Số mol trong \(100\left(g\right)\) hợp chất
\(n_C=\dfrac{81,81}{12}=6,82\left(mol\right)\)
\(n_H=\dfrac{18,19}{1}=18,19\left(mol\right)\)
\(n_C:n_H=6,82:18,19=1:2,66=3:8\)
Vậy \(CTPT\left(A\right):C_3H_8\)
\(CTCT\left(A\right):CH_3-CH_2-CH_3\left(Propan\right)\)