Question #79ec7

1 Answer
Feb 16, 2016

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"#.