A compound contains 40.0% C, 6.7% H 53.3% O. If the molecular mass of the compound is 180, its molecular formula is [C =12, H =1,O]

Answer Details

The first step to solving this problem is to determine the empirical formula of the compound. We assume that we have 100 grams of the compound, so we have 40.0 g of C, 6.7 g of H, and 53.3 g of O. To find the empirical formula, we need to convert these masses to moles. To do this, we divide each mass by the respective atomic weight: - C: 40.0 g / 12.01 g/mol = 3.33 mol - H: 6.7 g / 1.01 g/mol = 6.63 mol - O: 53.3 g / 16.00 g/mol = 3.33 mol The mole ratio of these elements can then be expressed as 1:2:1 (C:H:O). Therefore, the empirical formula is CH2O. To find the molecular formula, we need to know the molecular mass of the compound. In this case, it is given as 180 g/mol. We can calculate the molecular formula by dividing the molecular mass by the empirical formula mass and then multiplying each subscript by that factor. - Empirical formula mass: 12.01 + 2(1.01) + 16.00 = 30.03 g/mol - Factor: 180 g/mol ÷ 30.03 g/mol = 5.994 Multiplying each subscript by 5.994 gives us the molecular formula of the compound: C6H12O6, which is option (iv).