Lysine is an amino acid which has the following elemental composition: C, H, O, N. In one experiment, 2.175 g of lysine was combusted to produce 3.94 g of CO2 and 1.89 g H2O. In a separate experiment, 1.873 g of lysine was burned to produce 0.436 g of NH2. The molar mass of lysine is 150 g/mol. Determine the empirical and molecular formula of lysine.

Step One: determine the mass of each element present.

Carbon: 3.94 g x (12.011 / 44.0098) = 1.0753 g
Hydrogen: 1.89 g x (2.016 / 18.0152) = 0.2115 g
Nitrogen: 0.436 g x (14.007 / 16.023) = 0.38114 g
Oxygen: cannot yet be done

Why can't the oxygen be determined yet? It is because our C, H, and N data come from TWO different sources.

Step Two: Convert mass of each element to percentages.

Carbon: 1.0753 g ÷ 2.175 g = 49.44 %
Hydrogen: 0.2115 g ÷ 2.175 g = 9.72 %
Nitrogen: 0.38114 g ÷ 1.873 g = 19.17 %
Oxygen: 100 - (49.44 + 9.7 + 19.17) = 21.67 %

Step Three: Determine the moles of each element present in 100 g of lysine.

Carbon: 49.44 g ÷ 12.011 g/mol = 4.116 mol
Hydrogen: 9.72 g ÷ 1.008 g/mol = 9.643 mol
Nitrogen: 19.17 g ÷ 14.007 g/mol = 1.3686 mol
Oxygen: 21.67 g ÷ 15.9994 g/mol = 1.3544 mol

Step Four: find the ratio of molar amounts, expressed in smallest, whole numbers.

Carbon: 4.115 mol ÷ 1.3544 mol = 3.04
Hydrogen: 9.643 mol ÷ 1.3544 mol = 7.12
Nitrogen: 1.3686 mol ÷ 1.3544 mol = 1.01
Oxygen: 1.3544 mol ÷ 1.3544 mol = 1

The empirical formula is C3H7NO.

In order to determine the molecular formula, we need to know the "empirical formula weight." This value is 73.1.

We see that the approximate molecular weight is just about double this value, leading to the molecular formula of C6H14N2O2