Experiment 25 - Post-lab questions PDF

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This is the post lab for experiment 25 which was the calorimetry experiment for general chemistry 1 with lab....

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Calculations: A. Specific Heat of a Metal ~ Mass of water  (Mass of Calorimeter + water (g) – Mass of Calorimeter (g) ) 188.943 – 169.633 = 19.31 g ~ Temperature change of water, ΔT (°C)  (Max Temp of metal and water from graph (°C) – Temperature of water in calorimeter (°C) ) 22.405°C – 21.500°C = 0.905°C ~Heat gained by water (J)  (Mass of water x specific heat of water x Temperature change of water) (19.31) x (4.184) x (0.905°C) = 73.12 ~ Temperature change of metal, ΔT (°C)  (Max Temp of metal and water from graph – Temperature of metal (boiling water)

22.405°C – 60.5°C =  38.095°C ~ Specific heat of metal (J/g*°C)   (Specific heat capacity of water (J/g*°C) x Mass of water (g) x Temperature change of water, ΔT (°C)) / (Mass of Metal (g) x Temperature change of metal, ΔT (°C))  (4.184 J/g*°C x 19.31 x 0.905) / (1.580 x  38.095°C) = 1.2148 B. Enthalpy (Heat) of Solution for the Dissolution of a Salt ~ Moles of salt (mol)  (Mass of salt / Molar mass of Calcium Chloride (CaCl2)

5.14 / 110.98 = 0.046 mol ~ Mass of water (g)  (Mass of Calorimeter + water (g) – Mass of Calorimeter (g) ) 188.122 – 168.951 = 19.171 g ~ Temperature change of solution, ΔT (°C)  (Final temperature of mixture from graph (°C) – Initial temperature of water (°C) 34.890 – 21.500 = 13.39°C ~ Heat change of water (J)  (Mass of water x specific heat capacity of water x Temperature change of solution) (19.171 x 4.184 x 13.39) = 1,074.03 J ~ Heat change of salt (J) S.H. of  (Mass of salt x specific heat of CaCl2 x Temperature change of solution) (5.14 x 0.657 x 13.39) = 45.22 J ~ Specific heat of metal (J/g*°C)   (Specific heat capacity of water (J/g*°C) x Mass of water (g) x Temperature change of water, ΔT (°C)) / (Mass of Metal (g) x Temperature change of metal, ΔT (°C))  (4.184 x 19.171 x 13.39) / (1.580 x  38.095) = 17.844 J/g*°C ~ Total enthalpy change (J)  (Heat change of water (J) + Heat change of salt (J)) 1074.03 J + 45.22 J = 1,119.25 J ~ ΔH, (J/mol salt), equation 25.12  (  Specific heat of water x Mass of water x Temperature change of water) + (  Specific heat of salt x Mass of salt x Temperature change of salt) / (Moles of salt) (  4.184 x 19.171 x 13.39) + (  0.657 x 5.14 x 13.39) / (0.046) =  24,331.51 J/mol salt

Post-Lab Questions: 1. Part A.1. The 200-nm test tube also contained some water (besides the metal) that was subsequently added to the calorimeter (in Part A.4.). Considering a higher specific heat for water, will the temperature change in the calorimeter be higher, lower, or unaffected by this technique error? Explain. The temperature change in the calorimeter will be lower by this technique error because of the original specific heat of water. Considering a higher specific heat for water, this means that the water requires more energy to raise its temperature thus a lower change in temperature of the calorimeter is yielded. The test tube also contained some water aside from the metal which contributed to the mass of water in the calorimeter. A bigger mass means a higher heat required to heat it and result in a lower change in temperature. 2. Part A.5. In measuring the specific heat of a metal, Josh used the highest measured temperature for calculating the metal’s specific heat rather then the extrapolated temperature. Will this decision result in a higher or lower specific heat value for the metal? Explain. If Josh used the highest measured temperature for calculating the metal’s specific heat instead of the extrapolated temperature, then it would result in a lower specific heat value for the metal. By using the maximum temperature recorded, the change of temperature of the metal will be greater while the temperature change of water will be less. This can also be found using the formula Q ∁= where a greater change in temperature results in a greater denominator which will m∆T yield a lower specific heat. 7. Part C.3. If some of the salt remains adhered to the weighing paper (and therefore is not transferred to the calorimeter), will the enthalpy of solution for the salt be reported too high or too low? Explain. The enthalpy of solution for the salt will be reported as too low if some of the salt is not transferred to the calorimeter. This occurs because the enthalpy of the solution is being measured with a lower mass instead of the original, entire mass....