# Question #18199

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that the atomic mass that is listed in the Periodic Table is actually an **average atomic mass** because it is calculated by using the **atomic masses** of the stable isotopes of a given element and their respective abundances.

More specifically, each isotope will contribute to the average atomic mass of the element **proportionally** to its abundance.

For an element that has **stable isotopes**, you can say that

#"avg atomic mass" = sum_i "atomic mass isotope"_i xx "decimal abundance"_i#

Now, you know that chlorine has two stable isotopes, chlorine-35 and chlorine-37.

Since the problem doesn't provide you with the exact atomic masses of the two isotopes, you can say that you have

#"For chlorine-35: " "atomic mass " ~~ " 35 u"# #"For chlorine-37: " "atomic mass " ~~ " 37 u"#

Since there are only **two stable isotopes**, you can say that their percent abundances **must** add up to give **decimal abundances** must add up to give

If you take

This means that you have

#35.453 color(red)(cancel(color(black)("u"))) = overbrace(35 color(red)(cancel(color(black)("u"))) * x)^(color(blue)("the contribution of chlorine-35")) + overbrace(37 color(red)(cancel(color(black)("u"))) * (1-x))^(color(blue)("the contribution of chlorine-37"))#

Rearrange to find the value of

#35.453 = 35x + 37 - 37x#

#2z = 37 - 35.452 implies x = (37 - 35.453)/2 = 0.7735#

Consequently, you will have

#y = 1 - 0.7735 = 0.2265#

This means that the **percent abundances** of the two isotopes will be equal to

#"For chlorine-35: " 77.35%# #"For chlorine-37: " 22.65%#

I'll leave the two values rounded to four **sig figs**.