A bomb calorimeter is ideally suited for measuring the heat evolved during a combustion
reaction. A sample of sucrose (C12H22O11(s)) in a bomb calorimeter causes the temperature of
345 mL of water to rise from 24.9 o
C to 28.3 o
C. The heat capacitiy of the calorimeter assembly
is 4.90 kj/mol. What is the experimental molar enthalpy of combustion of sucrose in kj/mol?

To determine the experimental molar enthalpy of combustion of sucrose, we need to calculate the heat released by the combustion of sucrose and then relate this to the number of moles of sucrose combusted.

### Step 1: Calculate the heat absorbed by the water.
The heat absorbed by the water (\(q_{\text{water}}\)) can be calculated using the formula:

\[
q_{\text{water}} = m_{\text{water}} \cdot C_{\text{water}} \cdot \Delta T
\]

Where:
- \(m_{\text{water}}\) is the mass of the water,
- \(C_{\text{water}}\) is the specific heat capacity of water (\(4.18 \, \text{J/g}^\circ \text{C}\)),
- \(\Delta T\) is the change in temperature.

First, let's calculate \(\Delta T\):

\[
\Delta T = 28.3^\circ \text{C} - 24.9^\circ \text{C} = 3.4^\circ \text{C}
\]

Given that the density of water is \(1 \, \text{g/mL}\), the mass of the water is:

\[
m_{\text{water}} = 345 \, \text{mL} \times 1 \, \text{g/mL} = 345 \, \text{g}
\]

Now, calculate the heat absorbed by the water:

\[
q_{\text{water}} = 345 \, \text{g} \times 4.18 \, \text{J/g}^\circ \text{C} \times 3.4^\circ \text{C}
\]

\[
q_{\text{water}} = 4905.66 \, \text{J} = 4.91 \, \text{kJ}
\]

### Step 2: Calculate the total heat released by the combustion.
The total heat released (\(q_{\text{total}}\)) by the combustion is the sum of the heat absorbed by the water and the heat absorbed by the calorimeter assembly:

\[
q_{\text{total}} = q_{\text{water}} + C_{\text{calorimeter}} \times \Delta T
\]

Given the heat capacity of the calorimeter assembly:

\[
q_{\text{calorimeter}} = 4.90 \, \text{kJ/}^\circ \text{C} \times 3.4^\circ \text{C} = 16.66 \, \text{kJ}
\]

So the total heat released is:

\[
q_{\text{total}} = 4.91 \, \text{kJ} + 16.66 \, \text{kJ} = 21.57 \, \text{kJ}
\]

### Step 3: Determine the molar enthalpy of combustion.
Finally, to find the molar enthalpy of combustion (\(\Delta H_{\text{comb}}\)), we divide the total heat released by the number of moles of sucrose burned.

However, without the mass of sucrose burned, we cannot directly determine this. Please provide the mass of sucrose burned in the experiment to proceed with the calculation.

Would you like to proceed by providing the mass of sucrose or have any other questions?

Here are some related questions:
1. What is the relationship between heat capacity and specific heat?
2. How does a bomb calorimeter ensure accurate measurement of heat?
3. How can we correct for heat losses in a calorimetry experiment?
4. What is the significance of the molar enthalpy of combustion?
5. How does the density of water affect the calorimetry calculations?

**Tip:** In calorimetry, always ensure you account for all sources of heat absorption to get an accurate measure of the energy change.

Learn how to calculate the experimental molar enthalpy of combustion of sucrose using a bomb calorimeter. Includes step-by-step solution and formulas for heat absorption by water and calorimeter. Suitable for college-level students and researchers.

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