giả sử các h/c đều có số mol = 1
a/ %mS = \(\dfrac{32.100}{32+16.3}=40\%\)
=> %mO = 100% - 40% = 60%
b/ %mFe = \(\dfrac{56.2.100}{56.2+16.3}=70\%=>\%mO=100\%-70\%=30\%\)
c/ %mCa =\(\dfrac{40.3.100}{40.3+31.2+16.8}=38,7\%\)
%mP = \(\dfrac{31.2.100}{40.3+31.2+16.8}=20\%\)
%mO = 100%-38,7%-20%=41,3%
a) \(\%S=\dfrac{32}{32+16\times3}\times100\%=40\%\)
\(\Rightarrow\%O=100\%-40\%=60\%\)
b) \(\%Fe=\dfrac{56\times2}{56\times2+16\times3}\times100\%=70\%\)
\(\Rightarrow\%O=100\%-70\%=30\%\)
c) \(\%Ca=\dfrac{40\times3}{40\times3+2\times\left(31+16\times4\right)}\times100\%=38,71\%\)
\(\%P=\dfrac{31\times2}{40\times3+2\times\left(31+16\times4\right)}\times100\%=20\%\)
\(\Rightarrow\%O=100\%-38,71\%-20\%=41,29\%\)
