# Question #50a44

##### 1 Answer

#### Explanation:

The *average atomic mass* of an element is calculated by taking the **weighted average** of the atomic masses of its naturally occurring isotopes.

Simply put, each isotope will contribute to the average mass of the element **proportionally** to its *abundance*.

#color(blue)("avg. atomic mass" = sum_i ("isotope"_i xx "abundance"_i))#

In your case, you know that the average atomic mass of chlorine is equal to

chlorine-35#-> " 34.969 u"# and#75.77%# abundancechlorine-37#-> " ? u"# and#24.23%# abundance

Before doing any calculation, take a second to look at the values you have for the abundances of the two isotopes.

Notice that chjlorine-35 is considerably **more abundant** than chlorine-37, which means that the average atomic mass of the element will be *closer* in value to the atomic mass of chlorine-35.

Moreover, the atomic mass of chlorine-35 is **smaller** than the average atomic mass of chlorine, which means that you expect the atomic mass of chlorine-37 to be **bigger** than the average atomic mass of chlorine.

When calculating the average atomic mass of an element, it's useful to use **decimal abundances**, which are simply *percent abundances* divided by

So, if

#"35.446 u" = "34.969 u" xx 0.7577 + x * 0.2423#

Now all you have to do is solve this equation for

#x * 0.2423 = "35.446 u" - "34.969 u" xx 0.7577#

#x = color(green)("36.938 u")#

I'll leave the answer rounded to five sig figs, despite the fact that you only have four sig figs for the abundances of the two isotopes.