Question

1. Olive oil has a saponification value of 190 mg KOH per gram. This means that it takes 190 mg of potassium hydroxide to react completely with 1 g of olive oil.

(a) What is the mole ratio for the reaction of a triglyceride with KOH?

(b) Divide 0.190 g of KOH by its molar mass to calculate the number of moles of KOH that will react with 1 g of olive oil.

(c) Use the answers to (a) and (b) to determine the number of moles corresponding to 1 g of olive oil. Divide 1 g by this number of moles to calculate the average molar mass (g/mole) of olive oil.

Answer 1

Triglyceride have 3 ester groups. When these groups react with KOH, ester hydrolysis takes place and ester group is broken into alcohol group and carboxylate ion. This carboxylate ion forms a salt with potassium ions. This is what is termed as saponification. The resulting salts of carboxylate ion amd potassium ions are highly soluble in water.

a) As triglyceride have 3 ester groups, they require 3 moles of KOH for each mole of triglyceride. So the molar ratio of the reaction between triglyceride and KOH is 1:3 ( 1 mole of triglyceride and 3 moles of KOH)

b) Molar mass of KOH = 56.11 g/ mol

Mass of KOH = 0.190 g

Moles of KOH = mass/molar mass = 0.190g/ 56.11g/mol = 0.0034 mol

So, number of moles of KOH that will react with 1 g olive oil = 0.0034 mol = 3.4 x 10 -3 mol

c) as the mole ratio of the reaction is 1:3, we can calculate the moles of triglyceride.

3 moles of KOH will react with 1 mole of triglyceride

0.0034 mol of KOH will react with 0.0034/3 = 0.0011 mol of triglyceride. This is also equal to 1 g of olive oil

So we have the following information

Moles of olive oil = 0.0011 mol

Mass of olive oil = 1 g

Molar mass = mass of olive oil/ moles of olive oil = 1g / 0.0011 mol = 909.1 g/mol