Coconut oil contains a triester with three identical R groups - AQA - A-Level Chemistry - Question 1 - 2021 - Paper 2

The compound RCOOK is a potassium carboxylate salt, which is formed from the reaction of the triester with potassium hydroxide.

Potassium carboxylate salts can be used in the production of soap or as surfactants in various applications.

To find the value of n, we start with the total molecular mass of the triester:

$M = 638.0$ The formula for the triester is: $3 (C_nH_{2n+1}COO)$

e.g. the alkyl portion contributes:

$C + 2n + 1$ Substituting the molecular mass yields:

$638.0 = 3[(12 + 2n + 1) + (12)]$ Solving:

$3 (12n + 14) = 638.0$ $12n + 14 = 212.67$ Subtracting 14 yields: $12n = 198.67$ Dividing by 12 gives: $n = 16.6$ Since n must be an integer, it rounds to 16.

To calculate the percentage by mass of the triester in the coconut oil, the first step is to find the amount of KOH that reacted.

1. Initial amount of KOH: 0.421 g = 0.421 g / 56.11 g/mol = 0.00749 mol

2. Amount of HCl used: 15.65 cm³ of 0.100 mol/dm³ = 0.001565 mol

3. KOH remaining = 0.00749 mol - 0.001565 mol = 0.005925 mol

4. The moles of triester present: Since 1 mol of triester reacts with 3 mol of KOH, the moles of triester = (0.00749 - 0.001565) / 3 = 0.00214167 mol

5. Mass of triester in sample:

Mass = moles × molar mass = 0.00214167 mol × 638 g/mol = 1.367 g

6. Percentage by mass of the triester:

ext{Percentage by mass} = rac{ ext{mass of triester}}{ ext{initial mass of coconut oil}} imes 100

= rac{1.367}{1.450} imes 100 = 94.50 ext{ %}

Aqueous ethanol is a suitable solvent because it allows for better solubilization of both coconut oil and KOH, promoting effective reaction conditions for hydrolysis.

A safety precaution is to use a fume hood or ensure good ventilation. This is important because heating the mixture can release fumes that could be harmful if inhaled, and proper ventilation helps to mitigate this risk.