a, %Na= $\frac{23}{23+35,5}$ . $100$ =$39,32$ (%)
%Cl=$100-39,32=60,68$ (%)
b, %Fe= $\frac{56.2}{56.2+16.3}$ . $100$ =$70$ (%)
%O = $100-70=30$ (%)
c,%Cu= $\frac{64}{64+16}$ . $100$ =$80$ (%)
%O =$100-80=20$ (%)
d, %Na = $\frac{23}{23+14+16.3}$ . $100$ =$27,1$ (%)
%N = $\frac{14}{23+14+16.3}$ . $100$ =$16,5$ (%)
%O = $100-27,1-16,5=56,4$ (%)
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