4.8 Chemical analysis — notes on one page

Every AQA GCSE Chemistry 8462 specification point in 4.8, read straight through: 14 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against AQA 8462 section 4.8. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

How this checking works

4.8.1.1 · Pure substances

In chemistry, a pure substance contains one element or one compound and is not mixed with another substance.
Test purity by measuring a melting point or boiling point and comparing it with reliable reference data.
For example, a sample that melts sharply at the accepted melting point is consistent with a pure substance; a mixture usually changes the value and melts over a range.
A common error is to use the everyday meaning of pure, such as natural or unadulterated, instead of the chemical meaning.

Worked example

Reference data give the melting point of substance P as $64\,^\circ\mathrm{C}$ . Batch A melts sharply at $64\,^\circ\mathrm{C}$ , while batch B melts from $58$ to $61\,^\circ\mathrm{C}$ . Which batch is more likely to be pure? Explain.

1.Compare both results with the reference value. Batch A changes state at one temperature matching $64\,^\circ\mathrm{C}$ . Batch B changes state across a range below the reference value, which is evidence of a mixture.

Answer: Batch A; it melts at the accepted temperature and has a sharp melting point, whereas batch B melts over a different range.

Common mistakes

Don't fall into the trap of using the everyday meaning of pure, such as natural or unadulterated, instead of the chemical meaning.
Don't fall into the trap of calling a sample pure because it looks uniform, without comparing a measured melting or boiling point with reference data.

Exam tip. When data are supplied, compare both the measured value and whether melting occurs sharply or across a range.

4.8.1.2 · Formulations

A formulation is a mixture designed to be a useful product, with every component included for a particular purpose.
Make a formulation by measuring and mixing its components carefully so the final product has the required properties.
For example, a paint may contain a pigment for colour, a solvent to control flow and a binder that leaves a solid coating.
A common error is to call every mixture a formulation; a formulation must have a designed use and controlled composition.

Worked example

A $250\,\mathrm{g}$ fertiliser formulation is $18\%$ nitrogen compound, $12\%$ potassium compound and the remainder filler. Calculate the mass of filler and explain one purpose of measuring the components accurately.

1.The named components total $18+12=30\%$ , so filler is $70\%$ . Its mass is $0.70\times250=175\,\mathrm{g}$ . Careful measurement keeps each batch at the designed composition, so it performs consistently.

Answer: Mass of filler $=175\,\mathrm{g}$ ; accurate measurement gives the product its required composition or properties.

Common mistakes

Don't fall into the trap of calling every mixture a formulation; a formulation must have a designed use and controlled composition.
Don't fall into the trap of listing ingredients without linking each component or proportion to the product's required properties.

Exam tip. For an ‘identify a formulation’ question, state the designed purpose and the carefully controlled composition.

4.8.1.3 · Chromatography

Definition. Chromatography separates mixtures using a stationary phase and a mobile phase; separation depends on how each substance distributes itself between the two. The Rf value is the distance moved by the substance divided by the distance moved by the solvent.

In paper chromatography the paper is the stationary phase and the solvent is the mobile phase. A substance that is more soluble in the solvent and less strongly attracted to the paper travels further and has a higher Rf.
Because Rf is a ratio of two distances it has no units, is always between 0 and 1, and is measured to the centre of the spot from the start line.
The method is Required practical 6 and its marks are fixed: draw the start line in pencil, because pencil is insoluble and will not run as ink would;
spot the sample on the line; stand the paper in a suitable solvent with the solvent level below the start line; cover the container; let the solvent run until near the top; mark the solvent front immediately; dry the paper; then measure both distances and divide.
A pure compound gives a single spot in every solvent, whereas a mixture may give several.

A paper chromatogram showing the origin, separated spots and marked solvent front.

Worked example

A chromatogram has a solvent front $8.0\,\text{cm}$ above the origin. A spot centre is $5.6\,\text{cm}$ above the origin. Calculate its $R_f$ value.

1.Use $R_f=\dfrac{\text{distance moved by spot}}{\text{distance moved by solvent}}$ .
2.Substitute distances measured from the origin: $R_f=5.6\div8.0$ .
3.Evaluate the dimensionless ratio: $R_f=0.70$ .

Answer: $R_f=0.70$

Mark-scheme wording to use

“start line does not dissolve/run/smudge or pencil is insoluble” — AQA 8462, June 2025 Paper 2F, Q05.2
“some of the compounds are colourless (in solution) … dyes / compounds have the same Rf values” — AQA 8462, November 2020 Paper 2H, Q04.2
“pencil start line, spot, suitable solvent in beaker below line, lid, solvent travel, mark front, dry, measure both distances and calculate Rf” — AQA 8462, six-mark indicative content, November 2021 Paper 2F, Q10.1

What the examiner wants. The solvent level below the start line, and the reason for the pencil. Both are single marks that almost every candidate can score and many omit; the Rf itself must have no units.

Common mistakes

Don't fall into the trap of measuring from the paper edge or spot boundary instead of from the origin to the centre of the spot.
Don't fall into the trap of allowing the solvent to cover the origin line, so the samples dissolve directly into the solvent reservoir.

Exam tip. For an $R_f$ calculation, show both distances measured from the origin and give a ratio between $0$ and $1$ .

4.8.2.1 · Test for hydrogen

Hydrogen is identified by the characteristic pop produced when it burns rapidly. Hold a burning splint at the open end of the test tube containing the collected gas.
For example, a gas that gives a pop with a lit splint has a positive hydrogen-test result.
A common error is to insert a glowing splint; that is the test for oxygen, not hydrogen.
Only a small collected sample should be tested because hydrogen burns rapidly in oxygen.
The splint stays outside the test tube opening.

Worked example

Give the procedure and positive observation for confirming that gas collected from a metal-acid reaction is hydrogen.

1.Keep the flame at the open end of the container rather than pushing it deep inside. A pop sound is the required positive observation.

Answer: Place a burning splint at the tube opening; hydrogen burns with a pop.

Common mistakes

Don't fall into the trap of inserting a glowing splint; that is the test for oxygen, not hydrogen.
Don't fall into the trap of reporting only that the gas burns, rather than the characteristic squeaky pop at the mouth of the tube.

Exam tip. State ‘burning splint’ and ‘squeaky pop’ together; the observation earns the identification.

4.8.2.2 · Test for oxygen

Oxygen supports combustion and is identified when a glowing splint relights. Insert a glowing wooden splint into the test tube containing the gas.
For example, a splint with no flame that bursts back into flame gives a positive oxygen result.
A common error is to use a burning splint and look for a pop, which tests for hydrogen.
The test uses oxygen's ability to support combustion and should be performed on a collected gas sample.
Relighting is the required positive observation.

Worked example

Describe a test that distinguishes oxygen from a gas that does not support combustion. State the positive result.

1.First blow out the flame so the splint is glowing. Put it into the gas sample. Relighting is the diagnostic result for oxygen; merely remaining warm is not sufficient.

Answer: Insert a glowing splint; it relights in oxygen.

Common mistakes

Don't fall into the trap of using a burning splint and looking for a pop, which tests for hydrogen.
Don't fall into the trap of describing the splint as burning before insertion, rather than glowing with no visible flame.

Exam tip. The exact positive observation is that a glowing splint relights, not merely that it glows more brightly.

4.8.2.3 · Test for carbon dioxide

Carbon dioxide is identified because it forms a fine white precipitate in limewater, making the liquid look milky or cloudy.
Bubble the gas through, or shake it with, aqueous calcium hydroxide solution.
For example, clear limewater becoming cloudy is a positive result for carbon dioxide under this test.
A common error is to report that limewater becomes colourless; the required observation is that it turns milky or cloudy.
The cloudiness is caused by insoluble calcium carbonate, so ‘white precipitate forms’ is also a valid observation.

Worked example

State the reagent and the expected visible change when testing a gas sample for carbon dioxide.

1.Pass the gas through limewater or shake the two together. Name both the reagent and the formation of cloudiness for a complete answer.

Answer: Use limewater, aqueous calcium hydroxide; it changes from clear to milky or cloudy.

Common mistakes

Don't fall into the trap of reporting that limewater becomes colourless; the required observation is that it turns milky or cloudy.
Don't fall into the trap of naming calcium hydroxide but omitting the observation that a white precipitate makes the limewater cloudy.

Exam tip. Give reagent and observation as a pair: limewater changes from colourless to milky or cloudy.

4.8.2.4 · Test for chlorine

Chlorine gas bleaches damp litmus paper white. Expose damp litmus paper to the gas while using appropriate small-scale safety precautions because chlorine is toxic.
For example, litmus losing all its colour and becoming white is the positive chlorine result.
A common error is to use dry litmus paper; moisture is required for the specified test.
A positive result is bleaching rather than a permanent acid-base indicator colour, so the final white appearance is essential.
The gas must not be inhaled.

Worked example

Describe how litmus paper is used to test for chlorine and state the observation that confirms a positive result.

1.Moisten the litmus paper before exposing it to the gas. Record complete loss of colour to white, not simply a colour change between red and blue.

Answer: Use damp litmus paper; chlorine bleaches it white.

Common mistakes

Don't fall into the trap of using dry litmus paper; moisture is required for the specified test.
Don't fall into the trap of reporting only an initial colour change and omitting that the paper is ultimately bleached white.

Exam tip. Include both test conditions and result: damp litmus paper is bleached white by chlorine.

4.8.3.1 · Flame tests (chemistry only)

Flame tests identify some metal ions: Li⁺ crimson, Na⁺ yellow, K⁺ lilac, Ca²⁺ orange-red and Cu²⁺ green.
Place a small amount of a clean sample into a non-luminous flame and compare the observed colour with the known colours.
For example, a lilac flame indicates potassium ions, whereas an orange-red flame indicates calcium ions.
A common error is to treat a flame test as reliable for every mixture; a strong colour, especially sodium yellow, can mask another ion.
Contamination must be controlled by using clean apparatus because traces of sodium can produce an intense yellow flame.

Worked example

Samples M and N give a green flame and an orange-red flame respectively. Identify the metal ion in each sample.

1.Use the required colour associations. Green corresponds to copper compounds, and orange-red corresponds to calcium compounds.

Answer: M contains copper(II) ions, Cu²⁺. N contains calcium ions, Ca²⁺.

Common mistakes

Don't fall into the trap of treating a flame test as reliable for every mixture; a strong colour, especially sodium yellow, can mask another ion.
Don't fall into the trap of confusing lithium's crimson flame with calcium's orange-red flame when matching an unknown.

Exam tip. Record the observed flame colour first, then match it to one named ion from the specified colour list.

4.8.3.2 · Metal hydroxides (chemistry only)

With sodium hydroxide, Cu²⁺ gives a blue precipitate, Fe²⁺ green and Fe³⁺ brown.
Add sodium hydroxide solution dropwise, record any precipitate, then add excess when distinguishing aluminium from other white precipitates.
Al³⁺, Ca²⁺ and Mg²⁺ form white hydroxide precipitates, but only aluminium hydroxide dissolves in excess sodium hydroxide.
A common error is to claim every white precipitate dissolves in excess; calcium and magnesium hydroxides remain.
The coloured precipitates identify copper(II), iron(II) and iron(III) directly, whereas the three white precipitates require careful discrimination.

Worked example

Solutions A and B each form a white precipitate when a little sodium hydroxide is added. The precipitate from A dissolves after excess sodium hydroxide is added, but B's remains. What can be concluded about A, and why is B not fully identified?

1.All three candidate ions can first give a white precipitate. Only aluminium hydroxide dissolves in excess sodium hydroxide, so A is aluminium. An insoluble white precipitate is consistent with either calcium or magnesium, leaving B ambiguous.

Answer: A contains aluminium ions; B could contain calcium or magnesium ions, so the observations do not distinguish those two.

Common mistakes

Don't fall into the trap of claiming every white precipitate dissolves in excess; calcium and magnesium hydroxides remain.
Don't fall into the trap of identifying a white precipitate as aluminium without checking whether it dissolves in excess sodium hydroxide.

Exam tip. For a white precipitate, state the excess-sodium-hydroxide result before distinguishing aluminium from calcium or magnesium.

4.8.3.3 · Carbonates (chemistry only)

Carbonate ions are tested by adding a dilute acid; a carbonate reacts to release carbon dioxide gas.
Pass the gas produced into limewater to confirm that it is carbon dioxide.
For example, effervescence followed by limewater turning cloudy is a positive sequence for carbonate ions.
A common error is to identify a carbonate from fizzing alone; the gas should be confirmed with limewater.
The confirmatory gas test matters because effervescence alone can be produced by reactions involving gases other than carbon dioxide.

Worked example

Describe a complete chemical test for carbonate ions in an unknown powder, including how the gaseous product is identified.

1.Add a dilute acid to the powder and collect the gas released. Bubble it through aqueous calcium hydroxide. Fizzing plus cloudy limewater provides the two-stage positive result.

Answer: Add dilute acid, then pass the gas into limewater; effervescence occurs and the limewater turns milky or cloudy.

Common mistakes

Don't fall into the trap of identifying a carbonate from fizzing alone; the gas should be confirmed with limewater.
Don't fall into the trap of naming hydrogen as the gas from the acid-carbonate reaction instead of carbon dioxide.

Exam tip. A complete carbonate test gives both stages: effervescence with dilute acid, then limewater turning cloudy.

4.8.3.4 · Halides (chemistry only)

Halide ions are tested by acidifying the solution with dilute nitric acid and then adding silver nitrate solution.
Use clean samples and record the precipitate colour: chloride white, bromide cream and iodide yellow.
For example, a cream silver-halide precipitate identifies bromide ions under the specified test conditions.
A common error is to swap the cream and yellow results: silver bromide is cream, while silver iodide is yellow.
Acidification removes interfering ions, while nitric acid is chosen because nitrate ions do not form a precipitate with silver ions.

Worked example

Give the reagents, in order, used to test an aqueous sample for halide ions, and state the result for chloride ions.

1.First acidify the sample using dilute nitric acid. Then add aqueous silver nitrate. If chloride is present, insoluble silver chloride forms as a white precipitate.

Answer: Add dilute nitric acid, then silver nitrate solution; chloride ions give a white precipitate.

Common mistakes

Don't fall into the trap of swapping the cream and yellow results: silver bromide is cream, while silver iodide is yellow.
Don't fall into the trap of using hydrochloric acid to acidify, which introduces chloride ions and can create a misleading white precipitate.

Exam tip. Write the reagents in order—dilute nitric acid, then silver nitrate—before giving the precipitate colour.

4.8.3.5 · Sulfates (chemistry only)

Sulfate ions form a white precipitate when barium chloride solution is added under acidic conditions.
Acidify the sample with dilute hydrochloric acid, then add barium chloride solution and observe any precipitate.
For example, a white barium sulfate precipitate is the positive result for sulfate ions.
A common error is to report only that the mixture turns white without naming the formation of a precipitate.
The acidic conditions remove interfering carbonate ions before barium ions are added to test for sulfate.

Worked example

Describe the reagent sequence and positive observation for testing an unknown solution for sulfate ions.

1.Acidify the unknown with dilute hydrochloric acid before adding aqueous barium chloride. Record the formation of a white solid rather than merely a pale solution.

Answer: Add dilute hydrochloric acid, then barium chloride solution; a white precipitate forms if sulfate ions are present.

Common mistakes

Don't fall into the trap of reporting only that the mixture turns white without naming the formation of a precipitate.
Don't fall into the trap of using sulfuric acid to acidify, which adds sulfate ions and can produce a false positive.

Exam tip. State ‘dilute hydrochloric acid, then barium chloride’ and identify the white solid as a precipitate.

4.8.3.6 · Instrumental methods (chemistry only)

Instrumental methods detect and identify elements or compounds using measured signals rather than only visible chemical-test observations.
Choose an instrumental method when rapid analysis, high sensitivity or accurate measurement is important.
For example, a sensitive instrument can detect a component at a concentration too low to give a clear precipitate or colour change.
A common error is to list 'easy' as a specification advantage; the required comparison is that instrumental methods are accurate, sensitive and rapid.
Instrumental methods generate signals that can be compared with reference or calibration data to identify or measure substances.

Worked example

A laboratory must screen $180$ water samples in one day for a contaminant present at very low concentration. Give two reasons for choosing an instrumental method.

1.Link each circumstance to an advantage. The large number of samples requires speed, while the very small amount of contaminant requires sensitivity.

Answer: It is rapid enough for many samples and sensitive enough to detect a low concentration.

Common mistakes

Don't fall into the trap of listing 'easy' as a specification advantage; the required comparison is that instrumental methods are accurate, sensitive and rapid.
Don't fall into the trap of claiming instrumental analysis is always cheaper, even though the specification advantages are speed, sensitivity and accuracy.

Exam tip. When comparing methods, link ‘sensitive’ to small amounts and ‘accurate’ to reliable quantitative measurement.

4.8.3.7 · Flame emission spectroscopy (chemistry only)

In flame emission spectroscopy, a solution sample enters a flame and the emitted light passes through a spectroscope.
Identify metal ions by comparing the positions of lines in the sample spectrum with a reference set recorded in the same form.
Line intensity can be compared with calibration data to measure concentration; for example, an intensity halfway between two standards gives an intermediate concentration when the calibration is linear.
A common error is to use line brightness to identify the ion; line position identifies the ion, while intensity is used for concentration.
A sample can contain several metal ions because each element contributes its characteristic set of emission lines.

Worked example

A sodium calibration is linear through the origin. An intensity of $24$ units corresponds to $3.0\,\mathrm{mg\,dm^{-3}}$ . A sample gives $40$ units under identical conditions. Calculate its sodium-ion concentration.

1.For a linear calibration through the origin, concentration is proportional to intensity. Calculate $3.0\times\frac{40}{24}=5.0\,\mathrm{mg\,dm^{-3}}$ .

Answer: $5.0\,\mathrm{mg\,dm^{-3}}$

Common mistakes

Don't fall into the trap of using line brightness to identify the ion; line position identifies the ion, while intensity is used for concentration.
Don't fall into the trap of comparing a sample with reference spectra recorded on different axes or under unsuitable conditions.

Exam tip. Use line position for identity and a calibration relationship between intensity and concentration for quantity.