Lysine Amino Acid
Posted on 02 July 2010
What is the empirical formula of lysine?
6.Lysine, an essential amino acid in the human body, contains C, H, O and N. In one experiment, complete combustion of 2.175 g of lysine gave 3.94 g of CO2, and 1.89 g of H2O. In a separate experiment, the same sample mass of 0.506 g of lysine NH3. What is the empirical formula of lysine? Please write a detailed solution "steps of how you"
moles CO2 = 3.94 g / 44.009 g / mol = 0.0895 mol = 0.0895 mol mass = CC compounding x 12.011 g / mol = 1.07 g moles H2O = 1.89 g / 18.02 g / mol = 0.105 mol composed of H = 0.105 x 2 = 0.210 H = 0.210 mol x 1.008 g mass / g mol = 0.212 Y = 2.175 – (0.212 + 1.07) = 0.893 mol g O = 0.893 g / 15.999 g / mol = 0.0588 moles NH3 = 0.506 G/17.03 g / mol = 0.0297 the C: H, O and N 0.0895: 0.210: 0.0588: 0.0297 divided by the smallest number 0.0895 / 3 = 0.0297> C 0.210 / 0.0297 = 7 => H 0.0588 / 0.0297 = 2 => O 0.0297 / 0.0297 = 1 => N empirical formula C3H7O2N
Science Animation (amino acids)
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