Dumas Method. A method to determine the molar mass of a compound.

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Derick Craig
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1 Dumas Method A method to determine the molar mass of a compound.

2 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.

3 A flask is weighed and is determined to be g 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 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 ml. The atmospheric pressure for the day was torr. What is the molar mass of the liquid?

4 First, let s draw a schematic of this experiment.

5 A schematic of this experiment: weigh a flask empty: m MT

6 A schematic of this experiment: weigh a flask empty: m MT pour an excess of a volite liquid

7 A schematic of this experiment: weigh a flask empty: m MT pour an excess of a volite liquid heat the flask in a boiling water bath

8 A schematic of this experiment: Air weigh a flask empty: m MT pour an excess of a volite liquid heat the flask in a boiling water bath weigh a flask with condensate: m final

9 A flask is weighed and is determined to be g.

10 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. As long as an excess of liquid was added, this piece of information is not relevant.

11 The flask is heated in a boiling water bath at 100.0EC. until all the liquid is vaporized and excess liquid exits. T = 100.0EC of K

12 After this, the flask is cooled to room temperature and the weight is found to be g with the condensed liquid in it. Air

13 The volume of the flask up to the bottom of the stopper was measured by filling with water. It was found to be ml. V = 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.

14 It was found to be ml. The atmospheric pressure for the day was torr. V = ml P = 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.

15 What is the molar mass of the liquid? V = ml P = torr M =?

16 Solution: First calculate using PV = nrt the number of moles of the vapor in the flask at 100.0EC. R = L atm K -1 mol -1 V = ml P = torr M =? First convert pressure to atm: P = torr 1 atm 760 torr

17 Solution: First calculate using PV = nrt the number of moles of the vapor in the flask at 100.0EC. R = L atm K -1 mol -1 V = ml P = atm M =?

18 Solution: First calculate using PV = nrt the number of moles of the vapor in the flask at 100.0EC. R = L atm K -1 mol -1 V = ml P = atm M =? also converting ml to liters: V = ml / L

19 Solution: First calculate using PV = nrt the number of moles of the vapor in the flask at 100.0EC. R = L atm K -1 mol -1 V = L P = atm M =? and substitute into PV = nrt: ( atm)( L) = n( L atm K -1 mol -1 )(373.2 K)

20 Solution: First calculate using PV = nrt the number of moles of the vapor in the flask at 100.0EC. R = L atm K -1 mol -1 V = L P = atm M =? So, solving for n: n = mol

21 Solution: In order to substitute into the equation for the molar mass, that is m = Mn, the mass of the vapor is required. V = L n = mol P = atm M =? The mass of the vapor is given as the difference between the starting mass and final mass: m vapor = m final - m MT.

22 Solution: In order to substitute into the equation for the molar mass, that is m = Mn, the mass of the vapor is required. V = L n = mol P = atm M =? Substituting: m vapor = g g

23 Solution: In order to substitute into the equation for the molar mass, that is m = Mn, the mass of the vapor is required. V = L n = mol P = atm M =? or: m vapor = g

24 Solution: In order to substitute into the equation for the molar mass, that is m = Mn, the mass of the vapor is required. V = L n = mol P = atm m vapor = g M =? Substituting into m = Mn: ( g) = M( mol)

25 Solution: In order to substitute into the equation for the molar mass, that is m = Mn, the mass of the vapor is required. V = L n = mol P = atm m vapor = g M =? Solving for M: M = g mol -1

26 The End

**VII-1 C NC I-3 C NC II-2 C NC *VII-2 C NC I-4 C NC. CHEMISTRY 131 Quiz 5 Fall 2010 Form B

**VII-1 C NC I-3 C NC II-2 C NC *VII-2 C NC I-4 C NC. CHEMISTRY 131 Quiz 5 Fall 2010 Form B **VII-1 N I-3 N II-2 N *VII-2 N I-4 N EMISTRY 131 Quiz 5 Fall 2010 Form B NAME: Key hapter 11: States of Matter A. (2 pts) onsider the structures of the compounds shown below, and determine which intermolecular