Chapter 23, Problem 1P

Interpretation Introduction

Interpretation:

Calculate the available metabolic water from fatty acid oxidation for camel.

Introduction:

Water is released as a product in fatty acid oxidation (beta oxidation reactions). Therefore, beta oxidation provides a mean of water production for camels. Hump of the camel store a large amount of fatty acids and oxidizes releasing water.

Explanation of Solution

If the all the triglycerides are formed by palmitic acid, the palmitic acid percentage in triglycerides,

  $\text{Palmetic acid \%=}\frac{\text{3×Molecular}\text{weight of palmetic acid}}{\text{Molecular}\text{weight of triglyceride}}\text{×100 \%}$

Triglyceride is formed by one glycerol molecule with three palmetic acid molecules with the exclusion of three water molecules.

  $\begin{array}{l}\text{Palmetic acid \%=}\frac{\text{3}\times {\text{256 gmol}}^{\text{-1}}}{\text{3}\times \text{MWt of palmetic acid+Molecular Weight of glycerol-3}\times \text{MW of water}}\times \text{100\%}\\ \text{Palmetic acid \%=}\frac{\text{3}\times {\text{256 gmol}}^{\text{-1}}}{\text{3}\times {\text{256 gmol}}^{\text{-1}}{\text{+92 gmol}}^{\text{-1}}\text{-3}\times {\text{18 gmol}}^{\text{-1}}}\times \text{100\%}\\ \text{Palmetic acid \%=95 \%}\\ \text{30 pounds of triglycerides = }\frac{\text{30 lb}}{\text{1 g}}\times \text{453}\text{.6 g = 13608 g}\end{array}$

Where MW is molecular weight.

  $\text{Palmetic acid content = 13608 g}\times \text{95\% = 12928 g}$

  $\text{Palmetic acid moles = }\frac{\text{12928 g}}{256{\text{ gmol}}^{\text{-1}}}=50.5\text{ mol}$

Beta oxidation of one mole of palmitate produces 130 moles of water.

Therefore, the water moles produced in beta oxidation in this case = $50.5\text{ mol}\times \text{130=6565 moles}$

Produced water weight = $=6565\text{ mol}\times {\text{18 gmol}}^{-1}=118170\text{ g}$

Produced water volume = $=\frac{118170\text{ g}}{1000\text{ g/L}}=118.17\text{ L}$