Tính phần trăm khối lượng của các nguyên tố trong hợp chất: NaNO3; K2CO3 , Al(OH)3, SO2, SO3, NH4NO3 , Fe2O3

$M_{NaNO_3}=23+14+16.3=85(g/mol)\\ \%_{Na}=\dfrac{23}{85}.100\%=27,06\%\\ \%_N=\dfrac{14}{85}.100\%=16,47\%\\ \%_O=100\%-16,47\%-27,06\%=56,47\%\\ M_{K_2CO_3}=39.2+12+16.3=138(g/mol)\\ \%_K=\dfrac{39.2}{138}.100\%=56,52\%\\ \%_C=\dfrac{12}{138}.100\%=8,696\%\\ \%_O=100\%-8,696\%-56,52\%=34,784\%\\ M_{Al(OH)_3}=27+(16+1).3=78(g/mol)\\ \%_{Al}=\dfrac{27}{78}.100\%=34,62\%\\ \%_O=\dfrac{16.3}{78}.100\%=61,54\%\\ \%_H=100\%-61,54\%-34,62\%=3,84\%\\ M_{SO_2}=32+16.2=64(g/mol)\\ \%_S=\dfrac{32}{64}.100\%=50\%\\ \%_O=100\%-50\%=50\%\\ M_{SO_3}=32+16.3=80(g/mol)\\ \%_S=\dfrac{32}{80}.100\%=40\%\\ \%_O=100\%-40\%=60\% M_{NH_4NO_3}=14.2+4+16.3=80(g/mol)\\ \%_N=\dfrac{28}{80}.100\%=35\%\\ \%_H=\dfrac{4}{80}.100\%=5\%\\ \%_O=100\%-5\%-35\%=60\%\\ M_{Fe_2O_3}=2.56+3.16=160(g/mol)\\ \%_{Fe}=\dfrac{56.2}{160}.100\%=63,75\%\\ \%_O=100\%-63,75\%=36,25\%$