Elemental magnesium consists of three isotopes: 24mg with an accurate mass of 23.99 amu, 25mg with an accurate mass of 24.99 amu, and 26mg with an accurate mass of 25.99 amu. if 24mg is 79% of naturally occurring magnesium, what is the percent abundance of 26mg?

The relative abundances of the three isotopes must be equal to 1. Suppose x is the abundance of 25 mg, and y is the abundance of 26 mg, then,

x + y + 0.79 = 1 --> eqn 1

The average atomic weight of Magnesium is 24.305 amu. Thus,

23.99(0.79) + 24.99(x) + 25.99(y) = 24.305 --> eqn 2

Rearranging eqn 1: x = 1 - 0.79 - y = 0.21 - y. Substituting this to the second equation,

23.99(0.79) + 24.99(0.21 - y) + 25.99(y) = 24.305
Solving for y,
y = 0.105 or 10.5%

Thus, the percent abundance of 26 mg is y which is 10.5%.

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