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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 \(120 \mathrm{~g}\) 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 60 dietary Calories.

Step by step solution

01

Calculate Moles of Fructose

First, find the molar mass of fructose by adding the atomic masses of C, H, and O: \(6 \times 12.01 + 12 \times 1.01 + 6 \times 16.00 = 180.18 \, \text{g/mol}\).Now, calculate the moles of fructose in 16.0 g: \(\frac{16.0 \, \text{g}}{180.18 \, \text{g/mol}} = 0.0888 \, \text{mol}\).
02

Calculate Heat of Combustion for 16 g of Fructose

Using the heat of combustion value given, \(-2812 \, \mathrm{kJ/mol}\), multiply by the number of moles to find the total energy released:\(-2812 \, \mathrm{kJ/mol} \times 0.0888 \, \text{mol} = -249.7 \, \mathrm{kJ}\).
03

Convert Energy from Kilojoules to Calories

Knowing that 1 calorie equals 4.184 joules, convert kilojoules to calories:\(-249.7 \, \text{kJ} = -249700 \, \text{J}\).Now, convert to calories: \(\frac{-249700 \, \text{J}}{4.184 \, \text{J/cal}} = -59688.95 \, \text{cal}\).
04

Convert Calories to Dietary Calories

Understand that dietary Calories (uppercase C) are equivalent to kilocalories. Therefore, divide the calorie result by 1000 to convert:\(-59688.95 \, \text{cal} = -59.7 \, \text{kcal}\).In dietary terms, this means approximately 60 dietary Calories.

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Key Concepts

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

Understanding Heat of Combustion
The heat of combustion is a critical concept in thermochemistry that refers to the amount of heat energy released when a substance is completely burned in the presence of oxygen. This process is typically exothermic, meaning it releases energy.
For example, the heat of combustion for fructose, a simple sugar, is given as -2812 kJ/mol. The negative sign indicates that energy is released during the reaction.
It's important to remember that this value is specific to the substance being burned. Each different material, whether it be a fuel like gasoline or a nutrient like glucose, has a unique heat of combustion.
Ultimately, understanding the heat of combustion lets us predict how much energy we can extract from burning certain substances. In the context of food, it shows us how much energy, or calories, the body might gain from a specific amount of food.
Molar Mass Calculation
Calculating the molar mass is an essential step in performing stoichiometric calculations. It involves adding up the atomic masses of all atoms in a molecule.
Taking fructose as an example, it is composed of six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.
To find its molar mass, you simply multiply the number of each type of atom by the corresponding atomic mass and sum the total:
  • Carbon: 6 atoms × 12.01 g/mol
  • Hydrogen: 12 atoms × 1.01 g/mol
  • Oxygen: 6 atoms × 16.00 g/mol
This adds up to a molar mass of 180.18 g/mol for fructose.
Understanding molar mass is a fundamental part of converting grams of a substance to moles, which is crucial in calculating reaction yields, concentrations, and energy release.
Energy Conversion
Energy conversion is a key idea when working with different units of energy measurement. Often, calculations in chemistry require switching between joules, kilojoules, calories, and dietary calories.
The three steps to converting energy are:
  • First, determine the amount of energy in joules or kilojoules.
  • Next, convert from kilojoules to calories. Since 1 calorie = 4.184 joules, multiply or divide accordingly.
  • Finally, if needed, convert calories to dietary Calories. Note that 1 dietary Calorie (with uppercase C) is equivalent to 1 kilocalorie (kcal).
In the example involving fructose, the conversion from kilojoules to dietary Calories demonstrates how even small amounts of a substance can have measurable energy content.
Dietary Calories Explained
The concept of dietary Calories can often be confusing due to the way it is displayed on nutritional labels. Though often just written as "Calories," it actually refers to kilocalories (kcal).
When calculating the caloric content of food, the key is understanding the relationship between calories and dietary Calories.
1 kilocalorie equals 1000 regular calories, or the amount of energy needed to raise the temperature of 1 kg of water by 1 degree Celsius.
This conversion is particularly useful for assessing how food energy intake aligns with dietary needs. In the exercise example, fructose contributes to the total caloric intake of an apple, thus providing essential energy that the body can utilize. Overall, knowing how to convert and understand dietary Calories ensures you’re better equipped to assess the energy-providing capabilities of different foods.

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Most popular questions from this chapter

Identify the force present and explain whether work is being performed in the following cases: (a) You lift a book off the top of a desk. (b) Air is compressed in a bicycle pump.

The specific heat of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(I),\) is \(2.22 \mathrm{~J} / \mathrm{g}-\mathrm{K} .(\mathrm{a}) \mathrm{How}\) many J of heat are needed to raise the temperature of \(80.0 \mathrm{~g}\) of octane from 10.0 to \(25.0^{\circ} \mathrm{C} ?\) (b) Which will require more heat, increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(I)\) by a certain amount or increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(I)\) by the same amount?

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At the end of \(2012,\) global population was about 7.0 billion people. What mass of glucose in \(\mathrm{kg}\) would be needed to provide 1500 Cal/person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(I)\) according to the following thermochemical equation: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(I) \\ \Delta H^{\circ} &=-2803 \mathrm{~kJ} \end{aligned} $$

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