Describe the process of electrospray ionisation. Give an equation to represent the ionisation of P in this process. What is the relative molecular mass of P? A molecule Q is ionised by electron impact in a TOF mass spectrometer. The Q+ ion has a kinetic energy of 2.09 x 10^-1 J. This ion takes 1.23 x 10^-6 s to reach the detector. The length of the flight tube is 1.50 m. Calculate the relative molecular mass of Q.

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Describe the process of electrospray ionisation - AQA - A-Level Chemistry - Question 2 - 2019 - Paper 1

Question 2

Describe the process of electrospray ionisation. Give an equation to represent the ionisation of P in this process. What is the relative molecular mass of P? A mo... show full transcript

Worked Solution & Example Answer:Describe the process of electrospray ionisation - AQA - A-Level Chemistry - Question 2 - 2019 - Paper 1

Step 1

Describe the process of electrospray ionisation.

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Answer

Electrospray ionisation involves dissolving or placing the sample in a solvent, which is then sprayed through a fine needle under high voltage (around 4400 volts). This process creates charged droplets that shrink as the solvent evaporates, eventually resulting in the formation of gas-phase ions. In this method, the sample gains a proton, leading to the generation of a positively charged ion.

Step 2

Give an equation to represent the ionisation of P in this process.

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Answer

The ionisation of P can be represented as:

$ext{P} + ext{H}^+ \rightarrow ext{P}^+$

Step 3

What is the relative molecular mass of P?

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Answer

The relative molecular mass of P is 556.

Step 4

Calculate the relative molecular mass of Q.

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Answer

To find the relative molecular mass of Q, we can start by using the kinetic energy formula:

$KE = \frac{1}{2} mv^2$

Rearranging gives: $m = \frac{2 KE}{v^2}$

Where the time taken (t) to reach the detector is 1.23 x 10^-6 s, and the length of the flight tube (d) is 1.50 m. Thus, the speed v can be calculated as:

$v = \frac{d}{t} = \frac{1.50 \text{ m}}{1.23 x 10^{-6} \text{ s}} \approx 1.22 x 10^6 \text{ m/s}$

Substituting this into the kinetic energy equation:

$m = \frac{2 \times 2.09 \times 10^{-1} \text{ J}}{(1.22 \times 10^6)^2} \approx 2.81 \times 10^{-9} \text{ kg}$

To find the relative molecular mass (M) of Q:

$M = m \times L$

Where L = 6.022 x 10^{23} mol^{-1}. Thus:

$M = 2.81 \times 10^{-9} \text{ kg} \times 6.022 \times 10^{23} \text{ mol}^{-1} \approx 169 \text{ g/mol}$

Describe the process of electrospray ionisation - AQA - A-Level Chemistry - Question 2 - 2019 - Paper 1

Worked Solution & Example Answer:Describe the process of electrospray ionisation - AQA - A-Level Chemistry - Question 2 - 2019 - Paper 1

Describe the process of electrospray ionisation.

Give an equation to represent the ionisation of P in this process.

What is the relative molecular mass of P?

Calculate the relative molecular mass of Q.

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