Combustion analysis ....
This is more of a mathematical exercise than an actual useful tool in chemistry. There are more and better ways to determine the formulas of compounds that the archaic combustion analysis method. Therefore, don't think you are learning about something that modern chemists actually do.
Be that as it may, you can learn to do some mathemtaics which is related to chemistry. Determine both the mass and moles of C and H so that you can determine the mass and moles of O. The empirical formula is the simplest whole-number ratio of moles.
CxHyOz(g) + O2(g) --> CO2(g) + H2O(g)
1.130g ......................... 1.064g ...0.3631g
1.064g CO2 x (1 mol CO2 / 44.0g CO2) x (1 mol C / 1 mol CO2) = 0.0242 mol C, and 0.290g C
0.3631g H2O x (1 mol H2O / 18.0g H2O) x (2 mol H / 1 mol H2O) = 0.0403 mol H, and 0.0407g H
1.130g CxHyOz - 0.290g C - 0.0407g H = 0.799g O, and 0.0500 mol O
C..... 0.0242 mol C / 0.0242 = 1 mol C ......... x 3 = 3 mol C
H .... 0.0403 mol H / 0.0242 = 1.67 mol H .... x 3 = 5 mol H
O .... 0.0500 mol O / 0.0242 = 2 mol O ......... x 3 = 6 mol O
Unless I've done something remarkably bad and its not jumping out at me, the empirical formula seems to be heavy on oxygen and not on your list. In fact, none of these formulas when used to calculate the mass of CO2 will produce a mass of 1.064g. The masses of CO2 are are all in the range of 2.0g to 3.3g
1.130g C7H6O2 x (1 mol C7H6O2 / 122.0g C7H6O2) x (7 mol CO2 / 1 mol C7H6O2) x (44.0g CO2 / 1 mol CO2) = 2.852g CO2
1.130g C2H3O x (1 mol C2H3O / 43.0g C2H3O) x (2 mol CO2 / 1 mol C2H3O) x (44.0g CO2 / 1 mol CO2) = 2.313g CO2
1.130g C3H5O2 x (1 mol C3H5O2 / 73.0g C3H5O2) x (3 mol CO2 / 1 mol C3H5O2) x (44.0g CO2 / 1 mol CO2) = 2.043g CO2
1.130g C4H6O x (1 mol C4H6O / 60.0g C4H6O) x (4 mol CO2 / 1 mol C4H6O) x (44.0g CO2 / 1 mol CO2) = 3.315g CO2
1.130g C5H6O2 x (1 mol C5H6O2 / 98.0g C5H6O2) x (5 mol CO2 / 1 mol C5H6O2) x (44.0g CO2 / 1 mol CO2) = 2.537g CO2
Your question writer needs to go back to the drawing board.