Question

formula of oleate C₁₈H₃₄O₂

e. Calculate average change in oxidation number (+ or -) of carbon in each molecule as it is oxidized to carbon dioxide in aerobic respiration.

Fructose Average ΔO.N. of carbon = _____________ Oleate Average ΔO.N. of carbon = _____________

f. Calculate the total change in oxidation number (+ or -) over all carbons in each molecule. (i.e., you are calculating the total change in oxidation state per molecule as it goes onto become carbon dioxide.)

Fructose Total ΔO.N. per molecule = _____________ OleateTotal ΔO.N. per molecule = _____________

g. Calculate the total number of electrons that are lost in 1.00 g samples of fructose and oleate. (Note: the number of electrons lost per carbon atom is equivalent to its increase in oxidation state.)

Fructose No. e- lost in 1.00 g fructose = _______________ Oleate No. e- lost in 1.00 g oleate = _______________

h. If the loss of each electron represents the liberation of 1.875 x 10-19 J, calculate the number of kJ and Cal (kcal), of each substance in a 1.00 g sample. (Significant digits count.)

Fructose kJ = _____________ Cal = _____________ Oleate kJ = _____________ Cal = _____________

i. Comment on the relative values of the number of calories obtained from fats and sugars in a one gram sample. Do these relative values surprise you? Explain.

Answer 1

e. f. Average oxidation state of Carbons in Fructose is:

Oxidation state of C1= -2+1= -1
C2= +2
C3= +1-1=0
C4= +1-1=0
C5= +1-1=0
C6= -2+1=-1

Average oxidation state of 6 carbons in fructose = 0

Another way to calculate average oxidation number is

As Its a neutral molecule

Average oxidation state of C=0

In Carbon dioxide , oxidation state of Carbon = +4

Therefore change in fructose oxidation number

Oxidation number of Carbon in Oleate is

Average Oxidation state of C=

Oxidation number of C in

Change in oxidation number =

(g) The number of electrons lost per carbon atom is equivalent to its increase in oxidation state.

Moles in 1.0 g Fructose =

For 1 mol Fructose electrons lost = 4 mol

for 0.0056 moles of fructose electrons loss needed

Similarily for oleate

Number of moles in 1.0 gm oleate =

1 mole Oleate require loss of electrons = 5.67 moles

0.00354 moles lost electrons =