When a 1.50 g sample of aluminum metal is burned in an oxygen atmosphere, 2.83 g of aluminum oxide are producedHowever, the combustion of 1.50 g ultrafine aluminum in air results in 2.70 g of a product, which is a mixture of 80% aluminum oxide and 20 % aluminum nitride (% by mass)Use this information to determine the empirical formulas of aluminum oxide and aluminum nitride.
(1.50 g Al) / (26.9815 g Al/mol) 0.0556 mol Al (2.83 g - 1.50 g) / (15.9994 g O/mol) 0.0831 mol O Divide by the smaller number of moles: 0.0556 mol Al / 0.0556 1.000 0.0831 mol O / 0.0556 1.494 In order to make nearly whole numbers out of these molar ratios, multiply by 2 then round to the nearest whole number to find the empirical formula: Al2O3 (2.70 g) x (0.80) 2.16 g Al2O3 (2.16 g Al2O3) / (101.9614 g Al2O3/mol) x (53.9631 g Al2/mol) 1.14 g Al in Al2O3 1.50 g - 1.14 g 0.36 g Al in Al nitride (0.36 g Al) / (26.9815 g Al/mol) 0.01334 mol Al ((2.70 g) x (0.20) - 0.36 g Al) / (14.0067 g N/mol) 0.01285 mol N Divide by the smaller number of moles: 0.01334 mol Al / 0.01285 1.038 0.01285 mol N / 0.01285 1.000 Round to the nearest whole number to find the empirical formula: AlN
