A 2.20 g-sample of a compound containing carbon, hydrogen, and oxygen is burned and it produces 4.61 g CO2 and 0.94 g H2O. What is the empirical formula of this compound? (i.e., C2H4O)

I found the mass of Carbon
(4.61/44g)*(1 mol C/1 mol CO2)=.1047

I found the mass of Hydrogen
(.94/18g)*(2 mol H/1 mol H20)=.1044

2.20-.1047-.1044=1.99

Then I divided all the numbers by the lowest which is .1044. I keep getting CH019?
What am I doing everything wrong?

The problem is that you didn't quite follow my instructions.

I found the mass of Carbon
(4.61/44g)*(1 mol C/1 mol CO2)=.1047
What you have done here is to calculate MOLES C, not grams C. Multiply by 12 to get grams C.

I found the mass of Hydrogen
(.94/18g)*(2 mol H/1 mol H20)=.1044
Same thing here. You have calculated MOLES H. Convert this number to grams H atoms

2.20-.1047-.1044=1.99
Now you are subtracting 2.20 GRAMS - mols C - mols H and you can't do that. You should go back and calculate grams C and grams H so when you subtract from 2.20 g sample, you will get grams O instead of some other fictitious number.
Then I divided all the numbers by the lowest which is .1044. I keep getting CH019?
What am I doing everything wrong?

It seems like you made a small error during your calculations. Let's go through the steps again to find the correct empirical formula.

First, let's find the moles of carbon (C) and hydrogen (H) in the sample:

Moles of C = (mass of CO2 produced) / (molar mass of CO2)
= 4.61 g / 44 g/mol
= 0.1048 mol

Moles of H = (mass of H2O produced) / (molar mass of H2O)
= 0.94 g / 18 g/mol
= 0.0522 mol

Next, we need to find the moles of oxygen (O) by subtracting the moles of C and H from the total moles in the sample:

Moles of O = Total moles - Moles of C - Moles of H
= 0.1048 mol + 0.0522 mol
= 0.157 mol

Next, we need to find the whole number ratio of the elements by dividing the moles of each element by the smallest number of moles (in this case, H):

C: 0.1048 mol / 0.0522 mol = 2
H: 0.0522 mol / 0.0522 mol = 1
O: 0.157 mol / 0.0522 mol = 3

Finally, to get the empirical formula, we write the whole number ratio of atoms as subscripts:

C2H3O3

Therefore, the empirical formula of the compound is C2H3O3.