The element rubidium (atomic mass 85.5) consists of two isotopes, one mass number 85 (Rb-85) and the other mass number 87 (Rb-87). Which one of the following statements about the isotopes of rubidium is correct?

    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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