M\(Na_2O\)= 23.2 + 16 = 62(g/mol)
mNa trong \(Na_2O\)= 23.2 = 46(g)
%mNa trong \(Na_2O\)=\(\frac{46}{62}\).100% = 74,2%
mO trong \(Na_2O\)= 100% - 74,2% = 25,8%
\(M_{Na_2O}=23.2+16=62\left(g/mol\right)\)
\(\%m_{Na}=\dfrac{23.2}{62}.100\%\approx74,2\%\)
\(\%m_O=\%m_{Na_2O}-\%m_{Na}=100\%-74,2\%=25,8\%\)
M=23*2+16=62
%Na=23*2/62=74%
%O=100%-74%=26 %