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.
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