$\eqalign{
& a)N{H_4}N{O_3} \cr
& {M_{N{H_4}N{O_3}}} = 14 + 4.1 + 14 + 16.3 = 80 \cr
& \% N = {{14.2} \over {80}}.100\% = 35\% \cr
& \% H = {{4.1} \over {80}}.100\% = 5\% \cr
& \% O = 100\% - \% N - \% H = 60\% \cr
& b)CO{(N{H_2})_2} \cr
& {M_{CO{{(N{H_2})}_2}}} = 12 + 16 + (14 + 2).2 = 60 \cr
& \% C = {{12} \over {60}}.100\% = 20\% \cr
& \% O = {{16} \over {60}}.100\% = 26,67\% \cr
& \% H = {{2.2.1} \over {60}}.100\% = 6,67\% \cr
& \% N = 100\% - \% C - \% O - \% H = 46,67\% \cr
& c){H_3}P{O_4} \cr
& {M_{{H_3}P{O_4}}} = 3.1 + 31 + 16.4 = 98 \cr
& \% H = {{3.1} \over {98}}.100\% = 3,06\% \cr
& \% P = {{31} \over {98}}.100\% = 31,63\% \cr
& \% O = 100\% - \% H - \% P = 65,31\% \cr} $
