To find the abundances of the isotopes 10B and 11B in a natural sample of boron, we can use the concept of weighted averages based on their respective molar masses and the known average molar mass of the sample.
Let:
- x = abundance of 10B
- y = abundance of 11B
Since these are the only two isotopes considered, we can write:
- x + y = 1 (this is the total abundance)
Next, we can use the average molar mass formula:
- (10.013 g/mol) * x + (11.093 g/mol) * y = 10.81 g/mol
Now, substituting y = 1 – x into the average molar mass equation gives:
(10.013 g/mol) * x + (11.093 g/mol) * (1 – x) = 10.81 g/mol
Expanding this equation:
(10.013 g/mol) * x + (11.093 g/mol) - (11.093 g/mol) * x = 10.81 g/mol
We can combine like terms:
(10.013 - 11.093) * x + 11.093 = 10.81
Now, simplifying further:
-1.08 * x + 11.093 = 10.81
Subtracting 11.093 from both sides gives:
-1.08 * x = 10.81 - 11.093
-1.08 * x = -0.283
Now, dividing by -1.08:
x = 0.262
So, the abundance of the isotope 10B is approximately 26.2%. To find the abundance of 11B, we substitute back:
y = 1 - x = 1 - 0.262 = 0.738
This means the abundance of the isotope 11B is approximately 73.8%.
In summary, the abundances of the isotopes in the sample are:
- 10B: 26.2%
- 11B: 73.8%