$\eqalign{
& + NaOH \cr
& {M_{NaOH}} = 23.1 + 16.1 + 1.1 = 40 \cr
& \% Na = {{23.1} \over {40}}.100\% = 57,5\% \cr
& \% O = {{16} \over {40}}.100\% = 40\% \cr
& \% H = 100\% - \% Na - \% O = 2,5\% \cr
& + C{O_2} \cr
& {M_{C{O_2}}} = 12.1 + 16.2 = 44 \cr
& \% C = {{12.1} \over {44}}.100\% = 27,27\% \cr
& \% O = 100\% - 27,27\% = 72,73\% \cr
& + F{e_2}{\left( {S{O_4}} \right)_3} \cr
& {M_{F{e_2}{{\left( {S{O_4}} \right)}_3}}} = 56.2 + (32.1 + 16.4).3 = 400 \cr
& \% Fe = {{56.2} \over {400}}.100\% = 28\% \cr
& \% S = {{32.1.3} \over {400}}.100\% = 24\% \cr
& \% O = 100\% - \% Fe - \% S = 48\% \cr
& + NaN{O_3} \cr
& {M_{NaN{O_3}}} = 23.1 + 14.1 + 16.3 = 85 \cr
& \% Na = {{23.1} \over {85}}.100\% = 27,06\% \cr
& \% N = {{14.1} \over {85}}.100\% = 16,47\% \cr
& \% O = 100\% - \% Na - \% N = 56,47\% \cr} $
