# Question 2e766

Oct 2, 2015

90%

#### Explanation:

So, you know that you're dealing with two isotopes, let's say $\text{^20"X}$ and $\text{^22"X}$.

The relative atomic mass of the element will be determined by the atomic masses of the two isotopes in proportion to their respective abundances.

"relativ atomic mass" = sum_i ("isotope"""_i xx "abundance"""_i)

SInce you're only dealing with two isotopes, you can say that their abundances must add to give 100%.

If you take $x$ to be the decimal abundance, which is simply the percent abundance divided by $100$, of $\text{^20"X}$, the abundance of $\text{^22"X}$ will be $\left(1 - x\right)$.

This means that you can write

$\text{20 u" * x + "22 u" * (1-x) = "20.2 u}$

$20 x + 22 - 22 x = 20.2$

$2 x = 1.8 \implies x = \frac{1.8}{2} = 0.9$

The decimal abundance of $\text{^22"X}$ will thus be $\left(1 - 0.9\right) = 0.1$.

The percent abundances of the two isotopes are

""^20"X: " color(green)(90%)
""^22"X: " 10%#

The result makes sense because the relative atomic mass of the element is much closer to the atomic mass of $\text{^20"X}$ than it is to the atomic mass of $\text{^22"X}$, which can only imply that $\text{^20"X}$ has a significantly larger percent abundance that $\text{^22"X}$.
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