then 74.68% I --> 74.68 g I
divided by molar mass of 126.9 gives 0.58 moles of I
Now, of the three moles that you have, divide all three by the smallest number you have!
which is 0.58 (from the Iodine!)
so:
for C it's 1.766/0.58 = about 3
for H it's 4.12/ 0.58 = about 7
and for I it's 0.58/0.58 = 1
and thats your ratio so
empirical formula =
C3H7I !
and no i dont think the formula could contain oxygen because the oxygen in the product comes from the combustion of oxygen in the air!
ps. you may have noticed i didnt use the mass percentage of the iodine part, i think i just discovered i didnt need it.