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Dumas Method A method to determine the molar mass of a compound. The Dumas method uses the ideal gas law to determine the molar mass of a compound which is a liquid that has a boiling point between room temperature and 100EC. This method is best illustrated with an example. An example of the Dumas method. A flask is weighed and is determined to be 75.4142 g. 10.0 mL of a liquid is poured into this flask. The opening of the flask is stoppered with a small vapor escape hole in the stopper. The flask is heated in a boiling water bath at 100.0EC. until all the liquid is vaporized and excess liquid exits. After this, the flask is cooled to room temperature and the weight is found to be 77.7976 g with the condensed liquid in it. The volume of the flask up to the bottom of the stopper was measured by filling with water. It was found to be 265.0 mL. The atmospheric pressure for the day was 756.5 torr. What is the molar mass of the liquid? An example of the Dumas method. First, letâ€™s draw a schematic of this experiment. An example of the Dumas method. A schematic of this experiment: weigh a flask empty: mMT An example of the Dumas method. A schematic of this experiment: weigh a flask empty: mMT pour an excess of a volite liquid An example of the Dumas method. A schematic of this experiment: weigh a flask empty: mMT pour an excess of a volite liquid heat the flask in a boiling water bath An example of the Dumas method. A schematic of this experiment: Air weigh a flask empty: mMT pour an excess of a volite liquid heat the flask in a boiling water bath weigh a flask with condensate: mfinal An example of the Dumas method. A flask is weighed and is determined to be 75.4142 g. mMT = 75.4142 g. An example of the Dumas method. 10.0 mL of a liquid is poured into this flask. The opening of the flask is stoppered with a small vapor escape hole in the stopper. mMT = 75.4142 g. As long as an excess of liquid was added, this piece of information is not relevant. An example of the Dumas method. The flask is heated in a boiling water bath at 100.0EC. until all the liquid is vaporized and excess liquid exits. mMT = 75.4142 g. T = 100.0EC of 373.2 K An example of the Dumas method. After this, the flask is cooled to room temperature and the weight is found to be 77.7976 g with the condensed liquid in it. mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g Air An example of the Dumas method. The volume of the flask up to the bottom of the stopper was measured by filling with water. It was found to be 265.0 mL. mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 265.0 mL After the last step the volume of the flask is determined by pouring water into it and measuring how much is needed to fill up to the stopper. An example of the Dumas method. It was found to be 265.0 mL. The atmospheric pressure for the day was 756.5 torr. mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 265.0 mL P = 756.5 torr The pressure is normally measured with a barometer that is in the lab. This is the same pressure as the pressure inside the vapor filled flask at 100.0EC. An example of the Dumas method. What is the molar mass of the liquid? mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 265.0 mL P = 756.5 torr M=? An example of the Dumas method. Solution: First calculate using PV = nRT the number of moles of the vapor in the flask at 100.0EC. R = 0.08206 L atm K-1 mol-1 mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 265.0 mL P = 756.5 torr M=? First convert pressure to atm: 1 atm P = 756.5 torr × 760 torr An example of the Dumas method. Solution: First calculate using PV = nRT the number of moles of the vapor in the flask at 100.0EC. R = 0.08206 L atm K-1 mol-1 mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 265.0 mL P = 0.9954 atm M=? An example of the Dumas method. Solution: First calculate using PV = nRT the number of moles of the vapor in the flask at 100.0EC. R = 0.08206 L atm K-1 mol-1 mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 265.0 mL P = 0.9954 atm M=? also converting 265.0 mL to liters: V = 265.0 mL / 0.2650 L An example of the Dumas method. Solution: First calculate using PV = nRT the number of moles of the vapor in the flask at 100.0EC. R = 0.08206 L atm K-1 mol-1 mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 0.2650 L P = 0.9954 atm M=? and substitute into PV = nRT: (0.9954 atm)(0.2650 L) = n(0.08206 L atm K-1 mol-1)(373.2 K) An example of the Dumas method. Solution: First calculate using PV = nRT the number of moles of the vapor in the flask at 100.0EC. R = 0.08206 L atm K-1 mol-1 mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 0.2650 L P = 0.9954 atm M=? So, solving for n: n = 8.613 × 10-3 mol An example of the Dumas method. Solution: In order to substitute into the equation for the molar mass, that is m = Mn, the mass of the vapor is required. mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 0.2650 L n = 8.613 × 10-3 mol P = 0.9954 atm M=? The mass of the vapor is given as the difference between the starting mass and final mass: mvapor = mfinal - mMT. An example of the Dumas method. Solution: In order to substitute into the equation for the molar mass, that is m = Mn, the mass of the vapor is required. mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 0.2650 L n = 8.613 × 10-3 mol P = 0.9954 atm M=? Substituting: mvapor = 77.7976 g - 75.4142 g An example of the Dumas method. Solution: In order to substitute into the equation for the molar mass, that is m = Mn, the mass of the vapor is required. mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 0.2650 L n = 8.613 × 10-3 mol P = 0.9954 atm M=? or: mvapor = 2.3834 g An example of the Dumas method. Solution: In order to substitute into the equation for the molar mass, that is m = Mn, the mass of the vapor is required. mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 0.2650 L n = 8.613 × 10-3 mol P = 0.9954 atm mvapor = 2.3834 g M=? Substituting into m = Mn: (2.3834 g) = M(8.613 × 10-3 mol) An example of the Dumas method. Solution: In order to substitute into the equation for the molar mass, that is m = Mn, the mass of the vapor is required. mMT = 75.4142 g. T = 373.2 K mfinal = 77.7976 g V = 0.2650 L n = 8.613 × 10-3 mol P = 0.9954 atm mvapor = 2.3834 g M=? Solving for M: M = 276.7 g mol-1 The End