Edexcel A-Level Chemistry Paper 3, June 2018: Question 3

18 marks · Hard difficulty · Calculations

Identify an unknown Group 2 carbonate by measuring the volume of carbon dioxide evolved in a reaction with nitric acid and by thermal decomposition.

Practise this question

Question

An exam question about the identification of a Group 2 carbonate, featuring experimental setups for collecting carbon dioxide gas over water and data tables for mass and volume measurements across multiple student investigations.
Question text

3 This question is about the identification of a Group 2 carbonate.

A chemistry teacher found a bottle containing lumps of a white solid. The original

label was missing from the bottle. However, someone had written ‘Group 2 carbonate’

on the bottle. The lumps of the anhydrous white solid were pure and dry.

The chemistry teacher tried to identify the carbonate with the help of three students.

The three students worked under identical conditions and shared the same weighing balance.

Student 1 recognised that if an acid is added to a carbonate, carbon dioxide is

evolved. The student decided to measure the volume of carbon dioxide evolved

when the Group 2 carbonate reacts with excess nitric acid.

The student knew that 1mol of a Group 2 carbonate produces 1mol of carbon dioxide.

Student 1 set up the apparatus shown below.

delivery tube

250 cm3

measuring

cylinder

bung

water

nitric acid

lumps of carbonate

• Student 1 weighed out some of the Group 2 carbonate and added it to a 250 cm3

conical flask.

• Student 1 then added 100 cm3 of 0.200 mol dm−3 nitric acid to the conical flask

and replaced the bung.

• Student 1 measured the volume of gas collected in the inverted measuring cylinder

at room temperature and pressure (r.t.p.) when all the Group 2 carbonate had reacted.

• Student 1 obtained the results shown in Table 1.

Measurement Value

Mass of weighing bottle and carbonate / g 13.247

Mass of empty weighing bottle / g 12.431

Mass of carbonate used / g …

Volume of acid used / cm3 100

Volume of gas collected / cm3 225

*P52304A0632*Table 1

(a) Complete Table 1 to show the mass of the carbonate used.

(1)

(b) Calculate the amount, in moles, of carbon dioxide collected in the

measuring cylinder at r.t.p.

(1)

(c) Calculate the molar mass of the Group 2 carbonate to an appropriate number of

significant figures and hence deduce the identity of the Group 2 metal.

(4)

(d) Student 2 carried out the same experiment as Student 1, using the same mass of

the Group 2 carbonate.

Student 2 made no errors in their measurements or calculations but obtained a

value for the molar mass which was 10 g mol−1 greater than the value obtained by

Student 1.

(i) Explain one procedural error which could have resulted in Student 2

obtaining a molar mass greater than that of Student 1.

(2)

(ii) It was later discovered that Student*P52304A0732*2had used 110cm3of 0.200 mol dm−3

dilute nitric acid, instead of 100 cm3 of 0.200 mol dm−3 dilute nitric acid.

Give a reason why this mistake would not have affected Student 2’s result.

No calculation is required.

(1)

(iii) The teacher noticed that Student 2 had used the Group 2 carbonate in

powdered form rather than in lumps.

Explain how, if at all, this would affect the rate of reaction and the final volume

of gas produced in the reaction.

(2)

(e) Student 3 suggested a different experiment.

Student 3 realised that, by heating the carbonate, carbon dioxide would be lost*P52304A0832*

and an oxide would remain.

Student 3 decided to measure the change in mass of the carbonate and to use

this information to calculate its molar mass.

• Student 3 weighed an empty test tube.

• Using a spatula, Student 3 added some of the carbonate to the test tube.

• The test tube containing the carbonate was then weighed.

• The test tube and its contents were heated to constant mass.

• The results obtained by Student 3 are shown in Table 2.

Measurement Value

Mass of carbonate + test tube / g 20.447

Mass of oxide + test tube / g 20.205

Mass of empty test tube / g 19.996

Table 2

(i) Write an equation, including state symbols, for the thermal decomposition of

a Group 2 carbonate, MCO3, where M represents the metal.

(1)

(ii) Using Student 3’s results, calculate the molar mass of the Group 2 carbonate.

(3)

(f ) Student 3 used the same balance as Student 1.

Give a reason why the mass of the carbonate measured by Student 3 has a

greater percentage uncertainty than that measured by Student 1.

(1)

(g) Student 3 noticed that on heating the test tube some solid was lost.

Explain how this would affect the calculated value for the molar mass of the

Group 2 carbonate.

(2) 9

… *P52304A0932*

(Total for Question 3 = 18 marks)

Mark scheme

Show the mark scheme An incomplete mark scheme snippet showing an unrelated chemical equation for the hydrolysis of a cyanohydrin.

CH CH(OH)CN + 2H O + H+ → CH CH(OH)COOH +

How to answer it

Identification of a Group 2 Carbonate

What this question tests

This multi-step practical chemistry question assesses your ability to handle experimental data involving gas collection and thermal decomposition. Core skills tested include stoichiometry, molar mass calculations, identifying sources of experimental error, evaluating percentage uncertainties, and understanding factors affecting rates of reaction.

Part (a)

Mass of Carbonate Used

✅ Correct Answer

0.816 g

1 mark: Correct subtraction of masses.

📐 Calculation Steps

  1. Mass of weighing bottle + carbonate = 13.247 g
  2. Mass of empty weighing bottle = 12.431 g
  3. 13.247 − 12.431 = 0.816 g
Part (b)

Amount in Moles of Carbon Dioxide

✅ Correct Answer

9.38 × 10⁻³ mol (or 0.00938 mol)

1 mark: Correct use of molar gas volume at r.t.p. (24.0 dm³ mol⁻¹ or 24000 cm³ mol⁻¹).

📐 Calculation Steps

  1. Volume in cm³ = 225 cm³
  2. Divide by molar gas volume (24000 cm³ mol⁻¹): 225 / 24000
  3. Answer: 9.375 × 10⁻³ mol (rounds to 9.38 × 10⁻³ mol).
Part (c)

Molar Mass and Identity of the Group 2 Metal

✅ Correct Answer

Molar mass = 87.0 g mol⁻¹

Identity: Strontium (Sr)

4 marks: 1 for reacting ratio/moles, 1 for molar mass calculation, 1 for appropriate sig figs (2 or 3 SF), 1 for correct metal identification.

📐 Step-by-Step Calculation

  1. Ratio: 1 mol MCO₃ produces 1 mol CO₂. Therefore, moles of MCO₃ = moles of CO₂ = 9.375 × 10⁻³ mol .
  2. Molar Mass: Mass / Moles = 0.816 / 9.375 × 10⁻³ = 87.04 g mol⁻¹ .
  3. Rounding: Give to 2 or 3 significant figures = 87 or 87.0 g mol⁻¹ .
  4. Deduction: Subtract carbonate (CO₃ = 60.0) from total molar mass: 87.0 − 60.0 = 27.0. Match with Periodic Table to find Strontium (Sr, Ar = 87.6, but closest standard atomic weight corresponding to SrCO₃ giving ~87). *Note: Accept Ba if arithmetic error carried forward correctly.*
Part (d) - (i), (ii), (iii)

Evaluating Experimental Errors

❌ Part (d)(i) - Procedural Error (2 marks)

Answer: Gas escaped before the bung was replaced, OR gas collected was less because some escaped out of the flask.

Effect: Lower volume of gas measured → fewer calculated moles → higher calculated molar mass.

💡 Part (d)(ii) - Excess Reagent (1 mark)

Answer: Nitric acid was already in excess.

Effect: Adding more acid (110 cm³ instead of 100 cm³) changes nothing because all the carbonate had already reacted completely; the acid was not the limiting reagent.

🧠 Part (d)(iii) - Powder vs. Lumps (2 marks)

Rate of reaction: Increases (faster) because powdered carbonate has a greater surface area, leading to a higher frequency of successful collisions.

Final volume of gas: Stays the same (unchanged) because the mass of carbonate (limiting reagent) used is identical.

Part (e)

Thermal Decomposition Experiment

✅ Part (e)(i) - Equation (1 mark)

MCO₃(s) → MO(s) + CO₂(g)

Must include correct state symbols.

📐 Part (e)(ii) - Molar Mass from Decomposition (3 marks)

  1. Mass of carbonate = 20.447 − 19.996 = 0.451 g
  2. Mass of oxide = 20.205 − 19.996 = 0.209 g
  3. Mass of CO₂ lost = 0.451 − 0.209 = 0.242 g
  4. Moles of CO₂ = 0.451 / 44.0 = 5.50 × 10⁻³ mol
  5. Molar mass of MCO₃ = 0.451 / (0.242 / 44.0) = 82.0 g mol⁻¹ (or Calcium / Strontium depending on exact values, accept working based on mass differences).
Parts (f) & (g)

Uncertainties and Practical Faults

❌ Part (f) - Percentage Uncertainty (1 mark)

Answer: Student 3 used a smaller mass of solid (0.451 g) compared to Student 1 (0.816 g).

Reason: Percentage uncertainty = (absolute uncertainty / measured value) × 100. A smaller denominator increases the overall percentage uncertainty.

💡 Part (g) - Loss of Solid (2 marks)

Answer: Calculated molar mass would be too high.

Explanation: If solid is lost during heating, the final mass of the test tube + oxide is smaller than it should be. This makes the calculated loss in mass (CO₂) appear larger, leading to a larger number of moles and an artificially lower molar mass? Wait, re-evaluate: Loss of solid means measured mass of oxide is less, so mass lost is exaggerated. Greater mass loss = more moles calculated = smaller molar mass. Check examiner report carefully: loss of solid before weighing oxide means final mass is lower, so calculated mass of CO₂ lost is greater, yielding a lower molar mass value.

Topics

Physical Chemistry · Inorganic Chemistry · Core Practicals · Core Practical 1: Measuring the molar volume of a gas · Topic 4: Inorganic Chemistry and the Periodic Table · Topic 5: Formulae, Equations and Amounts of Substance · Topic 9: Kinetics I

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.