Question

2) When a molecule of fatty acid reacts with iodine (I2), the double bonds in a fatty acid are broken. The reaction is illustrated below.

ch2-hc=hc-ch2+I2 arrow ch2-h-i-c-h-i-c-ch2

a) The Iodine number (I#) is the mass of iodine that binds 100 g of fatty acid. An experiment shows that 1 g of the unknown fatty acid binds 0.98 g of I2. Calculate I# for the unknown fatty acid. Show your calculations. (2 pts)

b) Establish the mathematical equation that links MI (the mass of I2 bound to the unknown fatty acids), MWI2 (the molecular weight of I2) and mI (the number of moles contained in MI ). Show your reasoning. (2pts) MI = ?

c) Establish the mathematical formula that links I#, mI , MWI2, mFA, and MWFA. Show your reasoning. Remember the equation that you have establish in part 1)

d) Establish the mathematical equation between mI , mFA and p, where p is the number of double bonds contained in the unknown fatty acid. Show your reasoning. (2 pts) mI = ?

e) Based on the mathematical formula determined in parts c) and d), establish the mathematical formula that links I#, p, MWI2, and MWFA. Show your reasoning. (2 pts)

f) Knowing that the MWI2 = 254 g.mol-1, calculate how many double bonds are contained in the unknown fatty acid. Show your calculation. Note that you may have to round up the numerical value of p. (4 pts)

Answer 1

The balanced chemical equation is

-CH₂-CH=CH-CH₂- + I₂ -------> -CH₂-CH(I)-CH(I)-CH₂-

a) The iodine number I# = mass of iodine that binds to 100 g of the fatty acid (FA)

It is given that 1.00 g of an unknown FA binds 0.98 g I₂.

Therefore, I# = (100 g FA)*(0.98 g I₂/1.00 g FA) = 98 (ans).

b) Let mass of I₂ bound to an unknown FA = MI

Molecular weight of I₂ = MWI2

Therefore, we have 1 mole I₂ = MWI2 g I₂.

We have mI moles I₂; thus,

MI g I₂ = (MI g I₂)*(1 mole I₂/MWI2 g I₂) = MI/MWI2 mole.

By the problem.

mI = MI/MWI2

===> MI = (mI)*(MWI2) (ans)

c) We can see from the stoichiometric equation that

1 mole of I₂ reacts with 1 mole of FA.

Here, we must assume that the FA contains only one double bond.

Moles of FA, mFA = MFA/MWFA where mFA = moles of FA; MFA = mass of FA taken and MWFA = molecular weight of FA.

Assume a 1:1 reaction as above. We have defined I# as the mass of iodine that reacts with 100 g of FA.

Let MI be the mass of iodine that reacts with MFA g of the FA.

MI = (I#/100)*MFA

===> (mI)*(MWI2) = (I#/100)*MFA

===> (mI)*(MWI2) = (I#/100)*(mFA)*(MWFA) (ans)

d) We have assumed that 1 double bond reacts with 1 mole of I₂. If the FA contains 2 double bonds, then the number of moles of I₂ reacted is 2.

Assuming we have a single double bond, we must have

mFA = mI

However when p double bonds are present, then we must have

mFA = p*mI (ans)

e) From part (c) above, we have

mFA/mI = (100*MWI2)/(I#*MWFA)

Again, from part (d) above, mFA/mI = p.

Therefore,

p = (100*MWI2)/(I#*MWFA) (ans)

f) We put in I# = 98, MWI2 = 253.809 g/mol and MWFA = 254 g/mol; therefore,

p = (100*253.809 g/mol)/(98*254 g/mol) = 1.019 ≈ 1.00

The number of double bonds is 1 (ans).