Question #2e766

1 Answer
Oct 2, 2015

Answer:

#90%#

Explanation:

So, you know that you're dealing with two isotopes, let's say #""^20"X"# and #""^22"X"#.

The relative atomic mass of the element will be determined by the atomic masses of the two isotopes in proportion to their respective abundances.

#"relativ atomic mass" = sum_i ("isotope"""_i xx "abundance"""_i)#

SInce you're only dealing with two isotopes, you can say that their abundances must add to give 100%.

If you take #x# to be the decimal abundance, which is simply the percent abundance divided by #100#, of #""^20"X"#, the abundance of #""^22"X"# will be #(1 - x)#.

This means that you can write

#"20 u" * x + "22 u" * (1-x) = "20.2 u"#

#20x + 22 - 22x = 20.2#

#2x = 1.8 implies x = 1.8/2 = 0.9#

The decimal abundance of #""^22"X"# will thus be #(1- 0.9) = 0.1#.

The percent abundances of the two isotopes are

#""^20"X: " color(green)(90%)#
#""^22"X: " 10%#

The result makes sense because the relative atomic mass of the element is much closer to the atomic mass of #""^20"X"# than it is to the atomic mass of #""^22"X"#, which can only imply that #""^20"X"# has a significantly larger percent abundance that #""^22"X"#.