AQA A-Level Chemistry AS Paper 1, June 2025: Question 3

13 marks · Medium difficulty · State/Explain/Numerical

Answer questions on electron configurations, relative atomic mass definition, TOF mass spectrometry principles, calculating relative atomic mass from isotopic abundances, and determining the mass of an ion using time-of-flight data.

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Question

Question 03 containing parts 03.1 to 03.7. 03.1 asks for the full electron configuration of a fluoride ion (F-). 03.2 identifies element Q whose Q2+ ion has configuration 1s2 2s2 2p6 3s2 3p6 3d5. 03.3 asks for the meaning of relative atomic mass. 03.4 asks why the metal sample must be ionised, how the ion is detected, and how abundance is determined. Table 2 provides m/z values and percentage abundances: 50 (4.3%), 52 (82.8%), 53 (10.5%), 54 (2.4%). 03.5 asks for the m/z value of the ion reaching the detector last. 03.6 asks to calculate the relative atomic mass to 1 decimal place. 03.7 gives flight tube length 0.850 m, time of flight 6.41 x 10^-4 s, KE = 7.59 x 10^-20 J, and formula KE = 1/2 m v^2, asking to calculate the mass of one ion in grams.
Question text

03 This question is about atomic structure and time of flight (TOF) mass spectrometry.

03.1 Give the full electron configuration of a fluoride ion (F–).

[1 mark]

03.2 The Q2+ ion has electron configuration 1s2 2s2 2p6 3s2 3p6 3d5

Identify Q.

[1 mark]

03.3 Give the meaning of the term relative atomic mass.

[2 marks]

03.4 A TOF mass spectrometer is used to determine the relative atomic mass of a metal.

Explain why the sample of the metal must be ionised before passing into

the flight tube.

Describe how the ion is detected.

Describe how the abundance of each isotope is determined.

[3 marks]

Why sample must be ionised

How ion is detected

How abundance is determined

A sample of metal was analysed using a TOF mass spectrometer.

The mass spectrum showed four peaks.

Table 2 shows data about the four peaks in this spectrum.

*08* Table 2

m/z Percentage abundance

50 4.3

52 82.8

53 10.5

54 2.4

03.5 State the m/z value of the ion that would be the last to reach the detector.

[1 mark]

03.6 Calculate the relative atomic mass of this sample of metal.

Give your answer to 1 decimal place.

[2 marks]

10 Relative atomic mass

03.7 These are the data for one of the isotopes in this analysis.

Length of flight tube = 0.850 m

Time of flight = 6.41 × 10–4 s

KE = 7.59 × 10–20 J

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 g, of one ion of this isotope.

[3 marks]

Mass g

Mark scheme

Show the mark scheme Mark scheme for Question 03 showing marking guidance for parts 03.1 to 03.7. 03.1: 1s2 2s2 2p6. 03.2: Mn. 03.3: Average/mean mass of 1 atom of an element divided by 1/12 mass of one atom of 12C. 03.4: M1 to accelerate/detect; M2 ion gains electron; M3 abundance is proportional to current. 03.5: 54. 03.6: Sum of (m/z x abundance)/100 gives 52.1. 03.7: Calculates v = 1326 m/s, calculates mass in kg = 8.63 x 10^-26 kg, then converts to g giving 8.63 x 10^-23 g.

Question Marking guidance Additional Comments/Guidelines Mark

22 6 1

03.1 1s 2s 2p

(1 x AO1)

2+ – AS CHEMISTRY – 7404/1 – 1

03.2 Mn Mn

(1 x AO2)

Average / mean mass of 1 atom (of an element) If moles and atoms mixed, max = 1

1/12 mass of one atom of 12C

Mark top and bottom line independently. All key

OR terms must be present for each mark.

Average / mean mass of atoms of an element

1/12 mass of one atom of 12C

OR

Average / mean mass of atoms of an element × 12

mass of one atom of 12C

OR 2

03.3

(2 x AO1)

(Average) mass of one mole of atoms

1/12 mass of one mole of atoms of 12C

OR

(Weighted) average mass of all the isotopes (of an element)

1/12 mass of one atom of 12C

OR

Average mass of an atom/isotope compared to/relative to C−12 on a

scale in which an atom of C−12 has a mass of 12 – AS CHEMISTRY – 7404/1 –

This expression = 2 marks

M1 So it can be accelerated / so it can be detected 17

M2 Ion (hits negative plate or detector to) gain an electron M2 ion knocks out an electron into electron

03.4 multiplier

(3 x AO1)

M3 Abundance ∝ current flow / idea that abundance is proportional to

current. M3 signal from electron multiplier proportional to

number of ions

03.5 54

(1 x AO3)

(50 × 4.3) + (52 × 82.8) + (53 × 10.5) + (54 × 2.4)

100 2

03.6

– AS CHEMISTRY – 7404/1 – (2 x AO2)

52.1 (must be 1 dp)

M1: calculation of v

d 0.850 -1

v = = -4 = 1326 (m s )

t 6.41 ×10

M2: the correct calculation of mass in kg using their M1 M2: calculation of mass in kg

-20 2KE 2 ×7.59 ×10-20

2KE 2 × 7.59 ×10 -26

m = = m = 2 = 2 = 8.63 ×10 (kg)

v2 2 v (1326)

(M1)

M3: the correct calculation of mass in g using their M2

M3: calculation of mass in g

m = M2 × 1000

m = 8.63 × 10−26 × 1000 = 8.63 ×10-23 (g)

03.7

(3 x AO2)

Alternative approach

M1 expression

2KEd2

m = 2

t

M2 calculation of mass in kg

2 × 7.59 × 10-20× (0.850)2

m = = 8.63 ×10-26 (kg)

-4 2

(6.41 × 10 )

M3 answer in g

m = M2 × 1000

How to answer it

Atomic Structure & Time of Flight (TOF) Mass Spectrometry

What this question tests

This multi-part foundational question assesses core Physical and Inorganic Chemistry principles from AQA Section 3.1.1:

  • Electronic configurations: Writing configurations of negative ions and deducing transition metal identities from d-block cations.
  • Definitions: Recalling the precise, mark-scheme-standard definition of relative atomic mass (Ar).
  • Mass Spectrometry Mechanics: Understanding ionization, electric field acceleration, detector electron transfer, and ion abundance determination.
  • Quantitative TOF Calculations: Applying flight tube dynamics, relating kinetic energy ( KE = ½mv² ) to drift time ( t = d/v ), and managing unit conversions ( kg to g ).
Question 03.1 • 1 Mark

Full Electron Configuration of a Fluoride Ion (F⁻)

✅ Correct Answer

1s² 2s² 2p⁶

1 Mark: Full configuration must be shown. No noble gas shorthand allowed (e.g., [Ne] scores 0).

💡 Key Knowledge

  • Neutral fluorine has 9 electrons: 1s² 2s² 2p⁵ .
  • The fluoride ion (F⁻) gains 1 electron to complete its valence shell: 10 electrons total.
  • Subshells must be listed completely as requested by full electron configuration.
Question 03.2 • 1 Mark

Identifying Element Q from its Q²⁺ Electron Configuration

✅ Correct Answer

Mn (or Manganese)

1 Mark: Correct chemical symbol or full element name. Writing Mn²⁺ scores 0 because the question asks to identify Q.

🧠 Exam Technique: Transition Metal Ions

  • Q²⁺ configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ (23 electrons).
  • Neutral atom Q must have 23 + 2 = 25 electrons.
  • Atomic number 25 on the Periodic Table corresponds to Manganese (Mn).
  • Remember: 4s electrons are filled first, but they are also lost first upon ionization! Neutral Mn is [Ar] 4s² 3d⁵ .

❌ Common Errors

  • Identifying the element with 23 electrons (Vanadium, V) by forgetting to add back the two lost electrons.
  • Writing "Mn²⁺" instead of "Mn" — read carefully: Q is the neutral element.
Question 03.3 • 2 Marks

Meaning of the Term "Relative Atomic Mass"

✅ Correct Definition

M1: The average (or weighted mean) mass of 1 atom of an element

M2: Compared to (or divided by) 1/12th the mass of 1 atom of carbon-12 (¹²C)

2 Marks: Top and bottom lines marked independently. Either written out as words or expressed as an exact fraction.

🧠 Formulaic Expression

Examiners award both marks directly for this mathematical representation:

(Average mass of 1 atom of an element) / (¹/₁₂ mass of one atom of ¹²C)

❌ Major Mark Traps

  • Mixing atoms and moles: If you say "average mass of 1 mole" in the numerator, you MUST use "1/12th mass of 1 mole of ¹²C" in the denominator. Mixing atoms and moles caps marks at max 1.
  • Missing "average" or "mean": Stating merely "mass of an atom" loses M1 because it ignores isotopic abundance.
  • Omitting "¹²C" or "1/12th": Simply writing "carbon" without specifying the carbon-12 isotope loses M2.
Question 03.4 • 3 Marks

TOF Principles: Ionisation, Detection & Abundance

✅ Required Mark Points

  1. Why sample must be ionised: So it can be accelerated (by an electric field) OR so it can be detected.
  2. How the ion is detected: The positive ion hits the detector (negative plate) and gains an electron.
  3. How abundance is determined: The current generated is proportional to the abundance (number of ions hitting the plate).
3 Marks: 1 mark per point. Independent scoring.

💡 Detection Mechanism Explained

  • Positive ions are attracted to and strike the negatively charged detector plate.
  • When an ion hits the plate, it accepts electrons from the metal surface, which neutralises the ion.
  • This movement of electrons creates an electrical current.
  • A greater number of arriving ions draws more electrons, generating a larger current.

❌ Common Errors

  • Vague statement for ionisation like "so it can move through the tube" (must state accelerated or deflected/detected).
  • Saying ions "lose an electron" at the detector plate — positive ions need to gain electrons to be discharged.
Question 03.5 • 1 Mark

Deducing the Last Ion to Reach the Detector

✅ Correct Answer

54

1 Mark: The m/z value of the slowest ion.

💡 Physics of TOF Drift Region

  • All ions are given the same kinetic energy ( KE = ½mv² ) during acceleration.
  • Rearranging: v = √(2KE / m) . Ions with a larger mass ( m ) travel with a lower velocity.
  • Time of flight is given by t = d / v . Lower velocity means a longer time of flight.
  • Therefore, the isotope with the highest m/z ratio (54) travels slowest and hits the detector last.
Question 03.6 • 2 Marks

Calculating Relative Atomic Mass (Ar)

📐 Step-by-Step Calculation

  1. Multiply each m/z value by its percentage abundance:
    (50 × 4.3) + (52 × 82.8) + (53 × 10.5) + (54 × 2.4)
    = 215 + 4305.6 + 556.5 + 129.6 = 5206.7
  2. Divide by total abundance (100):
    Ar = 5206.7 / 100 = 52.067
  3. Round to 1 decimal place (as instructed):
    52.1
M1: Correct mathematical expression shown.
M2: 52.1 (Must strictly be 1 decimal place).

❌ Calculation Traps

  • Wrong rounding: Writing 52 or 52.07 loses the second mark. The question explicitly mandates 1 decimal place.
  • Sanity check: The answer (52.1) is very close to the most abundant isotope (52 at 82.8%), confirming the calculation is sensible. This also confirms the element is Chromium (Cr, Ar = 52.0).
Question 03.7 • 3 Marks

TOF Quantitative Calculation: Mass of an Ion in Grams

📐 Step-by-Step Method

Given:
Distance ( d ) = 0.850 m
Time ( t ) = 6.41 × 10⁻⁴ s
Kinetic Energy ( KE ) = 7.59 × 10⁻²⁰ J

  1. Find velocity (v):
    v = d / t = 0.850 / (6.41 × 10⁻⁴) = 1326 m s⁻¹
    M1: Calculation of velocity v = 1326 m s⁻¹ (or combining formulas).
  2. Rearrange KE = ½mv² to find mass in kg:
    m = 2KE / v²
    m = (2 × 7.59 × 10⁻²⁰) / (1326)² = 8.6327 × 10⁻²⁶ kg
    M2: Mass calculated in kg ( 8.63 × 10⁻²⁶ kg ).
  3. Convert mass from kg to grams (g):
    mass in g = mass in kg × 1000
    m = 8.6327 × 10⁻²⁶ × 1000 = 8.63 × 10⁻²³ g
    M3: Correct mass in grams ( 8.63 × 10⁻²³ g ).

🧠 Alternative Single-Step Algebra

Substitute v = d / t directly into m = 2KE / v² :

m = 2 · KE · t² / d²

m = [2 × (7.59 × 10⁻²⁰) × (6.41 × 10⁻⁴)²] / (0.850)²
m = 8.6327 × 10⁻²⁶ kg

This minimizes intermediate rounding errors on your calculator!

❌ Common Exam Traps in 03.7

  • Forgetting to convert kg to g: The SI formula gives mass in kg. Leaving the answer as 8.63 × 10⁻²⁶ loses the final mark.
  • Dividing by 1000 instead of multiplying: To convert kg to g, you must multiply by 1000.
  • Squaring errors: Forgetting to square v or squaring 2KE instead of just v .
  • Using Avogadro's number unnecessarily: The question asks for the mass of one ion, not a mole of ions. Do not divide by L .

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 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.