Question

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?

Answer 1

Answer:

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