(a) When 1.000 g of olive oil was completely burned in pure oxygen in a bomb calorimeter like the one shown in Figure 6.7, the temperature of the water bath increased from 22.000 °C to 26.049 °C. How many dietary Calories are in olive oil, per gram? The heat capacity of the calorimeter is 9.032 kJ/°C.
(b) Olive oil is almost pure glyceryl trioleate, C57H104O6. The equation for its combustion is
C57H104O6 + 80O2 → 57CO2 + 52H2O
What is the change in internal energy, ?E, for the combustion of one mole of glyceryl trioleate? Assume the olive oil burned in part (a) was pure glyceryl trioleate.
The Correct Answer and Explanation is:
The Correct Answer and Explanation is:
Sure! Let’s carefully solve both parts step-by-step, then provide a detailed explanation.
(a) Calories per gram in olive oil from calorimeter data
Given:
- Mass of olive oil burned, $m = 1.000 \, g$
- Initial temperature, $T_i = 22.000^\circ C$
- Final temperature, $T_f = 26.049^\circ C$
- Heat capacity of calorimeter, $C_{cal} = 9.032 \, kJ/^\circ C$
- Temperature change, $\Delta T = T_f – T_i = 26.049 – 22.000 = 4.049^\circ C$
Step 1: Calculate heat released, $q$, in kJ
$$
q = C_{cal} \times \Delta T = 9.032 \, \text{kJ/}^\circ C \times 4.049^\circ C = 36.58 \, \text{kJ}
$$
Since the olive oil burned completely, this is the energy released by 1.000 g of olive oil.
Step 2: Convert kJ to dietary Calories
1 dietary Calorie (Cal) = 1 kcal = 4.184 kJ
$$
\text{Calories per gram} = \frac{36.58 \, kJ}{4.184 \, kJ/Cal} = 8.75 \, \text{Cal/g}
$$
(b) Calculate the change in internal energy, $\Delta E$, per mole of glyceryl trioleate
Given combustion reaction:
$$
\mathrm{C}{57}\mathrm{H}{104}\mathrm{O}_6 + 80 \mathrm{O}_2 \rightarrow 57 \mathrm{CO}_2 + 52 \mathrm{H}_2\mathrm{O}
$$
Step 1: Calculate moles of olive oil burned in part (a)
Molecular weight of glyceryl trioleate:
- C: 57 atoms × 12.01 g/mol = 684.57 g/mol
- H: 104 atoms × 1.008 g/mol = 104.832 g/mol
- O: 6 atoms × 16.00 g/mol = 96.00 g/mol
Total molar mass:
$$
M = 684.57 + 104.832 + 96.00 = 885.402 \, g/mol
$$
Moles burned in part (a):
$$
n = \frac{1.000 \, g}{885.402 \, g/mol} = 0.00113 \, mol
$$
Step 2: Calculate energy released per mole
Energy released per mole:
$$
q_{\text{mole}} = \frac{36.58 \, kJ}{0.00113 \, mol} = 32,362 \, kJ/mol
$$
Since this is combustion, the change in internal energy $\Delta E$ is negative (energy released):
$$
\boxed{\Delta E = -32,362 \, kJ/mol}
$$
Detailed explanation (300+ words):
In part (a), we used data from a bomb calorimeter experiment to determine the energy content of olive oil per gram. A bomb calorimeter is a device used to measure the heat released during combustion of a sample. Here, 1.000 g of olive oil was burned in pure oxygen, causing the temperature of the surrounding water bath to rise. The temperature increase was 4.049 °C, and the heat capacity of the calorimeter system (which includes the water and container) is given as 9.032 kJ/°C. Multiplying these together gives the total energy released during combustion: about 36.58 kJ for 1 g of olive oil.
Dietary energy is usually measured in Calories (with a capital “C”), where 1 Calorie = 1 kcal = 4.184 kJ. Converting 36.58 kJ to Calories gives roughly 8.75 Cal/g, which matches well with typical values for fats (usually about 9 Cal/g). This means olive oil contains about 8.75 dietary Calories per gram, confirming its high energy density typical of fats.
In part (b), we examine the molecular-level combustion of olive oil, modeled as glyceryl trioleate ($C_{57}H_{104}O_6$). Using its molecular formula, we calculate its molar mass to be approximately 885.4 g/mol. Since only 1 g was burned in part (a), that corresponds to roughly 0.00113 moles. Dividing the total energy released (36.58 kJ) by this small mole quantity gives the energy released per mole, approximately -32,362 kJ/mol. The negative sign indicates that energy is released (exothermic reaction).
This large magnitude aligns with the fact that combustion of large hydrocarbon molecules releases substantial energy, which explains why fats like olive oil serve as dense energy sources in nutrition. The internal energy change $\Delta E$ here is nearly equivalent to the enthalpy change $\Delta H$ because the volume change in the sealed bomb calorimeter is minimal, so work done on the surroundings is negligible.
If you’d like, I can also help with enthalpy change calculations or explanations on how bomb calorimetry connects with thermodynamic quantities!
