\(\left\{{}\begin{matrix}n_H=2.\dfrac{5,4}{18}=0,6\left(mol\right)\\n_C=\dfrac{26,4}{44}=0,6\left(mol\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}m_H=0,6.1=0,6\left(g\right)\\m_C=0,6.12=7,2\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_H=\dfrac{0,6}{7,8}=7,7\%\\\%m_C=\dfrac{7,2}{7,8}=92,3\%\end{matrix}\right.\)\
Xét 0,6 + 7,2 = 7,8 => A chỉ có C và H
\(CTPT:C_aH_b\\ \rightarrow a:b=0,6:0,6=1:1\\ \rightarrow\left(CH\right)_n=2,69.29=78\\ \rightarrow n=6\\ CTPT:C_6H_6\left(benzen\right)\)