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Lysine, an essential amino acid in the human body, contains C,H,O, and N. In one experiment, the complete combustion of 2.175g of lysine gave 3.94g CO2 and 1.89g H2O. In a separate experiment, 1.873 g of lysine gave 0.436 g NH3.

(a) Calculate the empirical formula of lysine

(b) The approximate molar mass of lysine is 150g. What is the moleclar formula of the compound?

For (a) I think I found the # of each element except H.

For element C would this be right:
3.94g CO2 / 12 g C = 0.328 C

  • Chemistry - ,

    Also when I found the # of N
    I got it from 1.873g lysine and for the other elements C and O the lysine was 2.175g

    would i have to somehow add those number of lysine to find the empirical formula?

  • Chemistry - ,

    We can only calculate the empirical formula, which is the simplest ratio between the number of atoms of the constituents. This ratio may or may not be a sub-multiple of the actual formula.

    We will first decide on the processes.
    The complete combustion means that the limiting reagent will be the carbon and hydrogen in lysine, while oxygen is in excess supply.

    The second process is where nitrogen is the limiting reagent that produces ammonia.

    We will calculate the percentage weight of C, H and N, while that of O will be calculated by difference from 100%.

    Each percentage by weight will then be divided by the respective atomic mass to obtain the proportion of number of atoms. From there, we divide the HCF (approx.) of the numbers to obtain the ratio of number of atoms, in integers.

    Element, mass, %mass, #atoms, integer #atoms
    C: 3.94*12/44=1.075; 49.4%; 4.1167; 3
    H: 1.89*2/18=0.2100; 9.655%; 9.579; 7
    N: 0.436*14/17=0.3591; 19.17%; 1.3693; 1
    O: -; (100-49.4-9.655-19.17)=21.78%; 1.3609; 1

    From the last column, we conclude that the empirical formula is C3H7NO.

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