Isotopes and Mass Spectrometry for the ESAT

Defining Isotopes

All atoms belonging to a specific element possess the same number of protons in their nuclei, which is known as the atomic number. However, the number of neutrons in the nucleus of an atom of a particular element can vary. When atoms of the same element have different numbers of neutrons, they have different mass numbers. These varieties are called isotopes. The term originates from the Greek words 'isos', meaning equal, and 'topos', meaning place, reflecting the fact that isotopes occupy the same position in the Periodic Table.

A specific isotope is represented using standard notation that displays both the mass number and the atomic number alongside the element symbol.

For example, hydrogen exists in three isotopic forms: $^1_1H$, $^2_1H$, and $^3_1H$. Each of these contains exactly one proton. $^1_1H$ has zero neutrons, $^2_1H$ has one neutron, and $^3_1H$ has two neutrons.

Principles of Mass Spectrometry

To determine the number and relative abundance of different isotopes in a sample, chemists use a mass spectrometer. In this device, atoms or molecules are converted into positive ions through ionisation. These ions are then accelerated and separated according to their mass-to-charge ratio ($m/z$) as they move through the instrument. A detector records the number of ions arriving for each specific $m/z$ value.

The resulting mass spectrum is a graph plotting the relative number of ions on the $y$-axis against the $m/z$ values on the $x$-axis. The $y$-axis may use arbitrary units or percentage relative abundance. If relative abundance is used, the highest peak is usually set to 100 percent, and all other peaks are scaled relative to it.

Interpreting Mass Spectra

The number of peaks in a mass spectrum indicates the number of isotopes present in the sample. For instance, the mass spectrum of neon shows three distinct peaks, signifying that neon has three isotopes with mass numbers of 20, 21, and 22.

In the case of boron, the spectrum reveals two peaks at $m/z$ values of 10 and 11.

By comparing the heights of these peaks, we can determine the abundance ratio. For boron, the ratio of the peak at $m/z$ 10 to the peak at $m/z$ 11 is 1:4. This indicates that 80 percent of the atoms in this sample are boron-11 and 20 percent are boron-10.

Relative Atomic Mass Calculations

The relative atomic mass ($A_r$) is the weighted mean of the mass numbers of all isotopes of an element. This value is calculated relative to 1/12th of the mass of a carbon-12 atom. Because it is a ratio, $A_r$ does not have units.

When data is provided as percentages, use the following formula:

$A_r(X) = \frac{(\text{percentage } a \times \text{mass } q) + (\text{percentage } b \times \text{mass } r) + \dots}{100}$

If the data is provided as relative abundances from a mass spectrum, use the sum of the abundances in the denominator:

$A_r(X) = \frac{(\text{abundance } a \times \text{mass } q) + (\text{abundance } b \times \text{mass } r) + \dots}{\text{total abundance}}$

Worked Example: Chlorine

Consider a sample of chlorine where 75 percent of the atoms are $^{35}_{17}Cl$ and 25 percent are $^{37}_{17}Cl$. The calculation for the relative atomic mass is:

$A_r(Cl) = (\frac{75}{100} \times 35) + (\frac{25}{100} \times 37)$

$A_r(Cl) = 26.25 + 9.25 = 35.5$

Worked Example: Copper

Using the mass spectrum of copper below, we can calculate its relative atomic mass.

The $y$-axis shows 35 units for mass 63 and 15 units for mass 65. The total relative abundance is $35 + 15 = 50$.

$A_r(Cu) = \frac{(35 \times 63) + (15 \times 65)}{50}$

$A_r(Cu) = \frac{2205 + 975}{50} = \frac{3180}{50} = 63.6$

Analyzing Molecular Spectra

When dealing with diatomic molecules like bromine ($Br_2$), the mass spectrum is more complex. Single bromine atoms ($Br^+$) appear as peaks at $m/z$ 79 and 81 in equal heights, showing a 1:1 ratio. However, molecular ions ($Br_2^+$) appear at $m/z$ 158, 160, and 162. These result from different combinations of the two isotopes. Because there are two ways to form a molecule with a mass of 160 ($79+81$ and $81+79$), that peak is twice as high as the peaks for 158 ($79+79$) and 162 ($81+81$).