Question

A typical fat in the body is glyceryl trioleate, \(\mathrm{C}_{57} \mathrm{H}_{104} \mathrm{O}_{6}\) When it is metabolized in the body, it combines with oxygen to produce carbon dioxide, water, and \(3.022 \times 10^{4} \mathrm{~kJ}\) of heat per mole of fat. (a) Write a balanced thermochemical equation for the metabolism of fat. (b) How many kilojoules of energy must be evolved in the form of heat if you want to get rid of five pounds of this fat by combustion? (c) How many nutritional calories is this? (1 nutritional calorie \(=1 \times 10^{3}\) calories)

Short answer

Answer: The balanced thermochemical equation for the metabolism of glyceryl trioleate is: C57H104O6 + 80 O2 → 57 CO2 + 52 H2O + 3.022 x 10^4 kJ/mol. Combustion of 5 pounds of glyceryl trioleate releases 20,486 nutritional calories.

Step-by-step solution

(a) Balanced thermochemical equation

The unbalanced equation for the combustion of glyceryl trioleate is given as: $$\mathrm{C}_{57} \mathrm{H}_{104} \mathrm{O}_{6} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}$$ Balancing this equation, we get: $$\mathrm{C}_{57}\mathrm{H}_{104}\mathrm{O}_{6} + 80\mathrm{O}_{2} \rightarrow 57\mathrm{CO}_{2} + 52\mathrm{H}_{2}\mathrm{O}$$ Now, we also add the heat released (\(3.022\times 10^{4} \mathrm{kJ/mol}\)) in the thermochemical equation: $$\mathrm{C}_{57}\mathrm{H}_{104}\mathrm{O}_{6} + 80\mathrm{O}_{2} \rightarrow 57\mathrm{CO}_{2} + 52\mathrm{H}_{2}\mathrm{O} + 3.022\times 10^{4}\mathrm{~kJ/mol}$$

(b) Energy evolved from 5 pounds of fat

First, we need to convert 5 pounds of fat to moles. The molar mass of glyceryl trioleate is: $$57(12.01)+104(1.01)+6(16.00)=803.6 \mathrm{g/mol}$$ Now, convert 5 pounds to grams and then to moles: $$5 \mathrm{lb} \times \frac{453.6 \mathrm{g}}{1 \mathrm{lb}} \times \frac{1 \mathrm{mol}}{803.6 \mathrm{g}}=2.837 \mathrm{mol}$$ Using the thermochemical equation, we can find the energy evolved: $$2.837\mathrm{mol} \times 3.022\times 10^{4}\mathrm{~kJ/mol}=8.568\times 10^{4}\mathrm{~kJ}$$

(c) Conversion to nutritional calories

To convert the energy to nutritional calories, we'll use the conversion factor that 1 nutritional calorie \(= 1 \times 10^{3}\) calories and 1 calorie \(= 4.184 \mathrm{J}\): $$8.568\times 10^{4}\mathrm{~kJ} \times \frac{10^3\mathrm{~cal}}{4.184\mathrm{~kJ}} \times \frac{1\mathrm{~nutritional\,cal}}{10^3\mathrm{~cal}} = 20,486 \mathrm{~nutritional\,calories}$$ So, combustion of 5 pounds of glyceryl trioleate releases 20,486 nutritional calories.