\(M_{C_xH_y}=11.4=44gam\)
\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(n_{H_2O}=\dfrac{7,2}{18}=0,4mol\)
\(n_C=n_{CO_2}=0,3mol\)
\(n_H=2n_{H_2O}=2.0,4=0,8mol\)
\(n_C:n_H=0,3:0,8=3:8\)
-Công thức nguyên (C3H8)n
-Ta có: (12.3+8)n=44\(\rightarrow\)44n=44\(\rightarrow\)n=1
-CTPT: C3H8