The reaction for the metabolism of sucrose, C12H22O11, is the same as for its combustion in oxygen to yield CO2(g) and H2O(l). The standard heat of formation of sucrose is -2230 kJ mol-1. Based upon data in Table 6.2 and problem 6.75 the amount of energy (in kJ) released by metabolizing 4.17 oz (118 g) of sucrose is?
figure the moles of sucrose first.
Next,
energy=molessugar*2230Kj/mol
moles would be 118g divided by 342.3?
Well, isn't that sweet? We're talking about the metabolism of sucrose! I guess you could say that's some high-energy sweetness! Now, let's calculate the amount of energy released when metabolizing 4.17 oz (118 g) of sucrose!
First, we need to convert the mass of sucrose from grams to moles. The molar mass of sucrose is approximately 342.3 g/mol. So, 118 g of sucrose is roughly 0.345 mol.
Now, let's use the molar heat of formation of sucrose, which is -2230 kJ/mol, to calculate the energy released when metabolizing this amount of sucrose.
Using the equation:
Energy released (in kJ) = Molar heat of formation × Number of moles,
we can plug in the values:
Energy released = -2230 kJ/mol × 0.345 mol.
Calculating that out, we find that the amount of energy released by metabolizing 4.17 oz (118 g) of sucrose is approximately -769.3 kJ.
So, we've "sweetly" determined that about -769.3 kJ of energy is released when metabolizing that amount of sucrose. It seems like metabolism really knows how to burn through those calories!
To calculate the amount of energy released by metabolizing 4.17 ounces (118 grams) of sucrose, we need to determine the number of moles of sucrose and use the stoichiometry of the reaction to find the heat released.
1. Start by converting the mass of sucrose (118 grams) into moles. To do this, divide the mass by the molar mass of sucrose.
The molar mass of sucrose (C12H22O11) can be calculated by adding up the atomic masses of carbon (C), hydrogen (H), and oxygen (O).
(12*g/mol * 12) + (1*g/mol * 22) + (16*g/mol * 11) = 342.3 g/mol
Therefore, the number of moles of sucrose is: 118 g / 342.3 g/mol ≈ 0.345 mol
2. Now, we need to use the stoichiometry of the reaction to determine the amount of energy released per mole of sucrose. The balanced equation for the combustion of sucrose is:
C12H22O11(s) + 12O2(g) -> 12CO2(g) + 11H2O(l)
From the balanced equation, we can see that for every mole of sucrose metabolized, 2230 kJ of energy is released.
3. Finally, we can calculate the total amount of energy released by multiplying the number of moles of sucrose by the energy released per mole.
Total energy released = 0.345 mol * 2230 kJ/mol ≈ 768.35 kJ
Therefore, metabolizing 4.17 ounces (118 grams) of sucrose would release approximately 768.35 kJ of energy.