Boron has two isotopes with atomic masses 10 and 11. If its average atomic mass is 10.81, the abundance of the lighter isotope is:
Show Hint
For isotopic abundance problems, use the formula:
\[
\text{Average mass} = \frac{\text{(Mass of isotope 1) \(\times\) (abundance)} + \text{(Mass of isotope 2) \(\times\) (abundance)}}{100}.
\]
Simplify step-by-step for accurate results.
Let \( x \) be the percentage abundance of the lighter isotope (\( \mathrm{^{10}B} \)). The percentage abundance of the heavier isotope (\( \mathrm{^{11}B} \)) is then \( (100 - x)\%\).
The average atomic mass of boron is calculated as:
\[
\text{Average atomic mass} = \frac{(x \cdot 10) + ((100 - x) \cdot 11)}{100}.
\]
Given the average atomic mass is \( 10.81 \):
\[
10.81 = \frac{(x \cdot 10) + ((100 - x) \cdot 11)}{100}.
\]
Simplifying the equation:
\[
10.81 = \frac{10x + 1100 - 11x}{100}.
\]
Combining like terms:
\[
10.81 = \frac{1100 - x}{100}.
\]
Multiplying both sides by 100:
\[
1081 = 1100 - x.
\]
Solving for \( x \):
\[
x = 1100 - 1081 = 19.
\]
Therefore, the abundance of the lighter isotope is \( \mathbf{19\%} \).