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Identifying an inorganic unknown (Core Practical 15)

Edexcel IAL Chemistry · WCH16 Practical Skills in Chemistry II

Core Practical 15 asks you to identify unknown compounds from test results. In the exam you complete observation and inference tables and then put the ions together into a formula.

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A plan for an unknown salt

1. Look: a coloured solid or solution suggests a transition metal ion; crystals may be hydrated.

2. Cations: flame test; NaOH(aq) dropwise then excess (and warm, testing for ammonia with damp red litmus); NH3(aq) dropwise then excess.

3. Anions: dilute acid for carbonate; HCl then BaCl2 for sulfate; HNO3 then AgNO3 then ammonia for halides.

4. Extra clues: what happens on standing, with H2O2, or when two solutions are mixed (a brown colour with iodide means I- has been oxidised to I2).

5. Put it together: write a neutral formula by balancing charges, and add water of crystallisation if the solid is hydrated.

Writing observations and inferences

Double salts

Some unknowns contain two cations and one anion, such as ammonium iron(II) sulfate, (NH4)2Fe(SO4)2·6H2O (pale green crystals). Tests would show NH4+ (ammonia on warming with NaOH), Fe2+ (green precipitate turning brown) and SO42- (white precipitate with HCl and BaCl2). Charges: 2(+1) + (+2) = +4, balanced by 2SO42-.

Worked example

Green crystals X: warming with NaOH(aq) gives a gas that turns damp red litmus blue and a green precipitate that turns brown on standing. A solution of X with HCl and BaCl2 gives a white precipitate. Suggest a formula for X.

  1. Ammonia gas: NH4+.
  2. Green precipitate turning brown: Fe2+.
  3. White precipitate with BaCl2 in acid: SO42-.
  4. Balance charges: 2NH4+ + Fe2+ = +4, so 2SO42-. Crystals: hydrated.

Answer: (NH4)2Fe(SO4)2·6H2O, ammonium iron(II) sulfate

Key idea: Look, test cations, test anions, use extra clues, then balance charges. Observations are what you see; inferences name ions. Double salts have two cations.

Check yourself

  1. Which formula is correct for iron(III) sulfate?
    • FeSO4
    • Fe3(SO4)2
    • Fe(SO4)3
    • Fe2(SO4)3
    Show answer

    Fe2(SO4)3. 2 x (+3) = +6 balances 3 x (-2) = -6.

  2. Calculate the molar mass of ammonium iron(II) sulfate, (NH4)2Fe(SO4)2·6H2O. (H 1.0, N 14.0, O 16.0, S 32.1, Fe 55.8)
    Show answer

    392.0 g mol-1. 2(18.0) + 55.8 + 2(96.1) + 6(18.0) = 392.0 g mol-1.

  3. What is the percentage by mass of water in (NH4)2Fe(SO4)2·6H2O?
    Show answer

    27.6 %. 6 x 18.0 / 392.0 x 100 = 27.6%.

  4. When solution P is added to potassium iodide solution, the mixture turns brown. What does this show about P?
    • It contains an oxidising agent that oxidises iodide ions to iodine
    • It contains a reducing agent
    • It contains silver ions
    • It is a strong acid
    Show answer

    It contains an oxidising agent that oxidises iodide ions to iodine. Brown is iodine, formed by oxidising I-. The October 2024 report credited the redox point.

  5. Which is the best way to record the result when barium chloride is added to an acidified solution of a sulfate?
    • Barium sulfate
    • Sulfate present
    • White precipitate
    • White solid appears to be BaSO4
    Show answer

    White precipitate. The observation column needs what is seen.

  6. A solution gives a yellow precipitate with acidified silver nitrate, which does not dissolve in concentrated ammonia. Which ion is present?
    • I-
    • Br-
    • Cl-
    • SO42-
    Show answer

    I-. AgI is insoluble in concentrated ammonia.

  7. Warming a green solid with NaOH(aq) gives ammonia, and the solid's solution gives a green precipitate that turns brown in air. What does the solid contain?
    • Only Fe3+ ions
    • Cu2+ and NH4+ ions
    • NH4+ and Fe2+ ions
    • Only Ni2+ ions
    Show answer

    NH4+ and Fe2+ ions. Ammonia shows NH4+; green turning brown shows Fe2+.

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Last updated 8 November 2026. Spotted a mistake? Tell us.