AQA GCSE Chemistry Chemistry Paper 2 (Higher), November 2020: Question 4

8 marks · Standard Demand difficulty · Short Answer

Calculate the distance moved by the solvent from the Rf value and answer questions about paper chromatography and formulations.

Practise this question

Question

Question 4 displays a chromatography diagram (Figure 2) showing a paper strip with a start line, two spots close together labelled 'Yellow dye' and 'Blue dye', and a solvent front line at the top, marked not to scale. Five sub-questions follow: 04.1 asks candidates to calculate the distance moved by the solvent given an Rf value of 0.60 and dye distance of 5.7 cm (3 marks); 04.2 asks for a reason why only two spots appear if green ink has more than two compounds (1 mark); 04.3 asks to select two ways to increase separation between the spots from five options (2 marks); 04.4 asks why manufacturers use constant proportions of dyes (1 mark); and 04.5 asks candidates to tick which combination of solvent solubility and paper attraction definitely results in a smaller Rf value (1 mark).
Question text

04 This question is about ink.

A student investigated green ink using paper chromatography in a beaker.

The student used water as the solvent.

Figure 2 shows the chromatogram obtained.

Figure 2

Diagram not to scale

04.1 The Rf value of the yellow dye = 0.60

The distance moved by the yellow dye = 5.7 cm

Calculate the distance moved by the solvent.

[3 marks]

Distance moved by the solvent =11 cm

04.2 The green ink contains more than two compounds.

Suggest one reason why only two spots are seen on Figure 2.

[1 mark]

04.3 On the student’s chromatogram, the yellow and blue spots are very close together.

Which two ways could increase the distance between the spots?

[2 marks]

Tick ( ) two boxes.

Allow the solvent front to travel further.

Dry the chromatogram more slowly.

Use a different solvent.

Use a larger beaker.

Use a larger spot of green ink.

04.4 The manufacturers of the green ink always use the same proportions of yellow dye

and blue dye.

Suggest one reason why.

[1 mark]

04.5 The Rf value of a dye depends on:

• the solubility of the dye in the solvent

• the attraction of the dye to the paper.

Which will definitely produce a smaller Rf value if the solvent and paper are

both changed?

[1 mark]

Tick ( ) one box.

The dye is less soluble in the new solvent and

less attracted to the new paper.

The dye is less soluble in the new solvent and

more attracted to the new paper.

The dye is more soluble in the new solvent and

less attracted to the new paper.

The dye is more soluble in the new solvent and

more attracted to the new paper.

Mark scheme

Show the mark scheme Mark scheme for Question 4: 04.1 gives 1 mark for 0.60 = 5.7 / distance, 1 mark for rearranging to 5.7 / 0.60, and 1 mark for 9.5 (cm); 04.2 awards 1 mark for 'some compounds are colourless' or 'dyes have the same Rf values'; 04.3 awards 1 mark each for 'allow the solvent front to travel further' and 'use a different solvent'; 04.4 awards 1 mark for 'so that the (shade of) green is the same' or 'formulation'; 04.5 awards 1 mark for 'the dye is less soluble in the new solvent and more attracted to the new paper'. Total marks: 8.

Question 4

AO /

Question Answers Extra information Mark

Spec. Ref.

5.7

0.60 = 1

04.1 distance moved by solvent AO2

4.8.1.3

(distance moved by solvent =) 1 RPA6

5.7

0.60

= 9.5 (cm) 1

04.2 some of the compounds are allow there are only two 1 AO3

colourless (in solution) compounds that are coloured (in 4.8.1.3

solution) RPA6

or

dyes / compounds have the

same Rf values

04.3 allow the solvent front to travel 1 AO3

further 4.8.1.3

RPA6

use a different solvent 1

04.4 so that the (shade of) green is allow because the green ink is a 1 AO3

the same formulation 4.8.1.2

04.5 the dye is less soluble in the 1 AO3

new solvent and more attracted 4.8.1.3

to the new paper

Total 8

How to answer it

Paper Chromatography & Formulations: Analysis of Ink

📋 What this question tests

This question assesses your understanding of Required Practical 6 (Chromatography) and chemical formulations. Specifically: rearranging and calculating with the Rf formula, understanding factors affecting separation (solubility in mobile phase vs. attraction to stationary phase), interpreting chromatograms, and explaining why products like inks are manufactured to exact recipes.

Question 04.1: Calculating Distance Moved by the Solvent [3 marks]

Rearranging the Rf equation

📐 Step-by-Step Calculation

  1. State the formula:
    Rf = distance moved by substance ÷ distance moved by solvent
  2. Substitute the values given:
    0.60 = 5.7 ÷ distance moved by solvent [1 mark]
  3. Rearrange to make the unknown the subject:
    distance moved by solvent = 5.7 ÷ 0.60 [1 mark]
  4. Calculate the final value:
    = 9.5 cm [1 mark]

❌ Common Errors

  • Multiplying instead of dividing: Calculating 5.7 × 0.60 = 3.42 cm . Remember: the solvent front always travels further than any dye spot, so your answer must be greater than 5.7 cm!
  • Inverting the division: Calculating 0.60 ÷ 5.7 = 0.105 cm . Check your answer makes practical sense on the paper strip.

🧠 Exam Technique: Sanity Check

Because Rf values must always be between 0 and 1, the distance moved by the solvent must always be larger than the distance moved by any dye spot. If your calculated solvent distance is smaller than 5.7 cm, you know you have rearranged incorrectly!

Mark Scheme Breakdown:
• 1 mark for correct substitution: 0.60 = 5.7 / distance
• 1 mark for correct rearrangement: 5.7 / 0.60
• 1 mark for correct final value: 9.5 (cm)

Question 04.2: Explaining Missing Spots [1 mark]

Why only two spots appear when more than two compounds are present

✅ Acceptable Answers (Give ONE)

  • Some of the compounds are colourless (in solution).
  • There are only two coloured compounds (the others are colourless).
  • Two or more dyes/compounds have the same Rf value (they travel the same distance and overlap).
  • Some compounds did not dissolve or move from the pencil line.

💡 Key Knowledge

Chromatography separates mixtures based on relative attractions to the stationary and mobile phases. If two compounds have identical solubility and paper attraction, they share an Rf value and produce a single overlapping spot. Furthermore, non-coloured components (like binding agents or clear solvents) will not show up visually unless a locating agent is used.

Mark Scheme Breakdown: 1 mark for identifying either colourless compounds or overlapping / identical Rf values.

Question 04.3: Increasing Spot Separation [2 marks]

Improving chromatographic resolution

✅ Correct Checkboxes to Tick

  • ☑ Allow the solvent front to travel further. [1 mark]
  • ☑ Use a different solvent. [1 mark]

❌ Incorrect Distractors

  • Dry the chromatogram more slowly: Drying happens after chromatography is finished; it has zero impact on spot separation.
  • Use a larger beaker: The size of the container does not change how dyes interact with the paper or solvent.
  • Use a larger spot of green ink: This actually makes separation worse by causing large, smeared spots that overlap.

💡 Why These Work

  • Travelling further: Distance between spots is proportional to total solvent distance: Δd = (Rf1 - Rf2) × distancesolvent . Letting the solvent run higher expands the physical gap between them.
  • Different solvent: Different solvents have different polarities, changing the relative solubility of each dye and altering their Rf values relative to one another.
Mark Scheme Breakdown: 1 mark for each correctly selected box. Deduct marks if more than two boxes are ticked.

Question 04.4: Consistent Proportions in Inks [1 mark]

Formulations and quality control

✅ Acceptable Answers (Give ONE)

  • So that the (shade of) green is always the same / consistent.
  • Because the green ink is a formulation (a mixture designed with specific properties).

❌ Common Errors

  • Vague statements like "so it works" or "so it looks nice" without referring to shade, colour consistency, or the term formulation.
  • Confusing a formulation with a pure substance or a single chemical compound.
Mark Scheme Breakdown: 1 mark for mentioning colour/shade consistency OR stating that the ink is a formulation.

Question 04.5: Predicting Rf Value Changes [1 mark]

Solubility vs. Paper Attraction

✅ Correct Checkbox to Tick

☑ The dye is less soluble in the new solvent and more attracted to the new paper. [1 mark]

🧠 Exam Technique: Two Factors Pulling the Same Way

The question asks what will definitely produce a smaller Rf value:

  • Lower solubility in solvent: Moves more slowly / spends less time in mobile phase → decreases Rf.
  • Higher attraction to paper: Held back more firmly by stationary phase → decreases Rf.

Both changes work together in the same direction, guaranteeing a lower Rf value!

Mark Scheme Breakdown: 1 mark for ticking the 2nd checkbox (less soluble in new solvent AND more attracted to new paper).

Topics

Chemistry · Required Practicals · C8: Chemical Analysis · Required Practicals

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 2 (Higher), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.