Question

Question #79ec7

Answer 1

The atomic mass of `"Ag"` (silver; argentum) itself is the weighted average of the relative atomic masses for each isotope of `"Ag"`, `""^107 "Ag"` and `""^109 "Ag"`.

What we are asked to find is the `%""^107 "Ag"` and `%""^109 "Ag"`; that is, the contribution of each isotope's relative atomic mass to the overall atomic mass. Obviously they must add up to `100%`, since they belong to the only isotopes of `"Ag"`, and percent literally means "out of one hundred".

QUALITATIVE OBSERVATIONS

We can see how `107.870` is between `106.905` and `108.905`. If there were to be exactly `50%` of each one, then logically, the overall atomic mass would be exactly halfway in between `106.905` and `108.905`, which would be `107.905`.

Therefore, since `107.870 < 107.905`, there is slightly greater contribution from `""^107 "Ag"` than `""^109 "Ag"`, and `%""^107 "Ag" > 50%`. That is our initial guesstimate.

CALCULATION OF ISOTOPIC ABUNDANCE

To determine the exact percents, we look at the contribution by each given relative atomic mass:

`106.905x + 108.905(1-x) = 107.870`

`106.905x + 108.905 - 108.905x = 107.870`

`(106.905 - 108.905)x = 107.870 - 108.905`

`x = (107.870 - 108.905)/(106.905 - 108.905)`

`color(blue)(x) = 0.5175 -> color(blue)(51.75% ""^107 "Ag")`

`1 - x = color(blue)(48.25% ""^109 "Ag")`

And this gives a decent approximation to the actual isotopic abundances that we could have gotten if we had been given more accurate numbers to begin with:

`%""^107 "Ag" = 51.839%`
`%""^109 "Ag" = 48.161%`

More accurate numbers to use would be:

`M_(""^107 "Ag") = "106.905097 g/mol"`
`M_(""^109 "Ag") = "108.904752 g/mol"`
`M_"Ag" = "107.8682 g/mol"`

You could redo the calculations to get even more accurate results. I got `51.837% ""^107 "Ag"` and `48.163% ""^109 "Ag"`.