Answer
See the answer below.
Work Step by Step
Number of moles of water:
$0.09912\ g\div 18.015\ g/mol=5.502×10^{-3}\ mol$
$0.011\ mol\ H\rightarrow 0.0111\ g$
Number of moles of NaOH:
$0.3283\ M×28.81/1000=0.00946\ mol$
From stoichiometry:
$0.00473\ mol\ C\rightarrow 0.0568\ g$
Number of moles of nitrogen:
$n=(65.12/760×1\ atm)×0.225\ L/((0.082)(298\ K))$
$n=0.000788\ mol\ N_2$
$0.00158\ mol\ N\rightarrow 0.0221\ g$
Number of moles of oxygen:
$(0.1152-0.0111-0.0568-0.0221)\ g/(15.999\ g/mol)=0.00158\ mol$
H/O ratio: $0.0110/0.00158=7.0$
C/O ratio: $0.00473/0.00158=3.0$
N/O ratio: $0.00158/0.00158=1.0$
Empirical formula:
$C_4H_7NO$
Molecular weight of the empirical formula:
$85.11\ g/mol$
Molecular formula:
$C_8H_{14}O_2N_2$
