Dumas Method Molecular Weight Determination Example For Students

Dumas Method: Molecular Weight Determination 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: ï‚ · 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. ï‚ · ï‚ · Research Questions ï‚ · ï‚ · ï‚ · 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. The Dumas method is appropriate to determine the molecular weights of volatile organic substances that are liquids at room temperature. In the early 19th century, Jean-Baptiste Andrà © Dumas, a distinguished French chemist, created a relatively simple method for determination of the molecular weight of a substance. With this method, molecular weight is calculated by measuring the mass of a known volume of a vaporized liquid. Because the concept of the mole had not been developed in Dumas’ era, he computed relative molecular weights based on relative gas densities. Though Dumas got mixed results based on erroneous assumptions concerning elements in the gas phase, he is credited with establishing values for the molecular weights of thirty elements. In the modern version of the Dumas procedure, an Erlenmeyer flask is used rather than the glass bulb of Dumas’ day. The temperature, pressure and volume of the vapor are determined and the molar mass is found utilizing the Ideal Gas Law. The procedure entailed placing a small quantity of the unknown substance into a tared vessel of known volume. The vessel is then heated in a boiling water bath; all the air within the flask would be expelled, replaced by the vapor of the unknown substance. When no remaining liquid can be observed, the vessel may be sealed (e.g. with a flame), dried, and weighed. By subtracting the tare of the vessel, the actual mass of the unknown vapor within the vessel can be calculated. Assuming the unknown compound obeyed the ideal gas equation, the number of moles of the unknown compound, n, can be determined by PV = nRT where the pressure, P, is the atmospheric pressure, V is the measured volume of the vessel, T is the absolute temperature of the boiling water bath, and R is the ideal gas constant. By dividing the mass in grams of the vapor within the vessel by the calculated number of moles, the molecular weight may be obtained. Two major assumptions are used in this method: ï‚ · The compound vapor behaves ideally. Essay About MyselfV 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 1 + 18.810-5 (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) 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 1 + 18.810-5 (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