Ta có:
\(\%_O=100\%-\left(52,174\%-13,043\%\right)=60,869\%\)
\(X=\frac{12.x}{46}.100\%=52,174\%\)
\(\Leftrightarrow x=2\)
\(Y=\frac{1.y}{46}.100\%=13,043\%\)
\(\Leftrightarrow y=6\)
\(Z=\frac{16.z}{46}.100\%=60,069\%\)
\(\Leftrightarrow z=2\)
Vậy CTHH là C2H6O2