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?

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

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g