Qualitative Analysis of Inorganic and Organic Unknowns
Practical Skills: Qualitative Analysis identifies unknown organic liquids and inorganic solids from a fixed menu of test-tube reactions.
These are bromine water, warm acidified dichromate(VI), Fehling’s solution, hydrolysis then silver nitrate, and the acid, barium chloride, limewater and chlorine water tests. Every observation is written in the words an examiner credits.
Test Bench: Identify the Unknowns
Watch three colourless liquids and three white solids meet their test-tube reagents, then follow every observation to an inference and an identity.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
GCSE Recap: Testing for Ions
Four quick questions on the GCSE tests this practical builds on: flame colours, the carbonate test with limewater, the sulfate test and the silver nitrate colours.
What This Practical Is Testing
The evidence trail for the six unknowns on this page: every test, every observation and the inference each one supports.
Practical Skills: Qualitative Analysis asks you to identify several unknown substances, typically colourless organic liquids (here A, B and C) and white inorganic solids (here X, Y and Z), from a fixed menu of test-tube reactions.
Nothing here needs new theory: the tests come from the Group 17 chemistry, the tests for alkenes, alcohols, aldehydes and halogenoalkanes, and the ion tests you first met at GCSE.
What is new is the discipline: a small portion of the unknown for each test, the correct reagent in the correct order, and the right conditions (warming when the test needs it).
The last part of that discipline is an observation written in words an examiner will credit before any inference is made.
The logic is an evidence trail: unknown → test → observation → inference → identity. Negative results are part of the trail.
A liquid that turns acidified dichromate(VI) green but leaves Fehling’s solution blue is a primary or secondary alcohol; the negative Fehling’s result is what rules out an aldehyde, which dichromate(VI) would also have oxidised.
A solid that fizzes with acid is not shown to be a carbonate until the gas has turned limewater cloudy.
Students lose marks by jumping from one positive test to an identity and by describing a solution with no colour as “clear”.
Paper 3, a timed laboratory examination that assesses manipulation, observation, presentation of data, analysis and evaluation.
For this practical that means naming reagents and conditions precisely (warm in a water bath, dilute nitric acid before silver nitrate), giving observations in full (colour before and after, precipitate colour, gas test), and writing ionic equations with state symbols.
It also means explaining why a step is there: why the acid, why the warming, why the solid is dissolved first, why a second reagent is needed to confirm.
Key idea: An observation is what you see (orange to colourless, cream precipitate, effervescence). An inference is what it means (C=C present, bromide ions, carbonate ions). Write them separately and never skip the observation.
Organic Unknowns A, B and C
Bromine water added to A, B and C: only B decolourises it, from orange to colourless, which is the evidence for a C=C bond.
After warming in a water bath, only tube A has changed from orange to green: Cr₂O₇²⁻(aq) reduced to Cr³⁺(aq) as A is oxidised.
Fehling’s solution warmed with A, B and C: all three remain blue with no precipitate, so none of them is an aldehyde.
Hydrolysis, acidification with dilute nitric acid and then silver nitrate: only C gives a cream precipitate of silver bromide.
Four tests separate the organic liquids: bromine water for a C=C bond, warm acidified dichromate(VI) for anything oxidisable, warm Fehling’s solution (Fehling’s solution or Tollens’ reagent are accepted) for an aldehyde, and hydrolysis followed by silver nitrate for a halogenoalkane.
The table gives the sample observations and the inference each one allows. Notice that the inference column uses the negative results as much as the positive ones.
Three details carry marks. The dichromate(VI) and Fehling’s tubes are warmed in a water bath at about 60 °C; at room temperature both stay unchanged and the negative result is worthless.
Bromine water is decolourised, from orange to colourless, never “goes clear”.
The dichromate(VI) result alone does not identify A: dichromate(VI) would also oxidise an aldehyde, which is exactly why the negative Fehling’s result must be quoted in the inference.
Exam focus: Final identities: A is a primary or secondary alcohol (never “an oxidisable alcohol”), B is an alkene, C is a bromoalkane. Each identity must quote the test that supports it and the negative test that rules out the alternative.
Check: Three New Liquids
Propanal, propanone and but-1-ene meet the same three reagents. Predict every observation and decide which liquid is which.
Testing for an Aldehyde and the Rest of the Organic Menu
A positive Fehling’s result for reference: brick-red copper(I) oxide settles under the pale blue liquid after warming with an aldehyde. Not part of the sample data.
The sample set gives no positive Fehling’s result, so you must still know what one looks like.
On warming in a water bath an aldehyde reduces the deep blue copper(II) complex in Fehling’s solution to a brick-red precipitate of copper(I) oxide, Cu₂O.
Cu²⁺ is reduced to Cu⁺ while the aldehyde is oxidised to a carboxylic acid: RCHO + 2Cu²⁺ + 4OH⁻ → RCOOH + Cu₂O + 2H₂O.
A ketone has no hydrogen on its carbonyl carbon, so it cannot be oxidised under these conditions and the solution stays blue.
Tollens’ reagent (ammoniacal silver nitrate) is the accepted alternative: warmed with an aldehyde it gives a silver mirror on the inside of the tube as Ag⁺ is reduced to Ag.
The full organic test menu is longer than the four tests used on A, B and C, and any of it can appear in a question.
A carbonyl group (aldehyde or ketone) gives an orange precipitate with 2,4-dinitrophenylhydrazine (2,4-DNPH); Fehling’s or Tollens’ then decides between the two.
A carboxylic acid effervesces with sodium carbonate or sodium hydrogencarbonate solution, and the gas turns limewater cloudy; alcohols and phenols do not.
A tertiary alcohol is the one that leaves warm acidified dichromate(VI) orange but is still an alcohol, so it is identified by elimination once the other tests are negative.
The oxidation chemistry behind the dichromate(VI) test is the subject of Practical Skills: Distillation and Reflux (oxidation of an alcohol).
Two more organic tests from the notes
The tri-iodomethane test picks out one structural feature: a methyl group next to a carbonyl group, CH₃CO–, or next to a carbon carrying an OH group, CH₃CH(OH)–.
Warm the compound gently with iodine solution and sodium hydroxide solution: a pale yellow precipitate of CHI₃ with an antiseptic smell is a positive result, CH₃COR + 3I₂ + 4OH⁻ → RCOO⁻ + CHI₃(s) + 3I⁻ + 3H₂O(l).
Ethanol, ethanal, propan-2-ol and propanone give it; propan-1-ol, propanal and methanol do not, because the alkaline iodine first oxidises CH₃CH(OH)– to CH₃CO–.
Acidified potassium manganate(VII) turns from purple to colourless with any compound it can oxidise, including primary and secondary alcohols, aldehydes and alkenes, so it says only “can be oxidised”. Follow it with a more selective test to decide which group is present.
Key idea: Dichromate(VI) says “oxidisable”; Fehling’s or Tollens’ says “aldehyde”. You need both results to name a primary alcohol with confidence.
Check: Tri-iodomethane and Manganate(VII) Tests
Apply the alkaline iodine and acidified manganate(VII) tests to compounds that are not on this page.
Check: Aldehyde or Not?
Two questions on using a positive and a negative result together for liquids that are not on this page.
Testing C for the Halogen
The five steps for testing an organic liquid for a halogen: hydrolyse with NaOH(aq) and ethanol in a water bath, cool, acidify with dilute nitric acid, add silver nitrate, confirm with dilute then concentrated ammonia.
The halogenoalkane test has five steps and every one of them is examined. First, warm a few drops of C with sodium hydroxide solution and a little ethanol in a water bath.
The C–Br bond is covalent, so there are no bromide ions in the liquid until it has been hydrolysed: CH₃CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂OH + Br⁻ (1-bromopropane is used as the example).
Ethanol is the co-solvent that lets the organic liquid mix with the aqueous alkali, and it is flammable, so the tube is warmed in a water bath and never over a flame.
Second, cool the tube and acidify with dilute nitric acid. The alkali was added in excess, and Ag⁺ reacts with OH⁻ to give a brown precipitate of silver oxide, Ag₂O, which would hide the halide result:
2Ag⁺(aq) + 2OH⁻(aq) → Ag₂O(s) + H₂O(l).
Nitric acid is chosen because hydrochloric acid would add chloride ions and give a white precipitate of AgCl whatever the unknown, and sulfuric acid would add sulfate ions.
Third, add silver nitrate solution: Ag⁺(aq) + Br⁻(aq) → AgBr(s), a cream precipitate.
Fourth and fifth, add dilute ammonia solution (the cream precipitate stays) and then concentrated ammonia solution (it dissolves): AgBr(s) + 2NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Br⁻(aq).
Exam focus: Without the ammonia step an examiner can only accept “AgBr or AgI”, because cream and pale yellow are hard to tell apart in a small tube.
The same silver nitrate chemistry is used to compare the rates of hydrolysis of chloro-, bromo- and iodoalkanes in Practical Skills: Rates of Hydrolysis (hydrolysis of halogenoalkanes), where the time taken for the precipitate to appear is the measurement.
Here the precipitate is only the evidence for which halogen is present.
Exam trap: Adding silver nitrate straight to the organic liquid gives no precipitate and proves nothing: the halogen is covalently bonded. Hydrolyse, acidify with nitric acid, then test.
Inorganic Unknowns X, Y and Z
The silver halide test: acidify with dilute nitric acid, add silver nitrate, then use dilute and concentrated ammonia to tell white AgCl, cream AgBr and yellow AgI apart.
The sulfate test: acidify with dilute hydrochloric acid, then add barium chloride solution. A white precipitate of BaSO₄ shows sulfate; a carbonate fizzes with the acid instead.
The carbonate test done properly: acid added to the solid in a boiling tube, bung fitted at once, and the gas bubbled through limewater, which turns cloudy.
The anion tests on X, Y and Z side by side, with the orange bromine in the upper cyclohexane layer of the chlorine water tube for X.
The three white solids are tested for their anions in solution, so the first step is to dissolve a spatula tip of each in a little distilled water.
A dry solid cannot react with silver nitrate or barium chloride solution, and tap water would add chloride ions.
On this board the cation is identified separately, by the scheme in the cation card below; in this sample set X is a calcium salt, Y a sodium salt and Z a potassium salt.
The table gives the anion evidence with the inference beside each observation.
Identities: X is calcium bromide, CaBr₂; Y is sodium sulfate, Na₂SO₄; Z is potassium carbonate, K₂CO₃.
Each anion has one primary test and one confirmation: silver nitrate then ammonia for the halide, acid then barium chloride for the sulfate, acid then limewater for the carbonate.
The chlorine water result for X is a second, independent line of evidence for bromide, which is what makes the identification secure.
The acid before each precipitation test is not decoration. Carbonate ions give a white precipitate with silver ions (Ag₂CO₃) and with barium ions (BaCO₃), so a carbonate would give a false positive in both tests.
Dilute acid converts the carbonate to carbon dioxide and removes the problem.
The acid must add nothing that is being tested for: nitric acid before silver nitrate (hydrochloric acid would add Cl⁻), hydrochloric or nitric acid before barium chloride (sulfuric acid would add SO₄²⁻).
Exam wording: “Add dilute hydrochloric acid, then barium chloride solution: a white precipitate shows sulfate ions. The acid removes carbonate ions, which would also give a white precipitate.” Reagent, observation, inference and the reason for the acid.
Check: A New Solid, W
Follow the evidence trail for a solid that is not on this page: a yellow silver precipitate, a purple cyclohexane layer and a white precipitate with dilute sulfuric acid.
Identifying the Cation: Sodium Hydroxide and Ammonia Solutions
The ammonium ion test: warm the solid with sodium hydroxide solution and test the gas with damp red litmus paper, which turns blue.
On this board the metal ion is identified from the precipitate formed with sodium hydroxide solution and with ammonia solution, first with a few drops and then in excess.
The observations must be written in the standard wording: the colour of the precipitate, then whether it is “soluble in excess” or “insoluble in excess”, then the colour of any solution formed.
The table below is the one you need to know; the same table is printed in the practical paper, but you will answer faster and more accurately if you understand it.
Three white precipitates cause most of the lost marks. Zinc hydroxide dissolves in excess of both reagents; aluminium hydroxide dissolves in excess sodium hydroxide only; magnesium hydroxide dissolves in neither.
Calcium gives a white precipitate with sodium hydroxide unless its concentration is very low, barium gives a faint one unless its concentration is very low, and neither gives a precipitate with ammonia.
They are identified by the sulfate test on the anion side (barium sulfate is insoluble, calcium sulfate slightly soluble) or by a flame colour where the paper allows it.
For iron(II), the change to brown on standing is a mark point: Fe(OH)₂ is oxidised by air to Fe(OH)₃.
Two coloured hydroxides complete the table. Chromium(III) gives a grey-green precipitate that, like aluminium hydroxide, is amphoteric and dissolves in excess sodium hydroxide to give a dark green solution, but not in excess ammonia.
Cr³⁺(aq) + 3OH⁻(aq) → Cr(OH)₃(s), then Cr(OH)₃(s) + 3OH⁻(aq) → [Cr(OH)₆]³⁻(aq).
Manganese(II) gives an off-white precipitate of Mn(OH)₂ with either reagent, insoluble in excess, which turns brown rapidly as oxygen in the air oxidises it.
Iron(II) behaves in the same way but starts green, so the colour before it browns is what tells the two apart.
Equations with state symbols: Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s); Zn(OH)₂(s) + 2OH⁻(aq) → [Zn(OH)₄]²⁻(aq); Cu(OH)₂(s) + 4NH₃(aq) → [Cu(NH₃)₄]²⁺(aq) + 2OH⁻(aq).
Gas tests are also marked: carbon dioxide turns limewater cloudy, ammonia turns damp red litmus blue, chlorine bleaches damp litmus, and sulfur dioxide turns acidified potassium dichromate(VI) paper from orange to green.
The ammonium ion
Ammonium salts look like any other white solid, so the ammonium test is part of every scheme.
Warm a little of the solid (or its solution) with sodium hydroxide solution and hold a piece of damp red litmus paper at the mouth of the tube without letting it touch the liquid.
Ammonia gas is given off and the paper turns blue: NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l).
Students lose marks by writing “litmus turns blue” without saying the paper was damp and red, and by forgetting that the alkali itself would turn the paper blue if it splashed on it.
Exam wording: Always three parts: “white ppt, soluble in excess, colourless solution”. A precipitate that dissolves has not “disappeared”: it has formed a soluble complex.
Chlorine Water: a Stronger Oxidising Agent
Chlorine water added to chloride, bromide and iodide solutions: no change, orange bromine, brown iodine (purple when shaken with cyclohexane).
Chlorine water confirms bromide and iodide ions because chlorine is a stronger oxidising agent than bromine or iodine.
Reactivity falls down Group 17: the atoms get larger, the outer shell is further from the nucleus and more shielded, so the attraction for an extra electron weakens.
Chlorine therefore takes electrons from bromide and iodide ions, oxidising them to the element: 2Br⁻(aq) + Cl₂(aq) → Br₂(aq) + 2Cl⁻(aq) and 2I⁻(aq) + Cl₂(aq) → I₂(aq) + 2Cl⁻(aq).
Exam tip: Say “chlorine oxidises bromide ions to bromine”, never “chlorine displaces bromide”: what is displaced is the element bromine, and the mechanism is a transfer of electrons.
The colours in water are orange for bromine and brown for iodine, and dilute solutions of the two can look alike.
Shaking with a few drops of cyclohexane (or hexane) settles it: the halogen dissolves in the non-polar solvent, which is less dense than water and forms the upper layer, orange for bromine and purple (violet) for iodine, above a colourless aqueous layer.
Chloride ions give no change, because chlorine cannot oxidise chloride. With X the orange upper layer agrees with the cream AgBr precipitate: two independent tests, one inference.
Key idea: Oxidising power falls down Group 17. Chlorine oxidises Br⁻ and I⁻; bromine oxidises I⁻ only; iodine oxidises neither. The cyclohexane layer sits on top and shows the halogen colour clearly.
Check: Chlorine Water and Iodide Ions
Complete the ionic equation for chlorine water and potassium iodide solution, give the colours in water and in cyclohexane, and name what is oxidised.
Safety and Apparatus
Everything is done on a test-tube scale: a few drops of the liquid or a spatula tip of the solid.
Small quantities are a safety measure and a practical one, because a tube crowded with solid gives a cloudy mixture in which a precipitate cannot be seen.
Hazards and precautions: Bromine water and chlorine water: toxic and corrosive, use small volumes in a fume cupboard or a well-ventilated room, wear gloves.
Acidified potassium dichromate(VI): toxic and a suspected carcinogen, wear gloves, use small volumes and pour residues into the labelled waste bottle.
Sodium hydroxide solution: corrosive, eye protection throughout. Ethanol: flammable, so the hydrolysis tube is warmed in a water bath, never over a flame.
Silver nitrate: stains skin and clothing. Concentrated hydrochloric acid: corrosive, use a few drops only. Fehling’s solution is alkaline and contains copper: gloves and eye protection.
Method: Step by Step
The method below follows the sample set on this page: three colourless liquids A, B and C and three white solids X, Y and Z. Each test uses a fresh portion of the unknown, and every “why” is a mark-scheme point.
Exam wording: “Warm with acidified potassium dichromate(VI) in a water bath: orange to green.” “Add dilute nitric acid, then silver nitrate solution: cream precipitate, insoluble in dilute ammonia, soluble in concentrated ammonia.” Reagent, condition, observation.
More Anion Tests, Gas Tests and the Iodine Test
The qualitative analysis notes printed in the practical paper list more anions than the three in the sample set, and each has its own test.
Nitrate and nitrite are found by reduction to ammonia: warm the solution with sodium hydroxide solution and a small piece of aluminium foil, and test the gas with damp red litmus paper.
Because the test detects ammonia, warm the solution with sodium hydroxide alone first: if ammonia is given off before the foil is added, ammonium ions are present and must be driven off by boiling before the nitrate test means anything.
3NO₃⁻(aq) + 8Al(s) + 5OH⁻(aq) + 18H₂O(l) → 3NH₃(g) + 8[Al(OH)₄]⁻(aq)
Nitrate and nitrite give the same aluminium foil result, so a second test separates them.
Nitrite is a reducing agent: it is oxidised to nitrate and turns acidified potassium manganate(VII) from purple to colourless, 2MnO₄⁻(aq) + 5NO₂⁻(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 5NO₃⁻(aq) + 3H₂O(l).
Nitrate cannot be oxidised further, so it leaves the purple colour.
Sulfate and sulfite both give a white precipitate with barium ions, and the acid decides between them. Barium sulfate is insoluble in excess dilute strong acid; barium sulfite dissolves, BaSO₃(s) + 2H⁺(aq) → Ba²⁺(aq) + SO₂(g) + H₂O(l).
Sulfite is also a reducing agent and decolourises acidified manganate(VII), 2MnO₄⁻(aq) + 5SO₃²⁻(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 5SO₄²⁻(aq) + 3H₂O(l).
Calcium sulfate is only slightly soluble, which is why a solution with a high concentration of calcium ions also gives a white precipitate with sulfate.
Thiosulfate is recognised by the slow appearance of an off-white or pale yellow precipitate of sulfur when acid is added, S₂O₃²⁻(aq) + 2H⁺(aq) → S(s) + SO₂(g) + H₂O(l), the reaction timed in Practical Skills: Rates of Reaction.
Tests for gases and for iodine
| Gas or element | Test and test result |
|---|---|
| ammonia, NH₃ | turns damp red litmus paper blue |
| carbon dioxide, CO₂ | gives a white ppt with limewater |
| hydrogen, H₂ | “pops” with a lighted splint |
| oxygen, O₂ | relights a glowing splint |
| iodine, I₂ | gives blue-black colour on addition of starch solution |
Effervescence on its own is only an observation: the notes expect you to identify the gas with its test.
Hydrogen burns with a squeaky pop because it mixes with air in the tube, 2H₂(g) + O₂(g) → 2H₂O(l), and is given off when a reactive metal such as magnesium or zinc is added to dilute acid.
Oxygen relights a glowing splint because it supports combustion; it comes from decomposing hydrogen peroxide with a catalyst.
Starch is a very sensitive test for iodine, so a blue-black colour after an oxidising agent is added to iodide ions shows that iodine has been formed.
Exam wording: Nitrate and nitrite: “ammonia given off on warming with NaOH(aq) and Al foil turns damp red litmus paper blue”. Only nitrite, and sulfite, decolourise acidified KMnO₄(aq), because they are reducing agents.
Check: Ions and Gases From the Notes
Use the qualitative analysis notes to identify ions and gases in new unknowns.
Ionic Equations and Reagent Roles
Every precipitation and gas equation in this practical is expected as an ionic equation with state symbols; an equation without them loses the mark. The full set for the sample unknowns is below, together with the equations for the confirmatory and organic steps.
Three reagent roles are asked about again and again. Dilute nitric acid before silver nitrate has two jobs.
In the inorganic test it removes carbonate ions, which would give a white precipitate of Ag₂CO₃ and mask the halide. In the organic test it neutralises the excess sodium hydroxide from the hydrolysis, which would otherwise give brown Ag₂O.
Why nitric acid and not another acid: hydrochloric acid adds chloride ions (false white AgCl) and sulfuric acid adds sulfate ions (false white BaSO₄ in the sulfate test).
Dilute hydrochloric acid before barium chloride removes carbonate ions, which would give white BaCO₃; sulfuric acid can never be used here.
The ammonia steps are also reagent roles. Dilute ammonia dissolves AgCl by forming the soluble complex [Ag(NH₃)₂]⁺; AgBr is less soluble and needs concentrated ammonia; AgI is too insoluble to dissolve in either.
So the ammonia result is not a second test for the same thing, it is what separates the three halides once a precipitate has been seen.
Exam trap: Two different reasons for the same acid: nitric acid removes carbonate ions in the inorganic test and neutralises excess hydroxide ions in the organic test. Give the reason that fits the question.
Check: Diagnosing Test Results
Explain four results from halogenoalkane tests on compounds that are not on this page.
Exam-Style Data Practice
Two kinds of question follow this practical: an unknown-solution matrix, in which four labelled solutions are mixed pairwise and identified from the pattern, and a composition-and-spectra question that leads from percentage composition to a structure. Both are answered by the evidence-trail habit.
Unknown solution matrix
Four solutions are sodium carbonate, barium chloride, dilute hydrochloric acid and dilute sulfuric acid.
The trap is that barium chloride gives a white precipitate with sodium carbonate as well as with sulfuric acid: Ba²⁺(aq) + CO₃²⁻(aq) → BaCO₃(s). A student who forgets this misassigns C and D.
| Letter | Identity | Evidence from the pairwise mixing |
|---|---|---|
| A | sulfuric acid, H₂SO₄ | white precipitate with C (BaSO₄) and effervescence with D; no other change |
| B | hydrochloric acid, HCl | effervescence with D only; no precipitate with anything |
| C | barium chloride, BaCl₂ | white precipitate with A (BaSO₄, insoluble in acid) and with D (BaCO₃, which dissolves with effervescence when A or B is added) |
| D | sodium carbonate, Na₂CO₃ | effervescence with A and B; white precipitate with C |
The distinguishing observations are that A gives both a precipitate and a gas, B gives only a gas, C gives two precipitates and D gives two lots of gas and one precipitate.
Adding acid to the C + D precipitate dissolves it with effervescence, which tells BaCO₃ from BaSO₄.
Composition and spectra
An organic compound contains 38.7% carbon, 9.7% hydrogen and 51.6% oxygen by mass (the three values must add to 100%).
Its mass spectrum has a molecular ion peak at m/z 62 and a base peak at m/z 31; its infrared spectrum has a broad absorption at about 3400 cm⁻¹ and no absorption near 1700 cm⁻¹.
Two errors are common enough to name. Rounding the ratio to CH₂O gives C₂H₄O₂, which is ethanoic acid, a compound that would show a strong C=O absorption and no m/z 31 fragment; the data rule it out.
CHO⁺ has m/z 29, not 31: check every fragment by adding the atomic masses.
Key idea: Composition gives the empirical formula; the molecular ion gives Mr and so the molecular formula; the infrared spectrum names the functional groups; the fragments place them. Quote each piece of evidence for its own conclusion.
Check: Formula From Composition and Spectra
A different compound, Q: 52.2% carbon, 13.0% hydrogen, 34.8% oxygen, M⁺ at m/z 46, broad absorption at 3300 cm⁻¹ and a base peak at m/z 31. Work through the same steps.
Errors, Uncertainty and Improvements
A qualitative practical has no numerical result to compare with an accepted value, so the evaluation questions ask about false positives, false negatives and the steps that prevent them. The table lists the errors students actually make and what each does to the conclusion.
Where a quantity is measured, its percentage uncertainty still follows the usual rule.
If a method asks for “about 0.5 g” of solid and a 2 d.p. balance reads 0.50 g, the uncertainty is ±0.005 g, so the percentage uncertainty is (0.005) / (0.50) × 100 = 1.0%.
Measuring 5 cm³ of water with a 10 cm³ measuring cylinder (±0.5 cm³) gives (0.5) / (5) × 100 = 10%.
Neither matters here, because a precipitate forms or does not form whatever the exact mass or volume, and that is the answer to “why is ‘approximately’ acceptable in this method”: the result is qualitative, so the measurements need only be roughly right.
The composition data in the exam question are different: 38.7% quoted to 1 decimal place has an uncertainty of ±0.05%, about 0.1% of the value.
That is why the mole ratio comes out as 1.00 : 3.00 : 1.00 and not as something that needs rounding.
Exam wording: “Warming the tube in a water bath makes a negative result reliable.” “Dilute nitric acid is used because hydrochloric acid would add chloride ions and give a false positive.” State the error, its effect and the fix.
Common Mistakes
- Writing “clear” for a solution with no colour. Bromine water is decolourised; a solution with no colour is colourless; “clear” only means not cloudy.
- Not warming the dichromate(VI) and Fehling’s tubes, then reporting “no reaction”. A negative result at room temperature proves nothing.
- Calling A “an oxidisable alcohol”. The inference is “a primary or secondary alcohol”; the negative Fehling’s result is what rules out an aldehyde.
- Adding silver nitrate straight to the organic liquid. The halogen is covalently bonded; hydrolyse first, then acidify with nitric acid, then test.
- Testing a solid with silver nitrate or barium chloride solution without dissolving it in distilled water first.
- Giving “fizzes with acid” as proof of a carbonate. The gas must turn limewater cloudy.
- Stopping at “cream precipitate”. Without the dilute and concentrated ammonia results, AgBr has not been distinguished from AgI.
- Writing “chlorine displaces bromide ions”. Chlorine oxidises bromide ions to bromine; the element bromine is what is displaced.
- Ionic equations without state symbols, and minus signs missing from Br⁻ and SO₄²⁻.
- Rounding the mole ratio carelessly (CH₂O instead of CH₃O) and quoting a fragment whose mass does not add up (CHO⁺ is 29, CH₂OH⁺ is 31).
Check: Technique Check
Six sets of statements about the technique of this practical. Pick the accurate one in each set.
Common Exam Points
Say
“Warm with acidified potassium dichromate(VI) in a water bath: orange to green, so the liquid is oxidised.”
“Fehling’s solution remains blue on warming, so it is not an aldehyde; A is a primary or secondary alcohol.”
“Dissolve the solid in distilled water, add dilute nitric acid, then silver nitrate solution: cream precipitate, insoluble in dilute ammonia, soluble in concentrated ammonia, so bromide ions.”
“Effervescence; the gas turns limewater cloudy, so carbonate ions.”
“Chlorine oxidises bromide ions to bromine, which gives an orange upper layer in cyclohexane.”
Do not say
“Goes clear.” “Positive result.” “Oxidisable alcohol.” “Chlorine displaces bromide.” “Add acid” without saying which acid and why. “Fizzes, so carbonate” without the limewater step. Equations without state symbols. “No reaction” when a soluble product has formed.
Watch for
The order of the tests and the reason for it (acid before silver nitrate or barium chloride; carbonate removed first).
Which acid: nitric before silver nitrate, hydrochloric or nitric before barium chloride, never sulfuric. Water bath, not a naked flame, for every warming step.
Percentages in a composition question must add to 100%, and every fragment mass must add up.
FAQs
Quick answers to the questions students ask most about Practical Skills: Qualitative Analysis and the tests for unknown organic liquids and inorganic solids.
Why must the acidified dichromate(VI) and Fehling’s tubes be warmed?
Both oxidations are slow at room temperature.
Warming in a water bath at about 60 °C for a few minutes lets the reaction happen, so a tube that stays orange or blue after warming is a genuine negative result.
Without warming a negative result proves nothing, and the water bath keeps any naked flame away from flammable organic liquids.
Why is nitric acid, and not hydrochloric acid, added before silver nitrate?
Hydrochloric acid would add chloride ions and give a white precipitate of silver chloride in every tube, a false positive. Nitric acid adds only nitrate ions, which form no insoluble silver salt.
The acid removes carbonate ions in the inorganic test and neutralises excess sodium hydroxide after hydrolysing the organic liquid, which would otherwise give brown silver oxide.
How do I tell silver bromide from silver iodide?
Cream and pale yellow are easy to confuse in a small tube, so use ammonia: silver chloride dissolves in dilute ammonia, silver bromide dissolves only in concentrated ammonia, and silver iodide dissolves in neither. Quote the ammonia result as part of the inference.
Does fizzing with acid prove that a solid is a carbonate?
No. Effervescence suggests a carbonate (or hydrogencarbonate) but the gas must be identified: bubble it through limewater, which turns cloudy as a white precipitate of calcium carbonate forms. The observation and its confirmation are separate mark points.
What is the difference between “clear” and “colourless”?
Colourless means the solution has no colour; clear means it is not cloudy. Bromine water that has reacted with an alkene is decolourised, orange to colourless.
A blue Fehling’s solution is clear but not colourless. Examiners penalise “goes clear” because it does not describe a colour change.
Why shake the chlorine water mixture with cyclohexane?
Bromine and iodine dissolve much better in the non-polar solvent than in water, and cyclohexane floats as the upper layer, so the colour is concentrated and easy to read: orange for bromine, purple for iodine. Dilute aqueous solutions of the two can look alike.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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