SME
Organic Chemistry II
Biomolecules: carbohydrates, amino acids, lipids and nucleic acids
Use this for “Trilinolein has a molar mass of 879.4 g/mol and 6 C=C. Find its saponification and iodine values.”
Chart
Molar mass of the fat M = 885.4 g/mol
C=C per molecule ndb = 3
This is an example. Type your own values to start a new problem.
SV (mg KOH/g)
=
3
×
56.11
×
1000
M (g/mol)
IV (g I₂/100 g)
=
×
253.81
×
100
M (g/mol)
Formulas
SV
=
3
×
56.11
×
1,000
M
190.1
=
3
×
56.11
×
1,000
885.4
IV
=
ndb
×
253.81
×
100
M
86
=
3
×
253.81
×
100
885.4
Assumptions
•
The fat is a triglyceride: each molecule has three ester groups, so it takes 3 KOH (56.11 g/mol each) to saponify.
•
Each C=C in the fatty-acid chains adds one I₂ (253.81 g/mol); the ester C=O groups take none.
•
M is the average molar mass of the fat, so the values are for one gram (SV) or 100 g (IV) of the whole sample.
Step-by-step
WE KNOW
M = 885.4 g/mol, ndb = 3
FIND
saponification value (SV), iodine value (IV)
Step 1 · Find saponification value (SV)
Use SV = 3 × 56.11 × 1,000 ÷ M
One mole of fat takes 3 mol KOH, 168,330 mg. Divide by the grams in one mole.
SV
=
3
×
56.11
×
1,000
885.4
SV
=
168.33
×
1,000
885.4
SV
=
168,330
885.4
SV = 190.1 mg KOH/g
Step 2 · Find iodine value (IV)
Use IV = ndb × 253.81 × 100 ÷ M
Each C=C takes one I₂, 253.81 g. Divide by the grams in one mole, then scale to 100 g.
IV
=
3
×
253.81
×
100
885.4
IV
=
761.43
×
100
885.4
IV
=
76,143
885.4
IV
=
86
g
I₂
100
g
Check
Put the numbers back in. Both sides should match.
190.1
=
3
×
56.11
×
1,000
885.4
✓
86
=
3
×
253.81
×
100
885.4
✓
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