We're asked to find the percentage by mass of #"N"# in ammonium carbonate, #("NH"_4)_2"CO"_3#, given that its molar mass is #96.0"g"/"mol"#.
To find the mass of #"N"# in the compound, we multiply its molar mas, #14.01"g"/"mol"#, by however many atoms of #"N"# are in the compound, which is #2#:
#"mass of N" = (2)(14.01"g"/"mol") = 28.02"g"/"mol"#
Thus, in one mole of #("NH"_4)_2"CO"_3# (which has a mass of #96.0# #"g"#), there are #28.02# #"g"# of nitrogen. The percentage by mass of #"N"# is the mass of #"N"# divided by the total mass, then multiplied by #100#:
#% "N" = ("mass of N")/("mass of" ("NH"_4)_2"CO"_3) xx 100%#
#% "N" = (28.02"g N")/(96.0"g" ("NH"_4)_2"CO"_3) xx 100% = color(red)(29.2% "N"#
