# Question #85c8e

##### 1 Answer

#### Answer:

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 **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 **decimal abundance**, i.e. the *percent abundance* divided by

You can thus say that you have

#z = (z-1) * x + (z+2) * (1-x)#

This will be equivalent to

#color(red)(cancel(color(black)(z))) = color(red)(cancel(color(black)(zx))) - x + color(red)(cancel(color(black)(z)))-color(red)(cancel(color(black)(zx))) + 2 - 2x#

#3x = 2 implies x = 2/3#

Consequently, the decimal abundance of the second isotope will be

#1 - x = 1 - 2/3 = 1/3#

To convert these to **percent abundances**, simply multiply the two values by

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.