\(n_A=\dfrac{3,584}{22,4}=0,16\left(mol\right)\\ n_C=n_{CO_2}=\dfrac{14,08}{44}=0,32\left(mol\right)\\ n_H=2n_{H_2O}=2.\dfrac{8,64}{18}=0,96\left(mol\right)\)
Trong 1 mol A có: \(\left\{{}\begin{matrix}n_C=\dfrac{0,32}{0,16}=2\left(mol\right)\\n_H=\dfrac{0,96}{16}=6\left(mol\right)\end{matrix}\right.\)
=> CTHH của A là C2H6