What is the empirical formula for the compound?

Your sample definately contains Oxygen.

The law of conservation of mass tells us that there must be the same mass in the products as there were in the reactants.

The mass of C + mass H in you products = 2.202 g + 0.5509 g = 2.753 g
But you started with 4.226 g of compound, so the difference in mass must be Oxygen
mass O = 4.226 g - 2.202 - 0.5509 = 1.473 g Oxyygen in the compound

Now work out the moles of C, H and O that were in the sample

moles = mass / molar mass

moles C = 2.202 g / 12.01 g/mol = 0.1833 moles C
moles H = 0.5509 g / 1.008 g/mol = 0.5462 moles H
moles O = 1.473 g / 16.00 g/mol = 0.0921 moles O

Arrange the moles in a ratio format

moles C : moles H : moles O
= 0.1833 : 0.5462 : 0.0921

Divide each number by the smallest number in the ratio to get it into a whole number ratio.

C : H : O
(0.1833 / 0.0921) : (0.5462/0.0921) : (0.0921/0.0921)
= 2 : 6 : 1

Empirical formula
C2H6O