Bảo toàn C: \(n_C=n_{CO_2}=\dfrac{1,12}{22,4}=0,05mol\)
Bảo toàn H: \(n_H=2.n_{H_2O}=2.\dfrac{1,35}{18}=0,15mol\)
\(\rightarrow n_O=\dfrac{1,15-\left(0,05.12+0,15.1\right)}{16}=0,025mol\)
=> A gồm có C,H và O
\(C_xH_yO_z+\left(x+\dfrac{y}{4}+\dfrac{z}{2}\right)O_2\rightarrow\left(t^o\right)xCO_2+\dfrac{y}{2}H_2O\)
\(M=1,4375.32=46\) ( g/mol )
\(CTPT:C_xH_yOz\)
\(x:y:z=0,05:0,15:0,025=2:6:1\)
\(CTĐG:\left(C_2H_6O\right)n=46\)
\(\Leftrightarrow n=1\)
Vậy \(CTPT:C_2H_6O\)