Many organic compounds contain carbon and hydrogen and when burnt in air produce carbon dioxide and water.By calculating how many moles of carbon dioxide and water are formed, the moles of carbon and moles of hydrogen in the compound can be used to determine the empirical formula.The empirical formula is the simplest whole number ratio of moles of each element in a compound.The empirical formula of the compound can be found by determining the simplest whole number ratio of moles of carbon, hydrogen and any oxygen (found by subtraction).
A 1.87g sample of an organic compound completely burned in air to form 2.65g of carbon dioxide and 1.63g of waterAll the carbon in the carbon dioxide comes from the organic compound. The moles of carbon dioxide is 2.65÷44The moles of water is 1.63÷18There are two moles of hydrogen for each mole of water.Mass of carbon = 0.0602 × 12 = 0.723 g and mass of hydrogen = 0.181 × 1 = 0.181 g. These add up to 0.904 g, less than 1.87 g, so the compound also contains 1.87 - 0.904 = 0.966 g of oxygen, which is 0.966 ÷ 16 = 0.0604 mol.C : H : O = 0.0602 : 0.181 : 0.0604 = 1 : 3 : 1, so the empirical formula is CH3O.
A 1.87g sample of an organic compound completely burned in air to form 2.65g of carbon dioxide and 1.63g of waterAll the carbon in the carbon dioxide comes from the organic compound. The moles of carbon dioxide is 2.65÷44The moles of water is 1.63÷18There are two moles of hydrogen for each mole of water.Mass of carbon = 0.0602 × 12 = 0.723 g and mass of hydrogen = 0.181 × 1 = 0.181 g. These add up to 0.904 g, less than 1.87 g, so the compound also contains 1.87 - 0.904 = 0.966 g of oxygen, which is 0.966 ÷ 16 = 0.0604 mol.C : H : O = 0.0602 : 0.181 : 0.0604 = 1 : 3 : 1, so the empirical formula is CH3O.