Question #c6ffd

1 Answer
Oct 19, 2017

#"37.65 u"#


The idea here is that an isotope's contribution to the average atomic mass of the element will be proportional to its abundance.

Since this unknown element has three naturally occurring isotopes, you can say that the percent abundances of these isotopes must add up to give #100%#.

When working with decimal abundances, this is equivalent to saying that the decimal abundances of the isotopes must add up to give #1#.

This implies that the decimal abundance of the third isotope, let's say #x#, will be--remember, we convert a percent abundance to a decimal abundance by dividing it by #100%#.

#1 = 0.4557 + 0.3200 + x#

#x = 1 - (0.4557 + 0.3200)#

Now, the average atomic mass of the element is calculated by taking the weighted average of the atomic masses of its naturally occurring isotopes.

This means that you have

#"avg. atomic mass"#

#= overbrace("36.5681 u" * 0.4557)^(color(blue)("contribution from Lc-37")) + overbrace("38.105 u" * 0.3200)^(color(blue)("contribution of Lc-39")) + overbrace("39.2055 u" * [1 - (0.4557 + 0.32)])^(color(blue)("contribution of Lc-40"))#

You will end up with

#color(darkgreen)(ul(color(black)("avg atomic mass = 37.65 u")))#

The answer is rounded to four sig figs, the number of sig figs you have for the abundances of the first two isotopes.