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 that contains \(1.00 \mathrm{~L}\) of water \((d=1.00 \mathrm{~g} / \mathrm{mL})\). The heat capacity of the calorimeter is \(16.97 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter and water rise from \(23.49^{\circ} \mathrm{C}\) to \(27.72^{\circ} \mathrm{C}\) (a) What is \(q\) for the calorimeter? (b) What is \(q\) for water in the calorimeter? (c) What is \(q\) when \(4.50 \mathrm{~g}\) of fructose are burned in the calorimeter? (d) What is \(q\) for the combustion of one mole of fructose?

Short answer

Question: Calculate the heat generated for the combustion of one mole of fructose, given that a sample of 4.50 g of fructose is burned in a bomb calorimeter, which has a heat capacity of 16.97 kJ/°C. The initial temperature of the calorimeter is 23.49°C, and the final temperature is 27.72°C. There is 1.00 L of water inside the calorimeter. Answer: The heat generated for the combustion of one mole of fructose is -3576 kJ/mol.

Step-by-step solution

Calculate the heat gained by the calorimeter (q_calorimeter)

To find the heat gained by the calorimeter (q_calorimeter), we will use the formula: q = CΔT Where q is the heat, C is the heat capacity of the calorimeter (16.97 kJ/°C), and ΔT is the change in temperature (which is T_final - T_initial, and in this case, 27.72°C - 23.49°C). ΔT = 27.72 - 23.49 = 4.23°C Now, find q_calorimeter: q_calorimeter = CΔT = (16.97 kJ/°C) × 4.23°C = 71.7 kJ

Calculate the heat gained by water (q_water)

To find the heat gained by the water (q_water), we will use the specific heat capacity formula: q = mcΔT where m is the mass of the water, c is the specific heat capacity of the water (4.18 J/g°C), and ΔT is the same as calculated before, 4.23°C. First, we need to find the mass of the water. Since the density (d) = mass (m) / volume (V), then mass (m) = density (d) × volume (V). The density of water = 1.00 g/mL, and there's 1.00 L (1000 mL) of water. m_water = d × V = (1.00 g/mL) × (1000 mL) = 1000 g Now, calculate q_water: q_water = mcΔT = (1000 g) × (4.18 J/g°C) × (4.23°C) = 17708 J = 17.7 kJ (rounded to one decimal place)

Calculate the heat gained when 4.50 g of fructose are burned (q_combustion)

According to the law of conservation of energy, the heat lost during combustion is equal but opposite in sign to the heat gained by the calorimeter and water. q_combustion = - (q_calorimeter + q_water) = - (71.7 kJ + 17.7 kJ) = -89.4 kJ

Calculate the heat gained for the combustion of one mole of fructose (q_one_mole)

To calculate the heat generated for the combustion of one mole of fructose (q_one_mole), we need to find the molar mass of fructose (C₁H₁₂O₆). Molar mass of fructose = 6 × M_C + 12 × M_H + 6 × M_O = 6 × 12.01 g/mol + 12 × 1.01 g/mol + 6 × 16.00 g/mol = 72.06 g/mol + 12.12 g/mol + 96.00 g/mol = 180.18 g/mol Now, find out how many moles are in the given 4.50 g of fructose: 4.50 g / 180.18 g/mol ≈ 0.025 mol Next, divide the heat gained during combustion (q_combustion) by the number of moles to get q_one_mole: q_one_mole = q_combustion / 0.025 mol ≈ -89.4 kJ / 0.025 mol = -3576 kJ/mol