Question

A 3.159 g sample of an amino acid, CxHyOzNn, was analyzed by combustion reaction ad the following product data was obtained:

4.7885 g CO2 and 2.2868 g H2O
The nitrogen in the sample was converted to 1.2360 g NH3. The molecular mass of the compound was determined to be between 170 u and 200 u.

What is the (a) empirical formula and (b) molecular formula of this amino acid?

Answer 1

NO of mole of C = no of mole CO2 = 4.7885/44 = 0.11 mole

NO of mole of H = 2 * no of mole H2O = 2*(2.2868/18) = 0.254 mole

NO of mole of N = no of mole NH3 = 1.236/17 = 0.0727 mole

mass of O in sample = 3.159 - (0.11*12+0.254*1+0.0727*14) = 0.5672 g

no of mol of O = 0.5672/16 = 0.03545 mol

simplest ratio

C = 0.11/0.03545 = 3.1

H = 0.254/0.03545 = 7.16

N = 0.0727/0.03545 = 2

O = 0.03545/0.03545 = 1

whole number ratio

C = 3 , H = 7 , N = 2 , O = 1

empiricalformula = C3H7ON2

n = 2 ( because E.f.wt = 87 g/mol )

molecular formula = 2*(C3H7ON2)

   = C6H14O2N4