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Chapter 5: Problem 85

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 59.68 kcal to the caloric content of the apple.

Step by step solution

01

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. \(-2812 \mathrm{~kJ} / \mathrm{mol} \cdot 0.239\,\mathrm{kcal/kJ} = x\,\mathrm{kcal/mol}\)
02

Solve for x

Solve for x to find the heat of combustion in kcal/mol: \(x = -2812 \mathrm{~kJ} / \mathrm{mol} \cdot 0.239\,\mathrm{kcal/kJ}\) \(x \approx -672\,\mathrm{kcal/mol}\)
03

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. \(6 \cdot 12.01\,\mathrm{g/mol} + 12 \cdot 1.01\,\mathrm{g/mol} + 6 \cdot 16.00\,\mathrm{g/mol} = M\)
04

Solve for M

Solve for M to find the molar mass of fructose: \(M = 6 \cdot 12.01\,\mathrm{g/mol} + 12 \cdot 1.01\,\mathrm{g/mol} + 6 \cdot 16.00\,\mathrm{g/mol}\) \(M \approx 180.18\,\mathrm{g/mol}\)
05

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: \(-672\,\mathrm{kcal/mol} / 180.18\,\mathrm{g/mol} = y\,\mathrm{kcal/g}\)
06

Solve for y

Solve for y to find the heat of combustion per gram of fructose: \(y = -672\,\mathrm{kcal/mol} / 180.18\,\mathrm{g/mol}\) \(y \approx -3.73\,\mathrm{kcal/g}\)
07

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: \(16\,\mathrm{g} \cdot -3.73\,\mathrm{kcal/g} = E\)
08

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.

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Most popular questions from this chapter

(a) Calculate the kinetic energy in joules of a \(1200-\mathrm{kg}\) automobile moving at \(18 \mathrm{~m} / \mathrm{s}\). (b) Convert this energy to calories. (c) What happens to this energy when the automobile brakes to a stop?

Consider the decomposition of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) to gaseous acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g):\) $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \quad \Delta H=+630 \mathrm{~kJ} $$ (a) What is the enthalpy change for the reverse reaction? (b) What is \(\Delta H\) for the formation of 1 mol of acetylene? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) were consumed instead of \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

A pound of plain \(\mathrm{M\&M}\) candies contains \(96 \mathrm{~g}\) fat, \(320 \mathrm{~g}\) carbohydrate, and 21 g protein. What is the fuel value in \(\mathrm{kJ}\) in a \(42-\mathrm{g}\) (about 1.5 oz ) serving? How many Calories does it provide?

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{CO}_{2}(s)\) or \(1 \mathrm{~mol}\) \(\mathrm{CO}_{2}(g)\) at the same temperature, (b) 2 mol of hydrogen atoms or \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2},\) (c) \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) and \(0.5 \mathrm{~mol} \mathrm{O}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(g)\) at \(25^{\circ} \mathrm{C},\) (d) \(1 \mathrm{~mol} \mathrm{~N}_{2}(g)\) at \(100{ }^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol}\) \(\mathrm{N}_{2}(g)\) at \(300^{\circ} \mathrm{C}\)

Limestone stalactites and stalagmites are formed in caves by the following reaction: $$ \mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \longrightarrow \mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If \(1 \mathrm{~mol}\) of \(\mathrm{CaCO}_{3}\) forms at \(298 \mathrm{~K}\) under 1 atm pressure, the reaction performs \(2.47 \mathrm{~kJ}\) of \(P-V\) work, pushing back the atmosphere as the gaseous \(\mathrm{CO}_{2}\) forms. At the same time, \(38.95 \mathrm{~kJ}\) of heat is absorbed from the environment. What are the values of \(\Delta H\) and of \(\Delta E\) for this reaction?
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