Question

Tìm thành phần phần trăm các nguyên tố trong hợp chất. CU(NO³)²

CA³(PO4)²

Answer 1

\(M_{Cu(NO_3)_2}=188(g/mol)\\ \%_{Cu}=\dfrac{64}{188}.100\%=34,04\%\\ \%_N=\dfrac{28}{188}.100\%=14,89\%\\ \%_O=100\%-14,89\%-34,04\%=51,07\%\\ M_{Ca_3(PO_4)_2}=310(g/mol)\\ \%_{Ca}=\dfrac{120}{310}.100\%=38,71\%\\ \%_P=\dfrac{62}{310}.100\%=20\%\\ \%_O=100\%-20\%-38,71\%=41,29\%\)

Answer 2

\(\%Cu=\dfrac{64.1}{188}.100\%=34,04\%;\%N=\dfrac{14.2}{188}.100\%=14,9\%\)

\(\%O=\dfrac{16.6}{188}.100\%=51,06\%\)

\(\left\{{}\begin{matrix}\%Ca=\dfrac{40.3}{310}.100\%=38,7\%\\\%P=\dfrac{31.2}{310}.100\%=20\%\\\%O=\dfrac{16.8}{310}.100\%=41,3\%\end{matrix}\right.\)