Determining the Molar Mass of a Volatile Liquid
Determining the Molar Mass of a Volatile Liquid
Purpose: The purpose of this lab was to find the molar mass of a volatile liquid.
Data Table: Mass of Test Tube and Stopper (g) | 10.864 g | Barometric Pressure (mmHg) | 749.31 mmHg | Temperature of Boiling Water (C) | 97.1 C | Mass of Test Tube, Stopper, and Condensed Liquid (g) | 10.890 g | Volume of Flask (mL) | 9.90 mL |
Calculations:
749.31 mmHG*1 atm760 mmHg= .98593 atm
9.90 mL*1 L1000 mL= .00990 L
97.1 C+273=370.1 K
Substituting these values into the equation:
.98593 atm* .00990 L=n* .0821 atm L mol-1 K-1*370.1 K n= .000321 moles
Obtain the molar mass by dividing the moles of the liquid by the grams of the liquid:
.026 g.000321 mol=81.0gmol
Percent Error:
|81.0-46.07|46.07= .758*100%=75.8 %
There are many reasons why the percent error is so high. One reason could be that the liquid wasn’t all evaporated. This would lower the amount of condensated liquid. If the amount of condensated liquid was lowered, the mass would be lower and the molar mass would be smaller. This was unlikely though, since we had a higher molar mass than the accepted. The volume could have also been measured incorrectly. If the volume was lower than what is supposed to be, the molar mass would be higher than the accepted value. There is also a chance that the temperature was different than the value we recorded. If the tempereature was higher than what we recorded, the molar mass would be bigger as well. Like this, there are a lot of mistakes that could have been made which would skew the results of this
Purpose: The purpose of this lab was to find the molar mass of a volatile liquid.
Data Table: Mass of Test Tube and Stopper (g) | 10.864 g | Barometric Pressure (mmHg) | 749.31 mmHg | Temperature of Boiling Water (C) | 97.1 C | Mass of Test Tube, Stopper, and Condensed Liquid (g) | 10.890 g | Volume of Flask (mL) | 9.90 mL |
Calculations:
749.31 mmHG*1 atm760 mmHg= .98593 atm
9.90 mL*1 L1000 mL= .00990 L
97.1 C+273=370.1 K
Substituting these values into the equation:
.98593 atm* .00990 L=n* .0821 atm L mol-1 K-1*370.1 K n= .000321 moles
Obtain the molar mass by dividing the moles of the liquid by the grams of the liquid:
.026 g.000321 mol=81.0gmol
Percent Error:
|81.0-46.07|46.07= .758*100%=75.8 %
There are many reasons why the percent error is so high. One reason could be that the liquid wasn’t all evaporated. This would lower the amount of condensated liquid. If the amount of condensated liquid was lowered, the mass would be lower and the molar mass would be smaller. This was unlikely though, since we had a higher molar mass than the accepted. The volume could have also been measured incorrectly. If the volume was lower than what is supposed to be, the molar mass would be higher than the accepted value. There is also a chance that the temperature was different than the value we recorded. If the tempereature was higher than what we recorded, the molar mass would be bigger as well. Like this, there are a lot of mistakes that could have been made which would skew the results of this