A forensic scientist uses paper chromatography to identify an unknown ink found at a crime scene. The chromatography is carried out using ethanol as the solvent. The unknown ink produces three spots. The Rf values of the spots are 0.25, 0.51 and 0.78. Reference inks labelled P, Q and R each produce a single spot with Rf values of 0.51, 0.78 and 0.34 respectively. Explain what the results tell the scientist about the composition of the unknown ink, and explain why different compounds travel different distances up the chromatography paper.

OCR A-Level Chemistry B: Salters (H433) — 6.3 Analysis (chromatography, NMR) · Explain · 5 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

A forensic scientist uses paper chromatography to separate the components of an unknown ink. Ethanol is used as the solvent. The unknown ink produces three separate spots with Rf values of 0.25, 0.51 and 0.78. Reference inks P, Q and R produce single spots with Rf values of 0.51, 0.78 and 0.34 respectively.

Model answer (5 marks)

The unknown ink is a mixture of at least three different dyes or substances.

• The spot with Rf = 0.51 matches the reference ink P.
• The spot with Rf = 0.78 matches the reference ink Q.
• The spot with Rf = 0.25 does not match any of the reference inks (P, Q or R) and is therefore an unidentified component.

Different compounds travel different distances because they have different relative affinities for the stationary phase (the paper) and the mobile phase (ethanol). A compound that is more strongly attracted to the solvent will move further up the paper, whereas a compound that is more strongly attracted to the paper will remain closer to the origin.

Examiner tips

  • Use the Rf values to match spots to reference inks; state the unknown is a mixture; explain the movement in terms of affinity for paper vs solvent.
  • Show the logic of matching Rf values directly; mention the unidentified component.
  • Use the phrase "relative affinity" and "stationary vs mobile phase" to hit the mark.
  • Keep the answer concise – no extra wording beyond what is needed for the marks.

Common mistakes

  • Saying the unknown ink is a single component; ignoring the mixture requirement.
  • Confusing the direction of movement (more solvent attraction means further up, not lower).
  • Failing to mention the unidentified component with Rf = 0.25.

Mark scheme (5 marks)

  1. The unknown ink is a mixture (of at least three components / dyes / substances).
  2. One component of the unknown ink matches reference ink P (Rf = 0.51).
  3. One component of the unknown ink matches reference ink Q (Rf = 0.78).
  4. The component with Rf = 0.25 does not match any reference ink (P, Q or R) / is unidentified.
  5. Different compounds travel different distances because they have different attractions / interactions with the stationary phase (paper) and the mobile phase (solvent / ethanol); compounds more strongly attracted to the solvent travel further / compounds more strongly attracted to the paper travel less far.

Key terms in this question

chromatography · Rf value

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