Problem 107
Question
(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 \(0.53-\mathrm{g}\) 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?
Step-by-Step Solution
VerifiedKey Concepts
Heat of Combustion
To determine the heat of combustion of a sample, such as when using benzoic acid in a calorimeter, we calculate the total heat released. This is done using the formula:
- \[ q = m \cdot \Delta H \]
- where \( q \) represents the heat released, \( m \) is the mass of the substance, and \( \Delta H \) is the heat of combustion per gram.
Calorimeter Constant
To find the calorimeter constant, the equation used is:
- \[ C = \frac{q}{\Delta T} \]
- where \( q \) denotes the total heat transferred to the calorimeter, and \( \Delta T \) is the temperature change experienced.
Molar Mass
For a compound like caffeine, \( \text{C}_8\text{H}_{10}\text{N}_4\text{O}_2 \), calculating the molar mass involves summing the atomic masses of each atom present in the molecular formula. The calculation is straightforward:
- Carbon (C): 12.01 g/mol, 8 atoms => \( 8 \times 12.01 \) g/mol
- Hydrogen (H): 1.01 g/mol, 10 atoms => \( 10 \times 1.01 \) g/mol
- Nitrogen (N): 14.01 g/mol, 4 atoms => \( 4 \times 14.01 \) g/mol
- Oxygen (O): 16.00 g/mol, 2 atoms => \( 2 \times 16.00 \) g/mol