What is the empirical formula of a compound containing 40.00% carbon, 6.67% hydrogen, and 53.33% oxygen by mass?

To determine the empirical formula of a compound, we need to find the simplest whole-number ratio of the elements present in the compound. In this case, we need to find the ratio of carbon (C), hydrogen (H), and oxygen (O) in the compound. Given that the compound contains 40.00% carbon, 6.67% hydrogen, and 53.33% oxygen by mass, we can assume we have 100 grams of the compound. To find the number of moles of each element in 100 grams of the compound, we divide the mass of each element by its molar mass. The molar mass of carbon is 12.01 g/mol, so we have (40.00 g carbon) / (12.01 g/mol carbon) = 3.33 moles of carbon. The molar mass of hydrogen is 1.01 g/mol, so we have (6.67 g hydrogen) / (1.01 g/mol hydrogen) = 6.60 moles of hydrogen. The molar mass of oxygen is 16.00 g/mol, so we have (53.33 g oxygen) / (16.00 g/mol oxygen) = 3.33 moles of oxygen. Next, we need to find the simplest whole-number ratio of the elements. To do this, we divide the moles of each element by the smallest number of moles. The smallest number of moles is 3.33, which corresponds to both carbon and oxygen. Dividing the moles of each element by 3.33, we get: Carbon: 3.33 moles / 3.33 = 1 mole Hydrogen: 6.60 moles / 3.33 = 1.98 moles (approximated to 2 moles) Oxygen: 3.33 moles / 3.33 = 1 mole Therefore, the empirical formula of the compound is CH2O.