# Calorimitry Question

1. Jul 19, 2009

### webz

1. The problem statement, all variables and given/known data
20g of NH4NO3 (MM=80.05 g/mol) if dissolved in 125g of water in a coffee-cup colorimeter, the temperature falls from 23.5C to 13.4C. Determine 'q' for the dissolving of the compound. Is the process exothermic or endothermic?

2. Relevant equations
q(calorimeter)=C(calorimeter)*[Delta]T

3. The attempt at a solution
[Delta]T= -10.1
C(calorimeter)=~n(water)*C(water) = 125g H2O/18.016g/mol = 6.9382mol H2O
Ccal=~6.9382mol H2O* 75.291 J/mol C = 522.4 J/C
qcal=52.4J/C*-10.1C= -5.28kJ
q(calorimeter)= -q(chemical) = +5.28kJ

This answer makes sense, however, I don't understand why I am not taking into account the amount of moles NH4NO3. A larger amount of moles would make a larger deviation in T, which would change q.
Why are moles NH4NO3 not taken into account? Also, I'm using the heat capacity (C) of water, not of NH4NO3. The question is asking for 'q' for the dissolving compound. Is this correct?

2. Jul 21, 2009

### Staff: Mentor

To be precise you should take into account heat capacity of the solution - not water nor amonium nitrate. It is solution that changes temperature (well - this is more complicated - at first it is pure water that changes temperature, later it is solution, but it doesn't matter at the end). You are not given heat capacity of the solution, so you used heat capacity of water, which is the best approximation available to you.

You were asked to determine amount of heat needed for dissolving the compound - and looks like you did OK.

Number of moles will be required if the question asked for molar heat of dissolution - but it didn't.

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