Question

Vanadium has two naturally occurring isotopes, 50-V with an atomic mass of 49.9472 amu and 51-V with an atomic mass of 50.9440. The atomic weight of vanadium is 50.9415. What is the percent abundance of the vanadium isotopes?

Answer 1

`""^50"V: " 0.250803%`
`""^51"V: " 99.7492%`

Explanation:

The idea with elements that have two naturally occurring isotopes is that the percent abundances of those two isotopes must add up to give `100%`.

In calculations, it is often easier to work with decimal abundances, which are simply percent abundances divided by `100`.

So, if you're working with decimal abundances, you can say for sure that if `x` is the decimal abundance of the first isotope, then the decimal abundance of the second isotope will have to be `(1-x)`, since

`x + (1-x) = 1`

The decimal abundances of the two isotopes must add up to give `1`.

Now, an element's relative atomic mass is calculated by taking the weighted average of the atomic masses of its naturally occurring isotopes.

`color(blue)("relative atomic mass" = sum_i ("isotope"_i xx "abundance"_i))`

So, if we take `x` to be the decimal abundance of `""^50"V"` and `(1-x)` as the decimal abundance of `""^51"V"`, you can say that

`49.9472 color(red)(cancel(color(black)("u"))) xx x + 50.9440 color(red)(cancel(color(black)("u"))) xx (1-x) = 50.9415 color(red)(cancel(color(black)("u")))`

Solve this equation for `x` to get

`49.9472 * x + 50.9440 - 50.9440 * x = 50.9415`

`0.9968 * x = 0.0025 implies x = 0.0025/0.9968 = 0.00250803`

Since `x` represents the decimal abundance of `""^50"V"`, it follows that the decimal abundance of `""^51"V"` will be

`1 - x = 1 - 0.00250803 = 0.9974920`

The percent abundances of the two isotopes will be

• `""^50"V: " 0.002508 xx 100 = color(green)(0.250803%)`
• `""^51"V: " 0.9974920 xx 100 = color(green)(99.7492%)`

I'll leave the values rounded to six sig figs, the number of sig figs you have for the atomic masses of the isotopes and for the relative atomic mass of the element.