Determine the average atomic mass of helium, which has two isotopes having atomic masses 4.0026 amu and 3.0160 amu.

Question

The relative abundance of the two isotopes is 99.9999 percent and 0.0001 percent, respectively.

A. 

1.0066 amu

B. 

3.0160 amu

C. 

3.5903 amu

D. 

4.0026 amu

Answer 1

Sarah, thanks for the great question.

This is one of those classic chemistry questions where the math is pretty simple algebra that most students in chemistry can already do, but the wording of the problem is too intimidating for the student to see that. To help with that, we can break the word problem down into parts to turn complex chemistry into simple math.

The goal of the question is to find an average. You are given data about the set of numbers that form the average: "4.0026 amu and 3.0160 amu. The relative abundance of the two isotopes is 99.9999 percent and 0.0001 percent, respectively." In other words, you have this number set where 99.9999% of the numbers are "4.0026" and only 0.0001% are 3.0160.

Average = (sum of the numbers) / (the number of numbers)

We can't weigh every atom, so we can't get the sum of the numbers. But we do have the frequency that each number appears. We can't know the number of atoms, but we do know that the frequencies will all add to 100%

Average = (number*frequency)/ 100% + ... for each number and its frequency

Average = (4.0026)*(99.9999%)/(100%) + (3.0160)(0.0001%)/(100%)

Average = 4.0026 amu

The answer is D

This makes sense logically too, which is a great way to check our work. If the strong majority of atoms are 4.0026 (more than 99%!) it makes sense that the average atom (visualize scooping one atom up at random) weights 4.0026 amu.

Hope this helped! Feel free to message me any time you have other questions!