# Essay on Chem Lab Molar Heat Of Solution Of Ammo

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Pages: 3

Molar heat of solution of ammonium nitrate
The heat of solution of a solid compound can be easily measured in a calorimetric experiment. To calculate the molar heat of solutions by multiplying the number of joules of heat absorbed or evolved per gram of solute by the formula weight of the solute.
Apparatus and Materials
Styrofoam cup, Lid, 50 mL Beaker, Digital thermometer, Ammonium nitrate (NH4NO3­), Distilled water, Stir rod
Procedure
1. Measure and record a mass of 7.00 g ammonium nitrate into a 50 mL beaker.
2. Measure and record the mass of a Styrofoam cup.
3. Fill the cup with approximately 50 mL of distilled water.
4. Record the temperature of the water.
5. Record the mass of the water and the cup.
6. Add the ammonium nitrate to the cup, cover with lid. Record the temperature of the mixture once the temperature has stopped changing.
7. Wash cup out and repeat the trial once more.
Manipulative variable: Standardized Ammonium nitrate
Responding variable: Temperature of water in calorimeter
Controlled variable: Temperature and pressure of the room, cup mass and size, initial temperature of distilled water
Observations
Trial
Mass of NH4NO3 (g)
Mass of cup (g)
Mass of cup and water (g)
Mass of water (g)
Initial temperature (oC)
Final temperature (oC)
Δ Temperature (oC)
1st
7.04 g
4.52 g
49.39 g
44.87 g
19.4 oC
7.9 oC
-11.5 oC
2nd
14.04 g
4.52 g
46.98 g
42.46 g
19.4 oC
-0.2 oC
-19.6 oC

According to the CRC of Physics & Chemistry, the molar enthalpy of solution of ammonium nitrate is +25.7kJ/mol
Calculations
1. M NH4NO3= 80.06 g/mol, nΔrH = mc ΔT n = m / M = (7.04 g) / (80.06 g/mol) = 0.08793 mol Q = mc ΔT / (n) = (44.87 g)(4.184J/g* oC)(11.5 oC) / (0.08793 mol) ΔrH = 24553.22 J/mol = 24.55 kJ/mol
% Error = (Experimental value) / (Theoretical value) x 100% = (24.55 kJ/mol) / (25.7 kJ/mol) x 100% = 95.54 = 100 – 95.54 = 4.46%
2. n = m / M = (14.04 g) / (80.06 g/mol) = 0.17536 mol

Q = mc ΔT / (n) = (42.46 g)(4.184 J/g* oC)(19.6 oC) / (0.17536 mol)

ΔrH = 19855.28 J/mol = 19.86 kJ/mol
% Error = (Experimental value) / (Theoretical value) x 100% = (19.86 kJ/mol) / (25.7 kJ/mol) x 100% = 77.26 = 100 – 77.26 = 22.7 %

Error Analysis
One definite error reason is that the calorimeter wasn’t…