Question

Question #d4b30

Answer 1

Set up two simple equations then solve like you would any two-equation system.

Explanation:

You only have two isotopes, which means that the sum of their respective abundances will amount to 100%, or to 1 if you use fractional numbers.

Let's say the abundance of `""_3^6Li` is `"A"` and the abundance of `""_3^7Li` is `"B"`. To determine these abundances set up two simple equations (I'll use fractional numbers - you'll sometimes see this expressed as decimal abundance)

`A + B = 1`, and

`6.0151 * A + 7.0160 * B = 6.941`

Now solve this like you would any two-equation system.

`A = 1- B => 6.0151 * (1 - B) + 7.0160 * B = 6.941`

`6.0151 - 6.0151 * B + 7.0160 * B = 6.941`

This will get you

`1.0009 * B = 0.9259 => B = 0.92507`

Plug this into the first equation to get

`A = 1 - B = 1 - 0.92507 = 0.07493`

Expressed in percentages and rounded to four sig figs, these values will be

`A = 7.490%` - the abundance of `""_3^6Li`
`B = 92.51%` - the abundance of `""_3^7Li"`

Finally, check the result to see if it makes sense. Notice that the average atomic mass of the two isotopes is 6.941, which is closer to 7.0160, the atomic mass of the heavier isotope, `""_3^7Li`, than it is to 6.0151, the atomic mass of the lighter one.

This means that the heavier isotope is much more abundant than the other, which is confirmed by the abundaces we got - 92.5% for the heavier one and only 7.49% for the lighter isotope.