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
