\(M_{H_2O}=1.2+16=18\left(g/mol\right)\)
\(\%m_H=\dfrac{1.2}{18}.100\%=11\%\)
\(\%m_O=\dfrac{16}{18}.100\%=89\%\)
CTHH: \(H_2O_2\)
\(M_{H_2O_2}=1.2+16.2=34\left(g\right)\)
\(\%H=\dfrac{2.1}{34}.100\%\approx6\%\)
\(\%O=\dfrac{2.16}{34}.100\%\approx94\%\)