Question

A certain sample of rubidium has just two isotopes, 85Rb (mass = 84.911amu) and 87Rb (mass = 86.909amu). The atomic mass of this sample is 86.231 amu. What are the percentages of the isotopes in this sample?

Answer 1

Answer:

$\%_{Rb-85}=33.9\%$

$\%_{Rb-87}=66.1\%$

Explanation:

Hello,

In this case, for the natural occurring isotopes we equal the average atomic mass via:

$86.231=84.911*\%_{Rb-85}+86.909*\%_{Rb-87}$

Thus, since both percentages of abundance must turn out 100%, we can write:

$\%_{Rb-85}+\%_{Rb-87}=100\%\\\\\%_{Rb-85}=100\%-\%_{Rb-87}$

So we can write:

$86.231=84.911*(100\%-\%_{Rb-87})+86.909*\%_{Rb-87}$

Solving for the percentage of abundance of Rb-87:

$86.231=84911.00\%-84.911\%_{Rb-87}+86.909*\%_{Rb-87}\\\\\%_{Rb-87}=\frac{86.231-84.911}{-84.911+86.909}\\ \\\%_{Rb-87}=66.1\%$

Therefore, the percentage of abundance of Rb-85 turns out:

$\%_{Rb-85}=100\%-66.1\%\\\\\%_{Rb-85}=33.9\%$

Best regards.