The atomic weight of a newly discovered element is 110.352 amu. It has two naturally occurring isotopes. One has a mass of 111.624 amu. The other has an isotopic mass of 109.75 amu. What is the percent abundance of the last isotope (109.75 amu)?

Question

Peyton Adams · Accepted Answer

#67.876%#

The theory behind this is that, in proportion to their individual abundances, each isotope will add to the element's average atomic mass.

The secret to solving this puzzle now is to write down the abundances of the two isotopes.

Let's assume that the decimal abundance, which is simply the percent abundance divided by #100#, of the isotope that has an atomic mass of #"109.75 u"# is #x#.

Since you only have two isotopes, it follows that their decimal abundances must add up to give #1#. This means that the decimal abundance of the first isotope will be #(1-x)#.

One can compute the element's average atomic mass using

#color(blue)("avg. atomic mass" = sum_i("isotope"_i xx "abundance"_i))#

If it were you, you would have

#"110.352 u" = "111.624 u" xx (1-x) + "109.75 u" xx x#

This is the same as

#110.352 = 111.624 - 111.624 * x + 109.75 * x#

#1.874 * x = 1.272 implies x = 1.272/1.874 = 0.67876#

The two isotopes' percent abundances will be

Ellie Andrews · Answer

undefined Using the following formula, you can find the percent abundance of the isotope with a mass of 109.75 amu: percent abundance = (mass of isotope * abundance of isotope) / atomic weight of element.
Substituting the given values, you get: percent abundance = (109.75 amu * x) / 110.352 amu.
Solving for x, you get: x = (109.75 amu * 110.352 amu) / 111.624 amu
x ≈ 0.985.
This means that the isotope with a mass of 109.75 amu has a percent abundance of roughly 98.5%.