Question 79ec7

Feb 16, 2016

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

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 $\text{Ag}$, 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 $\text{^107 "Ag}$ than $\text{^109 "Ag}$, 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:

$106.905 x + 108.905 \left(1 - x\right) = 107.870$

$106.905 x + 108.905 - 108.905 x = 107.870$

$\left(106.905 - 108.905\right) x = 107.870 - 108.905$

$x = \frac{107.870 - 108.905}{106.905 - 108.905}$

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"
${M}_{\text{Ag" = "107.8682 g/mol}}$

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