If a mass of #"16.26 mg"# contains #1.66 xx 10^20# atoms, what is the molar mass of that atom? If you have #"1 million"# silver atoms constituting a total of #1.79 xx 10^(-16) "g"#, what is the molar mass? #"1 amu"# #=# #1.660599 xx 10^(-24) "g"#.
1 Answer
Both questions ask the same thing in opposite ways.
One gives you
#("16.26" cancel"mg")/(1.66xx10^20 cancel"atoms") xx "1 g"/("1000" cancel"mg") xx (6.0221413xx10^23 cancel"atoms")/"mol"#
#=# #color(blue)("58.988 g/mol")# ,
which is close enough to cobalt, whose accepted value is
The other gives you
#(1.79xx10^(-16) "g")/(10^6 cancel"Ag atoms")xx(6.0221413xx10^23 cancel"atoms")/"mol"#
#=# #color(green)("107.796 g/mol")#
which is close enough to the accepted value of
In either case you still get
WHAT IS AMU?
You are not required to use the conversion
#"1 g"/(6.0221413xx10^(23) "atoms") = "1 amu"#
#= 1.660599xx10^(-24) "g",#
unless you want to. It just depends on whether you are talking about
CONVERTING BETWEEN G/MOL and AMU
You can still interconvert between
#(58.988 cancel"g")/cancel"mol"xxcancel"1 mol"/(6.0221413xx10^23 cancel"atoms") xx "1 amu"/(1.660599xx10^(-24) cancel"g")xxcancel("1 atom")#
#~~ color(blue)(58.988)# #color(blue)("amu for one atom")#
i.e. The experimental atomic mass of
So,