Question

1. Carnivores may oxidize amino acids as a source of energy. Terrestrial vertebrates and sharks oxidize amino acids to urea (NH₂CONH₂), carbon dioxide and water - as an oxidation product, urea has a low toxicity, high solubility and easy to eliminate from the body. An example is provided by glycine, NH₂CH₂CO₂H. At 37 °C and 1.00 atm, what volume of carbon dioxide is produced for each microgram of glycine produced? Many aquatic venerates oxidize amino acids to ammonia, carbon dioxide and water. For each microgram of glycine oxidized, what volume of carbon dioxide is produced at 25 °C and 1.00 atm.

Answer 1

Write the balanced chemical equation as

2 NH₂CH₂CO₂H + 3 O₂ -----------> NH₂CONH₂ + 3 CO₂ + 3 H₂O

As per the stoichiometric equation,

2 moles glycine = 3 moles CO₂.

The atomic masses are

H: 1.008 g/mol.

C: 12.011 g/mol.

N: 14.007 g/mol.

O: 15.999 g/mol.

Molar mass of glycine, NH₂CH₂CO₂H = (1*14.007 + 5*1.008 + 2*12.011 + 2*15.999) g/mol

= 75.067 g/mol.

Mass of glycine = 1.0 µg

= (1.0 µg)*(1 g)/(1.0*10⁶ µg)

= 1.0*10^-6 g.

Mol(s) glycine corresponding to 1.0 µg glycine = (1.0*10^-6 g)/(75.067 g/mol)

= 1.332*10^-8 mol.

Mol(s) CO₂ corresponding to 1.0 µg glycine = (1.332*10^-8 mol glycine)*(3 mols CO₂)/(2 mols glycine)

= 1.998*10^-8 mol.

Assume the liberated CO₂ to behave as an ideal gas.

Pressure of the gas = 1.00 atm.

Temperature of the gas = 37ºC = (37 + 273) K = 310 K.

Use the ideal gas law to determine the volume of CO₂ liberated.

PV = nRT

where n denotes the number of moles of the gas.

Therefore,

V = nRT/P

Plug in values and get

V = (1.998*10^-8 mol)*(0.082 L-atm/mol.K)*(310 K)/(1.00 atm)

= 5.0789*10^-7 L

= (5.0789*10^-7 L)*(1.0*10⁶ µL)/(1 L)

= 0.50789 µL

≈ 0.51 µL.

The volume of CO₂ gas evolved per microgram of glycine oxidized = 0.51 µL (ans).