# Question 32c89

Oct 7, 2017

$\text{15.7 u}$

#### Explanation:

The idea here is that each isotope will contribute to the average atomic mass of the element in proportion to its abundance.

In other words, the more abundant an isotope is, the more its atomic mass will contribute to the average atomic mass of the element, i.e. the closer the average atomic mass of the element will be to the atomic mass of the isotope.

In this case, you know that element $\text{Z}$ has two stable isotopes

• $\text{Isotope 1: " "15.0 u", 30% = 3/10 color(white)(.)"abundance}$
• $\text{isotope 2: " "16.0 u", 70% = 7/10color(white)(.)"abundance}$

Even without doing any calculation, you should be able to say that average atomic mass of $\text{Z}$ will be closer in value to $\text{16.0 u}$ than to $\text{15.0 u}$ because the second isotope is more abundant than the first one.

To actually calculate the average atomic mass, use the atomic masses of the isotopes and their decimal abundances.

"avg. atomic mass" = overbrace("15.0 u" * 3/10)^(color(blue)("the contribution of isotope 1")) + overbrace("16.0 u" * 7/10)^(color(blue)("the contribution of isotope 2"))#

This will get you

$\text{avg. atomic mass" = "4.5 u" + "11.2 u}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{avg. atomic mass = 15.7 u}}}}$

As predicted, the average atomic mass of element $\text{Z}$ is closer in value to the atomic mass of the second isotope because this isotope is more abundant.