# Question 85c8e

Jun 22, 2017

Here's what I got.

#### Explanation:

The idea here is that each isotope will contribute to the average atomic mass of the element proportionally to their respective abundance.

In other words, the more abundant the isotope, the more its atomic mass will contribute to the average atomic mass of the element.

Now, you know that your element has an average atomic mass equal to $z$ and that its has $2$ naturally occurring isotopes that contribute to the average atomic mass.

Notice that the average atomic mass is closer in value to the atomic mass of the lighter isotope. This lets you know that the lighter isotope is more abundant than the heavier isotope.

If you take $x$ to be the decimal abundance, i.e. the percent abundance divided by 100%, of the isotope that has an atomic mass of $\left(z - 1\right)$, you can say that the percent abundance of the second isotope will be equal to $\left(1 - x\right)$.

You can thus say that you have

$z = \left(z - 1\right) \cdot x + \left(z + 2\right) \cdot \left(1 - x\right)$

This will be equivalent to

$\textcolor{red}{\cancel{\textcolor{b l a c k}{z}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{z x}}} - x + \textcolor{red}{\cancel{\textcolor{b l a c k}{z}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{z x}}} + 2 - 2 x$

$3 x = 2 \implies x = \frac{2}{3}$

Consequently, the decimal abundance of the second isotope will be

$1 - x = 1 - \frac{2}{3} = \frac{1}{3}$

To convert these to percent abundances, simply multiply the two values by 100%.

You will end up with

"For (z-1): " 2/3 * 100% = 66.7%

"For (z+2): " 1/3 * 100% = 33.3%#

As predicted, the lighter isotope is indeed more abundant than the heavier isotope.