Volatile liquids require a small amount of thermal energy to vaporize due to the fact that many contain nonpolar molecules. This property of volatile liquids was utilized in this experiment to determine the molar masses of unknown volatile liquids by employing the ideal gas law in the course of vaporization and condensation of the liquids. This was done by taking measurements of volume, temperature, mass, and pressure throughout various steps of the experiement. These findings were then applied to the ideal gas law and molar masses were obtained. The experimental data found from the calculations were similar to the theoretical values, showing that this experiment is practical in the laboratory setting.
INTRODUCTION:
Molar mass, which is …show more content…
At those given temperatures, the liquids started to receive a phase change from liquid to gas because there was enough thermal energy to overcome the attractive forces of the molecules. Additionally, when the test tubes were placed in the ice-cold beaker, the substances experienced another phase change: the change from a gas, or vapor, back into a liquid because enough thermal energy was absorbed by the ice water for the attractive forces to overcome the thermal energy.
In addition to these physical changes, it was also noticed that some of the substances took a longer period of time to vaporize than the others: Liquid A took the shortest time, while Liquid C took the longest time. Again, this is due to the amount of thermal energy required for vaporization to occur and overcome the attractive forces of the molecules were different and the different amount of substances heated, with more time representing more liquid and stronger attractive force.
Data Table Liquid A Liquid B Liquid …show more content…
Results
• Liquid A: 79.4 g/mol o Percent Error: 36.7%
• Liquid B: 49.5 g/mol o Percent Error: 7.4%
• Liquid C: 71.6 g/mol o Percent Error: 19.1%
Calculations
Mass of the Condensed Unknown
• (Mass of test tube, foil, and condensed gas) – (Mass of test tube and foil cover) o Liquid A: 0.08 g o Liquid B: 0.05 g o Liquid C: 0.07 g
Volume of the Test Tubes
• ((Mass of test tube, foil, and water) – (Mass of test and foil cover)) / (Density of water, 1.00 g/mL) o Test Tube A: 27.38 mL o Test Tube B: 27.42 mL o Test Tube C: 27.37 mL
Experimental Molar Mass
• PV=nRT, P= pressure, V= volume, n= number of moles or grams/(molar mass), R= ideal gas constant, T= temperature
• PV= (g/MM)RT [Plug in g/MM for n]
• (MM)PV=gRT [Multiply both sides by Molar Mass, MM]
• MM=gRT/PV [Divide both sides by PV] o Experimental MM A: 79.4 g/mol o Experimental MM B: 49.5 g/mol o Experimental MM C: 71.6
INTRODUCTION:
Molar mass, which is …show more content…
At those given temperatures, the liquids started to receive a phase change from liquid to gas because there was enough thermal energy to overcome the attractive forces of the molecules. Additionally, when the test tubes were placed in the ice-cold beaker, the substances experienced another phase change: the change from a gas, or vapor, back into a liquid because enough thermal energy was absorbed by the ice water for the attractive forces to overcome the thermal energy.
In addition to these physical changes, it was also noticed that some of the substances took a longer period of time to vaporize than the others: Liquid A took the shortest time, while Liquid C took the longest time. Again, this is due to the amount of thermal energy required for vaporization to occur and overcome the attractive forces of the molecules were different and the different amount of substances heated, with more time representing more liquid and stronger attractive force.
Data Table Liquid A Liquid B Liquid …show more content…
Results
• Liquid A: 79.4 g/mol o Percent Error: 36.7%
• Liquid B: 49.5 g/mol o Percent Error: 7.4%
• Liquid C: 71.6 g/mol o Percent Error: 19.1%
Calculations
Mass of the Condensed Unknown
• (Mass of test tube, foil, and condensed gas) – (Mass of test tube and foil cover) o Liquid A: 0.08 g o Liquid B: 0.05 g o Liquid C: 0.07 g
Volume of the Test Tubes
• ((Mass of test tube, foil, and water) – (Mass of test and foil cover)) / (Density of water, 1.00 g/mL) o Test Tube A: 27.38 mL o Test Tube B: 27.42 mL o Test Tube C: 27.37 mL
Experimental Molar Mass
• PV=nRT, P= pressure, V= volume, n= number of moles or grams/(molar mass), R= ideal gas constant, T= temperature
• PV= (g/MM)RT [Plug in g/MM for n]
• (MM)PV=gRT [Multiply both sides by Molar Mass, MM]
• MM=gRT/PV [Divide both sides by PV] o Experimental MM A: 79.4 g/mol o Experimental MM B: 49.5 g/mol o Experimental MM C: 71.6