An organic compound containing C, H, O, and S is subjected to two analytical procedures. In the first procedure a 9.33mg sample is burned which gives 19.50mg of CO2 and 3.99mg H2O. In a second procedure, a separate 11.05mg sample is fused with Na2O2 and the resulting sulfate is precipitated as BaSO4, which when dried weighs 20.4mg. The amount of oxygen in the original sample is obtained by the difference. Determine the empirical formula of this compound.

So what have you got so far ?

Not much, I set up some molar ratios and found the molar ratio between CO2 and H2o but I am not sure how to proceed from here

From the first analytical procedure, you can get the number of moles of C and H in the compound, then find the mass they occupy in the 9.33 mg sample.

I'm not sure why you found the mole ratio...

Anyway, next, using the mass of BaSO4, you can get the number of moles of S. This is the number of moles of S present in the original sample of 11.05 mg sample. Find the mass that S occupies in a 9.33 mg by using proportions.

Add up the masses of C, H and S that you have right now. This should give you a number below 9.33 mg. The number which you should add to the current sum to 9.33 mg is the mass of O.

Once you have the masses of C, H, S and O, the empirical formula should become a piece of cake.

PD: Are you sure it's Na2O2 and not Na2O? I have never come across that reaction before...