Question

Fructose is a sugar commonly found in fruit. A sample of fructose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), weighing \(4.50 \mathrm{~g}\) is burned in a bomb calorimeter. The heat capacity of the calorimeter is \(2.115 \times 10^{4} \mathrm{~J} /{ }^{\circ} \mathrm{C}\). The temperature in the calorimeter rises from \(23.49^{\circ} \mathrm{C}\) to \(27.71^{\circ} \mathrm{C}\). (a) What is \(q\) for the calorimeter? (b) What is \(q\) when \(4.50 \mathrm{~g}\) of fructose is burned? (c) What is \(q\) for the combustion of one mole of fructose?

Short answer

Answer: (a) The calorimeter absorbs 89141.3J, (b) the heat absorbed when 4.50g of fructose is burned is -89141.3J, and (c) the heat absorbed for the combustion of one mole of fructose is -3568000 J/mol.

Step-by-step solution

Calculate the temperature change

First, we need to find the temperature change in the calorimeter by subtracting the initial temperature from the final temperature: $$\Delta T = T_{final} - T_{initial} = 27.71^{\circ}C - 23.49^{\circ}C = 4.22^{\circ}C.$$

Calculate the heat absorbed by the calorimeter (\(q_{calorimeter}\))

The heat capacity of the calorimeter is given as \(2.115 \times 10^{4} J /{ }^{\circ}C\). Using the formula for heat absorbing capacity, \(q = C \times \Delta T\), we can find the heat absorbed by the calorimeter: $$q_{calorimeter} = (2.115 \times 10^{4}J /{ }^{\circ}C) \times (4.22^{\circ}C) = 89141.3J.$$

Calculate the heat absorbed by fructose (\(q_{fructose}\))

Since the system is adiabatic (no heat is exchanged with the surroundings), the heat absorbed by the calorimeter is equal to the heat released by the fructose in the reaction: $$q_{fructose} = -q_{calorimeter} = -89141.3J.$$

Calculate the molar mass of fructose

The molecular formula of fructose is \(C_6H_{12}O_6\). Using the molar masses of carbon (12.01g/mol), hydrogen (1.01g/mol), and oxygen (16.00g/mol), the molar mass of fructose can be calculated as follows: $$C_6H_{12}O_6: (6 \times 12.01g/mol) + (12 \times 1.01g/mol) + (6 \times 16.00g/mol) = 180.18 g/mol.$$

Calculate the heat absorbed per mole of fructose (\(q_{molar}\))

We know the heat absorbed by \(4.50g\) of fructose, so now we want to find the heat absorbed for one mole: $$q_{molar} = \frac{q_{fructose}}{(4.50g)(1 mol /180.18g)} = \frac{-89141.3J}{0.02498 mol} = -3568000\, J/mol.$$ So, the answers are: (a) The calorimeter absorbs \(89141.3J\), (b) The heat absorbed when \(4.50g\) of fructose is burned is \(-89141.3J\), and (c) The heat absorbed for the combustion of one mole of fructose is \(-3568000\, J/mol\).

Key concepts

Calorimetry

Calorimetry is an experimental technique that measures the amount of energy exchanged in a reaction in the form of heat. To visualize this, imagine a coffee cup calorimeter, a simple device where reactions occur inside a coffee cup surrounded by water. The water acts as the absorber or releaser of heat, depending on whether the process is exothermic or endothermic.

In more sophisticated setups like the bomb calorimeter described in our exercise, the sample is burnt in a sealed, oxygen-rich container submerged in water. Any heat released by the combustion causes the water temperature to rise, and by knowing the calorimeter's heat capacity, we can calculate the total energy change. This setup is ideal for studying combustion reactions because the sealed environment ensures that energy is not lost to the surroundings, thus adhering to the principles of an adiabatic process.

Heat Capacity

The concept of heat capacity is central to understanding how a substance responds to the addition or removal of heat. Defined as the amount of heat required to raise the temperature of an object by one degree Celsius, it provides insight into the substance's thermal 'inertia'.

Consider it like a thermal sponge - a high heat capacity means it can soak up a lot of heat without warming up much, ideal for a bomb calorimeter. In our solved problem, the calorimeter's capacity was key to determining the heat of the reaction. The larger the heat capacity, the more heat required to change the temperature, which explains why the bomb calorimeter can absorb the heat from the combustion without a significant change in its own temperature.

Enthalpy of Combustion

Enthalpy of combustion is the heat change when one mole of a substance burns completely in oxygen. This value is crucial for chemists as it's an indicator of the energy content within chemical bonds of the substance. In practical terms, it's a measure of a fuel's 'power'. Negative values, like the one calculated in our exercise (-3568000 J/mol), indicate that the reaction releases heat—an exothermic reaction.

The magnitude of this number for fructose enables us to compare it to other fuels and understand its behavior during combustion. In the context of the experiment, it corresponds to a significant release of energy, showcasing why carbohydrates like fructose are potent energy sources for both chemical and biological processes.