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, let’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