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.Order now
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 André 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: The compound vapor behaves ideally.
Either the volume of the vessel does not vary significantly between room and the working temperature, or the volume of the vessel may be accurately determined at the working temperature.
Significance In the experiment, mass and the number of moles of a particular vapor will be known which will help to identify the molecular weight of the substance. Design and Methods Reagents Benzene Unknown Liquid Distilled water Apparatus Erlenmeyer flasks Analytical balance Thermometer Beaker Clamp Iron stand Iron ring Wire gauze Bunsen burner Stopper Procedure The following procedure was done separately on benzene and the unknown liquid. A. Preparation of the Apparatus Both Erlenmeyer flasks were cleaned and were weighed using the other as a tare to minimize weighing errors due to the change in temperature and moisture forming on the surface of the flasks which will occur on both flasks, so that in weighing, the tare simply balances that of the weighed flasks. Ten mL of the test liquid was placed in one of the flask, and the flask was immersed in a beaker of water so that only the neck is above the surface. It was secured with a clamp and was tilted slightly so that the round bottom is the lowest part, and the liquid in the flask is clearly seen. B. Volatilization of the Liquid
The beaker of water was heated, keeping the flame of the burner constant. During the heating process, the flask was not covered with a stopper to avoid possible explosion and to allow the vapor to have similar pressure with the atmosphere. When the liquid started to vaporize, the temperature of the vapor was measured to identify the boiling point of the liquid. As soon as the liquid is gone, the flask was covered with a flask to ensure that the vapor was not replaced with air. The flask was cooled at room temperature, and its external part was dried. It was weighed accurately using the same tare as before. C. Determination of the Volume of the Vapor The volume of the vapor is the same as the volume of the flask. To determine this, the flask was filled with water and was stoppered, making sure that there was no bubbles present inside. The water was then decanted into the graduated cylinder, to measure the volume of the water which is equal to the volume of the flask and to the volume of the vapor. D. Computations The corrected pressure can be computed using the formula: Pcorr = Pb – Pb 1 + At Where A = 18.8 x 10-5 B = 1.84 x 10-5 Pb = barometric pressure t = room temperature (oC) ts = standard temperature (20 oC) The weight of the air is determined using the ideal gas equation: M = PVMW RT The weight of the vapor can be calculated using: Mvapor = (Mflask+stopper+vapor) – (Mflask+stopper+air) + Mair To calculate the MW of benzene, use the Berthelot’s equation: PV = nRT 128PcT T2 To calculate for the MW of the unknown liquid, use ideal gas equation (1.2). (1.4) (1.3) (1.2) (1.1)
To calculate for the percent error, use the equation: % error = I MWaccepted – MWcomputed I x 100% MWaccepted Set-up (1.5)
Figure No. 1 Volatilization of the Liquid III. Results and Interpretation of Data Table No. 1 Data and Results Benzene Troom Patm Pcorr Mflask+stopper+air Tvapor Mflask+stopper+vapor Vflask Mair Mvapor MW % error 77.2679 g 137 mL 0.0110609004 g 0.4141609004 84.9687 g/mol 8.9342 % 77.1055 g 137.5 mL 0.1305756683 g 0.4237756683 g 94.1180 g/mol 24.89 oC 24.94 inHg 24.82607485 inHg 76.8648 g Unknown Liquid 25.667 oC 25.14 inHg 24.02188244 inHg 76.8123 g
Based on the table above, the computed molecular weight of benzene is 84.9687 which has 8.9342 % marginal error from the accepted molecular weight of benzene which is 78. The computed molar mass of the unknown liquid based on ideal gas equation is 94.1180. Basing on the results of the experiment with benzene, 8 % error was made. This may result from errors on the weighing processes as well as the volatilization process. There is a possibility that air could have replaced some of the vaporized gas inside the flask during the heating process and during the cooling process, and this could result to errors on the mass of the actual vapor inside the flask. Thus, it had created error on the computed value of the molecular weight of benzene. Similar to the process of determining the molecular weight of benzene, there could also be possible errors in the computed molecular weight of the unknown liquid. If 10% should be accounted on this error, then the molecular weight of the unknown liquid could be possible to be 94 ± 10%. It could be ranging from 84.6 to 103.4. The unknown volatile liquid could possibly be hexane. Hexane has a molecular weight of 86 which is found in the range of probable molecular weight of the unknown liquid. Another source of correlation would be their boiling point. According to the Table 2-2 Physical Properties of Organic Compounds found in page 2-39 of the Perry’s Handbook, the boiling point of hexane is 69 oC which is also the measured boiling point of the unknown liquid. Organic compounds other than hexane which has molecular weight ranging from 84.6 – 103.4 have actually far higher and lower boiling point as compared to the measured boiling point of the unknown liquid. IV Conclusion and Recommendation Based on the interpretation of the data and results of the experiment, the following conclusions were drawn: The possible molecular weight of the unknown liquid ranges from 84.6 – 103.4. The unknown liquid has high probability to be hexane. Errors in the determination of the molecular weight of a particular volatile liquid could be due to 1. Inaccuracies in the weighing process 2. Possible replacement of the vapor with air 3. Inaccurate timing on stoppering the flask
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 + 18.8×10-5 (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 + 18.8×10-5 (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