# George Mason University General Chemistry Lab Report

Laboratory 3 Freezing Point DepressionColligative properties of solutions depend only on the concentration of the
solution. Freezing point depression is an example of a colligative property.
Drivers on icy streets depend on it. Salt placed on the road in the winter,
dissolves into the snow or ice and lowers the freezing point of water, preferably
Pure water freezes at 0o C. If one mole of dissolved particles is added to
one kilogram of water it dissolves at -1.86 oC. This gives the freezing point
constant for water, kfp = 1.86oC/m, where m is molality of the solution. Molality is
defined as the moles of solute per kilogram of solvent. When salt is dissolved in
water, two ions are present in solution, Na+ and Cl-, so the number of particles is
two, and if one mole of sodium chloride is dissolved in one kilogram of water the
freezing point of the water drops from 0 oC to -3.72 oC. If one mole of sugar is
dissolved in one kilogram of water the freezing point will be -1.86 oC because
sugar does not dissociate and gives only one particle in solution with water. So it
is clear why salt is used instead of sugar on icy roads.
This freezing point depression constant is only good for water. Each
solvent has a different kf.
ΔTfp = i kfp m
(1)
where ΔTfp is the freezing point change, kfp is the freezing point
depression constant and m is the molality of the solution, and i is the van Hoff’t
factor which tells the number of particles in solution. For NaCl this is 2. For
sugar this is 1. Molality is as the unit of concentration instead of molarity.
Molarity has the units moles solute/liter solution. Liquids tend to change volume
when the temperature changes therefore molality, which is a temperature
independent unit is used.
In this experiment first we will determine the freezing point of the pure
material, stearic acid. Then we will determine the freezing point depression
constant, kfp, of this material using a known substance, and finally we will
determine the molar mass of an unknown substance, by its freezing point
depression of stearic acid.
The normal freezing point is determined by removing heat from a
substance. In theory when heat is removed from a liquid the liquid will decrease
in temperature until it reaches the freezing point. The liquid then stays at the
freezing point until it becomes a solid, and then the temperature of the solid
decreases. The freezing point is determined to be the constant temperature or
the flat part of the cooling curve. In practice when you remove heat from a liquid,
21
there is a fast decrease in temperature for the liquid, followed by a much flatter,
but still sloped section where freezing occurs. The experimental cooling curve
for stearic acid is given in Figure 1.
Cooling Curve
95
90
85
Temperature (Celsius)
y = -0.105x + 89.378
80
Freezing Point is the intersection of the
two lines
75
70
y = -0.0085x + 66.2
65
60
55
0
100
200
300
400
500
Time (seconds)
For a pure substance typically a very nice cooling curve is obtained.
Solutions may exhibit more complicated cooling curves. An example of cooling
curve for a solution is shown below:
22
600
Cooling Curve 2 (Stearic Acid with Palmitic Acid)
85
80
y = -0.11x + 80.667
Temperature (Celisus)
75
Freezing point is the
intersection of the two lines
70
65
y = -0.0089x + 63.694
60
55
0
100
200
300
400
500
600
Time (seconds)
Because it is difficult to obtain a good cooling curve for the solutions, be
sure that they are well mixed and 100% dissolved, and measure them very
carefully. For both of the cooling curves given above, two straight lines have
been fit to the data, one for the portion where the liquid is cooling and one for the
portion where freezing occurred. A consistent method for determining the
freezing point is to use the temperature at which these lines intersect, which is
indicated by the vertical arrow on both plots. The minor grid lines have also been
included to make it easy to read the temperature.
Procedure: The freezing point of stearic acid
1. Prepare a hot water bath in a 250 mL beaker. Place it on the hot plate
and begin to heat it. Take a large test tube and add 3 grams of stearic acid.
Measure the mass exactly.
2. Using your test-tube holder, put the test tube into the hot water bath.
Put your thermometer into the test tube, not the water bath.
3. When the stearic acid has completely melted, remove it from the hot
water bath and insert the test-tube into a block of Styrofoam, which has a hole for
the test tube in it.
4. Record the temperature every 30 seconds until the stearic acid has
completely solidified.
23
700
Repeat the procedure using the same sample of stearic acid.
Plot the cooling curve for stearic acid, temperature (y axis) versus time for
both cooling curves. You can plot the data using EXCEL, but however you plot
the data you must read the value of the freezing point to one decimal point.
Directions for making a plot with EXCEL
First start the program, in a windows system usually you use a double
click.
The program opens to a fresh sheet. To move between sheets, click on
the tabs on the bottom of the page. The first step is to input some of your data.
The easiest method to keep everything straight is to use different sheets for your
different cooling curves. Use the first sheet to analyze your freezing point of pure
stearic acid. In the first column enter the time. Just type in the number and hit
return. In the second column, type the temperature. Now plot the line.
To make a graph in Excel 97, I have found the easiest method is to
make sure your x values are in one column and your y values are in the next
column. Using the mouse highlight your data, x and y for your standard curve.
Then click on the chart wizard icon, it looks like a bar graph. A series of 4 dialog
boxes open to make the chart.
Dialog Box 1: First the chart wizard asks what type of graph you want.
Click on xy plot, on the right side of the window it asks what type of xy plot you
want. Click on the style of graph you like. For this graph you will sketch your
own line so just use plot points without a line.
Dialog Box 2: Now the chart wizard asks if the data you want plotted is
A1:B5. If the range is what you want hit the next button, otherwise edit the range
of data, and then hit the next button.
Dialog Box 3: gives you the opportunity to add legends or labels if you
like. So you will be able to read the temperature values accurately, click on
gridlines and check the box for minor gridlines for the y-axis. When you are
happy with your graph, click on next.
Dialog Box 4: gives you a choice of placing the chart on the page with the
data or on a new page. Decide which you want, typically a new page is neater.
Click finish, and a graph of time versus temperature appears.
You should see two regions with two different slopes. The intersection of
these two lines gives the freezing point of pure stearic acid. Take the average of
the two freezing points to get the freezing point of pure stearic acid. If your two
values differ by more than 2o C take another cooling curve. You can compare
this value with a literature value from either the Aldrich catalogue or the CRC
Handbook.
24
Part 2 Freezing point depression constant of stearic acid
Weigh about 0.5 g of palmitic acid. Record the mass to the nearest
milligram. Add this to your test tube containing stearic acid. Calculate the
molality of the solution. Mix the solution thoroughly.
Using the same procedure as above, take a cooling curve for this solution.
Using your test tube holder put the test tube into the waterbath. As the solution
dissolves use a stirring rod to mix it thoroughly. Remove the liquid solution from
the hot water bath and place it in the insulated block. Again record the time and
temperature every thirty seconds until the solution is completely solid. Thaw the
solution and repeat the procedure. The freezing point of the solution is the
intersection of the two lines with different slopes. Plot both cooling curves and
determine the freezing point of the solution. Average the two measurements to
get the freezing point; remember it should be lower than that of pure stearic.
Once again if your two freezing points differ by more than 2o C, you should
repeat the measurement one more time, or if your average freezing point is
higher than that of pure stearic acid. Now we can use equation 1 to determine
the freezing point depression constant for stearic acid. Determine the molality of
the solution, and then the freezing point depression constant.
Discard the solution in the appropriate waste container.
Part 3 The molar mass of an unknown compound
Using a clean the test tube. Add 3 grams of pure stearic acid. Record the
unknown, recording the mass to the nearest milligram. Determine the freezing
point of your unknown solution by again recording two cooling curves. ΔTfp =
the freezing point of pure pure stearic acid minus the freezing point of the
unknown solution. Using the average of your two freezing points determine ΔTfp.
The molality of your solution, m = ΔTfp / kfp , where kfp was determined in part 2.
Determine the molality of your unknown solution. The molality is moles solute/kg
solvent. Determine the moles of unknown solute in the solution.
This is the number of moles present in your test tube. The molar mass of
your unknown is grams/mole. Divide the recorded mass of your unknown by
moles unknown to determine the molar mass.
Hints for organizing data and performing calculations:
You could set up one big data table with all the data you need to collect.
Pure Stearic Acid
MassInitial stearic acid =
Palmitic acid
25
Unknown
MassUnknown
Run 1
Time
Run 2
Temp
Time
Run 1
Temp
Time
Run 2
Temp
Time
Run 1
Temp
Time
Run 2
Temp
Time
Temp
Plot both pure runs on the same graph using different colors or x’s and o’s.
You want to do this while taking the data, or immediately after so you can see if
there is any problem. Determine the freezing points and average the two runs.
Plot your other runs in the same fashion. In your notebook write
stearic acid
acid
with Palmitic acid
with Unknown
Tf = Here put the values with the correct significant figures in each column.
For palmitic acid and your unknown:
i = 1, neither will dissociate in stearic acid
ΔTf = the difference of the freezing point of each solution and pure stearic
molality for palmitic solution is:
moles palmitic acid/kg stearic acid = mpalmitic acid
and now for stearic acid kfp , = ΔTfp palmitic acid /mpalmitic acid
Using this kfp, determine the molality of your unknown solution.
munknown = ΔTfp unknown / kfp
To get the moles of unknown from the molality of the unknown solution
remember m=moles solute/kg solvent. Just multiple the molality by the kg of
stearic acid used.
From this molar mass unknown, MM, is:
MM = massunknown / molesunknown
Question:
Based on the information in the introduction how much NaCl in grams is needed
to lower the freezing point of 1.2 liters of water 2.5oC? How many grams of
CaCl2 would be needed?
Safety and Disposal:
The stearic acid, palmitic acid and unknown solutions should be placed in the
waste container provided. To get the last bit out of the test tubes, rinse with hot
isopropanol.
26
Measurement of Physical Properties
In any measurement it is important to know the precision of the
measurement and also its accuracy. All physical measurements should be made
as precisely and accurately as possible. Maximizing the precision of a
measurement is accomplished by using the most precise equipment available,
and using it properly. If possible it is also wise to compare your result with either
a known or theoretical value.
Significant Figures
When calculating a result from more than one measurement it important to
retain the uncertainty information from all the measurements. There is an entire
field of mathematics devoted to this topic. In this course we use the relatively
simple method of significant figures. A summary of the rules with examples:
Addition and subtraction: line up the numbers to be added or subtracted;
the answer is truncated to the decimal place of the least precise number.
Ex. 12.1 + 2.345 = 14.4
15.678 – 2.2 = 13.5 (notice I rounded up)
Multiplication and Division: Significant Figures in the answer are equal to
the number of significant figures in the least precise number.
15.6 x 2.1 = 31
16.789 ⎟ 25.67432 = 0.65392
25.1 x 3.00 = 75.0
Note zeroes before another number as in 0.65392 do not count. In the
middle and the end they count.
Laboratory Notebooks
In this course you will be required to keep a laboratory notebook. A good
laboratory notebook is an accurate record of everything, which occurred in the
lab. In patent disputes a good lab book versus an inaccurate lab book can mean
millions of dollars. In this course it may mean hundreds of points. Before the lab
you will be required to prepare a lab report outline to be completed during the lab
session. Each lab report will contain:
7
Title and Purpose
1. Procedure and Observations
2. Data and Calculations
3. Results and Conclusions
4. Answers to questions in manual
An example of a lab report is shown below:
1. Title: Density of liquid and a solid.
Purpose to measure the density of a liquid and an unknown solid.
2. Procedure:
Observations
Part 1 Liquid
Unknown # 5 smells like gasoline
Weigh an empty 10.0 mL volumetric flask
Fill with unknown liquid.
Mass of empty flask = 12.032 grams
Mass of full flask = 18.685 grams
Part 2 Solid
Part 2 Unknown #12 Shiny orange color
water
Measure precisely volume of water.
Volume of water = 24.83 mL
Volume of water + metal = 28.53 mL
Mass of Dry metal = 46.409 grams
weigh dry solid sample
Data and Calculations:
Part 1 Liquid:
Density = Mass / volume
Mass of liquid = mass of liquid + flask – mass of flask = 18.685 grams – 12.032 grams = 6.652
grams.
Density = 6.652 grams / 10.000 mL = 0.6652 g/mL
Part 2 solid
volume of solid = volume of solid + water – volume water = 28.53 mL – 24.83 mL = 3.70 mL
density = 46.409 g/3.70 mL = 12.5 g/mL
Results and Conclusions:
The density of the liquid was determined to be 0.6652 g/mL by comparison with the
density table in the CRC it appears the sample could be hexane, which has a density of 0.660
g/mL
The density of solid was 12.5 grams / mL. The solid looked like copper, but the density of
copper from the CRC is: 8.94 g/mL, which is significantly less than my unknown sample.
Therefore although the sample looks like copper it must be something else.
8
In this example, the data is recorded in the section with the observations,
and the procedure is recorded in one column and the observations are recorded
in an adjoining column. This allows you to record your observations with the
correct section of the procedure. In some experiments, the type and volume of
data is better recorded in a table. In this case it should follow the procedure
section. You should still leave room in the procedure section for observations.
One of the objectives of this course is for students to learn how to determine
what data they need to collect, and how to organize it. For some experiments
explicit instructions for organizing the data and calculations will be given, but for
other experiments you will need to determine this for yourself before class. In the
case of repetitive calculations tables are necessary. A spreadsheet such as
Excel can be used, and instructions are included for the Reaction Rate
experiment. All your calculations must follow the rules for significant figures and
every value must have a correct unit. A spreadsheet or calculator will not
determine the correct number of significant figures; it is up to you.
When determining the results and conclusions, there are some things to
keep in mind. The results should relate back to the purpose. Address directly if
the purpose was fulfilled. If the result is a number clearly restate what it is and
the unit for the number. If possible compare your result with a literature value. If
you received no result or an unexpected result, give some scientific explanation
of this. Human error is not a good explanation, because the experiment or
section, which was in error, should be repeated. Thoroughness is important but it
is not necessary to write everything you know about density or volume etc.
To be ready to use all the lab time efficiently, before lab class you should
have completed the purpose, procedure and arranged the data table or written
down what you need to measure.
Lab Instructors may have additional report requirements.
9
Pure Stearic acid
Palmitic Acid
run1
run 2
time
temp
run 1
time
temp
run2
time
temp
time
30
77
30
77
30
79
30
60
72
60
72
60
74
60
90
63
90
64
90
65
90
120
62
120
63
120
62
120
150
61
150
62
150
61
150
180
60
180
60
180
60
180
80
80
75
75
Temprature (Celsius)
Temprature (Celsius)
Pure Stearic Acid
70
65
60
55
50
70
65
60
55
0
50
100
Time (Seconds)
150
200
0
Unknown
run 1
temp
run 2
time
temp
time
temp
77
30
74
30
76
71,5
60
70
60
72
64
90
69
90
70
63
120
67
120
68
62
150
63
150
64
60
180
61
180
60
Stearic Acid with Palmatic Acid
Temprature (Celsius)
80
75
70
65
60
55
50
100
Time (Seconds)
150
200
0
Steric Acid with Uknown
50
100
Time (Seconds)
150
200
Title: Lab 3: Freezing Point Depression
Purpose to determine the freezing point of stearic acid, then calculate the freezing point
depression constant of stearic acid. This will be used to determine the molar mass of an
unknown substance.
Procedure:
Part 1: Freezing point of stearic acid
• Prepare hot water bath in 250ml
beaker
• Add 3 grams of stearic acid to large
test tube
• Measure mass
• Add test tube to hot water bath
• Add thermometer to test tube
• Move melted stearic acid to test
tube block in Styrofoam
• Record temperature every 30
seconds until solidified
Part 2: Freezing point depression
constant of stearic acid
• Weight out .5 g palmitic acid
• Add palmitic acid to test tube with
stearic acid
• Add test tube to hot water bath
• Move melted solution to test tube
block in Styrofoam
• Record temperature every 30
seconds until solidified
Part 3: Molar mass of unknown
compound
• Add 3 grams stearic acid to clean
test tube
• Record mass to nearest milligram
• Add palmitic acid to test tube with
stearic acid
• Add test tube to hot water bath
• Move melted solution to test tube
block in Styrofoam
• Record temperature every 30
seconds until solidified
Observations:
Write any observation here
Write any observation here
Write any observation here
Data and Calculations:
Pure Stearic Acid
Mass =
Palmitic Acid
Mass =
Pure Stearic Acid
MassInitial stearic acid =
Run 1
Run 2
Time
Temp
Time
30
sec
1
min
1:30
min
2:00
min
2:30
min
3
min
77 c
30
sec
1
min
1:30
min
2:00
min
2:30
min
3
min
72
63
62
61
60
Unknown
Mass =
Palmitic acid
Run 1
Run 2
Temp
Time
Temp
77
30 sec 79
72
1 min
74
64
63
62
60
1:30
min
2:00
min
2:30
min
3 min
65
62
61
60
Time
30
sec
7 1
min
1:30
min
2:00
min
2:30
min
3
min
Temp
77
71
64
63
62
60
Unknown
MassUnknown
Run 1
Run 2
Time
Temp
Time
30
sec
1
min
1:30
min
2:00
min
2:30
min
3
min
74
70
69
67
63
60
30
sec
1
min
1:30
min
62:00
min
2:30
min
6 3
min
Temp
76
72
70
68
64
60
Molality palmitic acid is:
Moles palmitic acid/kg stearic acid = mpalmitic acid
Stearic acid Kfp = ΔTfp palmitic acid / mpalmitic acid
Molality unknown: Munknown = ΔTfp unknown / Kfp
Molar mass unknown: MM = massunknown / molesunknown
Results and Conclusions:
I need help doing the the result for the calculation
And write a conclusion
Questions:
How much NaCl in grams is needed to lower the freezing point of 1.2 liters of water 2.5
degrees. How many grams of CaCl2 would be needed? Please make sure the answer is
correct
ΔTfp = i Kfp m
NaCl = Na+ + Cli so i=2
Kf for water = 1.86
ΔT = 2.5
M=2.5/2×1.86 = .672 moles of NaCl / kg of water
.672 x 58.5 g/kg = 39.31 g of NaCl / kg of water
39.31 x 1.2 L = 47.2 g of NaCl
ΔTfp = i Kfp m
CaCl2= i=3
Kf for water = 1.86
ΔT = 2.5
M = 2.5/3×1.86 = .448 moles CaCl2 / kg of water
.448 x 111 g/kg = 49.73 g of CaCl2 / kg of water
49.73 x 1.2 L = 59.68 g of CaCl2
Graph Unt
Und
Materials
munk
ANSIS
The Ang
the follo
Find Tmolting pare Stories
2 Calculate molality palmitic / Stearie
Find (Tm palmitic & stearie
mp
Atmank = Tm-Tmunk
2
3
acid
4
-1
-1.562
3
Munka
at mank
ky
Atma
ATme
ko me
Munt X kg Stearic acid
= moles unt
m Munh=9/mole

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