Question

In a periodic table, the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are ${}_{12}{}^{24}\mathrm{Mg}$ (23.98504u), ${}_{12}{}^{25}\mathrm{Mg}$ (24.98584u), and ${}_{12}{}^{26}\mathrm{Mg}$ (25.98259u). The natural abundance of ${}_{12}{}^{24}\mathrm{Mg}$ is 78.99% by mass. Calculate the abundances of the other two isotopes.

Answer 1

Step 1: Assuming abundance of the other two isotopes.

Let the abundance of ${}_{12}{}^{25}\mathrm{Mg}$ is $\mathrm{x}\%$

The abundance of ${}_{12}{}^{24}\mathrm{Mg}$ is $78.99\%$ by mass [given]

So, the abundance of ${}_{12}{}^{26}\mathrm{Mg}$ by mass =

$(100-78.99-\mathrm{x})\%$

=$(21.01-\mathrm{x})\%$

Step 2: Finding the abundance of the other two isotopes.

The average atomic mass of Magnesium$\left(\mathrm{Mg}\right)$ is 24.312 =

$\left(23.98504\times \frac{78.99}{100}\right)+\left(24.98584\times \frac{\mathrm{x}}{100}\right)+\left(25.98259\times \frac{21.01-\mathrm{x}}{100}\right)$

By solving this, $\mathrm{x}=9.303\%\mathrm{for}{}_{12}{}^{25}\mathrm{Mg}$ by mass.

For ${}_{12}{}^{26}\mathrm{Mg}$ = $(21.01\%-\mathrm{x})\%=11.71\%$by mass.

Therefore, abundances for ${}_{12}{}^{25}\mathrm{Mg}\mathrm{and}{}_{12}{}^{26}\mathrm{Mg}\mathrm{are}9.303\%\mathrm{and}11.71\%$ by mass, respectively.