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
