AQA A-Level Chemistry Paper 1, 2023: Question 9

12 marks · Medium difficulty · State/Explain/Numerical

Answer questions on time of flight (TOF) mass spectrometry including ionisation equations, determining mass number from flight time, ion detection, isotopic abundance calculations, and electrospray ionisation.

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Question

Question 9 consists of five sub-questions on time of flight mass spectrometry. 09.1 asks for an equation with state symbols for the electron impact ionisation of Sr atoms. 09.2 gives kinetic energy of 7.02 x 10^-20 J, time 9.47 x 10^-4 s, flight tube length 95.0 cm, and Avogadro constant 6.022 x 10^23 mol^-1, asking to deduce the mass number of the ion. 09.3 asks how ions are detected and how relative abundance is determined. 09.4 provides isotopic percentages and ratios for 86Sr, 87Sr, and 88Sr to calculate the abundance of 87Sr and overall Ar to 1 decimal place. 09.5 asks why electrospray ionisation is preferred over electron impact for a protein.
Question text

09 This is a question about time of flight (TOF) mass spectrometry.

09.1 Give the equation, including state symbols, for the formation of Sr + ions from

Sr atoms by electron impact.

[1 mark]

09.2 A sample of strontium is analysed by TOF mass spectrometry.

The sample is ionised using electron impact.

The ions are accelerated to have a kinetic energy (KE) of 7.02 × 10–20 J

An ion takes 9.47 × 10–4 s to travel along a 95.0 cm flight tube.

KE = mv

where m = mass (kg) and v = speed (m s–1)

Use the information given to deduce the mass number of this ion.

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

[5 marks]

Mass number

09.3 Explain how the ions are detected in the TOF mass spectrometer.

State how the relative abundance of the ions is determined.

[2 marks]

*22* How ions are detected

How relative abundance is determined

09.4 A sample of strontium contains three isotopes, 86Sr, 87Sr and 88Sr

82% of the sample is 88Sr

The other isotopes are in a 1:2 ratio of 86Sr : 87Sr

Calculate the percentage abundance of 87Sr in this sample.

Use your answer to deduce the relative atomic mass (Ar) of the sample.

Give your answer to 1 decimal place.

[3 marks]

Abundance of 87Sr %

Ar

09.5 Electrospray ionisation is used instead of electron impact for the ionisation of a protein

in a mass spectrometry experiment.

Suggest why.

[1 mark]

Mark scheme

Show the mark scheme Mark scheme for Question 09: 09.1 accepts Sr(g) + e- -> Sr+(g) + 2e- or Sr(g) -> Sr+(g) + e-. 09.2 gives step-by-step working for v = d/t = 1003 m s^-1, m = 2KE/v^2 = 1.396 x 10^-25 kg, multiplying by 1000 and Avogadro constant to get 84.04, giving mass number = 84. 09.3 awards marks for ions gaining an electron at the detector causing a current, and current being proportional to abundance. 09.4 awards marks for calculating 87Sr abundance as 12% and Ar as 87.8. 09.5 awards a mark for stating the protein ion does not break up or fragment.

Question Answers Additional comments/Guidelines Mark

Sr(g) + e– → Sr +(g) + 2 e– Allow Sr(g) → Sr +(g) + e– 1

09.1

(AO1)

M1 V = (d÷t =) 0.950 ÷ 9.47 × 10–4 OR 1003 m s–1 Recall and conversion of d into metres

Allow 2 × 7.02 × 10–20 or 2KE t2

M2 m = 2KE or 2 × 7.02 × 10–20 (= 1.396 x 10-25 kg)

M12 d2

v2 10032

M3 = M2 x1000 5

M3 mass of ion = 1.396 × 10–22 (g)

09.2 (1 x AO1,

4 x AO2)

M4 = M3 x Avogadro’s number

M4 mass of one mol of ions in g =

–22 23 Conversion to g may be seen in M4

1.396 × 10 × 6.022 x 10 (= 84.04)

Answer as whole number

M5 mass number = 84

M1 (Ions hit a detector/electron multiplier and) each ion gains

09.3 an electron (generating a current)

(2 x AO1)

M2 current is proportional to abundance

Question Answers Additional comments/Guidelines Mark 27

M1 Abundance 87Sr = 2 × 18 ÷ 3 = 12(%)

09.4 M2 Ar = (82 × 88) + (12 × 87) + (6 × 86) Answer to 1 decimal place (2 x AO2,

100 1 x AO3)

M3 = 87.8

the protein (ion) does not break up/fragment 1

09.5

(AO3)

How to answer it

TOF Mass Spectrometry: Strontium Analysis & Calculations

📋 What This Question Tests

This comprehensive question assesses foundational and applied understanding of Time of Flight (TOF) Mass Spectrometry:

  • Writing state-specific ionisation equations for electron impact ionisation.
  • Multi-step mathematical derivations using kinetic energy ( KE = ½mv² ), velocity, mass conversions (kg to g), and Avogadro's constant ( L ) to determine isotope mass numbers.
  • Mechanisms of ion detection and how detector current directly relates to isotopic abundance.
  • Isotopic abundance calculations with proportional ratios and calculating relative atomic mass (Ar) to specified decimal places.
  • Comparison between electron impact and electrospray ionisation for biological macromolecules.
Question 09.1 [1 Mark]

Electron Impact Ionisation Equation for Strontium

Writing balanced equations with state symbols

✅ Correct Answer

Sr(g) + e⁻ → Sr⁺(g) + 2e⁻

OR alternatively:

Sr(g) → Sr⁺(g) + e⁻

Mark scheme: 1 mark for the correct balanced equation with gaseous state symbols (g) .

🧠 Exam Technique & Pitfalls

  • State symbols are mandatory: Always double-check that every strontium species has the gas state symbol (g) . Solids (s) score zero.
  • Single positive charge: Electron impact typically knocks out one electron to form a unipositive ion ( Sr⁺ ), not Sr²⁺ .
Question 09.2 [5 Marks]

Deducing Mass Number from Kinetic Energy & Flight Time

Five-step calculation combining physics formulas and molar conversions

📐 Step-by-Step Calculation

  1. Convert flight distance to metres & calculate velocity ( v ):
    d = 95.0 cm = 0.950 m
    v = d / t = 0.950 / (9.47 × 10⁻⁴ s) = 1003.17 m s⁻¹
    M1 awarded: Correct recall of v = d / t and unit conversion of distance (cm → m).
  2. Rearrange KE = ½mv² to find the mass of a single ion ( m ) in kg:
    m = 2(KE) / v²
    m = [2 × (7.02 × 10⁻²⁰)] / (1003.17)² = 1.3956 × 10⁻²⁵ kg
    M2 awarded: Correct rearrangement and substitution yielding ion mass in kg.
  3. Convert mass of one ion from kg to grams:
    m (g) = (1.3956 × 10⁻²⁵ kg) × 1000 = 1.3956 × 10⁻²² g
    M3 awarded: Correct multiplication by 1000 to convert kg into g.
  4. Calculate mass of one mole of these ions using Avogadro's constant ( L ):
    Molar Mass = (1.3956 × 10⁻²² g) × (6.022 × 10²³ mol⁻¹) = 84.04 g mol⁻¹
    M4 awarded: Multiplying single ion mass in grams by Avogadro's constant.
  5. Deduce mass number:
    Mass number must be an integer: 84
    M5 awarded: Mass number stated as the whole integer 84 (identifies the ⁸⁴Sr isotope).

❌ Common Errors to Avoid

  • Unit slip: Forgetting to convert 95.0 cm to 0.950 m causes a factor of 10⁴ error in the kinetic energy equation.
  • Forgetting kg to g: The mass from KE = ½mv² is strictly in kg. Forgetting to multiply by 1000 before (or dividing by 1000 after) Avogadro's constant leads to a value of 0.084.
  • Leaving as a decimal: Mass number is the sum of protons and neutrons; it must be rounded to a whole integer ( 84 ), not left as 84.04.

💡 Alternative Combined Formula Shortcut

You can substitute v = d / t directly into KE = ½m(d/t)² :

m = 2 × KE × t² / d²

m = 2 × (7.02 × 10⁻²⁰) × (9.47 × 10⁻⁴)² / (0.950)² = 1.396 × 10⁻²⁵ kg

This avoids rounding errors introduced by calculating intermediate velocities.

Question 09.3 [2 Marks]

Ion Detection & Relative Abundance

Describing the electron transfer and signal generation

✅ Correct Answer

How ions are detected:

Positive ions hit a negatively charged detector plate (electron multiplier) and gain an electron (which neutralises them and generates a flow of electrons / electric current).

How relative abundance is determined:

The size of the current produced is directly proportional to the abundance (number) of ions hitting the plate.

M1: Ion gains an electron (creating a current).
M2: The current generated is proportional to abundance.

❌ Common Misconceptions

  • Wrong electron direction: Stating the ions "lose an electron" or "transfer a charge" loses M1. Positive ions must gain an electron from the detector.
  • Vague electrical terms: Saying "the voltage" or "the charge is measured" will not score M2. The mark scheme explicitly requires the word current.
Question 09.4 [3 Marks]

Isotope Abundances & Relative Atomic Mass (Ar)

Ratio analysis and weighted average calculation

📐 Step-by-Step Calculation

  1. Determine abundance of ⁸⁷Sr:
    Total percentage of ⁸⁶Sr and ⁸⁷Sr = 100% − 82% = 18%
    The ratio is ⁸⁶Sr : ⁸⁷Sr = 1 : 2 (total parts = 1 + 2 = 3 parts).
    Abundance of ⁸⁷Sr = (2 / 3) × 18% = 12%
    (Consequently, abundance of ⁸⁶Sr = (1 / 3) × 18% = 6%)
    M1 awarded: Correct deduction that abundance of ⁸⁷Sr = 12%.
  2. Calculate Relative Atomic Mass (Ar):
    Ar = [(88 × 82) + (87 × 12) + (86 × 6)] / 100
    Ar = (7216 + 1044 + 516) / 100 = 8776 / 100 = 87.76
    M2 awarded: Correct weighted sum expression divided by 100.
  3. Round to specified precision:
    The question requests 1 decimal place: 87.76 rounds to 87.8
    M3 awarded: Correct final value to 1 d.p.

🧠 Exam Technique: Sanity Check

Your calculated value is 87.8. Strontium's major isotope is ⁸⁸Sr (82%), with small amounts of ⁸⁷Sr (12%) and ⁸⁶Sr (6%). The average must naturally lie just below 88. If your calculated answer is outside 86–88, check your arithmetic immediately!

❌ Common Errors

  • Inverting the ratio: Assigning 12% to ⁸⁶Sr and 6% to ⁸⁷Sr. Take care: ratio is ⁸⁶Sr : ⁸⁷Sr = 1 : 2.
  • Significant figures / Decimal places: Writing 87.76 or 88 instead of following the explicit command: "Give your answer to 1 decimal place" (87.8).
Question 09.5 [1 Mark]

Electrospray vs Electron Impact Ionisation

Ionisation method for large biomolecules (proteins)

✅ Correct Answer

The protein molecule (or ion) does not break up / does not fragment.

Mark scheme: 1 mark for stating that it prevents fragmentation / molecule stays intact.

💡 Key Knowledge: "Soft" vs "Hard" Ionisation

  • Electron Impact (Hard technique): High-energy electrons from an electron gun knock out an electron. This high energy causes covalent bonds to break, leading to extensive fragmentation.
  • Electrospray (Soft technique): The sample is dissolved in a volatile solvent and injected through a fine hypodermic needle at high voltage, gaining a proton ( H⁺ ). The mild conditions keep large, fragile biological macromolecules intact so their molecular ion can be observed.

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, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.