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To determine the percentage abundance of the isotopes of Boron with atomic masses 10 and 11, given that the average atomic mass of Boron is 10.80 u, follow these steps:
1. Identify the given data:
- The average atomic mass of Boron: [tex]\( 10.80 \)[/tex] u
- The atomic mass of isotope [tex]\(\text{Boron-10}\)[/tex]: [tex]\( 10 \)[/tex] u
- The atomic mass of isotope [tex]\(\text{Boron-11}\)[/tex]: [tex]\( 11 \)[/tex] u
2. Setup the equation for average atomic mass:
Let [tex]\( x \)[/tex] be the fractional abundance of isotope [tex]\(\text{Boron-10}\)[/tex]. Therefore, the fractional abundance of isotope [tex]\(\text{Boron-11}\)[/tex] will be [tex]\((1 - x)\)[/tex].
The average atomic mass of Boron can be expressed as:
[tex]\[ \text{Average atomic mass} = (\text{mass of Boron-10}) \times (\text{fractional abundance of Boron-10}) + (\text{mass of Boron-11}) \times (\text{fractional abundance of Boron-11}) \][/tex]
In terms of our variables:
[tex]\[ 10.80 = 10x + 11(1 - x) \][/tex]
3. Simplify the equation:
Substitute and distribute:
[tex]\[ 10.80 = 10x + 11 - 11x \][/tex]
Combine like terms:
[tex]\[ 10.80 = 11 - x \][/tex]
4. Solve for [tex]\( x \)[/tex]:
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ x = 11 - 10.80 \][/tex]
Calculate the value of [tex]\( x \)[/tex]:
[tex]\[ x = 0.20 \][/tex]
So, the fractional abundance of isotope [tex]\(\text{Boron-10}\)[/tex] is [tex]\( 0.20 \)[/tex] or 20%.
5. Calculate the fractional abundance of isotope [tex]\(\text{Boron-11}\)[/tex]:
[tex]\[ 1 - x = 1 - 0.20 = 0.80 \][/tex]
Thus, the fractional abundance of isotope [tex]\(\text{Boron-11}\)[/tex] is [tex]\( 0.80 \)[/tex] or 80%.
6. Convert fractional abundances to percentages:
- Percentage abundance of [tex]\(\text{Boron-10}\)[/tex]:
[tex]\[ 0.20 \times 100 = 20\% \][/tex]
- Percentage abundance of [tex]\(\text{Boron-11}\)[/tex]:
[tex]\[ 0.80 \times 100 = 80\% \][/tex]
Therefore, the percentage abundance of isotope [tex]\(\text{Boron-10}\)[/tex] is approximately [tex]\( 20\% \)[/tex] and the percentage abundance of isotope [tex]\(\text{Boron-11}\)[/tex] is approximately [tex]\( 80\% \)[/tex].
1. Identify the given data:
- The average atomic mass of Boron: [tex]\( 10.80 \)[/tex] u
- The atomic mass of isotope [tex]\(\text{Boron-10}\)[/tex]: [tex]\( 10 \)[/tex] u
- The atomic mass of isotope [tex]\(\text{Boron-11}\)[/tex]: [tex]\( 11 \)[/tex] u
2. Setup the equation for average atomic mass:
Let [tex]\( x \)[/tex] be the fractional abundance of isotope [tex]\(\text{Boron-10}\)[/tex]. Therefore, the fractional abundance of isotope [tex]\(\text{Boron-11}\)[/tex] will be [tex]\((1 - x)\)[/tex].
The average atomic mass of Boron can be expressed as:
[tex]\[ \text{Average atomic mass} = (\text{mass of Boron-10}) \times (\text{fractional abundance of Boron-10}) + (\text{mass of Boron-11}) \times (\text{fractional abundance of Boron-11}) \][/tex]
In terms of our variables:
[tex]\[ 10.80 = 10x + 11(1 - x) \][/tex]
3. Simplify the equation:
Substitute and distribute:
[tex]\[ 10.80 = 10x + 11 - 11x \][/tex]
Combine like terms:
[tex]\[ 10.80 = 11 - x \][/tex]
4. Solve for [tex]\( x \)[/tex]:
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ x = 11 - 10.80 \][/tex]
Calculate the value of [tex]\( x \)[/tex]:
[tex]\[ x = 0.20 \][/tex]
So, the fractional abundance of isotope [tex]\(\text{Boron-10}\)[/tex] is [tex]\( 0.20 \)[/tex] or 20%.
5. Calculate the fractional abundance of isotope [tex]\(\text{Boron-11}\)[/tex]:
[tex]\[ 1 - x = 1 - 0.20 = 0.80 \][/tex]
Thus, the fractional abundance of isotope [tex]\(\text{Boron-11}\)[/tex] is [tex]\( 0.80 \)[/tex] or 80%.
6. Convert fractional abundances to percentages:
- Percentage abundance of [tex]\(\text{Boron-10}\)[/tex]:
[tex]\[ 0.20 \times 100 = 20\% \][/tex]
- Percentage abundance of [tex]\(\text{Boron-11}\)[/tex]:
[tex]\[ 0.80 \times 100 = 80\% \][/tex]
Therefore, the percentage abundance of isotope [tex]\(\text{Boron-10}\)[/tex] is approximately [tex]\( 20\% \)[/tex] and the percentage abundance of isotope [tex]\(\text{Boron-11}\)[/tex] is approximately [tex]\( 80\% \)[/tex].
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