AQA A-Level Chemistry Paper 3, June 2022: Question 2

6 marks · Hard difficulty · Long Answer

Calculate the empirical formula and value of x for the hydrated mineral tschermigite, and describe chemical tests with ionic equations to confirm the presence of ammonium, aluminium, and sulfate ions.

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Question

Question 02 introduces tschermigite as a hydrated, water-soluble mineral with a relative formula mass of 453.2 and formula M.xH2O. Table 4 provides elemental composition by mass: N 3.09%, H 6.18%, Al 5.96%, S 14.16%, and O 70.61%. The mineral contains NH4+, Al3+, and SO4(2-) ions. Students are asked to calculate the empirical formula and the value of x, and describe chemical tests with results and ionic equations to confirm the identities of these ions for 6 marks.
Question text

02 Tschermigite is a hydrated, water-soluble mineral, with relative formula mass of 453.2

The formula of tschermigite can be represented as M.xH2O, where M represents all

the ions present.

Table 4 shows its composition by mass.

Table 4

Element % by mass

N 3.09

H 6.18

Al 5.96

S 14.16

O 70.61

In an analysis, it is found that the mineral contains the ions NH +, Al3+ and SO 2−

Calculate the empirical formula of tschermigite and the value of x in M.xH2O

Describe the tests, with their results, including ionic equations, that would confirm the

identities of the ions present.

[6 marks]

Mark scheme

Show the mark scheme Level of response mark scheme for Question 02 outlining 3 stages of chemistry content: Stage 1 involves calculating moles of each element to deduce empirical formula NH28AlS2O20 and finding x = 12. Stage 2 describes qualitative tests: NaOH and gentle heat testing for NH4+ (gas turns red litmus blue), acidified BaCl2 testing for SO4(2-) (white precipitate), and NaOH testing for Al3+ (white precipitate soluble in excess). Stage 3 gives balanced ionic equations for each test.

Question Answers Additional Comments/Guidelines Mark

This question is marked using levels of response. Refer to the Mark Indicative Chemistry content

Scheme Instructions for examiners for guidance on how to mark it.

Stage 1 Formula

Level 3 All stages are covered and the explanation of each (1a) divides % masses by Ar for each element

stage is correct and virtually complete (N = 0.221; H = 6.18; Al = 0.221; S = 0.441; O =4.41)

5-6 marks (1b) divides throughout by smallest and confirms

Answer communicates the whole explanation, formula as NH28AlS2O20

including equations, coherently and shows a Correct formula ticks 1a and 1b irrespective of method

logical progression through all three stages (1c) x = 12

Level 2 All stages are covered but the explanation of each

stage may be incomplete or may contain Stage 2 Ion ID

(2a) addition of NaOH/OH– and warming gives gas

3-4 marks inaccuracies

OR two stages covered and the explanations are that turns (damp) red litmus blue (= ammonia) 6

showing NH + (water bath = warm)

generally correct and virtually complete 4

(2 x AO2,

(2b) white ppt with acidified BaCl /Ba2+ = SO 2− 4 x AO3)

Answer is coherent and shows some progression 2 4

02 through all three stages. Some steps in each stage

(2c) addition of NaOH/OH– until in excess gives

may be incomplete 3+

white ppt that redissolves = Al

OR addition of carbonate giving white ppt and

Level 1 Two stages are covered but the explanation of effervescence/fizzing/bubbles/gas formed

each stage may be incomplete or may contain

1-2 marks inaccuracies Stage 3 Equations (Ignore state symbols)

OR only one stage is covered but the explanation (3a) NH + + OH− → NH + H O

43 2

is generally correct and virtually complete (3b) Ba2+ + SO 2− → BaSO

Answer shows some progression between two (3c) Al(H O) 3+ + 3 OH− → Al(H O) (OH) + 3 H O

26 2 3 3 2

stages Allow Al3+ + 3 OH− → Al(OH)

Level 0 Insufficient correct Chemistry to warrant a mark (3d) Al(H O) (OH) + OH− → Al(H O) (OH) − + H O

23 3 2 2 4 2

Allow Al(OH) + OH− → Al(OH) − etc.

0 marks OR 2Al(H2O)63+ + 3CO32– → 2Al(H2O)3(OH)3+ 3CO2 + 3H2O

Equation with CO 2– ‘ticks’ 3c AND 3d

How to answer it

Empirical Formula & Qualitative Inorganic Analysis of Tschermigite

📌 What this question tests

This 6-mark extended-response question assesses three core areas across physical and inorganic chemistry:

  • Empirical Formula & Water of Crystallisation: Calculating an empirical formula from percentage composition by mass and deducing the integer value of x in a hydrated salt formula M.xH₂O.
  • Inorganic Qualitative Analysis: Naming reagents, conditions, and observations for identifying ammonium ions (NH₄⁺), sulfate ions (SO₄²⁻), and aluminium ions (Al³⁺).
  • Ionic Equations: Writing precise ionic equations including precipitation, acid-base proton transfer, and amphoteric behaviour with excess hydroxide.
Stage 1

Empirical Formula & Value of x

Deducing NH₂₈AlS₂O₂₀ and determining x = 12

📐 Step-by-Step Calculation

  1. Find moles of each element (% mass ÷ Ar):
    • N: 3.09 ÷ 14.0 = 0.221 mol
    • H: 6.18 ÷ 1.0 = 6.18 mol
    • Al: 5.96 ÷ 27.0 = 0.221 mol
    • S: 14.16 ÷ 32.1 = 0.441 mol
    • O: 70.61 ÷ 16.0 = 4.41 mol
  2. Divide by the smallest value (0.221):
    • N = 1
    • H = 6.18 ÷ 0.221 = 27.96 ≈ 28
    • Al = 1
    • S = 0.441 ÷ 0.221 = 1.995 ≈ 2
    • O = 4.41 ÷ 0.221 = 19.95 ≈ 20
    Empirical Formula = NH₂₈AlS₂O₂₀
  3. Find x in M.xH₂O:
    The ions present are NH₄⁺, Al³⁺, and two SO₄²⁻ (overall neutral salt: NH₄Al(SO₄)₂).
    Anhydrous salt formula: 1 N, 4 H, 1 Al, 2 S, 8 O.
    Remaining atoms from empirical formula:
    H: 28 − 4 = 24 H atoms (which gives 24 ÷ 2 = 12 H₂O)
    O: 20 − 8 = 12 O atoms (which gives 12 H₂O)
    Alternative check via Mr:
    Mr[NH₄Al(SO₄)₂] = 14.0 + 4.0 + 27.0 + (2 × 32.1) + (8 × 16.0) = 237.2
    Mass of water = 453.2 − 237.2 = 216.0
    x = 216.0 ÷ 18.0 = 12

❌ Common Errors & Pitfalls

  • Premature Rounding: Rounding moles to 1 or 2 sig figs early on (e.g., 6.18 ÷ 0.22 instead of 0.221) leads to erroneous non-integer ratios for H.
  • Using Molecular Masses Instead of Atomic: Dividing H by 2.0 (for H₂) or N by 28.0 (for N₂) instead of their atomic masses (1.0 and 14.0).
  • Ignoring the Cation/Anion Balance: Forgetting to account for the 4 hydrogens already in NH₄⁺ when calculating x, mistakenly writing x = 14 (28 ÷ 2).
Marking breakdown: Stage 1 is fully achieved when moles are correctly evaluated, the empirical formula NH₂₈AlS₂O₂₀ is deduced, and x = 12 is stated.
Stage 2

Qualitative Tests & Observations

Identifying NH₄⁺, SO₄²⁻, and Al³⁺ ions

✅ Correct Reagents and Results

Ion Test / Reagent & Conditions Expected Observation
Ammonium (NH₄⁺) Add dilute aqueous NaOH (or OH⁻) and warm gently. Pungent gas produced (ammonia) that turns damp red litmus paper blue.
Sulfate (SO₄²⁻) Add acidified barium chloride solution (BaCl₂ acidified with HCl or HNO₃). A white precipitate forms (BaSO₄).
Aluminium (Al³⁺) Option A: Add dilute aqueous NaOH dropwise, then to excess.

OR Option B: Add sodium carbonate solution (Na₂CO₃). 		Option A: White precipitate forms initially, which redissolves in excess NaOH to give a colourless solution.

Option B: White precipitate and effervescence / bubbling.

🧠 Exam Technique: Securing Level 3

  • Always state warming for NH₄⁺: Adding NaOH alone is not sufficient; ammonia is soluble in water, so gentle heating is required to release the gas.
  • Litmus paper must be damp: NH₃ gas dissolves in water on the paper to produce OH⁻ ions which turn red litmus blue.
  • Specify "excess": Stating "add NaOH and get a white ppt" only proves the presence of a cation like Mg²⁺. Showing it redissolves in excess proves the amphoteric nature characteristic of Al³⁺.
Stage 3

Balanced Ionic Equations

Representing each analytical reaction

💡 Key Chemical Equations

1. Ammonium test:

NH₄⁺ + OH⁻ → NH₃ + H₂O

2. Sulfate test:

Ba²⁺ + SO₄²⁻ → BaSO₄

3. Aluminium test (using NaOH):

  • Formation of precipitate:
    [Al(H₂O)₆]³⁺ + 3OH⁻ → Al(H₂O)₃(OH)₃ + 3H₂O
    (Acceptable simplified form: Al³⁺ + 3OH⁻ → Al(OH)₃ )
  • Redissolving in excess:
    Al(H₂O)₃(OH)₃ + OH⁻ → [Al(H₂O)₂(OH)₄]⁻ + H₂O
    (Acceptable simplified form: Al(OH)₃ + OH⁻ → [Al(OH)₄]⁻ )

Alternative 3. Aluminium test (using Na₂CO₃):

2[Al(H₂O)₆]³⁺ + 3CO₃²⁻ → 2Al(H₂O)₃(OH)₃ + 3CO₂ + 3H₂O

Note: Providing this single carbonate equation earns credit for both precipitate formation and excess behaviour!

❌ Common Equation Blunders

  • Full molecular equations instead of ionic: Writing BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl when the question explicitly asked for ionic equations. Spectator ions must be cancelled out.
  • Unbalanced charge: Writing Al(OH)₃ + OH⁻ → Al(OH)₄ without the negative charge on the aluminate ion.
  • Forgetting the second equation for Al³⁺: Many students write the precipitation equation but omit the reaction where the precipitate dissolves in excess NaOH.
Levels of Response Summary:
  • Level 3 (5–6 marks): All three stages (Formula, Ion ID tests, Ionic Equations) are covered correctly and virtually completely with logical progression.
  • Level 2 (3–4 marks): Two stages fully correct OR all three stages covered with minor inaccuracies/omissions.
  • Level 1 (1–2 marks): Only one stage completely correct OR two stages partially attempted.

Topics

Physical Chemistry · Inorganic Chemistry · Required Practicals · 3.1.2 Amount of Substance · 3.2.6 Reactions of Ions in Aqueous Solution · Required Practical 4: Carry out simple test-tube reactions to identify Cations and Anions

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.