Question #7da3f

1 Answer
Oct 16, 2016

Here's how you can do that.


The idea here is that you can use the fact that rubidium, #"Rb"#, has two naturally occurring isotopes to figure out the percent abundance of the second isotope.

Since the element has two naturally occurring isotopes, their respective percent abundances must add up to give #100%#. This means that the percent abundance of the second isotope is

#100% - 72% = 28%#

Now, the average atomic mass of an element is calculated by taking the weighted average of the atomic mass of its stable isotopes.

Simply put, each isotope will contribute to the average mass of the element proportionally to its abundance.

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

Keep in mind that the above equation uses decimal abundances, which are simply percent abundances divided by #100%#.

So, you know that the average atomic mass of rubidium is #"85.47 u"# and that one of its isotopes has an atomic mass of #"85 u"#. This means that you can write

#85.47 color(red)(cancel(color(black)("u"))) = overbrace(85color(red)(cancel(color(black)("u"))) xx (72color(red)(cancel(color(black)(%))))/(100color(red)(cancel(color(black)(%)))))^(color(darkgreen)("contribution from first isotope")) + overbrace(?color(red)(cancel(color(black)("u"))) xx (28color(red)(cancel(color(black)(%))))/(100color(red)(cancel(color(black)(%)))))^(color(purple)("contribution from second isotope"))#

All you have to do now is find the value of #'?'#.

#85.47 = 85 xx 0.72 + ? xx 0.28#

This gets you

#? = (85.47 - 85 xx 0.72)/(0.28) = 86.68#

You should keep track of sig figs here and round the answer to #87#, i.e. rounded to two sig figs, but I'll leave it rounded to four sig figs.