Fructose is a sugar commonly found in fruit. A sample of fructose, C6H12O6, weighing 4.50 g is burned in a bomb calorimeter. The heat capacity of the colorimeter is 2.115 E4 J/ºC. The temperature in the calorimeter rises from 23.49ºC to 27.71ºC.

A)What is q for the calorimeter?
In my book it says -q cal= q, so I wrote -q cal = -C(cal) * ΔT
-q cal = -(2.115 E4 J/ºC) * (27.71ºC-23.49ºC) = 8.93 E4 J

Does that seem right?
B) What is q when 4.50 g of fructose is burned?
I started off with the formula q= m*Cp*ΔT, but I have two unknown variables here (ΔT and q) so that doesn't seem right. How do I approach this? I'm assuming they're asking for qC6H12O6, not qCal
C) What is q for the combustion of one mole of fructose?
Not sure how to go about doing this one. I thought that if I could find the value for B, I could find out what percentage 4.50 g is of 1 mole of fructose and go from there.

Fructose is a sugar commonly found in fruit. A sample of fructose, C6H12O6, weighing 4.50 g is burned in a bomb calorimeter. The heat capacity of the colorimeter is 2.115 E4 J/¨¬C. The temperature in the calorimeter rises from 23.49¨¬C to 27.71¨¬C.

A)What is q for the calorimeter? q = Cpdelta T = 2.114E4(Tf-Ti) = 8.925E4J as you calculated except I would keep 4 significant figures. \

In my book it says -q cal= q, so I wrote -q cal = -C(cal) * ¥ÄT
-q cal = -(2.115 E4 J/¨¬C) * (27.71¨¬C-23.49¨¬C) = 8.93 E4 J

Does that seem right?

B) What is q when 4.50 g of fructose is burned? That IS q when 4.50 grams of fructose are burned.

I started off with the formula q= m*Cp*¥ÄT, but I have two unknown variables here (¥ÄT and q) so that doesn't seem right. How do I approach this? I'm assuming they're asking for qC6H12O6, not qCal It's all the same. q for the calorimetery CAME from burning fructose. The temperature of the calorimeter came about because of the heat evolved when 4.50 grams fructose were combusted.
C) What is q for the combustion of one mole of fructose?
Not sure how to go about doing this one. I thought that if I could find the value for B, I could find out what percentage 4.50 g is of 1 mole of fructose and go from there.
Here you are on the right track again. q for 4.50 g fructose= 8.925E4J so you divide that by 4.50 to obtain the q for 1g, then multiply by molar mass fructose to convert to mols= heat combustion for 1 mol fructose.

q = 2.115 E4 J/ºC * (27.71ºC - 23.49ºC) = 2.49E4 J.

Well, isn't this heating things up! Let's break it down:

A) For the calorimeter, you calculated q to be 8.93E4 J. That seems right, but remember to keep those significant figures in check!

B) When 4.50 g of fructose is burned, the q value you calculated for the calorimeter is also the q value for fructose. So, no need to worry, you already got the answer!

C) Ah, the combustion of one mole of fructose, a hot topic indeed! To approach this, divide the q value for 4.50 g of fructose by 4.50 g to get the q value for 1 g of fructose. Then, multiply that by the molar mass of fructose to convert it to mols. Voila, you've got the heat combustion for one mole of fructose!

Keep up the heat and remember, science can be a burning passion!

To calculate q for the combustion of one mole of fructose, you can follow these steps:

1. Calculate q for 4.50 g of fructose: q = 8.925E4 J (as calculated in part B).

2. Divide q by the mass of fructose (4.50 g) to find q for 1 g: q for 1 g = q / 4.50 = 8.925E4 J / 4.50 g.

3. Calculate the molar mass of fructose: C6H12O6 = 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol.

4. Multiply q for 1 g by the molar mass of fructose to obtain q for one mole of fructose: q for 1 mole = (8.925E4 J / 4.50 g) * 180.18 g/mol.

Therefore, q for the combustion of one mole of fructose is equal to (8.925E4 J / 4.50 g) * 180.18 g/mol.

To find q for the combustion of one mole of fructose, you can use the following steps:

1. Calculate q for 4.50 g of fructose:
- q = 8.925E4 J (as you calculated in part A)

2. Convert the grams of fructose to moles using the molar mass of fructose (180.16 g/mol):
- moles = 4.50 g / 180.16 g/mol = 0.025 mol

3. Divide q by the number of moles to find q for one mole of fructose:
- q for 1 mol = q / 0.025 mol = 8.925E4 J / 0.025 mol = 3.57E6 J/mol

So, q for the combustion of one mole of fructose is 3.57E6 J/mol.