# Question ff210

Oct 20, 2017

$\text{185.0 u}$

#### Explanation:

The first thing that you need to do here is to use the percent abundance of rhenium-187 to calculate the percent abundance of rhenium-185.

To do that, use the fact that rhenium has only two naturally occurring isotopes. This implies that their percent abundances must add up to give 100%.

Alternatively, you can say that their decimal abundances must add up to $1$. This is the case because a decimal abundance is simply a percent abundance divided by 100%.

This means that you have

$\text{decimal abundance"color(white)(.)""^185"Re} = 1 - 0.6260$

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 can write

"186.207" color(red)(cancel(color(black)("u"))) = overbrace(186.956 color(red)(cancel(color(black)("u"))) * 0.6260)^(color(blue)("contribution from"color(white)(.)""^187"Re")) + overbrace(? color(red)(cancel(color(black)("u"))) * (1 - 0.6260))^(color(blue)("contribution from"color(white)(.)""^185"Re"))

Here ? represents the value of the atomic mass of rhenium-185. Rearrange the equation to solve for ?

? = (186.207 - 186.956 * 0.6260 )/(1 - 0.06260) = 184.953

Therefore, you can say that the atomic mass of rhenium-185 is equal to

"atomic mass"color(white)(.)""^185"Re" = color(darkgreen)(ul(color(black)("185.0 u")))#

The answer must be rounded to four sig figs, the number of sig figs you have for the percent abundance of rhenium-187.