When you combust that solvent, the general reaction is: (C-H-Cl) + O2 ==> CO2 + H2O + Cl2
All of the C in the sample ends up as CO2, all the H becomes H2O, and all Cl becomes Cl2 gas.
0.194 g CO2 x (1 mole CO2 / 44.0 g CO2) = 0.00441 moles CO2
0.00441 moles CO2 x (1 mole C / 1 mole CO2) = 0.00441 moles C in the solvent
0.00441 moles C x (12.0 g C / 1 mole C) = 0.0529 g C in the solvent
0.0263 g H2O x (1 mole H2O / 18.0 g H2O) = 0.00146 moles H2O
0.00146 moles H2O x (2 moles H / 1 mole H2O) = 0.00292 moles H in the solvent
0.00292 moles H x (1.01 g H / 1 mole H) = 0.0030 g H in the solvent
Since the original sample was 0.108 g, and the g C = 0.0529 and g H = 0.0030, then the rest must be Cl (0.108 - 0.0529 - 0.0030) = 0.0521 g Cl
0.0521 g Cl x (1 mole Cl / 35.5 g Cl) = 0.00147 moles Cl
So there are 0.00441 moles C, 0.00292 moles H, and 0.00147 moles Cl in the original sample. If we divide those by the smallest (0.00147) we get a C/H/Cl ratio of 3/2/1.
The empirical formula is C3H2Cl.