Boron has two naturally occurring isotopes, boron-10 with a mass of 10.0129 amu and boron-11 with a mass of 11.00931 amu. The atomic weight of boron is 10.811. What is the fractional abundance of the boron-11 isotope?
The Correct Answer and Explanation is :
To determine the fractional abundance of boron-11, we can use the formula for calculating the atomic weight (average atomic mass) of an element based on the masses and abundances of its isotopes:
[
\text{Atomic Weight} = (m_1 \times f_1) + (m_2 \times f_2)
]
Where:
- (m_1) and (m_2) are the masses of the isotopes (boron-10 and boron-11),
- (f_1) and (f_2) are their fractional abundances,
- The sum of the fractional abundances must equal 1: (f_1 + f_2 = 1).
Given:
- The mass of boron-10 ((m_1)) = 10.0129 amu,
- The mass of boron-11 ((m_2)) = 11.00931 amu,
- The atomic weight of boron = 10.811 amu.
Let (f_1) be the fractional abundance of boron-10 and (f_2) be the fractional abundance of boron-11. Since (f_1 + f_2 = 1), we can write (f_1 = 1 – f_2).
Substitute these values into the atomic weight equation:
[
10.811 = (10.0129 \times f_1) + (11.00931 \times f_2)
]
Now substitute (f_1 = 1 – f_2):
[
10.811 = (10.0129 \times (1 – f_2)) + (11.00931 \times f_2)
]
Simplify and solve for (f_2):
[
10.811 = 10.0129 – 10.0129 f_2 + 11.00931 f_2
]
[
10.811 – 10.0129 = (11.00931 – 10.0129) f_2
]
[
0.7981 = 0.99641 f_2
]
[
f_2 = \frac{0.7981}{0.99641} \approx 0.801
]
Thus, the fractional abundance of boron-11 is approximately 0.801 or 80.1%.
Explanation:
The fractional abundance of an isotope refers to the proportion of that isotope in a sample of the element. The atomic weight is the weighted average of the masses of the isotopes, taking into account their relative abundances. By solving for the unknown abundance of boron-11, we used the known atomic weight of boron (10.811) and the masses of the isotopes to set up a system of equations. This method allows us to accurately determine the fractional abundance of boron-11 in naturally occurring boron.