I offer an alternative method to that proposed by Erik D for solving this problem, not that there is anything wrong with the solution provided.
Assuming the unknown compound contains only C, H and O, we can proceed as follows:
moles of C: 1.3203 g CO2 x 1 mole CO2/44.00 g x 1 mole C/mole CO2 = 0.0300 moles C
mass of C: 0.0300 moles C x 12.01 g/mole = 0.3603 g C
moles H: 0.5405 g H2O x 1 mole/18 g x 2 mole H/mole H2O = 0.0600 moles H
mass H: 0.00600 moles H x 1.0079 g/mole = 0.06047 g H
To find mass of O and moles of O, we subtract the mass of C and H from the mass of the sample to get mass of O and then convert to moles of O.
0.7408 g - 0.3603 g - 0.06047 g = 0.3200 g O
moles O = 0.3200 g x 1 mole/16 g = 0.0200 moles O
Summarizing, we have...
0.0300 moles C
0.0600 moles H
0.0200 moles O
Dividing though by the smallest number (0.0200) in order to get whole numbers we obtain
1.5 moles C
3.0 moles H
1 mole O
and multiplying all by 2 to get whole numbers, we obtain
3 moles C
6 moles H
2 moles O
So empirical formula is C3H6O2