# Question c4218

Apr 6, 2016

$\text{12.667%}$

#### Explanation:

Right from the start, you can look at the atomic masses of the two isotopes and at the average atomic mass of the element and say that isotope $\text{^109"X}$ will be more abundant than the other isotope.

An element's average atomic mass is calculated by taking the weighted average of the atomic masses of its stable isotopes. In other words, each isotope will contribute to the average mass of the element in proportion to its abundance.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{avg. atomic mass" = sum_i "isotope"_i xx "abundance}}_{i} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Since the average atomic mass of $\text{106.33 u}$ is closer to the mass of the $\text{^106"X}$ isotope, it follows that this isotope is more abundant than the other.

If you take $x$ to be the decimal abundance of the first isotope, you can say that $y$, the decimal abundance of the $\text{^109"X}$ isotope, will be equal to

$y = 1 - x$

This happens because the decimal abundances of the two isotopes must be equal to $1$, since these are the only isotopes of element $\text{X}$.

You can thus say that the average atomic mass of element $\text{X}$ is equal to

"106.33" color(red)(cancel(color(black)(u))) = overbrace(x xx "105.95" color(red)(cancel(color(black)(u))))^(color(purple)("the contribution of" color(white)(a)""^106"X")) + overbrace((1-x) xx "108.95" color(red)(cancel(color(black)(u))))^(color(red)("the contribution of"color(white)(a)""^109"X"))

Solve this equation for $x$ to get

$106.33 - 108.95 = x \cdot \left(105.95 - 108.95\right)$

$x = \frac{2.62}{3} = 0.873333$

Since $x$ represents the decimal abundance of the lighter isotope, you can say that the decimal abundance of the heavier isotope will be equal to

$1 - 0.873333 = 0.126667$

The percent abundances of the two isotopes, which are simply the decimal abundances multiplied by $100$, will be

""^106"X: " color(green)(|bar(ul(color(white)(a/a)"87.333%"color(white)(a/a)|)))

""^109"X: " color(green)(|bar(ul(color(white)(a/a)"12.667%"color(white)(a/a)|)))#

As predicted, the lighter isotope contributes more to the average mass of the element because it is more abundant than the heavier isotope.