Edexcel A-Level Chemistry AS Paper 1, June 2018: Question 4

8 marks · Medium difficulty · Practical Techniques and Data Analysis

Identify an unknown hydrated ionic compound from chemical tests and flame tests, and calculate its water of crystallisation from thermal decomposition mass loss data.

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Question

Question text

4 An ionic compound contains a metal cation and a non-metal anion in a 1:1 ratio, and

water of crystallisation. The compound can be represented as MN.xH2O, where x is

the number of moles of water of crystallisation per mole of MN.

A sample of MN.xH2O was dissolved in distilled water to produce a colourless

solution, with a concentration of about 0.5 mol dm−3. 2 cm3 of the resulting solution

was transferred to each of two test tubes.

The following tests were carried out to identify the ions present.

(a) Test 1

(i) Addition of a few drops of a solution of barium chloride to one of the test

tubes gave a white precipitate.

Identify, by name or formula, two possible anions that would give this result.

(1)

(ii) Addition of 1 cm3 of dilute hydrochloric acid to the test tube in (a)(i) resulted

in no further change.

Give the formula of the anion.

(1)

(iii) What is the charge on the cation?

(1)

A +1

B −1

C +2

D −2

(b) Test 2

A flame test on a sample of solid MN.xH2O gave no change in the flame colour.

Give a possible identity of the cation, M.

(1)

solid MN by the following reaction:

MN.xH2O(s) → MN(s) + xH2O(g)

Heating a sample of the hydrated compound reduced the mass to 48.9% of its

original value.

Use this information and your answer to (a)(ii) and (b) to calculate the value of x.

[ Note: If you have been unable to identify*P51459A01024*MN,you may use this hydrated compound, ]

CoCl2.yH2O in which the sample reduced in mass to 54.6% of its original value.

Use this information to calculate the value of y.

(4)

Mark scheme

Show the mark scheme

Question

Acceptable Answer Additional Guidance Mark

Number

4(a)(i) An answer that makes reference to two of the Penalise lack of charge (1)

following:

sulfate / sulfate(VI) / SO4

sulfite / sulfate(IV) / SO3

carbonate / CO3

Question

Acceptable Answer Additional Guidance Mark

Number

4(a)(ii) 2- Ignore sulfate (ion) (1)

SO4

Only penalise lack of charge if not penalised in

4(a)(i)

Question

Acceptable Answer Mark

Number

4(a)(iii) The only correct answer is C (1)

A is not correct because the ratio is one-to-one

B is not correct because cations are positive

D is not correct because cations are positive

Question

Acceptable Answer Additional Guidance Mark

Number

4(b) Cation is Mg2+ / magnesium (ion) Do not award use of symbol just “Mg” (1)

Award Be2+/ beryllium (ion)

Question

Acceptable Answer Additional Guidance Mark

Number

4(c) In all cases correct answer with some correct (4)

working scores (4)

Method 1 Example of calculation:

Calculates Mr of MgSO4 (1) Mr of MgSO4 = 24.3 + 32.1 + (4 x 16) =

120.4

MgSO4 H2O

% 48.9 51.1

Divides percentage by relative formula mass (1)

Moles 48.9 / 120.4 = 51.1 / 18 =

Divides ratio by smallest (1) (÷ RFM) 0.406146170... 2.838888889...

Ratio

1 6.98982049

x = 7 (1) (÷ smallest)

Allow MgSO4.7H2O

For Alternative

Calculates Mr of CoCl2 (1)

Mr of CoCl2 = 58.9 + 2 x 35.5 = 129.9

CoCl2 H2O

Divides percentage by relative formula mass

% 54.6 45.4

(1)

Divides ratio by smallest Moles 54.6 / 129.9 = 45.4 / 18 =

(1) (÷ RFM) 0.42032 2.5222

Ratio

1 6.0007

y = 6 (1) (÷ smallest)

Allow CoCl2.6H2O

Method 2 Example of calculation:

Calculates Mr of MgSO4 (1) Mr of MgSO4 = 24.3 + 32.1 + (4 x 16) =

120.4

Forms algebraic equation for Mr of MgSO4.xH2O

(1) Mr of MgSO4.xH2O = 120.4 + 18x

Finds algebraic expression for ratio of MgSO4 to

hydrated MgSO4 (1) 120.4 = 48.9 %

120.4 + 18x

Solves for x (1)

x = 7

For Alternative

Calculates Mr of CoCl2 (1)

Mr of CoCl2 = 58.9 + 2 x 35.5 = 129.9

Forms algebraic equation for Mr of CoCl2.xH2O

(1) Mr of CoCl2.xH2O = 129.9 + 18x

Finds algebraic expression for ratio of CoCl2 to

hydrated CoCl2.xH2O (1) 129.9 = 54.6 %

129.9 + 18x

Solves for x (1)

x = 6

Method 3

Calculates Mr of MgSO4 (1) Mr of MgSO4 = 24.3 + 32.1 + (4 x 16.0) = 120.4

Calculates Mr of MgSO4.xH2O (1) Mr of MgSO4.xH2O = 120.4 x 100 = 246.2

48.9

Calculates mass of water in one mol (1) 246.2 – 120.4 = 125.8

Finds moles of water (1) 125.8 = 7

Method 3 for CoCl2.yH2O

Calculates Mr of for CoCl2. (1) Mr of CoCl2 = 58.9 + (2 x 35.5) = 129.9

Calculates Mr of CoCl2.yH2O Mr of CoCl2.yH2O = 129.9 x 100 = 237.9

(1) 54.6

237.9 – 129.9 = 108.0

Calculates mass of water in one mol (1)

108.0 = 6

Finds moles of water (1) 18

Use of Beryllium

Calculates Mr of BeSO4 = 105.1

Moles in 48.9% of 100g = 0.46527117

Ratio of BeSO4:H2O = 1:6.102

x=6

(Total for Question 4 = 8 marks)

How to answer it

Identification & Formula of a Hydrated Salt (MN·xH₂O)

📌 What this question tests

This question evaluates inorganic qualitative analysis and quantitative stoichiometry:

Qualitative Anion Tests: Precipitation reactions with barium chloride and differentiating precipitates using dilute hydrochloric acid.
Flame Tests & Cation Properties: Group 2 flame colours (identifying cations that impart no colour to a Bunsen flame).
Ionic Compound Deductions: Working out ion charges from stoichiometry (1:1 cation-to-anion ratio).
Empirical Formula & Water of Crystallisation: Determining moles of water ( x ) from mass loss percentages upon thermal decomposition.

Part (a)(i)

Test 1: Addition of Barium Chloride Solution

Identifying two anions forming a white precipitate with Ba²⁺(aq) [1 Mark]

✅ Correct Answers

Any two of the following (names or formulae):

Sulfate / sulfate(VI) / SO₄²⁻
Sulfite / sulfate(IV) / SO₃²⁻
Carbonate / CO₃²⁻

🧠 Exam Technique & Formula Rules

If you choose to give chemical formulae instead of names, you must include the correct charge (e.g., SO₄²⁻ , not SO₄ ). The mark scheme strictly penalises missing ionic charges.

Mark scheme guidance: 1 mark for any two correct names or formulae. Lack of charge on formulae results in 0 marks.

Part (a)(ii) & (a)(iii)

Test 1 Continued: Addition of Acid & Cation Charge

Deducing the specific anion and cation charge [2 Marks]

✅ Correct Answers

(a)(ii) Formula of anion: SO₄²⁻ (1 mark)

(a)(iii) Multiple Choice: C (+2) (1 mark)

💡 Key Knowledge

Why HCl is added: BaCO₃ and BaSO₃ react with acid and dissolve, releasing CO₂ and SO₂ respectively. Only BaSO₄ remains insoluble in dilute HCl.
Charge balance: The question states MN contains a cation and anion in a 1:1 ratio. Since the anion is SO₄²⁻ , the cation M must be +2 to ensure overall electrical neutrality.

❌ Common Errors

Ignoring the command word: Part (a)(ii) explicitly asked for the formula. Writing just the word "sulfate" risks losing the mark if not previously credited.
Sign errors: Confusing cations (positive) with anions (negative) leads to options B (-1) or D (-2).

Part (b)

Test 2: Flame Test for Cation M

Identifying a metal cation that gives no flame colour [1 Mark]

✅ Correct Answer

Mg²⁺ or magnesium ion (or Be²⁺ / beryllium ion)

🧠 Exam Technique

The question asks for the identity of the cation. Writing merely the elemental symbol Mg does not score the mark. You must write either the ion with its charge ( Mg²⁺ ) or state magnesium ion.

💡 Why does Magnesium show no flame colour?

In magnesium, the energy emitted when excited electrons return to lower energy levels falls within the ultraviolet region of the spectrum, which is outside the visible light range for the human eye.

Part (c)

Quantitative Analysis: Value of x in MN·xH₂O

Calculating water of crystallisation from thermal decomposition [4 Marks]

📐 Step-by-Step Calculation (Standard Route: MgSO₄·xH₂O)

From previous parts, MN is MgSO₄. Heating reduced the mass to 48.9% of its original value (meaning the anhydrous residue is 48.9%, and the lost water is 51.1%).

Substance	Anhydrous Salt: MgSO₄	Water: H₂O
Percentage Mass (%)	48.9%	100 - 48.9 = 51.1%
Molar Mass (M_r)	24.3 + 32.1 + (4 × 16.0) = 120.4 g mol⁻¹	(2 × 1.0) + 16.0 = 18.0 g mol⁻¹
Moles in 100 g	48.9 / 120.4 = 0.4061 mol	51.1 / 18.0 = 2.8389 mol
Mole Ratio (÷ smallest)	0.4061 / 0.4061 = 1	2.8389 / 0.4061 = 6.99 ≈ 7

Final Answer: x = 7 (Compound is MgSO₄·7H₂O )

📐 Alternative Route (If using CoCl₂·yH₂O)

The exam provides an alternative fallback compound if you could not identify MN:

M_r(CoCl₂): 58.9 + (2 × 35.5) = 129.9
% Mass: CoCl₂ = 54.6%, H₂O = 100 - 54.6 = 45.4%
Moles of CoCl₂: 54.6 / 129.9 = 0.4203 mol
Moles of H₂O: 45.4 / 18.0 = 2.5222 mol
Ratio: 2.5222 / 0.4203 = 6.00
Final Value: y = 6 (Compound is CoCl₂·6H₂O )

❌ Common Calculation Pitfalls

Inverting the percentages: The question states mass "reduced to 48.9%", meaning 48.9% is the residue (MgSO₄), NOT the water lost!
Rounding too early: Rounding intermediate mole values can skew the final ratio away from an integer. Keep at least 3-4 significant figures during calculations.
Non-integer final answers: Water of crystallisation ( x ) must be reported as a whole number.

Mark Breakdown (4 Marks total):
• Mark 1: Correct calculation of M_r of anhydrous salt (MgSO₄ = 120.4 or CoCl₂ = 129.9).
• Mark 2: Dividing respective mass percentages by M_r to find moles of both salt and H₂O.
• Mark 3: Dividing mole values by the smallest number of moles to establish ratio.
• Mark 4: Correct whole number value for x = 7 (or y = 6 ).

Topics

Inorganic Chemistry · Physical Chemistry · Topic 4: Inorganic Chemistry and the Periodic Table · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.