CHEM 108 Expt #8: Heat of Solution of CO(NH2)2
Sample Calculations:
mass of NaI = 7.89 g Mass of water = 35.2 g T initial = 22.8C Tfinal = 25.0C
Note: These are only a set of sample data. You will have different number of sig. fig. in your data.
Obviously you are not working of NaI but with urea.
1. Mass of solution = 7.89 g + 35.2 g = 43.09 g = 43.1 g soln
2. T of soln = Tfinal – Tinitial = 25.0C – 22.8C = + 2.2 C
3. Calc of qsoln
J
q soln = s m T = 4.18
x (43.1 g soln) x (2.2C) = 396 J
g.C
4. Convert to kJ
 1 kJ 
x kJ = 396 J  3  = 0.396 kJ = 0.40 kJ (2 sig. fig.)
 10 J 
5. qrxn = –qsoln = –0.40 kJ
6. Hrxn = –0.40 kJ
(q = H in an open container.)
7. Calc molar mass of NaI. MM = (22.99 + 126.9) g/mol = 149.9 g/mol
8. Calc # mol NaI
 1 mol NaI 
x mol NaI = 7.89 g NaI 

 149.9 g NaI 
= 0.0526 mol NaI (You may have more sig. fig. in your data.)
9. Calc molar Hrxn = q for 1 mole of NaI = q per mole of NaI
 0.40 kJ
=  7.6 kJ/mol NaI (2 sig. fig.)
0.0526 mol NaI
10. is this reaction endothermic or exothermic? It is exothermic. How do we tell??
11. Thermochemical equations:
NaI (s)  NaI (aq) + 7.6 kJ
(Incorrect to write 7.6 kJ/mol)
and
NaI (s)  NaI (aq)
H = –7.6 kJ
Molar H rxn =
We don’t write 7.6 kJ/mol because these thermochemical equations are written such that the amount of heat
corresponds to the coefficients in the equation.
Since the coefficient of NaI is “1” it is understood 7.6 kJ is for one mole.
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