An unknown compound contains only C, H, and O.

Combustion of 5.70g of this compound produced 11.4g of CO2 and 4.66g of H2O.

What is the empirical formula of the unknown compound? C(x)H(y)O(z)

the molar mass of H2O is 18.016 g/mol ((2 x 1.008) + 16.00)... you have used 8

From what I can see your method/reasonig is fine, so put in the correct molar mass of H2O all should be good.

Mass of C in 11.4g CO2 is (12/44)*11.4g= 3.109g C
Mass of H in 4.66g H2O is (2/18)*4.66g= 0.51778g H
Mass of O = 5.70g - (3.109 - 0.51778 ) which is 2.073 g O

Dividing each mass by the atomic number:
C=3.109/12 = .2591
H=0.51778/1 = 0.51778
O = 2.073/16 = 0.1296

Divide them all by 0.1296
since O is the smallest number

C= 0.2591/ 0.1296 = 2
H=0.51778/.1296 = 4
O= 0.1296/ 0.1296 = 1

C2H4O