The atomic masses of 6-Li and 7-Li are 6.0151 amu and 7.0160 amu, respectively. Calculate the natural abundances of these two isotopes.

Question

I understand the question, but I'm not sure how to solve this without an average atomic mass, it just feels like there's not enough information.

Answer 1

An element's atomic mass listed on the periodic table is a weighted average for each stable isotope of the element. Each isotope has its mass calculated and multiplied by it's abundance in nature (eg: 30% of all this element's isotopes). Lithium (Li) has two stable isotopes Li-6 and Li-7. Lithium's (Li) atomic mass is is 7.0 u. So in this case, the mass of isotope Li-6 times it's relative abundance (x) plus the average of the mass of isotope Li-7 times it's relative abundance (y) = lithium's atomic number:

6.0151(x) + 7.0160(y)= 7.0

Since there are only two isotopes, x + y= 100%.

Rearranging this gives us: y= 100% - x.

Simplifying yields: y= 1-x.

Substituting in the earlier formula yields:

6.0151x + 7.0160(1-x) = 7.0

Then it is an algebraic equation and we solve for x:

6.0151x + 7.0160(1-x) = 7.0

6.0151x + 7.0160 - 7.0160x = 7.0

6.0151x - 7.0160x = -0.0160

-1.0009x = -0.0160

x= 0.01599

x= 1.6%

since y=1-x ,

y=1-0.01599

y=0.98401

y=98.4%