Question

rubidium is soft, silvery-white metal that has two common isotopes, ^85 Rb and ^87 Rb. if the abundance of ^85 Rb is 80.2% and the abundance of ^87 Rb is 19.8%, what is the average atomic mass of rubidium?

Answer 1

To calculate the average atomic mass of rubidium based on the isotopes and their abundances, we can use the following formula:

\[
\text{Average atomic mass} = \left( \text{isotope mass} \times \text{abundance} \right) + \left( \text{isotope mass} \times \text{abundance} \right)
\]

Given the isotopes:
- \(^85 Rb\) with an abundance of 80.2%
- \(^87 Rb\) with an abundance of 19.8%

We'll use the approximate masses of the isotopes:
- Mass of \(^85 Rb\) = 84.911 amu
- Mass of \(^87 Rb\) = 86.909 amu

First, we convert the percentage abundances to decimal form:
- Abundance of \(^85 Rb\) = 80.2% = 0.802
- Abundance of \(^87 Rb\) = 19.8% = 0.198

Now we can calculate the average atomic mass:

\[
\text{Average atomic mass} = (84.911 \, \text{amu} \times 0.802) + (86.909 \, \text{amu} \times 0.198)
\]

Calculating each term:
1. \(84.911 \, \text{amu} \times 0.802 = 68.169822 \, \text{amu}\)
2. \(86.909 \, \text{amu} \times 0.198 = 17.233782 \, \text{amu}\)

Now, add these two results together:

\[
\text{Average atomic mass} = 68.169822 \, \text{amu} + 17.233782 \, \text{amu} = 85.403604 \, \text{amu}
\]

Thus, the average atomic mass of rubidium is approximately \(85.40 \, \text{amu}\).