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TRANSCRIPT
Burgin – Drylie
Introduction
Every year, somewhere around 3,400 people and countless numbers of
household pets die from ingesting ethylene glycol, the most common component
of antifreeze (Antifreeze Factsheet). Ethylene glycol is a sweet-tasting, odorless
chemical that can kill a grown man with as little as 120 milliliters. Victims that
consume ethylene glycol are afflicted by mental symptoms that mimic
drunkenness and physiological symptoms such as seizures, arrhythmias,
respiratory distress, and even heart failure or coma. If no medical help is sought,
the victim may suffer irreparable renal damage and will ultimately die.
An alternative to ethylene glycol is propylene glycol, a chemical commonly used
in food preservatives, but the use of propylene glycol is expensive, averaging
$100 dollars per gallon (“Propylene Glycol, 500 mL”), while ethylene glycol is
comparably cheaper, approximately $60 per gallon (“Ethylene Glycol, 500 mL”).
Another potential alternative for use in antifreeze is the chemical glycerin, a
byproduct of biodiesel.
Glycerin is a natural, non-toxic, easily accessible chemical. It was first
used in antifreeze as early as the 1900’s, but because biodiesel was uncommon,
it was very expensive and difficult to come by in its naturally occurring form.
Ethylene glycol was cheaper antifreeze that eventually made the use of glycerin
obsolete. However, since biodiesel production has been growing steadily for the
last three years, reaching 1.1 billion gallons annually in 2011 and 2012, there
now exists a surplus of glycerin (“Production Statistics”). Today, one gallon of
glycerin costs about $87 (“Glycerin, 500 mL”). Although more expensive then the
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toxic ethylene glycol, it is cheaper than the other nontoxic antifreeze, propylene
glycol.
This research intended to discover if a safer and less expensive antifreeze
than ethylene glycol or propylene glycol exists, concentrating on glycerin. The
purpose of this research was to determine if glycerin, now readily available, is as
effective an antifreeze as the commonly used chemicals ethylene and propylene
glycol. Glycerin as an antifreeze would be a safe, environmentally friendly
alternative.
Since the purpose of antifreeze is to lower the freezing point of water so
that fluid in car engines does not freeze in the winter, the experiment tested the
freezing point depression of different concentrations of solutions made with
ethylene glycol, propylene glycol, and glycerin. A control of pure water was also
tested. The experiment tried to discover which chemical depressed the freezing
point the most by finding which solution had the smallest change in temperature.
This was done by measuring the change in temperature after a certain amount of
time after the solutions were submersed in an ice-water bath. The chemical that
allowed the temperature of the solution to change by the smallest amount was
the most effective antifreeze. This is because the solution froze at a slower rate.
If it was determined that glycerin was as or more effective than the more
commonly used chemicals in antifreeze, it could potentially replace the harsh and
dangerous chemicals. Non-toxic antifreeze would be safer for humans, pets, and
the environment. Because glycerin is being made more abundant in recent times
do to its status as a byproduct of biodiesel production, this antifreeze that is safe
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to humans and to the environments is also cheaper than what is currently on the
market. If glycerin was proven effective, the results of this experiment could be
used in the future to employ glycerin as safer, inexpensive antifreeze to replace
the common, toxic chemicals used today.
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Review of Literature
Antifreeze is a substance that lowers the freezing temperature of a water
solution. There are numerous different antifreezes with different uses. Some
types are used in car engines while some are used in the environment. Road
salt, for instance is an antifreeze that allows ice to melt at lower temperatures by
adding ions that disrupt water molecules to make it harder for them to join
together and form ice, so that roads are not icy (Marder). Antifreeze agents can
also be used to keep brittle items that contain water from cracking in cold
conditions due to the expansion of water as it freezes. Antifreeze also elevates
the boiling point of a solution so that water will not boil until higher temperatures
are reached, so that engines in cars do not overheat (“How Does Antifreeze…”).
Perhaps the most well-known application of antifreeze chemicals is the use of a
mixture of antifreeze and water as coolant in automobile engines.
Figure 1. Diagram of Vehicle Engine (Jenkins)
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Figure 1above shows a diagram of a vehicle engine with components vital
to making the vehicle move labeled. In an engine, there are pistons situated
inside of cylinders in the chamber of the engine. These pistons are attached to
cams which are attached to cam shafts. Inside each of the chambers, there are
fuel injection systems. In a timed fashion, a small amount of gasoline is sprayed
into the cylinder. At the exact moment that the fuel is sprayed into the cylinder,
the spark plug sparks and a small explosion occurs. This explosion forces the
piston down into the cylinder head. When the piston is pushed down the cylinder,
the cam shaft rotates, and lobes on the cam shaft turn other pistons back up
toward the top. The cam shaft is connected to a drive shaft that turns the wheels,
and this process repeats at a fast pace to make the car move. However, the
burning gas in the combustion makes the engine get very hot, which can
potentially cause the engine to overheat (Brain).
Figure 2. Diagram of Vehicle Cooling System and Radiator (Dasan)
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When an engine overheats, the car does not function properly. Thus, a
cooling system must be used to keep the engine running at a safe temperature.
Such a system is shown above in Figure 2. This system involves a radiator that is
typically located in the front of the engine. Various tubes are attached to radiator
and go to other points in the engine that need to be cooled (Jennings). A mixture
of water and antifreeze circulates through the system to keep the engine cool, as
the heat from the system is transferred to the coolant, due to the Law of
Thermodynamics which states that energy cannot be created or destroyed.
Energy is always transferred, and is therefore changed and transferred in the
system. The coolant circulates through and around the head of engine to keep it
cool and eventually goes back to the radiator where it is cooled by fans and
circulated back through the system (Nice).
Water alone cannot be used as a coolant in the engine. Water alone
freezes at too high of a temperature, 0 °C, and boils at too low of a temperature,
100 °C, to keep from freezing or boiling inside the engine in extreme
temperatures. For this reason, it is necessary to use a solution of antifreeze and
water inside of the car to keep the coolant from boiling or freezing when
temperature extremes are reached because adding a solution to the water may
lower the freezing point or raise the boiling point of water due to the disruption of
molecules created by the addition of solution (“Antifreeze Factsheet”). As
temperatures inside the engine can rise above the boiling point of water, a
solution must be used to elevate the boiling point of the coolant so that it does
not boil inside the radiator. However, this solution also serves as a freezing point
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depressant in the winter when temperatures drop below freezing to keep the
coolant from freezing when the engine is not running. Antifreeze is therefore an
important feature in automobiles.
To understand how antifreeze works, one must understand why water
freezes when its temperature reaches freezing point, 0°C. Liquid water becomes
solid as the temperature decreases due to a loss of heat that slows the
molecules down. Bonds between molecules are formed as more energy is
released, and the water becomes solid. The molecules then form a crystalline
structure and freeze to form ice (Luedtke).
Figure 3. The Freezing of Pure Water vs. the Freezing of a Solution (Snelling)
Antifreeze changes the freezing point of water so that water does not
freeze until a lower temperature has been reached.. When the temperature of
pure water drops to 0ºC, the water molecules have slowed down enough for ice
to form. When a dissolvable substance, known as the solute, is added to water,
ions are introduced to the water that disrupt the placement of the molecules of
water and the formation of the crystalline structure of ice is impeded. The water
molecules are forced to move farther apart from each other, and more energy
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must be removed before the molecules slow down enough to form bonds and
become solid. Therefore, a lower temperature must be reached before water can
freeze when a solute is added (“Solutions and Colligative Properties:
Antifreeze”). The difference in how pure water and solutions change is shown in
the graphs in Figure 3. This difference is called freezing point depression, as the
freezing point of the solution is lower than the freezing point of water.
When water boils, it changes from its liquid phase to its gas phase, water
vapor. This occurs because when water reaches its boiling point of 100ºC, its
vapor pressure is equal to the vapor pressure of the outside air, and water vapor
escapes in the form of steam. This pressure is caused by the movement of
molecules, which increases as the molecules gain energy in the form of heat.
Figure 4. Boiling Point Elevation (Smith)
When solutes are added to the water, the vapor pressure of the water
decreases as additional ions are added to disrupt the water molecules so that it
takes a higher temperature for the liquid water to change to steam. This is called
boiling point elevation, as the boiling point of the solution is higher than the
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boiling point of water (Widom). This process is shown in Figure 4 above, as the
boiling point of the solution, the curve labeled 1, is higher than the boiling point of
water, the curve labeled 2.
Freezing point and boiling point of water are colligative properties. This
means that they depend on the concentration of the dissolved substance, the
antifreeze, and not on the chemical properties of the solute added. This is the
case because it depends on the number of ions introduced to displace the water
molecules, and not on the identity of said ions. However, not just any chemical
solutes can be put into cars and expected to be safe to use as antifreeze. The
chemicals used as antifreeze must be nonvolatile, non-corrosive, and safe to use
at the temperatures required in an engine. That is, they must not evaporate
under normal conditions, and they must not be damaging to the chemical
makeup of the engine of the car.
In automobiles today, the most common antifreeze used is composed of
ethylene glycol (C2H6O2), dyes, and corrosion inhibitors. This chemical is non-
corrosive and stable at high temperatures and is safe to use in cars.
Unfortunately, ethylene glycol is a toxic chemical, and ethylene glycol poisoning
is not uncommon. Ethylene glycol is sweet tasting substance that can be
accidentally ingested by animals and children. As the chemical metabolizes in
the body, it forms metabolites that are toxic to humans, and that inhibit many
systems that are necessary to keep the body functioning. As these toxins move
through the system, the body is subject to damage that is often irreparable.
Ingesting as little as 100 mL of ethylene glycol can prove to be fatal, and
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ingesting any of the substance can cause mental symptoms that mimic
drunkenness, and physiological symptoms such as seizures, arrhythmias, coma,
and renal damage that may take over a year to fully recover from (Bannerjee).
Using a nontoxic antifreeze in automobiles would therefore be a sensible choice.
Another chemical, propylene glycol (C3H8O2), is used as a “green,” safe,
antifreeze by some people, but it is more expensive than ethylene glycol. Today,
one gallon of pure ethylene glycol can be purchased for about $60 (“Ethylene
Glycol, 500 mL”) while one gallon of pure propylene glycol can be purchased for
about $100 (“Propylene Glycol, 500 mL”). Although ethylene glycol and
propylene glycol are generally the only antifreeze agents used in cars, another
substance, glycerin (C3H8O3), has being considered as an alternative antifreeze
in recent years.
In the past, glycerin was used as automobile antifreeze but it was not as
readily available as ethylene glycol. In recent times, however, glycerin is more
readily available, as it is a product of biofuel production that is becoming more
and more common (Treacy). If glycerin is as effective at lowering the freezing
point and raising the boiling point of water, it could be used as antifreeze today,
as a safer alternative to ethylene glycol.
To determine if glycerin was as effective an antifreeze as the more
commonly used chemicals, a freezing point depression laboratory experiment
was performed. A boiling point elevation procedure was planned, but could not
be carried out due to safety issues that arose after reviewing the Material Safety
Data Sheets (MSDS).
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One such freezing point depression experiment was published by a
Wisconsin AP Chemistry school teacher, Michael Nikson. In his experiment, a
solution made of a chemical and distilled water was placed into a test tube and
the tube was placed into an ice-salt bath inside of a beaker. The test tube
solution was stirred until ice crystals formed and the temperature at that point
was recorded to be the freezing point of the solution (Nikson). The methods used
in Nikson's experiment are similar to the methods used in this experiment in that
a chemical-water solution in a test tube is placed into an ice-water-salt bath in a
beaker. However, Nikson's methods differ from methods used in this experiment
because the solution does not get stirred. A temperature probe measures the
change in temperature as the solution remains in the ice water, and the freezing
point is calculated that way.
Another way the effectiveness of glycerin as an antifreeze could be tested
was through a boiling point elevation laboratory experiment. One experiment that
tested this was published by a teacher at a notable Korean school, Jeong S. Joo.
The boiling point was found using this experiment by recording the temperature
of each chemical solution as they were heated until the solution boiled. The
boiling point was reached when the temperature remained constant for three
readings at 30 second intervals (Joo). Unfortunately, due to laboratory limitations
and safety issues that were brought up after reviewing the MSDS, an experiment
comparing the boiling points of these chemicals could not be performed. The
hypothesis was tested by carrying out an experiment that compares freezing
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point. Glycerin’s ability to change the rate at which water freezes will be used as
judgment of how effective it is as an antifreeze.
Common antifreeze, although useful, is dangerous to humans, animals,
and the environment when it is made with ethylene glycol. Unfortunately,
ethylene glycol is cost-effective compared to a less harmful antifreeze chemical,
propylene glycol. As a byproduct of biodiesel production, glycerin could
potentially be a cost-effective, eco-friendly alternative to the chemicals most
commonly used in antifreeze.
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Problem Statement
Problem:
To determine if glycerin is an effective antifreeze by comparing the change
in temperature of glycerin solutions to solutions of ethylene glycol and propylene
glycol at three different concentrations.
Hypothesis:
There will be no statistical difference between the mean changes in
temperatures of the solutions, and glycerin will therefore be an effective
antifreeze.
Data Measured:
The independent variables in the experiment were the type of chemical,
either glycerin, ethylene glycol, or propylene glycol, and the concentration of the
solution which was measured in molarity (M). The dependent variable in the
experiment was the change in temperature of the solution, measured in °C. An
analysis of variance (ANOVA) test was carried out to determine if there was a
significant difference in the mean changes in temperature of each concentration
of each temperature.
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Experimental Design
Materials:
Ethylene Glycol, C2H6O2
Propylene Glycol, C3H8O2
Glycerin, C3H8O3
Graduated Cylinder, 10 mLGraduated Cylinder, 50 mL (2) Stir Rod(10) Beaker, 50 mL(1) Beaker, 600 mLRing Stand(2) Test Tube Clamp
(4) Test Tube, 50 mLTemperature ProbeLabQuestDistilled WaterIceRock Salt, NaClSpoon(3) Weighboat 1000 mL BeakerRefrigerator
Procedures:
1. Fill a 1000 mL beaker with water and place in refrigerator to keep the
water cold.
2. In the 50 mL beakers, prepare three solutions each of ethylene glycol,
propylene glycol and glycerin by using the graduated cylinders to measure
the necessary volume of the chemical with the necessary volume of
distilled water. Refer to Appendix A for detailed procedures on how the
solutions were made.
3. Fill the last 50 mL beaker with 30 mL of distilled water.
4. Setup the LabQuest by attaching the temperature probe and verifying that
the data collection is set to time based, and that the trial will run for 20
minutes, taking a data point every 0.1 minute.
5. Measure three 15 g samples of rock salt into three separate weigh boats.
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6. Fill the 600 mL beaker with approximately 200 mL of ice, and add water
from the refrigerated beaker until the ice water reaches the 200 mL line on
the beaker.
7. Set up the ring stand by placing the 600 mL beaker containing the ice
water on the stand and attaching the two test tube clamps so that they are
close together and nearly touching the top of the beaker.
8. Place 10 mL of the first solution in the test tube.
9. Place the test tube in the lower test tube clamp.
10. Place the temperature probe in the test tube, and secure by tightening the
top test tube clamp on the temperature probe.
11. Start data collection.
12. Quickly lower the test tube clamps so that the test tube with temperature
probe is inside the ice water bath.
13. When 0.5 minutes have passed since the data collection started, add one
of the 15 g samples of salt to the ice water and stir with a stirring rod until
the salt is no longer sitting on top of the ice.
14. After seven minutes have passed, add another 200 mL of ice and
refrigerated water and another 15 g of salt to the ice water bath and stir.
15. Repeat the process with a final 200 mL of ice water and the final weigh
boat of 15 g of salt after 14 minutes have passed.
16. When the 20 minutes are up, save the data and record the change in
temperature that occurred.
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17. Repeat for four trials each of the other solutions and the control of pure
distilled water.
Diagram:
Figure 5. Materials
Figure 5 shows all of the materials used in the experiment.
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Temperature Probe
Rock Salt
Ethylene Glycol Glycerin
Propylene Glycol
Ring Stand
Clamps
Distilled Water
Cold Water
Spoon
Graduated Cylinder600 mL Beaker
WeighBoats
LabQuest50 mL BeakerStir Rod
Ice
50 mL Test Tube
1000 mL Beaker
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Figure 6. Setup of Experiment
Figure 6 shows the setup of the experiment. The solution is inside the test
tube in the ice water bath and is being measured by the temperature probe.
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Clamps
50 mL beaker
600 mL beaker
Stir rod
Ring stand
Test tube
Temperature probe
Chemicals
Salt
Weigh boatLabQuest
1000 mL beaker
Burgin – Drylie
Data and ObservationsTable 1Average Temperature Change of 3M Solutions
3M Solutions of Ethylene Glycol, Glycerin, and Propylene Glycol
Chemical Trial
Initial Volume
of Solution
(mL)
Starting Temp. of Solution
(°C)
Ending Temp. of Solution
(°C)
Temp. Change
(°C)
Avg. Change in Temp.
(°C)
Ethylene Glycol
5 10.0 21.1 0.2 20.9
22.816 10.1 21.6 -0.6 22.229 10.0 23.0 -1.6 24.636 10.0 21.6 -2.0 23.6
Glycerin
2 10.0 20.1 -0.4 20.5
22.911 9.9 22.2 -2.4 24.630 10.0 21.9 -2.5 24.440 10.0 21.9 -0.3 22.2
Propylene Glycol
13 9.9 19.5 -3.4 22.9
23.923 10.0 23.9 -2.3 26.228 10.1 22.7 0.0 22.732 10.0 23.8 -0.1 23.9
Table 1 above shows the raw data from the trials that used 3M
concentrations of solutions. The trial numbers were randomized along with the
6M and 9M solutions and the control, and 40 total trials were carried out. The
Starting Temperature and Ending Temperature columns were found from directly
reading information from the LabQuest, and the Temperature Change was found
my subtracting the ending temperature from the starting temperature to find a
positive value for the decrease in temperature. The Initial Volume column is the
initial volume of solution that was measured in a graduated cylinder and placed
into the test tube. The average change in temperature for each chemical solution
was found by adding the changes in temperature for each solution and dividing
the result by four, the number of trials for each chemical.
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Table 2Average Temperature Change of 6M Solutions
6M Solutions of Ethylene Glycol, Glycerin, and Propylene Glycol
Chemical Trial
Initial Volume
of Solution
(mL)
Starting Temp. of Solution
(°C)
Ending Temp. of Solution
(°C)
Temp. Change
(°C)
Avg. Change in Temp.
(°C)
Ethylene Glycol
3 10.0 20.2 -0.8 21.0
24.110 9.9 21.2 -1.4 22.619 10.1 20.4 -3.8 24.235 10.0 27.0 -1.5 28.5
Glycerin
8 10.0 20.9 0.9 20.0
22.918 10.0 22.1 -0.9 23.021 10.0 21.8 -0.9 22.733 10.0 23.2 -2.6 25.8
Propylene Glycol
6 10.0 19.9 0.2 19.7
21.920 10.0 21.8 0.6 21.227 10.0 21.7 -0.5 22.238 10.0 22.3 -2.3 24.6
Table 2 above shows the raw data from the trials that used 6M
concentrations of solutions. It also shows the average change in temperature for
each chemical solution at 6M concentration.
Table 3
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Average Temperature Change of 9M Solutions9M Solutions of Ethylene Glycol, Glycerin, and Propylene Glycol
Chemical Trial
Initial Volume
of Solution
(mL)
Starting Temp. of Solution
(°C)
Ending Temp. of Solution
(°C)
Temp. Change
(°C)
Avg. Change in Temp.
(°C)
Ethylene Glycol
9 10.0 21.5 -0.9 22.4
22.612 10.0 22.9 0.3 22.617 10.0 21.9 -2.9 24.839 10.0 20.6 0.1 20.5
Glycerin
22 10.0 23.9 -1.4 25.3
25.924 10.1 22.1 -3.6 25.726 10.0 24.0 -1.1 25.131 10.0 25.0 -2.5 27.5
Propylene Glycol
1 10.0 22.0 0.5 21.5
22.34 10.1 20.8 -1.3 22.17 9.9 21.4 0.0 21.437 10.0 22.3 -1.8 24.1
Table 3 above shows the raw data from the trials that used 9M
concentrations of solutions. It also displays the average change in temperature
for each chemical solution at 9M concentration.
Table 4Control Data
Table 4 above shows the raw data from the control trials that used distilled
water instead of a chemical solution.
FinalTemperature=StartingTemperature−EndingTemperatureFigure 7. Final Temperature Formula
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Control, H20
Trial Initial Volume (mL)
Starting Temperatur
e (°C)
Ending Temperatur
e (°C)
Temperature Change
(°C)14 10.0 23.4 -1.6 25.015 10.0 22.8 -1.0 23.825 10.0 22.0 -2.5 24.534 10.1 24.1 -2.4 26.5
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Figure 7 shows the formula used to find the final temperature of the
solution. The ending temperature was subtracted from the starting temperature to
find the change in temperature as a positive value, although each temperature
change was a decrease in temperature.
Table 5
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ObservationsConcentratio
nTria
l Chemical Observations
3M
13 Propylene Glycol
LabQuest 1, Channel 2. Researcher 2. The second batch of new ice was added 0.5 minutes early, after 13.5 minutes had passed.
32 Propylene Glycol
LabQuest 1, Channel 2. Researcher 1. The first batch of new ice went in after 8.2 minutes, 1.2 minutes later than it should have been.
36 Ethylene Glycol
LabQuest 1, Channel 2. Researcher 2. This trial was redone because the temperature probe was never lowered into the solution the first time.
40 GlycerinLabQuest 1, Channel 2. Researcher 2. There was a little more than the 200 mL of ice water that there should have been at the start of the trial.
6M
3 Ethylene Glycol
LabQuest 2, Channel 1. Researcher 2. Temperature increased after first batch new of ice, the water added was too warm because it was early in the day and had not been refrigerator very long. Second batch of new ice went in at 14.5 minutes, 0.5 minutes late.
21 GlycerinLabQuest 1, Channel 2. Researcher 2. This trial had to be redone because of evaporation that affected the original solution concentration.
27 Propylene Glycol
LabQuest 2, Channel 1. Researcher 1. This trial had to be redone because of evaporation that affected the original solution concentration.
38 Propylene Glycol
LabQuest 2, Channel 1. Researcher 1. This trial was started before the other trials were ran at the same time because the LabQuest would not load.
9M
12 Ethylene Glycol
LabQuest 2, Channel 1. Researcher 1. The graph shown on the LabQuest was unusually wavy, but the trial was carried out in the same way as the other trials. Added ice water at 14.9 minutes, 0.9 minutes late.
17 Ethylene Glycol
LabQuest 2, Channel 1. Researcher 1. This trial was redone because the solution used the first time was affected by evaporation and the molarity may have been off.
31 Glycerin
LabQuest 1, Channel 1. Researcher 1. Approximately 1 mL of solution dripped down the side of the test tube before the trial started, so there was not as much solution in the test tube to be frozen. The first batch of new ice went in 1 minute late, after 8.0 minutes.
37 Propylene Glycol
LabQuest 1, Channel 1. Researcher 2. The second batch of new ice went in after 14.5 minutes, slightly
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later than it should have been.Table 5 shows the important and unusual observations that were taken
throughout the experiment. Note that Trials 17, 21, 27, and 36 were redone due
to evaporation that led to a possible change in concentration or other
experimental errors. Trials 3, 31, and 40 had errors that may have affected the
resulting temperature change. Also, many of the trials had batches of ice go in
slightly early or late, which may change the results of the experiment slightly.
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Data Analysis and Interpretation
The data was collected by reading the final temperature off of the
LabQuest after the solution had been in the ice water bath for the 20 minute trial.
This final temperature was subtracted from the initial temperature to find the
change in temperature. The final, lower temperature was subtracted from the
higher, initial temperature so that the change in temperature was a positive
value. To produce valid and viable data, the experiment had elements of control,
randomization, and repetition. These elements were used to ensure that the
effects of lurking variables were minimized and that the data can be trusted.
Trials with distilled water as opposed to the different concentrations of
solutions were run to act as controls in the experiment. These control trials were
used to see if the experiment was done correctly, because the data from the
controls should have been relatively horizontal.
0 1 2 3 4 5 60
2
4
6
8
10
12
Controls
Control Trial
Chan
ge in
Tem
pera
ture
(ºC)
Figure 8. Data From Control Trials
They acted as a valid control because lurking variables in the experiment
would have affected these trials in the same way they affected the experimental
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trials. However, as shown in Figure 8, the controls were not perfectly horizontal,
and the experiment may therefore have been affected by some lurking variables.
Randomization was used in the experiment as the trial numbers were
randomized to determine the order in which to test each chemical and each
concentration. The molarity and chemical of the solution used for each trial was
randomized to minimize the effect of lurking variables. Repetition was used in the
experiment to determine if the data from the same solutions had trends. Forty
trials were carried out, with four trials from each concentration of each solution,
and four control trials were carried out. There were an equal number of trials from
each concentration of each solution. The repetition of data resulted in less varied
results, according to the Law of Large Numbers.
In order to determine if there was in fact a difference in the mean
temperature changes for each of the concentrations, three Analysis of Variance
(ANOVA) tests were used. These tests were appropriate because the means of
three or more populations, in this case concentrations, were compared to one
another in each test. The tests were valid to use because of the ANOVA Rule of
Thumb: it is okay to use the test when the largest sample standard deviation is
no more than twice as large as the smallest standard deviation. For each test,
two times the smallest sample standard deviation was in fact larger than the
largest sample standard deviation, so all three tests would be valid.
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Figure 9. Comparative Boxplots of Solutions at 3 M
Figure 9 above shows all of the data for 3 M solutions graphed as different
boxplots. Glycerin has the highest spread among all of the different groups. The
upper and lower quartiles for that specific group are larger than any other group.
The data for 3 M ethylene glycol was most normally distributed.
The medians of all groups are very close to each other. The median of
propylene glycol is the largest at 23.4°C and the median of ethylene glycol is the
smallest at 22.9°C. Between the smallest and largest, there is only a difference of
0.5°C. This could indicate that there is not much of a difference in how the
different solutions affected the change in temperature. Although the boxplots
show similar trends in the data across all different groups, the ANOVA test was
used to see if there was a statistically significant difference among the obstacles.
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Figure 10. Comparative Boxplots of Solutions at 6 M
Figure 10 above shows all of the data for 3 M solutions graphed as
different boxplots. Glycerin and ethylene glycol both have a large amount of
spread. The upper quartile for ethylene glycol is fairly large compared to the
other groups. The data for 6 M glycerin, however, was most normally distributed
despite the spread.
The medians of all groups are again very close to each other. The median
of ethylene glycol is the largest at 23.4°C and the median of propylene glycol is
the smallest at 21.7°C. Between the smallest and largest, there is only a
difference of 1.7°C. Although this difference in temperature change is larger than
that of the 3 M solutions, this could still indicate that there is not much of a
difference in how the different solutions affected the change in temperature.
Although the boxplots show similar trends in the data across all different groups,
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Burgin – Drylie
the ANOVA test was used to see if there was a statistically significant difference
among the obstacles.
Figure 11. Comparative Boxplots of Solutions at 9 M
Figure 11 above shows all of the data for 9 M solutions graphed as
different boxplots. Ethylene glycol had the largest amount of spread but was also
the most normally distributed data among the other groups.
The medians of all groups are again very close to each other, although the
median for 9M glycerin is quite a bit higher in this grouping of solutions than it
had been in the others. The median of glycerin is the largest at 25.5°C and the
median of propylene glycol is the smallest at 21.8°C. Between the smallest and
largest, there is only a difference of 3.7°C. Because this difference is larger than
the other two concentrations, this could indicate that there may be a difference in
how the different solutions affected the change in temperature. To be sure,
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Burgin – Drylie
another ANOVA test was used to see if there was a statistically significant
difference among the obstacles.
Figure 12. Boxplot of Control Data
Figure 12 above shows a boxplot of data taken during the trials using a
control of distilled water. The median of the control data was 24.75°C. The data
appears to be relatively normally distributed. The control was used to ensure that
other trials were being done correctly.
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Table 6Table of Means, Sample Sizes, and Sample Standard Deviations
Chemical x̅ (°C) n s
3 MEthylene Glycol 22.8 4 1.617Propylene Glycol 23.9 4 1.605Glycerin 22.9 4 1.948
6 MEthylene Glycol 24.1 4 3.226Propylene Glycol 21.9 4 2.058Glycerin 22.9 4 2.371
9 MEthylene Glycol 22.6 4 1.760Propylene Glycol 22.8 4 1.255Glycerin 25.9 4 1.095
Table 6 above shows each chemical and concentration as well as its
sample mean, x̄, sample size, n, and sample standard deviation, s. This table
was used for easy reference while performing the three ANOVA tests.
Hypotheses for 3 M Solutions:
Ho: µeth3M = µprop3M = µglyc3M
Ha: Not all µeth3M, µprop3M, µglyc3M are equal
Hypotheses for 6 M Solutions:
Ho: µeth6M = µprop6M = µglyc6M
Ha: Not all µeth6M, µprop6M, µglyc6M are equal
Hypotheses for 9 M Solutions:
Ho: µeth9M = µprop9M = µglyc9M
Ha: Not all µeth9M, µprop9M, µglyc9M are equal
The null hypothesis of each test is that all sample means for all three
sample groups are equal to each other and that there is no difference between
solutions. The alternative hypothesis is that not all sample means are equal. This
would signify that there was a difference in how the different chemicals affected
the change in temperature of the solution.
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Assumptions:
i independent simple random samples
normal distribution in each population
same unknown standard deviation, σ, among sample groups
Most assumptions for the statistical test were met. Each sample i, where i is
any one of the chemicals, was randomly assigned to a trial. Because each trial
took 20 minutes to complete, the researchers were only able to complete 4 of
each sample with the time and resources available, so there is no way of
knowing if they were normally distributed. In each of the groups, standard
deviation, σ, was unknown. Even though one of the assumptions may not have
been met, the ANOVA tests were carried out anyway.
The F statistic of the test is the proportion of the variation among sample
means between each population to the variation among individuals in all the
samples within each population. In other words, the F statistic is the mean
square group, MSG, divided by the mean square error, MSE.
F= MSGMSE
Before this value could be determined, the weighted mean x̄ had to be found.
This is found by multiplying the sample size for each population by the
mean for each population, adding them together, and dividing by N, the total
number of trials in all samples combined. Refer to Appendix B for the formula
and calculations. After all values were input into the formula, the value of x̄ for 3
M solutions was found to be 23.225°C, 6 M solutions was found to be 22.958°C,
and 9 M solutions was found to be 23.583°C.
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Burgin – Drylie
Next, the mean square group, MSG, had to be calculated. This formula
takes the sample size of each population, multiplies it by the difference in sample
mean and weighted mean squared, adds them all together, and divides by one
less than I, the number of populations. Refer to Appendix B for the formula and
calculations. The value of 3 M solutions was found to be 1.480, 6 M solutions
was found to be 4.643, and 9 M solutions was found to be 16.191.
Finally, the mean square error, MSE, was calculated. This formula
uses the sample size of each population minus one and multiplies it by the
squared sample standard deviation. This is done for each population and they
are added together. Then the entire thing is divided by N – I, the total number of
populations subtracted from the total number of samples. Refer to Appendix B for
the formula and calculations. The value of 3 M solutions was found to be 2.996, 6
M solutions was found to be 6.756, and 9 M solutions was found to be 1.957.
Now that MSG and MSE had been determined for each of the tests, the F
statistic could be found. The F statistic is found by dividing MSG by MSE. See
Appendix B for the actual calculations.
The F statistic for the 3 M solutions was 0.494. This corresponds to a p-value
of 0.626. Because of this, it was concluded that the null hypothesis failed to be
rejected at α=0.05 significance level because the p-value is greater than the
alpha level. There is no significant evidence to suggest that the different
chemicals for 3 M solutions had an effect on the amount the temperature
changed. The p-value states that there is about a 62.6% chance that results this
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Burgin – Drylie
extreme were attained by chance alone if the null hypothesis was assumed to be
true.
The F statistic for the 6 M solutions was 0.687. This corresponds to a p-value
of 0.527. Because of this, it was concluded that the null hypothesis failed to be
rejected at α=0.05 significance level because the p-value is greater than the
alpha level. There is no significant evidence to suggest that the different
chemicals for 6 M solutions had an effect on the amount the temperature
changed. The p-value states that there is about a 52.7% chance that results this
extreme were attained by chance alone if the null hypothesis was assumed to be
true.
The F statistic for the 9 M solutions was 8.272. This corresponds to a p-value
of 0.009. Because of this, it was concluded that the null hypothesis was rejected
at α=0.05 significance level because the p-value is smaller than the alpha level.
There is significant evidence to suggest that the different chemicals for 9 M
solutions had an effect on the amount the temperature changed. The p-value
states that there is about a 0.900% chance that results this extreme were
attained by chance alone if the null hypothesis was assumed to be true.
There was no significant difference in temperature change for 3 M and 6 M
solutions, which means that glycerin was just as effective as ethylene and
propylene glycol as antifreeze. However, there was a difference in temperature
change in the 9 M solutions, which means that 9 M glycerin is not as effective as
9 M solutions of the other chemicals.
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Conclusion
The purpose of this experiment was to determine if glycerin is an effective
antifreeze by comparing the change in temperature in various concentrations of
glycerin solutions in an ice bath to the changes in temperature in various
concentrations of solutions made with commercially-used antifreezes. After
researching antifreeze and how it worked, an experiment was designed to
compare glycerin, ethylene glycol, and propylene glycol solutions of 3 M, 6 M,
and 9 M concentrations. Although the most common concentration of antifreeze
used in cars is a 50% solution of ethylene glycol, percent concentrations could
not be used in the experiment because the true molar concentrations of the
chemicals needed to be the same for the experiment to be valid.
The original hypothesis was that there would be no statistical difference
between the mean temperature depression, the changes in temperatures of the
solutions, and glycerin will therefore be effective antifreeze. After carrying out the
experiment and running ANOVA tests on the results, the hypothesis was
accepted because two of the three ANOVA tests that were carried out directly
support the hypothesis. The results of the third ANOVA test did not support the
hypothesis, but the results may be attributed to design flaws that occurred in the
experiment such as testing under non-ideal conditions.
A total of 40 randomized trials were carried out in the experiment,
including four control trials in which distilled water was used instead of a
concentrated solution. One solution with a designated concentration was used
per trial and placed into a test tube that was submerged in an ice bath. Salt was
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Burgin – Drylie
added to maximize temperature depression of the ice bath by depressing the
freezing point of the ice, and the temperature of the solution was recorded with a
LabQuest temperature probe for 20 minutes.
When the experiment was completed, the data was analyzed using three
ANOVA tests, one for each concentration. The ANOVA tests were used to
determine if there was a difference in the mean temperature changes for each of
the concentrations. For the 3 M solutions, the test resulted in a p-value of 0.626.
This means that the null hypothesis failed to be rejected at the alpha level α,
0.05, and there was no difference in temperature change between different
solutions at 3 M. In regards to the problem, this means that 3 M glycerin is as
effective an antifreeze as the standard antifreeze chemicals, ethylene and
propylene glycol.
For the 6 M solutions, the test resulted in a p-value of 0.527. This means
that the null hypothesis failed to be rejected at the alpha level α, 0.05, and there
was no difference in temperature change between different solutions at 6 M. In
regards to the problem, this means that 6 M glycerin is an effective an antifreeze,
like the standard antifreeze chemicals, ethylene and propylene glycol.
For the 9 M solutions, the test resulted in a p-value of 0.009. This means
that the null hypothesis was rejected at the alpha level α, 0.05, and there was a
difference in temperature change between different solutions at 9 M. In regards
to the problem, this means that 9 M glycerin was not as effective an antifreeze as
the standard antifreeze chemicals, ethylene and propylene glycol.
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The results of this experiment for the 3 M and 6 M solutions in which there
was no statistical difference in the mean change in temperature were supported
scientifically. Freezing point depression is a colligative property. When a
dissolvable substance, known as the solute, is added to water, ions are
introduced to the water that disrupt the placement of the molecules of water and
the formation of the crystalline structure of ice is impeded. The water molecules
are forced to move farther apart from each other, and more energy must be
removed before the molecules slow down enough to form bonds and solidify.
Because freezing point depression is a colligative property, it depends not on the
identity of the solute, but of the concentration of the solution. Thus, the number of
ions added is what matters, not what type of ions are added. Since each ANOVA
test compared the means of each chemical at the same molarity, each of the
solutions used in each group had the same ratio of solute to solvent on an atomic
level. The solutions therefore all had the same concentration of ions and there
should have been no difference amongst the results from each antifreeze used.
These results are supported by past findings related to freezing point
depression. In a lab that was written by Adrienne Oxley at Columbia College
discussed colligative properties of water and used the freezing point to find the
concentration of the solute, regardless of the identity of the dissolved substance
(Oxley). In a different lab performed at North Carolina State University, the
freezing point was predicted given the concentration of the solute (“Lab 3”).
These labs enforce that freezing point depression does not depend on the
identity of the solute, but rather on the concentration of said solute, because in
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Burgin – Drylie
each of these labs, the concentration of the solute was either used or found, but
its identity was not a factor in the results achieved. Since freezing point
depression is colligative, the identity of the solute did not matter for these labs,
assuming the concentration of ions in the solution was constant throughout.
The discrepancy between the results of this experiment and the colligative
property that occurred for the 9 M solutions could have been due to design flaws
and human error. Due to inadequate resources, many flaws occurred with the
experiment’s set up and trials. Perhaps the largest design flaw that occurred was
that the chemical solutions could not reach freezing point with the lab equipment
available, so the change in temperature was used instead. The statistical test
was run using the change in temperature, but there is no way to know if the same
results would have been gleaned from the experiment if the temperatures of
solutions were lowered all the way to freezing point.
An additional design flaw was that certain aspects of the experiment were
hard to control consistently. The times at which ice, water, and salt were added,
for instance, were often early or late because multiple trials were run at the same
time with the same requirements. If the salt was added to the ice bath at a later
time, the salt may not have lowered the temperature of the bath by the same
amount in each trial, and the temperature that the molecules in the solution were
exposed to may have varied from trial to trial. The temperature of water added
was also hard to control, because the earlier trials had warmer water that had
been cooling in freezer for very long. The temperature of the water that was
added to the ice bath may have affected the depression of freezing point
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Burgin – Drylie
because when the temperature of the water in the ice bath was lower, the
solution was given more of an opportunity to freeze as the molecules were at a
lower temperature and thus, lower energy. These lurking variables could not be
controlled, and the data obtained was therefore confounded.
The design flaws in the experiment are likely what led to the results of the
9 M ANOVA test that were not supported scientifically. Had the design errors not
occurred, it is possible that the 9 M solutions would also have been proven to
have no statistical difference and all of the antifreezes would have had the same
effectiveness. Testing whether or not 9 M solutions of the chemicals would be
effective could be something tested under ideal circumstances in further
research.
If the experiment were to be repeated under ideal circumstances, the
solutions would reach freezing point, and the freezing point and amount of time
passed before it reached freezing would be recorded. Additionally, the water
added would have been a more consistent temperature, and fewer trials would
be run at once, so that the ice, water, and salt could be added at the right time.
For further research, glycerin could be tested as antifreeze in an engine.
Since its effectiveness at depressing the freezing point has been proven, it
should hypothetically be capable of functioning as antifreeze. It must be tested in
a functioning engine, however, to ensure that there are no problems with its
performance in automobiles before people can start switching their ethylene
glycol-based antifreeze for glycerin.
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Another related experiment could be carried out to compare glycerin’s
effectiveness at elevating the boiling point of water. This would be a dangerous
experiment that could only be carried out with special lab equipment, but the
results could further scientific confidence in glycerin as an alternative to ethylene
glycol or propylene glycol.
The results of this experiment, along with the results of possible future
research can be used to employ glycerin as safer, greener antifreeze. Ethylene
glycol is a toxic chemical that should not be introduced to the environment
("Ethylene Glycol: Environmental Aspects"). Propylene glycol is safer for
humans, but can still be hazardous to the environment in extreme conditions, and
is expensive to produce and purchase. Glycerin, however, is nontoxic to humans
and is not harmful to the environment. Furthermore, glycerin is a byproduct of
biodiesel production, and is therefore abundant in recent times. Switching to
glycerin as antifreeze would be safer for all who use antifreeze, and would be
friendlier to the environment as a whole. Its abundance makes it an affordable,
sensible choice for those looking to use safer antifreeze.
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Acknowledgements
The researchers would like to express their sincerest thanks to the following
people for their continued support during the research process:
Mrs. Hilliard, for helping during experimentation and being all-around
supportive of the idea of the project.
Mr. Supal, for the tips he provided along the way.
The researchers would also like to extend their thanks to Mr. Darnell Jennings,
without whom they would lack knowledge of real-life applications of vehicle
engines and their components.
It is also important to mention the wonderful parents involved, who also provided
sufficient funding for the experiment:
Mr. and Mrs. Burgin
Mr. and Mrs. Drylie
Once again, to everybody involved in this entire process, the researchers
express their deepest thanks.
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Appendix A
To carry out the experiment, three different concentrations of solutions
were made for each chemical. The concentrations used were 3 M, 6 M, and 9 M,
which were roughly derived from 25%, 50%, and 75% solutions of the chemicals.
Thirty mL of each solution were made so that three trials of 10 mL each could be
carried out. Eventually, the process was repeated to make an additional 30 mL of
solution and to run the final trial and redo trials as necessary.
The molarity of a solution can be found by calculating the moles of the
substance, and dividing it by the volume of the final solution according to the
following equation:
M=molL
in which volume is in liters. Each concentration of solution had to be made for
each of glycerin, ethylene glycol, and propylene glycol.
Calculations for Necessary Volume of Each Chemical:
Table 7Molecular Weight and Density of Each Chemical
Molecular Weight (g/mol) Density (g/mL)Ethylene Glycol 62 1.11Glycerin 92 1.26Propylene Glycol 76 1.04
Table 7 above shows values that were used in calculating the volume of
chemical needed to make each solutions: the molecular weight and the density of
each chemical.
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Burgin – Drylie
M=molL
3= mol0.03
0.09=mol
0.09mol glycerin× 92g1molglycerin
=8.28 g
D=mV
1.26=8.28V
V =6.57mLglycerin
30−6.57=23.43mLwater
Figure 13. 3 M Glycerin
Figure 13 above shows the calculations used to find the amount of
glycerin and the amount of water needed to make 30 mL of the 3 M glycerin
solution. The desired molarity and the volume were substituted in for M and L in
the equation, and the moles were found to be 0.09.
This was multiplied by the molar mass of glycerin, 92 g, to find the grams
of glycerin needed. Since glycerin is a liquid, the liquid volume of glycerin was
found using the density of glycerin, 1.26 g/mL. The volume of glycerin required
was found to be 6.54 mL.
Since the total volume of is 30 mL, the volume of glycerin was subtracted
from 30 to find the necessary volume of water, 23.43 mL.
M=molL
6= mol0.03
0.18=mol
0.18mol glycerin× 92 g1molglycerin
=16.56 g
D=mV
1.26=16.56V
V=13.14 mLglycerin
30−13.14=16.86 mLwater
Figure 14. 6 M Glycerin
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Burgin – Drylie
Figure 14 above shows the calculations used to find the volume of glycerin
needed to make 30 mL of the 6 M glycerin solution.
M=molL
9= mol0.03
0.27=mol
0.27mol glycerin× 92 g1mol glycerin
=24.84 g
D=mV
1.26=24.84V
V =19.71mLglycerin
30−19.71=10.26mLwater
Figure 15. 9 M Glycerin
Figure 15 above shows the calculations used to find the volume of glycerin
needed to make 30 mL of the 9 M glycerin solution.
M=molL
3= mol0.03
0.09=mol
0.09molethylene glycol× 62g1molethylene glycol
=5.58g
D=mV
1.11=5.58V
V=5.03mLethylene glycol
30−5.03=24.97mLwater
Figure 16. 3 M Ethylene Glycol
Figure 16 above shows the calculations used to find the volume of
ethylene glycol needed to make 30 mL of the 3 M ethylene glycol solution.
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Burgin – Drylie
M=molL
6= mol0.03
0.18=mol
0.18molethylene glycol× 62g1mol ethylene glycol
=11.16 g
D=mV
1.11=11.16V
V =10.05mLethylene glycol
30−10.05=19.95mLwater
Figure 17. 6 M Ethylene Glycol
Figure 17 above shows the calculations used to find the volume of
ethylene glycol needed to make 30 mL of the 6 M ethylene glycol solution.
M=molL
9= mol0.03
0.27=mol
0.27mol ethylene glycol× 62g1mol ethylene glycol
=16.74 g
D=mV
1.11=16.74V
V =15.08mLethylene glycol
30−15.08=14.92mLwater
Figure 18. 9 M Ethylene Glycol
Figure 18 above shows the calculations used to find the volume of
ethylene glycol needed to make 30 mL of the 9 M ethylene glycol solution.
M=molL
3= mol0.03
0.09=mol
0.09mol propylene glycol× 76 g1mol propylene glycol
=6.84 g
D=mV
1.04=6.84V
V=6.58mL propylene glycol
30−6.58=23.42mLwater
Figure 19. 3 M Propylene Glycol
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Burgin – Drylie
Figure 19 above shows the calculations used to find the volume of
ethylene glycol needed to make 30 mL of the 3 M propylene glycol solution.
M=molL
6= mol0.03
0.18=mol
0.18mol propylene glycol× 76 g1mol propylene glycol
=13.68g
D=mV
1.04=13.68V
V =13.15mL propylene glycol
30−13.15=16.85mLwater
Figure 20. 6 M Propylene Glycol
Figure 20 above shows the calculations used to find the volume of
ethylene glycol needed to make 30 mL of the 6 M propylene glycol solution.
M=molL
9= mol0.03
0.27=mol
0.27mol propylene glycol× 76 g1mol propylene glycol
=20.52g
D=mV
1.04=20.52V
V=19.73mL propylene gl ycol
30−19.73=10.27mLwater
Figure 21. 9 M Propylene Glycol
Figure 21 above shows the calculations used to find the volume of
ethylene glycol needed to make 30 mL of the 9 M propylene glycol solution.
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Procedures for Making Solutions:
Materials:
50 mL Beaker (9)10 mL Graduated Cylinder (2)Stirring RodDistilled WaterGlycerin
Ethylene GlycolPropylene GlycolTapeMarkerPlastic Wrap
Procedures:
1. Use the graduated cylinder to measure the correct amount of the chemical
to make the desired solution.
2. Put the chemical into one of the 50 mL beakers.
3. Use the second graduated cylinder to measure the necessary distilled
water.
4. Pour the distilled water into the beaker with the chemical.
5. Use the stirring rod to mix the chemical and the distilled water to form the
solution.
6. Cover the beaker with plastic wrap.
7. Use the tape and marker to label the beaker.
8. Repeat procedures to make each solution.
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Appendix B
Calculations for the ANOVA Test for 3 M Solutions:
An ANOVA test was used to determine if there was a significant difference
between chemicals. All calculations for this test are shown here.
x̄=neth3M x̄eth3M+n¿ 3M x̄¿ 3M+nglyc3M x̄glyc3 M
N
Figure 22. Formula to find Weighted Mean
Figure 22 shows the formula used to find the weighted mean x̄ for the
experiment. This is found by multiplying the sample size for each population, n,
by the mean for each population, x̄, adding this value from each population
together, and dividing by N, the total number of trials in all samples combined.
x̄=4 (22.825)+4 (23.925)+4 (22.925)12
=23.225
Figure 23. Sample Equation Used to find Weighted Mean
Figure 23 above shows the formula to find the weighted mean x̄ when the
correct values are input. All sample sizes, n, consisted of 4 trials. The samples
means for each population were multiplied by n, and divided by 12, the total
number of trials. The value was found to be 23.225.
MSG=neth3 M ( x̄eth3M− x̄ )2+n¿3 M ( x̄¿ 3M− x̄ )2+nglyc 3M ( x̄glyc 3M− x̄ )2
I−1
Figure 24. Formula to find MSG
Next, the mean square group, MSG, had to be calculated. The formula for
this is shown in Figure 24 above. This formula takes the sample size of each
population, multiplies it by the difference in sample mean and weighted mean
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Burgin – Drylie
squared, adds this value for each population together, and divides by one less
than I, the number of populations.
MSG=4(22.825−23.225)2+4 (23.925−23.225)2+4 (22.925−23.225)2
3−1=1.48
Figure 25. Sample Equation Used to find MSG
Figure 25 above shows the formula to find the mean square group, MSG,
when the correct values are input. All sample sizes, n, consisted of 4 trials. The
weighted mean x̄ was subtracted from each sample mean for each population,
squared, and multiplied by n, then divided by 2, the total number of materials
minus one. The value was found to be 1.48.
MSE=(neth3M−1 ) seth3 M
2+ (n¿3 M−1 ) s¿3 M2+ (nglyc 3M−1 ) sglyc3 M
2
N−I
Figure 26. Formula to find MSE
Finally, the mean square error, MSE, was calculated. The formula to find
MSE is shown in Figure 26 above. This formula uses the sample size of each
population minus one and multiplies it by the squared sample standard deviation.
This is done for each population and they are added together. Then the
numerator is divided by N – I, the total number of populations subtracted from the
total number of samples.
MSE=(4−1 ) 1.617352+ (4−1 )1.604942+(4−1 ) 1.948292
12−3=2.99583
Figure 27. Sample Equation Used to find MSE
Figure 27 above shows the formula to find the mean square error, MSE,
when the correct values are input into the formula. All sample sizes, n, consisted
of 4 trials, so n-1 was 3 for each population. This value was then multiplied by the
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Burgin – Drylie
sample standard deviation squared and each population was added together.
Then it was divided by N-I, or 12-3, the total number of trials minus the number of
populations. The value was found to be 2.99583.
Now that MSG and MSE had been determined, the F statistic could be
found. The F statistic is found by dividing MSG by MSE.
F= MSGMSE
= 1.482.99583
=0.49402
After dividing MSG by MSE, the F statistic of the test was found to be 0.49402.
To calculate the p-value, the degrees of freedom had to be determined as
well. This value was calculated by dividing I-1 by N-I.
df = I−1N−I
= 3−112−3
=29
The degrees of freedom allowed for one to find the general interval in which the
p-value would fall. Using technology, the specific p-value was found to be
0.625792.
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Calculations for the ANOVA Test for 6 M Solutions:
An ANOVA test was used to determine if there was a significant difference
between chemicals. All calculations for this test are shown here.
x̄=neth6 M x̄eth6M+n¿ 6M x̄¿6M+nglyc6 M x̄ glyc6M
N
Figure 28. Formula to find Weighted Mean
Figure 28 shows the formula used to find the weighted mean x̄ for the
experiment. This is found by multiplying the sample size for each population, n,
by the mean for each population, x̄, adding this value from each population
together, and dividing by N, the total number of trials in all samples combined.
x̄=4 (24.075)+4 (21.925)+4 (22.875)12
=22.9583
Figure 29. Sample Equation Used to find Weighted Mean
Figure 29 above shows the formula to find the weighted mean x̄ when the
correct values are input. All sample sizes, n, consisted of 4 trials. The samples
means for each population were multiplied by n, and divided by 12, the total
number of trials. The value was found to be 22.9583.
MSG=neth6 M ( x̄eth6M− x̄)2+n¿6 M ( x̄¿6 M− x̄)2+nglyc6 M ( x̄glyc6 M− x̄)2
I−1
Figure 30. Formula to find MSG
Next, the mean square group, MSG, had to be calculated. The formula for
this is shown in Figure 30 above. This formula takes the sample size of each
population, multiplies it by the difference in sample mean and weighted mean
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Burgin – Drylie
squared, adds this value for each population together, and divides by one less
than I, the number of populations.
MSG=4(24.075−22.9583)2+4 (21.925−22.9583)2+4 (22.875−22.9583)2
3−1=4.64333
Figure 31. Sample Equation Used to find MSG
Figure 31 above shows the formula to find the mean square group, MSG,
when the correct values are input. All sample sizes, n, consisted of 4 trials. The
weighted mean x̄ was subtracted from each sample mean for each population,
squared, and multiplied by n, then divided by 2, the total number of materials
minus one. The value was found to be 4.64333.
MSE=(neth6M−1 ) seth6M
2+(n¿ 6M−1 ) s¿ 6M2+(nglyc6 M−1 ) sglyc 6M
2
N−I
Figure 32. Formula to find MSE
Finally, the mean square error, MSE, was calculated. The formula to find
MSE is shown in Figure 32 above. This formula uses the sample size of each
population minus one and multiplies it by the squared sample standard deviation.
This is done for each population and they are added together. Then the
numerator is divided by N – I, the total number of populations subtracted from the
total number of samples.
MSE=(4−1 ) 3.226322+( 4−1 )2.058112+ (4−1 )2.371182
12−3=6.75582
Figure 33. Sample Equation Used to find MSE
Figure 33 above shows the formula to find the mean square error, MSE,
when the correct values are input into the formula. All sample sizes, n, consisted
51
Burgin – Drylie
of 4 trials, so n-1 was 3 for each population. This value was then multiplied by the
sample standard deviation squared and each population was added together.
Then it was divided by N-I, or 12-3, the total number of trials minus the number of
populations. The value was found to be 6.75582.
Now that MSG and MSE had been determined, the F statistic could be
found. The F statistic is found by dividing MSG by MSE.
F= MSGMSE
=4.643336.75582
=0.687308
After dividing MSG by MSE, the F statistic of the test was found to be 0.687308.
To calculate the p-value, the degrees of freedom had to be determined as
well. This value was calculated by dividing I-1 by N-I.
df = I−1N−I
= 3−112−3
=29
The degrees of freedom allowed for one to find the general interval in which the
p-value would fall. Using technology, the specific p-value was found to be
0.527494.
52
Burgin – Drylie
Calculations for the ANOVA Test for 9 M Solutions:
An ANOVA test was used to determine if there was a significant difference
between chemicals. All calculations for this test are shown here.
x̄=neth9 M x̄eth9M+n¿ 9M x̄¿9M+nglyc 9M x̄ glyc9M
N
Figure 34. Formula to find Weighted Mean
Figure 34 shows the formula used to find the weighted mean x̄ for the
experiment. This is found by multiplying the sample size for each population, n,
by the mean for each population, x̄, adding this value from each population
together, and dividing by N, the total number of trials in all samples combined.
x̄=4 (22.575)+4 (22.275)+4 (25.9)12
=23.5833
Figure 35. Sample Equation Used to find Weighted Mean
Figure 35 above shows the formula to find the weighted mean x̄ when the
correct values are input. All sample sizes, n, consisted of 4 trials. The samples
means for each population were multiplied by n, and divided by 12, the total
number of trials. The value was found to be 23.5833.
MSG=neth 9M ( x̄eth9M− x̄)2+n¿ 9M ( x̄¿9 M− x̄)2+nglyc9 M ( x̄glyc9 M− x̄)2
I−1
Figure 36. Formula to find MSG
53
Burgin – Drylie
Next, the mean square group, MSG, had to be calculated. The formula for
this is shown in Figure 36 above. This formula takes the sample size of each
population, multiplies it by the difference in sample mean and weighted mean
squared, adds this value for each population together, and divides by one less
than I, the number of populations.
MSG=4(22.575−23.5833)2+4 (22.275−23.5833)2+4 (25.9−23.5833)2
3−1=16.1908
Figure 37. Sample Equation Used to find MSG
Figure 37 above shows the formula to find the mean square group, MSG,
when the correct values are input. All sample sizes, n, consisted of 4 trials. The
weighted mean x̄ was subtracted from each sample mean for each population,
squared, and multiplied by n, then divided by 2, the total number of materials
minus one. The value was found to be 16.1908.
MSE=(neth9M−1 ) seth9M
2+(n¿ 9M−1 ) s¿ 9M2+(nglyc9 M−1 ) sglyc 9M
2
N−I
Figure 38. Formula to find MSE
Finally, the mean square error, MSE, was calculated. The formula to find
MSE is shown in Figure 38 above. This formula uses the sample size of each
population minus one and multiplies it by the squared sample standard deviation.
This is done for each population and they are added together. Then the
numerator is divided by N – I, the total number of populations subtracted from the
total number of samples.
MSE=(4−1 ) 1.75952+ (4−1 )1.255322+ ( 4−1 )1.095452
12−3=1.95723
54
Burgin – Drylie
Figure 39. Sample Equation Used to find MSE
Figure 39 above shows the formula to find the mean square error, MSE,
when the correct values are input into the formula. All sample sizes, n, consisted
of 4 trials, so n-1 was 3 for each population. This value was then multiplied by the
sample standard deviation squared and each population was added together.
Then it was divided by N-I, or 12-3, the total number of trials minus the number of
populations. The value was found to be 1.95723.
Now that MSG and MSE had been determined, the F statistic could be
found. The F statistic is found by dividing MSG by MSE.
F= MSGMSE
=16.19081.95723
=8.27232
After dividing MSG by MSE, the F statistic of the test was found to be 8.27232.
To calculate the p-value, the degrees of freedom had to be determined as
well. This value was calculated by dividing I-1 by N-I.
df = I−1N−I
= 3−112−3
=29
The degrees of freedom allowed for one to find the general interval in which the
p-value would fall. Using technology, the specific p-value was found to be
0.009146.
55
Burgin – Drylie
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