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

The ionisation of P can be represented as:

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

The relative molecular mass of P is 556.

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:

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