Combustion Analysis help, please!!

To enter the AP course, a series of questions must be completed. I'm completely stuck on this one:

A compound produced as a by-product in an industrial synthesis of polymers was found to contain carbon, hydrogen, and iodine. A combustion analysis of 1.70g of the compound produced 1.32 g of CO2 and .631 g of H2O. The mass percentage of iodine in the compound was determined by converting the iodine in a .850-g sample of the compound into 2.31g of lead(II) iodide. What is the empirical formula of the compound? Could the compound also contain Oxygen?
Please show work and explain your answer :)

CHI (unknown) + O2 --> CO2 + H2O [formula]
you know that from the original compound, all the carbon went into the CO2.
So see how much C is in the CO2:
1.32 g CO2/ (44g/mol) =0.03 mol CO2
for every mol of CO2 there is 1 mole of C so 0.03 mol of CO2= 0.03 mol of C
0.03 mol C x 12 g/mol = 0.36 g of C
find the percent of C in the original compound:
0.36 g C/ 1.70 g CHI <-- unknown compound = 21.2 % C

okay now, do the same thing for Hydrogen using the amount given for water
take the 0.631 g H20 and convert to moles, then convert to Hydrogen moles which is two H for every 1 mol of H20 so be careful
and then change the moles of H's to grams
then make a percentage with the original compound like done above:
and I got : 0.07 g H/ 1.70 g CHI= 4.12% H

Now you have three components of a compound and two percentages so simply by subracting the percentages you already have from 100, you can find the percentage of Iodine.
100- (4.12% + 21.2 %) = 74.68% Iodine

Now take the percentages and pretend like they are grams of the element
so we had 21.2 % of C so make that
21.2 g C and do train tracks or mole map or whatever you learned to convert that into moles of C
which in this case you just divide by 12 g/mol and get 1.766 mol C

then 4.12%H --> to 4.12 g H which happens to be just 4.12 moles of H too