# In a periodic table the average atomic mass of magnesium is given as 24.312 u.

Question:

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

Solution:

Average atomic mass of magnesium, m = 24.312 u

Mass of magnesium isotope ${ }_{12}^{24} \mathrm{Mg}, m_{1}=23.98504 \mathrm{u}$

Mass of magnesium isotope ${ }_{12}^{25} \mathrm{Mg}, m_{2}=24.98584 \mathrm{u}$

Mass of magnesium isotope ${ }_{12}^{26} \mathrm{Mg}, m_{3}=25.98259 \mathrm{u}$

Abundance of ${ }_{12}^{24} \mathrm{Mg}, \eta_{1}=78.99 \%$

Abundance of ${ }_{12}^{25} \mathrm{Mg}, \eta_{2}=x \%$

Hence, abundance of ${ }_{12}^{26} \mathrm{Mg}, \eta_{3}=100-x-78.99 \%=(21.01-x) \%$

We have the relation for the average atomic mass as:

$m=\frac{m_{1} \eta_{1}+m_{2} \eta_{2}+m_{3} \eta_{3}}{\eta_{1}+\eta_{2}+\eta_{3}}$

$24.312=\frac{23.98504 \times 78.99+24.98584 \times x+25.98259 \times(21.01-x)}{100}$

$2431.2=1894.5783096+24.98584 x+545.8942159-25.98259 x$

$0.99675 x=9.2725255$

$\therefore x \approx 9.3 \%$

And $21.01-x=11.71 \%$

Hence, the abundance of ${ }_{12}^{25} \mathrm{Mg}$ is $9.3 \%$ and that of ${ }_{12}^{26} \mathrm{Mg}$ is $11.71 \%$.