Problem 83 - 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 caloric content of fructose in a golden delicious apple is approximately \( \left( \frac{16.0 g}{180.18 g/mol} * (-2812 kJ/mol) \right) \times \frac{1 cal}{4.184 kJ} \approx 67.9 \) cal.

Step 1: Convert grams of fructose to moles

To do this, we will use the molar mass of fructose, which is the sum of the molar masses of its constituent elements: C, H, and O. The molecular formula of fructose is C6H12O6. Molar mass of fructose = 6 * (molar mass of C) + 12 * (molar mass of H) + 6 * (molar mass of O) = 6 * 12.01 + 12 * 1.01 + 6 * 16.00 = 180.18 g/mol Now we'll divide the mass of fructose in the apple (16.0 g) by the molar mass of fructose (180.18 g/mol) to find the number of moles: Moles of fructose = \( \frac{16.0 g}{180.18 g/mol} \)

Step 2: Calculate the energy released during combustion in kJ

Now that we know the moles of fructose, we can use the heat of combustion value to find out how much energy is released during combustion. We are given that this value is -2812 kJ/mol: Energy released = Moles of fructose * Heat of combustion = \( \frac{16.0 g}{180.18 g/mol} * (-2812 kJ/mol) \)

Step 3: Convert the energy from kJ to cal

To find the caloric content of fructose in the apple, we have to convert the energy value we just found from kJ to cal. We know that 1 cal = 4.184 kJ. So, we will divide the energy released by this conversion factor: Caloric content of fructose = \( ( \frac{16.0 g}{180.18 g/mol} * (-2812 kJ/mol) ) \times \frac{1 cal}{4.184 kJ} \) Now, compute the value obtained in the above expression to find the caloric content of the fructose in a golden delicious apple.

Key Concepts

Stoichiometry

Stoichiometry is akin to a recipe in cooking. Just as a recipe specifies the amount of each ingredient needed, stoichiometry provides the quantitative relationships between reactants and products in chemical reactions. It explains how to calculate the amounts of substances consumed and produced, ensuring that atoms are conserved.

For the heat of combustion problem, we use stoichiometry to convert the mass of fructose into moles, a unit that represents the number of particles in a given substance. We can then relate the moles of fructose to the amount of energy released when it combusts. The key here is the mole concept, which allows us to bridge the gap between the macroscopic world of grams and the microscopic world of molecules and atoms.

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the energy changes during chemical reactions. Specifically, the term 'heat of combustion' refers to the amount of heat energy released when a substance fully combusts in the presence of oxygen.

In our exercise, we are focused on the combustion of fructose, which is an exothermic reaction—meaning it releases energy. Knowing the heat of combustion per mole of fructose, we can calculate the total energy released by the fructose present in an apple. Thermochemistry is vital for understanding energy flow in chemical processes and is an integral part of energy balance calculations.

Molar Mass

Molar mass is the mass of one mole of a substance and serves as a bridge between the mass of a material and the number of moles and molecules it contains. It is expressed in grams per mole (g/mol).

The molar mass is calculated by summing the masses of the individual elements contained in one molecule, adjusted by how many atoms of each element the molecule comprises. For instance, fructose (C₆H₁₂O₆) has a molar mass of 180.18 g/mol. We calculate it by multiplying the atomic masses of carbon (C), hydrogen (H), and oxygen (O) by the number of each atom in a molecule of fructose, and adding them together. The accurate determination of molar mass is crucial for stoichiometry calculations, as seen when converting grams of fructose to moles.

Caloric Content Calculation

Caloric content calculation enables us to understand how much energy we can obtain from consuming a particular amount of food. Specifically, it measures the energy that food provides to our bodies, usually expressed in calories or kilojoules.

From a chemical standpoint, the caloric content represents the energy released during the combustion of the food. For the apple exercise, after determining the amount of energy released by the fructose when it combusts, we convert this energy from kilojoules to calories because calories are a more familiar energy unit in the context of food. This conversion is essential for providing a practical understanding of the energy we gain from eating the apple, making the connection between thermochemical calculations and everyday nutritional values.