Question

Chlorine is made up of two isotopes, $\mathrm{Cl}-35(34.969\mathrm{u})$ and $\mathrm{Cl}-37(36.966\mathrm{u})$. Given chlorine's atomic weight is $35.453\mathrm{u}$, what is the percent abundance of each isotope?

Answer 1

Given:

Atomic mass of $\mathrm{Cl}-35=34.969\mathrm{u}$

Atomic mass of $\mathrm{Cl}-37=36.966\mathrm{u}$

Atomic weight of chlorine$=35.453\mathrm{u}$

Formula used:

$\begin{array}{rcl}\mathrm{Average}\mathrm{atomic}\mathrm{mass}& =& \frac{\mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}1\times \%\mathrm{of}\mathrm{occurence}+\mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}2\times \%\mathrm{of}\mathrm{occurence}}{100}\end{array}$

Calculate percent abundance of isotopes:

Let $\mathrm{x}$ be the abundance of the $\mathrm{Cl}-35$ isotope and $(1-\mathrm{x})$ be the abundance of the $\mathrm{Cl}-37$ isotope.

Thus,

$\mathrm{x}\%$ of $\mathrm{Cl}-35$ has atomic mass$=34.969\mathrm{u}$

$(1-\mathrm{x})\%$ of $\mathrm{Cl}-37$ has atomic mass$=36.966\mathrm{u}$

Thus,

$\begin{array}{rcl}\mathrm{Average}\mathrm{atomic}\mathrm{mass}& =& \frac{\mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}1\times \%\mathrm{of}\mathrm{occurence}+\mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}2\times \%\mathrm{of}\mathrm{occurence}}{100}\\ 35.453& =& \left(34.969\right)\left(\mathrm{x}\right)+\left(36.966\right)\left(1-\mathrm{x}\right)\\ 35.453& =& 34.969\mathrm{x}+36.966-36.99\mathrm{x}\\ 35.453-36.966& =& 34.969\mathrm{x}-36.966\mathrm{x}\\ 1.513& =& 1.997\mathrm{x}\\ \mathrm{x}& =& 0.76\end{array}$

On multiplying with $100$, percentage abundance will be obtained. Thus,

Percent abundance of $\mathrm{Cl}-35=\mathrm{x}\%$

Percent abundance of $\mathrm{Cl}-35=0.76\times 100\%$

Percent abundance of $\mathrm{Cl}-35=76\%$

Thus, the percent abundance of $\mathrm{Cl}-35$ isotope is $76\%$.

Percent abundance of $\mathrm{Cl}-37=(1-\mathrm{x})\%$

Percent abundance of $\mathrm{Cl}-37=(1-0.76)\times 100\%$

Percent abundance of $\mathrm{Cl}-37=(0.24)\times 100\%$

Percent abundance of $\mathrm{Cl}-37=24\%$

Thus, the percent abundance of $\mathrm{Cl}-37$ isotope is $24\%$.