# How do you calculate the average atomic mass of rubidium if 72.17% of its atoms have a mass of 84.91 amu and 27.83% of its atoms have a mass of 86.91 amu?

Mar 31, 2017

$\left(0.7217 \cdot 84.91\right) + \left(0.2783 \cdot 86.91\right) = 85.47$amu

#### Explanation:

The average atomic mass of some substance is measured indeed with the percentages of how much they exist in nature. An average is always calculated by adding up the numbers and divide that by the amount of number you had, so, for example, the average for the numbers 1, 3, and two times 7 is:

$\frac{1 + 3 + 7 + 7}{4} = 4.5$

If we now look at the percentage of the numbers encountered we see that we encounter
25% the number 1
25% the number 3
50% the number 7

Following this calculation, the average can also be determined by multiplying the numbers with the percentages and add them up.
$\left(0.25 \cdot 1\right) + \left(0.25 \cdot 3\right) + \left(0.50 \cdot 7\right) = 4.5$

For calculating the average atomic mass this is no different.
We know that
72.17% + 27.83%=100%

From the total of all rubidium, you will encounter 72.17% times the 84.91 amu one, and 27.83% times the 86.91 amu one.
We use exactly the same approach as in the example above:

$\left(0.7217 \cdot 84.91\right) + \left(0.2783 \cdot 86.91\right) = 85.47$amu