Question

Question #54d91

Answer 1

`"106.91 u"`

Explanation:

The average atomic mass of silver is given by the weighted average of the atomic masses of its two stable isotopes, silver-107 and silver-109.

This basically means that each isotope will contribute to the average mass of the element proportionally to its abundance.

Since you're only dealing with two stable isotopes here, you can say that their abundances must add up to give `100%`.

`100% = overbrace(48.161%)^(color(blue)("abundance of silver-109")) + "abundance of silver-107"`

This means that you have

`"abundance of silver-107" = 51.839%`

Next, convert the two percent abundances to decimal abundances by dividing both values by `100%`. You will end up with

`"For " ""^107"Ag: " (51.839 color(red)(cancel(color(black)(%))))/(100color(red)(cancel(color(black)(%)))) = 0.51839`

`"For " ""^109"Ag: " (48.161 color(red)(cancel(color(black)(%))))/(100color(red)(cancel(color(black)(%)))) = 0.48161`

Now, you know that

`"avg. atomic mass Ag"`

`= "mass """^107"Ag" * "abundance """^107"Ag" + "mass """^109"Ag" * "abundance """^109"Ag"`

This means that you have

`"mass """^107"Ag" = ("avg. mass of Ag " - " mass of """^109"Ag" * "abundance " ""^109"Ag")/("abundance """^107"Ag")`

Plug in your values to find

`"mass " ""^107"Ag" = ("107.8682 u " - " 108.9047 u" * 0.48161)/0.51839`

`"mass """^107"Ag" = color(darkgreen)(ul(color(black)("106.91 u")))`

The answer is rounded to five sig figs, the number of sig figs you have for the percent abundance of silver-109.