Câu 2: Tính % về khối lượng các nguyên tố trong các hợp chất sau: a. Ca(NO3)2 b. K2CO3 c. FeCl3 d. P2O5

$a,M_{Ca(NO_3)_2}=164(g/mol)\\ \%_{Ca}=\dfrac{40}{164}.100\%=24,39\%\\ \%_N=\dfrac{14.2}{164}.100\%=17,07\%\\ \%_O=100\%-24,39\%-17,07\%=58,54\%\\ b,M_{K_2CO_3}=138(g/mol)\\ \%_K=\dfrac{39.2}{138}.100\%=56,52\%\\ \%_C=\dfrac{12}{138}.100\%=8,7\%\\ \%_O=100\%-8,7\%-56,52\%=34,78\%\\ c,M_{FeCl_3}=162,5(g/mol)\\ \%_{Fe}=\dfrac{56}{162,5}.100\%=34,46\%\\ \%_{Cl}=100\%-34,46\%=65,54\%\\ d,M_{P_2O_5}=142(g/mol)\\ \%_P=\dfrac{31.2}{142}.100\%=43,66\%\\ \%_O=100\%-43,66\%=56,34\%$