Determining the Molar Mass of Volatile Liquid

Material:

Equipment:

• 1 x 600 cm3 beaker

• 1 x 250 cm3 Graduated cylinder (2 cm3 graduation ± 1.0 cm3)
• 1 x 10 cm3 Graduated Cylinder (1 graduation ±0.2 cm3)
• 2 x 250 cm3 Erlenmeyer flask

• 1 x Support stand
• 2 x Clamps and clamp holder
• 1 x Electronic balance (300 g, readability ±0.001 g)
• 1 x Precision Digital Barometer(range: 0~1100mbar abs, ±1 mbar)
• 1 x Precision-Digital Thermometer (range: -70°C~ +199.9  ±0.1°C)  
• 1x 5 cm3 Pipette ± 0.015 cm3
• 1x Cartridge Burner
• Tripod with wire gauze
• Boiling stones
• Aluminium foil
• Ice water bath

Chemicals:

• Water
• Volatile Liquid

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Procedure

1. The mass of a 250 cm3 Erlenmeyer flask together with a boiling stone and an aluminum foil cap was weighed and recorded accurately on the electronic balance to the nearest 0.001g
2. Small pinhole in the foil cap was made with a pin
3. About 5 cm3 of the volatile liquid was measured with a pipette and was added into the Erlenmeyer flask and then closed with the aluminum foil.
4. The Erlenmeyer flask was placed in the water bath of about 98°C in order to vaporize the liquid.
5. The water bath containing the Erlenmeyer with volatile liquid was heated gently until the liquid was vaporized at around 90°C for 3 minutes.
6. The water bath temperature and the atmospheric pressure were recorded.
7. The vapor in the Erlenmeyer was condensed by immersing in an ice bath
8. The Erlenmeyer flask was dried completely
9. The mass of the Erlenmeyer flask with the boiling stone, aluminum foil cap and condensed vapor was weighed and recorded accurately on the electronic balance to the nearest 0.001g
10. The room barometric pressure was recorded with Precision Digital Barometer
11. The foil was removed and the Erlenmeyer flask was rinse with water. Then it was fully filled with water. The volume of the flask was determined by pouring the water in the flask into a 250 cm3 graduated cylinder.
12. Procedure 1 to 11 were repeated 4 times

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Data Collection and Processing

Raw Data:

Observations: 

The volatile liquid is transparent in the beginning. When heated, the volatile gas evaporates to gas. When condensed, the volatile gas turns into liquid and is transparent.

Data processing:

Sample calculation for trial 1:

Conversion of mbar into Pascal:

Atmospheric pressure: P= 1015 mbar = 101500 Pa = 1.015 x 105 Pa  

 

Conversion of cm3 to m3 :

V= 338 cm3 = 3.38 x 10–4 m3

Conversion of Celsius into Kelvin :

T= 89.9 °C + 273.15 = 363.2 K

Calculation of the molar mass:

The ideal gas law:

Pv=nRT

n=

R= 8.31 J mol–1 K–1 

n = PV / RT = (1.015 x 105 x 3.38 x 10–4) / (8.31 x 363.2) = 1.14 x 10–2 mol

Uncertainty:

Volume: 1.0/338 x 100 = 0.30 %

Pressure: 1.0/1015 x 100 = 0.10%

Temperature: 0.1 / 89.9 x 100= 0.1 %

Total uncertainty: 0.30% + 0.10% +0.1% = 0.50%

Amount of volatile liquid = 1.14 x 10-2 ± 0.50 % moles

Calculation of the mass of the condensed gas:

Mass of condensed gas= mass of flask with condensed vapor- mass of empty flask

Mass of condensed gas = 98.083 g ±0.001g – 97.156 g ±0.001g

Mass of condensed gas = 0.927 g ±0.22 %

Mass of condensed gas = 0.927 ±0.22 %

Calculation of the molar mass of the volatile liquid:

M=m/n

Molar mass of volatile liquid = Mass of condensed gas / amount of volatile gas

M= 0.927/(1.14 x10-2)

M= 81.3 g mol–1

Uncertainty:

Mass of condensed gas = ±0.22 %

Amount of volatile liquid = 0.5 % moles

Total uncertainty:  ±0.22 % + 0.5% = 0.502% 0.72 %

Molar mass of volatile liquid = 81.3 g mol–1 ±0.72 %

Similar calculations have been carried out for Run 2, 3, 4 and 5.

Calculation of the average molar mass of volatile liquid:

Molar mass of volatile liquid = (81.3 + 83.1 + 84.6+ 82.9+ 86.4)/5

Molar mass of volatile liquid = 83.7 g mol–1 ± 0.73%

Conclusion:

        The purpose of this experiment was to determine the molar mass of a volatile liquid by evaporation of the liquid and condensation of a defined volume of its vapor.

The average experimental which is the determined molar mass is 83.7 g mol–1 ± 0.73%. We know that the volatile liquid was a hydrocarbon.

The nearest molar mass of a hydrocarbon to the experimental molar mass of the volatile liquid is hexane with 86.18 g mol–1. Compared to hexane and the experimental value, the %error would be:

%Error = (86.18 – 83.7) / 86.18 x 100 = 2.9 %

Therefore, the experiment was fairly accurate because the percentage error is only 2.9%.

Evaluation:

        To further improve the result, we can lessen the pin holes made because the air might have entered inside the Erlenmeyer flask containing volatile liquid. The use of the aluminum foil was to prevent the evaporation of the volatile liquid. Therefore, the volume of the gas collected might have been affected. When we weight the flask with condensed vapor, the total pressure in the flask is the combined pressure of the air and the vapor pressure of the volatile liquid. In addition, we can use more accurate gas syringe because its calibration has large intervals; thus less accurate.

Reference