A periodic table might list the average atomic mass of magnesium as being 24.312u, which the result of weighting the atomic masses of the magnesium isotopes is according to their natural abundances on Earth. The three isotopes and their masses are ${}^{\mathbf{24}}\mathbf{Mg}\mathbf{(}\mathbf{23}\mathbf{.}\mathbf{98504}\mathbf{u}\mathbf{)}$, ${}^{\mathbf{25}}\mathbf{Mg}\mathbf{(}\mathbf{23}\mathbf{.}\mathbf{98584}\mathbf{u}\mathbf{)}$, and${}^{\mathbf{26}}\mathbf{Mg}\mathbf{(}\mathbf{23}\mathbf{.}\mathbf{9859}\mathbf{u}\mathbf{)}$. The natural abundance of${}^{\mathbf{24}}\mathbf{Mg}$is 78.99%by mass (that is, 78.99%of the mass of a naturally occurring sample of magnesium is due to the presence of${}^{\mathbf{24}}\mathbf{Mg}$).What is the abundance of (a)${}^{\mathbf{25}}\mathit{M}\mathit{g}$and (b)${}^{\mathbf{26}}\mathit{M}\mathit{g}$?

1. Average atomic mass of magnesium,$M_{mg}=24.312u$
2. The mass of the isotope ${}^{24}\mathrm{Mg}$,$M_1=23.98504u$
3. The mass of the isotope ${}^{25}\mathrm{Mg}$,$M_2=24.98584u$
4. The mass of the isotope ${}^{26}\mathrm{Mg}$,$M_3=24.98259u$
5. The natural abundance of the isotope${}^{24}\mathrm{Mg}$ is 78.99% of its mass.

The abundance of the three isotopes represents their percent of mass percent in the environment. Considering that there are three isotopes of magnesium, we can get an equation of the total abundance of the three isotopes to be 100% of the natural magnesium substance. Thus, using this concept, we can calculate the abundance of the two other isotopes that are unknown.

Let $f_{24}$be the abundance of ${}^{24}\mathrm{Mg}$, let $f_{25}$be the abundance of ${}^{25}\mathrm{Mg}$, and let$f_{26}$ be the abundance of ${}^{26}\mathrm{Mg}$. Then, the entry in the periodic table for Mg is:

$24.312=23.98504f_{24}+24.98584f_{25}+25.98259f_{26}$ ….. (a)

Since, there are only three isotopes $f_{24}+f_{25}+f_{26}=1$. Solving for$f_{25}$ and $f_{26}$, we have the above equation as follows:

$f_{26}=1-f_{24}-f_{25}$ …… (b)

Now, substituting this above value and $f_{24}=0.7899$in equation (a), determine the abundance of isotope ${}^{25}\mathrm{Mg}$as follows:

role="math" localid="1661585620014" $$\begin{gathered} 24.312=23.98504\left(0.7899\right)+24.98259-24.98259\left(0.7899\right)-25.98259\left(f_{25}\right) \\ {f}_{25}=0.09303 \\ {f}_{25}=9.303\% \end{gathered}$$

Hence, the value of abundance is 9.303%.

Now, substitute the values$f_{24}=0.7899$ androle="math" localid="1661585674376" $f_{25}=0.09303$ in equation (b), solve to obtain theabundance of isotope${}^{26}Mg$ as follows:

$\begin{array}{l}{f}_{26}=1-0.7899-0.09303\\ =0.1171\\ =11.71\%\end{array}$

Hence, the value of abundance is 11.71%.