Lysine, an essential amino acid in the human body, contains C,H,O and N only. In an experiment, the complete combustion of 2.175 g lysine gave 3.94 g of CO2 and 1.89 g of H2O. In another reaction, the nitrogen content of 1.873 g of lysine is quantitatively converted to 0.436 g of NH3.
i) Calculate the empirical formula of lysine.
To calculate the empirical formula of lysine, we need to determine the relative number of atoms of each element present in the compound.
Let's start by calculating the number of moles for each component from the given masses.
1) CO2 (carbon dioxide):
Mass of CO2 = 3.94 g
Molar mass of CO2 = 12.01 g/mol (carbon) + 2(16.00 g/mol) (oxygen) = 44.01 g/mol
Number of moles of CO2 = Mass of CO2 / Molar mass of CO2 = 3.94 g / 44.01 g/mol ≈ 0.0895 mol
2) H2O (water):
Mass of H2O = 1.89 g
Molar mass of H2O = 2(1.01 g/mol) (hydrogen) + 16.00 g/mol (oxygen) = 18.02 g/mol
Number of moles of H2O = Mass of H2O / Molar mass of H2O = 1.89 g / 18.02 g/mol ≈ 0.105 mol
3) NH3 (ammonia):
Mass of NH3 = 0.436 g
Molar mass of NH3 = 14.01 g/mol (nitrogen) + 3(1.01 g/mol) (hydrogen) = 17.04 g/mol
Number of moles of NH3 = Mass of NH3 / Molar mass of NH3 = 0.436 g / 17.04 g/mol ≈ 0.0256 mol
Next, we need to determine the mole ratio of each element. We divide the number of moles of each element by the smallest number of moles.
For carbon (C):
Mole ratio of C = Number of moles of CO2 / Number of moles of NH3
= 0.0895 mol / 0.0256 mol ≈ 3.49
For hydrogen (H):
Mole ratio of H = Number of moles of H2O / Number of moles of NH3
= 0.105 mol / 0.0256 mol ≈ 4.10
For nitrogen (N):
Mole ratio of N = Number of moles of NH3 / Number of moles of NH3 (It is the same element, so the ratio is 1)
Now, we can use the mole ratios as the subscripts in the empirical formula.
Empirical formula of lysine: C3.49H4.10N
To obtain whole numbers, we can round these subscripts to the nearest whole number:
Empirical formula of lysine: C3H4N