Question

Question #0f03b

Answer 1

The answer is indeed (b).

Explanation:

The thing to remember about isotopes is that they contribute to the average atomic mass of the element in proportion to their abundance.

That is, the average mass of the element will be closer to the atomic mass of the most abundant isotope.

So right from the start, you can look at the values given to you and say that `""^85"Rb"` is more abundant than `""^87"Rb"` because the average atomic mass of rubidium, `"85.5 u"`, is closer to the atomic mass of `""^85"Rb"`, which is `"85 u"`.

I'll try to use a more intuitive approach here.

If the two isotopes were to have a `50%-50%` split, then the average atomic mass would be equal to

`"avg. atomic mass" = ("85 u " + " 87 u")/2 = "86 u"`

Now, notice that the difference between the atomic masses of the two isotopes is equal to

`Delta_"mass" = "87 u" - "85 u" = "2 u" = color(red)(4) * "0.5 u"`

This means that for every

`(100%)/color(red)(4) = 25%`

increase in abundance from `50%`, the average atomic mass becomes closer to the average atomic mass of the more abundant isotope by a margin of `"0.5 u"`.

In your case, the average atomic mass is equal to `"85.5 u"`. This is equivalent to

`"85.5 u" = overbrace("86 u")^(color(purple)("50% - 50% split")) - "0.5 u"`

So the average atomic mass is now `"0.5 u"` closer to the atomic mass of `""^85"Rb"`, which means that the abundance of this isotope is `25%` higher than `50%`.

This means that you have

`"Abundance """^85"Rb" = 50% + 25% = color(green)(75%)`

The abundance of `""^87"Rb"` must decrease because the two abundances must be equal to `100%`

`"Abundance """^87"Rb" = 50% - 25% = color(green)(25%)`

Therefore, `""^85"Rb"` is about three times as abundant as `""^87"Rb"`.

Answer 2

Alternative approach.

Explanation:

Here's another approach to use in order to find the abundances of the two isotopes.

As you know, each isotope will contribute to the average atomic mass of rubidium in proportion to their abundance.

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

Here `i` represents the atomic mass of an isotope `i`. This equation uses decimal abundance, which is simply percent abundance divided by `100`

`color(blue)(|bar(ul(color(white)(a/a)"decimal abundance" = "percent abundance"/100color(white)(a/a)|)))`

For example, if an isotopes has a `13%` percent abundance, it will have a

`"decimal abundance" = 13/100 = 0.13`

So, you know that your element has two stable isotopes, `""^85"Rb"` and `""^87"Rb"`. If you take `x` to be the decimal abundance of `""^85"Rb"`, you can say that the decimal abundance of `""^87"Rb"` will be `1-x`.

This is the case because the abundances of the two elements must add up to give `100%`, or `1` as a decimal abundance.

You know that the average atomic mass of rubidium is `"85.5 u"`, and that the two isotopes have atomic masses equal to `"85 u"` and `"87 u"`, respectively.

The equation will thus take the form

`85.5 color(red)(cancel(color(black)("u"))) = 85color(red)(cancel(color(black)("u"))) xx x + 87color(red)(cancel(color(black)("u"))) xx (1-x)`

`85.5 = 85x + 87 - 87x`

Rearrange to solve for `x`

`87x - 85x = 87 - 85.5`

`2x = 1.5 implies x = 0.75`

So, the abundances of the two isotopes will be

`"For """^85"Rb: " overbrace(0.75)^(color(purple)("decimal abundance")) = overbrace(75%)^(color(green)("percent abundance"))`

`"For """^87"Rb: " 1 - 0.75 = overbrace(0.25)^(color(purple)("decimal abundance")) = overbrace(25%)^(color(green)("percent abundance"))`

Once again, this shows that `""^85"Rb"` is three times as abundant as `""^87"Rb"`.