Q.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 $_{12}^{24}$Mg (23.98504 u), $_{12}^{25}$Mg (24.98584 u) and $_{12}^{26}$Mg (25.98259 u). The natural abundance of $_{12}^{24}$Mg is 78.99% by mass. Calculate the abundances of other two isotopes.

[!TLDR] With $x+y=21.01\%$ (abundances of Mg-25, Mg-26) and the weighted-average equation $0.7899(23.98504)+x(24.98584)+y(25.98259)=24.312$, solving gives $x\approx9.2\%$ (Mg-25) and $y\approx11.8\%$ (Mg-26). [ …

Let x be the fractional abundance of $^{25}Mg$ and y that of $^{26}Mg$, with the remaining abundance being $^{24}Mg$ at 78.99%. Since the three abundances must sum to 100%, $x+y=100\%-78.99\%=21.01\%=0.2101$. The average atomic mass is the abundance-weighted sum of the three isotope masses: $0.7899(23.98504)+x(24.98584)+y(25.98259)=24.312$. Substituting $y=0.2101-x$: $18.9448+24.98584x+25.98259(0.2101-x)=24.312$, which simplifies to $18.9448+5.4590+(24.98584-25.98259)x=24.312$, i.e. $24.4038-0.99675x=24.312$, giving $x=\frac{24.4038-24.312}{0.99675}=\frac{0.0918}{0.99675}\approx0.0921$, so $^{25}Mg$ …

Set up two simultaneous equations -- the abundances summing to the remaining percentage, and the abundance-weighted masses summing to the given a …

Treating the given 78.99% $^{24}Mg$ abundance as leaving the other two isotopes EQUALLY split by default, instead of setting up and solving the weighted-average eq …