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13 August, 13:41

Lysine, which is an essential amino acid, contains only C, H, N, and O. In one

experiment, the complete combustion of 2.175 g of lysine produces 3.93 g of CO2

and 1.87 g of H2O. In a separate experiment, 1.873 g of lysine produces 0.436 g of

NH3. What is the empirical formula of lysine?

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Answers (1)
  1. 13 August, 13:47
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    The empirical formula is C3H7NO

    Explanation:

    Step 1: Data given

    Mass of the lysine sample in the first experiment = 2.175 grams

    Mass of CO2 produced = 3.93 grams

    Mass of H2O produced = 1.87 grams

    Molar mass CO2 = 44.01 g/mol

    Molar mass H2O = 18.02 g/mol

    Atomic mass of C = 12.01 g/mol

    Atomic mass O = 16.0 g/mol

    Atomic mass H = 1.01 g/mol

    In a separate experiment, 1.873 g of lysine produces 0.436 g of NH3

    Molar mass NH3 = 17.03 g/mol

    Step 2: Calculate moles CO2

    Moles CO2 = mass CO2 / molar mass CO2

    Moles CO2 = 3.93 grams / 44.01 g/mol

    Moles CO2 = 0.0893 moles

    Step 3: Calculate moles C

    For 1 mol CO2 we have 1 mol C

    For 0.0893 moles CO2 we have 0.0893 moles C

    Step 4: Calculate mass C

    Mass C = 0.0893 moles * 12.01 g/mol

    Mass C = 1.072 grams

    Step 5: Calculate moles H2O

    Moles H2O = 1.87 grams / 18.02 g/mol

    Moles H2O = 0.104 moles

    Step 6: Calculate moles H

    For 1 mol H2O we have 2 moles H

    For 0.104 moles H2O we have 0.208 moles H

    Step 7: Calculate mass H

    Mass H = 0.208 moles * 1.01 g/mol

    Mass H = 0.210 grams

    Step 8: Calculate mass %

    Mass % C = (1.072 grams / 2.175 grams) * 100%

    MAss % C = 49.3 %

    Mass % H = (0.210 grams / 2.175 grams) * 100 %

    Mass % H = 9.66 %

    Step 9: Calculate moles NH3

    Moles NH3 = 0.436 grams / 17.03 g/mol

    Moles NH3 = 0.0256 moles

    Step 10: Calculate moles N

    For 1 mol NH3 we have 1 mol N

    For 0.0256 moles NH3 we have 0.0256 moles N

    Step 11: Calculate mass N

    Mass N = 0.0256 grams * 14.0 g/mol

    Mass N = 0.358 grams

    Step 12: Calculate mass C in Lysine

    MAss C = 0.493 * 1.873 grams

    Mass C = 0.923 grams

    Step 13: Calculate mass H

    Mass H = 0.0966 * 1.873 grams

    Mass H = 0.181 grams

    Step 14: Calculate mass O

    Mass O = 1.873 grams - 0.923 grams - 0.181 grams - 0.358 grams

    MAss O = 0.411 grams

    Step 15: Calculate moles in the compound

    Moles C = 0.923 grams / 12.01 g/mol

    Moles C = 0.07685 moles

    Step 16: Calculate moles H

    Moles H = 0.181 grams / 1.01 g/mol

    Moles H = 0.1792 moles

    Step 17: Calculate moles N

    Moles N = 0.358 grams / 14.0 g/mol

    Moles N = 0.02557 moles

    Step 18: Calculate moles O

    Moles O = 0.411 grams / 16.0 g/mol

    Moles O = 0.02569 moles

    Step 19: Calculate the mol ratio

    We divide by the smallest amount of moles

    C: 0.07685 moles / 0.02557 moles = 3

    H: 0.1792 moles / 0.02557 moles = 7

    N: 0.02557 moles / 0.02557 moles = 1

    O: 0.02569 moles / 0.02557 = 1

    The empirical formula is C3H7NO
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