# Naturally occurring boron is 80.20% boron-11 (atomic mass of 11.01 amu) and 19.80% of some other isotopic form of boron. What must the atomic mass of this second isotope be in order to account for the 10.81 a average atomic mass of boron?

Sep 30, 2016

9.99 amu

#### Explanation:

Background Info
Most elements have different isotopes, or atoms with the same number of protons, but different numbers of neutrons. Hence, it is possible to have two atoms that are the of the same element, but with different masses. Also, for a variety of reasons, some isotopes are more common than others. For example, Carbon-12 is much more common than Carbon-13.

Now, how is this taken into account in the listed mass on the periodic table? Well, we use what is called a weighted average, which takes into account the percentage abundances of each isotopes when taking the average atomic mass of the element.

Actual Solution
So this problem has given you the percent abundances, and the final average atomic mass. Let's go ahead and set this up as an equation:

$\left(\frac{80.2}{100}\right) \left(B 1\right) + \left(\frac{19.80}{100}\right) \left(B 2\right) = 10.81$ amu

B1 and B2 are the atomic masses of each isotope. So is there any additional information we can use? Well, we have the atomic mass of the more abundant isotope, Boron-11. Let's plug that in for B1:

$\left(\frac{80.2}{100}\right) \left(11.01\right) + \left(\frac{19.80}{100}\right) \left(B 2\right) = 10.81$ amu

And that's honestly the most confusing part. From here on, it's just a simple process of algebra as we solve for B2:

$\implies \left(\frac{19.80}{100}\right) \left(B 2\right) = 10.81 - \left(\frac{80.2}{100}\right) \left(11.01\right)$
$\implies \left(B 2\right) = \frac{10.81 - \left(\frac{80.2}{100}\right) \left(11.01\right)}{\frac{19.80}{100}} = 9.99$ amu

Here's some videos by Tyler DeWitt that can help you some more:

Conceptual:
Video 1
Video 2

Examples:
Video 3
Video 4

Hope that helped :)