Calculate the relative atomic mass of an element from the relative masses and abundances of its isotopes

Atoms, Elements and Compounds – Isotopes

Learning Objective

By the end of this lesson you will be able to calculate the relative atomic mass (Ar) of an element from the relative masses and natural abundances of its isotopes.

Key Terms (AO1)

• Isotope – atoms of the same element that have the same number of protons (Z) but different numbers of neutrons.
• Mass number (A) – total number of protons + neutrons in an isotope (a whole number).
• Atomic number (Z) – number of protons; defines the element.
• Relative atomic mass (Ar) – weighted average of the masses of all naturally occurring isotopes of an element, expressed in atomic mass units (u).
• Natural abundance – percentage of each isotope present in a typical sample of the element on Earth.

What are Isotopes? (2.3 Isotopes – Core)

• Isotopes have the same Z (hence the same electron configuration) and therefore exhibit the same chemical behaviour and identical chemical properties.
• They differ only in the number of neutrons, giving different mass numbers (A) and different relative masses.
• Notation: ^A/_Z X.

  
  Example: ³⁵Cl, ²⁴Mg.

Why Do We Need a Weighted Average? (2.3 Isotopes – Supplement)

Because a bulk sample contains a mixture of isotopes, the mass that we measure on a balance (or that appears on the periodic table) is not the mass of a single isotope but the average mass of the mixture. The average is calculated by weighting each isotope’s mass by its fractional natural abundance.

Formula for Relative Atomic Mass

\[

Ar = \sum{i} \bigl( mi \times fi \bigr)

\]

where

\(m_i\) = relative mass of isotope i (in u)

\(f_i\) = fractional abundance of isotope i (percentage ÷ 100).

Step‑by‑Step Procedure (AO2)

1. List each isotope’s relative mass (u).
2. List the natural abundance of each isotope as a percentage.
3. Convert each percentage to a decimal fraction (divide by 100).
4. Multiply the relative mass of each isotope by its decimal abundance.
5. Add all the products – the sum is the element’s relative atomic mass (Ar).
6. Round the final answer only after the calculation; keep at least three significant figures throughout.

Common Pitfalls (AO2)

• Using the mass number (A) instead of the measured relative mass (e.g., 35 u for ³⁵Cl) – the relative mass is slightly different.
• Adding percentages instead of converting to fractions before multiplication.
• Rounding intermediate results; keep extra significant figures until the final step.
• Forgetting that the total fractional abundance should be 1 (or 100 %).

Worked Example – Chlorine

Calculation:

\[

\begin{aligned}

A_r(\text{Cl}) &= 34.969 \times \frac{75.78}{100}

+ 36.966 \times \frac{24.22}{100} \\[4pt]

&= 34.969 \times 0.7578

+ 36.966 \times 0.2422 \\[4pt]

&= 26.49 + 8.95 \\[4pt]

&= 35.44\;\text{u (rounded to 2 d.p.)}

\end{aligned}

\]

The tabulated value is 35.45 u, confirming the method.

Worked Example – Magnesium

Calculation:

\[

\begin{aligned}

A_r(\text{Mg}) &= 23.985 \times 0.7899

+ 24.986 \times 0.1000

+ 25.983 \times 0.1101 \\[4pt]

&= 18.94 + 2.50 + 2.86 \\[4pt]

&= 24.30\;\text{u}

\end{aligned}

\]

The accepted value is 24.305 u – the small difference is due to rounding.

Worked Practice Question – Bromine

Data

Solution

\[

\begin{aligned}

A_r(\text{Br}) &= 78.918 \times \frac{50.69}{100}

+ 80.916 \times \frac{49.31}{100} \\[4pt]

&= 78.918 \times 0.5069

+ 80.916 \times 0.4931 \\[4pt]

&= 40.00 + 39.84 \\[4pt]

&= 79.84\;\text{u (to 2 d.p.)}

\end{aligned}

\]

The periodic‑table value is 79.904 u; the slight difference arises from the rounding of the abundances to two decimal places.

Practice Questions (Unsolved)

1. Natural bromine consists of ⁷⁹Br (mass = 78.918 u, abundance = 50.69 %) and ⁸¹Br (mass = 80.916 u, abundance = 49.31 %). Calculate the relative atomic mass of bromine. (Solution given above)
2. The element X has two isotopes: ^AX (mass = 55.934 u, abundance = 60 %) and ^BX (mass = 56.935 u, abundance = 40 %). Determine Ar for element X.
3. Given the data for element Y, verify that the tabulated relative atomic mass of 63.55 u is correct.
   

   Isotope	Relative mass (u)	Abundance (%)
   58Y	57.933	68.1
   60Y	59.933	31.9

Link‑on (AO3 – Application)

The same weighted‑average method is used later when you calculate:

• Empirical formulae from percent composition.
• Average molar masses of compounds containing more than one element.

Understanding isotopic weighting therefore underpins many later topics in the syllabus.

Summary

• Isotopes share the same number of electrons, giving identical chemical properties.
• The relative atomic mass (Ar) is a weighted average of the relative masses of all naturally occurring isotopes, using fractional natural abundances as the weights.
• Use the formula Ar = Σ (relative mass × fractional abundance), keep extra significant figures during calculation, and round only at the final step.
• Experimental isotopic data come from mass‑spectrometry and are essential for later quantitative chemistry work.