4.8 Chemical analysis — revision question pack

14 specification points · notes, questions, answers and worked methods

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

Loading your tier…

Answer all questions in the spaces provided.

4.8.1.1 · Pure substances

Explanation

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.

Tier 1 · Easy

A sealed sample contains only solid sulfur. Decide whether it is a pure substance in the chemical sense and justify your decision.
[2 marks]
Total for this question: 2
A bottle contains distilled water only. State whether its contents are chemically pure and give one reason.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A sample contains exactly two compounds and no elements. A student says it is chemically pure because it contains only compounds. Evaluate this statement.
[2 marks]
Total for this question: 2
The accepted boiling point of cyclohexane is $81\,^\circ\mathrm{C}$ . Sample P boils steadily at $81\,^\circ\mathrm{C}$ , while sample Q boils over the range $78$ to $83\,^\circ\mathrm{C}$ . Identify the sample that is more likely to be pure and explain your choice.
[3 marks]
Total for this question: 3
Explain why a solid made from one compound can be chemically pure even though each particle of the compound contains three different elements.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A drink is advertised as 'pure fruit juice'. Analysis shows water, sugars, acids and flavour compounds. Explain why the label may be reasonable in everyday language but the drink is not pure in chemistry.
[4 marks]
Total for this question: 4
A thermometer reads $2\,^\circ\mathrm{C}$ below the actual temperature. A reference substance melts at $153\,^\circ\mathrm{C}$ . Sample R melts sharply at a displayed $151\,^\circ\mathrm{C}$ ; sample S melts from a displayed $151$ to $153\,^\circ\mathrm{C}$ . Use the corrected data to decide which sample is stronger evidence of purity.
[4 marks]
Total for this question: 4
Evaluate a student's claim that a liquid is pure because it starts boiling at the accepted boiling point of $76\,^\circ\mathrm{C}$ . During continued heating, the liquid boils from $76$ to $82\,^\circ\mathrm{C}$ .
[4 marks]
Total for this question: 4
Candidate R melts at $48\,^\circ\mathrm{C}$ and boils at $93\,^\circ\mathrm{C}$ ; candidate S melts at $71\,^\circ\mathrm{C}$ and boils at $126\,^\circ\mathrm{C}$ . An unknown sample melts sharply at $48\,^\circ\mathrm{C}$ and boils sharply at $126\,^\circ\mathrm{C}$ . Evaluate the claim that the sample is pure candidate R.
[4 marks]
Total for this question: 4
A liquid begins boiling at $64\,^\circ\mathrm{C}$ , but its temperature rises steadily to $91\,^\circ\mathrm{C}$ while boiling continues. A fraction collected near $91\,^\circ\mathrm{C}$ is retested and boils sharply at $91\,^\circ\mathrm{C}$ , the accepted boiling point of compound Z. Evaluate the purity of the original liquid and of the collected fraction.
[5 marks]
Total for this question: 5

4.8.1.2 · Formulations

Explanation

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.

Tier 1 · Easy

A cleaning spray is made from measured amounts of solvent, detergent, fragrance and dye, each with a stated purpose. State why the spray is a formulation.
[2 marks]
Total for this question: 2
An alloy is produced by mixing measured masses of metals so that it has the required strength. State why the alloy is a formulation.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A paint contains carefully measured pigment, solvent and binder. Explain why the paint is a formulation and give the purpose of each named component.
[4 marks]
Total for this question: 4
Mixture A is a breakfast cereal made with controlled amounts of grain, vitamins and flavouring. Mixture B is sand accidentally spilled into water. Identify the formulation and explain your decision.
[3 marks]
Total for this question: 3
Explain why a medicine tablet containing a small measured mass of active drug and a much larger measured mass of filler can still be a formulation.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Two batches of a medical cream contain the same chemicals. In batch X the components were weighed precisely; in batch Y their proportions varied. Evaluate which batch better fits the definition of a formulation.
[4 marks]
Total for this question: 4
A $500\,\mathrm{g}$ cleaning-gel formulation should contain $8\%$ surfactant, $2\%$ fragrance and the remainder water. A batch contains $40\,\mathrm{g}$ surfactant, $15\,\mathrm{g}$ fragrance and $445\,\mathrm{g}$ water. Calculate the required masses of fragrance and water, then evaluate the batch.
[4 marks]
Total for this question: 4
Evaluate a proposed cleaning-agent formulation containing $72\%$ solvent, $25\%$ detergent and $6\%$ fragrance. Calculate the corrected solvent percentage if the detergent and fragrance percentages stay fixed, and explain why the original recipe is unsuitable.
[4 marks]
Total for this question: 4
A $480\,\mathrm{g}$ batch of skin lotion contains $36\,\mathrm{g}$ of active ingredient. The designed formulation must contain $6.0\%$ active ingredient. Calculate the mass of filler that must be added without removing any material, then explain why the adjusted batch fits the formulation more closely.
[5 marks]
Total for this question: 5
A wall coating must spread evenly, hide the surface and leave a firmly attached layer. Trial P spreads easily but dries translucent and peels away. Trial Q spreads evenly, dries opaque and remains attached. Choose the better formulation and explain how the trial evidence guides changes to the pigment, solvent and binder in trial P.
[5 marks]
Total for this question: 5

4.8.1.3 · Chromatography

Explanation

Chromatography separates a mixture because its substances distribute differently between a stationary phase and a mobile phase; a mixture may give several spots, while a pure compound gives one spot in every solvent.
For paper chromatography, draw a pencil origin line, add small sample spots, keep the solvent below the line, allow the solvent to rise, then mark the solvent front.
Calculate $R_f=\frac{\text{distance moved by the centre of the spot}}{\text{distance moved by the solvent front}}$; for example, $3.6/6.0=0.60$ .
A common error is to measure from the paper edge or spot boundary instead of from the origin to the centre of the spot.
Separation also depends on solubility in the mobile phase and attraction to the stationary phase, so changing the solvent can change the pattern.

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: $R_f=0.70$

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 $1$ .

Tier 1 · Easy

On a chromatogram, a pigment centre is $4.2\,\mathrm{cm}$ above the origin and the solvent front is $7.0\,\mathrm{cm}$ above the origin. Calculate the pigment's $R_f$ value.
[2 marks]
Total for this question: 2
In a paper-chromatography setup, the solvent level starts above the sample spots on the origin line. Explain why this setup is unsuitable.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Describe how to use paper chromatography to find out whether a purple pen ink contains more than one soluble dye. Include two setup details that prevent misleading results.
[5 marks]
Total for this question: 5
A spot moves $3.3\,\mathrm{cm}$ from the origin while the solvent front moves $7.5\,\mathrm{cm}$ . Calculate the $R_f$ value and identify the spot from references M $0.36$ , N $0.44$ and P $0.58$ in the same solvent.
[3 marks]
Total for this question: 3
Calculate the $R_f$ value of a spot whose centre is $5.9\,\mathrm{cm}$ from the bottom edge of the paper. The origin is $1.5\,\mathrm{cm}$ from the bottom edge and the solvent front is $9.5\,\mathrm{cm}$ from that edge.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A solvent front moves $8.0\,\mathrm{cm}$ . An unknown gives spots at $2.4\,\mathrm{cm}$ and $5.6\,\mathrm{cm}$ . Reference $R_f$ values in this solvent are: J $0.30$ , K $0.55$ , L $0.70$ . Identify the substances present and state why this evidence does not prove the sample contains only those substances.
[5 marks]
Total for this question: 5
A colourless sample produces one spot in solvent A but two separated spots in solvent B. Evaluate the claim that the sample is pure, and explain why the two solvents give different evidence.
[4 marks]
Total for this question: 4
Evaluate an identification made from a broad chromatogram spot extending from $3.9$ to $5.1\,\mathrm{cm}$ above the origin. The solvent front moved $7.5\,\mathrm{cm}$ . A student used the upper edge and matched reference B, $R_f=0.68$ ; reference A has $R_f=0.60$ .
[4 marks]
Total for this question: 4
On a chromatogram, a reference spot with $R_f=0.65$ moved $5.2\,\mathrm{cm}$ . On the same run an unknown spot moved $3.6\,\mathrm{cm}$ . Calculate the solvent-front distance and the unknown's $R_f$ , then identify the unknown from Q $0.36$ , R $0.45$ and S $0.72$ .
[4 marks]
Total for this question: 4
Ink A gives spots $2.4$ and $4.2\,\mathrm{cm}$ above the origin when its solvent front moves $6.0\,\mathrm{cm}$ . Ink B, run in the same solvent on a longer paper, gives spots $3.6$ , $6.3$ and $7.2\,\mathrm{cm}$ above the origin when its solvent front moves $9.0\,\mathrm{cm}$ . Use $R_f$ values to evaluate the claim that the inks contain exactly the same dyes.
[5 marks]
Total for this question: 5

4.8.2.1 · Test for hydrogen

Explanation

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.

Tier 1 · Easy

A colourless gas makes a pop when tested at the mouth of its tube with a lit splint. Identify the gas.
[1 mark]
Total for this question: 1
Describe the splint test that identifies hydrogen, including where the splint is held and the sound heard.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A student tests a tube before it has filled reasonably fully with the collected gas. Explain why this could give an unreliable negative result in the hydrogen test.
[2 marks]
Total for this question: 2
A learner pushes a burning splint deep into a test tube of collected gas. State the procedural correction for a hydrogen test and explain why only a small sample should be tested.
[3 marks]
Total for this question: 3
Explain why a gas sample collected immediately after a reaction starts may give no pop, while a later sample from the same apparatus gives a pop with a burning splint.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A student tests a gas using a glowing splint and sees no change, then concludes that hydrogen is absent. Evaluate the conclusion and describe the correct confirmation test.
[3 marks]
Total for this question: 3
A collected gas gives a squeaky pop with a burning splint. A student concludes that the tube contains pure hydrogen. Evaluate the conclusion.
[3 marks]
Total for this question: 3
Choose the report that gives a positive hydrogen result and explain the other two observations: A, a glowing splint relights; B, a burning splint at the tube opening causes a pop; C, damp litmus paper is bleached white.
[4 marks]
Total for this question: 4
A known hydrogen control and an unknown gas both fail to produce a pop when tested with the same charred, unlit splint. Evaluate the conclusion that the unknown is not hydrogen, and give a valid repeat procedure.
[4 marks]
Total for this question: 4
Tubes P and Q contain hydrogen and oxygen, one gas in each tube. Devise a test sequence that assigns both labels using fresh wooden splints, and state the positive observation required for each gas.
[5 marks]
Total for this question: 5

4.8.2.2 · Test for oxygen

Explanation

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.

Tier 1 · Easy

A glowing splint returns to flame inside a jar of colourless gas. Name the gas.
[1 mark]
Total for this question: 1
State the condition of the wooden splint used to test for oxygen and give the positive observation.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A gas causes a glowing splint to burst back into flame. Identify the gas and state the property demonstrated by this test.
[2 marks]
Total for this question: 2
A burning splint continues to burn in an unknown gas. Evaluate whether this observation alone confirms oxygen and state the required positive result.
[3 marks]
Total for this question: 3
Evaluate the statement that oxygen has been confirmed because a glowing splint became brighter but did not relight.
[2 marks]
Total for this question: 2

Tier 3 · Hard

Three observations are reported for one gas: a lit splint gives no pop, limewater stays clear, and a glowing splint relights. Use all three results to identify the gas and explain which observation is decisive.
[4 marks]
Total for this question: 4
Explain why a glowing wooden splint relights in oxygen but this does not mean that oxygen itself is a fuel.
[3 marks]
Total for this question: 3
Evaluate a negative oxygen test in which a student blows out a splint, waits until it is no longer glowing, and then inserts it into the gas. Describe the valid repeat test.
[3 marks]
Total for this question: 3
A gas collected immediately after an apparatus is assembled does not relight a glowing splint. After the gas has flowed for a while, a fresh sample relights a fresh glowing splint. Evaluate what the two results show about oxygen in the samples.
[4 marks]
Total for this question: 4
A fresh glowing splint relights in jar P. The student immediately transfers the now-burning splint into jar Q; it continues burning, so the student reports oxygen in both jars. Evaluate both identifications and describe the test needed for Q.
[5 marks]
Total for this question: 5

4.8.2.3 · Test for carbon dioxide

Explanation

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.

Tier 1 · Easy

An unknown gas is shaken with clear limewater, which becomes cloudy. Identify the gas.
[1 mark]
Total for this question: 1
Name the reagent used to test for carbon dioxide and state its positive appearance.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Carbon dioxide makes clear limewater turn cloudy. Explain what causes the cloudiness and name the solid formed.
[2 marks]
Total for this question: 2
A student bubbles an unknown gas through pure water, sees no visible change and rules out carbon dioxide. Evaluate this decision and describe the correct test.
[3 marks]
Total for this question: 3
Evaluate a carbon-dioxide test in which the limewater is cloudy before any gas is passed through it, and state how to improve the test.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A learner writes: 'Carbon dioxide is confirmed because it extinguishes a flame.' Improve this claim to give the specification test, its reagent and its positive observation.
[3 marks]
Total for this question: 3
Carbon dioxide is bubbled through limewater. Write the balanced symbol equation, including state symbols, using $\mathrm{CO_2}$ and $\mathrm{Ca(OH)_2}$ , and link the visible result to one product.
[4 marks]
Total for this question: 4
Determine what can be concluded from these valid tests: gas X turns fresh limewater cloudy; gas Y leaves fresh limewater clear but extinguishes a burning splint. Explain why gas Y cannot be identified from the information given.
[4 marks]
Total for this question: 4
Fresh limewater stays clear when air from an empty delivery tube is passed through it, turns cloudy with a known carbon dioxide control, and turns cloudy with unknown gas U. Evaluate the evidence for identifying U and state one conclusion that the test cannot support.
[5 marks]
Total for this question: 5
A gas mixture is divided into two fresh portions. One portion turns clear limewater cloudy; a glowing splint relights in the other portion. Evaluate the claim that the original sample contained carbon dioxide only.
[4 marks]
Total for this question: 4

4.8.2.4 · Test for chlorine

Explanation

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.

Tier 1 · Easy

Damp litmus paper loses its colour and becomes white in an unknown gas. Identify the gas.
[1 mark]
Total for this question: 1
State the paper condition needed for the chlorine test and the final positive observation.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A student says a gas must be chlorine because damp litmus paper turned white. Another student says any acidic gas would do this. Explain which student is correct.
[2 marks]
Total for this question: 2
Damp blue litmus paper turns red and then becomes white in an unknown gas. Identify the gas and state which observation provides the positive identification.
[3 marks]
Total for this question: 3
Explain why damp red litmus paper can still give a positive chlorine test even though it cannot show an initial change from blue to red.
[2 marks]
Total for this question: 2

Tier 3 · Hard

Two students test the same gas. One uses dry blue litmus and sees no change; the other uses damp litmus and it turns white. Explain which result should be used and what conclusion follows.
[3 marks]
Total for this question: 3
Three gases are tested with damp blue litmus. Gas A leaves it blue, gas B turns it red and the colour remains, and gas C removes the colour completely. Identify the gas that has tested positive for chlorine and explain why B is not a positive chlorine result.
[4 marks]
Total for this question: 4
Evaluate a student's plan to identify chlorine by smelling an unknown gas, and describe the specified test and positive result.
[3 marks]
Total for this question: 3
Damp blue litmus paper turns red as soon as it enters an unknown gas. The observer removes it immediately and reports an acidic gas but rules out chlorine. Evaluate this conclusion and describe the observation needed before chlorine can be ruled out.
[4 marks]
Total for this question: 4
A damp litmus blank left in clean air becomes white. A second strip becomes white in a known chlorine control, and a third becomes white in unknown gas V. Evaluate whether V has been identified, diagnose the problem and describe a reliable repeat.
[5 marks]
Total for this question: 5

4.8.3.1 · Flame tests (chemistry only)

Explanation

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.

Tier 1 · Easy

A compound produces a crimson flame. Identify the metal ion present.
[1 mark]
Total for this question: 1
A metal compound gives an orange-red flame. Identify the metal ion in the compound.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A pure potassium salt unexpectedly gives an intense yellow flame instead of a clear lilac flame. Suggest a likely cause and one improvement to the procedure.
[3 marks]
Total for this question: 3
Sample U gives a green flame and sample V gives a lilac flame. Identify the metal ion in each sample.
[2 marks]
Total for this question: 2
Predict the flame-test results for separate samples of sodium chloride and sodium carbonate, and explain what the test can and cannot identify.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A mixture known to contain two metal ions gives only an intense yellow flame. State one ion supported by the observation and explain why the second ion cannot be identified confidently from this test alone.
[3 marks]
Total for this question: 3
An ionic compound produces no distinctive flame colour. Explain why this result does not prove that the compound contains no metal ions.
[3 marks]
Total for this question: 3
Explain why a mixture containing sodium ions and calcium ions may show only a yellow flame, and suggest how flame emission spectroscopy could provide evidence for both ions.
[4 marks]
Total for this question: 4
An unclean wire makes a potassium reference appear yellow. After the wire is cleaned, the reference gives a lilac flame. The wire is cleaned again before unknown X is tested and gives an orange-red flame. Determine the ions supported by the valid observations and explain the anomalous yellow result.
[5 marks]
Total for this question: 5
A solution tested on a clean wire gives an intense yellow flame. A blank gives no flame colour. A separate portion of the solution forms a blue precipitate with sodium hydroxide solution. The solution may contain more than one metal ion. Use all the evidence to identify the ions supported and explain why the flame test alone was incomplete.
[5 marks]
Total for this question: 5

4.8.3.2 · Metal hydroxides (chemistry only)

Explanation

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.

Tier 1 · Easy

Adding sodium hydroxide solution to an unknown ionic solution forms a brown precipitate. Identify the metal ion.
[1 mark]
Total for this question: 1
Sodium hydroxide solution forms a blue precipitate with an unknown solution. Identify the metal ion.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Sodium hydroxide solution gives a green precipitate with sample P and a brown precipitate with sample Q. Identify the metal ion in each sample.
[2 marks]
Total for this question: 2
Solutions J and K both form white precipitates when a little sodium hydroxide solution is added. J's precipitate dissolves in excess sodium hydroxide, but K's remains. Identify the ion in J and state what can be concluded about K from this test alone.
[4 marks]
Total for this question: 4
Predict the observations when sodium hydroxide solution is added first in a small amount and then in excess to separate solutions containing copper(II) ions and aluminium ions.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Three solutions are tested with sodium hydroxide. W gives a blue precipitate, X gives a green precipitate, and Y gives a white precipitate that dissolves in excess sodium hydroxide. Identify the metal ion in W, X and Y.
[3 marks]
Total for this question: 3
Iron(III) ions react with hydroxide ions to form a brown precipitate. Write the balanced ionic equation, including state symbols, and name the precipitate.
[4 marks]
Total for this question: 4
Determine the ion in each of three solutions known to contain aluminium, calcium or magnesium ions. All form white precipitates with a little sodium hydroxide. A's precipitate dissolves in excess; B's remains and B gives an orange-red flame; C's remains and gives no distinctive flame colour.
[5 marks]
Total for this question: 5
A student pours a large excess of sodium hydroxide solution directly into an aluminium-ion sample. The final mixture is colourless with no solid, so the student concludes that aluminium ions are absent. Evaluate the conclusion and give a procedure that would reveal the diagnostic changes.
[5 marks]
Total for this question: 5
A solution is known to contain copper(II) ions, aluminium ions, or both. A little sodium hydroxide produces a blue precipitate. After excess sodium hydroxide is added, a blue precipitate remains. Determine what is established about the ions and explain the limitation of the test for this mixture.
[5 marks]
Total for this question: 5

4.8.3.3 · Carbonates (chemistry only)

Explanation

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.

Tier 1 · Easy

A solid fizzes when dilute acid is added, and the gas makes limewater cloudy. Name the ion detected in the solid.
[1 mark]
Total for this question: 1
Dilute acid is added to a carbonate and a gas is released. Name the gas and state how to confirm it.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An unknown powder and a known carbonate are tested in parallel with dilute acid and limewater. Neither test makes the limewater cloudy. Explain why the unknown result cannot be used to rule out carbonate ions, and state what should be done next.
[2 marks]
Total for this question: 2
Describe a complete test for carbonate ions in an unknown solid, including the reagent added to the solid and the test applied to the gas.
[3 marks]
Total for this question: 3
Describe how gas tests distinguish a carbonate reacting with dilute acid from a metal reacting with dilute acid.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Powders C and D both fizz when dilute acid is added. The gas from C turns limewater cloudy, but the gas from D leaves limewater clear. Explain which powder has tested positive for carbonate ions and why fizzing alone is insufficient evidence.
[4 marks]
Total for this question: 4
Aqueous sodium carbonate reacts with dilute hydrochloric acid. Write the balanced symbol equation with state symbols, then state the observation used to confirm the gaseous product.
[4 marks]
Total for this question: 4
Identify the cation, anion and compound in an unknown solid that gives an orange-red flame and releases a gas with dilute acid that turns limewater cloudy.
[4 marks]
Total for this question: 4
A tablet contains an active carbonate and an insoluble binder. Dilute acid makes the crushed tablet fizz, and the gas turns fresh limewater cloudy, but some solid remains after the reaction. Evaluate the claim that the remaining solid disproves the presence of carbonate ions.
[4 marks]
Total for this question: 4
An unknown water-soluble powder may contain both carbonate ions and chloride ions. Describe how to use separate portions to test for both ions, including the reagent sequence and positive observation for each portion.
[6 marks]
Total for this question: 6

4.8.3.4 · Halides (chemistry only)

Explanation

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.

Tier 1 · Easy

After dilute nitric acid and silver nitrate are added to a solution, a yellow precipitate forms. Identify the halide ion.
[1 mark]
Total for this question: 1
A correctly acidified solution forms a cream precipitate with silver nitrate solution. Identify the halide ion.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A student acidifies an unknown with dilute hydrochloric acid before adding silver nitrate solution. Explain why this can give a false positive for chloride ions and name the correct acid.
[3 marks]
Total for this question: 3
An unknown solution is treated with the correct acid and then silver nitrate solution. A cream solid forms. Name the acid, identify the ion and name the precipitate.
[4 marks]
Total for this question: 4
Explain why an unknown solution is acidified with dilute nitric acid before silver nitrate solution is added in a halide test.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Samples R, S and T form white, cream and yellow precipitates respectively after the correct halide test. Identify all three ions and state the two reagents that must have been added, in order.
[5 marks]
Total for this question: 5
A deionised-water blank produces a white precipitate during a halide test. The unknown sample produces a pale cream precipitate in the same apparatus. Evaluate the sample result and state how to obtain reliable evidence.
[4 marks]
Total for this question: 4
Determine the ions and formula of a compound that gives a green flame and, after dilute nitric acid followed by silver nitrate solution, forms a cream precipitate.
[5 marks]
Total for this question: 5
An unknown solution may contain carbonate ions as well as a halide. Silver nitrate added directly to one portion gives a pale yellow precipitate. A second portion is first acidified correctly and then gives a cream precipitate with silver nitrate. Evaluate the two results and identify the halide supported by the valid test.
[5 marks]
Total for this question: 5
After dilute nitric acid and silver nitrate are added correctly, an unknown solution forms a very pale cream precipitate. A student reports that bromide ions are present and chloride ions are absent. Evaluate both parts of the report if the solution could contain a mixture of halide ions.
[4 marks]
Total for this question: 4

4.8.3.5 · Sulfates (chemistry only)

Explanation

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.

Tier 1 · Easy

An acidified solution forms a white precipitate when barium chloride solution is added. Identify the ion being tested.
[1 mark]
Total for this question: 1
State the two reagents used, in order, to test a solution for sulfate ions and give the positive result.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A solution contains carbonate ions but no sulfate ions. Explain why adding barium chloride directly could be misleading and how acidifying first prevents this problem.
[3 marks]
Total for this question: 3
An unknown solution is acidified with dilute hydrochloric acid and then gives a white precipitate with barium chloride solution. Identify the ion and name the precipitate.
[3 marks]
Total for this question: 3
Evaluate a plan to acidify an unknown solution with dilute sulfuric acid before adding barium chloride solution in a sulfate test. Give the correct acid.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A student adds barium chloride directly to an unknown solution, sees a white precipitate and reports sulfate ions. Evaluate the procedure and give the complete test needed before accepting the identification.
[4 marks]
Total for this question: 4
Aqueous sodium sulfate is acidified correctly and reacts with barium chloride solution. Write the balanced symbol equation with state symbols and explain the visible change.
[4 marks]
Total for this question: 4
Describe how to use separate portions of an unknown solution to show that both sulfate ions and chloride ions are present. Give each reagent sequence and positive observation.
[6 marks]
Total for this question: 6
Dilute hydrochloric acid is added to an unknown solution and the released gas turns limewater cloudy. When fizzing has stopped, barium chloride solution is added to the acidified liquid and a white precipitate forms. Use the complete sequence to determine the two anions supported and explain the purpose of waiting for the fizzing to stop.
[5 marks]
Total for this question: 5
An acidified deionised-water blank and an acidified unknown both form white precipitates when the same bottle of barium chloride is added. A fresh bottle gives no precipitate with a new blank but gives a white precipitate with a new portion of the unknown. Evaluate the first result and the repeated evidence.
[5 marks]
Total for this question: 5

4.8.3.6 · Instrumental methods (chemistry only)

Explanation

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.

Tier 1 · Easy

State one advantage of an instrumental method over a chemical test based on a visible colour change.
[1 mark]
Total for this question: 1
A contaminant is present at a concentration too small to produce a visible chemical-test result. State the instrumental-method advantage needed to detect it.
[1 mark]
Total for this question: 1

Tier 2 · Standard

An instrument reads $5.02\,\mathrm{mg\,dm^{-3}}$ for a certified $5.00\,\mathrm{mg\,dm^{-3}}$ standard and can detect $0.001\,\mathrm{mg\,dm^{-3}}$ . State which result demonstrates accuracy and which demonstrates sensitivity.
[2 marks]
Total for this question: 2
A laboratory must screen $200$ water samples in one day for a trace pollutant. Explain two advantages of choosing a suitable instrumental method instead of a visible chemical test.
[3 marks]
Total for this question: 3
Evaluate the claim that instrumental methods are better than chemical tests because they are always cheaper and easier. State two specification advantages instead.
[3 marks]
Total for this question: 3

Tier 3 · Hard

For a certified $10.0\,\mathrm{mg\,dm^{-3}}$ standard, method A reports $12\,\mathrm{mg\,dm^{-3}}$ after $25$ minutes and misses very dilute samples. Method B reports $10.1\,\mathrm{mg\,dm^{-3}}$ after $40$ seconds and detects every sample. Compare the methods using the three specification advantages of instrumental analysis.
[4 marks]
Total for this question: 4
A certified solution contains $8.00\,\mathrm{mg\,dm^{-3}}$ of an ion. Instrument X reports $7.842\,\mathrm{mg\,dm^{-3}}$ and detects down to $0.0001\,\mathrm{mg\,dm^{-3}}$ . Instrument Y reports $8.01\,\mathrm{mg\,dm^{-3}}$ and detects down to $0.01\,\mathrm{mg\,dm^{-3}}$ . Evaluate the claim that X is more accurate because it displays more decimal places.
[4 marks]
Total for this question: 4
Calculate the total time saved when $120$ samples are analysed sequentially by an instrumental method taking $15$ seconds per sample instead of a chemical test taking $4$ minutes per sample. The detection limits are $0.002\,\mathrm{mg\,dm^{-3}}$ and $0.5\,\mathrm{mg\,dm^{-3}}$ respectively. Explain two advantages shown by the data.
[5 marks]
Total for this question: 5
An instrument processes a large batch rapidly and detects a trace control that a visible test misses. However, on every repeat it reports a certified standard above the accepted value. Evaluate the claim that repeatability makes the instrument suitable for reporting the samples immediately.
[5 marks]
Total for this question: 5
Before analysing an unknown, an instrumental method gives the analyte's signal when a blank containing no analyte is measured. The unknown gives the same signal. Evaluate whether the analyte has been detected and set out the corrective chain before the unknown is retested.
[5 marks]
Total for this question: 5

4.8.3.7 · Flame emission spectroscopy (chemistry only)

Explanation

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.

Tier 1 · Easy

Reference lines occur at $589\,\mathrm{nm}$ for sodium and $671\,\mathrm{nm}$ for lithium. A sample has one line at $671\,\mathrm{nm}$ . Identify the metal ion.
[1 mark]
Total for this question: 1
State what the positions of lines in a flame emission spectrum can reveal about a solution.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A flame emission spectrum contains lines at the reference positions for lithium and sodium. The sodium line is brighter. State what can be concluded about the ions present and why the brighter line alone does not prove that sodium has the higher concentration.
[3 marks]
Total for this question: 3
A reference set lists calcium lines at $423$ and $616\,\mathrm{nm}$ and copper lines at $510$ and $578\,\mathrm{nm}$ . A sample produces all four lines. Identify the ions and explain how the spectrum supports the identification.
[3 marks]
Total for this question: 3
Describe how flame emission spectroscopy produces information that can be used to identify metal ions in a solution.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A solution produces emission lines at $589\,\mathrm{nm}$ and $766\,\mathrm{nm}$ . References assign these to sodium and potassium. At $766\,\mathrm{nm}$ , standards of $2.0$ and $6.0\,\mathrm{mg\,dm^{-3}}$ give intensities $15$ and $47$ units; the sample gives $31$ units. Identify both ions and estimate the potassium-ion concentration, assuming a linear response.
[5 marks]
Total for this question: 5
A solution is diluted by transferring $10.0\,\mathrm{cm^3}$ into a $50.0\,\mathrm{cm^3}$ final volume. Its flame-emission intensity is $25$ units. Standards of $2.0$ , $4.0$ and $6.0\,\mathrm{mg\,dm^{-3}}$ give intensities $10$ , $20$ and $30$ units. Determine the ion concentration in the original solution, assuming a linear response.
[5 marks]
Total for this question: 5
Evaluate an attempt to identify an ion by comparing line distances measured in centimetres on a sample printout with reference wavelengths listed in nanometres when no conversion scale is provided. State what reference evidence is needed.
[3 marks]
Total for this question: 3
A sample has lines at $671\,\mathrm{nm}$ and $589\,\mathrm{nm}$ , identifying lithium and sodium respectively. Both calibrations are linear through the origin. A $2.0\,\mathrm{mg\,dm^{-3}}$ lithium standard gives $12$ intensity units and the sample's lithium line gives $30$ units. A $5.0\,\mathrm{mg\,dm^{-3}}$ sodium standard gives $20$ units and the sample's sodium line gives $24$ units. Calculate both concentrations and evaluate which ion is more concentrated.
[5 marks]
Total for this question: 5
A line at $423\,\mathrm{nm}$ identifies calcium. The instrument gives $4$ intensity units for a blank, $16$ units for a $2.0\,\mathrm{mg\,dm^{-3}}$ calcium standard and $40$ units for a $6.0\,\mathrm{mg\,dm^{-3}}$ standard. A sample gives $31$ units. Correct for the blank and determine the calcium-ion concentration.
[4 marks]
Total for this question: 4

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.8.1.1 · Pure substances

Tier 1 · Easy

Mark scheme for 4.8.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	It is pure because it contains a single element and no other substance.	Apply the chemical definition rather than the everyday meaning. Sulfur is one element, and the sample contains nothing else, so it is a pure substance.	2
Total Question 1			2
02.1	The contents are chemically pure because water is one compound and no other substance is present.	Use the chemical definition. A sample containing a single compound, with nothing mixed into it, is a pure substance.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.8.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The statement is incorrect. A chemically pure substance contains only one element or one compound, so the sample is a mixture of two compounds.	Count distinct substances, not whether they are elements or compounds. Two compounds means two substances are mixed, so the sample is not pure.	2
Total Question 1			2
02.1	Sample P is more likely to be pure because it boils at the accepted temperature and does not boil over a range.	Compare both the value and the spread. P has the specific accepted boiling point expected for a pure compound, whereas Q changes state across several temperatures, which indicates a mixture.	3
Total Question 2			3
03.1	A pure substance may be a single compound; containing different elements within that compound does not make it a mixture.	Count the substances present rather than the types of atom in each particle. If every particle belongs to the same compound and nothing else is mixed in, the solid is chemically pure.	2
Total Question 3			2

Tier 3 · Hard

Mark scheme for 4.8.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Everyday 'pure' can mean nothing unwanted has been added; chemically the drink is a mixture because it contains several substances, not one element or compound.	Separate the two meanings. The advertising claim can describe an unadulterated natural product. The composition list, however, contains several different compounds, so the chemical definition classifies the drink as a mixture.	4
Total Question 1			4
02.1	R is stronger evidence of purity: its corrected melting point is a sharp $153\,^\circ\mathrm{C}$ , matching the reference. S has a corrected melting range of $153$ to $155\,^\circ\mathrm{C}$ , so it is more likely to be a mixture.	Add $2\,^\circ\mathrm{C}$ to every displayed value. R becomes one sharp value at the reference point. Although S's range begins at the reference point, melting across a range is evidence against purity.	4
Total Question 2			4
03.1	The claim is not supported because the liquid boils over a range rather than at one specific temperature. Starting at the accepted value is insufficient evidence; the full boiling range indicates that the liquid is a mixture.	Consider the complete change-of-state record. A pure substance boils at a specific temperature, whereas this liquid continues boiling across a $6\,^\circ\mathrm{C}$ interval. The first reading alone therefore gives a misleading impression of purity.	4
Total Question 3			4
04.1	The sharp melting and boiling points are evidence that the sample may be a pure substance. Its melting point matches candidate R. Its boiling point does not match R; it matches candidate S instead. The two results do not support identification as pure R, so the reference data, sample label or measurements should be checked.	Use both physical properties rather than accepting the first match. A pure sample should have a consistent set of characteristic temperatures. Here each measurement points to a different candidate, so the identity claim is not supported even though both changes of state are sharp.	4
Total Question 4			4
05.1	The original liquid is not pure because it boils across a range of temperatures. The rising boiling temperature is evidence that the composition changes during boiling. The collected fraction boils at one sharp temperature. That temperature matches the accepted boiling point of compound Z. The fraction is therefore consistent with pure Z, although the result is evidence rather than absolute proof.	Judge the original liquid and the separated fraction independently. A changing boiling temperature shows that the starting liquid contains more than one substance. The fraction gives much stronger evidence of purity because its sharp boiling point also agrees with reliable reference data for Z.	5
Total Question 5			5

4.8.1.2 · Formulations

Tier 1 · Easy

Mark scheme for 4.8.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	It is a useful mixture whose components are present in measured quantities for particular purposes.	Check both parts of the definition: the product is a mixture designed for cleaning, and its components have specified functions and quantities. It is therefore a formulation.	2
Total Question 1			2
02.1	It is a useful mixture whose components are combined in controlled quantities to give a required property.	Identify the designed use and the controlled composition. Both features are needed for the mixture to be classed as a formulation.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.8.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	It is a formulation because it is a mixture designed for a useful purpose with controlled amounts of its components. The pigment gives colour, the solvent controls flow or spreading, and the binder forms the solid coating.	A formulation is more than any mixture: its composition is chosen to produce required properties. Link each ingredient to a distinct function in the finished paint.	4
Total Question 1			4
02.1	Mixture A is the formulation because it is designed as a useful product and its components are present in controlled amounts for particular purposes. Mixture B is an accidental mixture rather than a designed product.	Do not label every mixture as a formulation. A is deliberately designed and measured to have required properties; B has neither a designed use nor controlled composition.	3
Total Question 2			3
03.1	The tablet is a useful product made from a mixture whose components have particular purposes. The quantities are controlled to give the required dose and tablet properties; the active ingredient does not need to be the largest component.	Apply the definition rather than comparing component masses. The active drug supplies the medical effect, while the filler helps form a usable tablet, and both are deliberately measured.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.8.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Batch X fits better because a formulation has carefully measured components chosen to give required properties; variable proportions in Y may change how the cream works.	Having several chemicals is not sufficient. A formulation is deliberately composed in controlled quantities. Batch X meets this requirement, whereas batch Y lacks the reliable composition needed to ensure the designed properties.	4
Total Question 1			4
02.1	Required fragrance mass $=10\,\mathrm{g}$ and required water mass $=450\,\mathrm{g}$ . The batch does not match the formulation: it has $5\,\mathrm{g}$ too much fragrance and $5\,\mathrm{g}$ too little water; only the surfactant mass is correct.	Calculate $0.02\times500=10\,\mathrm{g}$ fragrance. Water is $90\%$ , so $0.90\times500=450\,\mathrm{g}$ . The stated $40\,\mathrm{g}$ surfactant is $8\%$ of $500\,\mathrm{g}$ , but the other two quantities do not meet the controlled recipe.	4
Total Question 2			4
03.1	The listed percentages total $103\%$ , so the recipe cannot describe one complete mixture. The corrected solvent percentage is $100-25-6=69\%$ . A formulation needs carefully controlled quantities, so an impossible composition would not reliably give the required properties.	Check that all component percentages form one whole. The two fixed components use $31\%$ , leaving $69\%$ for the solvent; the original $72\%$ makes the total exceed $100\%$ .	4
Total Question 3			4
04.1	The current active-ingredient percentage is $36/480\times100=7.5\%$ . For $36\,\mathrm{g}$ to be $6.0\%$ , the final batch mass must be $36/0.060=600\,\mathrm{g}$ . The mass of filler to add is $600-480=120\,\mathrm{g}$ . Adding filler changes the proportion without changing the active-ingredient mass. The adjusted batch has the controlled composition required for the lotion's designed dose and properties.	First check the existing percentage. Then work backwards from the required percentage: the unchanged $36\,\mathrm{g}$ must be $0.060$ of the new total mass. Subtract the original mass from that total to find the filler addition, and link the corrected proportion to the purpose of a formulation.	5
Total Question 4			5
05.1	Trial Q is the better formulation because it meets all three required properties. P's translucent coating indicates that its pigment amount or proportion is unsuitable for hiding the surface. P's peeling indicates that its binder amount or proportion is unsuitable for forming an attached layer. Because P already spreads easily, any change to its solvent must preserve the flow needed for even spreading. The components should be remeasured in controlled proportions and the adjusted coating retested.	Start with the product requirements, not merely the ingredient list. Match opacity to pigment, adhesion to binder and flow to solvent. Q meets the complete design brief, while P needs controlled changes followed by testing rather than an arbitrary addition of one component.	5
Total Question 5			5

4.8.1.3 · Chromatography

Tier 1 · Easy

Mark scheme for 4.8.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$R_f=0.60$	Use $R_f=\frac{\text{spot distance}}{\text{solvent distance}}=\frac{4.2}{7.0}=0.60$ . The ratio has no unit.	2
Total Question 1			2
02.1	The samples can dissolve directly into the solvent reservoir instead of being carried up the paper, so they may not separate into reliable spots.	The solvent must begin below the origin. Otherwise the sample can wash off the paper into the liquid before the mobile phase travels through the stationary phase.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.8.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Use a pencil origin, place a small ink spot on it, stand the paper in a suitable solvent below the origin, let the solvent rise, remove and mark the solvent front; more than one separated spot shows more than one dye.	Draw the start line in pencil so it does not dissolve. Add a small concentrated ink spot and place the paper in a suitable solvent with the liquid level below the spot. Cover if appropriate, allow separation, then remove the paper and mark the solvent front before it evaporates. Count the separated spots.	5
Total Question 1			5
02.1	$R_f=0.44$ ; the spot matches substance N.	Calculate $R_f=3.3/7.5=0.44$ . Identification requires reference data obtained with the same solvent, and $0.44$ matches N.	3
Total Question 2			3
03.1	$R_f=0.55$	Measure both movements from the origin: the spot moves $5.9-1.5=4.4\,\mathrm{cm}$ and the solvent moves $9.5-1.5=8.0\,\mathrm{cm}$ . Therefore $R_f=4.4/8.0=0.55$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.8.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	J and L are present; the values are $0.30$ and $0.70$ . Another substance could have the same $R_f$ in this solvent, so a second solvent or other evidence is needed.	Calculate both ratios: $2.4/8.0=0.30$ and $5.6/8.0=0.70$ , matching J and L. An $R_f$ match is conditional on the solvent and stationary phase and is not unique proof; repeat with another solvent to strengthen the identification.	5
Total Question 1			5
02.1	The sample is impure because solvent B separates it into two spots, showing at least two substances. One spot in solvent A is not proof of purity because the substances may have moved together in that solvent; changing the solvent changes their distribution between the mobile and stationary phases.	Use all the chromatograms, not just the first result. Compounds can have the same or very similar movement in one solvent but separate in another because their solubilities and attractions to the paper differ.	4
Total Question 2			4
03.1	The spot centre is $(3.9+5.1)/2=4.5\,\mathrm{cm}$ , so the correct value is $R_f=4.5/7.5=0.60$ . The result matches reference A, not B; measuring to the upper edge produced the incorrect value $5.1/7.5=0.68$ .	The specified distance is measured to the centre of a spot. Find the midpoint of its lower and upper edges, divide that distance by the solvent-front distance, and then compare the result with the references.	4
Total Question 3			4
04.1	The solvent-front distance is $5.2/0.65=8.0\,\mathrm{cm}$ . The unknown's $R_f$ is $3.6/8.0=0.45$ . The unknown matches reference R. The comparison is valid because the reference and unknown were run under the same conditions.	Rearrange $R_f=\text{distance moved by substance}/\text{distance moved by solvent}$ to recover the solvent-front distance from the reference. Use that common distance for the unknown, then compare the calculated ratio with the listed references.	4
Total Question 4			4
05.1	Ink A's two $R_f$ values are $2.4/6.0=0.40$ and $4.2/6.0=0.70$ . Ink B has matching values $3.6/9.0=0.40$ and $6.3/9.0=0.70$ . Ink B also has a third value, $7.2/9.0=0.80$ . The matching values support the conclusion that the inks share two dyes. They do not contain exactly the same dyes because B has evidence of one additional component.	Direct spot distances cannot be compared because the solvent fronts travelled different distances. Convert every spot distance to an $R_f$ value. Two ratios coincide across the runs, while B has an extra ratio, so the evidence supports partial rather than complete compositional agreement.	5
Total Question 5			5

4.8.2.1 · Test for hydrogen

Tier 1 · Easy

Mark scheme for 4.8.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Hydrogen	Recall the diagnostic observation: rapid burning with a pop is the positive test for hydrogen.	1
Total Question 1			1
02.1	Hold a burning splint at the open end of the tube; hydrogen gives a pop sound.	Use a burning, not glowing, splint and keep it at the tube opening. Rapid combustion of hydrogen produces the diagnostic pop.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.8.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Too little hydrogen has collected to give an audible pop. So failing to hear a pop does not show that hydrogen is absent; the result is inconclusive rather than negative.	The squeaky pop needs enough hydrogen present at the tube opening to ignite audibly. Let the tube fill reasonably fully before testing, and treat a silent test on a nearly empty tube as inconclusive rather than as evidence that no hydrogen is being produced.	2
Total Question 1			2
02.1	Hold the burning splint at the open end of the test tube rather than pushing it inside. Use only a small gas sample because hydrogen burns rapidly when it reacts with oxygen.	Match the prescribed position exactly: the flame is presented at the mouth of the tube. Limiting the sample reduces the amount of gas involved in the rapid combustion.	3
Total Question 2			3
03.1	The first sample may contain mostly air or too little hydrogen because the gas has not yet displaced the air in the apparatus. The later pop is a positive result showing that hydrogen is present.	Account for the gas initially occupying the apparatus. Early collection can dilute the hydrogen with air, whereas a later sample can contain enough hydrogen at the tube opening to give the specified pop.	2
Total Question 3			2

Tier 3 · Hard

Mark scheme for 4.8.2.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The conclusion is not valid because a glowing splint is not the hydrogen test; use a burning splint at the opening and listen for a pop.	A negative result from the wrong test cannot exclude hydrogen. Replace the glowing splint with a burning splint and test at the mouth of the tube. A pop confirms hydrogen.	3
Total Question 1			3
02.1	The pop confirms that hydrogen is present, but it does not show that the gas is pure. Other gases, including air or oxygen needed for the hydrogen to burn, may also be in the tube.	Treat the gas test as an identification test, not a purity test. A positive pop establishes the presence of hydrogen; it cannot exclude additional gases in the collected sample.	3
Total Question 2			3
03.1	Report B is the positive hydrogen result because hydrogen burns with a pop when tested with a burning splint at the opening. A is the positive observation for oxygen, while C is the positive observation for chlorine.	Match both the test condition and observation. Hydrogen requires the burning-splint pop; relighting a glowing splint and bleaching damp litmus belong to different gas tests.	4
Total Question 3			4
04.1	The conclusion is invalid because the known hydrogen control also gave a negative result. The splint was not burning, so the specified hydrogen test was not performed. Use a fresh burning splint at the opening of a tube containing a small sample of gas. A pop or squeaky pop in a valid repeat is the positive result for hydrogen.	Use the control to test whether the procedure can produce a positive result. Its failure exposes the unlit splint as the procedural fault, so the unknown's negative result carries no valid identification evidence. Repeat with the specified burning-splint test.	4
Total Question 4			4
05.1	Test a small sample from each tube with a separate fresh glowing splint. The tube that relights the glowing splint contains oxygen. Test a small sample from the other tube with a fresh burning splint at its opening. The tube that gives a pop contains hydrogen. Fresh splints and separate samples prevent the first test from creating a misleading observation in the second tube.	Use a positive identification test for each gas rather than assigning a label from a negative result alone. A glowing splint identifies oxygen by relighting, while a burning splint identifies hydrogen by a pop. Keep the samples and splints separate so each observation belongs to one valid test.	5
Total Question 5			5

4.8.2.2 · Test for oxygen

Tier 1 · Easy

Mark scheme for 4.8.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Oxygen	The relighting of a glowing splint is the characteristic positive observation for oxygen.	1
Total Question 1			1
02.1	Use a glowing splint; it relights in oxygen.	The splint must have no flame before insertion but must still be glowing. Returning to flame is the specified positive result.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.8.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The gas is oxygen; the result shows that oxygen supports combustion. Accept ‘supports burning’ or ‘is needed for burning’.	The defining positive result in the oxygen test is relighting a glowing splint. This happens because oxygen supports combustion or burning and is needed for burning.	2
Total Question 1			2
02.1	The observation does not confirm oxygen because the splint was already burning. The valid test uses a glowing splint, which must relight in oxygen.	A test is diagnostic only when its starting condition and observation match the specification. Extinguish the flame first, insert the glowing splint, and look for relighting.	3
Total Question 2			3
03.1	The statement is not justified because the specified positive result is that the glowing splint relights; becoming brighter without relighting is not the required observation.	Compare the report with the exact diagnostic observation. Only a return to flame counts as the specified positive oxygen result.	2
Total Question 3			2

Tier 3 · Hard

Mark scheme for 4.8.2.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The gas is oxygen; relighting a glowing splint is decisive, while no pop and clear limewater rule out positive results for hydrogen and carbon dioxide.	Match each observation to a gas test. No pop is not a positive hydrogen result, and unchanged limewater is not a positive carbon-dioxide result. The glowing splint relighting is the specific positive oxygen result.	4
Total Question 1			4
02.1	Oxygen supports combustion, so the hot wood reacts faster and reaches a temperature at which it flames again. The wood is the fuel; oxygen supports its burning rather than burning as the fuel.	Separate the roles in combustion. The glowing wood is the substance being oxidised, while oxygen enables the combustion reaction to continue rapidly enough for a flame to reappear.	3
Total Question 2			3
03.1	The negative result is invalid because the splint was no longer glowing when it entered the gas. Repeat with a glowing splint inserted into the collected gas; relighting is the positive result for oxygen.	The starting condition is part of the test. Extinguish the flame but insert the wood while its end is still glowing, then observe whether it returns to flame.	3
Total Question 3			3
04.1	The later positive result confirms that the later sample contains oxygen. The early negative result does not prove that oxygen was absent. Air already in the apparatus could have diluted the early sample so that the splint did not relight. Neither observation alone proves that either sample is pure oxygen.	Give more weight to the valid positive observation than to an early negative result from an unflushed apparatus. Dilution can prevent the characteristic relighting, while relighting supports the presence of oxygen. A gas test identifies a gas present; it does not establish purity.	4
Total Question 4			4
05.1	Jar P has given the specified positive oxygen result because it relit a glowing splint. The result for Q is not a valid positive oxygen test. The splint entered Q already burning, so continuing to burn is not the same as relighting. Prepare a separate fresh glowing splint and insert it into Q. Report oxygen in Q only if that glowing splint relights.	Track the state of the splint at the start of each test. P changes a glowing splint back into flame and is therefore positive. Q was never tested with a glowing splint, so it needs an independent repeat before any identification is justified.	5
Total Question 5			5

4.8.2.3 · Test for carbon dioxide

Tier 1 · Easy

Mark scheme for 4.8.2.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Carbon dioxide	Cloudiness in limewater is the positive observation used to identify carbon dioxide.	1
Total Question 1			1
02.1	Use limewater; it turns milky or cloudy.	Limewater is aqueous calcium hydroxide. Carbon dioxide produces the specified milky or cloudy result.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.8.2.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	An insoluble white precipitate forms. The solid is calcium carbonate.	Carbon dioxide reacts with calcium hydroxide in limewater. Insoluble calcium carbonate is produced as fine solid particles, making the liquid appear cloudy.	2
Total Question 1			2
02.1	The decision is invalid because pure water is not the specified reagent and a lack of visible change is inconclusive. Bubble or shake the gas with limewater; carbon dioxide turns it milky or cloudy.	A negative result from an unsuitable reagent cannot identify the gas. Repeat with aqueous calcium hydroxide and use formation of cloudiness as the positive observation.	3
Total Question 2			3
03.1	The result is unreliable because any cloudiness after the test cannot be attributed to the gas sample. Repeat with fresh, clear limewater and look for it turning milky or cloudy when the gas is bubbled through or shaken with it.	Check the reagent before exposure to the unknown. The limewater must start clear so that formation of cloudiness can be observed as a change caused by the gas.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.8.2.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Bubble or shake the gas with limewater; the limewater turns milky or cloudy.	Extinguishing a flame is not the specified identification test and is not sufficiently diagnostic. Use aqueous calcium hydroxide and report the visible cloudiness produced by carbon dioxide.	3
Total Question 1			3
02.1	$\mathrm{CO_2(g)+Ca(OH)_2(aq)\rightarrow CaCO_3(s)+H_2O(l)}$ The insoluble $\mathrm{CaCO_3(s)}$ forms a precipitate that makes the limewater milky or cloudy.	The equation is already balanced with one carbon, one calcium, two hydrogen and four oxygen atoms on each side. The state symbol $(s)$ identifies calcium carbonate as the solid responsible for the observation.	4
Total Question 2			4
03.1	Gas X contains carbon dioxide because it gives the positive limewater result. Gas Y has not tested positive for carbon dioxide, but extinguishing a flame is not a specific identification test, so the gas cannot be identified from these observations alone.	Give priority to a specified positive test. Cloudy limewater identifies carbon dioxide in X. A flame going out can occur in more than one gas, so it does not provide a unique identity for Y.	4
Total Question 3			4
04.1	The clear blank shows that the apparatus or air supply does not itself make the limewater cloudy. The known carbon dioxide control shows that the limewater can give the expected positive result. Gas U gives the same positive cloudy result as the control. The evidence therefore supports the presence of carbon dioxide in U. The test cannot show that U is pure carbon dioxide or that no other gas is present.	Use the blank to check for a false positive and the known control to check that the reagent works. With both checks behaving correctly, U's cloudiness is valid evidence for carbon dioxide. The test identifies a component, but it does not establish the complete composition of the sample.	5
Total Question 4			5
05.1	The cloudy limewater is a positive test for carbon dioxide. The relighting glowing splint is a positive test for oxygen. The original sample therefore contains carbon dioxide and enough oxygen to give its positive test. The claim that it contained carbon dioxide only is incorrect.	Interpret each positive result with its own specified gas test, then combine the evidence because both portions represent the same mixture. Limewater identifies carbon dioxide, while relighting identifies oxygen, so a single-gas conclusion is inconsistent with the complete data.	4
Total Question 5			4

4.8.2.4 · Test for chlorine

Tier 1 · Easy

Mark scheme for 4.8.2.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Chlorine	Bleaching damp litmus paper to white is the diagnostic observation for chlorine gas.	1
Total Question 1			1
02.1	Use damp litmus paper; chlorine bleaches it white.	Moisture must be present on the litmus paper. Loss of the litmus colour, leaving white paper, is the diagnostic result.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.8.2.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The first student is correct: chlorine bleaches damp litmus paper, turning it white. An acidic gas turns damp litmus red rather than bleaching it white, so turning white is not something any acidic gas would do.	Separate a colour change from bleaching. An acidic gas dissolves in the water on the paper and turns blue litmus red; chlorine removes the colour altogether, leaving the paper white. The white result is therefore the specific positive test for chlorine.	2
Total Question 1			2
02.1	The gas is chlorine; bleaching the damp litmus paper white is the identifying observation. Turning blue litmus red shows acidity but is not by itself specific to chlorine.	Use the final loss of colour as the diagnostic evidence. The earlier red colour is consistent with an acidic gas but cannot uniquely identify chlorine.	3
Total Question 2			3
03.1	Chlorine bleaches damp litmus paper white, so complete loss of the red colour is still the positive result.	The diagnostic observation is bleaching rather than the earlier acidic colour change. Red litmus can therefore identify chlorine if it loses its colour and becomes white.	2
Total Question 3			2

Tier 3 · Hard

Mark scheme for 4.8.2.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Use the damp-paper result because moisture is needed; bleaching to white shows that the gas is chlorine.	The dry-paper procedure does not meet the specified test conditions, so its negative result is not reliable. The valid damp litmus test gives bleaching, which identifies chlorine.	3
Total Question 1			3
02.1	Gas C has tested positive for chlorine because it bleaches the damp litmus paper white. Gas B gives only an acidic colour change; without bleaching, the specified chlorine result has not occurred.	Distinguish a change to red from removal of the indicator colour. Only C produces bleaching, so only C matches the chlorine identification test.	4
Total Question 2			4
03.1	The gas should not be smelled because chlorine is toxic and smell is not the specified identification test. Expose damp litmus paper to a small sample; chlorine bleaches the paper white.	Reject direct inhalation and replace it with the specified indicator test. Moisture must be present, and the identifying observation is complete bleaching to white.	3
Total Question 3			3
04.1	Turning damp blue litmus red shows acidic behaviour but does not by itself rule out chlorine. Chlorine can turn damp blue litmus red before bleaching it. The paper was removed too soon to observe the complete response. Repeat safely with damp litmus and observe whether it is bleached white, the positive chlorine result.	Judge the full specified observation rather than only its first stage. Chlorine's positive result is bleaching of damp litmus paper. An early red colour is therefore incomplete evidence, so the test must continue long enough to check for loss of colour.	4
Total Question 4			4
05.1	V has not been identified because the blank also lost its colour without chlorine. The litmus strips may already be contaminated with a bleaching substance. The known control cannot rescue the result because contaminated paper would also turn white in chlorine. Repeat with fresh uncontaminated damp litmus and a blank that remains coloured in clean air. V is positive for chlorine only if the valid test paper is bleached white.	The blank tests whether the paper changes without the gas being investigated. Because it bleaches in clean air, every result from that batch of paper is unreliable. Replace the paper, establish a valid blank, and then compare the unknown with the specified positive observation.	5
Total Question 5			5

4.8.3.1 · Flame tests (chemistry only)

Tier 1 · Easy

Mark scheme for 4.8.3.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Lithium ions, Li⁺	Match the observed crimson colour to the specified flame-test table: lithium compounds give crimson.	1
Total Question 1			1
02.1	Calcium ions, Ca²⁺.	Match the observed flame colour to the required set: calcium compounds produce an orange-red flame.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.8.3.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The sample or apparatus is probably contaminated with sodium ions, whose intense yellow flame can mask the lilac colour. Use clean apparatus and a clean, uncontaminated sample.	A yellow flame indicates sodium, but the sample is stated to be a pure potassium salt. Trace sodium contamination is therefore the likely source; cleaning prevents one sample carrying ions into the next test.	3
Total Question 1			3
02.1	U contains copper ions; V contains potassium ions, K⁺.	Use the specified colour matches independently: copper compounds give green, while potassium compounds give lilac.	2
Total Question 2			2
03.1	Both samples give a yellow flame because both contain sodium ions, Na⁺. The flame test identifies the metal ion but cannot distinguish the chloride ion from the carbonate ion.	Base the flame colour on the cation. Changing the anion does not change the specified sodium-yellow result, so another chemical test is required to identify the anion.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.8.3.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Sodium ions are supported; the strong sodium-yellow colour may mask the flame colour of the other ion.	Yellow is the specified flame colour for sodium. Because the sample is a mixture, one intense emission can conceal another colour, so a separate instrumental or chemical result is needed for the second ion.	3
Total Question 1			3
02.1	Flame tests identify only some metal ions by distinctive colours. A metal ion outside the specified lithium, sodium, potassium, calcium and copper set may give no useful colour, so another test is needed.	Treat the required flame-colour list as a limited identification tool. A negative or unclear result cannot rule out every possible metal ion.	3
Total Question 2			3
03.1	The intense yellow sodium flame can mask the orange-red calcium flame. Flame emission spectroscopy produces a line spectrum; matching sample line positions with sodium and calcium reference lines can show that both ions are present.	Treat the visible flame as a limited mixture test. The stronger sodium colour may conceal calcium, whereas separate characteristic line positions can be compared with a reference set to identify both metal ions.	4
Total Question 3			4
04.1	The valid lilac reference flame supports potassium ions in the reference. The orange-red flame supports calcium ions in X. The first yellow result should not be used because the wire was unclean. A small amount of sodium contamination can produce an intense yellow flame and mask the expected colour. Cleaning the wire between samples is needed to prevent carry-over and obtain reliable identifications.	Separate the invalid trial from the observations made after cleaning. Match lilac to potassium and orange-red to calcium. The changed reference result shows that the first yellow flame came from contamination rather than from the labelled potassium sample alone.	5
Total Question 4			5
05.1	The yellow flame supports the presence of sodium ions. The colourless blank makes contamination from the wire or blank less likely. The blue precipitate supports the presence of copper(II) ions. The intense sodium flame can mask the green flame expected from copper compounds. Together the independent tests support both sodium and copper(II) ions, whereas the flame test alone revealed only sodium.	Interpret the controlled flame result first, then combine it with the metal-hydroxide observation from a separate portion. Sodium accounts for the strong yellow flame, while copper(II) accounts for the blue precipitate. Masking explains why one test did not display both ions.	5
Total Question 5			5

4.8.3.2 · Metal hydroxides (chemistry only)

Tier 1 · Easy

Mark scheme for 4.8.3.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Iron(III) ions, Fe³⁺	The required observation table assigns a brown hydroxide precipitate to iron(III) ions.	1
Total Question 1			1
02.1	Copper(II) ions, Cu²⁺.	A blue hydroxide precipitate is the specified observation for copper(II) ions.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.8.3.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	P contains iron(II) ions, Fe²⁺; Q contains iron(III) ions, Fe³⁺.	Match the hydroxide precipitate colours: iron(II) hydroxide is green, whereas iron(III) hydroxide is brown.	2
Total Question 1			2
02.1	J contains aluminium ions, Al³⁺, because its white precipitate dissolves in excess sodium hydroxide. K could contain calcium ions or magnesium ions; this test alone does not distinguish between them.	Aluminium, calcium and magnesium ions all give white precipitates initially. Only aluminium hydroxide dissolves in excess sodium hydroxide, leaving calcium and magnesium unresolved by this test.	4
Total Question 2			4
03.1	Copper(II) ions form a blue precipitate, which remains in excess sodium hydroxide. Aluminium ions form a white precipitate, which dissolves in excess sodium hydroxide.	State both the initial precipitate and the excess-reagent result for each ion. Colour identifies copper(II), while dissolving in excess distinguishes aluminium from the other listed white precipitates.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.8.3.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	W contains copper(II) ions, Cu²⁺. X contains iron(II) ions, Fe²⁺. Y contains aluminium ions, Al³⁺.	Match each observation to the sodium-hydroxide results: blue identifies copper(II), green identifies iron(II), and a white precipitate that dissolves in excess uniquely identifies aluminium among the listed white precipitates.	3
Total Question 1			3
02.1	$\mathrm{Fe^{3+}(aq)+3OH^-(aq)\rightarrow Fe(OH)_3(s)}$ The precipitate is iron(III) hydroxide.	Three hydroxide ions supply the three $\mathrm{OH}$ groups and total charge $-3$ , balancing one $\mathrm{Fe^{3+}}$ . The product is neutral and solid, so both atoms and charge balance.	4
Total Question 2			4
03.1	A contains aluminium ions, Al³⁺, because its precipitate dissolves in excess sodium hydroxide. B contains calcium ions, Ca²⁺, because calcium gives an orange-red flame. C contains magnesium ions, Mg²⁺, by elimination; its white precipitate remains in excess and magnesium has no required distinctive flame colour.	First use the excess-sodium-hydroxide result to identify aluminium. Then use the calcium flame colour to identify B. The stated three-ion candidate set leaves magnesium for C, consistent with its insoluble white precipitate.	5
Total Question 3			5
04.1	The conclusion is incorrect because aluminium hydroxide dissolves in excess sodium hydroxide. A small amount of sodium hydroxide would first form a white precipitate. Adding a large excess immediately can make that intermediate precipitate easy to miss. Repeat by adding sodium hydroxide dropwise and record the white precipitate before adding more. Then add excess and observe the precipitate dissolving, which supports aluminium ions.	The order and amount of reagent matter. Aluminium ions first produce white aluminium hydroxide, but this solid dissolves when sodium hydroxide is in excess. A stepwise repeat exposes both diagnostic observations instead of recording only the final clear mixture.	5
Total Question 4			5
05.1	The blue precipitate establishes that copper(II) ions are present. Copper(II) hydroxide remains insoluble in excess sodium hydroxide, so the remaining blue solid is expected. Any white aluminium hydroxide could have been hidden by the blue precipitate when little reagent was added. Any aluminium hydroxide present would dissolve in the excess reagent. The observations therefore do not show whether aluminium ions are also present; this test cannot fully resolve the mixture.	Identify the evidence that is unambiguous before considering what could be masked. Blue copper(II) hydroxide confirms copper(II). The two characteristic aluminium observations can both be concealed in a mixture, first by colour and then by dissolution, so aluminium cannot be confirmed or excluded from these results.	5
Total Question 5			5

4.8.3.3 · Carbonates (chemistry only)

Tier 1 · Easy

Mark scheme for 4.8.3.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Carbonate ions, CO₃²⁻	Dilute acid releases carbon dioxide from a carbonate. The cloudy limewater confirms the gas, so the original solid contains carbonate ions.	1
Total Question 1			1
02.1	The gas is carbon dioxide; bubble or shake it with limewater, which turns milky or cloudy.	Carbonates release carbon dioxide when they react with dilute acids. Use the separate limewater result to confirm the gas rather than relying on bubbles alone.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.8.3.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The known carbonate is a positive control and should give a positive result, so its failure shows that the reagents, apparatus or method did not work. Repeat both tests with a verified working setup before interpreting the unknown.	Use the known carbonate to validate the procedure. When the positive control fails, the unknown's negative result is inconclusive rather than evidence that carbonate ions are absent.	2
Total Question 1			2
02.1	Add dilute acid to the solid, then bubble or shake the gas produced with limewater; a milky or cloudy result confirms carbonate ions.	Give both linked stages. Dilute acid releases carbon dioxide from a carbonate, and the positive limewater observation identifies that gas.	3
Total Question 2			3
03.1	Test the gas from the carbonate reaction with limewater; carbon dioxide turns it milky or cloudy. Test the gas from the metal reaction with a burning splint at the tube opening; hydrogen gives a pop.	Do not rely on effervescence because both reactions release a gas. Apply the appropriate confirmatory test to each collected gas and compare the specified positive observations.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.8.3.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	C has tested positive because acid released carbon dioxide, confirmed by cloudy limewater. Fizzing alone is insufficient because another gas-producing reaction could cause bubbles; D's gas was not confirmed as carbon dioxide.	Use both stages of the test. Acid causes gas production, but only C's gas gives the positive carbon-dioxide result with limewater. Therefore C is the carbonate result, while D shows why effervescence by itself is not diagnostic.	4
Total Question 1			4
02.1	$\mathrm{Na_2CO_3(aq)+2HCl(aq)\rightarrow2NaCl(aq)+H_2O(l)+CO_2(g)}$ The gas turns limewater milky or cloudy.	Two hydrochloric acid formula units provide two chlorides for $2\mathrm{NaCl}$ and two hydrogens for water. The equation has two sodium, one carbon, three oxygen, two hydrogen and two chlorine atoms on each side; limewater then confirms $\mathrm{CO_2}$ .	4
Total Question 2			4
03.1	The orange-red flame identifies calcium ions, Ca²⁺. The acid and limewater results identify carbonate ions, CO₃²⁻. The compound is calcium carbonate, CaCO₃.	Combine independent tests on the two ions. Calcium gives the orange-red flame, while carbonate releases carbon dioxide that clouds limewater. Equal and opposite ion charges give the formula $\mathrm{CaCO_3}$ .	4
Total Question 3			4
04.1	The claim is incorrect because the cloudy limewater confirms carbon dioxide from the acid reaction. Producing carbon dioxide with dilute acid is a positive test for carbonate ions in the tablet. The carbonate component can react completely while the insoluble binder remains. The remaining solid shows that the tablet is a mixture; it does not cancel the positive carbonate result.	Follow the component that reacts and test its gaseous product. The confirmed carbon dioxide supplies the evidence for carbonate ions. A different, insoluble component can remain because the tablet is a formulation rather than a single pure substance.	4
Total Question 4			4
05.1	Add dilute acid to the first portion of the powder. Pass any gas produced through fresh limewater. Fizzing with a gas that turns limewater cloudy is positive evidence for carbonate ions. Dissolve a separate portion in water, acidify it with dilute nitric acid and allow any fizzing to stop. Add silver nitrate solution to the acidified second portion. A white precipitate is positive evidence for chloride ions.	Use separate portions so the chloride test does not depend on material already consumed in the carbonate test. Confirm carbonate by generating carbon dioxide and testing that gas. For chloride, remove interfering carbonate with dilute nitric acid before adding silver nitrate and looking for white silver chloride.	6
Total Question 5			6

4.8.3.4 · Halides (chemistry only)

Tier 1 · Easy

Mark scheme for 4.8.3.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Iodide ions, I⁻	Match the silver-halide colour to the required table: yellow silver iodide indicates iodide ions.	1
Total Question 1			1
02.1	Bromide ions, Br⁻.	A cream silver-halide precipitate identifies bromide ions.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.8.3.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Hydrochloric acid introduces chloride ions, which can form a white silver chloride precipitate even if the original sample contained no chloride. Use dilute nitric acid.	The acid must not add the ion being tested. Nitrate ions do not precipitate with silver ions, so nitric acid removes interferences without creating a chloride result.	3
Total Question 1			3
02.1	Use dilute nitric acid; the ion is bromide, Br⁻, and the precipitate is silver bromide, AgBr.	The specified acid is dilute nitric acid. Cream corresponds to silver bromide, so the original solution contains bromide ions.	4
Total Question 2			4
03.1	Acidifying first removes interfering ions such as carbonate ions that could otherwise form a precipitate with silver ions. Nitric acid is used because nitrate ions do not form a precipitate with silver ions.	Preserve the reagent order: remove possible interferences before adding silver ions. The acid must not introduce a halide, so dilute nitric acid is suitable.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.8.3.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	R is chloride, S is bromide and T is iodide. Dilute nitric acid, followed by silver nitrate solution.	Use the colour sequence: white identifies chloride, cream identifies bromide and yellow identifies iodide. The specified procedure is to acidify first with dilute nitric acid and then add silver nitrate solution.	5
Total Question 1			5
02.1	The sample result is unreliable because the blank should give no precipitate; the white solid shows chloride contamination in a reagent or the apparatus. Repeat the blank and sample with clean apparatus and uncontaminated dilute nitric acid and silver nitrate, accepting the sample only when the blank stays clear.	A positive blank reveals that the procedure itself introduces halide ions. Contamination could alter the sample's observed colour, so clean the setup and verify a negative blank before identifying the unknown.	4
Total Question 2			4
03.1	The green flame identifies copper(II) ions, Cu²⁺. The cream precipitate identifies bromide ions, Br⁻. The compound is copper(II) bromide, CuBr₂.	Combine the cation and anion tests. One Cu²⁺ ion requires two Br⁻ ions for an overall neutral compound, giving $\mathrm{CuBr_2}$ .	5
Total Question 3			5
04.1	The direct test is unreliable because carbonate ions can also form a precipitate with silver ions. The second portion was correctly acidified to remove interfering carbonate ions. Dilute nitric acid is the correct acid because it does not introduce halide ions. A cream precipitate in the correctly acidified portion identifies bromide ions. The valid evidence is therefore the second result, not the untreated portion's pale yellow solid.	Decide which portion followed the complete halide procedure. The unacidified result can include silver carbonate and cannot be interpreted securely. Nitric acid removes carbonate interference without adding a halide, so the subsequent cream silver bromide precipitate supports bromide.	5
Total Question 4			5
05.1	A cream colour supports the presence of bromide ions. Silver bromide is cream, whereas silver chloride is white. A mixture of white silver chloride and cream silver bromide could appear very pale cream. The observation therefore does not prove that chloride ions are absent or that bromide is the only halide.	Treat the precipitate colour as evidence of what is present, not proof that every other ion is absent. Bromide can account for cream, but white silver chloride could be hidden within a mixed precipitate, so the second part of the student's report overstates the test.	4
Total Question 5			4

4.8.3.5 · Sulfates (chemistry only)

Tier 1 · Easy

Mark scheme for 4.8.3.5 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Sulfate ions, SO₄²⁻	A white precipitate with barium chloride in acidified solution is the specified positive test for sulfate ions.	1
Total Question 1			1
02.1	Add dilute hydrochloric acid followed by barium chloride solution; a white precipitate is positive.	Acidify the solution first, then add the barium reagent. Formation of the white insoluble solid is the specified observation.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.8.3.5 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Carbonate ions can form a white barium carbonate precipitate, which could be mistaken for barium sulfate. Dilute hydrochloric acid removes the carbonate interference before barium chloride is added.	A white barium precipitate is not unique to sulfate in an untreated sample. Acid reacts with carbonate ions first, so a later white precipitate after adding barium chloride is valid evidence for sulfate.	3
Total Question 1			3
02.1	The solution contains sulfate ions, SO₄²⁻; the precipitate is barium sulfate, BaSO₄.	Because the test was performed under the specified acidic conditions, the white barium sulfate precipitate is evidence for sulfate ions.	3
Total Question 2			3
03.1	The plan can give a false positive because sulfuric acid introduces sulfate ions, which form a white precipitate with barium ions. Use dilute hydrochloric acid before adding barium chloride solution.	The acid used in a test must not add the ion being investigated. Acidify with dilute hydrochloric acid, then add the barium reagent.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.8.3.5 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The procedure is incomplete because the sample was not first acidified. Add dilute hydrochloric acid, then barium chloride solution; a white precipitate is the positive sulfate result.	Do not accept a result obtained under incomplete conditions. Repeat the test by acidifying the unknown with dilute hydrochloric acid before adding barium chloride. Only then use formation of a white precipitate as the specified positive result.	4
Total Question 1			4
02.1	$\mathrm{Na_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+2NaCl(aq)}$ Insoluble barium sulfate forms as a white precipitate.	The equation has two sodium, one sulfate group, one barium and two chlorine atoms on each side. The sulfate product has state $(s)$ because it is insoluble and appears as the white solid.	4
Total Question 2			4
03.1	For sulfate ions, acidify one portion with dilute hydrochloric acid, then add barium chloride solution; a white precipitate is positive. For chloride ions, acidify a fresh portion with dilute nitric acid, then add silver nitrate solution; a white precipitate is positive.	Keep the two tests independent. Perform each acidification before adding its precipitating reagent, and use a fresh portion for the chloride test so that its white silver chloride result cannot come from acid added during the sulfate test.	6
Total Question 3			6
04.1	The gas turning limewater cloudy identifies carbon dioxide. Carbon dioxide released by dilute acid supports carbonate ions in the original solution. Waiting until fizzing stops lets the acid remove the interfering carbonate as carbon dioxide before the barium reagent is used. A white precipitate with barium chloride in the acidified liquid is positive evidence for sulfate ions. The original solution is therefore supported to contain both carbonate and sulfate ions.	Interpret the stages in order. The first stage detects and removes carbonate as carbon dioxide. Once that reaction has finished, carbonate should no longer create a misleading barium precipitate, so the later white solid can be used as evidence for sulfate ions.	5
Total Question 4			5
05.1	The first unknown result is invalid because the blank also formed a precipitate. The original barium chloride or apparatus may have been contaminated with sulfate ions. The clear new blank shows that the fresh reagent and apparatus do not give that false positive. The repeated unknown result is a valid white-precipitate result after acidification. The repeated evidence therefore supports sulfate ions in the unknown.	Use the blank to decide whether a precipitate is caused by the unknown. The failed first blank makes that pair of results unusable. Fresh reagent restores a clear blank, so the precipitate produced only by the new unknown portion supplies reliable sulfate evidence.	5
Total Question 5			5

4.8.3.6 · Instrumental methods (chemistry only)

Tier 1 · Easy

Mark scheme for 4.8.3.6 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	It can be more accurate, more sensitive or more rapid.	Select one of the three specification advantages: accurate, sensitive or rapid. One precise comparison earns the mark.	1
Total Question 1			1
02.1	High sensitivity (or that instrumental methods are sensitive).	Sensitivity is the ability to detect a very small amount or concentration of a substance.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.8.3.6 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The reading close to $5.00\,\mathrm{mg\,dm^{-3}}$ demonstrates accuracy; detecting $0.001\,\mathrm{mg\,dm^{-3}}$ demonstrates sensitivity.	Accuracy is closeness to the accepted value. Sensitivity is the ability to detect a very small amount or concentration.	2
Total Question 1			2
02.1	An instrumental method can be rapid, allowing many samples to be tested in the available time. It can be sensitive, allowing a very small pollutant concentration to be detected. Accept accuracy as one advantage if linked to reliable results.	Link each specification advantage to the context. Speed addresses the large sample number, while sensitivity addresses the trace concentration.	3
Total Question 2			3
03.1	The claim is not supported because instrumental methods are not necessarily cheaper or easier. Valid advantages include being more accurate, more sensitive and more rapid; any two are required.	Judge the claim using the stated comparison rather than assuming a cost benefit. Replace the unsupported properties with two of the specified advantages: accuracy, sensitivity and speed.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.8.3.6 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Method B is more accurate, more rapid and more sensitive than method A.	Method B is closer to the certified value: its error is $\|10.1-10.0\|=0.1\,\mathrm{mg\,dm^{-3}}$ , compared with $\|12-10.0\|=2\,\mathrm{mg\,dm^{-3}}$ for A. Forty seconds rather than $25$ minutes shows greater speed, and detecting the dilute samples shows greater sensitivity.	4
Total Question 1			4
02.1	The claim is incorrect: Y is more accurate because $8.01\,\mathrm{mg\,dm^{-3}}$ is closer to the certified $8.00\,\mathrm{mg\,dm^{-3}}$ than X's $7.842\,\mathrm{mg\,dm^{-3}}$ . X is more sensitive because its detection limit is smaller; the number of displayed decimal places alone does not establish accuracy.	Compare absolute errors: X differs by $0.158\,\mathrm{mg\,dm^{-3}}$ , whereas Y differs by $0.01\,\mathrm{mg\,dm^{-3}}$ . Then compare the minimum detectable concentrations separately to judge sensitivity.	4
Total Question 2			4
03.1	Instrumental time $=120\times15=1800$ seconds $=30$ minutes; chemical-test time $=120\times4=480$ minutes; time saved $=450$ minutes. The instrumental method is more rapid and more sensitive because it takes less time and has the lower detection limit.	Convert both totals to minutes before subtracting: $1800/60=30$ and $480-30=450$ . Then link the shorter analysis time to speed and the smaller detectable concentration to sensitivity.	5
Total Question 3			5
04.1	The rapid batch processing demonstrates the speed advantage of the instrumental method. Detecting the trace control demonstrates high sensitivity. Consistently reporting the certified standard too high shows a systematic error and poor accuracy. Repeatability does not make a consistently biased result accurate. The instrument should be recalibrated and the standard rechecked before sample results are reported.	Evaluate each claimed advantage separately. The instrument is rapid and sensitive, but the certified standard exposes an accuracy fault. Repeating the same biased value only shows consistency, so calibration must be corrected before the sample measurements can be trusted.	5
Total Question 4			5
05.1	The analyte has not been shown to be present because the blank gives the same signal. The signal may come from contamination, an incorrect zero or an instrument fault. Clean the apparatus or replace contaminated materials and obtain a blank with no analyte signal. Check or recalibrate the instrument using suitable standards. Retest a fresh portion of the unknown and compare its result with the valid blank and standards.	A blank establishes the background response of the method. When the unknown is indistinguishable from that background, the result cannot identify the analyte. Remove the source of the blank signal, verify calibration, and only then repeat the sample measurement.	5
Total Question 5			5

4.8.3.7 · Flame emission spectroscopy (chemistry only)

Tier 1 · Easy

Mark scheme for 4.8.3.7 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Lithium ions, Li⁺	Compare the sample line position with the reference set. The $671\,\mathrm{nm}$ match identifies lithium ions.	1
Total Question 1			1
02.1	They can identify the metal ions present in the solution.	Compare the line positions with a reference spectrum. Matching positions identify the corresponding metal ion.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.8.3.7 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Both lithium and sodium ions are present because identification uses line positions. Their concentrations cannot be compared from brightness alone without suitable calibration data for each ion.	Use wavelength or line position to identify an ion. Intensity is used for concentration only by comparison with the relevant calibration, so raw brightness from different ions is not a concentration comparison.	3
Total Question 1			3
02.1	Calcium and copper ions are present because the sample's line positions match both reference line sets.	Match positions rather than line brightness. Both calcium reference positions and both copper reference positions occur in the sample, supporting both identifications.	3
Total Question 2			3
03.1	Put the solution sample into a flame and pass the emitted light through a spectroscope. The output is a line spectrum; compare its line positions with a reference set in the same form to identify the metal ions.	Follow the signal from sample to conclusion: the flame excites the sample, the spectroscope separates the emitted light into lines, and matching line positions with reference data gives the identities.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.8.3.7 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Sodium and potassium ions are present. Potassium-ion concentration $=4.0\,\mathrm{mg\,dm^{-3}}$ .	Match the two line positions to sodium and potassium. The sample intensity $31$ is $16$ above $15$ , exactly half of the $32$ -unit rise from $15$ to $47$ . Move halfway from $2.0$ to $6.0\,\mathrm{mg\,dm^{-3}}$ : $2.0+0.5(4.0)=4.0\,\mathrm{mg\,dm^{-3}}$ .	5
Total Question 1			5
02.1	Original concentration $=25\,\mathrm{mg\,dm^{-3}}$ .	An intensity of $25$ is halfway between $20$ and $30$ , so the diluted solution is halfway between $4.0$ and $6.0\,\mathrm{mg\,dm^{-3}}$ : $5.0\,\mathrm{mg\,dm^{-3}}$ . The final volume is five times the transferred volume, so dilution reduced concentration by a factor of $5$ . Therefore the original concentration was $5\times5.0=25\,\mathrm{mg\,dm^{-3}}$ .	5
Total Question 2			5
03.1	The identification is not valid because the sample and reference line positions are not given in the same form and cannot be matched without a scale. A reference spectrum or table in the same form and on the same position scale as the sample is needed.	Line positions identify ions only through a direct comparison with appropriate reference data. Centimetres on an unscaled printout cannot be matched numerically with wavelengths in nanometres.	3
Total Question 3			3
04.1	For lithium, the sample-to-standard intensity factor is $30/12=2.5$ . The lithium-ion concentration is $2.5\times2.0=5.0\,\mathrm{mg\,dm^{-3}}$ . For sodium, the sample-to-standard intensity factor is $24/20=1.2$ . The sodium-ion concentration is $1.2\times5.0=6.0\,\mathrm{mg\,dm^{-3}}$ . Sodium is more concentrated even though its line is less intense, because the two ions have different calibrations.	Use wavelength to identify each ion, then apply its own calibration rather than comparing the two raw intensities. With a linear response through the origin, multiply each standard concentration by the corresponding intensity ratio. The calculated concentrations, not line brightness across different ions, determine the comparison.	5
Total Question 4			5
05.1	Blank-corrected standard intensities are $16-4=12$ and $40-4=36$ units. The calibration response is $12/2.0=36/6.0=6$ intensity units per $\mathrm{mg\,dm^{-3}}$ . The sample's blank-corrected intensity is $31-4=27$ units. The calcium-ion concentration is $27/6=4.5\,\mathrm{mg\,dm^{-3}}$ .	Subtract the blank response from every reading before using the calibration. Both corrected standards give the same response per concentration unit, confirming linearity. Divide the corrected sample intensity by that response to obtain the concentration.	4
Total Question 5			4