\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{H_2O}=\dfrac{5,4}{18}=0,3\left(mol\right)\)
Bảo toán C: nC = 0,3 (mol)
Bảo toàn H: nH = 0,6 (mol)
\(n_O=\dfrac{5,8-0,3.12-0,6.1}{16}=0,1\left(mol\right)\)
nC : nH : nO = 0,3 : 0,6 : 0,1= 3:6:1
=> CTPT: (C3H6O)n
Mà MB = 29.2 = 58 (g/mol)
=> n = 1
=> CTPT: C3H6O