a, $M_{NaCl}$ = 23 + 35,5 = 58,5
%$m_{Na}$ = $\frac{23}{58,5}$ ~ 39,31%
%$m_{Cl}$ = 100% - 39,31% = 60,68%
b, $M_{Fe_{2}O_{3}}$ = 56.2 + 16.3 = 160
%$m_{Fe}$ = $\frac{112}{160}$ = 70%
%$m_{O}$ = 100% - 70% = 30%
c, $M_{CuO}$ = 64.1 + 16.1 = 80
%$m_{Cu}$ = $\frac{64}{80}$ = 80%
%$m_{O}$ = 100% - 80% = 20%
d, $M_{NaNO_{3}}$ = 23 + 14 + 3.16 = 85
%$m_{Na}$ = $\frac{23}{85}$ = 27,05%
%$m_{N}$ = $\frac{14}{85}$ ~ 16,47%
%$m_{O}$ = 100% - 27,05% - 16,47% = 56,48%
