AQA A-Level Chemistry AS Paper 2, June 2025: Question 5

7 marks · Medium difficulty · Practical Techniques & Data Analysis

Explain how high resolution mass spectrometry distinguishes between compounds and calculate the relative molecular mass of a volatile liquid using ideal gas data.

Practise this question

Question

Question 05 consists of two parts. Part 05.1 states that propanone, prop-2-en-1-ol, and butane each have an Mr of 58.0, and asks why high resolution mass spectrometry can distinguish between propanone and butane, but cannot distinguish between propanone and prop-2-en-1-ol (2 marks). Part 05.2 describes a method to determine the Mr of a volatile liquid X by vaporising it in a conical flask submerged in boiling water and weighing the condensed liquid. Figure 1 shows the experimental setup with a clamped conical flask having a foil top with a pinhole, submerged in a beaker of boiling water. Table 1 provides: volume of conical flask = 272 cm³, temperature of boiling water = 100 °C, pressure = 102 000 Pa, mass of flask + foil in step 2 = 105.872 g, mass of flask + foil + X in step 7 = 106.633 g. The student must calculate the Mr of X (5 marks), given R = 8.31 J K⁻¹ mol⁻¹.
Question text

05 This question is about determining the relative molecular mass (Mr) of

different compounds.

05.1 The Mr of a compound can be determined by mass spectrometry.

Propanone, prop-2-en-1-ol and butane each have Mr = 58.0

Give the reason why high resolution mass spectrometry can distinguish between

propanone and butane, but cannot distinguish between propanone and

prop-2-en-1-ol.

[2 marks]

05.2 A student does an experiment to determine the Mr of a volatile liquid X.

In this experiment a conical flask of known volume is filled with vaporised X which is

then condensed back into a liquid in the flask.

Method

1. Place a piece of aluminium foil over the opening of a conical flask.

2. Record the mass of the conical flask and foil.

3. Remove the foil and add 5 cm3 of liquid X to the flask.

4. Replace the foil and use a pin to make a tiny hole in the middle of the foil.

5. Place the conical flask in a large beaker of boiling water (see Figure 1) until all of

liquid X has vaporised. The flask is now full of vaporised X (the air and some of X

escaped through the tiny hole).

6. Allow the conical flask to cool so that the vaporised X that was in the flask cools

and condenses (air returns to the flask).

7. Record the mass of the conical flask with foil and condensed X.

Figure 1

Table 1 shows the student’s results.

Table 1

Volume of conical flask / cm3 272

Temperature of boiling water / °C 100

Pressure / Pa 102 000

Mass of flask + foil in step 2 / g 105.872

Mass of flask + foil + X in step 7 / g 106.633

Calculate the Mr of X.

Assume that the mass of air in the flask is the same in steps 2 and 7.

*12Gas constant,* R = 8.31 J K–1 mol–1

[5 marks]

Mr

Mark scheme

Show the mark scheme Mark scheme for Question 05. For 05.1: M1 is butane and propanone have different Mr to several decimal places; M2 is prop-2-en-1-ol and propanone have the same molecular formula or are isomers (2 marks). For 05.2: M1 is n = PV / RT; M2 is converting V to 272 × 10⁻⁶ m³ and T to 373 K; M3 is calculating n = (102000 × 272 × 10⁻⁶) / (8.31 × 373) = 0.00895 mol; M4 is mass of X = 106.633 - 105.872 = 0.761 g; M5 is Mr = mass / n = 85.0 (at least 2 sf). Alternative method uses Mr = mRT / PV (5 marks).

Question Marking guidance Additional Comments/Guidelines Mark

M1 butane (and propanone) have different Mr to several/more M1 ignore reference to fragmentation and/or

decimal places same molecular formula

05.1

M2 prop-2-en-1-ol (and propanone) have the same molecular (2 x AO2)

formula or are isomers

PV If expression not written out, M1 could score from a

M1 n =

RT substituted correct expression later on (even if any

unit conversions are incorrect)

M2 converting V to 272 × 10–6 and T to 373

Allow ECF from M1 to M3, M2 to M3; M3 and/or M4

102000 × 272 × 10–6

to M5

M3 n = (= 0.00895)

8.31 × 373

If 373.15 used as T, then Mr = 85.055 (scores 5) 5

05.2 M4 mass of X = 106.633 – 105.872 (=0.761 (g)) If 100 used as T, then Mr = 22.79…. (scores 4) (1 x AO1,

4 x AO3)

M4 Alternative method

M5 Mr = = 85.0 (at least 2 sf)

M3

mRT

M3 Mr = PV

0.761 × 8.31 × 373

M5 Mr = ( –6) = 85.0 (at least 2 sf)

102000 ×272 ×10

How to answer it

Determining Molecular Mass via High-Resolution MS & the Ideal Gas Equation

WHAT THIS QUESTION TESTS

This question assesses core physical and analytical chemistry principles from AS/A-Level physical and organic chemistry:

  • High-Resolution Mass Spectrometry: The difference between integer (nominal) molecular mass and accurate mass measured to several decimal places, and why isomers cannot be separated by precise mass alone.
  • Ideal Gas Equation (pV = nRT): Rearranging the gas law to determine molar mass ( M r) of an unknown volatile liquid from experimental Dumas-method data.
  • Unit Conversions: Converting temperature from °C to K and volume from cm³ to m³.
  • Significant Figures & Practical Analysis: Calculating accurate mass changes and quoting values to appropriate precision.

Question 05.1

High-Resolution Mass Spectrometry Distinction [2 Marks]

✅ Mark Scheme Model Answer

  • Mark 1: Butane and propanone have different accurate molecular masses ( M r) to several decimal places (or more decimal places).
  • Mark 2: Prop-2-en-1-ol and propanone have the same molecular formula (C₃H₆O) or are isomers (so have identical precise masses).

💡 Key Knowledge

  • Nominal Mass: At low resolution, all three have integer masses:
    Propanone (C₃H₆O) = (3 × 12.0) + (6 × 1.0) + 16.0 = 58.0
    Butane (C₄H₁₀) = (4 × 12.0) + (10 × 1.0) = 58.0
    Prop-2-en-1-ol (C₃H₆O) = 58.0
  • High Resolution: Measures accurate masses using exact isotopic values (¹H = 1.0078, ¹²C = 12.0000, ¹⁶O = 15.9949):
    C₃H₆O = 58.0417
    C₄H₁₀ = 58.0780
  • Because propanone and prop-2-en-1-ol share the identical molecular formula, even infinite resolution cannot distinguish them by mass alone.

🧠 Exam Technique & Top Tips

Always split your answer into two clear sentences:

  1. Explain why butane can be distinguished: emphasize "different accurate mass to several decimal places" or "different precise atomic masses".
  2. Explain why prop-2-en-1-ol cannot: state clearly that they have the "same molecular formula" or are "structural isomers".

❌ Common Errors to Avoid

  • Discussing fragmentation patterns: The question asks about determining M r via mass spectrometry, not fingerprint fragmentation. Examiner guidance explicitly states: "ignore reference to fragmentation".
  • Vague statements about mass: Simply stating "they have different masses" is not enough; at 1 decimal place they are all 58.0. You must state to several / more decimal places.
Mark Breakdown: 2 marks total (AO2). 1 mark for identifying different accurate decimal M r between butane and propanone; 1 mark for stating propanone and prop-2-en-1-ol are isomers / have the same molecular formula.

Question 05.2

Calculation of M r Using Experimental Gas Data [5 Marks]

📐 Step-by-Step Calculation

Step 1: Calculate the mass of condensed liquid X (m) [M4]
m = 106.633 g - 105.872 g = 0.761 g
Step 2: Convert all given quantities to SI units [M2]
  • Pressure, P = 102 000 Pa (already in SI units)
  • Volume, V = 272 cm³ = 272 × 10⁻⁶ m³ (divide by 1 000 000)
  • Temperature, T = 100 °C = 100 + 273 = 373 K (or 373.15 K)
Step 3: State the Ideal Gas Equation and rearrange for moles (n) [M1]
pV = nRT ==> n = pV / RT
Step 4: Calculate the moles of vaporised liquid X [M3]
n = (102 000 × 272 × 10⁻⁶) / (8.31 × 373)
n = 27.744 / 3099.63 = 0.0089507... mol
Step 5: Calculate the relative molecular mass ( M r) [M5]
Mr = mass / moles = m / n
Mr = 0.761 / 0.0089507... = 85.02...
Final answer: 85.0 (allow 85 to 85.1; or 85.06 if using 373.15 K; minimum 2 significant figures).

✅ Mark Scheme Breakdown

  • M1: Rearranging expression: n = pV / RT (or correct substitution).
  • M2: Unit conversions: V = 272 × 10⁻⁶ m³ and T = 373 K .
  • M3: Calculating moles: n = 0.00895 mol .
  • M4: Mass of X: 106.633 - 105.872 = 0.761 g .
  • M5: Final M r = 85.0 (at least 2 sig figs).

❌ Common Calculation Traps

  • Forgetting to convert cm³ to m³: Multiplying by 10⁻³ instead of 10⁻⁶ is the most common blunder. 1 m³ = 100 cm × 100 cm × 100 cm = 10⁶ cm³.
  • Leaving temperature in °C: Using 100 instead of 373 K leads to M r = 22.8 (which caps marks at 4 due to ECF).
  • Premature rounding: Rounding intermediate values (e.g., using 0.009 mol) creates rounding drift in the final answer. Keep values stored in calculator memory!

🧠 Top Tip: Direct One-Step Formula

You can substitute n = m / Mr directly into pV = nRT :

Mr = (m × R × T) / (p × V)

Mr = (0.761 × 8.31 × 373) / (102 000 × 272 × 10⁻⁶) = 85.0

This avoids rounding errors completely and directly scores M3 and M5 together if executed correctly.

💡 Why the Air Assumption Matters

The prompt mentions: "Assume that the mass of air in the flask is the same in steps 2 and 7."

Because the flask cools down and air refills it in step 6, subtracting step 2 mass from step 7 mass cleanly cancels out the mass of the flask, foil, and enclosed air, leaving only the mass of condensed liquid X.

Mark Breakdown: 5 marks total (1 × AO1, 4 × AO3). Full Error Carried Forward (ECF) applies if unit conversions or intermediate moles are incorrectly calculated.

Topics

Physical Chemistry · Organic Chemistry · 3.1.2 Amount of Substance · 3.3.6 Organic Analysis

Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.