# Question #fd4b1

##### 1 Answer

Here's what's going on here.

#### Explanation:

I think that you're referring to a problem Tyler solved in one of his YouTube videos. The problem starts at the 8:03 minute mark.

The idea here is that the average atomic mass of an element is calculated by taking the **weighted average** of the atomic masses of its stable isotopes.

Now, *weighted average* means that each isotope contributes to the average atomic mass **in proportion** to its abundance.

In this case, lithium is said to have two stable isotopes. Right from the start, you can say that the percent abundances of the two isotopes **must** add up to give

This is equivalent to saying that their *decimal abundances*, which are simply the percent abundances divided by **must** add up to give

Why is that the case?

Because lithium has only two stable isotopes, which implies that there are only two isotopes that contribute to its average atomic mass.

So if we take

#overbrace(" " color(red)(cancel(color(black)(x))) " ")^(color(blue)("decimal abundance """^6"Li")) + overbrace(" " 1 - color(red)(cancel(color(black)(x)))" ")^(color(blue)("decimal abundance """^7"Li")) = 1#

Now, the average atomic mass for lithium will be equal to

#"avg. atomic mass Li" = "atomic mass"color(white)(.)""^6"Li" xx "decimal abundance"color(white)(.)""^6"Li" + "atomic mass"color(white)(.)""^7"Li" xx "decimal abundance"color(white)(.)""^7"Li"#

Since you know that lithium has an average atomic mass of

#6.941 = 6.015 * x + 7.016 * (1-x)#

You would then go on to solve this equation to get

#x = 0.075#

This is the decimal abundance of

#1 - 0.075 = 0.925#

To convert these to *percent abundances*, multiply them by

#"For """^6"Li: " 0.075 * 100% = 7.5%#

#"For " ""^7"Li: " 0.925 * 100% = 92.5%#