# To use the Ideal Gas Law to calculate the molar mass

Please take a look at the attached lab report on vapor density. Please see the ** items regarding percentage error and calculation of R. I can't figure out how to do the percentage error of my trials and the calculations for the R values are confusing. Thanks for your help.

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Vapor Density and the Ideal Gas Law
Lab Conducted: October 10, 2005

Purpose: To use the Ideal Gas Law (PV=nRT) to calculate the molar mass (M) of an unknown real vapor and to determine its identity as either Methanol, Ethanol, Acetone, Pentane or Cyclohexane.

Procedure: Boiled 0.5L of water and inserted a flask of an unknown liquid which was covered with a foil cap that had a small hole in it. The unknown substance was heated in three separate trials to determine the volume of gas that evaporated through the hole in the foil cap.

Calculations:

Pressure (P) = 753 mmHg/760 mmHg = 0.990789 atm
Temperature (T) = 99.74&#730;Celcius + 273.15 = 373.89&#730;Kelvin
Constant R = 0.082057

Trial #1
Volume of flask = 262 mL = 0.262 liters

Mass of flask before heating = 123.610 grams
Mass of flask after heating = 124.590 grams
Difference = mass of substance = 0.980 grams

n = PV = (0.990789)(0.262)_ = 0.0084610 moles
RT (0.082057)(373.89)

Molar Mass = 0.980 grams/.0084610 = 115.826 grams/mole

Trial #2
Volume of flask = 298 mL = 0.298 liters

Mass of flask before heating = 113.487 grams
Mass of flask after heating = 114.344 grams
Difference = mass of substance = 0.857 grams

n = PV = (0.990789)(0.298)_ = 0.0096236 moles
RT (0.082057)(373.89)

Molar Mass = 0.857 grams/.0096236 = 89.052 grams/mole

Trial #3
Volume of flask = 262 mL = 0.262 liters

Mass of flask before heating = 123.610 grams
Mass of flask after heating = 124.410 grams
Difference = mass of substance = 0.800 grams

n = PV = (0.990789)(0.262)_ = 0.008461 moles
RT (0.082057)(373.89)

Molar Mass = 0.800 grams/.008461 = 94.5515 grams/mole

Molar Masses of Possible Substances
Methanol CH3OH 32 g/mol
Ethenol C2H5OH 36 g/mol
Acetone CH3CH2CO 57 g/mol
Pentane C5H12 72 g/mol
Cyclohexane C6H12 84 g/mol

Conclusion/Discussion

It is likely that the unknown substance is Cyclohexane because the average of molar mass from the three trials is 99.8 g/mol [(94.5515+89.052+115.826)/3] and the largest molar mass of the possible substances is 84 g/mol.

**Percentage Error of molar mass calculations:

**Maximum Value, Minimum Value and Average value of R with uncertainty range to the average value

R = MP/T
m/V

Assume uncertainties are as follows: Pressure = +/- 0.1 cmHg, Temperature = +/-0.3&#730;C, Volume = +/-0.0004 L, mass = +/- 0.001

Lab Book Questions

1. Weighing the flask with the foil cap makes the measurement more accurate as the foil cap has mass.

2. It will create error in the calculation of the mass of the substance.

3. The temperature value from the table is for pure water. We actually used tap water in the experiment which will change the boiling point of water because of other substances in the non-pure water.

4a. 1.0261 g

4b. 250 mL

4c. n = PV = (0.99408)(0.250)_ = 0.00813 moles
RT (0.082057)(372.59)

Molar Mass = 1.0261 grams/.00813 = 126.2 grams/mole

5. I believe the signs would be opposite which would cancel the error arising from the neglect of the vapor.
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• Vapor Density.doc

Vapor Density and the Ideal Gas Law
Lab Conducted: October 10, 2005

Purpose: To use the Ideal Gas Law (PV=nRT) to calculate the molar mass (M) of an
unknown real vapor and to determine its identity as either Methanol, Ethanol, Acetone,
Pentane or Cyclohexane.

Procedure: Boiled 0.5L of water and inserted a flask of an unknown liquid which was
covered with a foil cap that had a small hole in it. The unknown substance was heated in
three separate trials to determine the volume of gas that evaporated through the hole in
the foil cap.

Calculations:

Pressure (P) = 753 mmHg/760 mmHg = 0.990789 atm
Temperature (T) = 99.74˚Celcius + 273.15 = 373.89˚Kelvin
Constant R = 0.082057

Trial #1
Volume of flask = 262 mL = 0.262 liters

Mass of flask before heating = 123.610 grams
Mass of flask after heating = 124.590 grams
Difference = mass of substance = 0.980 grams

n = PV = (0.990789)(0.262)_ = 0.0084610 moles
RT (0.082057)(373.89)

Molar Mass = 0.980 grams/.0084610 = 115.826 grams/mole

Trial #2
Volume of flask = 298 mL = 0.298 liters

Mass of flask before heating = 113.487 grams
Mass of flask after heating = 114.344 grams
Difference = mass of substance = 0.857 grams

n = PV = (0.990789)(0.298)_ = 0.0096236 moles
RT (0.082057)(373.89)

Molar Mass = 0.857 grams/.0096236 = 89.052 grams/mole
Trial #3
Volume of flask = 262 mL = 0.262 liters

Mass of flask before heating = 123.610 grams
Mass of flask after heating = 124.410 grams
Difference = mass of substance = 0.800 grams

n = PV = (0.990789)(0.262)_ = 0.008461 moles
RT (0.082057)(373.89)

Molar Mass = 0.800 grams/.008461 = 94.5515 grams/mole

Molar Masses of Possible Substances
Methanol CH3OH 32 g/mol
Ethenol C2H5OH 36 g/mol
Acetone CH3CH2CO 57 g/mol
Pentane C5H12 72 g/mol
Cyclohexane C6H12 84 g/mol

Conclusion/Discussion

It is likely that the unknown substance is Cyclohexane because the average of molar mass
from the three trials is 99.8 g/mol [(94.5515+89.052+115.826)/3] and the largest molar
mass of the possible substances is 84 g/mol.

**Percentage Error of molar mass calculations:

**Maximum Value, Minimum Value and Average value of R with uncertainty range
to the average value

R = MP/T
m/V

Assume uncertainties are as follows: Pressure = +/- 0.1 cmHg, Temperature = +/-0.3˚C,
Volume = +/-0.0004 L, mass = +/- 0.001
Lab Book Questions

1. Weighing the flask with the foil cap makes the measurement more accurate as the
foil cap has mass.

2. It will create error in the calculation of the mass of the substance.

3. The temperature value from the table is for pure water. We actually used tap
water in the experiment which will change the boiling point of water because of
other substances in the non-pure water.

4a. 1.0261 g

4b. 250 mL

4c. n = PV = (0.99408)(0.250)_ = 0.00813 moles
RT (0.082057)(372.59)

Molar Mass = 1.0261 grams/.00813 = 126.2 grams/mole

5. I believe the signs would be opposite which would cancel the error arising from
the neglect of the vapor.