OCR GCSE Chemistry B (Twenty First Century)

OCR GCSE Chemistry B C5 Chemical Analysis Exam Questions

Free OCR GCSE Chemistry B revision for C5 Chemical Analysis: what you need to know, common mistakes and exam-style questions with answers. Covers purity, formulations, chromatography and Rf values, separation methods, ion tests, conservation of mass, moles, reacting masses, concentration and titration.

Practise C5 Chemical Analysis with past papers

Once you have learnt this topic, find the C5 Chemical Analysis questions in real OCR GCSE Chemistry B past papers and mark them with the mark scheme. Doing lots of real questions is the quickest way to learn how examiners ask about each idea.

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About OCR GCSE Chemistry B C5 Chemical Analysis

Chapter C5 is the quantitative heart of OCR Chemistry B. You learn what chemists mean by a pure substance, how melting points and chromatography test purity, and which separation technique suits a mixture. Separate chemistry adds flame tests, precipitate tests for ions and instrumental methods. The chapter then moves to amounts: conservation of mass, relative formula mass, the mole and the Avogadro constant, reacting masses, solution concentrations and titrations. These skills are reused in C6 when you calculate yields. Examiners use C5 for many of the calculation marks, so expect Rf values, mole calculations and concentration questions with unit conversions.

What you need to know

  • C5.1 Purity and formulations A pure substance contains only one element or compound and has a sharp melting point; impurities lower and broaden it. Formulations, such as paints and medicines, are mixtures made in exact proportions for a purpose.
  • C5.1 Chromatography and separation Chromatography has a stationary phase and a mobile phase; Rf = distance moved by spot / distance moved by solvent. Filtration, crystallisation, simple distillation and fractional distillation are chosen depending on the substances in the mixture.
  • C5.2 Identifying ions Separate chemistry only: flame tests identify lithium (red), sodium (yellow), potassium (lilac), calcium (orange red) and copper (blue green). Precipitate and anion tests identify other ions, and instrumental methods such as emission spectroscopy are fast, sensitive and accurate.
  • C5.3 Moles and masses Mass is conserved in reactions; an apparent change in an open container is due to gases escaping or being gained. Moles = mass / Mr, and one mole contains 6.02 x 10^23 particles. Balanced equations give mole ratios for reacting mass calculations.
  • C5.4 Concentration and titration Concentration can be in g/dm3 or mol/dm3 (moles / volume in dm3). In a titration, a burette adds a solution of known concentration to a measured volume until the indicator changes colour, and concordant results are averaged.

Common mistakes in C5 Chemical Analysis

  • Measuring the Rf distance to the edge of the spot or from the bottom of the paper. Measure from the baseline to the centre of the spot and to the solvent front.
  • Forgetting to convert cm3 to dm3. Divide by 1000 before calculating concentration.
  • Using the wrong mole ratio in reacting mass questions. Use the balanced equation to scale moles before converting back to mass.
  • Saying mass is lost in a reaction. Mass is conserved; an open container loses mass because a gas escapes.
  • Drawing the chromatography baseline in ink. Use pencil, because ink would dissolve and move with the solvent.

Try three C5 Chemical Analysis exam-style questions

1. A spot moved 3.6 cm and the solvent front moved 8.0 cm. Calculate the Rf value. (2 marks)

Show the mark scheme answer

Rf = 3.6 / 8.0; = 0.45.

2. Calcium carbonate decomposes: CaCO3 to CaO + CO2. Calculate the mass of carbon dioxide made from 10.0 g of calcium carbonate. (Ar: Ca = 40, C = 12, O = 16) (3 marks)

Show the mark scheme answer

Mr CaCO3 = 100, so moles = 10.0 / 100 = 0.100 mol; 1 : 1 ratio so 0.100 mol CO2; mass = 0.100 x 44 = 4.40 g.

3. 2.0 g of sodium hydroxide (Mr = 40) is dissolved to make 250 cm3 of solution. Calculate the concentration in g/dm3 and in mol/dm3. (3 marks)

Show the mark scheme answer

250 cm3 = 0.250 dm3; 2.0 / 0.250 = 8.0 g/dm3; moles = 2.0 / 40 = 0.050 mol, so 0.050 / 0.250 = 0.20 mol/dm3.

C5 Chemical Analysis FAQs

What does OCR Chemistry B C5 Chemical Analysis cover?

It covers purity, formulations, chromatography, separation techniques, conservation of mass, moles, reacting masses, concentrations and titration. Separate chemistry adds tests to identify ions and instrumental analysis.

Which paper is it examined in?

OCR Chemistry B has two papers at each tier, Breadth in chemistry and Depth in chemistry, and both can assess content from every chapter, so C5 can appear on either paper.

What equations do I need for C5?

Learn Rf = distance moved by substance / distance moved by solvent, moles = mass / Mr, and concentration = amount / volume in dm3, with the mole ratio method for reacting masses.