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Time of flight (TOF) mass spectrometry can be used to analyse large molecules such as the pentapeptide, leucine enkephalin (P) - AQA - A-Level Chemistry - Question 2 - 2019 - Paper 1

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Time of flight (TOF) mass spectrometry can be used to analyse large molecules such as the pentapeptide, leucine enkephalin (P). P is ionised by electrospray ionisat... show full transcript

Worked Solution & Example Answer:Time of flight (TOF) mass spectrometry can be used to analyse large molecules such as the pentapeptide, leucine enkephalin (P) - AQA - A-Level Chemistry - Question 2 - 2019 - Paper 1

Step 1

Describe the process of electrospray ionisation.

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Answer

Electrospray ionisation is a technique where a sample solution is pushed through a fine needle at high voltage (typically around 4000 volts). This creates an aerosol of charged droplets, which evaporate, leaving charged ions in the gas phase. In this process, molecules gain protons, leading to ion formation.

Equation:

ightarrow ext{P}H^+ $$

Step 2

What is the relative molecular mass of P?

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Answer

The relative molecular mass of P is 556.

Step 3

Calculate the relative molecular mass of Q.

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Answer

To find the mass of the ion Q, we first need to use the kinetic energy equation:

KE=12mv2KE = \frac{1}{2} mv^2

Given that:

  • KE=2.09×1018 JKE = 2.09 \times 10^{-18} \text{ J}
  • The time taken to reach the detector is t=1.23×106 st = 1.23 \times 10^{-6} \text{ s}
  • The length of the flight tube is d=1.50 md = 1.50 \text{ m}

We can find the speed using: v=dt=1.50 m1.23×106 s1.22×106 m s1v = \frac{d}{t} = \frac{1.50 \text{ m}}{1.23 \times 10^{-6} \text{ s}} \approx 1.22 \times 10^6 \text{ m s}^{-1}

Plugging into the kinetic energy equation: KE=12mv2KE = \frac{1}{2} m v^2

Rearranging gives: m=2×KEv22×2.09×1018 J(1.22×106)22.81×1018 kgm = \frac{2 \times KE}{v^2} \approx \frac{2 \times 2.09 \times 10^{-18} \text{ J}}{(1.22 \times 10^6)^2} \approx 2.81 \times 10^{-18} \text{ kg}

To convert to relative molecular mass: mrelative=m×Lm_{relative} = m \times L where L=6.022×1023 mol1L = 6.022 \times 10^{23} \text{ mol}^{-1}.

Calculating gives: mrelative=2.81×1018 kg×6.022×1023 mol1169extgmol1m_{relative} = 2.81 \times 10^{-18} \text{ kg} \times 6.022 \times 10^{23} \text{ mol}^{-1} \approx 169 ext{ g mol}^{-1}

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