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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

Expert verified
The fructose in the apple contributes approximately 59.7 calories to the apple's total caloric content.

Step by step solution

01

Convert heat of combustion to caloric content

To convert the heat of combustion of fructose to caloric content, we will use the fact that 1 cal is equal to 4.184 J. So, we have to divide the given heat of combustion by 4.184. Heat of combustion in calories/mol = \(\frac{-2812 \: kJ}{4.184 \: J/cal} = -672\: cal/mol\)
02

Calculate the number of moles of fructose

Now, we need to find the number of moles of fructose in the 16.0 g of fructose contained in the apple. The molecular weight of fructose is \(C_6H_{12}O_6\), which is equal to 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol. Number of moles of fructose = \(\frac{16.0 \: g}{180.18 \: g/mol} = 0.0888 \: mol\)
03

Calculate the caloric content contributed by the fructose

Finally, we can calculate the total caloric content provided by the fructose using the heat of combustion (in calories/mol) and the number of moles. Caloric content of fructose = Heat of combustion (cal/mol) x Number of moles Caloric content of fructose = \(-672 \: cal/mol \times 0.0888 \: mol = -59.7 \: cal\) Note that the caloric content is a negative value because heat is released during combustion, which is an exothermic process. Thus, the fructose in the apple contributes approximately 59.7 calories to the apple's total caloric content.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fructose
Fructose is a simple sugar, or monosaccharide, found in many plants. It is often referred to as fruit sugar due to its high concentration in fruits. Its chemical formula is \(\text{C}_6\text{H}_{12}\text{O}_6\). While similar to glucose, fructose has a slightly different molecular structure, which influences its sweetness and metabolic processing.Fructose is naturally found in a variety of foods including:
  • Fruits such as apples, pears, and berries
  • Some vegetables like sweet potatoes and onions
  • Honey
In the human body, fructose is metabolized mostly in the liver, unlike glucose, which is utilized by almost all body cells. Extensive consumption of fructose can lead to health issues like insulin resistance and fatty liver disease. Despite its potential drawbacks, fructose is a key component of many foods, contributing to their natural sweetness.
Caloric Content
The caloric content is a measure of how much energy food provides when consumed. It is typically expressed in calories (cal) or kilocalories (kcal), where 1 kcal is equivalent to 1,000 cal.In the context of chemical reactions, such as combustion, the caloric content refers to the energy released or absorbed. When talking about foods, it reflects the energy we obtain from digestion and metabolism.Here are the basic steps to calculate the caloric content of a substance:
  • Convert the substance's energy value (e.g., heat of combustion) from kilojoules to calories, using the conversion factor \(1 \text{ cal} = 4.184 \text{ J}\).
  • Determine the number of moles based on the substance's mass and molar mass.
  • Multiply the energy per mole by the total number of moles to get the total caloric content.
For the apple example, the fructose provides around 59.7 cal, calculated from its heat of combustion and molar amount.
Exothermic Process
An exothermic process refers to a chemical reaction that releases energy in the form of heat. When the total energy of the products is lower than that of the reactants, the excess energy is released, often as heat. This is typical of combustion reactions, such as the burning of fuel.Key features of exothermic reactions include:
  • Release of heat
  • Negative change in enthalpy (\(\Delta H\))
  • Increases in temperature of the surroundings
In the context of combustion, the process involves the breakdown of molecules like fructose, releasing stored chemical energy. This is why the caloric content linked to combustion is often negative; it's an indication of energy release.Understanding exothermic processes is crucial in fields like chemistry and engineering, where managing heat release is fundamental. In our everyday diet, the energy we obtain from food is a result of metabolic reactions that are exothermic by nature.

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Most popular questions from this chapter

Can you use an approach similar to Hess's law to calculate the change in internal energy, \(\Delta E,\) for an overall reaction by summing the \(\Delta E\) values of individual reactions that add up to give the desired overall reaction?

(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) \(H\) is a state function, but \(q\) is not a state function. Explain. (c) For a given process at constant pressure, \(\Delta H\) is positive. Is the process endothermic or exothermic?

The decomposition of \(\mathrm{Ca}(\mathrm{OH})_{2}(s)\) into \(\mathrm{CaO}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) at constant pressure requires the addition of 109 \(\mathrm{kJ}\) of heat per mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) . (a) Write a balanced thermochemical equation for the reaction. (b) Draw an enthalpy diagram for the reaction.

Given the data $$\begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) & \Delta H=+180.7 \mathrm{kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \Delta H=-113.1 \mathrm{kJ} \\ 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-163.2 \mathrm{kJ} \end{aligned}$$ use Hess's law to calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g)$$

Diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l),\) a flammable compound that was once used as a surgical anesthetic, has the structure $$\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$$ The complete combustion of 1 mol of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) yields \(\Delta H^{\circ}=-2723.7 \mathrm{kJ}\) . (a) Write a balanced equation for the combustion of 1 \(\mathrm{mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) .\) (b) By using the information in this problem and data in Table \(5.3,\) calculate \(\Delta H_{f}^{\circ}\) for diethyl ether.

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