Figure 1. Palmitic acid, CH3(CH2)14COOH, is an example of a saturated long-chain fatty acid with 16 carbons.
Figure 2. Oleic acid, CH3(CH2)7CH=CH(CH2)7COOH, is an example of an unsaturated long-chain fatty acid with 18 carbons.
Figure 3. A triglyceride is formed from one molecule of glycerol and three fatty acids in a condensation reaction.
Figure 4. Saturated fatty acids result in a higher melting point, while unsaturated fatty acids result in a lower melting point.
Figure 5. Iodine reacts with unsaturated carbon-carbon double bonds. Oils that have more unsaturation 'absorb' more iodine, resulting in a greater iodine number. Linoleic acid is polyunsaturated, with two C=C bonds, absorbing two moles of iodine.
Table 1. Typical iodine numbers from some fats and oils.
18.5 g of lipid reacts completely with 11.5 g of iodine. Calculate the iodine number of the lipid.
Iodine number=(mass of iodine reacted/mass of lipid)×100
Iodine number=(11.5 g/18.5 g)×100
Iodine number=0.622×100
Iodine number=62.2
Note that there are no units for iodine number, since it represents a ratio.
A long-chain carboxylic acid molecule has a Mr = 302.45 g mol-1 and an iodine number of 420. Calculate the number of carbon-carbon double bonds in a molecule of the acid.
Assume a 1 gram sample of the lipid to calculate the mass of iodine reacted:
Iodine number=(mass of iodine reacted/mass of lipid)×100
420=(mass of iodine reacted/1 gram of lipid)×100
4.20 g=mass of iodine reacted
Convert grams of iodine into moles of iodine:
4.20 g ×(1mol I2/253.8 g)=0.0165 mol I2
Calculate the number of moles of the lipid:
1 g ×(1mol lipid/302.45 g)=0.00331 mol lipid
Compare the mole ratio of lipid:iodine
0.0165 mol I2:0.00331 mol lipid
To examine the mole ratio more simply, divide by the smallest number of moles:
0.0165 mol I2÷0.00331 mol : 0.00311 mol lipid÷0.00331 mol
4.99 mol I2: 1.00 mol lipid
5 mol I2: 1 mol lipid
Therefore, there are five carbon-carbon double bonds per molecule in the unsaturated lipid.