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ScotlandChemistrySyllabus dot point

How do chemists separate, identify and measure the substances in a sample?

The principles of chromatography for separating mixtures, the use of volumetric (titration) analysis to find an unknown concentration, the idea of standard solutions, and how to interpret analytical data to identify and quantify substances.

An SQA Higher Chemistry answer on chemical analysis, covering the principles of chromatography for separating mixtures, volumetric (titration) analysis to find an unknown concentration, standard solutions, and how analytical data is interpreted to identify and quantify substances.

Generated by Claude Opus 4.811 min answer

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  1. What this key area is asking
  2. Chromatography
  3. Volumetric analysis
  4. The key relationship
  5. Worked example: a standardised titration
  6. Reliability of results
  7. Examples in context
  8. Try this

What this key area is asking

The SQA wants you to explain how chromatography separates the components of a mixture, describe volumetric (titration) analysis to find an unknown concentration, explain the role of a standard solution, and interpret analytical data to identify and quantify substances. The titration calculation is one of the most reliable sources of marks in the whole course, so the n=CVn = CV method here must be secure.

Chromatography

The distance each component moves can be used to identify it by comparison with known substances run under the same conditions. If a single spot of a sample separates into several spots, the sample is a mixture.

Volumetric analysis

A standard solution is one of accurately known concentration, prepared by dissolving a known mass of a pure solute and making it up to a precise volume in a volumetric flask.

The key relationship

Worked example: a standardised titration

Reliability of results

To be reliable, repeated titres should be concordant (within about 0.2 cm30.2 \text{ cm}^3 of each other), and the average of the concordant titres is used. A rough first titration is ignored when averaging. Accurate technique, reading the burette to the bottom of the meniscus, and swirling near the end point all improve reliability.

Examples in context

Volumetric analysis is the backbone of routine quality control. A brewery uses an acid-base titration to check the acidity of beer, and a swimming-pool technician titrates to monitor chlorine levels. In environmental monitoring, the concentration of dissolved oxygen in a river is found by a redox titration, where the same n=CVn = CV logic applied above turns a titre volume into a concentration that tells ecologists whether fish can survive. Chromatography, meanwhile, is used by forensic scientists to separate the dyes in an ink left at a crime scene and match them to a suspect's pen, and by anti-doping laboratories to separate and identify banned substances in an athlete's sample.

Try this

Q1. Explain how chromatography separates the components of a mixture. [2 marks]

  • Cue. Components are separated by how strongly each is attracted to the stationary phase relative to the mobile phase; those held less strongly travel further.

Q2. What is a standard solution? [1 mark]

  • Cue. A solution of accurately known concentration, made from a known mass of pure solute and a precise volume.

Q3. 25.0 cm325.0 \text{ cm}^3 of NaOHNaOH needs 22.0 cm322.0 \text{ cm}^3 of 0.100 mol l10.100 \text{ mol l}^{-1} HClHCl (1:11 : 1 ratio). Calculate the concentration of the NaOHNaOH. [3 marks]

  • Cue. n(HCl)=0.100×0.0220=2.20×103n(HCl) = 0.100 \times 0.0220 = 2.20 \times 10^{-3} mol =n(NaOH)= n(NaOH); C=2.20×103/0.0250=0.0880 mol l1C = 2.20 \times 10^{-3} / 0.0250 = 0.0880 \text{ mol l}^{-1}.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20194 marksA 25.0 cm325.0 \text{ cm}^3 sample of sodium hydroxide solution was titrated against 0.100 mol l10.100 \text{ mol l}^{-1} hydrochloric acid. The concordant titres were 24.824.8, 24.924.9 and 25.0 cm325.0 \text{ cm}^3. Calculate the concentration of the sodium hydroxide solution. The equation is NaOH+HClNaCl+H2ONaOH + HCl \rightarrow NaCl + H_2O.
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Markers reward averaging only the concordant titres, the moles of acid, the 1:11 : 1 ratio, and the final concentration with a unit.

Average concordant titre:

24.8+24.9+25.03=24.9 cm3\frac{24.8 + 24.9 + 25.0}{3} = 24.9 \text{ cm}^3

Moles of acid:

n(HCl)=CV=0.100×0.0249=2.49×103 moln(HCl) = CV = 0.100 \times 0.0249 = 2.49 \times 10^{-3} \text{ mol}

The equation shows a 1:11 : 1 ratio, so n(NaOH)=2.49×103 moln(NaOH) = 2.49 \times 10^{-3} \text{ mol}.

Concentration of the alkali:

C=nV=2.49×1030.0250=0.0996 mol l1C = \frac{n}{V} = \frac{2.49 \times 10^{-3}}{0.0250} = 0.0996 \text{ mol l}^{-1}

A common loss is including a rough titre in the average or forgetting the unit.

SQA Higher 20223 marksExplain how paper chromatography can be used to show that an unknown food dye is a mixture, and describe how a component can be identified by comparison with reference dyes.
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A 3 mark answer needs the separation principle, the evidence of a mixture, and the identification by comparison.

A spot of the dye is placed on the paper (the stationary phase) and a solvent (the mobile phase) moves up through it. Each component is separated according to how strongly it is attracted to the stationary phase compared with the mobile phase: a component held weakly travels further.

If the single spot of the unknown dye separates into two or more spots, this shows it is a mixture rather than a pure substance.

A component can be identified by running reference dyes alongside it: a component that travels the same distance as a known reference (the same RfR_f value, under the same conditions) is likely to be that substance.

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