Page 1
NAMES: Mary-Kate McGlinchey
MISE - Physical Basis of Chemistry
Third Experiment
Molecular Weight of a Volatile Liquid
Lab Report – Submit electronically (digital drop box) by Sunday, November 6, 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. By vaporizing a measured amount of the
sample, and determining the pressure, volume, and temperature of the sample,
the approximate molecular weight may be determined by presuming
ideal gas behavior.
Background:
•
# 2 - Pressure, Gas Laws, Ideal Gas Law, Partial Pressures - Some Examples
•
Gases - Experiments and Relationships - PowerPoint Summary
(PowerPoint summary of our discussion of gases.)
These handouts are on Bb.
Reference:
For some procedural aspects of this lab:
Catalyst - The Prentice-Hall Custom Laboratory Program for Chem 53 at the
University of Pennsylvania Department of Chemistry, Pearson Custom
Publishing, 2005, pages 57-65.
Introduction:
In Experiment # 2, you had the opportunity to determine the empirical formula of
a binary compound. In order to determine the molecular formula of a compound
- given the empirical formula - you need the molecular weight of the compound
(in addition to the empirical formula weight).
In this experiment, you will utilize one of many experimental procedures to determine the
molecular weight of a compound. This method, first developed by Dumas, exploits the
ideal gas law in order to determine the molecular weight of a compound that can be
vaporized (without significant decomposition). Specifically - via the ideal gas law - you
will determine the molecular weight of a substance by determining the pressure exerted,
volume occupied, temperature, and mass of a sample of its vapor. The example below
will hopefully enable you to understand how this is accomplished.
Presuming that the vaporized volatile liquid follows the ideal gas law (PV = nRT),
we may determine the moles of the contained vapor via:
Page 2
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 by the calculated number of moles.
g
g
gRT
MW = n = PV =
.
PV
 
RT
 
Before you carry out the determination of the molecular weight (MW) of the volatile liquid,
a worked example may help. Please refer to the Appendix at the end of this experiment.
DATA/CALCULATIONS
Note: Data and Calculations may be completed on an Excel spreadsheet.
Unknown Letter: _______A______
Molecular Weight of a Volatile Liquid
(If you performed more than one trial, adjust your data table accordingly.)
Weight of flask + foil cap + rubber band
81.44
g
Temperature of water bath when (excess) vapor ceased to escape
94.0
Weight of flask + foil cap + rubber band + condensed vapor
Weight of condensed vapor
0.44
ºC
81.88
g
g
Determination of full volume of flask:
Weight of Erlenmeyer flask filled with water
227.4
Weight of clean, dry, and empty Erlenmeyer flask
g
80.98
g
Weight of water that completely filled Erlenmeyer flask
146.42
Temperature of water in the filled Erlenmeyer flask
24
Calculated volume of Erlenmeyer flask _____0.147______ L
Barometric pressure
752
mm Hg (xxx.x mm Hg)
g
ºC
Page 3
Before you carry out the determination of the molecular weight (MW) of the volatile liquid,
a worked example may help. Please refer to the Appendix at the end of this experiment.
Table 1: (Revised**) Density of pure liquid water at various temperatures:
Temperature (ºC)
1ºC
2ºC
3ºC
4ºC
5ºC
6ºC
7ºC
8ºC
9ºC
10ºC
11ºC
12ºC
13ºC
14ºC
density (d) in g/mL
0.999890
0.999940
0.999960
0.999970
0.999970
0.999950
0.999910
0.999860
0.999790
0.999720
0.999620
0.999520
0.999400
0.999270
Temperature (ºC)
15ºC
16ºC
17ºC
18ºC
19ºC
20ºC
21ºC
22ºC
23ºC
24ºC
25ºC
26ºC
27ºC
28ºC
density (d) in g/mL
0.999099
0.998943
0.998774
0.998595
0.998405
0.998203
0.997992
0.997770
0.997538
0.997296
0.997044
0.996783
0.996512
0.996232
** Density data from 1ºC to 14ºC (at 0.1013 Mbar (1 atm)) source:
NIST - National Institute of Standards & Technology website: http://webbook.nist.gov/
Determination of the volume occupied by the vapor in the Erlenmeyer flask:
(Using the mass of water that completely filled the Erlenmeyer flask and the above
table of water densities as needed – determine the volume (in L) occupied by the vapor
in the Erlenmeyer flask. Also, explain your reasoning. Then, place your result
in the data table at the appropriate spot. Show work below.
Volume of vapor
0.147
L
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.
Page 4
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.
(Average) Molecular Weight
90.41
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 %
Page 5
Empirical Formula
CH2
Molecular Formula
C6H12
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.
Initially, the pressure inside the flask was greater than the barometric pressure which was outside the
flask. As the vapor escaped, due to heating, both pressures equalized. The experiment ensured this due
to the minute pinhole in the foil on top of the flask. This allowed substance “A’s” vapor to escape until
there was no more vapor visible, which meant the pressures had equalized. Therefore, I could infer that
the pressure inside the flask (the pressure of the vapor) could be considered the barometric pressure.
2. Why isn't it necessary to weigh the amount of liquid initially put into the flask?
Explain.
It isn’t necessary to measure the weight of the liquid that is initially put into the flask because some
of the liquid will be changed into a vapor, which will escape through the pinhole. Therefore, for
experimental purposes, some of the substance is lost when it changes from a liquid to a gas. Much of
the liquid is vaporized throughout the experiment anyway.
Extra –
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
by the calculated number of moles.
g
g
gRT
MW = n = PV =
. In other words, if the gas did not behave ideally, then the
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”?
Page 6
Appendix - A worked Example:
A certain volatile hydrocarbon (a binary compound of carbon and hydrogen) is found to be
92.3 % carbon, by mass. In a separate experiment, utilizing the Dumas method, a
4.00 mL pure liquid sample of this hydrocarbon is vaporized in a 125 mL Erlenmeyer flask
when the barometric pressure is 768.0 torr. The empty flask - fitted with a foil cap pierced with
a pinhole - weighs 25.3478 g. After the excess gas escapes, the temperature is measured as
98.0oC. The flask and contents are subsequently cooled to 25oC and the vapor condenses to
a liquid. The weight of the flask and contents is found to be 25.6803 g. The flask is then
emptied, cleaned, and filled with water. When weighed on a triple-beam balance, the
difference in weight between the flask filled to the brim with water and the dry empty
flask – at 25ºC - is 128.12 g. Please determine the following:
• The empirical formula of this hydrocarbon.
• The volume that the vapor occupied in the Erlenmeyer flask (in L).
• The molecular formula of this hydrocarbon.
Solution to Example:
The ideal gas law in terms of mass (g = grams), molecular weight (MW),
temperature (T), volume (V), and pressure (P).
mass
g
PV = nRT and
n = moles = molecular weight =
.
MW
g
PV
gRT
n = MW = RT
Solving for MW:
MW =
.
PV
The first equation above may be easily used to find the molecular weight of the
L atm.
hydrocarbon. Note that the ideal gas constant R = 0.0821 mole K .
Also note that 760 torr = 1 atm. The kelvin (K) temperature is obtained by adding 273 to
the temperature in oC.
The first part of this problem, obtaining the empirical formula. A hydrocarbon contains only
carbon and hydrogen. Thus, the mass % of H = 100 % - 92.3 % = 7.70 % H. We assume
100 g of the hydrocarbon. Then, there are 92.3 g of C and 7.70 g of H present (relatively).
Determining moles of each element:
 1 mole C 
moles C = (92.3 g C)•12.01 g C
= 7.69 moles C.


moles H =
1 mole H
(7.70 g H)• 1.01 g H 


Divide to obtain the mole ratio:
= 7.62 moles H.
Page 7
moles C
7.69 moles
1.01
1
moles H = 7.62 moles = 1 = 1 so, the Empirical Formula = CH.
As was presumed when we carried out this experiment, the vapor completely fills
the flask - once the excess escapes (also flushing out the air). The volume of water that
fills the flask equals the full volume of the flask and the volume occupied by the gas in the
ideal gas law (V).
V = volume of vapor = full volume of the flask = volume of water that fills it to the brim.
V = Vwater filling flask =
mass of water filling flask
.
dwater
Recall that density is temperature -dependent. So, we must use the density of water at
its temperature when it filled the flask. By the info in the problem, this temperature
is 25ºC. Referring to the table of densities (Table 1), the density of liquid water at 25ºC
is listed as: 0.997044 g/mL.
128.12 g
V = Vwater filling flask = 0.997044 g/mL = 128.50 mL = 0.12850 L
To determine the molecular formula, we need to find the molecular weight. The most useful
form of the ideal gas law - considering the given information - is:
MW =
gRT
; T = 98 + 273 = 371 K and V = 0.1285 L.
PV
mass of condensed vapor = mass of gas in flask = 25.6803 g - 25.3478 g = 0.3325 g.
MW =
L atm. 

(0.3325 g) 0.0821 mole K (371 K)


gRT
= 78.0 g/mole.
PV =
768.0


 760  (0.12850 L)


The EFW = Empirical Formula Weight for CH = 12.01 + 1.01 = 13.02 g/mole.
N = # of empirical formula units =
78.0 g/mole
MW
= 13.02 g/mole = 5.99 = 6.
EFW
Thus, in this case, since N = 6, the molecular formula is 6 x Empirical Formula.
Thus, the Molecular Formula = C6H6 . (The compound is benzene).