Dumas Method: Molecular Weight Determination
PHYSICAL CHEMISTRY LABORATORY REPORT (CHE 414L) Department of Chemical Engineering School of Engineering and Architecture Saint Louis University Group No. 12 Members: DEMOT, Judy Ann SOMERA, Randolf P. TIPAYNO, Samantha TORZAR, Precious Imee I. Signatures: ____________________ ____________________ ____________________ ____________________
Title: Determination of Molecular Weight of a Volatile Liquid By Vaporization: Dumas Method Background of the Study Problem In this experiment, an unknown liquid is in need to be identified and one of the key factors to identify it is to determine its molecular weight. In order to determine the molecular weight of a particular substance which in this experiment’s case is a volatile liquid, the need to convert the liquid into a gas arises. The relation among the pressure, volume, temperature and the number of moles of a gas will be an important key in the conclusion of its molecular weight. However, in the process of vaporization, there is a possibility that some of the vaporized liquid will escape the flask and will be replaced with air. If this happens, it will incorporate error in the measurement of the mass of the vaporized liquid, thus contributing error in the computations for its molecular weight. Objectives The experiment aims to: Determine the molecular weight of benzene and an unknown liquid by applying the simple variation of the Dumas Method which is an appropriate process in the determination of organic volatile substances that are liquid at room temperature. Utilize the Ideal Gas Law and Berthelot’s equation in connection with the experiment Get a hint in the identity of the unknown volatile liquid.
II.
Research Questions
What is the molecular weight of Benzene? What is the molar mass of the unknown liquid? What is the unknown liquid?
Conceptual Framework The Dumas method of molecular weight determination was historically a procedure used to determine the molecular weight of an unknown substance.
Title: Determination of Molecular Weight of a Volatile Liquid By Vaporization: Dumas Method Background of the Study Problem In this experiment, an unknown liquid is in need to be identified and one of the key factors to identify it is to determine its molecular weight. In order to determine the molecular weight of a particular substance which in this experiment’s case is a volatile liquid, the need to convert the liquid into a gas arises. The relation among the pressure, volume, temperature and the number of moles of a gas will be an important key in the conclusion of its molecular weight. However, in the process of vaporization, there is a possibility that some of the vaporized liquid will escape the flask and will be replaced with air. If this happens, it will incorporate error in the measurement of the mass of the vaporized liquid, thus contributing error in the computations for its molecular weight. Objectives The experiment aims to: Determine the molecular weight of benzene and an unknown liquid by applying the simple variation of the Dumas Method which is an appropriate process in the determination of organic volatile substances that are liquid at room temperature. Utilize the Ideal Gas Law and Berthelot’s equation in connection with the experiment Get a hint in the identity of the unknown volatile liquid.
II.
Research Questions
What is the molecular weight of Benzene? What is the molar mass of the unknown liquid? What is the unknown liquid?
Conceptual Framework The Dumas method of molecular weight determination was historically a procedure used to determine the molecular weight of an unknown substance.
References: Grider, Douglas J., Tobiason, Joseph D., Tobiason, Fred L. (1988). Molecular Weight Determination by an Improved Temperature Monitored Vapor Density Method. Journal of Chemical Education, 65 (7), 641. Kaya, Julie J., Campbell, J. Arthur (1967). Molecular Weights from Dumas Bulb Experiments. Journal of Chemical Education, 44 (7), 394. APPENDIX A Computations of Data and Results A. Benzene Pcorr = 24.94 inHg – 24.94 inHg [18.8x10-5 (24.89) – 1.84x10-5 (24.89 – 20)] 1 + 18.8x10-5 (24.89) Pcorr = 24.82607485 inHg Mair = (24.82607485 inHg) (1 atm / 29.92 inHg) (0.137 L) (29g/mol) (0.08205 L-atm/K-mol) (298.04 K) Mair = 0.0110609004 g Mvapor = 77.2679 g – 76.8648 g + 0.0110609004 g Mvapor = 0.4141609004 g MW = (0.4141609004g)(0.08205)(298.04K)[ 1+ 9(0.829748491)(562.05) (1 – 6(562.052)] 128(48.30989391)(298.04) (298.04 )2 MWbenzene = 84.9687g/mol % error = I 78 – 84.9687I x 100% 78 % error = 8.9342 % B. Unknown Liquid Pcorr = 25.14 inHg – 25.14 inHg [18.8x10-5 (25.667) – 1.84x10-5 (25.667 – 20)] 1 + 18.8x10-5 (25.667) Pcorr = 24.02188244 inHg Mair = (24.02188244 inHg) (1 atm / 29.92 inHg) (0.1375 L) (29g/mol) (0.08205 L-atm/K-mol) (298.817 K) Mair = 0.1305756683 g Mvapor = 77.2679 g – 76.8648 g + 0.1305756683 g Mvapor = 0.4237756683 g MWunknown liquid = (0.4237756683 g)(0.08205 L-atm/K-mol)(298.817 K) (24.02188244 inHg)(1 atm/ 29.92 inHg)(0.1375 L) MWunknown liquid = 94.1180 g/mol