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Chapter 5: Problem 107

(a) When a 0.47-g sample of benzoic acid is combusted in a bomb calorimeter (Figure 5.19), the temperature rises by \(3.284^{\circ} \mathrm{C}\). When a 0.53 -g sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\), is burned, the temperature rises by \(3.05^{\circ} \mathrm{C}\). Using the value of \(26.38 \mathrm{~kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to \(0.001 \mathrm{~g},\) what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

Short Answer

Expert verified
The heat of combustion per mole of caffeine is calculated to be 13771 kJ/mol, and the estimated uncertainty in this value is approximately 16.66 kJ/mol.

Step by step solution

01

Calculate the heat released by benzoic acid

First we need to find out the heat released by the benzoic acid when it is combusted. We are given the temperature rise and the mass of the benzoic acid sample and we will use the heat of combustion of benzoic acid to find the heat released using the following equation: Heat released (Q) = Mass of sample 脳 Heat of combustion 脳 Temperature rise Q = 0.47 g 脳 26.38 kJ/g 脳 3.284 掳C Q = 40.61 kJ
02

Calculate the heat released by caffeine

Now, using the temperature rise and the mass of the caffeine sample, we will find the heat released by the caffeine using the information from the benzoic acid combustion. Heat released by caffeine, \( Q_{caffeine} \) = Heat released by benzoic acid 脳 (Temperature rise of caffeine / Temperature rise of benzoic acid) \( Q_{caffeine} \) = 40.61 kJ 脳 (3.05 掳C/3.284 掳C) \( Q_{caffeine} \) = 37.64 kJ
03

Calculate the moles of caffeine

Now we need to find the number of moles of caffeine in the sample. We are given the mass of the caffeine sample and the molecular formula C8H10N4O2. Using the molar mass of each element, we can calculate the molar mass of caffeine: Molar mass of C = 12.01 g/mol Molar mass of H = 1.01 g/mol Molar mass of N = 14.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of caffeine = 8 脳 12.01 + 10 脳 1.01 + 4 脳 14.01 + 2 脳 16.00 Molar mass of caffeine = 194.19 g/mol Now, we will find the moles of caffeine sample: Moles of caffeine = mass of caffeine / molar mass of caffeine Moles of caffeine = 0.53 g / 194.19 g/mol Moles of caffeine = 0.00273 mol
04

Calculate the heat of combustion per mole of caffeine

To find the heat of combustion per mole of caffeine, we will divide the heat released by the sample by the moles of caffeine: Heat of combustion = Heat released by caffeine / moles of caffeine Heat of combustion = 37.64 kJ / 0.00273 mol Heat of combustion = 13771 kJ/mol Therefore, the heat of combustion per mole of caffeine is 13771 kJ/mol.
05

Calculate the uncertainty of the heat of combustion

We are given the uncertainties in the temperature readings and the mass of the samples. We can use these uncertainties to estimate the uncertainty in the heat of combustion per mole of caffeine. In general, the uncertainty in the heat of combustion, 未Q = Q 脳 (未T / T + 未M / M) Where 未T and 未M are the uncertainties in the temperature readings and mass of the samples, respectively, and T and M are the actual temperature readings and mass of the samples. For benzoic acid, 未Q = Q 脳 (未T / T + 未M / M) 未Q = 40.61 kJ 脳 (0.002 掳C / 3.284 掳C + 0.001 g / 0.47 g) 未Q = 40.61 kJ 脳 0.00123 未Q = 0.0499 kJ For caffeine, 未Q = Q 脳 (未T / T + 未M / M) 未Q = 37.64 kJ 脳 (0.002 掳C / 3.05 掳C + 0.001 g / 0.53 g) 未Q = 37.64 kJ 脳 0.00121 未Q = 0.0455 kJ Now, we will propagate the uncertainties to find the uncertainty in the heat of combustion per mole: 未(Heat of combustion) = Heat of combustion 脳 (未Q / Q) 未(Heat of combustion) = 13771 kJ/mol 脳 (0.0455 kJ / 37.64 kJ) 未(Heat of combustion) = 13771 kJ/mol 脳 0.00121 未(Heat of combustion) = 16.66 kJ/mol Therefore, the estimated uncertainty in the heat of combustion per mole of caffeine is approximately 16.66 kJ/mol.

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Key Concepts

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

Heat of Combustion
The heat of combustion refers to the energy released when a substance undergoes complete combustion in the presence of an oxidizer, typically oxygen, at constant volume. This process is essential for understanding how much energy can be obtained from burning fuels or specific substances. In our case, we calculate the heat of combustion for both benzoic acid and caffeine using a bomb calorimeter. The device measures how much the temperature rises when the sample burns, providing a basis to calculate the energy released.
It's important to note the heat of combustion is expressed per gram or per mole, allowing us to compare the energy yields from different compounds. In this exercise, we used the heat of combustion per gram for benzoic acid to find the heat released, which assisted in calculating caffeine's heat of combustion per mole.
Benzoic Acid
Benzoic acid plays a crucial role as a calibration substance in bomb calorimetry. It has a known and consistent heat of combustion, making it an ideal standard for calibrating the calorimeter. When we combust 0.47 g of benzoic acid, the energy released causes a measurable temperature rise, used to calculate the calorimeter's heat capacity.
This step is vital because it sets the baseline for future experiments by creating a relationship between the temperature change and the heat released. In this specific exercise, the heat liberated by burning benzoic acid allowed us to determine the heat capacity of the bomb calorimeter, which was subsequently used to calculate the energy released by caffeine.
Caffeine Combustion
Caffeine is an organic compound whose heat of combustion we aim to determine. After determining the calorimeter's heat capacity using benzoic acid, the next step is to combust caffeine and note the temperature rise, which was 3.05掳C. By comparing this temperature increase with that from benzoic acid, we calculate how much energy caffeine releases.
The formula for calculating this takes into account the mass of caffeine and applies the heat capacity derived from benzoic acid combustion. From there, dividing the total energy by the number of moles of caffeine gives us the heat of combustion per mole, a critical measure for understanding its energy potential. The final value calculated was 13771 kJ/mol for caffeine.
Uncertainty Calculation
Understanding uncertainty is crucial for generating reliable experimentation results. It involves quantifying the possible deviations in measurement from their exact value. In this exercise, we consider uncertainties in temperature readings and mass measurements.
For each component鈥攂enzoic acid and caffeine鈥攚e calculate their respective uncertainties using the given margin of error. These are then combined to find the overall uncertainty in the heat of combustion calculation. For caffeine, the estimated uncertainty in the measurement was about 16.66 kJ/mol.
  • Temperature uncertainty affects how accurately we can link the temperature rise to energy released.
  • Mass uncertainty influences precise determination of how much sample was burned.
Calculating these helps assess the reliability of the heat of combustion value obtained.

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Most popular questions from this chapter

A magnesium ion, \(\mathrm{Mg}^{2+}\), with a charge of \(3.2 \times 10^{-19} \mathrm{C}\) and an oxide ion, \(\mathrm{O}^{2-},\) with a charge of \(-3.2 \times 10^{-19} \mathrm{C},\) are separated by a distance of \(0.35 \mathrm{nm}\). How much work would be required to increase the separation of the two ions to an infinite distance?

A coffee-cup calorimeter of the type shown in Figure 5.18 contains \(150.0 \mathrm{~g}\) of water at \(25.2^{\circ} \mathrm{C}\). A \(200-\mathrm{g}\) block of silver metal is heated to \(100.5^{\circ} \mathrm{C}\) by putting it in a beaker of boiling water. The specific heat of \(\mathrm{Ag}(s)\) is \(0.233 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K})\). The \(\mathrm{Ag}\) is added to the calorimeter, and after some time the contents of the cup reach a constant temperature of \(30.2^{\circ} \mathrm{C} .(\mathbf{a})\) Determine the amount of heat, in J, lost by the silver block. (b) Determine the amount of heat gained by the water. The specific heat of water is \(4.184 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K}) .(\mathbf{c})\) The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam \(^{\circ}\) cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by \(1 \mathrm{~K} .\) Calculate the heat capacity of the calorimeter in \(\mathrm{J} / \mathrm{K}\). (d) What would be the final temperature of the system if all the heat lost by the silver block were absorbed by the water in the calorimeter?

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?

Under constant-volume conditions, the heat of combustion of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) is \(40.18 \mathrm{~kJ} / \mathrm{g}\). A 2.50 -g sample of naphthalene is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.50 to \(28.83^{\circ} \mathrm{C}\). (a) What is the total heat capacity of the calorimeter? (b) A 1.50-g sample of a new organic substance is combusted in the same calorimeter. The temperature of the calorimeter increases from 21.14 to \(25.08^{\circ} \mathrm{C}\). What is the heat of combustion per gram of the new substance? (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

At the end of 2012, global population was about 7.0 billion people. What mass of glucose in 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 \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) according to the following thermochemical equation: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ} &=-2803 \mathrm{~kJ} \end{aligned} $$
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