Question

Question #ff210

Answer 1

`"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

`"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.