I need help writing a lab report
the format should be like this
Title and Purpose
Procedure and Observations
Data and Calculations
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.
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.
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:
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.
Part 1 Liquid
Unknown # 5 smells like gasoline
Weigh an empty 10.0 mL volumetric flask
Fill with unknown liquid.
Weigh filled volumetric flask
Mass of empty flask = 12.032 grams
Mass of full flask = 18.685 grams
Part 2 Solid
Part 2 Unknown #12 Shiny orange color
Fill a graduated cylinder with about 25 mL of
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
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
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.
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.
Laboratory 2 Preparation of Soap
All compounds are divided into one of three types, ionic, polar covalent,
and nonpolar covalent. In general, intermolecular interactions are strongest
between two different molecules when they belong to the same type. For
example, HCl interacts very strongly with water, because both are polar covalent,
however, HCl interacts very weakly with hexane, because hexane is nonpolar.
In the attempt to clean, humans have long been interested in improving
the interactions between water and nonpolar compounds. As early as the first
century AD they had developed soaps. Soaps improve the ability of water to
dissolve nonpolar compounds. Since the advent of chemistry we now
understand that they do this by having a polar end on a large nonpolar
compound. The nonpolar end dissolves nonpolar compounds, and the polar end
in turn is dissolved in water.
Figure 1 illustrates how soaps can dissolve nonpolar compounds in water.
The nonpolar ends of many soap molecules surround the nonpolar compound,
leaving the ionic tails pointing toward the water. The suspended particle is called
Picture a drop of water on a newly waxed car. The water forms a
spherical shape. Because of the strong water-water intermolecular interaction,
each water molecule tries to stay as close as possible to other water molecules.
This property is called surface tension. Soaps decrease the surface tension of
water and are also called surface-active compounds or surfactants.
The simplest soaps are made by the reaction of fat with base. Fats are
also frequently called triglycerides. Triglycerides are esters of glycerol and three
long chain carboxylic acids.
R is CH3(CH2)n
fatty acids R , R’ and R” may be the
same or different
If the triglyceride is mixed with a strong base the following reaction occurs:
O- Na+or K+
+ NaOH or KOH
O- Na+or K+
O- Na+or K+
This converts the nonpolar fat molecule to carbonate ions and glycerol.
This process is called saponification. All anions of fatty acids will act as
surfactants and form micelles. However, different fatty acids produce other
effects, which may or may not be desirable. A personal products company may
want to produce a hard soap bar with a stable lather. The fatty acid anion,
laurate, will make a hard soap bar, which cleanses, but the lather is not stable
and laurate is also somewhat drying for skin. The fatty acid anion, ricinoleate,
will be moisturizing and have a stable lather, but will not produce a hard bar or do
much cleansing. For this reason commercial soaps and detergents contain more
than just one saponicated fatty acid. The area of chemistry, which is concerned
with achieving all the desired characteristics in a product is formulations
chemistry. Table 1 gives the more common, natural fatty acids and the soap
characteristics they produce. As can be seen in Table 1, some fatty acids
contain carbon-carbon double bonds, and these are the unsaturated fats.
Table 1: Fatty Acids
Many soaps are produced from natural fats. These natural fats, aside
from appealing to consumers, are an inexpensive source of fatty acids. Table 2
gives some common natural fats and oils and their content of the common fatty
Table 2. Percent of Fatty Acids in Fats and Oils
Fat or Oil
As can be seen from Table 2, coconut oil and palm oil both contain large
amounts of lauric acid or palmitic acid. These are commonly used to provide
cleansing. Just as no one fatty acid produces a perfect bar soap, neither does
just one natural fat. Too much coconut and palm oils gives a crumbly bar. All
commercial bar-soaps use a mixture of natural fats and oils. Other ingredients
are also added. Essential oils or perfumes may be added for fragrance.
Beeswax may be added to harden soap. Milk is sometimes used in place of
water as an emollient or skin softener. Used early in the saponification process it
can cause the soap to be tan or brown in color, because the excess base
denatures the protein in the milk. In a white bar soap it is added after the initial
saponification reaction. All commercial formulations include a 1-7% excess of fat
in the final soap, this prevents the soap from drying the skin, and the soap from
being too basic. Other materials may be added to help the soap work in hard
water. The calcium and magnesium ions in hard water form a precipitate with the
carbonate ion in natural soap. This is commonly called soap scum. If the
carbonate group is changed to a sulfate group, the precipitate does not form.
These modified anions are called detergents instead of soaps.
In today’s experiment you will design and make one bar of your own
useable soap. To help you select the amount and types of fat and oils to use
Table 3 lists the maximum amount of each natural fat, which may be used in a
successful bar of soap. It also gives the amount of NaOH in mg needed to
saponificate 1.0 grams of that fat or oil. First you will decide which fats and oils
to use, then you will calculate the amount of NaOH needed to give the proper
saponification, select the fragrance and produce the soap. The soap will need to
harden for one week before it is useable.
To make one bar of soap use a total of 50 grams of fat, 19 grams of water
or cream (in place of milk) and .7 grams of scent. You will need to calculate the
amount of NaOH needed by using Table 3.
Table 3 % Maximum Usage and Saponification Values for Fats and Oils
Fat or Oil
Value (g NaOH/ 1.0 gram
Example: Design of Soap
To make a hard soap with conditioning properties, one could use a
combination of coconut oil, castor oil, olive oil and sunflower oil. The maximum
amount of castor oil is 10%, coconut oil is 50%, olive oil is 50% and sunflower oil
is 15%. To be sure it is hard I want include 50% coconut oil, and for the
conditioning 10% castor oil. If I use 10% sunflower oil that leaves 30% for olive
oil. The total amount of each oil will be:
0.50 x 50 grams = 25 grams coconut oil
0.10 x 50 grams = 5 grams castor oil
0.10 x 50 grams = 5 grams sunflower oil
0.30 x 50 grams = 15 grams olive oil
The amount of NaOH needed is
25 grams coconut oil x 0.183 g NaOH/gram coconut oil = 4.575 g
5 grams castor oil x 0.129 g NaOH/gram castor oil = 0.645 g
5 grams sunflower oil x 0.134 grams NaOH/gram sunflower oil = 0.670 g
15 grams olive oil x 0.137 grams NaOH/gram olive oil = 2.055 g
for a total of 7.945 grams NaOH. This is the total saponification value.
The amount we will use is 95% of this or 7.548 grams.
Preparation of Soap
Be sure to record all data, including the exact masses of all materials used,
and any observations in your lab notebook.
1. Design your soap. Determine which oil to use, which scents to use and
whether or not cream will be used. Calculate the amount of NaOH needed to
provide 100% saponification and then multiple this by 0.95 to insure a 5% excess
2. Weigh out the amount of NaOH needed and place it in a 100 mL
beaker. Weigh the fats and oils and combine them in a 250 mL beaker. If you
are using cream (do not use more than 9 grams of cream) in your formulation,
subtract the amount of cream in grams from the amount of distilled water, 19
grams total, you are using. Add the water to the NaOH in the 100 mL beaker and
dissolve it. Set it aside to cool.
3. Place the beaker containing the fats and oils on the hot plate. While
stirring with a glass rod, gently heat the fats and oils until they melt. If they burn
discard them in the waste container provided and start again. Once they are
warm (about 40oC) turn off the hot plate and leave the beaker on it.
4. Weigh a total of .7 grams of the essentials oils to be used to fragrance
the soap. If cream is to used, weigh the amount of cream to be used and have
5. While gently stirring the fats and oils, slowly add the NaOH solution
from step 2 into the fats and oils. Stir the mixture with the glass stir rod and/or a
magnetic stir bar until the mixture reaches trace. (Test for this by pulling the stir
rod out of the mixture. Move the rod about. As the soap falls on the surface if it
traces a visible pattern on the surface before it sinks, the soap has traced.)
6. Once trace is reached, add the fragrance compounds and any cream
to be used. Stir until trace is reached again.
7. Pore the soap into a 3”x 3” labeled weigh boat. Place it in the insulated
container to harden for one week.
Questions for discussion:
1. The fatty acids, which produce hard bars, have something in common. What
2. Saturated fats are less healthy in the diet than unsaturated fats. Based on
this information determine which fats or oils listed in table 2 are healthy and
which are not.
3. Did your bar soap have the properties you desired?
Safety and Disposal:
All waste and products can be placed in the nonhazardous waste
container or flushed down the drain.
Laboratory 3 Freezing Point Depression
Colligative 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
below the road’s temperature.
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.
The general equation for freezing point depression is:
ΔTfp = i kfp m
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,
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.
y = -0.105x + 89.378
Freezing Point is the intersection of the
y = -0.0085x + 66.2
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:
Cooling Curve 2 (Stearic Acid with Palmitic Acid)
y = -0.11x + 80.667
Freezing point is the
intersection of the two lines
y = -0.0089x + 63.694
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
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
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
Amount of NaOH needed:
25 g Coconut Oil * 0.183 g NaOH =
20 g Olive Oil * 0.137 g NaOH = 2.74 g
2.5 g Sunflower Oil * 0.134 g NaOH =
2.5 g Castor Oil * 0.129 g NaOH = 0.3225 g
Total Mass Amount of NaOH: 7.9725 grams
Total Mass Amount of 95% Saponification: 0.95 *7.9725 g = 7.574 grams
Results and Conclusions:
I predict that the
The oils were teat& Cocount oil was the only one that was slid at
* Olive, caster, and Sunflower oil were liquid in room temprature
* The oils melted and became a
* When taken off heat, mixture started to get thicker, that’s due to the
Data and Calculations:
Amount of Oils needed:
50% Coconut Oil
40% Olive Oil
S% Sunflower Oil →
5% Castor Oil
0.50 * 50g= 25$
0.40 * 50 g = 205 – 20-64
0.05 * 50g= 258
0.05 * 50g= 25$
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