Example Question - 2

0.10000 g of an amino acid produced 0.11710 g of carbon dioxide and 0.05992 g of water on complete combustion.In a second reaction, 0.10000 g of the amino acid produced 0.02264 g of ammonia.Determine the empirical formula of the amino acid.First find the mass of each element. All the carbon ends up in CO₂: moles of C = 0.11710 ÷ 44.0 = 0.002661 mol, so mass of C = 0.002661 × 12.0 = 0.03194 g. All the hydrogen ends up in H₂O: moles of H = 2 × (0.05992 ÷ 18.0) = 0.006658 mol, so mass of H = 0.00666 g. All the nitrogen ends up in NH₃: moles of N = 0.02264 ÷ 17.0 = 0.001332 mol, so mass of N = 0.01865 g.Oxygen is found by difference: 0.10000 - 0.03194 - 0.00666 - 0.01865 = 0.04275 g, so moles of O = 0.04275 ÷ 16.0 = 0.002672 mol.Divide each amount by the smallest (0.001332 mol): C = 2.00, H = 5.00, N = 1.00, O = 2.01. The empirical formula is C₂H₅NO₂, which is the amino acid glycine.