Writing the qualitative analysis report
The structure that maps onto the criteria
Write your report in these sections and every criterion has somewhere to live.
- Purpose and context — the sample, the question, and why it matters.
- Procedure — the identification procedure you were given, and any modifications.
- Results — a table of tests, reagents and observations.
- Identification — which species, and the reasoning from data to conclusion.
- Equations — for every change, including redissolutions.
- Justification — the secondary data and chemical principles supporting each identification.
- Significance — of at least one identified species, in your context.
The results table
- One row per test, not per conclusion. A test that had two stages gets two rows.
- Columns: test number, reagent added, observation.
- Record negative results. "No precipitate formed" eliminates possibilities and is evidence.
- Use precise colour words: white, cream, pale yellow, pale blue, green, red-brown, deep blue. Not "a coloured precipitate".
Turning observations into an identification
- Move from data to conclusion explicitly, in writing.
- Weak: "Test 3 showed Cu2+."
- Strong: "In test 3, adding NH3 in excess dissolved the pale blue precipitate to give a deep blue solution. The complex ion table identifies this as , so the cation is Cu2+."
- Say what each result eliminates as well as what it indicates. An identification is as much about ruling things out as ruling them in.
Equations for all the changes
The most commonly incomplete Merit requirement. For a single cation test with copper you need three:
- Write ionic equations with state symbols — the is what shows a precipitate formed.
- For organic compounds, use structural formulae, not molecular ones.
Linking species to the compound
- Achieved requires you to link the chemical species to the compound present in the sample.
- Combine your cation and anion with the correct charge balance:
- Cu2+ and SO42− → CuSO4
- Mg2+ and Br− → MgBr2
- Al3+ and Cl− → AlCl3
- State the compound's name and formula, and check the charges cancel.
Justifying with secondary data
- Name the secondary data you used and quote what it says.
- "The solubility rules state that all hydroxides are insoluble except those of Group 1 metals, so adding OH− to any solution containing a transition metal ion must give a precipitate."
- "The table of complex ions gives and , both soluble, which accounts for the precipitate dissolving in both excess reagents."
- Then connect it to your data. Secondary data listed but not used earns nothing.
Worked ExampleA complete identification write-up
A student is given a white solid, dissolves it in water to give a colourless solution, and follows the identification procedure. Their results are below. Write the identification section of the report.
| Test | Reagent added | Observation |
|---|---|---|
| 1 | dilute HNO3 | no effervescence |
| 2 | NaOH(aq) dropwise | white precipitate |
| 3 | NaOH(aq) in excess | precipitate dissolved, colourless solution |
| 4 | NH3(aq) in excess (fresh portion) | white precipitate, dissolved in excess |
| 5 | acidified with HNO3, then AgNO3 | white precipitate |
| 6 | dilute NH3 added to the precipitate from test 5 | precipitate dissolved |
Identifying the cation
Test 1 produced no effervescence with dilute acid, so no carbonate is present. This also means the white precipitates in later tests cannot be carbonates.
Test 2 gave a white precipitate with hydroxide. The solubility rules state that hydroxides are insoluble except those of Group 1, so a metal hydroxide has formed. The white colour eliminates Cu2+ (pale blue), Fe2+ (green) and Fe3+ (red-brown), leaving Al3+, Zn2+, Mg2+ and Ca2+.
Test 3 showed the precipitate dissolving in excess sodium hydroxide, so the hydroxide is amphoteric. Only Al(OH)3 and Zn(OH)2 are amphoteric, so Mg2+ and Ca2+ are eliminated.
Test 4 showed the precipitate also dissolving in excess ammonia. Zinc forms the soluble tetraamminezinc(II) complex; aluminium does not form an ammine complex and would have remained as a white solid. Al3+ is eliminated.
The cation is Zn2+, identified by the combination of an amphoteric white hydroxide that also forms an ammine complex — a pattern unique to zinc in this scheme.
Identifying the anion
Test 5, on a portion acidified with nitric acid (so that no chloride was introduced by the acid), gave a white precipitate with silver nitrate. The silver halides are white (Cl), cream (Br) and yellow (I), so this indicates chloride, but cream and white can be confused.
Test 6 resolved this: the precipitate dissolved in dilute ammonia. AgCl is the only silver halide that dissolves in dilute ammonia — AgBr requires concentrated ammonia and AgI does not dissolve at all.
The anion is Cl−.
The compound
Zn2+ carries a 2+ charge and Cl− a 1− charge, so two chloride ions are needed per zinc ion for the formula to be neutral.
The compound is zinc chloride, ZnCl2.
Equations for all the changes
Precipitation of the hydroxide (tests 2 and 4):
Ammonia acting as a weak base to supply that hydroxide in test 4:
Dissolution in excess sodium hydroxide (test 3):
Dissolution in excess ammonia (test 4):
Precipitation of the silver halide (test 5):
Dissolution in dilute ammonia (test 6):
Answer: the sample is zinc chloride, ZnCl2 — Zn2+ identified by an amphoteric white hydroxide that dissolves in both excess NaOH and excess NH3, and Cl− identified by a white silver precipitate soluble in dilute ammonia.