Đáp án:
$* Fe3O4$
$MFe3O4 = 3.56 + 4.16 = 232 (g/mol)$
%$mFe = $$\frac{3.56}{232}$$.100 = 72,41$%
%$mO = 100 - 72,41 = 27,59$%
$* CO2$
$MCO2 = 12 + 2.16 = 44 (g/mol)$
%$mC = $$\frac{12}{44}$$.100 = 27,27$%
%$mO = 100 - 27,27 = 72,73$%
$* NaCl$
$MNaCl = 23 + 35,5 = 58,5 (g/mol)$
%$mNa = $$\frac{23}{58,5}$$.100 = 39,32$%
%$mCl = 100 - 39,32 = 60,68$%