Question

Question #6aa96

Answer 1

`"114.9 u"`

Explanation:

The problem tells you that indium has two stable isotopes, so right from the start, you know that their percent abundances must add up to give `100%`.

This means that the percent abundance of the indium-115 isotope is equal to

`100% - 4.305% = 95.695%`

Before moving forward, convert the two percent abundances to decimal abundances by dividing them by `100%`.

You will have

• `""^113"In: " (4.305 color(red)(cancel(color(black)(%))))/(100color(red)(cancel(color(black)(%)))) = 0.04305`

• `""^115"In: " (95.695 color(red)(cancel(color(black)(%))))/(100color(red)(cancel(color(black)(%)))) = 0.95695`

Now, the average atomic mass of indium is given by the weighted average of the isotopic masses of its stable isotopes.

If you take `x` `"u"` to be the isotopic mass of indium-115, you can say that you have

`overbrace(114.8 color(red)(cancel(color(black)("u"))))^(color(blue)("avg. atomic mass")) = overbrace(112.90 color(red)(cancel(color(black)("u"))) * 0.04305)^(color(blue)("contribution from"""^113"In")) + overbrace(xcolor(red)(cancel(color(black)("u"))) * 0.95695)^(color(blue)("contribution from"""^115"In"))`

Rearrange to solve for `x`

`x = (114.8 - 112.90 * 0.04305)/0.95695 = 114.9`

You can thus say that indium-115 has an isotopic mass equal to

`color(darkgreen)(ul(color(black)("isotopic mass"color(white)(.)""^115"In = 114.9 u")))`

The answer is rounded to four sig figs, the number of sig figs you have for the average atomic mass of the element.