Đáp án:

$C_4H_6O_2Cl_2$

Giải thích các bước giải:

$\text{Gọi CTTQ là }\ C_xH_yO_zCl_t\\ x:y:z:t=\dfrac{\%C}{M_C}:\dfrac{\%H}{M_H}:\dfrac{\%O}{M_O}:\dfrac{\%Cl}{M_{Cl}}=\dfrac{30,6}{12}:\dfrac{3,85}{1}:\dfrac{20,39}{16}:\dfrac{45,16}{35,5}=2:3:1:1\\ \to CTĐG:\ C_2H_3OCl\\ \text{A có dạng}\ (C_2H_3OCl)_n\ (n\in N^*)\\ 120<M_A<190\\ \Rightarrow 120<78,5n<190\\ \Rightarrow 1,5<n<2,42\\ Mà \ n\in \ N^*\\ \Rightarrow n=2\\ \to CTPT:\ C_4H_6O_2Cl_2$

Bạn tham khảo:

$C_xH_yO_zCl_t$

$x:y:z:t=\dfrac{30,6}{12} : \dfrac{3,85}{1} : \dfrac{20,39}{16}: \dfrac{45,16}{35,5}$

$x:y:z:t=2,55 : 3,85 : 1,27 : 1,27$

$x:y:z:t=2:3:1:1$

$\to (C_2H_3OCl)_n$

$120 \leq M_A=(C_2H_3OCl)_n \leq 190$

$\to 120 \leq 78,5n \leq 190$

$1,52 \leq n \leq 2,42$

$\to n=2$

$\to C_4H_6O_2Cl_2$