Question

1) Consider the oxidation of Stearic acid (18 carbon fatty acid, fully saturated) to CO2, NADH and FADH2

a) How many moles of electrons are removed from a molecule of stearic acid during the oxidation process?

b) How many moles of electrons are removed during the oxidation of a mole of glucose?

c) How many moles of electrons are removed during the oxidation of 100 g of stearic acid?

d) How many moles of electrons are removed during the oxidation of 100 g or glucose?

e) Write a balanced equation for the oxidation of stearic acid (C18H36O2) to CO2, NADH and FADH2

Please show work for a-e thank you.

Answer 1

Ans. #a. 1 NAD⁺ accepts 2 electron whereas 1 FAD accepts 1 electron during their reduction.

So,

Moles of e^- released per mol stearic acid upon complete oxidation

= (2e / NADH) x moles of NADH formed

+ (1e / FAD) x Moles of FADH₂ formed

                                    = (2e / NADH) x 35 NADH + (1e / FADH₂) x 17 FADH₂

                                    = 70 mol e + 17 mol e

                                    = 87 mol e

#b. Moles of e^- released per mol glucose upon complete oxidation

= (2e / NADH) x moles of NADH formed

+ (1e / FAD) x Moles of FADH₂ formed

                                    = (2e / NADH) x 10 NADH + (1e / FADH₂) x 2 FADH₂

                                    = 22 mol e

#c. Moles of stearic acid in 100.0 g sample = Mass / Molar mass

                                    = 100.0 g / (284.48) g

                                    = 0.35152 mol

#from #a, electron yield of stearic acid = 87 mol e per mol stearic acid

Now,

            Moles of e^- = electron yield of stearic acid x moles of stearic acid in 100 g

                                    = (87 mol e / mol stearic acid) x 0.35152 mol stearic acid

                                    = 30.582 mol e

#d. Moles of glucose in 100.0 g sample = 100.0 g / (180.156) g = 0.55507 mol

#from #b, electron yield of glucose = 22 mol e per mol glucose

Now,

            Moles of e^- = electron yield of glucose x moles of glucose in 100 g

                                    = (22 mol e / mol glucose) x 0.55507 mol glucose

                                    = 12.212 mol e

#e. Balanced reaction for oxidation of stearic acid-

C₁₇H₃₅COOH + 26 O₂ + 35 NAD⁺ + 17 FAD -------->

18 CO₂ + 18H₂O + 35 NADH + 17 FADH2