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Determining the Molar Mass of Volatile Liquid

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Introduction

Determining the Molar Mass of Volatile Liquid Aim: In this experiment the molar mass of an unknown volatile hydrocarbon liquid was determined by evaporation of the liquid and condensation of a defined volume of its vapor. Background: The molar mass of the volatile liquid was calculated from the number of moles and the mass of the sample. In order to get the number of mole of the volatile liquid, the ideal gas law was used. The moles of an ideal gas can be calculated with the ideal gas law: PV = nRT R represents the gas constant (8.314 J mol-1 K-1). n= PV / RT With this formula, we can determine the moles of the gas in a given volume of the glass ware (V). The molar mass can then be calculated with the moles of the gas and the molar mass formula. M = m / n Variables: Controlled Volume of volatile liquid Pressure of the room Room temperature Independent Temperature of hot water bath Dependent Mass of the condensed vapor of the volatile liquid Method for controlling variables: * The volume of the Erlenmeyer flasks used for the trials was determined by filling it with water to the overflow and measuring the volume of the water in a 250 cm3 graduated cylinder. ...read more.

Middle

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 Data Collection and Processing Raw Data: Run 1 Run 2 Run 3 Run 4 Run 5 Mass of empty flask /g � 0.001 g 97.156 95.108 99.452 96.887 97.987 Mass of flask with condensed vapor /g � 0.001 g 98.083 96.064 100.374 97.815 98.981 Volume of flask / cm3 � 1.0 cm3 338 340 324 334 340 Temperature of water �C ?0.1�C 89.9 88.0 89.5 90.5 87.5 Room pressure /mbar �1.0 mbar 1015 1015 1015 1015 1015 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: ...read more.

Conclusion

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 http://www.translatorscafe.com/cafe/units-converter/pressure/calculator/millibar-%5Bmbar%5D-to-newton-per-square-meter-%5BN/m%5E2%5D/ ?? ?? ?? ?? Stephany NISHIKAWA Date: October 21, 2008 IB Chemistry SL Lab 4 GESM- International Baccalaureate Page 1 of 3 ...read more.

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