# Question #79ec7

##### 1 Answer

The **atomic mass** of *argentum*) itself is the weighted average of the *relative atomic masses* for each isotope of

What we are asked to find is the ** contribution** of each isotope's relative atomic mass to the overall atomic mass. Obviously they must add up to

**QUALITATIVE OBSERVATIONS**

We can see how *exactly* halfway in between

Therefore, since

**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