Question

The heat of combustion of fructose, ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$, is $-2812\mathrm{kJ}/\mathrm{mol}$. If a fresh golden delicious apple weighing $4.23$ oz $(120\mathrm{g})$ contains $16.0\mathrm{g}$ of fructose, what caloric content does the fructose contribute to the apple?

Short answer

The short answer to the problem is as follows: 1. Convert the heat of combustion of fructose to Calories: $-2812\frac{kJ}{mol}\times \frac{1Cal}{4.184kJ}=-672\frac{Cal}{mol}$ 2. Calculate the moles of fructose in the apple: $\frac{16.0g}{180.16\frac{g}{mol}}=0.089\mathrm{mol}$ 3. Calculate the caloric content contributed by fructose: $0.089\mathrm{mol}\times -672\frac{Cal}{mol}=-59.8\mathrm{Cal}$ The fructose contributes approximately -59.8 Calories to the apple.

Step-by-step solution

Convert the heat of combustion of fructose to calories

We are given the heat of combustion of fructose as -2812 kJ/mol. To convert it into Calories, we use the conversion factor 1 Calorie = 4184 J: Heat of combustion (Cal/mol) = $-2812\frac{kJ}{mol}\times \frac{1Cal}{4184J}$ Heat of combustion (Cal/mol) = $-2812\frac{kJ}{mol}\times \frac{1Cal}{4.184kJ}$ Now, calculate the heat of combustion in Calories per mole.

Calculate the moles of fructose in the apple

To calculate the moles of fructose, use the given mass of fructose (16.0 g) and the molar mass of fructose, C6H12O6 (180.16 g/mol): Moles of fructose = $\frac{16.0g}{180.16\frac{g}{mol}}$ Now, calculate the actual moles of fructose in the apple.

Calculate the caloric content contributed by fructose

To find the caloric content contributed by fructose, multiply the moles of fructose (from step 2) by the heat of combustion (from step 1): Caloric content (Cal) = Moles of fructose × Heat of combustion (Cal/mol) Now, calculate the caloric content contributed by fructose in the apple in Calories.