Posted by **Chris** on Sunday, January 28, 2007 at 9:01pm.

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.
**

## Answer This Question

## Related Questions

- chemistry - A solution of fructose, C6H12O6, a sugar found in many fruits, is ...
- Chemistry - A sample of benzene, C6H6, weighing 3.51 g was burned in an excess ...
- Chemistry - A 1.50 g sample of glucose (C6H12O6) is burned in a bomb calorimeter...
- Chemistry - A sample of ethanol, C2H5OH, weighing 2.84 g was burned in an excess...
- chemistry - A hydrocarbon sample was burned in a bomb calorimeter. The ...
- Chemistry - A 1.20g sample of benzoic acid is burned in excess of )2 in a bomb ...
- CHEMISTRY - A hydrocarbon sample was burned in a bomb calorimeter. The ...
- chemistry - An alternative approach to bomb calorimetry is toestablish the heat ...
- Algebra 2 - Ordinary corn syrup contains dextrose sugar. A process called enzyme...
- chemistry - A 0.167 g sample of Mg(s) is burned in a bomb calorimeter which has ...

More Related Questions