4.8 Chemical analysis — coverage pack

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

4.8.1.1 · Pure substances

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

Tier 1 · Easy

1. A sealed sample contains only solid sulfur. Decide whether it is a pure substance in the chemical sense and justify your decision.[2 marks]
Answer
- It is pure because it contains a single element and no other substance.
Method: 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.

Tier 2 · Standard

1. 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.[3 marks]
Answer
- Batch A; it melts at the accepted temperature and has a sharp melting point, whereas batch B melts over a different range.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.8.1.2 · Formulations

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

Tier 1 · Easy

1. 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]
Answer
- It is a useful mixture whose components are present in measured quantities for particular purposes.
Method: 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.

Tier 2 · Standard

1. 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.[4 marks]
Answer
- Mass of filler $=175\,\mathrm{g}$ ; accurate measurement gives the product its required composition or properties.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.8.1.3 · Chromatography

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.

Tier 1 · Easy

1. 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]
Answer
- $R_f=0.60$
Method: Use $R_f=\frac{\text{spot distance}}{\text{solvent distance}}=\frac{4.2}{7.0}=0.60$ . The ratio has no unit.

Tier 2 · Standard

1. 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]
Answer
- 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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.8.2.1 · Test for hydrogen

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

Tier 1 · Easy

1. A colourless gas makes a pop when tested at the mouth of its tube with a lit splint. Identify the gas.[1 mark]
Answer
- Hydrogen
Method: Recall the diagnostic observation: rapid burning with a pop is the positive test for hydrogen.

Tier 2 · Standard

1. Give the procedure and positive observation for confirming that gas collected from a metal-acid reaction is hydrogen.[2 marks]
Answer
- Place a burning splint at the tube opening; hydrogen burns with a pop.
Method: Keep the flame at the open end of the container rather than pushing it deep inside. A pop sound is the required positive observation.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.8.2.2 · Test for oxygen

Oxygen supports combustion and is identified when a glowing splint relights.
Insert a glowing wooden splint into the test tube containing the gas.
For example, a splint with no flame that bursts back into flame gives a positive oxygen result.
A common error is to use a burning splint and look for a pop, which tests for hydrogen.

Tier 1 · Easy

1. A glowing splint returns to flame inside a jar of colourless gas. Name the gas.[1 mark]
Answer
- Oxygen
Method: The relighting of a glowing splint is the characteristic positive observation for oxygen.

Tier 2 · Standard

1. Describe a test that distinguishes oxygen from a gas that does not support combustion. State the positive result.[2 marks]
Answer
- Insert a glowing splint; it relights in oxygen.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.8.2.3 · Test for carbon dioxide

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

Tier 1 · Easy

1. An unknown gas is shaken with clear limewater, which becomes cloudy. Identify the gas.[1 mark]
Answer
- Carbon dioxide
Method: Cloudiness in limewater is the positive observation used to identify carbon dioxide.

Tier 2 · Standard

1. State the reagent and the expected visible change when testing a gas sample for carbon dioxide.[2 marks]
Answer
- Use limewater, aqueous calcium hydroxide; it changes from clear to milky or cloudy.
Method: Pass the gas through limewater or shake the two together. Name both the reagent and the formation of cloudiness for a complete answer.

Tier 3 · Hard

1. 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]
Answer
- Bubble or shake the gas with limewater; the limewater turns milky or cloudy.
Method: 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.

4.8.2.4 · Test for chlorine

Chlorine gas bleaches damp litmus paper white.
Expose damp litmus paper to the gas while using appropriate small-scale safety precautions because chlorine is toxic.
For example, litmus losing all its colour and becoming white is the positive chlorine result.
A common error is to use dry litmus paper; moisture is required for the specified test.

Tier 1 · Easy

1. Damp litmus paper loses its colour and becomes white in an unknown gas. Identify the gas.[1 mark]
Answer
- Chlorine
Method: Bleaching damp litmus paper to white is the diagnostic observation for chlorine gas.

Tier 2 · Standard

1. Describe how litmus paper is used to test for chlorine and state the observation that confirms a positive result.[2 marks]
Answer
- Use damp litmus paper; chlorine bleaches it white.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- Use the damp-paper result because moisture is needed; bleaching to white shows that the gas is chlorine.
Method: 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.

4.8.3.1 · Flame tests (chemistry only)

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

Tier 1 · Easy

1. A compound produces a crimson flame. Identify the metal ion present.[1 mark]
Answer
- Lithium ions, Li+
Method: Match the observed crimson colour to the specified flame-test table: lithium compounds give crimson.

Tier 2 · Standard

1. Samples M and N give a green flame and an orange-red flame respectively. Identify the metal ion in each sample.[2 marks]
Answer
- M contains copper(II) ions, Cu2+.
- N contains calcium ions, Ca2+.
Method: Use the required colour associations. Green corresponds to copper compounds, and orange-red corresponds to calcium compounds.

Tier 3 · Hard

1. 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]
Answer
- Sodium ions are supported; the strong sodium-yellow colour may mask the flame colour of the other ion.
Method: 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.

4.8.3.2 · Metal hydroxides (chemistry only)

With sodium hydroxide, Cu2+ gives a blue precipitate, Fe2+ green and Fe3+ brown.
Add sodium hydroxide solution dropwise, record any precipitate, then add excess when distinguishing aluminium from other white precipitates.
Al3+, Ca2+ and Mg2+ form white hydroxide precipitates, but only aluminium hydroxide dissolves in excess sodium hydroxide.
A common error is to claim every white precipitate dissolves in excess; calcium and magnesium hydroxides remain.

Tier 1 · Easy

1. Adding sodium hydroxide solution to an unknown ionic solution forms a brown precipitate. Identify the metal ion.[1 mark]
Answer
- Iron(III) ions, Fe3+
Method: The required observation table assigns a brown hydroxide precipitate to iron(III) ions.

Tier 2 · Standard

1. 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?[4 marks]
Answer
- A contains aluminium ions; B could contain calcium or magnesium ions, so the observations do not distinguish those two.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- W contains copper(II) ions, Cu2+.
- X contains iron(II) ions, Fe2+.
- Y contains aluminium ions, Al3+.
Method: 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.

4.8.3.3 · Carbonates (chemistry only)

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

Tier 1 · Easy

1. A solid fizzes when dilute acid is added, and the gas makes limewater cloudy. Name the ion detected in the solid.[1 mark]
Answer
- Carbonate ions, CO32−
Method: Dilute acid releases carbon dioxide from a carbonate. The cloudy limewater confirms the gas, so the original solid contains carbonate ions.

Tier 2 · Standard

1. Describe a complete chemical test for carbonate ions in an unknown powder, including how the gaseous product is identified.[3 marks]
Answer
- Add dilute acid, then pass the gas into limewater; effervescence occurs and the limewater turns milky or cloudy.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.8.3.4 · Halides (chemistry only)

Halide ions are tested by acidifying the solution with dilute nitric acid and then adding silver nitrate solution.
Use clean samples and record the precipitate colour: chloride white, bromide cream and iodide yellow.
For example, a cream silver-halide precipitate identifies bromide ions under the specified test conditions.
A common error is to swap the cream and yellow results: silver bromide is cream, while silver iodide is yellow.

Tier 1 · Easy

1. After dilute nitric acid and silver nitrate are added to a solution, a yellow precipitate forms. Identify the halide ion.[1 mark]
Answer
- Iodide ions, I−
Method: Match the silver-halide colour to the required table: yellow silver iodide indicates iodide ions.

Tier 2 · Standard

1. Give the reagents, in order, used to test an aqueous sample for halide ions, and state the result for chloride ions.[3 marks]
Answer
- Add dilute nitric acid, then silver nitrate solution; chloride ions give a white precipitate.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- R is chloride, S is bromide and T is iodide.
- Dilute nitric acid, followed by silver nitrate solution.
Method: 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.

4.8.3.5 · Sulfates (chemistry only)

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

Tier 1 · Easy

1. An acidified solution forms a white precipitate when barium chloride solution is added. Identify the ion being tested.[1 mark]
Answer
- Sulfate ions, SO42−
Method: A white precipitate with barium chloride in acidified solution is the specified positive test for sulfate ions.

Tier 2 · Standard

1. Describe the reagent sequence and positive observation for testing an unknown solution for sulfate ions.[3 marks]
Answer
- Add dilute hydrochloric acid, then barium chloride solution; a white precipitate forms if sulfate ions are present.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.8.3.6 · Instrumental methods (chemistry only)

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

Tier 1 · Easy

1. State one advantage of an instrumental method over a chemical test based on a visible colour change.[1 mark]
Answer
- It can be more accurate, more sensitive or more rapid.
Method: Select one of the three specification advantages: accurate, sensitive or rapid. One precise comparison earns the mark.

Tier 2 · Standard

1. 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.[2 marks]
Answer
- It is rapid enough for many samples and sensitive enough to detect a low concentration.
Method: Link each circumstance to an advantage. The large number of samples requires speed, while the very small amount of contaminant requires sensitivity.

Tier 3 · Hard

1. 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]
Answer
- Method B is more accurate, more rapid and more sensitive than method A.
Method: 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.

4.8.3.7 · Flame emission spectroscopy (chemistry only)

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

Tier 1 · Easy

1. 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]
Answer
- Lithium ions, Li+
Method: Compare the sample line position with the reference set. The $671\,\mathrm{nm}$ match identifies lithium ions.

Tier 2 · Standard

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

Tier 3 · Hard

1. 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]
Answer
- Sodium and potassium ions are present.
- Potassium-ion concentration $=4.0\,\mathrm{mg\,dm^{-3}}$ .
Method: 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 $6.0\,\mathrm{mg\,dm^{-3}}$ : $2.0+0.5(4.0)=4.0\,\mathrm{mg\,dm^{-3}}$ .