The combustion of 0.1584 g benzoic acid increases the temperature of a bomb calorimeter by \(2.54^{\circ} \mathrm{C}\) . Calculate the heat capacity of this calorimeter. (The energy released by combustion of benzoic acid is 26.42 \(\mathrm{kJ} / \mathrm{g} .\) ) A \(0.2130-\mathrm{g}\) sample of vanillin \(\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) is then burned in the same calorimeter, and the temperature increases by $3.25^{\circ} \mathrm{C}$ . What is the energy of combustion per gram of vanillin? Per mole of vanillin?

Calculate the heat released by the combustion of 0.1584 g of benzoic acid. We are given that the heat of combustion for benzoic acid is 26.42 kJ/g. We can find the heat released by using the formula: Heat released (\(q\)) = mass (\(m\)) × heat of combustion (\(H_c\)) So, \(q_{benzoic \space acid} = 0.1584 \space \mathrm{g} \times 26.42 \frac{\mathrm{kJ}}{\mathrm{g}} = 4.184 \space \mathrm{kJ}\).

To find the heat capacity of the calorimeter, we need to divide the heat released by benzoic acid by the increase in temperature caused by the combustion. The formula for this is: Heat capacity (\(C_{cal}\)) = \(q_{benzoic \space acid} \div \Delta T\) Where \(\Delta T\) is the change in temperature. In this case, \(\Delta T = 2.54 \degree \mathrm{C}\). So, \(C_{cal} = \frac{4.184 \space \mathrm{kJ}}{2.54 \degree \mathrm{C}} = 1.647 \frac{\mathrm{kJ}}{\degree \mathrm{C}}\).

Since the heat capacity of the calorimeter is now known, we can use it to calculate the heat released by the combustion of vanillin. The formula for this is: Heat released (\(q\)) = heat capacity (\(C_{cal}\)) × increase in temperature (\(\Delta T\)) In this case, \(\Delta T = 3.25 \degree \mathrm{C}\). So, \(q_{vanillin} = 1.647 \frac{\mathrm{kJ}}{\degree \mathrm{C}} \times 3.25 \degree \mathrm{C} = 5.353 \space \mathrm{kJ}\).

To determine the energy of combustion per gram of vanillin, we need to divide the heat released by vanillin by the mass of vanillin: Energy of combustion (\(H_c\)) = \(\frac{q_{vanillin}}{m}\) Where \(m = 0.2130 \space \mathrm{g}\) So, \(H_c = \frac{5.353 \space \mathrm{kJ}}{0.2130 \space \mathrm{g}} = 25.13 \frac{\mathrm{kJ}}{\mathrm{g}}\).

Lastly, we need to convert the energy of combustion per gram of vanillin to energy of combustion per mole of vanillin. First, determine the molar mass of vanillin: Molar mass of vanillin = C\(_8\)H\(_8\)O\(_3\) = \(8 \times 12.01 \space \mathrm{g} + 8 \times 1.01 \space \mathrm{g} + 3 \times 16.00 \space \mathrm{g} = 152.15 \space \mathrm{g/mol}\) Now, convert the energy of combustion per gram of vanillin to per mole: Energy of combustion per mole (\(H_c' \space \mathrm{kJ/mol}\)) = energy of combustion per gram (\(H_c \space \mathrm{kJ/g}\)) × molar mass of vanillin (\(\frac{\mathrm{g}}{\mathrm{mol}}\)) \(H_c' = 25.13 \frac{\mathrm{kJ}}{\mathrm{g}} \times 152.15 \frac{\mathrm{g}}{\mathrm{mol}} = 3826.75 \frac{\mathrm{kJ}}{\mathrm{mol}}\) Therefore, the energy of combustion per gram of vanillin is \(25.13 \frac{\mathrm{kJ}}{\mathrm{g}}\) and per mole is \(3826.75 \frac{\mathrm{kJ}}{\mathrm{mol}}\).