Question

Bài 1. Tính % khối lượng các nguyên tố có trong các phân tử sau:

a)    CO; MgCl₂; C₆H₆.

b)    FeO; Fe₃O₄;

c)     CuSO₄; CaCO₃

Answer 1

Làm một cái rồi tương tự nhé

\(a,\%C=\dfrac{12}{44}=27,27\%\\ \%O=100\%-27,27\%=72,73\%\)

Answer 2

\(a,CO\\ \%m_C=\dfrac{M_C}{M_C+M_O}.100\%=\dfrac{12}{12+16}.100\approx42,857\%\\ \Rightarrow\%m_O\approx100\%-42,857\%\approx57,143\%\\ MgCl_2\\ \%m_{Mg}=\dfrac{M_{Mg}}{M_{Mg}+2.M_{Cl}}.100\%=\dfrac{24}{24+2.35,5}.100\approx25,263\%\\ \Rightarrow\%m_{Cl}\approx100\%-25,263\%\approx74,737\%\\ C_6H_6\\ \%m_C=\dfrac{6.M_C}{6.M_C+6.M_H}.100\%=\dfrac{6.12}{6.12+6.1}.100\approx92,308\%\\ \Rightarrow\%m_H\approx100\%-92,308\%\approx7,692\%\)

Answer 3

\(b,FeO\\ \%m_{Fe}=\dfrac{M_{Fe}}{M_{Fe}+M_O}.100\%=\dfrac{56}{56+16}.100\approx77,778\%\\ \Rightarrow\%m_O\approx100\%-77,778\%=22,222\%\\ Fe_3O_4\\ \%m_{Fe}=\dfrac{3.M_{Fe}}{3.M_{Fe}+4.M_O}.100\%=\dfrac{3.56}{3.56+4.16}.100\approx72,414\%\\ \Rightarrow\%m_O\approx100\%-72,414\%\approx27,586\%\)

\(c,CuSO_4\\ \%m_{Cu}=\dfrac{M_{Cu}}{M_{Cu}+M_S+4.M_O}.100\%=\dfrac{64}{64+32+4.16}.100=40\%\\ \%m_S=\dfrac{M_S}{M_{Cu}+M_S+4.M_O}.100\%=\dfrac{32}{64+32+4.16}.100=20\%\\ \Rightarrow\%m_O=100\%-\left(\%m_{Cu}+\%m_S\right)=100\%-\left(40\%+20\%\right)=40\%\\ CaCO_3\\ \%m_{Ca}=\dfrac{M_{Ca}}{M_{Ca}+M_C+3.M_O}.100\%=\dfrac{40}{40+12+3.16}.100=40\%\\ \%m_C=\dfrac{M_C}{M_{Ca}+M_C+3.M_O}.100\%=\dfrac{12}{40+12+3.16}.100=12\%\\ \Rightarrow\text{ }\%m_O=100\%-\left(40\%+12\%\right)=48\%\)

Answer 4

\(a,\%Mg=\dfrac{24}{95}=25.26\%\\ \%Cl=100\%-25.26\%=74.74\%\\ \%C=\dfrac{72}{78}=92.3\%\\ \%H=100\%-92.3\%=7.7\%\)

\(b,\%Fe=\dfrac{56}{72}=77.77\%\\ \%O=100\%-77.77\%=22.23\%\\ \%Fe=\dfrac{168}{232}=72.41\%\\ \%O=100\%-72.41\%=27.59\%\)

\(c,\%Cu=\dfrac{64}{160}=40\%\\ \%S=\dfrac{32}{160}=20\%\\ \%O=100\%-40\%-20\%=40\%\\ \%Ca=\dfrac{40}{100}=40\%\\ \%C=\dfrac{12}{100}=12\%\\ \%O=100\%-40\%-12\%=48\%\)