (polar and polar). Iodine dissolves in hexane (non-polar and non-polar). Oil does not dissolve in water because water molecules attract each other (hydrogen bonds) much more strongly than they attract oil molecules, so the oil is left out.

1.6 Measuring and graph skills

Use a ruler marked in millimetres and read to the nearest 0.1 cm. A ratio of two distances in the same unit has no units. Give answers to a sensible number of decimal places (usually 2 for a ratio, or 3 significant figures for a concentration).

A calibration graph plots a measured reading (y-axis) against a known amount (x-axis). When a line goes through the origin, y = mx, where m is the gradient. You can read a value across from the y-axis to the line and then down to the x-axis.

1.7 Safe laboratory habits

Wear safety glasses. Many solvents (ethanol, propanone, hexane) are flammable: keep them away from flames and use a fume cupboard or good ventilation. Never look straight at an ultraviolet lamp. Wash your hands after practical work and tell your teacher about spills straight away.

2. Topic Explanation: Investigating solubility Part-2 (Chromatography)

What is chromatography?

Chromatography is a family of techniques that separate the substances in a mixture and help us identify them. It works even when the substances are very similar or when only a tiny sample is available. All forms use two phases: a stationary phase that stays still and a mobile phase that moves through it and carries the sample.

Chromatography answers three kinds of questions. What is in this sample? (qualitative analysis). Is this sample pure? (purity test). How much of each substance is there? (quantitative analysis). In this topic you will learn simple forms (paper and thin-layer), column and gas chromatography, and high-performance liquid chromatography (HPLC), and you will calculate Rf values and concentrations.

2.1 How chromatography separates substances

Two phases

The stationary phase is a solid, or a liquid held on a solid, that does not move. The mobile phase is a liquid or gas that flows past it. The sample is placed at the start, called the origin, and the mobile phase carries the sample particles along.

TechniqueStationary phaseMobile phase
Paper chromatographyWater held on the cellulose fibres of the paperA solvent (water or another solvent) rising up the paper
Thin-layer chromatography (TLC)A thin layer of solid such as silica gel on a plateA solvent rising up the plate
Column chromatographySolid packing in a glass tubeA solvent flowing down the tube
HPLCSmall solid particles packed in a steel columnA liquid pushed through at high pressure
Gas chromatography (GC)A liquid coated inside a long columnAn inert carrier gas

The tug-of-war: adsorption and desorption

Each substance in the mixture is attracted to the stationary phase and also to the mobile phase. The particles of a substance keep adsorbing (sticking to the surface of the stationary phase) and desorbing (breaking away and returning to the mobile phase). While stuck, a particle does not move. While in the mobile phase, it is carried forward.

A substance that is strongly attracted to the stationary phase spends more time stuck and travels a short distance. A substance that is weakly attracted spends more time in the mobile phase and travels a long distance. Because different substances have different attractions, they travel at different speeds and separate into bands or spots.

Why polarity matters

The attractions that matter are the intermolecular forces from section 1.4, and these depend on polarity. Paper and silica gel are polar stationary phases, so they attract polar substances strongly (by hydrogen bonds and dipole–dipole forces).

  • A polar solute sticks well to a polar stationary phase, so it moves less (a low Rf).
  • A non-polar solute is attracted only weakly to the stationary phase and dissolves well in a non-polar solvent, so it moves further (a high Rf).
  • A more polar solvent competes better for the polar solutes, so all spots move further, especially the polar ones.

Common mistake: students say the substance “is heavier so it moves less”. Separation depends on attractions (polarity), not on weight.

Extra Knowledge

In adsorption chromatography (TLC on silica) solutes stick to the solid surface. In partition chromatography (paper) solutes spread between the water held on the fibres and the moving solvent. Chemists describe both with an equilibrium: solute (mobile) ⇌ solute (stationary). The further the equilibrium lies to the right, the slower the solute moves.

Some plates use a non-polar stationary phase (reversed-phase). Then the order of travel is reversed: non-polar solutes stick and move less, and polar solutes move more.

Worked Example 1

Three dyes X (non-polar), Y (slightly polar) and Z (very polar) are separated on paper using a fairly non-polar solvent. Predict the order from nearest the origin to furthest, and explain.

Step 1. Identify the phases. The paper is a polar stationary phase and the solvent is the less polar mobile phase.
Step 2. Z is very polar, so it forms strong hydrogen bonds and dipole–dipole attractions with the paper. It sticks most strongly and moves least.
Step 3. X is non-polar. It is only weakly attracted to the paper and dissolves well in the non-polar solvent. It moves furthest.
Step 4. Y is in between.
Answer: Z (nearest the origin), then Y, then X (furthest). So Rf(Z) < Rf(Y) < Rf(X).

Worked Example 2

A student repeats the experiment of Worked Example 1 but uses a much more polar solvent. What happens to the Rf value of the polar dye Z, and why?

Step 1. A more polar solvent forms stronger attractions with polar dye Z.
Step 2. Z now competes between two polar phases, and it spends more time dissolved in the mobile phase than before.
Step 3. More time in the mobile phase means a longer distance travelled.
Answer: The Rf value of Z increases.

2.2 Paper chromatography and thin-layer chromatography

Paper chromatography: how to do it

In paper chromatography a strip of paper is held in a container so that only its bottom edge touches a shallow layer of solvent. The solvent rises up the paper by capillary action and carries the sample with it.

  1. Draw a pencil line about 2 cm from the bottom of the strip. This line is the origin.
  2. Use a fine capillary tube to place a small spot of each sample on the line. Let it dry. Add more drops on the same spot if the colour is faint.
  3. Pour solvent into the container to a depth of about 1 cm. The solvent level must be below the origin line.
  4. Hang the strip so its bottom edge dips in the solvent. Put on the lid.
  5. When the solvent is near the top, take the strip out and immediately mark the solvent front with a pencil, then let the paper dry.

Why each rule matters: pencil because graphite does not dissolve, but pen ink contains dyes that would also travel and be mistaken for sample; solvent below the origin so the spot is not washed into the beaker; lid on so the air stays full of solvent vapour and the solvent does not evaporate unevenly from the paper; small spots so bands do not overlap.

Thin-layer chromatography (TLC)

In TLC the stationary phase is a thin layer of a fine polar solid, such as silica gel, coated on a glass, plastic or aluminium plate. The method is the same as for paper: pencil origin, small spots, solvent below the origin, lid on, mark the solvent front.

FeaturePaper chromatographyTLC
Stationary phaseWater on cellulose fibresThin layer of silica gel or alumina
SpeedSlowerFaster
SeparationGood for simple mixturesUsually sharper, better for similar substances
Sample sizeSmallVery small
Making spots visibleLimited choice of methodsUV lamp, iodine, sprays (the plate is strong)
CostVery cheap and simpleCheap, but plates cost more than paper

Making colourless spots visible

Many substances are colourless. We then develop the chromatogram by adding something that makes the spots visible. Common ways are: viewing the plate under an ultraviolet (UV) lamp (TLC plates often contain a fluorescent powder, so spots show up as dark patches); exposing the plate to iodine vapour (spots turn brown); or spraying with a locating agent such as ninhydrin, which turns amino acids purple.

Common mistake: marking the position of a colourless spot after the plate dries without developing it first. Also, mark the solvent front straight away, because the solvent evaporates and the front becomes invisible.

Extra Knowledge

In two-way chromatography the plate is run in one solvent, dried, turned through 90° and run again in a second solvent. Substances that overlap in the first solvent often separate in the second. Chemists also check identity by co-spotting: the unknown and a standard are spotted on the same origin. If only one spot forms, the two are probably the same substance.

Worked Example 3

A student draws the origin line with a ballpoint pen and runs a sample that contains two dyes. Four bands appear. Explain the problem.

Step 1. Ballpoint ink is itself a mixture of dyes dissolved in a solvent.
Step 2. When the solvent reaches the line, the dyes in the pen ink dissolve and are carried up with the sample.
Step 3. Extra bands from the pen line mix with the bands from the sample, so we cannot tell which bands belong to the sample.
Answer: The line should be drawn in pencil because graphite does not dissolve or move.

Worked Example 4

A forensic scientist has a tiny smear of a suspected drug and needs a sharp separation from three similar substances. Should she use paper chromatography or TLC? Justify.

Step 1. The sample is very small, and the substances are similar.
Step 2. TLC needs very small samples and usually gives sharper, better-resolved spots because the silica layer is fine and even.
Step 3. TLC is also faster, and the plate allows visualisation under UV light for colourless substances.
Answer: She should use TLC.

2.3 Rf values: identifying substances and testing purity

The Rf value

To compare spots from different runs, we calculate the Rf value (retardation factor). It is the fraction of the way to the solvent front that a substance has travelled.

Rf = distance moved by the solute ÷ distance moved by the solvent

  • Measure both distances from the origin (the pencil line), not from the bottom edge of the paper.
  • Measure to the centre of the spot, and to the solvent front line you marked.
  • An Rf value is always between 0 and 1 and has no units, because the solute can never travel further than the solvent.
  • An Rf value is only useful when the conditions are identical: same stationary phase, same solvent, same temperature.

Using Rf values to identify a substance

Run the unknown mixture and known substances (standards) under identical conditions, then calculate the Rf value of each spot. If a spot in the mixture has the same Rf value (and colour) as a standard, that substance is probably present.

Example: a band in an ink has Rf = 0.50 and the standard dye E also has Rf = 0.50 under the same conditions, so the ink probably contains dye E. Careful: a match shows the substance is probably the same. To be more certain, repeat in a second solvent or use another technique.

Using chromatography to test purity

A pure substance should give one single spot. Extra spots show impurities: the sample is a mixture. A big, dark spot means a lot of that substance. Pharmaceutical chemists use this to check tablets for unwanted by-products, and food testers use it to check dyes.

Exception: one spot in one solvent does not prove a sample is pure. Two different substances can have the same Rf value in a given solvent. A pure substance gives one spot in several different solvents.

Extra Knowledge

Chemists try to choose a solvent that gives Rf values between about 0.2 and 0.8. Spots very near the origin or the front are hard to compare. In more advanced work the retention factor is defined as k = (1 − Rf) ÷ Rf, which is the ratio of time a substance spends in the stationary phase to the time it spends in the mobile phase. A large k means a strongly held substance. This links the Rf value directly to the tug-of-war from section 2.1.

Worked Example 5

A TLC run of a mixture gives three spots at 2.8 cm, 4.2 cm and 6.9 cm from the origin. The solvent front is 8.4 cm from the origin. A standard of substance A, run under identical conditions, has a spot 5.0 cm from the origin and a front 10.0 cm from the origin. Which spot is most likely substance A?

Step 1. Write the formula: Rf = distance of solute ÷ distance of solvent
Step 2. Spot 1: 2.8 ÷ 8.4 = 0.33
Step 3. Spot 2: 4.2 ÷ 8.4 = 0.50
Step 4. Spot 3: 6.9 ÷ 8.4 = 0.82
Step 5. Standard A: 5.0 ÷ 10.0 = 0.50. Use each run’s own solvent front.
Answer: Spot 2 (Rf = 0.50) matches substance A.

Worked Example 6

(a) Substance P has Rf = 0.62 in a solvent that rose 9.5 cm from the origin. How far did P move? (b) In another run, substance Q moved 3.2 cm and has Rf = 0.40. How far did the solvent front move?

(a) Step 1. Rearrange: distance of solute = Rf × distance of solvent.
(a) Step 2. 0.62 × 9.5 = 5.9 cm.
(b) Step 3. Rearrange: distance of solvent = distance of solute ÷ Rf.
(b) Step 4. 3.2 ÷ 0.40 = 8.0 cm.
Answer: (a) 5.9 cm (b) 8.0 cm

2.4 Column chromatography and gas chromatography

Column chromatography

In column chromatography a glass tube is packed with a solid stationary phase, such as silica gel. The mixture is placed at the top (the origin) and solvent is added continuously. The solvent flows down and the components move at different speeds, forming bands. Each band reaches the bottom at a different time and is collected in a separate flask as a fraction. This washing out of components with solvent is called elution.

With a polar stationary phase and a less polar solvent, the least polar component is held least and leaves the column first. Column chromatography is often used to purify larger amounts of a substance, not only to analyse it.

Gas chromatography (GC)

Gas chromatography separates substances that can be turned into vapour without breaking down, such as petrol components or solvents. The mobile phase is an inert carrier gas (for example helium or nitrogen). The stationary phase is a liquid coated on the inside of a long, thin column coiled in a heated oven. A sample is injected, vaporised, and carried through the column. Substances that are more strongly attracted to the stationary phase (and less volatile) take longer to travel through. A detector records a signal, and the time each substance takes is its retention time.

TechniqueBest forGives numbers (quantitative)?
Paper chromatographySimple, cheap tests (dyes, inks)No (only Rf values)
TLCQuick purity checks and identificationMostly no
Column chromatographySeparating and collecting larger amountsNot usually
GCVolatile substancesYes (peak areas)
HPLCNon-volatile or heat-sensitive substancesYes (peak areas)

Extra Knowledge

In GC, retention depends on both volatility (boiling point) and attraction to the stationary phase. Ovens are often heated gradually (temperature programming) so that low-boiling substances leave early and high-boiling ones leave in a reasonable time. Many laboratories join a gas chromatograph to a mass spectrometer (GC–MS): the chromatograph separates the substances and the spectrometer identifies each one from its molecular fragments.

Worked Example 7

A mixture of a non-polar oil (N) and a polar compound (P) is placed on a silica gel column and washed with a non-polar solvent. Which leaves the column first?

Step 1. Silica gel is a polar stationary phase.
Step 2. P is polar, so it forms strong attractions with silica and is held back.
Step 3. N is non-polar, is only weakly held, and dissolves in the non-polar solvent.
Answer: N leaves first, then P.

Worked Example 8

Which technique would you use (a) to detect volatile solvents in paint, and (b) to measure the amount of a large, heat-sensitive vitamin in a drink?

(a) Volatile substances can be vaporised without decomposing: use gas chromatography.
(b) A large, heat-sensitive vitamin would break down when heated and is not volatile: use HPLC, because HPLC works with liquids at near room temperature and gives peak areas for quantitative analysis.

2.5 High-performance liquid chromatography (HPLC) and quantitative analysis

How HPLC works

High-performance liquid chromatography (HPLC) is a sensitive, instrumental form of column chromatography. A pump pushes the liquid mobile phase through a steel column packed with very small particles of stationary phase at high pressure. A small volume of sample is injected at the top. Components move at different speeds and leave the column at different times. A detector at the end (often measuring absorbed ultraviolet or visible light) sends a signal to a recorder.

The recorder draws a graph called a chromatogram: signal (absorbance) on the y-axis against time on the x-axis. Each substance appears as a peak. The time at which a peak appears is the retention time, Rt. Under identical conditions, the same substance always has the same retention time. So retention time identifies a substance in the way an Rf value does for paper and TLC.

Why use instruments?

Instrumental methods are fast, accurate, need very small samples, leave less room for human error, and can measure amounts. They can detect chemicals at very low concentrations. This is why news stories about traces of pesticides or banned drugs are more common today: the instruments are better, not necessarily the world dirtier.

Quantitative analysis with a calibration curve

The area under a peak is proportional to the amount of that substance that passed the detector. To turn an area into a concentration, we first calibrate the instrument:

  1. Prepare standards of known concentration of the substance.
  2. Run each standard and note its retention time.
  3. Measure the peak area for each standard.
  4. Plot a calibration curve: peak area (y-axis) against concentration (x-axis), and draw the line of best fit.
  5. Run the sample under the same conditions. Find the peak with the same retention time and measure its area.
  6. Read the area across to the line and then down to find the concentration.
  7. If the sample was diluted, multiply by the dilution factor to get the original concentration.

Common mistake: confusing the calibration curve (area against concentration) with the chromatogram (signal against time). Also, never extend your conclusion beyond the standards: if the sample’s area is higher than that of your strongest standard, dilute the sample and run it again.

Extra Knowledge

Most HPLC analyses in industry use reversed-phase columns: the stationary phase is non-polar (long hydrocarbon chains bonded to silica) and the mobile phase is polar (water mixed with methanol or acetonitrile). The elution order is the reverse of silica TLC: polar substances come out first and non-polar substances later. To improve accuracy, chemists often add an internal standard, a known amount of a different substance added to every solution, and compare peak areas to it. This reduces errors from small differences in injection volume.

Worked Example 9

Standards of a dye give these peak areas: 5 mg L⁻¹ → 40, 10 mg L⁻¹ → 80, 15 mg L⁻¹ → 120, 20 mg L⁻¹ → 160 (arbitrary units). A sample run under the same conditions has a peak with the same retention time and an area of 100. Find its concentration.

Step 1. Plot area (y) against concentration (x). The points lie on a straight line through the origin, so area = 8 × concentration (gradient = 160 ÷ 20 = 8 per mg L⁻¹).
Step 2. The area 100 lies between the standards 80 and 120, so the sample is inside the calibrated range.
Step 3. Concentration = area ÷ gradient = 100 ÷ 8 = 12.5.
Answer: 12.5 mg L⁻¹ (read from the graph, the area 100 gives 12.5 on the x-axis).

Worked Example 10

To analyse a sports drink, 25.0 mL of the drink was diluted to 250.0 mL. The diluted solution gave a concentration of 12.5 mg L⁻¹ of a vitamin. Calculate the concentration in the original drink and the mass of vitamin in a 600 mL bottle.

Step 1. Dilution factor = 250.0 ÷ 25.0 = 10.
Step 2. Original concentration = 12.5 × 10 = 125 mg L⁻¹. Multiply by the dilution factor.
Step 3. Volume = 600 mL = 0.600 L.
Step 4. Mass = concentration × volume = 125 × 0.600 = 75 mg.
Answer: 125 mg L⁻¹ in the drink; 75 mg in the bottle.

Exam Style Questions

These questions combine ideas from more than one subtopic. The marks are shown in brackets. Work through each answer step by step, and notice how the marks are earned.

Exam Style Question 1 (10 marks)

Investigators compared the ink on a threatening note (N) with the inks of three pens (P1, P2, P3) using paper chromatography with water as the solvent. The solvent front moved 8.0 cm from the origin. shows the positions of the spots.

(a) Calculate the Rf value of each spot in the note ink N. (3 marks)

(b) Which pen most likely wrote the note? Justify your answer using Rf values. (2 marks)

(c) Explain why the origin line was drawn in pencil and why distances were measured from the origin. (2 marks)

(d) The dye with the highest Rf value moved much further than the dye with the lowest Rf value. Explain, in terms of forces and polarity, how this occurs. (2 marks)

(e) Suggest one further test that would increase confidence in the identification. (1 mark)

Worked answer

(a) Rf = distance of spot ÷ distance of solvent front (1 mark for the correct method).
2.0 ÷ 8.0 = 0.25; 4.4 ÷ 8.0 = 0.55; 6.8 ÷ 8.0 = 0.85 (2 marks: all three correct; 1 mark if two are correct).
(b) P2 has spots at the same distances (2.0, 4.4 and 6.8 cm), so the same Rf values 0.25, 0.55 and 0.85 (1 mark). P1 has two spots (0.30 and 0.70) and P3 has one (0.40), so they do not match (1 mark for justification). Answer: P2.
(c) Pencil (graphite) does not dissolve or move, but pen ink would dissolve and its dyes would form extra spots (1 mark). The solvent and solute both start moving from the origin, so only distances measured from there are comparable. Measuring from the paper’s edge would give wrong Rf values (1 mark).
(d) Paper is a polar stationary phase. The dye with the high Rf is weakly attracted to the paper (weaker hydrogen bonds or dipole–dipole forces) and spends more time dissolved in the mobile phase (1 mark). The dye with the low Rf is strongly attracted to the paper, so it is adsorbed for longer and moves less (1 mark).
(e) Repeat in a second solvent (or by co-spotting, or by analysing the inks with HPLC) and compare again (1 mark).

Exam Style Question 2 (9 marks)

A manufacturer claims a tablet extract contains only substance Z. A chemist ran the extract on a silica TLC plate beside standards of Z and two possible impurities, M and N. After development under UV light, the solvent front was 9.0 cm from the origin. shows the plate.

(a) Calculate the Rf values of Z, M and N. (3 marks)

(b) State which of Z, M and N are present in the extract. (2 marks)

(c) Evaluate the manufacturer’s claim. (2 marks)

(d) The chemist repeats the run with a less polar solvent. Predict how the Rf values will change and explain. (2 marks)

Worked answer

(a) Z: 5.4 ÷ 9.0 = 0.60. M: 3.6 ÷ 9.0 = 0.40. N: 7.2 ÷ 9.0 = 0.80. (1 mark for each correct value.)
(b) The extract shows spots at 5.4 cm and 3.6 cm, which match Z and M (1 mark). There is no spot at 7.2 cm, so N is absent (1 mark).
(c) The extract gives two spots, so it is a mixture of Z and the impurity M (1 mark). The claim that it contains only Z is not supported. (1 mark)
(d) Silica is polar. A less polar solvent competes less well for the polar substances, so they stay adsorbed longer and all Rf values decrease (1 mark for the prediction, 1 mark for the explanation).

Exam Style Question 3 (10 marks)

An analyst measured caffeine in an energy drink by HPLC. Five standards of caffeine were run, and each gave a peak at a retention time of 3.2 minutes. The results are in the table. The drink (5.00 mL) was diluted to 50.0 mL with water. Its chromatogram showed several peaks. The peak at 3.2 minutes had an area of 150.

Concentration of standard (mg L⁻¹)Peak area (arbitrary units)
1061
20119
30182
40241
50300

(a) Describe how to draw the calibration curve from this data. (2 marks)

(b) Determine the concentration of caffeine in the diluted solution. (2 marks)

(c) Calculate the concentration of caffeine in the energy drink. (1 mark)

(d) Calculate the mass of caffeine in a 250 mL can. (2 marks)

(e) Explain why the analyst used the retention time and not just the largest peak. (2 marks)

(f) A second diluted sample gave an area of 420. What should the analyst do next? (1 mark)

Worked answer

(a) Plot peak area on the y-axis and concentration (mg L⁻¹) on the x-axis with suitable scales, using labelled axes and units (1 mark). Plot the five points and draw a straight line of best fit through them (and through the origin, since no caffeine gives no peak) (1 mark).
(b) The gradient is about 300 ÷ 50 = 6.0 area units per mg L⁻¹. The area 150 gives 150 ÷ 6.0 = 25 mg L⁻¹ (1 mark for reading or calculating, 1 mark for the correct value, accept 24 to 26).
(c) Dilution factor = 50.0 ÷ 5.00 = 10, so concentration = 25 × 10 = 250 mg L⁻¹ (1 mark).
(d) 250 mL = 0.250 L (1 mark). Mass = 250 mg L⁻¹ × 0.250 L = 62.5 mg (1 mark).
(e) The drink contains other substances that also give peaks (1 mark). Under identical conditions each substance has its own retention time, so only the peak at 3.2 minutes is caffeine (1 mark).
(f) An area of 420 is above the highest standard (300), so it is outside the calibrated range. Dilute the sample further and run it again; do not extrapolate. (1 mark)

Exam Style Question 4 (12 marks)

A soft drink is coloured with a mixture of two red food dyes. Paper chromatography was used first. The solvent front moved 10.0 cm from the origin. The drink’s dye mixture gave spots at 7.2 cm and 3.1 cm. Standards of dye E, dye F and dye G, run on the same strip under the same conditions, gave spots at 7.2 cm, 3.1 cm and 5.0 cm respectively.

The amount of dye E was then found by HPLC. Four standards of dye E, at 2.0, 4.0, 6.0 and 8.0 mg L⁻¹, gave peak areas of 36, 72, 108 and 144. A 10.0 mL sample of the drink was diluted to 50.0 mL. Its dye E peak (with the same retention time as the standards) had an area of 90.

(a) Calculate the Rf values of the two spots in the drink and identify the dyes. (3 marks)

(b) Explain why the drink and the standards had to be run under the same conditions. (1 mark)

(c) Determine the concentration of dye E in the diluted sample. (2 marks)

(d) Calculate the concentration of dye E in the drink. (2 marks)

(e) Explain two reasons why HPLC is better than paper chromatography for finding the amount of dye E. (4 marks)

Worked answer

(a) 7.2 ÷ 10.0 = 0.72 and 3.1 ÷ 10.0 = 0.31 (1 mark each). Dye E (0.72) and dye F (0.31) match the standards; dye G (0.50) is absent (1 mark).
(b) Rf values depend on the solvent, stationary phase and temperature, so they can only be compared when these are identical (1 mark).
(c) Gradient = 144 ÷ 8.0 = 18 area units per mg L⁻¹ (the points lie on a line through the origin) (1 mark). Concentration = 90 ÷ 18 = 5.0 mg L⁻¹ (1 mark).
(d) Dilution factor = 50.0 ÷ 10.0 = 5 (1 mark). Original concentration = 5.0 × 5 = 25 mg L⁻¹ (1 mark).
(e) Reason 1: HPLC gives peak areas that are proportional to the amount, so a calibration curve can turn an area into a concentration, but paper chromatography gives only positions (Rf values), not amounts (2 marks: reason and explanation). Reason 2: HPLC is more sensitive and accurate and there is less human error because the instrument measures the signal, while judging the size of a spot by eye is subjective (2 marks). (Accept speed, small samples or reproducible retention times as valid reasons.)

Higher-Order Worked Examples

Higher-Order Example 1: reversing the order

A student separates a black ink on (i) paper with a polar stationary phase and a non-polar solvent, then on (ii) a reversed-phase TLC plate (a non-polar stationary phase) with a polar solvent. The ink contains a polar dye P and a non-polar dye N. Predict the order of the Rf values in each case and explain the difference.

Step 1: case (i). The paper is polar. P forms strong hydrogen bonds or dipole–dipole attractions with it and is poorly soluble in the non-polar solvent. P sticks and moves little. N is weakly attracted to the paper and dissolves well in the non-polar solvent, so N moves far. Rf(N) > Rf(P).
Step 2: case (ii). The stationary phase is now non-polar, and the solvent is polar. N forms dispersion forces with the non-polar surface, and it is poorly soluble in the polar solvent (the solvent molecules prefer to hold on to each other). N sticks and moves little. P is strongly attracted to the polar solvent, which carries it far. Rf(P) > Rf(N).
Step 3: generalise. The separation depends on the balance of attractions to the two phases. The substance more like the stationary phase moves less, and the substance more like the mobile phase moves more.
Answer: The Rf order is reversed when the polarity of the two phases is swapped: N moves further in (i) and P moves further in (ii).
Why it matters: Reversed-phase HPLC uses this idea, so polar substances leave the column first.

Higher-Order Example 2: does one spot prove purity?

A chemist says: “My sample gave a single spot at Rf = 0.45, the same as standard S, so it is pure S.” Then the sample is run in a more polar solvent. It now shows two spots at Rf = 0.70 and Rf = 0.62, while standard S gives one spot at 0.70. Evaluate the chemist’s claim and plan the next steps.

Step 1. In solvent 1, the sample and S had the same Rf (0.45). This is consistent with pure S, but it does not prove purity, because two different substances can happen to have the same Rf in one solvent.
Step 2. In solvent 2 (more polar), one spot at 0.70 matches S, and a second spot appears at 0.62. A new spot shows a second substance with a different polarity: it interacts differently with the new solvent, so the two substances separate.
Step 3. Conclusion: the sample is a mixture. It contains S and an unknown impurity that co-moved with S in solvent 1. The chemist’s claim is not supported.
Step 4: next steps. (i) Co-spot the sample with S in several solvents. (ii) Use HPLC to measure the area of the impurity peak, so its percentage can be estimated. (iii) Collect the impurity by column chromatography and identify it.
Why it matters: A reliable purity test uses more than one solvent (or more than one technique), because one single test can hide an impurity.

3. Tips, Tricks and Exceptions

Tips and tricks

TipWhy it works
Pencil for the origin, always.Graphite does not dissolve or move.
Solvent level below the origin.Spots stay on the paper and are not washed away.
Mark the solvent front the moment you remove the strip.The solvent evaporates and the front disappears.
Measure from the origin to the centre of each spot.Spots are spread out. The centre is the best single position.
Rf must be between 0 and 1. If you get more than 1, you divided the wrong way round.The solute cannot go further than the solvent.
Polar spots stay low on polar paper or silica.Stronger attractions to the stationary phase.
Calibration graphs: read across, then down.y (area) is the measured value, x (concentration) is the answer.
Check the units before you finish: mg L⁻¹ × L = mg.Keeps mass and concentration calculations consistent.

Exceptions and traps

  • One spot is not proof of purity. Different substances can share an Rf in one solvent (see Higher-Order Example 2).
  • Rf values can vary slightly with temperature, the batch of paper or plate, and the exact solvent mixture. Always run standards on the same sheet.
  • A spot that stays on the origin may be insoluble in the solvent or extremely strongly adsorbed. Try a more polar solvent.
  • Streaky or “tailing” spots often mean the sample was too concentrated. Use a smaller spot.
  • Colourless substances will not be seen without developing (UV lamp, iodine, ninhydrin).
  • Calibration lines can curve at high concentration. Do not use values beyond your highest standard.
  • In reversed-phase HPLC the order is reversed: polar substances leave first.
  • Peak height is not always proportional to amount. Peak area is the reliable measure.

4. Application and Further Reading

Real-life applications

  • Forensic science: comparing inks on documents (paper chromatography or TLC) and dyes from fibres.
  • Medicine and pharmacy: TLC and HPLC test tablets for the right amount of active ingredient and for impurities.
  • Food safety: HPLC finds banned dyes, preservatives and pesticide residues in food and drinks. GC finds volatile flavour compounds.
  • Sport: samples from athletes (and racehorses) are screened for banned drugs by GC–MS and LC–MS.
  • Environmental science: pollutants such as oil components, pesticides and organic chemicals in water are measured by GC and HPLC.
  • Hospitals: HPLC measures substances in blood, such as glucose or the level of medicines.

Advanced ideas and related topics

  • Resolution and efficiency: columns with smaller particles give narrower peaks, so closely spaced substances can be separated.
  • Ion-exchange chromatography: separates charged particles using a stationary phase with fixed charges.
  • Size-exclusion chromatography: separates large and small molecules by size.
  • Mass spectrometry linked to chromatography (GC–MS, LC–MS): separates and identifies at the same time.
  • Spectroscopy (UV–visible, infrared, NMR): other instrumental methods used to identify substances.
  • Solubility and equilibrium: the “tug-of-war” in chromatography is a dynamic equilibrium, and it links to your work on solubility and intermolecular forces.

5. Difficult Words and Vocabulary