Question

@shiraz14

benzoic acid sucrose
mass of sample 0.9488 0.9953
initial temperature 25.001 25.009
final temperature 30.411 28.264

Answer 1

1. Calculate DeltaT for the water by DeltaT = |Tf - Ti|
DeltaT = |Tf - Ti| = |28.264 – 25.009| = 3.255

2. Calculate the moles of sucrose in the sample (MWsucrose = 342.3g/mol)
Number of moles of sucrose = m/MW = 0.9953/342.3 = 0.002908 mol = 2.908 mmol

3. DeltaHcomb for sucrose can be calculated by DeltaHcomb = (CsystemDeltaT)/n, where n is the moles of sucrose in the sample and Csystem is the heat capacity of the calorimetric system. Using the value you calculated for Csystem, calculate the heat of combustion for sucrose.
(Csystem * DeltaT)/n = DeltaH
(4632.679113 * 3.255)/0.002908 = DeltaH
DeltaH = 5185478.168

Do these look correct? I tried.
@shiraz14

Answer 2

@shiraz14
are you still there?

Answer 3

@JJ_Stark : You missed out the units for Q3 - otherwise, it's fine.

Answer 4

(Csystem * DeltaT)/n = DeltaH
(4632.679113 * 3.255)/0.002908 = DeltaH
DeltaH = 5185478.168 J/mol = 5185.478168 kJ/mol
Would those be the correct units for DeltaH? Do I keep J or kJ?

Answer 5

Yes, this would be correct. Leave the answer in kJ - the magnitude of the value of ΔH looks too large if left in J.

Answer 6

Okay thank you! Last one:
4. If the accepted value for the heat of combustion for sugar is 5639 kJ/mol, calculate the percent error.
% Error = |your answer - accepted answer|/accepted answer
* 100

Answer 7

% Error = (|5185.478168 – 5639|/5639) * 100

Answer 8

|-453.5218322|/5639 =
-0.08042593229 * 100
-8.042593229

Answer 9

Is that what I should put for the percent error? Or is there something I'm missing because somehow that seems like a weird percent error to me. I don't know...

Answer 10

You forgot that |x| is always >=0, even if x<0. Remove the negative sign in your answer, and it should be fine.

Answer 11

Ohhhh, that makes sense. I forgot about that. Thank you! So:
|-453.5218322|/5639 =
|453.5218322|/5639 =
0.08042593229 * 100 =
8.042593229 % error =

Answer 12

@JJ_Stark : Yes, your last post is correct.

Answer 13

Oh my god, yay. Thank you so much! I really, really appreciate all of the help that you've given me. Like so, so much. Thank you!

Answer 14

You're welcome. :)