A compound 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