Earlier in this lesson, you learned that the atomic mass of an element shown on the periodic table is a weighted average that accounts for the abundance of each isotope in nature. You can calculate the atomic mass of any element if you know its number of naturally occurring isotopes, their masses, and their percent abundances. It is calculated using an equation.
Equation for Calculating Atomic Mass
\(\text{Average Atomic Mass} = \) \(\left( \frac{\text{% isotope 1}}{100} \times \text{mass of isotope 1} \right) + \) \(\left( \frac{\text{% isotope 2}}{100}\times \text{mass of isotope 2} \right) + \ \ldots \)
Let's Watch
In the video you will learn how to calculate the atomic mass of an element by seeing two example problems. As you watch the video, pay attention to how the percent abundance for each isotope is used to calculate the average as a weighted average.
You may want to use the study guide to follow along. If so, click below to download the study guide.
The first problem reads:
Chlorine has two stable isotopes, chlorine-35 and chlorine-37. Chlorine-35, which has a mass of 34.97 u, makes up 75.76 percent of the naturally occurring chlorine. Chlorine-37, which has a mass of 36.97 u, makes up 24.24 percent. What is the average atomic mass of chlorine?
And there’s the equation that we’re going to use to calculate this. Alright, so let’s plug in the values that we know, and calculate from there. Average atomic mass is going to be equal to the percentage of isotope 1 divided by 100. So percentage of isotope 1, that’s 75.75 percent, divided by 100, times the mass of isotope 1, which is 34.97 unified atomic mass units. And we’re going to add to that the percent frequency of isotope 2, chlorine-37, which is 24.24 percent, divided by 100, times the mass of chlorine 37, which is 36.97 unified atomic mass units. If we simplify each of these terms, the first term multiplies out to 26.49 u, plus the second term is 8.96 u. If we add those two terms together, then we find that the average atomic mass is equal to 35.45 unified atomic mass units.
And you see, if we were to just take the mean of the two atomic masses, we would get a number much closer to 36. But because this is a weighted average, and because the lighter isotope is more common, we got a number much closer to the lighter isotope value, about 35.45 unified atomic mass units. Alright, let’s look at another problem, and I’m going to have you work this one on your own.
The problem reads:
Boron has two stable isotopes, boron-10 and boron-11. Boron-10, which has a mass of 10.01 u, makes up 19.9 percent of the naturally occurring boron, while boron-11, which has a mass of 11.01 u, makes up 80.1 percent. What is the average atomic mass of boron?
Well, our equation is right there, so let’s enter in the values that we know, and calculate from there. The percentage concentration of isotope 1, boron-10, is 19.9 percent. We’re going to divide that by 100, and multiply it by the mass of boron-10, which is 10.01 units. And we’re going to add to that the percentage concentration of boron-11, which is 80.1 percent, divide it by 100, multiplied by the mass of boron-11, which is 11.01 units.
If you multiply out each of these terms, for the first term you’ll get 1.992 unified atomic mass units, plus the second term, which is 8.819 atomic mass units.
Add these together, and you’ll find that the average atomic mass of boron is 10.81 unified atomic mass units. And again, if we had just taken the arithmetic mean of the two atomic masses, it’s clear to see we would have gotten 10.51 u. But because this is weighted more towards the heavier isotope, which is more common, we got a slightly heavier number, 10.81 u.
Question
What values for each isotope of an element are required to calculate the weighted average atomic mass?
percent abundance and atomic mass
