Question

Question #a254f

Answer 1

`"6.945 u"`

Explanation:

The average atomic mass of an element is simply the weighted average of the atomic masses of its naturally occurring isotopes.

`color(blue)(bar(ul(|color(white)(a/a)"avg. atomic mass" = sum_i "isotope"_i xx "abundance"_icolor(white)(a/a)|)))`

Each naturally occurring isotope contributes to the average atomic mass of the element proportionally to its abundance.

Now, the problem gives you percent abundance, but the above equation uses decimal abundances. To get from one to the other, simply divide a percent abundance by `100%`.

This means that you have

• `"isotope no. 1: " "6.017 u", (7.30 color(red)(cancel(color(black)(%))))/(100color(red)(cancel(color(black)(%)))) = 0.0730`

• `"isotope no. 2: " "7.018 u", (92.7color(red)(cancel(color(black)(%))))/(100color(red)(cancel(color(black)(%)))) = 0.927`

Notice that the percent abundances must add up to give `100%`, which implies that the decimal abundances must add up to give `1`.

Also, notice that the second isotope is much more abundant than the first. This tells you that the average atomic mass of the element will be closer in value to the atomic mass of the second isotope.

The average atomic mass of lithium will be

`"avg. atomic mass" = overbrace("6.017 u" xx 0.0730)^(color(darkgreen)("contribution from isotope 1")) + overbrace("7.018 u" xx 0.927)^(color(purple)("contribution from isotope 2"))`

This gets you

`"avg. atomic mass " = " 6.945 u"`

The answer is rounded to three sig figs.

As you can see, the average atomic mass of the element is indeed closer in value to the mass of the more abundant isotope.