Q. how to calculate percent abundance of isotopes

We will explain, step by step, how to calculate percent abundance of isotopes for the common two‑isotope case. The same ideas extend to three or more isotopes, and I will give a numerical example (chlorine) to show the arithmetic.

Step 1. Define variables. Let \(m_{1}\) and \(m_{2}\) be the masses (in atomic mass units) of the two isotopes. Let \(x\) be the fractional abundance of isotope 1, and therefore \(1-x\) is the fractional abundance of isotope 2. Let \(M\) be the measured average atomic mass shown on the periodic table. The weighted average relation is

\[
x\,m_{1} + (1 – x)\,m_{2} = M
\]

Step 2. Expand and solve algebraically for \(x\). First expand the left hand side:

\[
x\,m_{1} + m_{2} – x\,m_{2} = M
\]

Collect the terms containing \(x\):

\[
x\,(m_{1} – m_{2}) + m_{2} = M
\]

Isolate the \(x\) term:

\[
x\,(m_{1} – m_{2}) = M – m_{2}
\]

Finally solve for \(x\):

\[
x = \frac{M – m_{2}}{m_{1} – m_{2}}
\]

Step 3. Convert the fractional abundance \(x\) to a percent abundance by multiplying by \(100\%\). The percent abundance of isotope 1 is \(x \times 100\%\). The percent abundance of isotope 2 is \((1 – x)\times 100\%\).

Step 4. Numerical example: chlorine. The common isotope masses and atomic mass are approximately \(m_{1}=34.96885\) for \(\text{Cl-35}\), \(m_{2}=36.96590\) for \(\text{Cl-37}\), and \(M=35.453\) for the average atomic mass of chlorine. Substitute these values into the formula:

\[
x = \frac{35.453 – 36.96590}{34.96885 – 36.96590}
\]

Compute numerator and denominator separately:

\[
35.453 – 36.96590 = -1.51290
\]

\[
34.96885 – 36.96590 = -1.99705
\]

Divide to obtain the fractional abundance:

\[
x = \frac{-1.51290}{-1.99705} \approx 0.75765
\]

Convert to percent:

\[
\%\,\text{Cl-35} = 0.75765 \times 100\% \approx 75.77\%
\]

The other isotope abundance is

\[
\%\,\text{Cl-37} = (1 – 0.75765)\times 100\% \approx 24.23\%
\]

Summary. To find percent abundances for two isotopes, set up the weighted average equation \(x\,m_{1} + (1-x)\,m_{2} = M\), solve for \(x\) with \(x = \dfrac{M – m_{2}}{m_{1} – m_{2}}\), and multiply the resulting fraction by \(100\%\) to get percent abundance. For chlorine this gives about \(75.77\%\) \(\text{Cl-35}\) and \(24.23\%\) \(\text{Cl-37}\).