Question

Lysine is an amino acid which has the following elemental composition: C, H, O, N. In one experiment, 3.263 g of lysine was combusted to produce 5.910 g of CO2 and 2.835 g of H2O. Separate analysis determined that lysine is 19.17% N. The molar mass of lysine is 150 g/mol. Determine the empirical and molecular formulas of lysine.

Answer 1

Given that  lysine has 19.17% N.

mass of Nitrogen = mass of lysine x % N =  3.263 g x  19.17% = 0.6255 g

3.263 g of lysine was combusted to produce 5.910 g of CO2 and 2.835 g of H2O.

Mass of carbon present in 5.91 g of CO2 = 5.91 g x molar mass of carbon/ molar mass of CO2

= 5.91g x 12 (g/mol) / 44 (g/mol)

= 1.611 g

Mass of hydrogen present in 2.835 g of H2O = 2.835 g x molar mass of hydrogen/ molar mass of H2O

= 2.835 g x2 (g/mol) / 18 (g/mol)

= 0.315 g

Mass of C + Mass of H + Mass of N + Mass of O = Mass of Lysine

1.611 g + 0.315 g + 0.6255 g + Mass of N = 3.263 g

Then, Mass of O = 0.7115 g

Dividing each mass by elemental molar mass tells us how many moles of each element present in the sample of butyric acid.

For C, 1.611 g/ 12 (g/mol) = 0.1342 mol C

For H, 0.315 g/ 1 (g/mol) = 0.315 mol H

For O, 0.7115 g/ 16 (g/mol) = 0.0444 mol O

For N, 0.6255 / 14 (g/mol) = 0.0446

Divide each by smallest among them i.e. 0.0444

For C, 0.1342/ 0.0444 = 3.02 mol C

For H, 0.315/ 0.0444 = 7.09 mol H

For O, 0.0444/ 0.0444 = 1 mol O

For N, 0.0446/ 0.0444 = 1 mol N

Round each value to the nearest integer,

C = 3

H = 7

O =1

N= 1

Therefore, empirical formula of Lysine = C₃H₇NO

molecular formula of lysine = (C₃H₇NO)n

empirical formula mass = 73 g/mol

Given that molar mass of lysine = 150 g/mol

n =  molar mass of lysine/   empirical formula mass = 150 g/mol/ 73 g/mol = 2

Therefore,

molecular formula of lysine = (C₃H₇NO)n =  (C₃H₇NO)₂ = C₆H₁₄N₂O₂