Atomic Structure and Relative Atomic Mass

Isotopes and Atomic Notation

Atoms of the same element always possess the same number of protons in their nuclei, which defines their atomic number. However, atoms of the same element can have different numbers of neutrons. These variations are known as isotopes. The term isotope is derived from the Greek roots 'isos', meaning equal or same, and 'topos', meaning place. This reflects that isotopes occupy the same position in the Periodic Table because they are the same chemical element.

Since neutrons contribute to an atom's mass, isotopes have different mass numbers. A specific isotope is identified using standard notation, which displays the mass number ($A$) above the atomic number ($Z$) next to the element symbol ($X$).

For example, hydrogen has three isotopes: protium ($^1_1H$), deuterium ($^2_1H$), and tritium ($^3_1H$). While each has 1 proton, they have 0, 1, and 2 neutrons respectively.

Mass Spectrometry and Abundance

The number and relative abundances of an element's isotopes are determined using a mass spectrometer. In this device, atoms are ionised, accelerated, and then separated based on their mass-to-charge ratio ($m/z$) as they drift through the machine. The detected ions produce a mass spectrum.

The mass spectrum is a graph plotting the $m/z$ ratio on the x-axis against the number of ions (abundance) on the y-axis. The y-axis can show 'relative abundance', where the most abundant ion is set to 100% and others are scaled accordingly, or it may use 'arbitrary units'.

In the mass spectrum of neon shown above, there are three peaks. This indicates neon has three isotopes with mass numbers of 20, 21, and 22. In the case of boron, two peaks at $m/z$ 10 and 11 indicate two isotopes. If the ratio of these peaks is 1 to 4, it means there are four times as many boron-11 atoms as boron-10 atoms, corresponding to 80% and 20% abundance respectively.

The Concept of Relative Atomic Mass ($A_r$)

The relative atomic mass, $A_r$, is not a simple average of isotopic masses but a weighted mean. This means it accounts for how common each isotope is. The term 'relative' signifies that these masses are compared to a standard: one twelfth of the mass of a carbon-12 atom ($^{12}_6C$).

Calculation from Percentage Data

If isotopic data is provided as percentages (e.g., $a\%$ of isotope $q$ and $b\%$ of isotope $r$), the general formula is:

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

Worked Example: Chlorine

A sample of chlorine contains 75% $^{35}_{17}Cl$ and 25% $^{37}_{17}Cl$. To find the $A_r$:

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

$A_r(Cl) = \frac{2625 + 925}{100} = \frac{3550}{100} = 35.5$

Calculation from Mass Spectra (Relative Abundance)

When data is presented on a mass spectrum with relative units rather than percentages, we divide by the total sum of the peak heights. If the values on the y-axis are $a$ for mass $q$ and $b$ for mass $r$, the formula is:

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

Worked Example: Copper

Consider a mass spectrum for copper with peaks at $m/z$ 63 (height 35) and $m/z$ 65 (height 15).

1. Calculate the total number of atoms: $35 + 15 = 50$
2. Calculate the weighted sum of masses: $(35 \times 63) + (15 \times 65) = 2205 + 975 = 3180$
3. Divide the sum by the total abundance: $A_r(Cu) = \frac{3180}{50} = 63.6$

Worked Example: Boron

In a sample showing 80% $^{11}B$ and 20% $^{10}B$:

$A_r(B) = (\frac{80}{100} \times 11) + (\frac{20}{100} \times 10) = 8.8 + 2.0 = 10.8$