Atomic Structure and Relative Atomic Mass for the ESAT

Atomic Number and Mass Number

All atoms of a specific element contain the same number of protons in their nuclei. This value is known as the atomic number. While the atomic number identifies the element, the mass of atoms within that element can vary. This variation is due to the presence of different numbers of neutrons. The mass number is the sum of the protons and neutrons in an atom's nucleus.

To identify a particular atom or ion, we use a standard notation where the mass number is written as a superscript and the atomic number as a subscript to the left of the chemical symbol.

Calculating Subatomic Particles

In a neutral atom, the number of electrons is equal to the atomic number (protons). In an ion, the number of electrons changes based on the charge.

1. Protons: Equal to the atomic number.
2. Neutrons: Calculated as $\text{Mass Number} - \text{Atomic Number}$.
3. Electrons: In a neutral atom, this equals the protons. For an ion, electrons = $\text{Atomic Number} - \text{Charge}$.

Worked Example: Calcium Ion For the ion ${}^{41}_{20}\text{Ca}^{2+}$:

• The atomic number is 20, so there are 20 protons.
• The mass number is 41. Neutrons = $41 - 20 = 21$.
• The charge is $2+$. Electrons = $20 - 2 = 18$.

Isotopes

Isotopes are defined as atoms of the same element that have the same number of protons but different numbers of neutrons. This means they have the same atomic number but different mass numbers. The term comes from the Greek roots isos (equal) and topos (place), referring to the fact that isotopes occupy the same position in the Periodic Table.

Hydrogen provides a classic example with three isotopes:

• Protium (${}^{1}_{1}\text{H}$): 1 proton, 0 neutrons.
• Deuterium (${}^{2}_{1}\text{H}$): 1 proton, 1 neutron.
• Tritium (${}^{3}_{1}\text{H}$): 1 proton, 2 neutrons.

Mass Spectrometry

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

The x-axis shows the $m/z$ value (effectively the mass of the ion if the charge is $+1$), while the y-axis shows the relative abundance or the number of ions detected. In some spectra, the most abundant peak is set to 100 percent, and others are scaled accordingly.

Interpreting Spectra: Neon and Boron In the neon spectrum above, there are three peaks, meaning neon has three isotopes with masses 20, 21, and 22. For boron, the spectrum shows two peaks at $m/z$ 10 and 11.

The ratio of the peaks for boron is 1:4. This indicates that for every five atoms, four have a mass of 11 and one has a mass of 10. Thus, boron-11 (${}^{11}_{5}\text{B}$) has an abundance of 80 percent and boron-10 (${}^{10}_{5}\text{B}$) has an abundance of 20 percent.

Diatomic Elements: Bromine Bromine exists as $\text{Br}_2$ molecules. Its mass spectrum shows peaks for individual atoms (${}^{79}\text{Br}$ and ${}^{81}\text{Br}$) and for molecular ions (${}^{158}\text{Br}_2$, ${}^{160}\text{Br}_2$, and ${}^{162}\text{Br}_2$).

Because the ${}^{79}\text{Br}$ and ${}^{81}\text{Br}$ peaks are equal in height, the isotopes are 50 percent abundant each. The molecular peaks occur in a 1:2:1 ratio because there are two ways to form a mass 160 molecule (${}^{79}\text{Br}-{}^{81}\text{Br}$ and ${}^{81}\text{Br}-{}^{79}\text{Br}$).

Relative Atomic Mass ($A_r$)

The relative atomic mass ($A_r$) is the weighted mean of the mass numbers of the isotopes of an element. The term 'relative' signifies that these masses are compared to a standard: $1/12$ the mass of an atom of carbon-12.

To calculate $A_r$ from percentage data: $A_r(X) = \frac{(a \times q) + (b \times r) + ...}{100}$ Where $a, b$ are percentages and $q, r$ are isotopic masses.

Worked Example: Chlorine Chlorine is 75 percent ${}^{35}\text{Cl}$ and 25 percent ${}^{37}\text{Cl}$. $A_r(\text{Cl}) = (\frac{75}{100} \times 35) + (\frac{25}{100} \times 37) = 26.25 + 9.25 = 35.5$

If the data is provided as relative abundances (peak heights) rather than percentages, use the following formula: $A_r(X) = \frac{(a \times q) + (b \times r) + ...}{(a + b + ...)}$

Worked Example: Copper Using the mass spectrum for copper:

• Mass 63 has a height of 35.
• Mass 65 has a height of 15.
• Total abundance = $35 + 15 = 50$. $A_r(\text{Cu}) = \frac{(35 \times 63) + (15 \times 65)}{50} = \frac{2205 + 975}{50} = \frac{3180}{50} = 63.6$