Edexcel A-Level Chemistry Paper 2, June 2023: Question 3

7 marks · Medium difficulty · Calculations

Calculate and determine the molar masses of three organic compounds using high-resolution relative atomic masses, room temperature gas volumes, and the ideal gas equation.

Practise this question

Question

A three-part chemistry question about the molar masses of organic compounds X, Y, and Z. Part (a) provides a table of accurate relative atomic masses for H, C, N, and O, and asks to identify compound X given its molecular ion peak at m/z = 60.0323 from four multiple-choice options. Part (b) gives mass and volume data for gaseous compound Y at r.t.p. to calculate its molar mass. Part (c) provides mass, volume, pressure, and temperature data for vaporised volatile liquid Z to calculate its molar mass using the ideal gas equation, specifying appropriate significant figures.
Question text

3 This question is about the molar masses of three organic compounds, X, Y and Z.

(a) The accurate relative atomic masses, Ar, of four of the elements that could be

present in an organic compound are shown.

Element Ar

hydrogen, H 1.0078

carbon, C 12.0000

nitrogen, N 14.0031

oxygen, O 15.9949

The mass spectrum of organic compound X gives a molecular ion peak

at m/z = 60.0323

What is compound X?

(1)

A ethanamide, CH3CONH2

B ethanoic acid, CH3COOH

C trimethylamine, (CH3)3N

D urea, CO(NH2)2

(b) 9.90 g of a gaseous organic compound, Y, occupies a volume of 5.40 dm3 at

room temperature and pressure (r.t.p.).

Calculate the molar mass of the compound Y.

[molar gas volume at r.t.p. = 24.0 dm3 mol–1]

(2)

(c) A quantity of a volatile organic liquid, Z, is placed in a 60.0 cm3 flask and heated to

95.0°C. When all the liquid has vaporised, the flask is sealed.

Mass of vapour = 0.170g

Pressure = 100.6kPa

Gas constant (R) = 8.31 J mol–1 K–1

Calculate the molar mass of compound Z, giving your answer to an appropriate

number of significant figures.

Assume there was no air left in the flask once the liquid Z had vaporised.

(4)

*P71913A0628*

(Total for Question 3 = 7 marks)

Mark scheme

Show the mark scheme The mark scheme provides the answers for parts (a), (b), and (c). Part (a) indicates D is the correct answer with explanations for incorrect options. Part (b) awards 2 marks for calculating the moles of gas and the molar mass. Part (c) awards 4 marks for converting units, rearranging the ideal gas equation, calculating moles, and determining the molar mass to 2 or 3 significant figures with working examples.

Question Answer Mark

number

3(a) (1)

The only correct answer is D (urea, CO(NH2)2)

A is incorrect because ethanamide has Mr of 59.037

B is incorrect because ethanoic acid has Mr of 60.021

C is incorrect because trimethylamine has Mr of 59.0733

Question

Answer Additional Guidance Mark

Number

3(b) Example of calculation (2)

• calculation of mol of gas (1) mol of gas = 5.40 = 0.225 (mol)

24.0

molar mass = 9.90 = 44.0 / 44 (g mol−1)

• calculation of molar mass (1)

0.225

TE on mol gas

Ignore SF except 1 SF

Ignore units

Correct answer with no working scores (2)

Question

Answer Additional Guidance Mark

Number

3(c) Example of calculation (4)

• conversion of pressure, temperature and volume (1) pressure = 100.6 × 103 = 1.006 × 105 Pa

and

temperature = 273 + 95.0 = 368 K

and

volume = 60.0 × 10−6 = 6.00 × 10−5 m3

• rearrangement of ideal gas equation (1) n = pV

RT

Allow values correctly substituted into pV = nRT

• calculation of n (1)

n = 1.006 × 105 × 6.00 × 10−5 = 1.9738 × 10−3

8.31 × 368

TE on p, V and T

Molar mass = 0.170 = (86.129) = 86 / 86.1(g mol−1)

• calculation of molar mass and answer to 2 / 3 SF (1)

1.9738 × 10−3

TE on n

Ignore units.

If their final answer is less than 1 do not award the final mark

(Total for Question 3 = 7 marks)

How to answer it

Molar Masses of Organic Compounds

What this question tests

This question assesses your understanding of accurate relative atomic masses, high-resolution mass spectrometry (molecular ion peaks), gas volumes at room temperature and pressure (r.t.p.), and the ideal gas equation ( pV = nRT ). You will need to demonstrate unit conversions, significant figure handling, and multi-step molar mass calculations.

Part (a) — Identifying Compound X from Mass Spectrometry

Determine compound X given its molecular ion peak at m/z = 60.0323

✅ Correct Answer

D — urea, CO(NH₂)₂

Mark: 1 / 1

💡 Key Knowledge

  • The molecular ion peak ( m/z ) corresponds to the exact relative molecular mass of the intact molecule.
  • High-resolution mass spectrometry uses accurate Aᵣ values (e.g., H = 1.0078, C = 12.0000, N = 14.0031, O = 15.9949) to distinguish between compounds with the same integer nominal mass.

❌ Common Errors & Distractors

  • A (ethanamide): Mr = 59.037
  • B (ethanoic acid): Mr = 60.021 (Nominal mass is 60, but accurate high-res mass does not match 60.0323).
  • C (trimethylamine): Mr = 59.0733

Part (b) — Gas Volumes at r.t.p.

Calculate the molar mass of gaseous organic compound Y

📐 Step-by-Step Calculation

  • Step 1: Calculate moles of gas Y
    n = Volume / Molar Gas Volume
    n = 5.40 dm³ / 24.0 dm³ mol⁻¹ = 0.225 mol
  • Step 2: Calculate molar mass
    Molar Mass = Mass / Moles
    M = 9.90 g / 0.225 mol = 44.0 g mol⁻¹
Marks: 2 / 2 (1 for moles of gas, 1 for molar mass)

🧠 Exam Technique & Guidance

  • Correct working: A correct final answer with no working scores full marks (2 marks).
  • Error carried forward (TE): If you miscalculate moles in step 1, consequential marking applies if you divide mass correctly by your moles value.
  • Units: Units are not strictly penalized here, but expressing molar mass in g mol⁻¹ is standard best practice.

Part (c) — The Ideal Gas Equation

Calculate the molar mass of volatile liquid Z using PV = nRT

📐 Step-by-Step Calculation

  • Step 1: Convert units for PV = nRT
    Pressure ( p ) = 100.6 kPa = 100.6 × 10³ Pa (or 1.006 × 10⁵ Pa)
    Temperature ( T ) = 95.0 °C + 273 = 368 K
    Volume ( V ) = 60.0 cm³ = 60.0 × 10⁻⁶ m³ (or 6.00 × 10⁻⁵ m³)
  • Step 2: Rearrange the ideal gas equation
    n = pV / RT
  • Step 3: Calculate moles (n)
    n = (1.006 × 10⁵ × 6.00 × 10⁻⁵) / (8.31 × 368) = 1.9738 × 10⁻³ mol
  • Step 4: Calculate molar mass and apply SF
    Molar Mass = Mass / n = 0.170 g / 1.9738 × 10⁻³ mol = 86.129... = 86 g mol⁻¹ (or 86.1 g mol⁻¹ for 3 SF)
Marks: 4 / 4 (1 for conversions, 1 for rearranging PV=nRT, 1 for calculating n, 1 for molar mass & 2/3 SF)

❌ Common Calculation Traps

  • Temperature trap: Forgetting to convert Celsius to Kelvin (+273). Using 95.0 instead of 368 K loses marks instantly.
  • Volume conversion trap: Failing to convert cm³ to m³ properly ( ×10⁻⁶ instead of ×10⁻³ ).
  • Pressure conversion trap: Forgetting to convert kPa to Pa ( ×10³ ).
  • Significant figures: The question asks for an appropriate number of significant figures. Give your answer to either 2 or 3 significant figures (86 or 86.1). Do not write excessive decimal places.

Topics

Organic Chemistry · Physical Chemistry · Topic 7: Modern Analytical Techniques I · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.