+ CuSO4
Ta có: MCuSO4 = 64 + 32 + 16 x 4 = 160 (g/mol)
=> \(\%_{Cu}=\frac{64}{160}.100\%=40\%\)
\(\%_S=\frac{32}{160}.100\%=20\%\)
\(\%_O=100\%-40\%-20\%=40\%\)
+ Mg(NO3)2
Ta có: MMg(NO3)2 = 24 + 14 x 2 + 16 x 6 = 152 (g/mol)
=> \(\%_{Mg}=\frac{24}{152}.100\%=15,8\%\)
\(\%_N=\frac{14.2}{152}.100\%=18,4\%\)
\(\%_O=100\%-15,8\%-18,4\%=65,8\%\)
a) CuSO4
MCuSO4 = 64 + 32 + 4 . 16 = 160 g
%mCu = \(\frac{64.100}{160}\)= 40%
%mS = \(\frac{32.100}{160}\)= 20%
%mO= \(\frac{16.4.100}{160}\)= 40%
b) MMg(NO3)2= 24 + 14 . 2 + 6 . 16 = 148 g
%mMg = \(\frac{24.100}{148}\)= 16,2 %
%mN = \(\frac{14.2.100}{148}\)= 19 %
%mO= \(\frac{6.16.100}{148}\)= 64,8 %
MCuSO4 = 64 + 32 + 16.4 = 160 (g/mol)
nCu = 1 mol
nS = 1 mol
nO = 4 mol
mCu = 1.64 = 64 (g)
mS = 1.32 = 32 (g)
mO = 4.16 = 64 (g)
\(\%m_{Cu}=\frac{64}{160}.100\%=40\%\)
\(\%m_S=\frac{32}{160}.100=20\%\)
\(\%m_O=100\%-40\%-20\%=40\%\)
Bài còn lại tương tự cậu tự làm nhé.
