WagenborgMWVOLGASLab Page 1
Name:Bill Wagenborg
MISE - Physical Basis of Chemistry
Third Experiment
Molecular Weight of a Volatile Liquid
Lab Report – Submit electronically (digital drop box) by Sunday, November 18, 6:01PM
Note: When submitting to digital Drop Box label you files with your name first and then a
brief description of what the document is.
Purpose: The Purpose of this experiment is to determine the molecular weight of a compound
that is a volatile liquid.
DATA/CALCULATIONS
Note: Data and Calculations may be completed on an Excel spreadsheet.
Unknown Letter: ____C_________
Molecular Weight of a Volatile Liquid
(If you performed more than one trial, adjust your data table accordingly.)
1. Determination of mass of vapor:
Weight of flask + foil cap + rubber band
Trial 1
81.3358 g
Temperature of water bath when (excess)
vapor ceased to escape
Weight of flask + foil cap + rubber band +
condensed vapor
Weight of condensed vapor
89 0 C
2. Determination of full volume of flask:
Weight of Erlenmeyer flask filled with water
Weight of clean, dry, and empty Erlenmeyer
flask
Weight of water that completely filled
Erlenmeyer flask
Temperature of water in the filled Erlenmeyer
flask
Calculated volume of Erlenmeyer flask (in
liters)
Show work here:Density of water at 21 C =
.997992
D=M/V V=M/D = 146.7/.997992=
80.8894 g
.4464 g
227.8 g
81.1 g
146.7 g
21 0 C
.147 L
Trial 2
81.3413
g
90 0 C
80.9104
g
.4309 g
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146.995 ml
(146.995 ml)(1 L/ 1000ml) =.1469951663 L
Barometric pressure :
Temperature (Kelvin):
770.9mmHg(1.01Atm)
294 0 K
Determination of molecular weight of the volatile liquid:
(Assuming ideal gas behavior, use the pressure of the vapor (P), its volume (V), and
its temperature (T) to determine the moles of vapor in the Erlenmeyer flask.
Then, using the mass of vapor which occupied the Erlenmeyer flask, determine the
molecular weight (MW in g/mole) of the volatile liquid. If you performed more than
one trial, show all work for each trial and then compute the average molecular weight.) Show
work below.
PV=NRT N= PV/RT
N= (1.01)(.147)/(.0821)(362) = .0049955922 MOLES
MW(1) =
MW(2)=
gRT (.4464)(.0821)(362)

 89.60562215 g / mole
PV
(101
. )(.147)
gRT (.4309)(.0821)(363)

 86.49431582 g / mole
PV
(101
. )(.147)
(89.60562215 + 86.49431582) / 2 = 88.04996899
(Average) Molecular Weight
88.05
g/mole
Determination of Empirical and Molecular Formula of volatile liquid:
The table below lists the pure volatile liquids that were distributed in lab. The first column gives
the letter designation of the compound and the remaining columns give the appropriate elemental
mass percents. Using the data for your particular volatile liquid, please determine the empirical
formula and the molecular formula for your particular sample. Show all work.
Sample Letter
mass % carbon (C)
mass % hydrogen (H)
mass % oxygen (O)
“A” (cyclohexane)
85.60 %
14.40 %
None
“B” (ethyl acetate)
54.52 %
9.17 %
36.31 %
“C” (2-propanol)
59.94 %
13.44 %
26.62 %
1mole
 4.99 moles of C
12.01 g
1mole
 13.31moles of H
Moles Of H = (13.44 g)
101
. g
Moles of C = (59.94 g)
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Moles of oxygen = (26.62 g)
1mole
 166
. moles O
16 g
Moles C/ moles O = 4.99/1.66 = 3
Moles H/moles O = 13.31 / 1.66 = 8
MW/EFW = N
EFW = (12.01 * 3) + (1.01 * 8) + (16 *1) =60.11
88.05/60.11 = 1.46= 1
C = 1 *3 = 3 H=1 * 8 = 8 O= 1 * 1 = 1
(1 * THE EMPIRICAL FORMULA)
Empirical Formula
C3 H 8 O
Molecular Formula C3 H8 O
CONCLUSION QUESTIONS
1.Why is the barometric (i.e., atmospheric) pressure considered to be the pressure of the vapor, i.e.,
how does the experimental procedure ensure this? Explain carefully.
We know that the pressure on the outside (barometric pressure) is equal to the pressure on the
inside (the vapor) because we do not see any vapor escaping through the pinhole after awhile. For
this to occur, both pressures must be equal. This experimental procedure points to the pressure
being equal.
2. Why isn't it necessary to weigh the amount of liquid initially put into the flask?
Explain.
We do not have to weigh the amount of liquid in the flask because it ( the liquid’s weight) does not play
play a part in the final outcome. During the heating excess vapor escaped through the
pinhole. The weight of the vapor in the flask after the evaporation is what we are
concerned with because that is crucial to our calculations. We need that along with the
pressure, temperature and the volume of the gas to plug into the ideal Gas formula in order
to find the molecular weight. Since we are using the barometric pressure we need to use the
weight of the vapor when the vapor’s pressure matches it (the barometric pressure).
Extra Credit–
It is sometimes stated that the above method (Dumas method) relies on the presumption that the
investigated gas follows the ideal gas law (PV = nRT). This allows one to determine the moles
of the contained vapor via:
PV
n = moles of contained vapor = RT , i.e., its measured pressure (P), volume (V),
and temperature (T). Of course, the gas constant (R) is tabulated and equals
0.0821 L• atm•mol-1•K-1. Then, the molecular weight of the gas (MW in g/mol)
can be determined from the measured mass of the contained vapor (g in grams) divided
WagenborgMWVOLGASLab Page 4
by the calculated number of moles.
g
g
gRT
MW =
=
=
. In other words, if the gas did not behave ideally, then the
n
PV
PV
RT
 
calculated molecular weight would be a crude approximation at best. Please explain
why this is commonly not a big problem, i.e., why an approximate value of the
molecular weight is often sufficient when determining the actual chemical formula
of a volatile compound. What other information is commonly obtained for a compound
in the process of determining its chemical formula such that an approximate molecular
weight is often “good enough”?
It is good enough because along with the molecular weight, we use the empirical formula weight
to find the chemical (molecular formula). We divide the molecular weight by the empirical
formula weight to find an integer to multiply the empirical formula by. If the molecular weight
is not exact, it does not matter because we will round the answer we get to a whole number to
multiply the empirical formula by. In the end the chemical formula will be correct.