AQA A-Level Chemistry AS Paper 1, June 2023: Question 7
10 marks · Hard difficulty · State/Explain/Numerical
Calculate the molar mass of compound X using TOF mass spectrometry data, and explain the principles of ionisation and detection.
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Question text
07 This question is about time of flight (TOF) mass spectrometry.
07.1 Compound X is dissolved in a polar, volatile solvent and is ionised by
electrospray ionisation.
Each ion is accelerated so that it has a kinetic energy of 1.36 × 10 –16 J
The kinetic energy of an ion is given by the equation KE = mv
where:
KE = kinetic energy / J
m = mass / kg
v = speed / m s–1
The time of flight along the 0.750 m flight tube is 2.48 × 10 –5 s
Determine the mass, in g, of one mole of X.
The Avogadro constant, L = 6.022 × 1023 mol–1
[5 marks]
Mass of one mole of X g
A mixture of gases is analysed using TOF mass spectrometry.
The mixture contains argon, carbon dioxide, nitrogen and oxygen.
The mixture is ionised by electron impact.
*0127.*2 State the meaning of the term electron impact ionisation.
[1 mark]
07.3 Identify the ion formed from this mixture that reaches the detector last.
Justify your answer.
[2 marks]
Ion that reaches detector last
Justification
07.4 State how the ions are detected, and how the abundance of each ion is measured, in
a TOF mass spectrometer.
[2 marks]
How ions are detected
How abundance is measured
Mark scheme
Show the mark scheme
Question Marking guidance Additional Comments/Guidelines Mark
M1 M1 Calculation of v
d 0.750 –1
v = = −5 = 30241.9 m s
t 2.48 × 10
M2 Calculation of m (in kg)
M2
2 ke 2 × 1.36 × 10−16
−16 m = = = 2.974 × 10–25 kg
2 ke 2 × 1.36 × 10 2 2
m = = v (30241.9)
v2 (ans to M1)2
M3 M3 calculation of m (in g)
07.1 (4 x AO2,
m = (ans to M2) × 1000 m = 2.974 × 10–25 × 1000 = 2.974 × 10–22 g
1 x AO3)
M4 M4 calculation of mass of one mole of ions
mass = (ans to M3) × 6.022 × 1023
mass = 2.974 × 10–22 × 6.022 × 1023 = 179(.1)
M5
M5 subtracts 1 for mass of H+
Mass of one mole = (ans to M4) – 1 = 178(.1) Mass of one mole = 179.1 – 1 = 178(.1)
(High energy) electrons (from an electron gun) are used to knock out 1
07.2
an electron (from each molecule or atom.) (AO1)
20 Ion that reaches detector last: CO +
07.3
Justification: Has the highest mass (to charge ratio) (so will travel the (2 x AO3)
slowest)
M1 (ions hit a detector and) each ion gains an electron (generating a Allow the use of electron multiplier to amplify the
current) current 2
07.4
(2 x AO1)
M2 (the abundance is) proportional to (the size of) the current
How to answer it
TOF Mass Spectrometry Calculations & Principles
This question examines core aspects of Time of Flight (TOF) mass spectrometry: rearranging and applying the kinetic energy equation ( KE = ½mv² ) and velocity formula ( v = d / t ), accounting for added protons ( H⁺ ) in electrospray ionisation, defining electron impact ionisation, predicting flight times based on m/z values, and describing the operation of the ion detector.
Calculation of Molar Mass via Electrospray Ionisation
Determining the mass in grams of one mole of compound X
📐 Step-by-Step Calculation
- Velocity of the ion ( v ):
v = d / t = 0.750 / (2.48 × 10⁻⁵) = 30241.9 m s⁻¹ - Mass of one ion in kg ( m ):
KE = ½mv² ⇒ m = 2(KE) / v²
m = [2 × (1.36 × 10⁻¹⁶)] / (30241.9)² = 2.974 × 10⁻²⁵ kg - Convert mass of one ion to g:
m = 2.974 × 10⁻²⁵ × 1000 = 2.974 × 10⁻²² g - Mass of one mole of ions:
mass = (2.974 × 10⁻²²) × (6.022 × 10²³) = 179.1 g - Subtract mass of added H⁺ (electrospray ionisation):
Molar mass of X = 179.1 − 1.0 = 178.1 g (or 178 g)
✅ Correct Answer
178 g (or 178.1 g)
❌ Common Traps & Mistakes
- Forgetting to subtract 1: Electrospray adds an H⁺ ion ( XH⁺ ). Failing to subtract 1 at the end loses mark M5.
- Unit conversion errors: KE is in Joules, so the mass calculated via KE = ½mv² is always in kg, not grams. You must multiply by 1000 to get grams.
- Premature rounding: Rounding intermediate values like v too early introduces severe rounding errors in the final answer. Keep full values in calculator memory.
• M1: Correct calculation of velocity v = 30241.9 m s⁻¹
• M2: Correct calculation of single ion mass in kg m = 2.974 × 10⁻²⁵ kg
• M3: Converting kg to g ( × 1000 ) to give 2.974 × 10⁻²² g
• M4: Multiplying by Avogadro constant ( L ) to obtain mass of 1 mole of ions ( 179.1 g )
• M5: Subtracting 1 for H⁺ to find molar mass of X ( 178 g or 178.1 g )
Electron Impact Ionisation
Definition of the ionisation process
✅ Correct Answer
High-energy electrons (fired from an electron gun) knock out / displace an electron from each molecule or atom.
🧠 Exam Technique
- Mention high-energy electrons (or electrons from an electron gun).
- Clearly state the outcome: knocks out / removes an electron (resulting in a 1+ ion: M + e⁻ → M⁺ + 2e⁻ ).
• 1 Mark: High-energy electrons knock out an electron from each molecule/atom.
Flight Time & Detection Order
Argon, Carbon Dioxide, Nitrogen, and Oxygen mixture
✅ Correct Answer
Ion that reaches detector last: CO₂⁺
Justification: It has the highest mass-to-charge ratio ( m/z ) / largest mass, and therefore travels with the lowest velocity / slowest.
💡 Key Knowledge
- Under electron impact, gases form 1+ ions:
• Nitrogen: N₂⁺ ( m/z = 28 )
• Oxygen: O₂⁺ ( m/z = 32 )
• Argon: Ar⁺ ( m/z = 40 )
• Carbon dioxide: CO₂⁺ ( m/z = 44 ) - All ions have identical kinetic energy ( KE ). Since v = √(2KE/m) , the heaviest ion moves the slowest and takes the longest time to travel down the flight tube.
❌ Common Errors
- Forgetting that nitrogen and oxygen are diatomic ( N₂ and O₂ ), leading to incorrect mass comparisons.
- Forgetting to write the charge: answering CO₂ instead of the ion CO₂⁺ loses the identification mark.
• 1 Mark: CO₂⁺ correctly identified (must include charge).
• 1 Mark: Justification stating it has the highest m/z ratio (or largest mass) so it moves slowest / takes the longest time.
Ion Detection & Current Measurement
How the detector functions in TOF mass spectrometry
✅ Correct Answer
How ions are detected: Positive ions hit the (negatively charged) detector plate and gain an electron, which generates / induces an electric current.
How abundance is measured: The abundance of an ion is proportional to the size of the current generated.
🧠 Exam Technique
- Always link detection to electron transfer: ions hit the plate, gain an electron, causing a flow of charge (current).
- Always link abundance directly to the size / magnitude of the electric current (more ions = more electrons transferred = greater current).
- Note: The mark scheme allows mention of an electron multiplier used to amplify the current.
• M1: Positive ions hit the detector and gain an electron, generating an electric current.
• M2: Abundance is proportional to the size/magnitude of this current.
Topics
Physical Chemistry · 3.1.1 Atomic Structure · 3.1.2 Amount of Substance
Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.