Aniline, a starting material for urethane plastic foams, consists of c, h, and n. combustion of such compounds yields co2, h2o, and n2 as products. if the combustion of 9.71 g of aniline yields 6.63 g h2o and 1.46 g n2, what is its empirical formula?

Since we're going to need to calculate molar masses, let's start with looking up the atomic weights of the associated elements. Atomic weight carbon = 12.0107 Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol Molar mass N2 = 2 * 14.0067 = 28.0134 g/mol Now calculate the number of moles of H2O and N2 we have Moles H2O = 6.63 g / 18.01488 g/mol = 0.368029096 moles Moles N2 = 1.46 g / 28.0134 g/mol = 0.052117915 moles Since there's 2 hydrogen atoms per H2O molecule, we have 2 * 0.368029096 mol = 0.736058192 moles of hydrogen. The mass of that hydrogen is 0.736058192 mol * 1.00794 g/mol = 0.741902494 g Since there's 2 nitrogen atoms per N2 molecule, we have 2 * 0.052117915 mol = 0.10423583 moles We already know the mass of nitrogen we have. Now we need to determine the mass of carbon we have and the number of moles of carbon. Subtract the know mass of hydrogen and nitrogen from the total mass of Aniline. So 9.71 g - 0.741902494 g - 1.46 g = 7.508097506 g Moles carbon = 7.508097506 g / 12.0107 g/mol = 0.625117396 mol We now have the ratio of carbon : hydrogen : nitrogen of 0.625117396 : 0.736058192 : 0.10423583 We need to get a ratio of small integers that closely approximates the above ratio. Start by dividing all the numbers by the smallest of the numbers. Giving: 5.997145089 : 7.061470053 : 1 All three of those values are acceptably close to integers with the ratio of 6:7:1 So the empirical formula of Aniline is C6H7N

Forgot the question hahaha