Answer:
[tex]\%_{Rb-85}=33.9\%[/tex]
[tex]\%_{Rb-87}=66.1\%[/tex]
Explanation:
Hello,
In this case, for the natural occurring isotopes we equal the average atomic mass via:
[tex]86.231=84.911*\%_{Rb-85}+86.909*\%_{Rb-87}[/tex]
Thus, since both percentages of abundance must turn out 100%, we can write:
[tex]\%_{Rb-85}+\%_{Rb-87}=100\%\\\\\%_{Rb-85}=100\%-\%_{Rb-87}[/tex]
So we can write:
[tex]86.231=84.911*(100\%-\%_{Rb-87})+86.909*\%_{Rb-87}[/tex]
Solving for the percentage of abundance of Rb-87:
[tex]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\%[/tex]
Therefore, the percentage of abundance of Rb-85 turns out:
[tex]\%_{Rb-85}=100\%-66.1\%\\\\\%_{Rb-85}=33.9\%[/tex]
Best regards.
Considering the definition of atomic mass, isotopes and atomic mass of an element, the percentages of the isotopes in this sample are:
First of all, the atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.
The same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element.
On the other hand, the atomic mass of an element is the weighted average mass of its natural isotopes. In other words, the atomic masses of chemical elements are usually calculated as the weighted average of the masses of the different isotopes of each element, taking into account the relative abundance of each of them.
In this case, a certain sample of rubidium has just two isotopes, 85Rb (mass = 84.911amu) and 87Rb (mass = 86.909amu). Then, the average mass of lithium can be calculated as:
84.911× percent of Rb-85 + 86.909× percent of Rb-87= 86.231
Since both abundance percentages must be 100%, you can write:
percent of Rb-85 + percent of Rb-87= 100
Then:
percent of Rb-85 = 100% - percent of Rb-87
So, replacing this expression in the first equation:
84.911× (100% - percent of Rb-87) + 86.909× percent of Rb-87= 86.231
Solving:
84.911×100% - 84.911× percent of Rb-87 + 86.909× percent of Rb-87= 86.231
8491.1% -84.911× percent of Rb-87 + 86.909× percent of Rb-87= 86.231
-84.911× percent of Rb-87 + 86.909× percent of Rb-87= 86.231 - 8491.1%
-84.911× percent of Rb-87 + 86.909× percent of Rb-87= 86.231 - 84.911
1.998× percent of Rb-87= 86.231 - 84.911
1.998× percent of Rb-87= 1.32
percent of Rb-87= 1.32÷ 1.998
percent of Rb-87= 0.6607
percent of Rb-87= 66.07%
Therefore, the percentage of abundance of Rb-85 is:
percent of Rb-85 = 100% - percent of Rb-87
percent of Rb-85 = 100% - 66.07%
percent of Rb-85 = 33.93%
Finally, the percentages of the isotopes in this sample are:
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