The naturally occurring isotopes of Li are Li-6 of mass 6.015121 amu and Li-7 of mass 7.016003 amu. The atomic mass of Li is 6.9409 amu, the percent abundance can be calculated using the following relation.
Atomic mass=m(Li-6 )×%(Li-6 )+m(Li-7 )×%(Li-7 )
Let the percent abundance of Li-6 be X thus, that of Li-7 will be 1-X, putting the values,
[tex]6.9409 amu=6.015121 amu\times X+7.016003 amu(1-X)[/tex]
Or,
[tex]6.9409 =6.015121 X+7.016003-7.016003 X[/tex]
Or,
X=0.075
Thus, [tex]1-X=1-0.075=0.925[/tex]
Thus, percent abundance of Li-6 is 0.075 or 7.5 % and that of Li-7 is 0.925 or 92.5%.
7.5%
Given:
Lithium has two naturally occurring isotopes:
Lithium has an atomic mass of 6.9409 amu.
Question:
What is the percent abundance of Lithium-6?
The Process:
The average atomic mass is the weighted average of atomic masses of isotopes, weighted by their respective fractional abundance.
[tex]\boxed{ \ Average \ atomic \ mass = \sum_{i = 1}^{n} (isotopic \ mass_{(i)})(abundance_{(i)}) \ }[/tex]
Let the percent abundance of Lithium-6 as X and Lithium-7 as (1 - X). Keep in mind, lithium has two naturally occurring isotopes only.
mass the percent abundance (%)
Lithium has an atomic mass of 6.9409 amu.
Let us find out the percent abundance of Lithium-6.
[tex]\boxed{ \ 6.9409 = (6.015121)X + (7.016003)(1 - X) \ }[/tex]
[tex]\boxed{ \ 6.9409 = (6.015121)X - (7.016003)X + 7.016003 \ }[/tex]
[tex]\boxed{ \ (7.016003)X - (6.015121)X = 7.016003 - 6.9409 \ }[/tex]
[tex]\boxed{ \ (1.000882)X = 0.075103 \ }[/tex]
[tex]\boxed{ \ X = \frac{0.075103}{1.000882} \ }[/tex]
[tex]\boxed{ \ X = 0.075 \ }[/tex] [tex]\ \rightarrow \boxed{ \ X = 7.5 \% \ }[/tex]
Thus, the percent abundance of Lithium-6 is 7.5%.
One more thing, what is the abundance of Lithium-7? 1 - 0.075 = 0.925, i.e., 92.5%
