Đáp án:

$C_3H_6O_2$

Giải thích các bước giải:

$n_{C_xH_yO_z}=\dfrac{3,584}{22,4}=0,16(mol)$

$n_{C_2H_4}=\dfrac{4,032}{22,4}=0,18(mol)$

$m_{C_xH_yO_z}+m_{C_2H_4}=16,88$

$\Rightarrow m_{C_xH_yO_z}=16,88-0,18.28=11,84(g)$

$\Rightarrow M_{C_xH_yO_z}=\dfrac{11,84}{0,16}=74(g/mol)$

$n_{H_2O}=\dfrac{15,12}{18}=0,84(mol)$

$\Rightarrow n_{H(\text{trong hh})}=2n_{H_2O}=1,68(mol)$

$n_{CO_2}=\dfrac{18,816}{22,4}=0,84(mol)$

$\Rightarrow n_{C(\text{trong hh})}=n_{CO_2}=0,84(mol)$

Mặt khác:

$n_{C(\text{trong }C_2H_4)}=2n_{C_2H_4}=0,36(mol)$

$n_{H(\text{trong }C_2H_4)}=4n_{C_2H_4}=0,72(mol)$

$\Rightarrow n_{C(C_xH_yO_z)}=0,84-0,36=0,48(mol)$

$n_{H(C_xH_yO_z)}=1,68-0,72=0,96(mol)$

$\Rightarrow n_{O(C_xH_yO_z)}=\dfrac{11,84-0,48.12-0,96}{16}=0,32(mol)$

$\Rightarrow x:y:z=0,48:0,96:0,32=3:6:2$

$\Rightarrow C_xH_yO_z$ có dạng $(C_3H_6O_2)_n$

Mà $M_{C_xH_yO_z}=74$

$\Rightarrow (12.3+6+16.2).n=74$

$\Rightarrow n=1$

Vậy $C_xH_yO_z$ là $C_3H_6O_2$