# chemistry

posted by
**Chris**
.

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.
**