$ \%Fe = \dfrac{M_{Fe}}{M_{Fe_3O_4}}.100\% = \dfrac{3.56}{232} .100 = 72,41\% $
$\%O = 100\% - \%Fe = 100 - 72,41 = 27,59 \% $
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$ \%Na = \dfrac{M_{Na}}{M_{Na_2SO_4}}.100\% = \dfrac{2.23}{142} .100 = 32,39\% $
$ \%S = \dfrac{M_{S}}{M_{Na_2SO_4}}.100\% = \dfrac{32}{142} .100 = 22,54\%$
$ \%O = 100\% - \%Na - \%S = 100 - 32,39 - 22,54 = 45,07 \% $