AQA A-Level Chemistry Paper 1, June 2024: Question 1

13 marks · Medium difficulty · State/Explain/Numerical

Answer questions on atomic structure, relative atomic mass, electron impact ionisation, and time of flight (TOF) mass spectrometry calculations.

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Question

Question 1 covers atomic structure in five parts. Part 1.1 displays a diagram of Thomson's plum pudding model containing 5 electrons in a sphere of positive charge, asking for the atom's identity and two differences from the modern model. Part 1.2 asks to define relative atomic mass and justify tellurium's placement before iodine. Part 1.3 asks for the electron impact ionisation equation for tellurium. Part 1.4 provides flight tube length 1.25 m, KE 1.88 times 10 to the minus 12 J, and time 3.00 times 10 to the minus 7 s to calculate the mass of 1 mole of ions in grams and determine mass number y. Part 1.5 asks to compare the kinetic energy of two tellurium isotopes in a TOF mass spectrometer.
Question text

01 This question is about atomic structure.

01.1 In 1897 JJ Thomson discovered the electron. He suggested that atoms were

positively charged spheres with electrons embedded within them.

Figure 1 represents an atom using Thomson’s model.

Figure 1

Suggest the identity of this atom.

Give two differences between the modern model of an atom and the Thomson model

of an atom.

[3 marks]

Identity

Difference 1

Difference 2

01.2 Tellurium has a relative atomic mass of 127.6

Iodine has a relative atomic mass of 126.9

Define relative atomic mass.

Suggest one property of tellurium that justifies its position before iodine in the

modern Periodic Table.

[3 marks]

Definition

Justification

01.3 A sample of tellurium is analysed in a time of flight (TOF) mass spectrometer using

electron impact ionisation.

Give an equation, including state symbols, for this ionisation.

[1 mark]

01.4 In the TOF mass spectrometer an ion of an isotope of tellurium, with mass number y,

travels along a 1.25 m flight tube with a kinetic energy of 1.88 x 10–12 J

The ion takes 3.00 x 10–7 s to reach the detector.

KE = mv

KE = kinetic energy / J

m = mass / kg

v = speed / m s–1

Calculate the mass, in g, of 1 mole of these tellurium ions.

Use your answer to suggest the mass number y of the tellurium isotope.

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

[5 marks]

Mass g

5 Mass number y

01.5 Tellurium has several other isotopes.

Two of these isotopes are 126Te and 124Te

A different sample of tellurium is analysed using a TOF mass spectrometer.

Which statement about kinetic energy (KE) is correct?

[1 mark]

*04* Tick ( ) one box.

The KE of 126Te+ is greater than the KE of 124Te+

The KE of 126Te+ is the same as the KE of 124Te+

The KE of 126Te+ is less than the KE of 124Te+

Mark scheme

Show the mark scheme Mark scheme for Question 1: 01.1 awards 1 mark for Boron/B and 2 marks for differences between modern and Thomson models (protons/neutrons in nucleus, electrons in shells, atom mostly empty space). 01.2 awards 2 marks for defining relative atomic mass based on 1/12 mass of an atom of carbon-12 and 1 mark for justifying placement based on atomic number or valence electrons. 01.3 gives Te(g) forms Te+(g) + e-. 01.4 details 5-step TOF calculation yielding v = 4.17 times 10^6 m/s, m = 2.16 times 10^-25 kg, molar mass = 130.4 g/mol, giving mass number y = 130 or 131. 01.5 awards 1 mark for the KE of both isotopes being the same.

Question Answers Additional comments/Guidelines Mark

B/ Boron

any 2 from:

protons in the centre of the atom/nucleus

01.1 electrons are in shells/energy levels (around the nucleus)

(3 x AO1)

neutrons in the centre of the atom/nucleus

most of the atom is empty space/most of mass in nucleus

Definition

Average / mean mass of 1 atom (of an element) (1)

1/12 mass of one atom of 12C (1)

01.2 Or

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

(Average) mass of one mole of atoms

1/12 mass of one mole of 12C

Or

(Weighted) average mass of all the isotopes

1/12 mass of one atom of 12C

Or

01.2 Average mass of an atom/isotope

(cont) compared to/relative to C−12 on a scale in which an atom of C−12 (2 x AO1,

has a mass of 12 1 x AO3)

Justification

Tellurium has Z = 52 but iodine has Z = 53

Or

Te has one fewer proton than I / I has one more proton

Or

Tellurium has 6 outer shell electrons/valence electrons but iodine

has 7

Or

Te has similar chemistry/chemical properties to other Group 6

elements

Or

I has similar chemistry/chemical properties to other Group 7

elements

Te(g) + e– → Te+(g) + 2 e–

01.3 Or

(1 x AO1)

Te(g) → Te+(g) + e–

d 6 –1

M1 v = t = 4.17 × 10 (m s )

2KE×t2 2KE 2 𝑥𝑥 1.88 𝑥𝑥 10−12

M2 m = d2 or m = 2 or 6)2

v (4.17 𝑥𝑥 10

01.4 M3 m = 2.16 x 10-25 to 2.17 × 10–25 (kg) (4 x AO2,

1 x AO3)

M4 mass of 1 mole of ions = L × 1000 × M3 = 130.4 (g) M4 Allow 130 to 131 (3 or more significant

figures)

M5 y = 130 or 131 M5 Must be an integer

126 + 124 + 1 13

01.5 The KE of Te is the same as the KE of Te

(1 x AO1)

How to answer it

Atomic Structure, the Periodic Table & TOF Mass Spectrometry

📌 What this question tests

This question examines foundational atomic physics and physical chemistry principles from AQA Section 3.1.1:

  • Historical vs. modern atomic models: Contrasting Thomson’s plum pudding model with Rutherford-Bohr nuclear theory.
  • Periodic Table fundamentals: The formal definition of relative atomic mass (Ar) and ordering by atomic number (Z) rather than atomic mass.
  • Ionisation in TOF MS: Writing correct equations and state symbols for electron impact ionisation.
  • Time of Flight (TOF) mathematics: Rearranging kinetic energy and velocity formulas, unit conversions (kg to g), and working via the Avogadro constant to find isotopic mass numbers.
  • Kinetic energy principles: Understanding that ions of equal charge receive identical kinetic energy in an electrostatic field.
Question 01.1 • 3 Marks

Thomson Model vs. Modern Atomic Structure

Identifying an atom from electron count & describing structural differences

✅ Correct Answers

Identity: Boron (or B) [1 mark]

Differences (any two) [2 marks]:

  • In the modern model, protons are in a central nucleus (Thomson: positive charge is a diffuse sphere).
  • Neutrons are present in the nucleus (Thomson: no neutrons).
  • Electrons are in energy levels / shells / orbitals orbiting the nucleus (Thomson: embedded within the sphere).
  • Most of the atom is empty space / mass is concentrated in the nucleus (Thomson: mass/charge is spread out).

🧠 Exam Technique & Insight

  • Counting electrons: Figure 1 clearly shows 5 embedded dots (electrons). Because an atom is electrically neutral, it must have 5 protons. Element 5 is Boron.
  • Structure your comparison: Always explicitly reference the modern model feature (e.g., "electrons orbit in shells" or "protons and neutrons reside in a dense nucleus"). AQA credits modern model descriptions directly.

❌ Common Errors

Giving vague statements like "Thomson has no nucleus" without explaining where the particles actually are in the modern atom. Ensure you mention protons, neutrons, or electron shells specifically.

Question 01.2 • 3 Marks

Relative Atomic Mass Definition & Periodic Ordering

Defining Ar and explaining anomalous atomic mass positioning

✅ Correct Answers

Definition of Ar [2 marks]:

The average (or mean) mass of an atom of an element ÷ (1/12th the mass of an atom of ¹²C)

• Mark 1: Average / weighted mean mass of an atom of an element
• Mark 2: Relative to 1/12th of the mass of a carbon-12 (¹²C) atom

Justification [1 mark] (any one):

  • Tellurium has a lower atomic number than iodine (Te has Z = 52, I has Z = 53).
  • Te has one fewer proton than I (or I has one more proton).
  • Te has 6 outer/valence electrons, whereas I has 7.
  • Te has chemical properties matching Group 6 (or I matches Group 7).

💡 Key Knowledge

  • Mendeleev vs. Modern Table: Mendeleev initially arranged elements by atomic weight and had to place Te before I based on chemical properties. Today, the table is strictly arranged by increasing atomic number (number of protons).
  • Strict Ar wording: AQA insists on the word "average" or "mean". Omitting it will cap you at 1/2 for the definition.

❌ Common Errors

  • Forgetting the word "average" or "weighted mean" before mass of an atom.
  • Saying "mass of a carbon atom" instead of specifying ¹²C or carbon-12.
  • For the justification, stating that "Te has a smaller mass" — that is factually false (Te is 127.6, I is 126.9). The reason it sits first is strictly its proton number (Z).
Question 01.3 • 1 Mark

Electron Impact Ionisation Equation

Formulating the equation with state symbols

✅ Correct Answer

Either of the following standard representations:

Te(g) + e⁻ → Te⁺(g) + 2e⁻

OR

Te(g) → Te⁺(g) + e⁻

🧠 Exam Technique

  • Always include state symbols: In TOF mass spectrometry, samples are first vaporised. The reactant MUST have the gas symbol (g) .
  • Charge conservation: Make sure the positive ion Te⁺ is shown with a single positive charge (electron impact removes one electron).
Question 01.4 • 5 Marks

TOF Mass Spectrometer Calculation

Determining isotopic mass number from kinetic energy and drift time

📐 Step-by-Step Calculation Breakdown

1 Calculate velocity of the ion (v):

v = d ÷ t = 1.25 m ÷ (3.00 × 10⁻⁷ s) = 4.167 × 10⁶ m s⁻¹

[M1 awarded for calculating velocity]

2 Rearrange KE formula for the mass of one single ion (m):

KE = ½ m v² ⇒ m = 2(KE) ÷ v²

[M2 awarded for correct algebraic rearrangement]

3 Calculate mass of ONE ion in kilograms (kg):

m = [2 × (1.88 × 10⁻¹² J)] ÷ (4.167 × 10⁶ m s⁻¹)²

m = 3.76 × 10⁻¹² ÷ 1.736 × 10¹³ = 2.166 × 10⁻²⁵ kg

[M3 awarded for m between 2.16 × 10⁻²⁵ and 2.17 × 10⁻²⁵ kg]

4 Calculate the mass of 1 mole of ions in grams (g):

Multiply mass of one ion (in kg) by Avogadro's constant (L), then convert kg to g (× 1000):

Mass of 1 mole = m × L × 1000

Mass = (2.166 × 10⁻²⁵ kg) × (6.022 × 10²³ mol⁻¹) × 1000 g kg⁻¹ = 130.4 g

[M4 awarded for mass of 130 to 131 g]

5 State the mass number (y):

Mass number y = 130 (or 131)

[M5 awarded: Mass number must be given as an integer]

❌ Common Calculation Traps

  • Forgetting to square velocity: A classic calculator error when dividing by v² . Use brackets carefully: 2 × KE / (v)² .
  • Unit conversion oversight: The mass from KE = ½ m v² is in kg. Forgetting to multiply by 1000 to convert to grams results in a molar mass of 0.13 g mol⁻¹.
  • Non-integer mass number: Mass number is the sum of protons and neutrons — it must be an integer. Writing y = 130.4 will forfeit the final mark.
Question 01.5 • 1 Mark

Kinetic Energy Comparison of Isotopes

Understanding electrostatic acceleration in TOF MS

✅ Correct Choice

Tick Box 2:

The KE of ¹²⁶Te⁺ is the same as the KE of ¹²⁴Te⁺ [✓]

💡 Underlying Principle

During the acceleration phase of TOF mass spectrometry, ions pass through an electric field with a fixed potential difference ( V ).

KE = q × V

Because both ions have an identical +1 charge ( q ), they are accelerated by the exact same voltage and therefore gain the exact same kinetic energy.

🧠 Examiner Insight

Students frequently confuse kinetic energy with speed:

  • Kinetic energy: Identical for all 1+ ions.
  • Velocity: Different! Because ¹²⁴Te⁺ has a smaller mass, it travels faster and reaches the detector in a shorter time than ¹²⁶Te⁺ ( t ∝ √m ).

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, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.