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To use the Ideal Gas Law to calculate the molar mass

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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.

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.


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


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

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.

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See Also This Related BrainMass Solution

Calculations for decompositions of a solid using gas laws under STP. How to use molecular masses and the given equation to find out the solution.

What volume of carbon dioxide at STP is produced by the complete decomposition of 24 g of CaCO3--2.69 dm3, 4.64 dm3,5.37 dm3, 7.68 dm3

To solve the question we write down the equation of the reaction:

CaCO3 -----> CO2 + CaO

This equation is balanced out. The number of atoms of every element is the same on the left and right side.
What we can see is that 1 mole CaCO3 gives 1 mole CO2.

To find out the correct volume of CO2 we have to do the following steps:

1) Transforming the given mass [g] of CaCO3 into mole gives us the number of moles of CO2.

2) We need to calculate the molar mass of both CaCO3 and CO2. To do that we need the atomic mass of every atom of CaCO3 and CO2.

in a table of elements we find:

Ca : 40.078 u
C : 12.011 u
O : 15.999 u

1 mole of CaCO3 has the molar mass : 40.078 + 12.011 + (3 * 15.999) = 100.086 g

In the same way we get the molar mass of CO2 ( 40.009 g )

Now, we can see in case of CaCO3:

100.086 g = 1 mole
24 g = 100.086 / 24 = 0.239 mole

Based on our equation we can see that 1 mole CaCO3 gives 1 mole of CO2.
That means that 0.239 mole CaCO3 give 0.239 mole CO2

2) Using the relation ' 1 mole of a gaseous compound has a volume of 22.41 dm3 ' gives us the volume of the formed CO2 .

We find:

1 mole = 22.41 dm3
0.239 mole = 0.239 * 22.41 = 5.37 dm3

=> you have calculated the correct value :)
Thanks for using BrainMass.com help service

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