Problem 85 - Chemistry The Central Science

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?

Step-by-Step Solution

Answer

The fructose in the apple contributes approximately 59.68 kcal to the caloric content of the apple.

Step 1: Convert heat of combustion to kcal/mol

First, we need to convert the heat of combustion of fructose from kJ/mol to kcal/mol. There are approximately 0.239 kcal in 1 kJ. Therefore, the heat of combustion of fructose in kcal/mol can be obtained by multiplying -2812 kJ/mol by the conversion factor 0.239 kcal/kJ. \(-2812 \mathrm{~kJ} / \mathrm{mol} \cdot 0.239\,\mathrm{kcal/kJ} = x\,\mathrm{kcal/mol}\)

Step 2: Solve for x

Solve for x to find the heat of combustion in kcal/mol: \(x = -2812 \mathrm{~kJ} / \mathrm{mol} \cdot 0.239\,\mathrm{kcal/kJ}\) \(x \approx -672\,\mathrm{kcal/mol}\)

Step 3: Determine the molar mass of fructose

Find the molar mass of fructose (\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)) by adding together the molar masses of its constituent atoms. The molar mass of each atom is: C = 12.01 g/mol, H = 1.01 g/mol, and O = 16.00 g/mol. \(6 \cdot 12.01\,\mathrm{g/mol} + 12 \cdot 1.01\,\mathrm{g/mol} + 6 \cdot 16.00\,\mathrm{g/mol} = M\)

Step 4: Solve for M

Solve for M to find the molar mass of fructose: \(M = 6 \cdot 12.01\,\mathrm{g/mol} + 12 \cdot 1.01\,\mathrm{g/mol} + 6 \cdot 16.00\,\mathrm{g/mol}\) \(M \approx 180.18\,\mathrm{g/mol}\)

Step 5: Convert heat of combustion to kcal/g

Divide the heat of combustion in kcal/mol by the molar mass of fructose to convert it to kcal/g: \(-672\,\mathrm{kcal/mol} / 180.18\,\mathrm{g/mol} = y\,\mathrm{kcal/g}\)

Step 6: Solve for y

Solve for y to find the heat of combustion per gram of fructose: \(y = -672\,\mathrm{kcal/mol} / 180.18\,\mathrm{g/mol}\) \(y \approx -3.73\,\mathrm{kcal/g}\)

Step 7: Calculate the caloric content of fructose in the apple

Multiply the mass of fructose in the apple (16 g) by the heat of combustion per gram of fructose (-3.73 kcal/g) to find the total energy released from the fructose in the apple: \(16\,\mathrm{g} \cdot -3.73\,\mathrm{kcal/g} = E\)

Step 8: Solve for E

Solve for E to find the caloric content of the fructose in the apple: \(E = 16\,\mathrm{g} \cdot -3.73\,\mathrm{kcal/g}\) \(E \approx -59.68\,\mathrm{kcal}\) The fructose in the apple contributes approximately 59.68 kcal to the caloric content of the apple.

Key Concepts

Heat of Combustion

The heat of combustion is the amount of energy released when a substance completely burns in oxygen. For fructose, it's given as \(-2812 \, \mathrm{kJ/mol}\). This negative sign indicates that energy is released during the combustion process. To better understand energy values in food and nutrition, we often convert this value to \(\mathrm{kcal/mol}\), since calories are a more familiar unit for measuring energy. By knowing the heat of combustion, we can calculate how much energy a food provides when metabolized. It’s important because it directly relates to the caloric content, which indicates how much energy the body can derive from the food item.

Molar Mass

The concept of molar mass helps us understand how much one mole of a substance weighs. For fructose \((\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6})\), we calculate it by adding the molar masses of all the atoms present:

• Carbon (\(C\)) has a molar mass of \(12.01 \, \mathrm{g/mol}\), with 6 atoms in fructose.
• Hydrogen (\(H\)) is \(1.01 \, \mathrm{g/mol}\), with 12 atoms.
• Oxygen (\(O\)) is \(16.00 \, \mathrm{g/mol}\), with 6 atoms.

Adding these gives us a molar mass for fructose of approximately \(180.18 \, \mathrm{g/mol}\). This value is key when converting between moles and grams.

Conversion Factors

Conversion factors are essential for translating between different units. In this case, converting the heat of combustion from \(\mathrm{kJ/mol}\) to \(\mathrm{kcal/mol}\) makes the information more practical for dietary needs. We use the conversion factor 1 \(\mathrm{kJ} = 0.239 \, \mathrm{kcal}\). Therefore, multiplying the energy in \(\mathrm{kJ/mol}\) by \(0.239 \, \mathrm{kcal/kJ}\) allows us to find that the heat of combustion for fructose is approximately \(-672 \, \mathrm{kcal/mol}\). Understanding conversion factors helps in making such calculations smooth and error-free, crucial for accurate nutritional analysis.

Caloric Content

Caloric content tells us the amount of energy food provides. For fructose in an apple, we calculate it by first finding the heat of combustion per gram \((-3.73 \, \mathrm{kcal/g})\). Knowing the mass of fructose in the apple (16 g), we can calculate the total energy contribution:

• Multiply the mass of fructose by the heat of combustion per gram.

This gives approximately \(59.68 \, \mathrm{kcal}\), indicating how much energy the fructose in the apple can offer. Caloric content is a vital aspect of food energy evaluation, helping us understand the potential energy a body can use.