Rhenium has an atomic number of 75 - AQA - A-Level Chemistry - Question 2 - 2022 - Paper 1

Given the relative isotopic mass of 185 with an abundance of 10 and the unknown isotopic mass M with an abundance of 17,

The formula for the relative atomic mass is: $\text{Relative atomic mass} = \frac{(185 \times 10) + (M \times 17)}{10 + 17}$ Substituting the values: $186.3 = \frac{(1850 + 17M)}{27}$ Multiply by 27: $186.3 \times 27 = 1850 + 17M$ $5030.1 = 1850 + 17M$ $17M = 5030.1 - 1850$ $17M = 3180.1$ So, $M = \frac{3180.1}{17} \approx 186.0$

The isotopes of rhenium have the same chemical properties because they have the same number of electrons and thus the same electron configuration.

First, rearrange the kinetic energy formula to find velocity: $KE = \frac{1}{2} mv^2$ So, v = \sqrt{\frac{2 \times KE}{m}}.

Calculating mass (m) for 117Re: $m = \frac{185}{6.022 \times 10^{23}} \times 10^{-3} \text{ kg} = 3.072 \times 10^{-25} \text{ kg}$ Substituting in the kinetic energy: $v = \sqrt{\frac{2 \times 1.153 \times 10^{-3}}{3.072 \times 10^{-25}}} \approx 8.66 \times 10^6 m s^{-1}$ Next, using the flight tube length (d = 1.450 m) to calculate the time (t): $t = \frac{d}{v} = \frac{1.450}{8.66 \times 10^6} \approx 1.67 \times 10^{-7} s$

The relative abundance is determined by measuring the intensity of the detected ions which corresponds to the number of ions hitting the detector; more ions will generate a larger signal.