# Question 74c1d

Feb 15, 2016

$\text{7.00157 u}$

#### Explanation:

It's important to realize that the problem is asking for the average atomic mass of this sample of lithium, not of lithium in general.

There is a distinction between the two because this sample is depleted in the lighter isotope, which means that it contains less $\text{^6"Li}$ that a regular sample of lithium would contain.

So right from the start, you should expect the average atomic weight of lithium in this sample to be bigger than lithium's average atomic weight.

So, an element's average atomic weight is calculated by taking the weighted average of the atomic weights of its stable isotopes.

Each isotope will contribute to the average atomic weight proportionally to its abundance.

$\textcolor{b l u e}{{\text{avg. atomic weight" = sum_i "isotope"_i xx "abundance}}_{i}}$

Now, to make calculations easier, you can use decimal abundances, which are simply percent abundances divided by $100$.

So, you know that this sample contains 1.442% (according to the image you added) $\text{^6"Li}$. Since lithium only has two stable isotopes, it follows that it also contains

100% - 1.442% = "98.558% """^7"Li"

The decimal abundances for the two isotopes will be

"For """^6"Li: " (1.442%)/100 = 0.01442

"For """^7"Li: " (98.558%)/100 = 0.98558

The average atomic weight of the sample will thus be

$\text{avg. atomic weight" = "6.0151122 u" xx 0.01442 + "7.016004 u} \times 0.98558$

"avg. atomic weight " = color(green)(" 7.00157 u")#

So, is the average atomic weight bigger in this sample than what you would expect to see in a typical sample of lithium?

Indeed it is!

The average atomic weight for lithium varies between $6.9387$ and $\text{6.9959 u}$. The two isotopes have the following abundances

As you can see, the sample depleted in the lighter isotope really does contain less $\text{^6"Li}$ than you would normally see in a sample of lithium.