First we need to determine the average atomic mass of these isotopes.
The atomic masses of Lithium-6 and Lithium-7 are 6.0151amu and 7.0160amu respectively.
Now that we have that, we need to set up the relative abundance problem.
We can do that using the formula (M1)(x) + (M2)(1-x) = M(E)
Because you asked for details, I'll explain what each variable stands for-
M1 = the mass of one isotope
x = the relative abundance
M2 = the mass of the second isotope
M(E) = the atomic mass of the element from the periodic table
Therefor all we do now is plug in our respective variable.
M1 = 6.0151amu (Lithium-6)
x = unknown relative abundance
M2 = 7.0160amu (Lithium-7)
M(E) = 6.94004amu
Now the formula plugged in will look like this:
6.0151x + 7.0160(1-x) = 6.94004
Now doing some quick math, we can determine that x=0.07589
For the last step we need to find the percent abundance of each element, which is quite simple. We take X and multiply it by 100 to find Lithium-6, so 0.076 x 100 = 7.6%
and since then we use 1 - X, so 1 - 0.076 = 0.924
We take that and also multiply it by 100, so 0.924 x 100 = 92.4%
Conclusion:
The abundance of the Lithium-6 isotope is 7.6%, and the abundance of the Lithium-7 isotope is 92.4%.
Please ask if you have any questions and I hope this was helpful!
