Acompound used to test for the presence of ozone in the stratosphere contains 96.2 percent thallium and 3.77 percent oxygen what is its empirical formula?

When ozone = O3. and we have 96.2% thallium of the compound. and 3.77% oxygen of the compound. We need to get the no.of moles of thallium & oxygen to get the empirical formula.
So we can assume that:
96.2% thallium = 96.2 grams of thallium
3.77% oxygen = 3.77 grams of oxygen
as each 100 gram of the compound contains 3.77 grams of O2 & 96.2 grams of Ti
no.of moles of thallium = 96.2 gm/molar mass of thallium 
                                       = 96.2 gm / 204 = 0.5 mol
no.of moles of oxygen = 3.77 gm / molar mass of oxygen 
                                      = 3.77 gm / 16
                                      = 0.25 mol 
to get the empirical formula we need to make this numbers a whole numbers so we multiply by 4 by oxygen and thallium to get the lowest whole numbers that we can get.
So, after multiple:
∴ no.of moles of oxygen = 1 
and no.of moles of thallium = 2 
So the empirical formula is 
Ti2O1 
→
Ti2O

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