Question
Combustion analysis of a 18.0-mg sample of common headaheremedy which contains carbon, hydrogen, nitrogen, and oxygen gave42.0 mg of carbon dioxide and 9.72 mg of water. In a separateanalysis, a 35.0-mg sample gave 3.941 mg of ammonia. The molar masswas also determined to be 302 g/mole. What is the molecular formula of thiscompound? Combustion analysis of a 18.0-mg sample of common headaheremedy which contains carbon, hydrogen, nitrogen, and oxygen gave42.0 mg of carbon dioxide and 9.72 mg of water. In a separateanalysis, a 35.0-mg sample gave 3.941 mg of ammonia. The molar masswas also determined to be 302 g/mole. What is the molecular formula of thiscompound?
Explanation / Answer
We Know that : The number of moles of CO2 obtained = 42.0*10-3 g / 44 g / mol = 0.9545 * 10-3 mole The number of moles of Carbonin the compound = 0.9545 * 10-3 mole * 1 mole C / 1 mole CO2 The number of moles of Water formed = 9.72 * 10-3g / 18 g / mol = 5.4 * 10-4 mole The number of moles of Hydrogen in the compound = 5.4 *10 -4 mol H2O * 2mol H / 1 mol H2O = 1.08 * 10-3 mol The number of moles of NH3 = 3.491 * 10-3 g / 17 g / mol = 2.05 * 10 -4 mole The number of moles ofNitrogen in the sample = 2.05 * 10-4 mole For 35.0 * 10-3 g of the sampleconsist of 2.05 * 10-4 mole For 18.0 * 10-3 g of the sample consistof 1.05 * 10-4 mol For thenumber of moles of Oxygen = 18 - ( 11.54 +1.08 + 1.47 ) = 3.91 g number of moles of Oxygen in the compound is = 3.91 g / 32 g / mol = 0.122 *10-3 mol So theempirical formula of the compound can be written as : C - 0.9545 * 10-3 mole / 1.05 * 10-4 mol H - 1.08 *10 -3 / 1.05 * 10-4 mol N - 1.05 * 10-4 / 1.05* 10-4 mol O - 0.122 * 10-3 mol / 1.05 * 10-4 mol C9 H10N1O1 empirical formula x n = molecular formula n = molecular weight / empirical formulaweight = 302 / 148 = 2 Molecularformula of the compound is : C18 H20 N2O2 The number of moles of Water formed = 9.72 * 10-3g / 18 g / mol = 5.4 * 10-4 mole The number of moles of Hydrogen in the compound = 5.4 *10 -4 mol H2O * 2mol H / 1 mol H2O = 1.08 * 10-3 mol The number of moles of NH3 = 3.491 * 10-3 g / 17 g / mol = 2.05 * 10 -4 mole The number of moles ofNitrogen in the sample = 2.05 * 10-4 mole For 35.0 * 10-3 g of the sampleconsist of 2.05 * 10-4 mole For 18.0 * 10-3 g of the sample consistof 1.05 * 10-4 mol For thenumber of moles of Oxygen = 18 - ( 11.54 +1.08 + 1.47 ) = 3.91 g number of moles of Oxygen in the compound is = 3.91 g / 32 g / mol = 0.122 *10-3 mol So theempirical formula of the compound can be written as : C - 0.9545 * 10-3 mole / 1.05 * 10-4 mol H - 1.08 *10 -3 / 1.05 * 10-4 mol N - 1.05 * 10-4 / 1.05* 10-4 mol O - 0.122 * 10-3 mol / 1.05 * 10-4 mol C9 H10N1O1 empirical formula x n = molecular formula n = molecular weight / empirical formulaweight = 302 / 148 = 2 Molecularformula of the compound is : C18 H20 N2O2