A technician is investigating three paint samples — paint P, paint Q and paint R. He carries out paper chromatography on each sample using the same solvent. Paint P produces a single spot that travels 4.8 cm, whilst the solvent front travels 6.0 cm. Paint Q produces three separate spots. Paint R produces a single spot that travels 3.6 cm, whilst the solvent front also travels 6.0 cm. The technician wants to determine whether any of the paints are pure substances and to identify what paints P and R might be by comparing their Rf values with a reference table of known pigments. Explain how the technician can use the chromatography results to determine which paints are pure substances, and calculate the Rf values of the spots in paints P and R, stating what these Rf values tell him about the identity of these paints.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Reference Rf values (solvent used: ethanol) • Pigment A: Rf = 0.80 • Pigment B: Rf = 0.60 • Pigment C: Rf = 0.45 • Pigment D: Rf = 0.30
Model answer (5 marks)
A pure substance gives only one spot on a chromatogram.
Paint P gives a single spot, so it is a pure substance. Paint Q gives three spots, so it is a mixture. Paint R also gives a single spot, so it is a pure substance.
Rf for paint P = 4.8 cm ÷ 6.0 cm = 0.80.
Rf for paint R = 3.6 cm ÷ 6.0 cm = 0.60.
The Rf of 0.80 matches Pigment A in the reference table, so paint P contains Pigment A. The Rf of 0.60 matches Pigment B, so paint R contains Pigment B.
Paint P gives a single spot, so it is a pure substance. Paint Q gives three spots, so it is a mixture. Paint R also gives a single spot, so it is a pure substance.
Rf for paint P = 4.8 cm ÷ 6.0 cm = 0.80.
Rf for paint R = 3.6 cm ÷ 6.0 cm = 0.60.
The Rf of 0.80 matches Pigment A in the reference table, so paint P contains Pigment A. The Rf of 0.60 matches Pigment B, so paint R contains Pigment B.
Examiner tips
- Use the command word ‘explain’ to describe the reasoning behind purity and identification.
- Show the calculation of Rf with units and compare directly to the reference values.
- Mention that a single spot indicates a pure substance, multiple spots a mixture.
Common mistakes
- Calculating Rf incorrectly (e.g. using 6.0/4.8).
- Failing to state that a single spot means purity.
- Not matching the calculated Rf to the correct pigment in the table.
Mark scheme (5 marks)
- A pure substance produces a single spot in chromatography
- Paints P and R are pure substances because each produces only one spot; paint Q is a mixture because it produces more than one spot
- Rf value for paint P = 4.8 ÷ 6.0 = 0.80
- Rf value for paint R = 3.6 ÷ 6.0 = 0.60
- Paint P matches Pigment A (Rf = 0.80) and paint R matches Pigment B (Rf = 0.60), so the Rf values can be used to identify the pigments in P and R
Key terms in this question
pure substance · Rf value · paper chromatography
Related
- All WJEC A-Level Chemistry (Wales) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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