# Question #c6ffd

##### 1 Answer

#### Answer:

#### Explanation:

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

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

This implies that the decimal abundance of the third isotope, let's say

#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.