# Question 50a44

Dec 6, 2015

$\text{36.938 u}$

#### 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 $\text{35.446 u}$, and that two isotopes contribute to this value

• chlorine-35 $\to \text{ 34.969 u}$ and 75.77% abundance
• chlorine-37 $\to \text{ ? u}$ 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 $100$.

So, if $x$ represents the atomic mass of chlorine-37, you can say that

$\text{35.446 u" = "34.969 u} \times 0.7577 + x \cdot 0.2423$

Now all you have to do is solve this equation for $x$

$x \cdot 0.2423 = \text{35.446 u" - "34.969 u} \times 0.7577$

$x = \textcolor{g r e e n}{\text{36.938 u}}$

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.