AQA A-Level Chemistry AS Paper 1, June 2022: Question 5

5 marks · Hard difficulty · State/Explain/Numerical

Calculate the mass of a 121Sb+ ion in kg and determine the time of flight for a 123Sb+ ion in a time of flight mass spectrometer.

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Question

Question 05 states that a sample of antimony containing isotopes 121Sb and 123Sb is analysed in a TOF mass spectrometer. Ions are accelerated to identical kinetic energy and travel through a 1.05 m flight tube, with 121Sb+ taking 5.93 × 10⁻⁴ s. The kinetic energy equation KE = 1/2 m v² is provided, along with the Avogadro constant L = 6.022 × 10²³ mol⁻¹. Candidates are asked to calculate the mass in kg of one 121Sb+ ion and the time taken for a 123Sb+ ion to travel through the tube.
Question text

05 A sample of antimony is analysed in a time of flight (TOF) mass spectrometer

and is found to contain two isotopes,121Sb and 123Sb

After electron impact ionisation, all of the ions are accelerated to the same

kinetic energy (KE) and then travel through a flight tube that is 1.05 m long.

A 121Sb+ ion takes 5.93 × 10–4 s to travel through the flight tube.

The kinetic energy of an ion is given by the equation KE = mv

KE = kinetic energy / J

m = mass / kg

v = speed / m s–1

Calculate the mass, in kg, of one 121Sb+ ion.

Calculate the time taken for a 123Sb+ ion to travel through the same flight tube.

The Avogadro constant, L = 6.022 × 1023 mol–1

[5 marks]

Mass of one 121Sb+ ion kg

Time taken by a 123Sb+ ion s

Mark scheme

Show the mark scheme Mark scheme for Question 05 detailing 5 marks: M1 calculates mass of 121Sb+ as 121 / (1000 × 6.022 × 10²³) = 2.009 × 10⁻²⁵ kg; M2 finds velocity v = d/t = 1770.658 m s⁻¹; M3 calculates KE = 3.1493 × 10⁻¹⁹ J; M4 determines velocity of 123Sb+ as 1756.07 m s⁻¹; M5 finds time t = 5.98 × 10⁻⁴ s. An alternative method relating m/t² is also accepted.

Question Marking guidance Additional Comments/Guidelines Mark

Mass of one ion of 121Sb+ =121 / (1000 × 6.022 × 1023) 1

05 –25

= 2.009 × 10 kg

Alternative method 1

V = d/t

KE = ½ m d2/t2

= 1.050 / 5.93 × 10–4

= 1770.658 (m s–1)

m /t 2 = m /t 2 1

2 121 121 123 123

KE = ½ m v

= ½ × 2.009 × 10–25 × (M2)2 (or = ½ × M1 × (M2)2 )

= 3.1493 × 10–19 (J)

t 2 = 123/121 x t 2 1

123 121

2KE = 3.57 x 10–7 (s2)

V123 = √ ( )

m

= √ [2(M3) / 2.0425 × 10–25]

= √ 3083769.889

= 1756.07 (m s–1)

t123 = √M4 1

t = d / v

(5 x AO2)

= 1.050 / (M4)

= 5.98 × 10–4 s

How to answer it

TOF Mass Spectrometry: Ion Velocity & Flight Time

What this question tests

This 5-mark calculation tests your ability to link physical principles to chemical analysis in a Time of Flight (TOF) mass spectrometer:

  • Calculating the mass of a single ion in kilograms (kg) using isotopic mass and the Avogadro constant.
  • Applying the equations for velocity ( v = d / t ) and kinetic energy ( KE = ½mv² ).
  • Understanding that all ions are accelerated to the same kinetic energy regardless of mass.
  • Rearranging formulas or applying direct proportional reasoning ( t ∝ √m ) to find the flight time of a second isotope.

Question 05 (5 Marks)

Calculation of single ion mass and flight time for antimony isotopes

✅ Correct Answers & Target Values

  • Mass of one ¹²¹Sb⁺ ion: 2.009 × 10⁻²⁵ kg (or 2.01 × 10⁻²⁵ kg )
  • Time taken by a ¹²³Sb⁺ ion: 5.98 × 10⁻⁴ s (3 significant figures)

💡 Key Knowledge

  • Molar Mass vs. Particle Mass: Isotopic mass (121) is in g mol⁻¹ . To convert to kg per ion , divide by both L ( 6.022 × 10²³ ) and 1000 .
  • Equal Kinetic Energy: In TOF, all ions with the same charge receive the same KE . Lighter ions travel faster; heavier ions travel slower.
  • Flight time relationship: Since KE = ½m(d/t)² , time is directly proportional to the square root of mass: t ∝ √m .

🧠 Exam Technique

  • Store values in your calculator: Avoid intermediate rounding. Carry all digits in your calculator memory (e.g., using ANS or sto recall) to avoid losing the final accuracy mark.
  • Sanity check your answer: ¹²³Sb⁺ is heavier than ¹²¹Sb⁺. Therefore, it must travel slower and take longer than 5.93 × 10⁻⁴ s . ( 5.98 × 10⁻⁴ s is indeed greater).
  • Match precision: Give the final answer to 3 significant figures, matching the precision given in the question ( 5.93 × 10⁻⁴ s and 1.05 m ).

📐 Step-by-Step Calculations

Step 1: Calculate the mass of one ¹²¹Sb⁺ ion in kg (Mark 1)

m(¹²¹Sb⁺) = 121 / (1000 × 6.022 × 10²³) = 2.0093 × 10⁻²⁵ kg

Step 2: Calculate the velocity of the ¹²¹Sb⁺ ion (Mark 2)

v₁₂₁ = d / t = 1.050 / (5.93 × 10⁻⁴) = 1770.66 m s⁻¹

Step 3: Calculate the kinetic energy (KE) of the ions (Mark 3)

KE = ½ m v² = 0.5 × (2.0093 × 10⁻²⁵) × (1770.66)² = 3.1493 × 10⁻¹⁹ J

Step 4: Calculate the velocity of the ¹²³Sb⁺ ion (Mark 4)

First find mass of ¹²³Sb⁺: m(¹²³Sb⁺) = 123 / (1000 × 6.022 × 10²³) = 2.0425 × 10⁻²⁵ kg

v₁₂₃ = √(2 × KE / m) = √(2 × 3.1493 × 10⁻¹⁹ / 2.0425 × 10⁻²⁵) = √3083770 = 1756.07 m s⁻¹

Step 5: Calculate the flight time for the ¹²³Sb⁺ ion (Mark 5)

t₁₂₃ = d / v₁₂₃ = 1.050 / 1756.07 = 5.98 × 10⁻⁴ s

⚡ Alternative Shortcut Method (Marks 2–5):

Since KE₁₂₁ = KE₁₂₃ and d is constant, m₁ / (t₁)² = m₂ / (t₂)² :

(t₂)² = (m₂ / m₁) × (t₁)² = (123 / 121) × (5.93 × 10⁻⁴)² = 3.5746 × 10⁻⁷ s²

t₂ = √(3.5746 × 10⁻⁷) = 5.98 × 10⁻⁴ s

❌ Common Errors & Traps

  • Forgetting to divide by 1000: Calculating the mass as 121 / (6.022 × 10²³) = 2.01 × 10⁻²² gives grams, NOT kg. This leads to invalid kinetic energy values.
  • Premature rounding: Rounding v₁₂₁ to 1770 or KE to 3.15 × 10⁻¹⁹ introduces rounding propagation errors, yielding an inaccurate final time (e.g., 6.0 × 10⁻⁴ s ).
  • Inverting the ratio in the shortcut: Multiplying by 121 / 123 instead of 123 / 121 . Remember, heavier ions must take more time!
  • Forgetting to square or square-root: Forgetting that v² is in the KE equation, or forgetting to take the square root when solving for v .
Examiner Mark Breakdown:
• M1: Mass of one ion correctly determined in kg ( 2.009 × 10⁻²⁵ kg ).
• M2: Velocity of ¹²¹Sb⁺ correctly found ( 1771 m s⁻¹ ).
• M3: Kinetic energy calculated ( 3.15 × 10⁻¹⁹ J ).
• M4: Velocity of ¹²³Sb⁺ calculated ( 1756 m s⁻¹ ) OR ratio expression set up ( (t₂)² = 3.57 × 10⁻⁷ ).
• M5: Final time for ¹²³Sb⁺ calculated accurately to at least 3 sig figs ( 5.98 × 10⁻⁴ s ).

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 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.