Alternative approach.
Here’s another approach to use in order to find the abundances of the two .
As you know, each isotope will contribute to the average 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 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”##.
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