# Molar Enthalpy question with Combustion

1. Jun 27, 2009

### staka

1. The problem statement, all variables and given/known data
Using the data for the combustion of 3-pentanol calculate the molar enthalpy for the alcohol.

Initial mass of alcohol: 27.35g
Final mass of alcohol: 26.41g
Change in temperature: 21.0C
Volume of water: 100.0mL (100.0g)
Specific heat capacity of water: 4.18 J/gC

2. Relevant equations
Q=mcΔT

3. The attempt at a solution
3-pentanol: C5H12O (88.14818 g/mol)
27.35g-26.41g=0.94g
0.94g/88.14818g/mol=0.010663861mol

Q=mcΔT
=(100)(4.18)(21.0)
=8778J
ΔH=-8778J
ΔH/n=-8778J/0.0106638161mol
=-823154.0152J/mol
=-8.23x10^2kJ/mol ←is this the correct answer? (I don't have an answer key.)

2. Jun 29, 2009

### chemisttree

It looks good. The only complicating factor is to determine if the 3-pentanol underwent an increase in temperature as well. You used about a gram of the alcohol and about 26 grams are left. Did that remaining alcohol undergo an increase in temperature of 21oK?

3. Jun 29, 2009

### staka

I believe the remaining alcohol increased 21 Celcius.
I just want to know if I determined the molar enthalpy of alcohol in the end.

4. Jun 29, 2009

### chemisttree

If that is the case you will need to revise (very slightly) your answer to account for the heat required to heat 26.41 g of the alcohol through 21o. It's probably small enough to neglect.