To find the percentage abundances of the two isotopes of boron, we can use the concept of weighted averages. Here’s a detailed step-by-step solution:
1. Define the variables and given data:
- Let the fractional abundance of the isotope [tex]${}^{10}$[/tex]B be [tex]\( x \)[/tex].
- The fractional abundance of the isotope [tex]${}^{11}$[/tex]B would therefore be [tex]\( (1 - x) \)[/tex].
- The masses of the isotopes are:
- [tex]${}^{10}$[/tex]B has a mass of 10.
- [tex]${}^{11}$[/tex]B has a mass of 11.
- The average atomic mass of boron is given as 10.80.
2. Set up the weighted average equation:
The weighted average of the atomic masses, taking into account their abundances, will be equal to the average atomic mass. Hence, we write:
[tex]\[
10x + 11(1 - x) = 10.80
\][/tex]
3. Solve the equation for [tex]\( x \)[/tex]:
- Expand and simplify the equation:
[tex]\[
10x + 11 - 11x = 10.80
\][/tex]
- Combine like terms:
[tex]\[
10x - 11x + 11 = 10.80
\][/tex]
[tex]\[
-x + 11 = 10.80
\][/tex]
- Isolate [tex]\( x \)[/tex]:
[tex]\[
-x = 10.80 - 11
\][/tex]
[tex]\[
-x = -0.20
\][/tex]
[tex]\[
x = 0.20
\][/tex]
So, the fractional abundance of [tex]${}^{10}$[/tex]B is 0.20.
4. Find the fractional abundance of [tex]${}^{11}$[/tex]B:
[tex]\[
1 - x = 1 - 0.20 = 0.80
\][/tex]
So, the fractional abundance of [tex]${}^{11}$[/tex]B is 0.80.
5. Convert fractional abundances to percentages:
- For [tex]${}^{10}$[/tex]B:
[tex]\[
0.20 \times 100 = 20\%
\][/tex]
- For [tex]${}^{11}$[/tex]B:
[tex]\[
0.80 \times 100 = 80\%
\][/tex]
Therefore, the percentage abundances of the boron isotopes are:
- [tex]${}^{10}$[/tex]B: 20%
- [tex]${}^{11}$[/tex]B: 80%
