To determine the natural abundances of lithium-6 (^6Li) and lithium-7 (^7Li), we can use the given atomic masses and the average atomic mass of lithium.
Let's denote the natural abundance of ^6Li as x and the natural abundance of ^7Li as y.
The average atomic mass can be expressed as:
Average Atomic Mass = (Mass of ^6Li x Abundance of ^6Li) + (Mass of ^7Li x Abundance of ^7Li)
Using the given values:
6.941 amu = (6.0151 amu x x) + (7.0160 amu x y)
Now, we can solve this system of equations:
x + y = 1 (The sum of the natural abundances is equal to 1)
6.0151x + 7.0160y = 6.941
From the first equation, we can isolate x as x = 1 - y.
Substituting this value into the second equation:
6.0151(1 - y) + 7.0160y = 6.941
6.0151 - 6.0151y + 7.0160y = 6.941
y - 6.0151y + 7.0160y = 6.941 - 6.0151
2.0009y = 0.9259
y = 0.9259 / 2.0009
y ≈ 0.4623
Substituting this value back into x = 1 - y:
x = 1 - 0.4623
x ≈ 0.5377
Therefore, the natural abundance of lithium-6 is approximately 53.77%, and the natural abundance of lithium-7 is approximately 46.23%.