CHEM108 Lab Manual
Includes Lab Reports and
Pre-Lab Assignments
Required for all Saddleback CHEM108 Classes
!
NAME: ____________________
Pre-lab #1: Introduction to Lab Techniques
Introduction to Measurements
There are numerous aspects to chemistry, but a common thread between them all is the process of
collecting data and observations through field studies or in a laboratory. It is critical to understand how to
make and read measurements, and we will focus on the basic types in this experiment: mass, length,
volume and temperature.
All quantitative measurements (involving a numerical value) are meaningless without an associated
unit. It is necessary to always include a unit with a number, and that unit is determined by the measuring
device. Since there are many different units for mass, volume, length, etc., we will look at the relation
between some common units from the English system and the metric system during lab.
The other major goal is to learn how to use, read and interpolate the basic measuring devices found in
the laboratory, such as the balances, graduated cylinders, rulers and thermometers. The concept of
significant figures will be used to aid in understanding the limitations inherent in any measurement and
later calculations involving that value.
Points to keep in Mind:
There are many basic concepts that will be carried throughout the entire semester in lab which are
briefly covered below. More details will be given as the semester progresses.
1) Safety: All safety rules must be followed at all times, with specific ones covered in the prelab lecture
for each class. Make sure you are on time! Safety goggles are required to be worn at all times while
doing experimental work in the lab. No flip-flops or open-toe shoes are to be worn in the lab. Shoes
must cover the tops of your feet.
2) Pre-lab Assignments: These are due at the beginning of lab, no exceptions. You may not perform
the lab if the prelab is not complete. You can download these from the course website and have an entire
week to ask questions if you are unsure of anything. Be familiar with the proper instructions and
calculations for the lab before you try anything!
4) Problem Solving and Calculations: All work must be shown for mathematical calculations, and all
unit conversions should be set up as unit line equations for clarity. Be neat! Always include units for any
number.
For example, to calculate the density (D) of a liquid with a measured volume (V) of 9.83 mL and a
mass (M) of 12.90 g, show the formula, insert the primary data with units, show the unrounded answer,
and then box the rounded answer to the appropriate number of significant figures with the correct units.
D = M =
V
12.90 g =
9.83 mL
1.31¦23 g
mL
------>
1.31 g/mL
If this value was to be converted into units of grams per Liter (L), use the unrounded answer in the Unit
Line Equation with the conversion factor to change the units of mL to L:
1.31¦23 g
mL
1000 mL
L
=
131¦2.3 g ------>
L
1310 g/L
or
1.31 x 103 g/L
in scientific notation
Making Measurements:
Each measuring device has its own limitations and techniques. For digital readings,
there is no estimation involved and all digits shown are recorded except for any zero
preceding a non – zero number (i.e.: 3.20 g instead of 03.20 g).
For non – digital instruments, the first step is to determine the value represented by
the increment, or the space between the smallest divisions (lines) on the measuring
device (represented by dashed lines in the figure to the right). This is most easily done
by dividing the difference between two labeled lines by the number of lines between them.
For the example shown to the right (assume it is a graduated cylinder), the difference
between the labeled lines is: 50 mL – 40 mL = 10 mL. There are 10 divisions between
these labeled lines, and so the value of the increment is 10 mL ÷ 10 = 1 mL.
50
40
WHEN USING NON-DIGITAL MEASURING DEVICES, REPORT MEASURED VALUES ONE DEMICAL
PLACE MORE PRECISE THAN THE INCREMENT.
Metric Rulers: A portion of a metric ruler is shown on the right.
The smallest division (the increment) is a tenth (0.1) of
a centimeter (cm). You must estimate the reading between these
lines to hundredth (0.01) of a cm in order to report
the measured value one decimal place more precise than the
increment. The position marked by the dashed line is read as
5.38 cm (or 5.39 cm).
5
Thermometers: All temperature measurements are made in degrees Celsius (oC)
for the thermometers used in this lab. You do not need to shake the thermometer
before use, and do not hold the bulb in your hand when making any measurements.
If the red liquid in the thermometer is segmented or the thermometer broken, turn
it in to your instructor to be fixed and get a new one.
A portion of a thermometer is shown to the right in units of oC. Note the
(increment) is one degree, and the measurement is estimated to a tenth of a
degree (10 mental divisions), which is one decimal place more precise than the
increment.
• The position marked by the dashed line is read as 33.5 oC.
Graduated Cylinders: There are two main factors to watch out for when reading
a graduated cylinder: reading from the meniscus and avoiding parallax error.
Since the liquid surface is not level (concave or convex), then a standard
must by agreed upon for consistency. This point is at the bottom of the curve,
called the meniscus. Line up a burette reading card behind the cylinder so that
the upper edge of the black line almost touches the bottom of the meniscus.
This is shown on the figure to the right.
Parallax error involves orienting your line of sight even with the meniscus.
For very large cylinders, leave them on the counter and adjust your height to
even your line of sight. For smaller cylinders, you can pick them up and hold
them like a plumb bob between your fingers in line with your eyes.
There are some tricks involved with determining whether you are oriented
properly which will be covered in lab as a demonstration.
For the example on the right, the increment is 0.1 mL (3 mL – 2 mL = 1 mL divided by
10 divisions). Measurements in this cylinder is are estimated to the nearest hundredth
of a mL (± 0.01 mL), (one decimal place more precise than the
increment). Since the meniscus falls exactly on the line, the value would be 2.60 mL.
(You must include the last zero since this cylinder is estimated to 2 decimal places).
6
40
30
3
2
Balances: We use top loader digital balances in this lab, and there are some important rules to follow so
as to not damage them. Unless instructed otherwise, report all of the numbers displayed in the digital
readout.
a) Never move a balance or turn them off.
b) Never put chemicals directly on a balance (use weighing paper or a container).
c) Clean up any spilled chemicals immediately!
Practice Problems:
For the following scales (no units), determine the size of the increment, and then estimate the
readings at the lines given to the appropriate number of digits.
Answers to the Practice problems:
1) Increment: _1 __
Reading at X: 67.4
Reading at Y: 61.0
2)
Increment: _0.1
Reading at X: 7.74
Reading at Y: 7.10
Density
Introduction:
Matter is anything that has a mass associated with it, as well as encompassing a particular volume.
The more matter there is in a certain volume (i.e.: the more mass within that volume), the more dense
that matter is. So the density of matter, D, is the mass of that matter, M, divided by its volume, V.
D = M
V
Solids normally have densities recorded in units of g/cm3 whereas liquids are g/mL. Gases are much
less dense than liquids or solids and so typically have densities given in units of g/L.
Density is constant for a particular species, unless the temperature changes. Under thermal expansion,
the volume of a species increases when it is heated or contracts when cooled. If the mass remains
constant while the volume changes upon heating, then the density changes based on the equation above.
In this experiment, the densities of various solids and liquids will be measured with the particular
techniques used explained in each individual section (i.e.: regularly shaped vs. irregularly shaped solids
vs. liquids). You will also be given an unknown liquid and are to determine its density as accurately as
possible.
The Density of a Liquid:
In determining the density of any liquid, the volume is easily measured directly from a graduated cylinder or
burette. The mass, though, must be measured indirectly through a method called “measuring by difference”.
Using this method, the mass of a dry graduated cylinder is subtracted from the mass of that cylinder filled with
a measured volume of liquid. This gives the liquid’s mass, and when divided by its volume, its density.
The Density of a Regularly Shaped Solid:
If a solid is square or rectangular, the volume can be directly measured using a ruler or other similar
measuring device. Knowing that the volume of a square or rectangle is length (l) x width (w) x height (h), and
with the mass being directly measured on a balance, then the density is easily obtained.
Other solid volumes can be determined if their equation for volume is known (i.e.: Cones, cylinders,
spheres).
The Density of an Irregularly Shaped Solid:
It can be very difficult or impossible to directly measure the volume of an irregularly shaped object (like an
egg or tooth). A common method used is to measure the volume by water displacement. In this method, the
solid object of known mass is placed in a graduated cylinder or other volume measuring device with a prerecorded volume of water in it. The increase in volume due to addition of the solid is the volume of the solid.
Prelab Questions:
These are due at the beginning of lab and must be completed to start the lab.
1) For the following scales (no units), determine the size of the increment, and then estimate
the readings at the lines given to the appropriate number of digits.
100
2) If a brick has a length of 13.77 cm, a width of 8.50 cm, and a height of 5.12 cm:
a) What is the volume of the brick?
b) If the brick has a mass of 895.3 g, what is its density?
3) If 8.15 mL of water is placed in a graduated cylinder (dry mass = 56.98 g), and the combined
mass of both the cylinder + water is 65.11 g:
a) What is the mass of the water:
b) What is the density of the water:
4) If a non-polar organic liquid with a density of 0.88 g/mL were mixed with the water from
question #3, would the water or the organic liquid float on top?
Chemistry 108 Lab #1
!
Name_____________________________
Lab #1: Introduction to Lab Techniques
INTRODUCTION
Our goals in this experiment are (1) to make some measurements using a metric ruler, (2) to learn
how to determine volumes with a graduated cylinder, and (3) to determine the density of an
unknown liquid and an unknown solid.
PROCEDURE
A. The Metric Ruler
i) On the image of the metric ruler below, draw a small arrow pointing to 2.00 cm and write “A”,
do the same and write “B" at 12.00 cm, "C" at 3.50 cm, "D" at 8.55 cm, and “E" at 10.60 cm.
NOTE: The numbers that are displayed are cm, the increment is 0.1 cm.
ii) Use a metric ruler to determine the length and width of this piece of paper. Metric rulers
are on the counter in the front of the lab. Use the correct number of decimal places base on
the ruler increments (as you did in the pre-lab).
Increment on metric ruler = ______________ cm
Length of the page = ___________cm
Do calculations to convert the length of the page to mm and meter
(show your calculations for full credit)
= ______mm
= _______m.
Width of the page = ___________cm
Do calculations to convert the width of the page to mm and meter
(show your calculations for full credit)
= ______mm
= _______m.
1!
!
Chemistry 108 Lab #1
!
B. The Graduated Cylinder
There are 3 graduated cylinders set up in the lab. Each line on the 1000mL graduated
cylinder represents ten milliliters (note that 1000 mL = 1 L). Each line on the 100mL
graduated cylinder represents one milliliter. Each line on the 10 mL graduated cylinder
represents 0.1 milliliters.
Observe the top of the liquid in the 100 mL cylinder. Note that the liquid surface is curved,
not level. The curved surface is called the meniscus. The volume is always read at the lowest
point of the meniscus. Hold the graduated cylinder so the meniscus is exactly at eye level.
Now raise and lower the graduated cylinder and observe that the volume reading changes as
the cylinder is raised and lowered. Only when your eye is at exactly the same level as the
bottom of the meniscus can you obtain an accurate volume reading. (The error introduced if
your eye is high or low is called parallax.)
Using the correct number of decimal places (as you did in the pre-lab), determine the
volume of liquid in each of the 3 cylinders and record the data below.
Graduate Cylinder
Increment of
Graduated Cylinder
Volume of Liquid
in Cylinder
1000 mL (= 1L)
100 mL
10 mL
Remember to use units whenever you write a number.
2!
!
Chemistry 108 Lab #1
!
C. Measuring the Density of a Solid and of a Liquid
i) Density of a metal cylinder
1) Each student will be given a metal cylinder, record the unknown number in the table below.
2) Weigh the metal cylinder, record the mass (in table below) to at least three places past the
decimal.
3) Determine the increment of the 100 mL graduated cylinder and record it in the table below.
4) Place about 30 ml of water in your 100 mL graduated cylinder, record the volume to one
decimal place more precise than the increment (as done in pre-lab) in the table below.
5) While holding the graduated cylinder at an angle, carefully slide your metal slug into the
graduated cylinder. The metal cylinder must be completely submerged. Record the new volume to
one decimal place more precise than the increment in the table below.
6) Calculate the volume of your metal cylinder.
7) Go to page 5 and calculate the density of your metal cylinder. Show your calculations, using
the correct number of significant figurers in the appropriate box on page 5.
Metal Cylinder: Unknown number: ……………...…._______________________
Mass of metal……………………………..…….…_____________ g
Increment on a 100 ml graduated cylinder………….. ___________mL
(Figure it out just as you did on the prelab or ruler)
Volume before submersion of metal ….…..….….______________mL
(report volume to one decimal place more precise than the increment)
Volume after submersion of metal ….…..……….______________mL
(report volume to one decimal place more precise than the increment)
Calculate the volume of the metal
(Use the data above and the “volume by displacement method” as
discussed in the powerpoint prelab introduction to calculate the
volume of the metal)
= _____________mL
3!
!
Chemistry 108 Lab #1
!
ii) Density of a liquid
1) Obtain an unknown liquid from the instructor and record the number of the unknown on the data
table below.
2) Determine the increment of the 10 mL graduated cylinder and record it in the table below. Note
that the increment is different for the 10ml graduated cylinder than it was for the 100 ml graduated
cylinder that was used for the metal slug.
3) Measure and record, in the table below, the mass of a clean, dry 10 mL graduated cylinder using
a balance. Record the mass to at least three places past the decimal.
4) Pour between 5 and 10 mL of the unknown liquid in the 10 mL graduated cylinder. Re-weigh and
record the mass, to at least three places past the decimal, in the table below.
5) Next, record the volume of the unknown liquid in the 10 mL cylinder to one decimal place
more precise than the increment. If you need to redo the density of your unknown liquid, the
cylinder must be completely dry before you weigh it. The graduated cylinder can be dried using a
rolled-up paper towel.
6) Calculate the mass of your unknown liquid.
7) Go to page 5 and calculate the density of your unknown liquid. Show your calculations, using
the correct number of significant figurers in the appropriate box on page 5.
Unknown liquid: Unknown number: …………………….______________________
Increment on a 10 ml graduated cylinder………….. ___________mL
(Figure it out just as you did on the prelab or ruler)
Mass of 10 mL graduated cylinder…..………………____________g
Mass of 10 mL graduated cylinder + liquid…………____________g
Volume of liquid ……………………………….…_____________mL
(Report volume to one decimal place more precise than the increment.)
Calculate the mass of the liquid
(Calculated using “weight by difference” method)
=____________ g
4!
!
Chemistry 108 Lab #1
!
DENSITY CALCULATIONS
In the area below, calculate the densities of the metal cylinder and the unknown liquid.
Remember: Every number in a measurement must have a number and a unit)
i) Calculation of the unknown solid’s density
Unknown solid number_______
Density of your unknown solid ___________________
Did you use the correct number of significant figures?
ii) Calculation of the unknown liquid’s density
Unknown liquid number__________
Density of your unknown liquid_______________
Did you use the correct number of significant figures?
5!
!
!
Chemistry 108 prelab
Name_________________________
Prelab #2
Coffee Cup Calorimetry
Heat is a form of energy, sometimes called thermal energy that will pass spontaneously from an
object at a high temperature to an object at a lower temperature. If the two objects are in contact
they will, given sufficient time, both reach the same temperature. Heat flow is ordinarily
measured in a device called a calorimeter. A calorimeter is simply a container with insulating
walls, made so that essentially no heat is exchanged between the contents of the calorimeter and
the surroundings. A coffee cup or thermos is an example of containers designed to prevent heat
from escaping its contents. Within the calorimeter chemical reactions may occur or heat may
pass from one part of the contents to another, but no heat flows into or out of the calorimeter
from or to the surroundings.
A. Specific Heat.
When heat energy flows into a substance, the temperature of that substance will increase. The
quantity of heat energy (Q) required to cause a temperature change in any substance is equal to
the specific heat (S) of that particular substance times the mass (m) of the substance times
temperature change (∆T), as shown in Equation 1.
Q = S x m x (∆T)
Equation (1)
The specific heat can be considered to be the amount of heat required to raise the temperature
of one gram of the substance by one degree Celsius. Amounts of heat energy are measured in
either joules or calories. To raise the temperature of 1 gram of water by 1 degree Celsius, 4.184
joules of heat must be added to the water. The specific heat of water is therefore 4.184
joules/goC. Since 4.184 joules equals one calorie, we can also say that the specific heat of water
is 1 calorie/goC. That is not a coincidence, this is how the calorie was originally defined- the
amount of energy needed to raise 1g of water by 1oC!
B. Calorimetry
The specific heat of a metal can readily be measured in a calorimeter. A calorimeter is a
container with a known heat capacity. When a heating or cooling process in a calorimeter
occurs, the amount of energy (heat) gained or lost in the process is absorbed by the calorimeter.
This energy can be calculated from equation (1) if the calorimeter’s change in temperature is
measured, since its heat capacity (S) and mass (m) are known. Very often, a calorimeter is
composed of pure water in an insulating container. If one assumes that a negligible amount of
heat escapes the insulated container and there is a negligible change in temperature of the
container during the process, then all the energy released during the process of interest goes in to
the water. Since we know the heat capacity and the mass of the water used in calorimeter and
can measure the temperature change of the water, then we can use equation (1) to calculate the
energy gained or lost by the water. The key point is that the energy gained (or lost) by the water
is equal to the energy lost (or gained) in the process (such as cooling of the metal) that occurs in
the calorimeter.
Chemistry 108
Heat Capacity Prelab
A pre-weighed amount of metal is heated to some known temperature and is then quickly
transferred into a calorimeter that contains a measured amount of water at a known temperature.
Energy (heat) flows from the metal to the water, and the two eventually equilibrate (come to the
same temperature) at some temperature between the initial temperatures of the water and metal.
Assuming that no heat is lost from the calorimeter to the surroundings, and that a negligible
amount of energy is absorbed by the calorimeter walls, the amount of energy that flows from the
metal as it cools is equal to the amount of energy absorbed by the water. In other words, the
energy that the metal loses is equal to the energy that the water gains. Since the metal loses
energy (Tfinal is less than Tinitial; therefore ∆T is negative) Qmetal is negative. The water in the
calorimeter gains energy (Tfinal is greater than Tinitial; therefore ∆T is positive) and Qwater is
positive. Since the total energy is always conserved (cannot disappear), we can write equation
(2):
Qmetal + Qwater = 0
equation (2)
Rearranging equation (2) gives (note the negative sign):
Qmetal = - Qwater
equation (3)
In this experiment we measure the mass of water in the calorimeter and mass of the unknown
metal and their initial and final temperatures. Using equation (1), the heat energy gained by the
water and lost by the metal can be written:
Qwater = Swaterx mwater x ∆Twater
equation (4)
Qmetal = Smetalx mmetal x ∆Tmetal
equation (5)
(Note that ∆Tmetal< 0 and ∆Twater > 0, since ∆T = Tfinal - Tinitial)
We can calculate Qwater from the experimental data; we measured mwater, ∆Twater, and we know
Swater= 4.184 J/goC. We know that the heat energy lost by the metal is equal (but opposite sign)
as the heat energy gained by the water. To determine the specific heat of your unknown metal
(Smetal), substitute Qmetal (in equation (5)) with – Qwater:
-Qwater = Smetalx mmetal x ∆Tmetal
equation (6)
In this equation, we know 3 of the 4 variables: we calculated Qwater from experimental data; we
measured mmetal and ∆Tmetal. We can now solve equation (6) for Smetal. Equation (6) can be
rearranged to give Smetal as a function of the known variables (Qwater, mmetal, and ∆Tmetal):
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!"#$% = !
!!!"#$%
!"#$%!× ∆!!"#$%
equation (7)
You will use this procedure to obtain the specific heat of an unknown metal. You will then
compare the specific heat of your unknown metal to a table containing values of specific heats
for several metals in order to determine the identity of your metal.
2
Chemistry 108
Heat Capacity Prelab
Prelab Questions:
1) Why can we substitute Qmetal (in equation (5)) with – Qwater?
2) Why is ∆Tmetal< 0?
3) Why is ∆Twater > 0 ?
4) Should Qmetal be positive or negative? Why?
5) Should Qwater be positive or negative? Why?
6) A metal sample weighing 45.2 g and at a temperature of 100.0°C was placed in 38.6 g of
water in a calorimeter at 25.2°C. At equilibrium the temperature of the water and metal was
32.4oC.
a. What was ∆T for the water? (∆T = Tfinal - Tinitial)
____ °C
b. What was ∆T for the metal?
_____oC
c. Using the specific heat of water (4.184 J/goC), calculate how much heat flowed into the water?
_____joules
d. Calculate the specific heat of the metal, using equation (7).
_____________joules/goC
3
Chemistry 108
Heat Capacity Prelab
4
Chemistry 108 lab
Name_________________________
Lab #2: Coffee Cup Calorimetry
INTRODUCTION
In this experiment, you will determine the specific heat for an unknown metal. The metal sample
will be heated to a high temperature (100oC) then placed into a coffee cup calorimeter containing
a known amount of water. If you can find out how much heat was gained by the water in the
calorimeter then you will know how much heat was lost by the metal. You can then calculate
and compare the specific heat of your unknown metal to known values of metal specific heats
and identify your metal.
EXPERIMENTAL
1) Set up a calorimeter. The instructor will give each student a sample of an unknown
metal. Each student must do their own unknown. Record your unknown number in
the DATA section below. (The thermometer is very expensive, so be careful when
handling it.) The calorimeter consists of two polystyrene coffee cups fitted with a
styrofoam cover (placed in a 400 ml beaker for balance). There are two holes in the cover
for a thermometer and a glass stirring rod. Do not fill the calorimeter with water yet.
Assemble the experimental setup as shown in the figure below.
400 ml beaker
2) One of the two lab partners do the following: Fill a 600 ml beaker two-thirds full of water,
add a couple of boiling chips, and begin heating it to boiling on a hotplate. Hotplates are located
in the common drawers. This boiling water will be shared by both lab partners.
3) Each person will weigh an empty, extra large test tube and stopper. Record the mass of the
empty, stoppered tube your data table (use all of the digits displayed on the balance).
Chemistry 108
Heat Capacity Lab
4) Transfer your unknown metal to the extra large test tube that you just weighed and replace
the stopper. (the boiling water bath would remove the labels from the tubes your metals came in).
Weigh your sample of unknown metal in the extra large, stoppered test tube. Record the mass of
the stoppered tube and metal in your data table. Record all of the units displayed on the
balance read-out.
• You will return your metal back to the original tube at the end of the experiment, so keep
track of your own tube and do not mix it up with your partner’s.
5) Place the loosely stoppered tube with the metal into the boiling water in the beaker. Do not
put water in the tube with the metal! Do not put the stopper tightly in to the tube or the tube
may explode as the air in the tube is heated. The water level in the beaker should be high
enough so that the top of the metal is below the water surface. Continue heating the metal in the
water for at least 15 minutes after the water begins to boil to ensure that the metal attains the
temperature of the boiling water (100.0oC). Add water to the beaker as necessary to maintain the
water level.
6) While the water is boiling, weigh the empty calorimeter (both styrofoam cups and cover only;
no water, no stirring rod, no thermometer, no 400ml beaker). Record the mass of the empty
calorimeter in your data table, record this mass using three numbers after the decimal
point.
7) Place about 30 ml of tap water in the calorimeter and weigh it again. Record the mass of the
calorimeter and water in your data table. (styrofoam cups, water, and cover only; no stirring
rod, no thermometer, no 400ml beaker) Do not be concerned that the last digit on the balance is
not stable, you are seeing the water evaporate! Because of evaporation occurring, record this
mass using three numbers after the decimal point.
8) Measure the initial temperature of the water contained in calorimeter. Note that you will need
to hold the calorimeter at an angle so that the thermometer bulb is completely under the water.
Record, to 0.1°C (one place to the right of the decimal), the temperature of the water (Tinitial
water) in the calorimeter.
9) Insert the stirrer and thermometer into the calorimeter through the 2 holes in the cover.
10) Take the test tube out of the beaker of boiling water, remove the stopper, and pour the metal
into the water in the calorimeter. Replace the calorimeter cover and stir the water/metal mixture
as best you can with the glass stirrer. Record, to 0.loC, the maximum temperature reached by
the water in the calorimeter (this is the Tfinal of the water and the metal).
11) OPTIONAL
If you have 40 more minutes before the end of lab time, if you wish, you can repeat the
experiment a second time (trial 2). Be sure to dry your metal before reusing it; this can be done
using several paper towels. Be sure to dry the metal completely or you will
introduce error to your measurements.
2
Chemistry 108
Heat Capacity Lab
WHEN FINISHED: The metal used in this experiment is to be dried with paper towels and
returned to the front counter in the test tube in which you obtained it.
DATA
Your unknown number ________________________
Trial 1
Trial 2
Mass of empty test tube and stopper
g
g
g
g
g
g
g
g
°C
°C
100.0 °C
100.0 °C
°C
°C
Mass of stoppered test tube plus metal
Mass of empty calorimeter
Mass of calorimeter and water
Initial temperature of water in calorimeter
(Tinitial water)
Initial temperature of metal (assume 100.0°C)
(Tinitial metal)
Equilibrium temperature of metal and water in the calorimeter
(Tfinal water = Tfinal metal)
CALCULATIONS
Be sure to use the correct number of significant figures and to use units with every
number you write!!!!
1) Calculate ∆Tmetal
Trial 1
Trial 2
Trial 1
Trial 2
2) Calculate ∆Twater
3
Chemistry 108
Heat Capacity Lab
3) Calculate the mass of the water in the calorimeter (mwater).
(Think about how you get this from the data table values….you have the mass of the calorimeter
with the water in it and the mass of the empty calorimeter……)
Trial 1
Trial 2 (OPTINAL)
4) Calculate the mass of the metal (mmetal).
(Think about how you get this from the data table values….you have the mass of the metal in the
tube and the mass of the empty tube……)
Trial 1
Trial 2
5) Calculate the heat energy gained by the water (Qwater).
Trial 1
Trial 2
Did you use the correct number of sig. figs.?
6) Knowing that the heat gained by the water is equal and opposite the heat lost by the metal,
use equation (6) or (7) in the prelab to calculate the specific heat of your metal.
Trial 1
Trial 2
Did you use the correct number of sig. figs.?
7) If you did two trials, take the average of the specific heats calculated in the two trials (above).
4
Chemistry 108
Heat Capacity Lab
Average specific heat of metal (if you did 2 trials only)……..________________________
(Did you use the correct number of significant figures and the correct units?)
CONCLUSION
Re-write your unknown number here: ____________________
Your unknown metal is one of the metals in the table below. Use this table of specific heats to
determine the identity of your metal. Choose the metal with the specific heat that is closest to
your experimental value.
Metal
Aluminum
Bismuth
Tin
Nickel
Specific Heat
(J/goC)
0.900
0.126
0.226
0.443
My metal is (circle one):
Aluminum
Bismuth
Tin
Nickel
5
!
Chemistry 108 Lab #3
Name_____________________________
Prelab #3 Gases:
Percent Yield of Hydrogen Gas from Magnesium
and Hydrochloric Acid
Introduction
For chemical reactions involving gases, gas volume measurements provide a convenient means of
determining stoichiometric relationships. A gaseous product is collected in a long, thin graduated glass
tube, called a eudiometer, by displacement of a liquid, usually water. Magnesium reacts with hydrochloric
acid, producing hydrogen gas:
Mg (s) + 2HCl (aq)
MgCl2 (aq) + H2 (g)
Note: for every mole of Mg (s) that is reacted, one mole of H2(g) is produced. If we know the
mass of Mg(s) we can convert to moles of Mg(s). Then, since we get 1 mole of H2(g) for every mole of
Mg(s), we can predict how many moles of H2(g) could be made (theoretical yield). We use an excess of
HCl so that we would react all the Mg(s) before we used all of the HCl.
Eudiometer Tube Set-up
Reaction Set-up (after inverting eudiometer tube)
1
Chemistry 108 Lab #3
When the magnesium reacts with the acid, the evolved hydrogen gas is collected by water
displacement and its volume is measured. The temperature of the gas is taken to be the same as the
temperature of the water it is in contact with because, given a sufficient amount of time, the two will reach
thermal equilibrium. The level of water in the eudiometer is adjusted so that it is equal to the level of water
outside the eudiometer. This insures that the pressure in the eudiometer is equal to the prevailing
atmospheric pressure. Because the hydrogen gas was collected above water, and water has a significant
vapor pressure, to get the pressure of pure hydrogen (dry hydrogen), we must subtract the vapor pressure of
water. The pressure of the dry hydrogen gas (PH2) is calculated from Dalton's Law of Partial Pressures:
Ptotal = PH2 + PH2O
so
PH2 = Ptotal - PH2O
where Ptotal (the pressure in the eudiometer) is atmospheric pressure, and PH2O (the water vapor
pressure) is the pressure exerted by water vapor that has evaporated into the eudiometer. We will get the
vapor pressure of water from the table below of vapor pressure vs. temperature.
The number of moles of hydrogen gas collected can then be calculated from the ideal gas law:
(n = # moles H2)
n = PV
RT
(Use PH2 here, not Ptotal)
This will give you the experimental # moles of hydrogen gas collected. The theoretical # of moles
of H2(g) can be calculated by converting the mass of Mg to moles Mg, and understanding that we get 1
mole of H2 from every mole of Mg(s). From the theoretical yield and the experimental yield, one can
calculate the percent yield:
Percent Yield = experimental # moles H2
theoretical #mole H2
100%
2
Chemistry 108 Lab #3
PRELAB QUESTIONS:
1. Suppose you did this experiment and obtained the following data. The initial mass of magnesium metal
was 0.0426 g. The volume of the gas produced in the eudiometer was 41.75 mL, the atmospheric pressure
is 763.2 torr, and the water temperature is 24.1 C. As if you carried out this experiment and obtained the
following data, fill in the data table and do the calculations below.
DATA TABLE:
Mass of Magnesium Metal
.
g
Volume of Gas
.
mL
Temperature of Gas (assumed to be the same temp. as the water)
.
C
Atmospheric Pressure
.
torr
Water Vapor Pressure
(at the above temperature, see table on last page)
.
torr
CALCULATIONS:
1. Theoretical (calculated) yield of H2 gas (# moles H2).
a) convert mass Mg to #moles Mg
b) Convert # moles Mg reacted to moles of H2 that could be produced. (1 mole H2 is produced for every
1 mole Mg reacted- this comes from the balanced chemical equation):
Mg (s) + 2HCl (aq)
MgCl2 (aq) + H2 (g)
# moles H2 = _____________________________________ (theoretical yield of H2)
2. Experimental yield of H2 gas (# moles H2).
a) Determine pressure of dry H2 (PH2) by subtracting the vapor pressure of water from the total
pressure,
Ptotal = PH2 + PH2O
so
PH2 = Ptotal - PH2O
where Ptotal (the pressure in the eudiometer) is atmospheric pressure, and PH2O (the water
vapor pressure) is the pressure exerted by water vapor that has evaporated into the
eudiometer. Use the table provided on the previous page to find the vapor pressure of water
as a function of temperature.
PH2 = _____________________torr
b) Convert this pressure from torr to atm (760.0 torr = 1atm)
PH2 = _____________________atm
3
Chemistry 108 Lab #3
c) Use the Ideal Gas Equation to calculate the number of moles (n) of H2 that you produced
in your experiment (experimental yield). Make sure to use the correct units so that they
match the units in the gas constant (R).
3. Calculate the Percent Yield.
% Yield = experimental # moles H2
theoretical #mole H2
100%
% Yield =______________________________
OTHER PRE-LAB
QUESTIONS
1. When the volume of gas is measured in a eudiometer, the water levels are the same on the
inside and outside of the eudiometer. Why do we need to do this?
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
2. If a gas is collected by water displacement in a eudiometer, the atmospheric pressure is
known, and the water vapor pressure is known, what equation will allow you to calculate the
pressure of the dry gas?
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
4
Chemistry 108 Lab #3
Name_____________________________
Lab # 3: Gases
Percent Yield of Hydrogen Gas from
Magnesium and Hydrochloric Acid
Introduction
For chemical reactions involving gases, gas volume measurements provide a convenient means of
determining stoichiometric relationships. A gaseous product is collected in a long, thin graduated glass
tube, called a eudiometer, by displacement of a liquid, usually water. Magnesium reacts with hydrochloric
acid, producing hydrogen gas:
Mg (s) + 2HCl (aq)
MgCl2 (aq) + H2 (g)
Note: for every mole of Mg (s) that is reacted, one mole of H2(g) is produced. If we know the
mass of Mg(s) we can convert to moles of Mg(s). Then, since we get 1 mole of H2(g) for every mole of
Mg(s), we can predict how many moles of H2(g) could be made (theoretical yield). We use an excess of
HCl so that we would react all the Mg(s) before we used all of the HCl.
When the magnesium reacts with the acid, the evolved hydrogen gas is collected by water
displacement and its volume is measured. The temperature of the gas is taken to be the same as the
temperature of the water it is in contact with because, given a sufficient amount of time, the two will reach
thermal equilibrium. The level of water in the eudiometer is adjusted so that it is equal to the level of water
outside the eudiometer. This insures that the pressure in the eudiometer is equal to the prevailing
atmospheric pressure. Because the hydrogen gas was collected above water, and water has a significant
vapor pressure, to get the pressure of pure hydrogen (dry hydrogen), we must subtract the vapor pressure of
water. The pressure of the dry hydrogen gas (PH2) is calculated from Dalton's Law of Partial Pressures:
Ptotal = PH2 + PH2O
so
PH2 = Ptotal - PH2O
where Ptotal (the pressure in the eudiometer) is atmospheric pressure, and PH2O (the water vapor
pressure) is the pressure exerted by water vapor that has evaporated into the eudiometer. We will get the
vapor pressure of water from the table below of vapor pressure vs. temperature.
1
Chemistry 108 Lab #3
The number of moles of hydrogen gas collected can then be calculated from the ideal gas law:
(n= # moles H2)
n = PV
RT
(Use PH2 here, not Ptotal)
This will give you the experimental # moles of hydrogen gas collected. The theoretical # of moles
of H2(g) can be calculated by converting the mass of Mg to moles Mg, and understanding that we get 1
mole of H2 from every mole of Mg(s). From the theoretical yield and the experimental yield, one can
calculate the percent yield:
Percent Yield = experimental # moles H2
theoretical #mole H2
100%
PROCEDURE:
1.Fill the largest beaker in your drawer (400 ml or a 600mL beaker) about 2/3
full of water and allow it to sit on the base of a ring stand so that the
temperature of the water may adjust to room temperature. Place a double buret
clamp on the ring stand well above the beaker.
2.Obtain a 4-5 cm length of magnesium ribbon from the back counter of the
lab room. With sand paper, sand the magnesium ribbon until it is shiny.
Determine its mass on the side-loading balance, and record its mass in your
data table. Your magnesium should have a mass no larger than 0.0450 g. Roll
the magnesium ribbon into a loose coil. Obtain a piece of thread 25 cm in
length, and tie it to one end of the magnesium ribbon in such a way that all the
loops of coil are tied together (see Figure 1).
3.Obtain a eudiometer from the stockroom. Always
carry a eudiometer in a vertical position. The eudiometer
will contain water to keep dust out. Empty out the water
Figure 1
into your sink and temporarily attach it to the buret
clamp, open end up.
4.Measure out 10 mL of hydrochloric acid in a graduated cylinder and pour it into
your eudiometer. Remove the eudiometer from the buret clamp, hold it on a slight
slant, and add enough water to the eudiometer to fill it completely. Try to mix the
water and the acid as little as possible. Reattach the eudiometer to the buret clamp,
open end up (see Figure 2).
5.Obtain a one-hole rubber stopper from the back counter. Take your magnesium coil
and lower it into the water of the eudiometer to a depth of about 5 cm. Have the
thread attached to the coil hang over the lip and out of the eudiometer. Insert the onehole rubber stopper into the eudiometer so the thread is held firmly against the edge,
and when water squirts out of the hole in the stopper, cover the hole firmly with your
thumb (see Figure 1).
Figure 2
2
Chemistry 108 Lab #3
6. PUT ON RUBBER GLOVES. Taking care that no air enters, remove the eudiometer from the buret
clamp, invert it, and place its open end underwater in the beaker. Re-clamp the eudiometer to the buret
clamp so that the bottom of the eudiometer is about 1 cm below the surface of the water in the beaker. The
acid will flow down the eudiometer and react with the magnesium.
7.When the magnesium has disappeared
entirely and the reaction has stopped,
tap the tube with your finger to dislodge
any bubbles you see attached to the side
of the eudiometer. Measure the
temperature of the water in your beaker;
this will be the temperature of the
hydrogen gas in the eudiometer. Record
this value, to the nearest 0.1oC, in
your data table. Because your
thermometer reads to a tenth of a degree
Celsius, add 273.2 when converting to
Kelvin.
8.Place your finger over the hole in the
stopper and remove the eudiometer
from the beaker. Lower the eudiometer
into the leveling tank and remove your
finger. Raise or lower the eudiometer
until the water level inside the
eudiometer is the same as the water
level in the leveling tank. Read the
volume of gas in the eudiometer and
record it in your data table. Record
the volume to the nearest 0.01 mL.
Figure 3. Inverted eudiometer illustrating gas
collection and water displacement.
9.Your instructor will write the
barometer for today’s atmospheric
pressure on the chalk board. Record this
value in your data table. The water
vapor pressure can be found in the table
on page 2. Record this value for the
vapor pressure of water in your data
table.
10.When finished with the experiment, clean the eudiometer with water, dry the outside, fill it with
deionized water, and return it to the stockroom.
3
Chemistry 108 Lab #3
DATA TABLE:
Mass of Magnesium Metal
(must be less than 0.0450 g)
=
g
Volume of Gas
=
mL
Temperature of Gas = Temperature of the water =
=
C
Atmospheric Pressure
=
torr
Water Vapor Pressure
(at the above temperature, see table on last page)
=
torr
CALCULATIONS:
1. Theoretical (calculated) yield of H2 gas (# moles H2).
a) Convert mass Mg to #moles Mg
b) Convert # moles Mg reacted to moles of H2 that could be produced (1 mole H2 is produced for every 1
mole Mg reacted- this comes from the balanced chemical equation):
Mg (s) + 2HCl (aq)
MgCl2 (aq) + H2 (g)
# moles H2 = _____________________________________ (theoretical yield of H2)
2. Experimental yield of H2 gas (# moles H2).
a) Determine pressure of dry H2 (PH2) by subtracting the vapor pressure of water from the total
pressure,
Ptotal = PH2 + PH2O
so
PH2 = Ptotal - PH2O
where Ptotal (the pressure in the eudiometer) is atmospheric pressure, and PH2O (the water
vapor pressure) is the pressure exerted by water vapor that has evaporated into the
eudiometer. Use the table provided on the last page to find the vapor pressure of water as a
function of temperature.
PH2 = _____________________torr
b) Convert this pressure from torr to atm (760.0 torr = 1atm)
PH2 = _____________________atm
4
Chemistry 108 Lab #3
c) Use the Ideal Gas Equation to calculate the number of moles (n) of H2 that you produced
in your experiment (experimental yield). Make sure to use the correct units so that they
match the units in the gas constant (R).
3. Calculate the Percent Yield.
% Yield = experimental # moles H2
theoretical #mole H2
100%
% Yield =______________________________
4. What are the possible sources of error in this experiment?
5
!
Chemistry 108
Chemical Reactions
Prelab 4
Pre-Lab #4: Chemical Reactions
Many chemical reactions can be placed into one of two categories: oxidation-reduction reactions and
double-replacement reactions.
Oxidation-reduction reactions are ones in which electrons are transferred from one species to another.
There are four types of oxidation-reduction reactions that we will investigate: synthesis, decomposition,
single-replacement, and combustion.
A) SYNTHESIS REACTIONS
A synthesis reaction is one in which a single compound is formed from two or more substances:
A + B → AB
An example of a synthesis reaction (one type of oxidation-reduction reaction) is the reaction that occurs
between sodium metal and oxygen gas (O2):
Na(s) + O2(g) → Na2O(s)
The equation above is not balanced. Balance this equation:
_______Na(s) +
________O2(g)
→
_________Na2O(s)
As stated earlier, oxidation-reduction reactions are ones in which electrons are transferred from one
species to another. Let’s see how this transfer of electrons occurred in our reaction of sodium with
oxygen. First, we must remember that the total charge of any pure element or compound is always
ZERO! This fact will help you to determine the charge of each atom in a compound. Na(s) and O2(g)
are pure elements (not combined with other elements) therefore the charge of each of the atoms in a piece
of pure sodium metal and a sample of pure oxygen gas is equal to ZERO. Next, let’s consider the
charges on the Na and O in the Na2O(s) product. We once again must understand that the total charge of
any pure element or compound is always ZERO! Since sodium oxide (Na2O(s)) is a compound, it has a
total charge = ZERO. We also know that sodium oxide is an ionic compound, and that although the total
charge is ZERO, the sodium cations and oxygen anions are charged particles.
What is the charge of an oxide ion? _________________
What is the charge of a sodium ion? ________________
Note that ions combine in a ratio such that the total charge of a compound = ZERO, that is why sodium
oxide has the formula Na2O. This discussion of charge can be found in detail in chapter 3, it was
repeated here as a review.
Now, back to our discussion as to why this synthesis reaction is classified as an oxidation-reduction
reaction. Let us consider how electrons are removed from one reactant and transferred to the other. We
will do so by considering what happens to the charge of sodium and oxygen as they are converted from
reactants to products.
1
Chemistry 108
Chemical Reactions
Prelab 4
Let’s consider Na first. In the reactant, Na exists as Na(s) and has a charge of ZERO (charges are shown
below as superscripts). In the product, Na has a charge of 1+.
Na0 (sodium has 0 charge in Na2O) → Na+ (sodium has a +1 charge in Na2O)
In this chemical reaction, sodium went from ZERO charge to 1+ charge…. THIS IS ONLY DONE BY
LOSING AN ELECTRON! When a species loses electron(s), we call that oxidation.
Na0 → Na+ + eWhere does the electron go? It must get transferred to the oxygen! Let’s consider the charge on oxygen
as it is converted from reactant to product. In the reactant, oxygen exists as O2 and has a charge of ZERO
(charges are shown below as superscripts). In the product, O has a charge of 2-.
.
O0 (oxygen has 0 charge in O2) → O2- (oxygen has a 2- charge in Na2O)
We see in this chemical reaction, oxygen went from ZERO charge to 2-…. THIS IS ONLY DONE BY
GAINING ELECTRONS! When a species gains electron(s), we call that reduction.
O0 + 2e- → O2The fact that oxygen gains 2 electrons and sodium only loses 1 electron is accounted for in the balanced
chemical equation, note that 4 sodium atoms react with 2 oxygen atoms (O2)….making the electron
donation ration equal 2:1, two sodium atoms are required to reduce each oxygen atom.
Practice Problems:
1) Predict the product and balance the following synthesis reaction:
______ Li(s) +
_______ O2(g)
→
_________________________
What is the charge of the lithium in the reactant Li(s)? ________________
What is the charge of the lithium ion in the product? ________________
Did lithium gain or lose electron(s) in this reaction: _______________ if so, how many? ______
Was lithium oxidized or reduced? ________________
What is the charge of the oxygen in the reactant O2(g) ? _________________
What is the charge of the oxide ions in the product? _________________
Did oxygen gain or lose electron(s) in this reaction: ______________ if so, how many? __________
Was oxygen oxidized or reduced? ________________
2
Chemistry 108
Chemical Reactions
Prelab 4
2) Predict the product and balance the following synthesis reaction:
Mg(s) +
Cl2(g)
→
_________________________
What is the charge of the magnesium in the reactant Mg(s)? ________________
What is the charge of the magnesium ion in the product? ________________
Did magnesium gain or lose electron(s) in this reaction: _______________ if so, how many? _____
Was magnesium oxidized or reduced? ________________
What is the charge of the chlorine in the reactant Cl2(g) ? _________________
What is the charge of the chloride ion in the product? _________________
Did chlorine gain or lose electron(s) in this reaction: _______________ if so, how many? ________
Was chlorine oxidized or reduced? ________________
B) DECOMPOSITON REACTIONS
A decomposition reaction is an oxidation-reduction reaction in which a single reactant breaks down
into two or more substances:
AB → A + B
An example of a decomposition reaction is the thermal (heat induced) decomposition of mercury(II)
oxide:
HgO (s) → Hg (l) + O2 (g)
(not balanced, you will balance this reaction in the problem below)
Note that the key to identifying a decomposition reaction is that one reactant species is converted to two
or more products species. In our example, we start the reaction with just one compound present, HgO (s),
after the reaction there are two different chemical species present, Hg (l) and O2 (g).
3
Chemistry 108
Chemical Reactions
Prelab 4
Practice Problems:
3) Balance our example decomposition reaction:
________ HgO (s) → _______ Hg (l) + __________ O2 (g)
What is the charge of the mercury ion in the reactant (HgO (s)) ? ________________
What is the charge of the mercury in the product (Hg (l))? ________________
Did mercury gain or lose electron(s) in this reaction: _______________ if so, how many? ______
Was mercury oxidized or reduced? ________________
What is the charge of the oxygen ion in the reactant (HgO (s))? _________________
What is the charge of the oxygen in the product (O2 (g))? _________________
Did oxygen gain or lose electron(s) in this reaction: ______________ if so, how many? __________
Was oxygen oxidized or reduced? ________________
C) SINGLE-REPLACEMENT REACTIONS
A single-replacement reaction is an oxidation-reduction reaction in which an element replaces another
element from a compound:
A + BX → AX + B
For example, copper metal (Cu(s)) will replace the silver ion (Ag) in silver nitrate:
Cu(s) + AgNO3 (aq) → Cu(NO3)2 (aq) + Ag (s)
(not balanced, you will balance this reaction in the problem below)
Practice Problems:
4) Balance our example single-replacement reaction:
______ Cu(s) + _____AgNO3 (aq) → _____Cu(NO3)2 (aq) + _____Ag(s)
What is the charge of the copper in the reactant Cu(s) ? ________________
What is the charge of the copper in the product (Cu(NO3)2 )? ________________
Did copper gain or lose electron(s) in this reaction: _______________ if so, how many? ______
4
Chemistry 108
Chemical Reactions
Prelab 4
Was copper oxidized or reduced? ________________
What is the charge of the silver in the reactant (AgNO3)? _________________
What is the charge of the silver in the product (Ag(s)) ? _________________
Did silver gain or lose electron(s) in this reaction: ______________ if so, how many? __________
Was silver oxidized or reduced? ________________
D) COMBUSTION REACTIONS
A combustion reaction is an oxidation reduction reaction in which a compound reacts with oxygen. In
the case of organic compounds, complete combustion will always result in the formation of CO2 (g) and
H2O (g). Incomplete combustion of organic molecules often results in the formation of C(s), often in the
form of small particles of pure carbon called “soot” and other small organic molecules like carbon
monoxide. Soot can be thought of as tiny pieces of charcoal. In chemistry 108, we only focus on
complete combustion of organic molecules to form the products CO2 (g) and H2O (g). As an example, let
us consider the combustion of pentane:
Write in the products and balance the reaction for the complete combustion of pentane:
C5H12 (l) + O2(g)
→
__________
+ ____________
In part A of the prelab, we learned how to identify an oxidation reduction reaction by understanding how
electrons are transferred from one species to another. We were able to see which species was oxidized
and which species was reduced by keeping track of the charges of atoms in reactants and products.
Seeing how the charges of atoms changed in an inorganic chemical reaction allows us to determine how
electrons were transferred, which species gained electrons, and which species lost electrons. In the case
of the oxidation and reduction of organic compounds, that process is not as easily applied, so we learned
a couple of rules (in chapter 6) to determine if a species is oxidized or reduced:
Oxidation and Reduction of Organic Molecules
Method for determining oxidation or reduction in organic compounds:
An atom in a molecule is oxidized if it:
• gains oxygen (is attached to more oxygen atoms in the product than in the reactant) OR
• loses hydrogen (is attached to fewer hydrogen atoms in the product than in the reactant)
An atom in a molecule is reduced if it:
• loses oxygen (is attached to fewer oxygen atoms) OR
• gains hydrogen (is attached to more hydrogen atoms)
Let’s apply these rules to our example of the combustion of pentane by asking: is carbon oxidized or is
it reduced in this reaction?
5
Chemistry 108
•
Chemical Reactions
Prelab 4
When the carbon in C5H12 is converted to CO2, carbon is oxidized– not only does it lose hydrogen but it
also gains oxygen. Carbons are bound to hydrogens in pentane (reactant) and are bound to oxygen in
CO2 (product).
What is reduced? Where did the electrons that carbon lost go to?
•
O2 is reduced – each oxygen in O2“lost” an oxygen in the formation of water. The two oxygen atoms in
O2 that were bound to each other are no longer bound to an oxygen in the product. In addition, each
oxygen gained 2 hydrogens in the formation of water.
Practice Problems:
5) In aerobic organisms, in the absence of sufficient oxygen, the pyruvate made in glycolysis is not able
to enter the citric acid cycle and is converted to lactate (using hydrogen supplied by a molecule called
nicotinamide adenine dinucleotide, NADH):
Is this an oxidation reduction reaction? _____________, if so,
is pyruvate being oxidized or reduced in the reaction? ________________
Is this a combustion reaction?__________________
E) DOUBLE REPLACEMENT REACTIONS
Double replacement reactions are ones in which two substances in solution switch partners. They are
not oxidation-reduction reactions as were the first 4 reactions discussed. There are two types of double
replacement reactions that we will investigate: precipitation reactions and gas forming reactions.
A precipitation reaction is one in which, when the partners of the two reactant compounds are switched,
one of the products is insoluble in water:
AX + BY → AY + BX
(one of the products, AY or BX is insoluble in water = solid = precipitate)
6
Chemistry 108
Chemical Reactions
Prelab 4
A gas forming reaction is one in which, when the partners of the two reactant compounds are switched,
the products include water, an ionic compound, and a gas. The only gas forming reaction that we are
concerned with in chemistry 108 is the following reaction:
HCl (aq) + NaHCO3 → H2O(l) + CO2(g) + NaCl(aq)
•
See your chapter 7 notes for more information on precipitation reactions and gas forming reactions.
Practice Problems:
6) For each double replacement reaction below, determine if and what type of reaction will occur
(precipitation, gas forming, or no reaction), determine the appropriate products for the reaction and
then write a complete balanced chemical equation. For those which do not react, indicate this by
writing "No Reaction." Hint: See chapter 7 lecture notes on double replacement reactions.
REACTION:
REACTION TYPE
Sodium bromide + silver nitrate…………………………………………____________________
Iron (III) chloride + sodium carbonate……………………………….______________________
Copper (II) sulfate + ammonium hydroxide…………………………______________________
Sodium bicarbonate + hydrochloric acid (HCl)……………………._______________________
In this experiment each of these six types of reactions will be performed. The evidence that reactions are
occurring may be (1) a gas is produced (2) a precipitate (a solid produced when two solutions are mixed)
is formed, (3) a color change is observed, (4) a temperature change is noted, or (5) a flame is produced.
7
Chemistry 108
Chemical Reactions
Prelab 4
8
Chemistry 108
Chemical Reactions
lab
Name__________________________________
Lab #4: Chemical Reactions
Many chemical reactions can be placed into one of two categories: oxidation-reduction reactions and
double replacement reactions.
Oxidation-reduction reactions are ones in which electrons are transferred from one species to another.
There are four types of oxidation-reduction reactions that we will investigate: synthesis, decomposition,
single replacement, and combustion. There are two types of double replacement reactions that we will
investigate: precipitation reactions and gas forming reactions. In this experiment, these six types of
reactions will be performed. The evidence that reactions are occurring may be (1) a gas is produced (2) a
precipitate (a solid produced when two solutions are mixed) is formed, (3) a color change is observed, (4) a
temperature change is noted, or (5) a flame is produced.
The purpose of this lab is to give the learner hands-on experience in classification, observation, and
analysis of various types of chemical reactions.
PROCEDURE
SAFETY NOTE: WEAR GLOVES AND GOGGLES FOR THE ENTIRE EXPERIMENT.
PART A - SYNTHESIS REACTIONS
Obtain a piece of magnesium ribbon. If the ribbon is dull, sand it with a piece of sand paper. In order not to
damage the bench top, perform the next step of the experiment over your evaporating dish (an evaporating
dish looks like a small, white ceramic bowl and is located in your locker drawer). Holding one end of the
magnesium ribbon with your tongs, ignite the other end in the burner flame. Look for just a moment at
the burning magnesium; do not stare at the magnesium while it is burning because of the intensity of the
light. Record any evidence of chemical change in the Part A Observation Data Table. NOTE: The
evidence that reactions are occurring may be (1) a gas is produced (2) a precipitate (a solid produced when
two solutions are mixed) is formed, (3) a color change is observed, (4) a temperature change is noted, or (5)
a flame is produced. Dispose of the ash from this part of the experiment in the trashcan.
PART B - DECOMPOSITION REACTIONS
Obtain a plastic cup with two platinum wires fused into it and fill the container with the catalystcontaining water (the water used for this reaction contains an acid catalyst) until the water is about 1cm
from the top of the container. Next, fill an extra-small test tube completely with the catalyst-containing
water. Wearing a disposable glove, cover the top of the test tube with your finger, so that there are no
bubbles in it, invert it into the water in the container, and position it over one of the platinum wires. The
extra-small test tubes are the smallest test tubes in your drawer. Repeat the same procedure with a second
extra-small test tube, positioning it over the second platinum wire. Obtain a 9-volt battery, and connect it to
the battery clip on the wires sticking out of the bottom of the container. Keep the battery connected for two
minutes. Record any evidence of chemical change in the Part B Observation Data Table. The reactant
is water; think about what the identity of the two products could be. If you need more information, review
the decomposition reaction discussion in your prelab. Dispose of the solution from this part of the
experiment down the sink.
1
Chemistry 108
Chemical Reactions
lab
PART C – SINGLE REPLACEMENT REACTIONS
Part C-1. Place 3 mL of silver nitrate solution in a medium test tube. Add a piece of copper metal and
observe the test tubes for several minutes. Record any evidence of chemical change in the Part C-1
Observation Data Table.
IMPORTANT: Pour the solution into the largest beaker from your lab drawer. This beaker will be used
for other chemical waste throughout the lab. Keep the waste beaker on your lab bench. At the end of the
experiment, the entire contents of your waste beaker will be disposed of in the waste bottle in the fume
hood near the instructor’s desk. Clean the test tube.
Part C-2. Place 3 mL of 6M hydrochloric acid (HCl) into a medium test tube. Add a small piece of
magnesium metal (labeled "magnesium turnings") to the test tube. Record any evidence of chemical
change in the Part C-2 Observation Data Table. If a gas is produced, hold your thumb over the mouth of
the test tube for about one minute, light a wooden splint with a Bunsen burner, then remove your thumb
and hold the burning splint to the mouth of the test tube. H2 a combustible gas, is produced in the reaction.
In this reaction, magnesium replaces the hydrogen in HCl. Place the solution in the waste beaker on your
lab bench. Clean the test tube.
Part C-3. Place 3 mL of deionized water into a medium test tube, which is resting in your test tube rack.
Using the forceps provided, add a small piece of calcium metal to the test tube.
CAUTION: Do not touch calcium metal, and handle with care.
Record any evidence of chemical change in the Part C-3 Observation Data Table. Obtain a piece of
red litmus paper. Dip a stirring rod in the solution and touching it to the piece of red litmus paper. Red
litmus paper turning blue indicates that a product of the chemical reaction is an ionic compound where the
anion is hydroxide (OH-). In this reaction, the calcium replaces one of the hydrogens in H2O. Place the
solution in a waste beaker on your lab bench. Clean the test tube.
PART D – COMBUSTION REACTIONS
Light your laboratory burner to burn the methane gas coming from the gas outlet at your lab station.
Remove any flammable material from your lab bench. Holding the burner by its base, turn the burner over
and have the flame come in contact with the lab bench for 2 or 3 seconds. Place the burner back on the lab
bench and turn it off. Record your observations of the product seen on the lab bench in the Part D
Observation Data Table.
2
Chemistry 108
Chemical Reactions
lab
PARTS E and F : DOUBLE REPLACEMENT REACTIONS
PART E – PRECIPITATION REACTIONS
Place about 2 mL of mercury(II) nitrate solution into a clean, medium test tube, and place about 2 mL of
aluminum nitrate solution into a second clean, medium test tube. Add about 2 mL of potassium iodide
solution to both of the test tubes. Record any evidence of chemical change in the Part E-1 and E-2
Observation Data Tables. If a precipitate forms, write the balanced equations in the Part E–1 or Part
E–2 sections of the Reaction Table. If no reactions occurred, write NR (NR means No Reaction). Place both
solutions in the waste beaker on your lab bench.
Place about 2 mL of mercury(II) nitrate solution into a clean, medium test tube, and place about 2 mL of
aluminum nitrate solution into a second clean, medium test tube. Add about 2 mL of sodium phosphate
solution to both. Record any evidence of chemical change in the Part E-3 and E-4 Observation Data
Tables. If a precipitate forms, write the balanced equations in the Part E–3 or Part E–4 sections of the
Reaction Table. If no reactions occurred, write NR. Place both solutions in a waste beaker on your lab
bench.
PART F – GAS FORMING REACTIONS
Place about 0.50 grams of solid sodium bicarbonate (NaHCO3) in the bottom of a large test tube. Add
several drops of 6M HCl. Record any evidence of chemical change in the Part F Observation Data
Table. The contents of the tube can be washed down the sink with water.
Empty the entire contents of your waste beaker in the waste bottle in the fume hood near the instructor’s
desk.
COMPLETE THE REACTION TABLE ON THE LAST PAGE OF THE HANDOUT.
Write the balanced equations for each of these reactions in the Reaction Table. Remember to write the
phase [(g),(l),(s), or (aq)] after each reactant or product for all of the chemical equations in your
Reaction Table.
THE REACTION TABLE CAN BE DONE AT HOME. TURN THE LAB AND PRELAB IN AT THE
BEGINNING OF THE NEXT LAB PERIOD.
3
Chemistry 108
Chemical Reactions
lab
DATA TABLE
PART A OBSERVATIONS
EVIDENCE OF REACTION
Magnesium + Oxygen
PART B OBSERVATIONS
EVIDENCE OF REACTION
Water
PART C-1 OBSERVATIONS
EVIDENCE OF REACTION
Silver Nitrate + Copper
PART C-2 OBSERVATIONS
EVIDENCE OF REACTION
Hydrochloric Acid + Magnesium
PART C-3 OBSERVATIONS
EVIDENCE OF REACTION
Water + Calcium
PART D OBSERVATIONS
EVIDENCE OF REACTION
4
Chemistry 108
Chemical Reactions
lab
Methane + Oxygen
PART E-1 OBSERVATIONS
EVIDENCE OF REACTION
Mercury (II) Nitrate + Potassium Iodide
PART E-2 OBSERVATIONS
EVIDENCE OF REACTION
Aluminum Nitrate + Potassium Iodide
PART E-3 OBSERVATIONS
EVIDENCE OF REACTION
Mercury (II) Nitrate + Sodium Phosphate
PART E-4 OBSERVATIONS
EVIDENCE OF REACTION
Aluminum Nitrate + Sodium Phosphate
PART F OBSERVATIONS
EVIDENCE OF REACTION
Hydrochloric Acid (HCl) + Sodium Bicarbonate
5
Chemistry 108
Chemical Reactions
lab
REACTION TABLE:
Write a balanced chemical equation for each of these reactions. Remember to write the phase
[(g),(l),(s), or (aq)] after each reactant and product.
PART A
Hint: One of the reactants is O2, and the product is an ionic compound.
PART B
Note: The only reactant is water; the acid is just a catalyst and not a reactant or product. Since the only
reactant is water, think about what the identity of the two products are for a decomposition reaction.
PART C-1
Hints: (1) See the single replacement discussion and examples in your pre-lab, and (2) the copper is being
oxidized to Cu2+.
PART C-2
PART C-3
In this reaction, the calcium replaces one of the hydrogens in H2O.
PART D
PART E-1 Mercury (II) Nitrate + Potassium Iodide
PART E-2 Aluminum Nitrate + Potassium Iodide
PART E-3 Mercury (II) Nitrate + Sodium Phosphate
PART E-4 Aluminum Nitrate + Sodium Phosphate
PART F HCl + Sodium Bicarbonate
6
Chemistry 108
Name_______________________________
Lab #5 Prelab:
EXTRACTION AND SEPARATION OF PLANT PIGMENTS
Purpose of the lab:
The purpose of this lab activity is for the student to learn about extraction and chemical separation
technology. Specifically, the student will learn how to do a liquid phase-extraction and Thin Layer
Chromatography in order to separate a mixture of molecules.
Method:
In the first step of this lab, the student will remove some of the water from spinach using a
methanol rinse. In the second step, liquid-phase extraction will be used to extract pigment molecules from
dried, pulverized spinach. In the third and final step, the pigment molecules will be separated from each
other using Thin Layer Chromatography.
Introduction
Chromatography is a useful analytical technique that allows various components of a mixture to be
separated based on their polarity and/or size. There many different types of chromatographic separation
apparatuses and methods. We will use a technique called thin-layer chromatography (TLC). The thinlayer chromatographic surfaces that we will use in this experiment are plastic slides (TLC slides) with a
coating of very small porous silica (SiO2) particles. The particles are so small that when they are not
adhered to a surface, they behave like dust.
A small drop of the mixture to be separated is applied (spotted) near one end of the TLC slide. The
spotted end of the TLC slide is then dipped into a developing solvent (see figure 1a), called the mobile
phase¸ which flows up the TLC surface by capillary action. As the developing solvent flows up the TLC
surface, it can carry along the components of the mixture. Each component of the spotted mixture will
move upward at a different rate.
The fact that different molecules move along the chromatographic surface/solvent interface at
different rates is the key feature of chromatographic separation. Think of the component molecules as
constantly moving back-and-forth from being adsorbed to the TLC surface to being carried upward with
the mobile phase solvent. In the case of TLC, the more soluble the component is in the solvent, the faster
Chemistry 108
Plant Pigments
it travels up the surface. If it is not very soluble, it will remain adsorbed on the TLC surface (stationary
phase) longer and therefore travel at a slower rate. In addition to the component’s affinity for the solvent,
the component’s affinity for the stationary phase also plays a part in the rate at which the component
moves. If there is a strong affinity between a component molecule and the stationary phase surface, the
molecule will move upward more slowly. The affinity to the surface is determined by (1) non covalent
interactions such as dipole-dipole forces and (2), in cases of porous stationary phases, the size of the
molecule. If a molecule has a size such that it is easily “stuck” in a pore (hole), it will move upward more
slowly since it takes more time to become “un-stuck”.
The rates of flow are measure in terms of Rf (retardation factor) values. An Rf is the relative
distance that a sample component has moved relative to the distance moved by the mobile phase solvent.
The following illustration will demonstrate how the Rf is calculated. Rf is measured by dividing the
distance the component traveled by the distance the solvent traveled. Therefore, an Rf value can never be
greater than 1. The Rf value for a particular component is characteristic for that component in that
particular solvent. Therefore, it will always be the same (considering that the mobile and stationary phases
are the same) and can be identified in other mixtures
Rf = Distance component traveled
Distance solvent traveled
Example: Calculation of TLC Rf values for a sample
containing 4 components:
Solute
Distance
Traveled
(cm)
Rf value
A
2.0 cm
2.0 cm = 0.33
6.0 cm
B
3.0 cm
3.0 cm = 0.50
6.0 cm
3.5cm
3.5 cm = 0.58
6.0 cm
5.5cm
5.5 cm = 0.92
6.0 cm
C
D
Chemistry 108
Plant Pigments
Prelab Questions
1) Consider the following TLC chromatograph. Calculate the Rf factors for each of the
corresponding spots.
Solute
Distance
Traveled
(cm)
A
3.10 cm
B
2.58 cm
C
4.20 cm
D
5.08 cm
Rf value
2) In the first step of this lab you will crush the wash the spinach with methanol. Crushing
breaks the plant cells open and the methanol wash removes much of the water from the
spinach. Make a drawing showing how methanol molecules (CH3OH, methyl alcohol) are
attracted to water molecules (and vice versa). Name the attraction and explain why it is a
relatively strong intermolecular attraction.
Chemistry 108
Plant Pigment Lab
In the second step of the lab, we will extract the pigment molecules in a technique called
liquid-phase extraction. In this step you will separate the hydrophobic plant pigment
molecules from other hydrophilic component molecules and solids. This is done by placing
the pulverized spinach in a flask that contains an extremely hydrophobic liquid called
“petroleum ether” and a hydrophilic liquid. These two liquids will not mix and form two
layers because one is hydrophobic and the other hydrophilic. When the flask is shaken
vigorously, then left to settle so that the two liquids separate into layers again, the spinach
pigment molecules will be extracted into the hydrophobic petroleum ether layer. These
various pigment molecules that are extracted into the petroleum ether layer will be separated
from each other in a final TLC Chromatography step of the lab described earlier.
3) Petroleum ether contains molecules like CH3CH2CH2CH2CH2CH3 and
CH3C(CH3)2CH2CH(CH3)CH2CH3. (Note: Although the name implies that petroleum ether
is in the ether family, it is not; it is a mixture of non-aromatic hydrocarbons. The name is
historical from when “ether” was the same as “spirits” and had to do with the fact that the
hydrocarbon mixture had a high vapor pressure.)
Look at the structure of β-carotene (one of our plant pigments) below and
thoroughly explain why β-carotene is much more soluble in petroleum ether than in water.
Do not simply say that “like dissolves like” (although it is true), name and explain the
nature of the intermolecular force that is involved here.
4
Chemistry*108*lab**
Name_________________________*
Lab #5:
EXTRACTION AND SEPARATION OF PLANT PIGMENTS
Purpose of the lab:
The purpose of this lab activity is for the student to learn about extraction and chemical separation
technology. Specifically, the student will learn how to do a liquid phase-extraction and Thin Layer
Chromatography in order to separate a mixture of molecules.
Method:
In the first step of this lab, the student will remove some of the water from spinach using a
methanol rinse. In the second step, liquid-phase extraction will be used to extract pigment molecules
from the partially dehydrated pulverized spinach. In the third and final step, the pigment molecules will
be separated from each other using thin layer chromatography.
PROCEDURE
1. Vigorously crush about 10 spinach leafs with a mortar and pestle for 30 seconds.
2. Add 30 mL of methanol and continue to crush the spinach-methanol mixture for 30 seconds. Pour off
as much of the methanol extract as possible and discard it in the waste container in the hood. It is ok
if a small amount of crushed spinach is lost in the waste.
3. Locate the nearest fire extinguisher. There is a fire extinguisher near three of the four corners of the
lab, so one should be nearby. Pick it up, study it, and decide exactly what you would do if you needed
to use it. If you have any questions, ask the instructor. Replace the fire extinguisher on its mount on
the wall.
4. Working under your bench-top hood to contain the vapors, add 25 mL of methanol and 35 mL of
petroleum ether to the spinach in the mortar (Caution: petroleum ether is extremely flammable).
5. Crush for two minutes, and filter the
liquid with vacuum as follows:
• Use a 250 ml side-arm flask,
Buchner funnel, and a piece of #3filter paper in the funnel. Clamp the
vacuum flask to a ring-stand.
Clamps are in the common drawers.
• Turn on the vacuum then pour the
spinach-liquid mixture from the
mortar into the funnel.
Buchner Funnel
Line to
vacuum
6. Place the liquid that you collected in to a
clean 125 mL Erlenmeyer flask.
250 ml
side-arm
flask
Figure 1. Filtration apparatus.
Chemistry*108*
Plant*Pigment*Lab*
*
7. Add 25 mL of water to the liquid in the 125 ml Erlenmeyer flask from step 6. Stopper, then shake
slowly (try not to get foaming) for 30 seconds, and then allow the two liquid layers to separate. At
this point you should have a dark, intensely green, clear layer on top and a lighter green, cloudy layer
on the bottom. (If you don't have a dark green layer on top, add 10 mL of petroleum ether and
stopper, then shake again. If that doesn't work, consult with the instructor.)
8. Use a Pasteur pipette (located on the counter) to transfer about 1 ml of the dark green, top layer to
your smallest, Erlenmeyer flask. Be sure the flask is dry; do not pre-rinse it.
9. Obtain two chromatography slides. You will be working in pairs. Each student should spot one
slide. Spot using the capillary tubes provided and develop the slides in the capped jars as instructed
below.
9.1. The spots must be small and dark. Spot the slide about 3 times (in the same place): ¾ of an inch
from the bottom of the slide, waiting about 30 seconds between each spotting so the previous
spot dries. DRAW A LINE NEXT TO YOUR SPOT (NOT TOUCHING THE SPOT) SO YOU
WILL BE ABLE TO CALCULATE THE Rf VALUES LATER.
9.2. Use the developing troughs at the end of the lab benches. Place both partners’ slides in the
trough, the developing solvent will work best if you keep the lid on trough. Note your slot
number so you do not exchange your TLC slide with someone else’s.
9.3. Stop the development when the solvent comes to within 1.5 inches of the top of the TLC slide or
when you can clearly see good separation of 5 spots. Immediately draw a line near the top of
the slide where the solvent reached for calculation of Rf values.
DISCUSSION/DATA
The visible spots are, in order of decreasing
Rf values, due to β-carotene (orange to
orange-yellow with the highest Rf value),
pheophytin a & b (grey, the pheopthytin b
spot is often too weak to be seen),
chlorophyll a & b (green to blue-green), and
the xanthophylls (yellow with the lowest Rf
value). Xanthophyll is a general term that
encompasses many different compounds.
They are molecules of !-carotene that have
been oxidized, so they contain many OH
groups. Pheophytin a & b are identical to
chlorophyll a & b except the Mg ions are
replaced by hydrogen ions.
Chlorophyll(a(
Chlorophyll(b((
2*
*
Chemistry*108*
Plant*Pigment*Lab*
*
Measure the distance traveled by each spot and by the solvent. START ALL MEASUREMENT
FROM WHERE THE SLIDE WAS ORIGNALLEY SPOTTED. Make a drawing in the box (below and
to right) of your TLC slide, identify the spots, and record the distance traveled for the solvent and for
each pigment molecule in the data table below. The pigment molecules are ordered from the top of
your slide to the bottom, with spot #1 being the spot that travels the furthest (β-carotene). Note that you
may not be able to see spot # 3, #7, #8 because of the small amounts present in most spinach samples.
DATA:
If you do not see all these spots, just write “not visible “for those not seen. Start all distance
measurements, including the distance the solvent traveled from your original spot.
Spot
solvent
1
top spot,
highest
on slide
2
3
4
5
6
7
8
Molecule
distance the solvent traveled
from your original spot
β-carotene
color = yellow/orange
Distance
traveled(cm)
Start your Rf measurements where you
originally spotted the slide. Then
measure to the center of each spot.
Drawing*of*developed*slide*goes*here:*
Pheophytin a
color = grey
Pheophytin b
color = grey if visible
Chlorophyll a
color = blue/green
Chlorophyll b
color = green
less intense than chlor. a
Xanthophyll l
color = yellow
Xanthophyll 2
color = yellow if visible
Xanthophyll 3
color = yellow if visible
3*
*
Chemistry*108*
Plant*Pigment*Lab*
*
Calculate the Rf values:
Spot
Molecule
1
top spot,
highest
on slide
β-carotene
color = yellow/orange
Rf value
note: pay attention to significant
figures and units
2
Pheophytin a
color = grey
3
Pheophytin b
color = grey if visible
4
Chlorophyll a
color = blue/green
5
Calculation
Chlorophyll b
color = green
less intense than chlor. a
6
Xanthophyll l
color = yellow
7
Xanthophyll 2
color = yellow if visible
8
Xanthophyll 3
color = yellow if visible
QUESTION
1) Silica gel (the coating on your TLC slide) is very polar and the developing solvent is quite non-polar.
Based on your TLC slide data (Rf values), which of the plant pigments that you see on your slide is the
most polar? Explain.
4*
*
Chemistry 108
Carboxylic Acids Pre-lab
Lab#6 Prelab
CARBOXYLIC ACIDS
Background:
O
||
Carboxylic acids contain the carboxyl functional group:
* carboxyl + hydroxyl groups.
–C–O–H
Also written as – COOH
These are found in amino acids, proteins, fatty acids and oils, and as intermediates of carbohydrate
metabolism. Four to twenty C-atom-chain carboxylic acids are obtained from animal and plant fats or oils and
so are often called fatty acids.
Nomenclature:
IUPAC Names:
• Take the base name from longest C atom chain containing the carboxyl group, with that
acting as C #1. Drop the "e" ending from the parent alkane name and add the suffix "–oic
acid".
O
||
ex:
ethanoic acid :
CH3 — C — OH
Common Names:
• Use the suffix " - oic acid" added to the Latin/Greek name origin.
Examples: formic acid (methanoic acid) and acetic acid (ethanoic acid).
Acidity of Carboxylic Acids:
• The hydroxyl H atom is weakly acidic (ionizes ~ 0.5 %).
–
•
•
RCOOH
+ H2O (l)
RCOO (aq) + H3O+ (aq)
RCOO– is the carboxylate ion, and is much more soluble than the parent acid because it
has a charge (it is an ion).
• Nomenclature of carboxylate ions is done by changing the "–ic acid" suffix of the parent
acid to "–ate ion".
Note that the reaction is reversible, the carboxylate ion can be converted to the carboxylic
acid form under acidic conditions (neutralization of a base with an acid):
–
RCOO (aq) +
–
H3O+ (aq)
RCOOH
+ H2O (l)
Or
RCOO (aq) + HCl (aq)
RCOOH
+ Cl- (aq)
• Likewise, The acid form can be converted to the base form (carboxylate ion) under basic
conditions (neutralization of an acid with a base):
RCOOH
+ OH- (aq)
–
RCOO (aq) +
H2O (l)
1
Chemistry 108
Carboxylic Acids Pre-lab
Preparation (synthesis) of Carboxylic Acids:
1. Oxidation of an Aldehyde: Addition of an oxygen atom to an aldehyde creates a carboxylic
acid.
O
O
||
||
R–C –H +
O
R–C–O–H
• The source of the oxygen atom can be from oxygen gas (O2), or another oxygen containing
compound such as the permanganate ion (MnO4-).
2. Oxidation of a primary (1o) alcohol: Oxidation of a primary alcohol creates a carboxylic acid.
R – C– O – H
•
O
||
R–C–O–H
+ O
The source of the oxygen atom is from a strong oxidizing agent such as the dichromate ion
(Cr2O72-).
Carboxylic Acid Reactions:
1) Neutralization: A carboxylate salt is created upon addition of a strong base to a carboxylic.
• Note: neutralization of a long chain carboxylic acid (fatty acid) is called Saponification.
RCOOH + NaOH
RCOO– Na+ +
H2O
2) Esterification: Addition of an alcohol with a strong acid catalyst creates an ester.
O
||
R–C–O–H
H+
+ H – O – R'
O
||
R – C – O – R' + H2O
3) Amide Formation: Amides are formed by reaction of an amine with a carboxylic acid. This is an
endothermic reaction (requires energy). In biological systems, where it is not possible to use heat, a
specific enzyme is needed
a) Reaction of carboxylic acids and ammonia (NH3)
O
O
||
||
R’ – C – OH + H – N– H
R’ – C – NH2
|
H
+ H2O
2
Chemistry 108
Carboxylic Acids Pre-lab
b) Reaction of carboxylic acids and primary (1o) amines (NH2R)
O
||
R’ – C – OH + H – N – R
|
H
O
||
R’ – C – N–R
|
H
+ H2O
c) Reaction of carboxylic acids and secondary (2o) amines (NHR2)
O
||
R’ – C – OH + H – N– R’
|
R”
O
||
R’ – C – N –R’
|
R”
+ H2O
d) Reaction of carboxylic acids and tertiary (3o) amines (HR3)
NO REACTION occurs with tertiary (3o) amines
Solubility of Carboxylic Acids and their Conjugates Bases:
Acid Form
Base Form
O
||
R – C – OH
Carboxylic Acid
O
||
R – C – OCarboxylate ion
If R has less than 5 carbons:
Soluble (aq)
If R has less than 12 carbons:
Soluble (aq)
If R has 5 or more carbons:
Insoluble (s)
If R has 12 or more carbons:
Amphipathic
Forms monolayers or micelles
For the acid (HA) and its conjugate base (A-),
HA + H2O
A- + H3O+
• There is more HA when the pH is lower than the pKa.
• There is more A- when the pH is higher than the pKa.
• There are equal amounts of HA and A- when the pH = pKa.
3
Chemistry 108
Carboxylic Acids Pre-lab
Questions:
1) Name the following carboxylic acids:
a) CH3CH2CH2COOH:
IUPAC: ___________________________
b) CH3CHCH2CH2COOH:
|
CH3
IUPAC: ___________________________
2) Draw line bond structures for the following carboxylic acids:
a) 2, 3 – dibromohexanoic acid
b) 3 – methylpentanoic acid:
3) Write the reaction when acetic acid reacts with 1-butanol (CH3CH2 CH2 CH2―OH)
4) Write the reaction when propanoic acid reacts with ethanamine:
5) Write the reaction when the butanoate ion is neutralized by hydrochloric acid:
4
Chemistry 108 lab
Name_________________________
Lab #6: CARBOXYLIC ACIDS LAB
PART I: Preparation of Carboxylic Acids
(a) Oxidation of an Aldehyde by Oxygen from the Air:
Benzaldehyde is an aromatic aldehyde with a familiar odor. On a clean, dry watch glass, place 3
or 4 drops of the pure liquid benzaldehyde (this is 100% benzaldehyde, it is not a solution). Spread
the benzaldehyde out on the glass so as to expose more of it to the oxygen in the air and allow it to
remain for most of the lab period. Near the end of the period, examine the material on the watch
glass. Record your observations describing fully the starting material and any changes that occur.
Write a balanced equation for the reaction of benzaldehyde and oxygen gas to produce the
carboxylic acid product:
Move on to part (b) and the rest of the experiment, then come back to this question:
Based your observation of the product, how do you know a chemical reaction occurred?
______________________________________________________________________________
______________________________________________________________________________
(b) Oxidation of an Aldehyde by Potassium Permanganate Solution:
Place about 2 drops of benzaldehyde in a small test tube and add about 4 drops of 6 M sodium
hydroxide solution and about 12 drops of 0.1 M potassium permanganate (KMnO4) solution. MIX
contents of the test tube. Allow the mixture to react for 10 minutes. (You can do a model from Part
3 while you are waiting.)
Describe your observations:
Note: The oxidizing agent is the permanganate ion (MnO4-). MnO4- is reduced to MnO2, a brown
precipitate, in the reaction.
Chemistry 108
Carboxylic Acids Lab
Does the acid-form (benzoic acid) or the base-form
(benzoate ion) dominate the equilibrium in this tube?
Since the solution is very basic, the pH is much
greater than 7 and the pKa of benzoic acid is about 5,
you should know if the acid- or base-form dominates
based on what we learned in lecture (chapter 9 and 10,
comparing pH and pKa). Draw the line bond
structure of that predominant form in the box to
the right.
Obtain a piece of filter paper from the supplied for the experiment. Folding it properly, first, place
it in your funnel and then moisten the filter paper with DI water. Put your funnel in a clean large
test tube and pour the brown suspension from the small test tube into the filter/funnel. Wait until
at least a few drops of liquid pass through the funnel and into the large test tube. Then, REMOVE
THE FUNNEL, and add 4 drops of 12 M HCl to the filtrate (liquid that went through the filter
paper and is now in the large test tube). The HCl will make the solution acidic. Describe what you
see. If you do not see a change occur when you add the HCl, see the instructor.
Observations:
Think about what you observed based on the solubility of carboxylic acids vs. carboxylate ions!
Write a balanced chemical equation for the reaction that occurred when you added the HCl.
• By adding the acid, you are converting your product formed (you drew it in the box in the
upper right hand corner of this page) in the oxidation reaction to its acid form (reacting with
the HCl, HINT: see the prelab neutralization equations).
Equation:
2
Chemistry 108
Carboxylic Acids Lab
Part II: Solubility of a Carboxylic Acid and Its Salt in Water:
Neutralization of a Carboxylic Acid:
The solubility of carboxylic acids in water is very low when they contain more than 5 or 6
carbon atoms. Knowing this, predict the water solubility of benzoic acid and explain your
answer.
Circle one:
predicted to be soluble
or
predicted to be insoluble
Explain_______________________________________________________________________
Now test your prediction by adding benzoic acid crystals (0.01g to 0.02 g) to 2 ml of water in a
large test tube. Shake well and record your observations (what observation tells you if it was
soluble or not):
Test the pH of the mixture in your test tube by touching the tip of your stirring rod to the suspension
in the test tube and then touching a spot of liquid from the tip of the stirring rod onto a piece of red
litmus paper, and then place another spot of the liquid onto blue litmus paper. The spot on the
red litmus paper should look red (indicating that the solution is acidic, pH<7). The spot on the blue
litmus paper should look also look red (indicating that the solution is acidic, pH<7).
Record the spot color on the red litmus paper: __________________________________
Record the spot color on the blue litmus paper: _________________________________
Note that although carboxylic acids are “weak acids”, they do produce enough H3O+ to make the
solution acid.
Write a balanced chemical equation for the acid reacting with H2O (see prelab):
•
Is the acid form or the base form predominant at pH < 5? __________________.
Now add, one drop at a time, 1 M NaOH solution, stirring the contents of the test tube after each
addition of the NaOH and testing the pH of the solution using the red litmus paper.
Continue adding NaOH solution in this manner until the solution pH causes red litmus paper to turn
blue, indicating that the pH >7.
3
Chemistry 108
Carboxylic Acids Lab
Check the contents of the test tube and record your observations (is the solid still present?):
Write the chemical equation describing this reaction (see prelab the reaction of a carboxylic acid and
a base):
Explain why you no longer see a solid present.
What kind of reaction is this (see prelab)? _____________________________ .
4
Chemistry 108
Carboxylic Acids Lab
Part III: Models
Examine the numbered models set out by the instructor in the balance room. Draw condensed
structural formulas for the compounds represented. Write the organic family name of the molecule
(for example: alkane, alkene, amine, carboxylic acid etc.….) Also write the systematic name for each
molecule. Atoms Colors: Black = carbon, White = hydrogen, Red = oxygen, Blue = nitrogen
MODEL #
CONDENSED STRUCTURAL FORMULA
FAMILY
SYSTEMATIC NAME
1
2
3
4
For this model, do not draw the condensed form
here; instead, draw a side view. See you Chapter 4
Lecture Notes if you do not recall side-view
representations.
5
6
7
5
Chemistry 108
Carboxylic Acids Lab
Post-Lab Questions: Draw the products for the following reactions:
O
H2
C
+
C
H3C
Acid
Catalyst
CH3
OH
HO
CH
C
H2
CH3
Hint: Check your pre- lab
CH3CH2CH2COOH + NaOH
CH3CH2CH2COO- + HCl
6
NAME: ____________________
Lab #7: Carbohydrates Prelab
Introduction:
Of the three main sources of energy and raw materials for cellular function, carbohydrates are
normally ingested in the largest portions, typically 60 to 70% of our diet. They are also the primary fuel,
or source of energy, for the cells in our body. The main categories of carbohydrates are sugars, starches
and cellulose.
Carbohydrates were originally thought to be hydrates (containing water) of carbon, and this was the
derivation of the name. It turned out that they were not hydrates, but did contain the elements C, H and
O expressed in the ratio: Cx(H2O)y. They are now usually classified as polyhydroxy (many OH groups)
aldehydes or ketones.
The simple sugars are the monosaccharides, with the three main dietary ones being glucose (blood
sugar), galactose and fructose (fruit sugar). They all have the general formula C6H12O6, and are isomers
that exist in both open and closed forms. The open chain forms are regularly shown as Fischer
projections and the closed ring forms as Haworth diagrams. These are shown below for the
monosaccharides.
H
O
C
HC
HO
CH2OH
OH
O
CH
HC
OH
H2C
OH
H
OH
OH
OH
α-D-Glucose
H
OH
CH2OH
Glucose
D-Glucose
aldohexose)
(an(an
aldohexose)
HC
HO
HO
CH2OH
C
HO
CH2OH
HC
OH
H2C
OH
CH2OH
Fructose
D-Fructose
ketohexose)
(an(aketohexose)
OH
O
OH
3
4
OH
Β-D-Fructose
beta-D-Fructose
2
CH2OH
1
OH
CH2OH
OH
O
CH
OH
OH
CH
H2C
O
CH
O
C
H
OH
CH2OH
Galactose
D-Galactose
(an aldohexose)
OH
β-D-Galactose
The position of the OH group on the anomeric carbon on the far right of the ring determines whether
the sugar is an alpha (α) or a beta (β) sugar. The OH group points down below the ring in α sugars and
up above the ring in β sugars. These have important consequences later for digestion purposes.
The other sugars are the disaccharides, formed when two monosaccharides join together, releasing a
molecule of water in the process. If the O-link (glycosidic bond) between the monosaccharide units is
in the down position, it is an α O-bridge, or a β O-bridge if pointing up. The three main dietary
disaccharides are maltose, sucrose and lactose. Maltose (malt sugar) forms when two glucose units join
with an α O-bridge. Sucrose (table sugar) is a glucose with fructose, and lactose (milk sugar) is glucose
with a galactose. To digest these disaccharides, the necessary digestive enzymes, maltase, sucrose, and
lactase, must be present in the small intestine.
If many glucose monosaccharides are joined together, then a polysaccharide forms. Sometimes
called complex carbohydrates, the polysaccharides include plant and animal starch, as well as cellulose.
Starch is the main energy storage form for plants, usually in the seeds. The glucose units are joined
by α O-bridges and are therefore digestible by humans with the aid of the amylase enzyme in saliva and
pancreatic juices. Starch consists of about 20% amylose and 80 % amylopectin. Amylose is linear and
hundreds to several thousand glucose units long, whereas amylopectin is branched and up to 100,000
glucose units.
Glycogen is the main energy storage form for animals, mainly in their liver and muscles. It is often
called animal starch. It consists of up to 1 million glucose units joined by α O-bridges and is even more
highly branched than amylopectin.
Cellulose is the main structural component of plant cell walls. It consists of several thousand glucose
units in a linear fashion just like amylose, but has β O-bridges and therefore is not digestible by humans.
Some ruminant animals contain special bacteria in their stomachs that allow digestion of cellulose β
bridges. For humans, it acts as dietary fiber, or “roughage”, and cleans out the digestive tract.
In this experiment we will become acquainted with some of the common tests/reactions for
carbohydrates. We will look at the reactions of fructose, sucrose, cellulose, lactose, and starch. We
will use these reactions to determine which of the five sugars is present in an unknown.
Part I: Benedict's Test
Benedict’s reagent is a copper compound that will oxidize only aldehyde groups (aldoses) and not
alcohols. If you consider cyclic forms of carbohydrates, hemiacetals give positive tests while acetals
give negative tests. The reason for this is that the cyclic form interconverts (is in equilibrium) with the
linear form that contains an aldehyde!
A sugar that reacts with Benedict’s
solution is called a reducing sugar
since it reduces the ion Cu2+ ! Cu+
Part II: Seliwanoff’s Test
The Seliwanoff reagent contains hydrochloric acids which converts fructose to Shydroxymethylfurfural. The reagent also contains resorcinol which reacts with the Shydroxymethylfurfural to give a red color. Complex carbohydrates which contain fructose units can also
give a positive test. Aldohexoses react similarly, but more slowly.
Part III: Iodine test
One of the two molecular components of starch are shaped like very long spiral staircases, inside of
which is just enough space to accommodate iodine molecules. The blue color arises when the electrons
of the entrapped iodine molecules interact with the electrons of the starch molecule and the resulting
complex absorbs visible light (appears dark).
Questions:
1) Why do some disaccharides give positive Benedict’s tests while other disaccharides give negative
tests?
2) Fructose is not an aldose. Why does it give a positive Benedict's test?
3) Is the aldehyde group on a sugar oxidized or reduced in the Benedict’s test? Explain.
4) A carbohydrate that gives a positive Benedict's test is said to be a reducing sugar. Why? What's being
reduced?
5) Cu2+ in Benedicts tests is a weak oxidizing agent. Why don't we use a stronger oxidizing agent?
6) Starch is made up of two components.
a) What are the two components, and how do the two components differ from each other?
b) Which of the two components of starch is similar to glycogen? How does the component differ from
glycogen?
c) Draw a section of a amylose molecule containing two glucose units. Label the linkages (for example
β (1→4)) and place an asterisk (*) next a carbon where branching would occur if it was a amylopectin
molecule.
Chemistry 108 lab
Name_________________________
Lab #7: CARBOHYDRATES LAB
INTRODUCTION
In this experiment we will become acquainted with some of the common carbohydrate reactions of
fructose, sucrose, cellulose, lactose, and starch. We will use these reactions to determine which of the
five is present in an unknown.
PROCEDURE
Part I: Benedict's Test
Benedict’s reagent is a copper compound that will oxidize only aldehyde groups (aldoses) and not
alcohols. If you consider cyclic forms of carbohydrates, hemiacetals give positive tests while acetals give
negative tests. The reason for this is that the cyclic form interconverts (is in equilibrium) with the linear
form that contains an aldehyde, for example:
1) Add about 1-2 inches of water and 3 boiling chips to a 600 mL beaker (it is the largest beaker in your
drawer). Place the beaker and water on a hot plate, turn the hot plate on, and begin to boil the water.
2) Label 7 medium size test tubes (from your drawer) for each of the 5 sugars, your unknown, and
your partner’s unknown. DO NOT ADD THE SUGAR SOLUTIONS UNTIL LATER
3) Using the metering dispenser, place 2 mL of Benedict's solution in EACH OF THE 7 labeled,
medium size test tubes and heat in gently-boiling water bath for 2 minutes. NOTE: groups of 3 students
will have 8 tubes.
4) Remove the tube that is labeled for the fructose solution. Add 10 drops of the 5% fructose solution to
the Benedict's solution, mix thoroughly, and set the tube in the boiling water bath for 60 more seconds.
Remove the tube from the water bath and check for a color change. A color change from clear blue to
cloudy green, yellow or to brick red indicates the carbohydrate is a reducing sugar. The brick green,
yellow, or red color is copper(I) oxide.
5) Repeat step 4 for your other 6 sample test tubes, using sucrose, cellulose, starch, lactose and each
partner’s unknown instead of fructose. When using dropper to add your unknown solutions to the tubes,
be sure to rinse the droppers after each partner’s unknown to prevent cross-contamination of
unknowns.
Record your results in the DATA TABLE.
Chemistry 108
Carbohydrates Lab
Part II: Seliwanoff’s Test
The Seliwanoff reagent contains hydrochloric acids which converts fructose to S-hydroxymethylfurfural.
The reagent also contains resorcinol which reacts with the S-hydroxymethylfurfural to give a red color.
Complex carbohydrates which contain fructose units can also give a positive test. Aldohexoses react
similarly, but more slowly.
1) Maintain the boiling water bath used in Part II, you may need to add more water because of
evaporation.
2) Put 5 mL of freshly prepared Seliwanoff’s reagent in a medium test tube, add 10 drops of 5%
fructose solution and place in a boiling water bath. (Caution: This is a strongly acidic solution.) A red
color within two minutes is indicative of a ketohexose.
• Record your results in the DATA TABLE.
3) Repeat step 2 for each of your other six carbohydrate solutions (instead of fructose).
Part III: Iodine test
Some of the starch molecules are shaped like very long spiral staircases, inside of which is just enough
space to accommodate iodine molecules. The blue color arises when the electrons of the entrapped iodine
molecules interact with the electrons of the starch molecule and the resulting complex absorbs visible
light (appears dark). Be sure to rinse your graduated cylinder between sugar additions so you do not
cross contaminate your samples!
1) Add 2 mL of the 5% fructose solution to a small test tube (from your drawer), add two drops of
iodine solution.
• A deep blue color is a positive test for starch.
• Record your results in the DATA TABLE.
2) Repeat for the other 6 carbohydrate solutions (instead of fructose).
2
Chemistry 108
Carbohydrates Lab
DATA TABLE (write (+) for a positive test and (–) for a negative test)
Benedict’s
Test
Seliwanoff’s
Test
Iodine
Test
Fructose
Sucrose
Cellulose
Starch
Lactose
Unknown #
Conclusion: Unknown Number _________________
is __________________________.
3
Chemistry 108
Carbohydrates Lab
Part IV: Representing 3D structures with two-dimensional drawings.
(a) Fischer projection formulas of open-chain monosaccharides:
•
Select model #1 of an open-chain monosaccharide. If the model comes apart when you are
manipulating it and you are not sure how to put it back together, bring it to the instructor!
Does the model represent an aldose or ketose? ______________
Does the model represent a triose, tetrose, etc.?______________
Does the model represent a L- or a D- sugar?______________
Classify this monosaccharide as a combination of the previous 3 answers ___________________.
Determine and draw the Fischer projection for this model below, being sure to view each C atom so that the
–H and – OH groups are facing up towards you (above the plane).
Replace model #1 and repeat the same procedure for models #2-4.
MODEL# 2
MODEL #3
Classification: ________________
Classification: ________________
Fischer Projection
Fischer Projection
MODEL#4
Classification: ________________
Fischer Projection
(b) Haworth projection formulas of cyclic monosaccharides.
View model #5 of the cyclic monosaccharide and answer the following:
•
Is the ring structure a pyranose (6 sides) or a furanose (5 sides): ___________________
•
Is the structure α or β?__________________
Draw the Haworth projection for the compound represented by this model.
4
Chemistry 108 lab
Name_________________________
Lab #8: Protein, Triglycerides, and Esters Lab
Purposes of the Lab:
In Part 1, you make cheese by separating the caseins from the other components of milk. In Part 2, you
will make soap using the saponification of triglycerides reaction. In Part 3, you will synthesize two esters
that smell very good!
Safety Precautions:
1. Use goggles at all times.
2. Wear gloves at all times.
3. Never put room temperature glass on a hot hotplate surface.
4. Do not eat the cheese that you make; you did not use sanitary glassware!
5. Do not use the soap that you make; it was made in lab glassware that may have contained other
chemical residues.
Part 1 –Cheese Making: Separating the Caseins from Milk
Procedure:
1. Pour 30 mL of nonfat milk into your 100mL beaker.
2. Place the 100 mL beaker on your hotplate. Turn the heat to the 50% (half way) position.
3. Monitor the temperature until the milk reaches 40oC. Swirl the milk every few minutes. See the
instructor’s apparatus for proper thermometer installation.
4. When the temperature of the milk has reached 40oC, TURN OFF THE HOTPLATE. Remove the
beaker from the hotplate. Add glacial acetic acid drop-by-drop to the warm milk, while slowly stirring
with your glass stir-rod, until the caseins flocculate (become solid), then add two more drops of glacial
acetic acid. The caseins may solidify on your stir-rod, that is ok.
5. Obtain a piece of cheesecloth. Fold the cheesecloth in half two times so that it is 4 layers thick. You
will filter the flocculate using cheesecloth. While one lab partner holds the cheesecloth above the sink,
the other partner pours the cheese and liquid onto the cheesecloth. Next, rinse the cheese by running tap
water over cheese while it is still on the cheesecloth.
6. After rinsing, wrap the cheesecloth around the cheese then squeeze any excess liquid into the sink.
7. Remove the cheesecloth to examine the cheese you made.
8. Dispose of the cheese and cheesecloth using any trashcans in the room
•
You can make cheese at home in the same manner using some vinegar or lemon juice as a
substitute for the glacial acetic acid.
Chemistry 108
Protein, Triglycerides, and Ester Lab
Part 2: Synthesis of Triglycerides: Soap Making
•
SAFETY WARNING: YOU WILL BE USING A VERY STRONG BASE (NaOH)
SOLUTION. STRONG BASES CAN BE JUST AS DANGEROUS AS STRONG ACIDS.
PROCEDURE:
1. Using the graduated cylinder that is located next to the olive oil, measure 20.0 mL of olive oil into
your 100mL beaker.
2. OPTIONAL: Add 6 drops of any fragrant oil extract or one drop of perfume/cologne to your 100
mL beaker that contains the olive oil.
3. Using the graduated cylinder that is provided and labeled for the NaOH solution, measure 5.0 mL
of the NaOH solution. SLOWLY add it to the 100ml beaker containing your olive oil while
stirring with your glass stir-rod.
4. Obtain a magnetic stir-bar and add it to the beaker containing the olive oil and NaOH.
5. Place the beaker on the stir-plate (it is the hotplate that you used for heating in previous labs). If
your hotplate surface is still hot from Part 1, use another hotplate (there are two hotplates at each
lab station). Set the stir speed to ‘medium’ and stir for 5 minutes. SAFETY ALERT: BE SURE
TO USE THE STIR CONTROL KNOB AND NOT THE HEAT CONTROL KNOB!!!
6. While the mixture is being stirred, get a paper cup (for use as a soap mold) and write your initials
on the cup.
7. After 5 minute of stirring, turn off the stir-plate. While wearing gloves, remove the stir-bar.
8. Pour the mixture of NaOH and oil into your paper cup mold.
9. Place your mold with the soap in the area designated by the instructor for storage.
10. OPTIONAL: If both partners want to make a bar of soap, using gloves and paper towels, wipe
the excess oil from your 100 mL beaker and stir-bar then repeat the procedure for lab partner #2.
You can put the used paper towels into any of the trashcans in the lab.
11. OUR LEGAL DEPARTMENT FORBIDS THE USE OF THE SOAP MADE IN LAB.
12. Use soap and water to thoroughly clean your 100 mL beaker and stir bar.
13. Store you soap in your lab locker until we meet for the lab locker checkout, it will be solidified
by then.
Part 3: That Smells Good: Esterification Reactions
Esterification: Addition of an alcohol with a strong acid catalyst creates an ester.
O
||
R–C–O–H
H3 O
+ H – O – R'
+
O
||
R – C – O – R' + H2O
(a) Esterification of Acetic Acid with 1-pentanol:
1) Start a boiling hot water bath by adding about 2 inches of water and a few boiling chips to your 600
mL beaker. Place the beaker on a hotplate and set the temperature to the highest setting until the water
begins to boil. (Note, after the water bath begins to boil, turn the heat control down to about half-power
to maintain a gentle boil). Go on to step #2 while waiting for the water to boil.
2) While the water is coming to a boil, wearing protective gloves, add about 15 drops of glacial acetic
acid and about 6 drops of 1-pentanol to a large dry test tube. (Glacial acetic acid is the name given to
pure acetic acid, which is a liquid above 16.6 °C). Mix the contents of the tube, and then add 5 drops of
concentrated sulfuric acid, one drop at a time, with mixing after each drop is added.
2
Chemistry 108
Protein, Triglycerides, and Ester Lab
3) After the H2SO4 has been added, heat the contents by placing the test tube in the beaker of boiling
water for 3 minutes.
4) Now pour the contents of the test tube into a small beaker that is half-filled with tap water (not the
beaker that you used for boiling).
Does there appear to be any "oil" on top of the water? ________________. (The liquid ester appears like
oil does on water because it is insoluble.)
Cautiously determine/describe the odor of the product. (Can you identify the aroma?):
Write the chemical equation that describes the formation of this ester:
Name this ester: _______________________________________________ (see your chapter 10 lecture
notes for naming esters)
(b) Esterification of Salicylic Acid with Methanol:
1) Place about 0.5 grams of salicylic acid (2-hydroxybenzoic acid) in a clean dry test tube.
2) Add 3 ml of methanol to the salicylic acid.
3) When the entire solid has dissolved, wearing protective gloves, slowly add 10 drops of concentrated
sulfuric acid, mixing the contents after each addition.
4) After all of the sulfuric acid has been added, heat the test tube in the beaker of boiling water for 3
minutes. Next, pour the contents of the test tube into a small beaker of tap water (not the beaker that
you used for boiling). You should see a solid form in the cool water. The solid is the ester that you
synthesized.
Cautiously determine/describe the odor of the product:
Write the chemical equation that describes the formation of this ester.
Name this ester: _______________________________________________
3
Chemistry 108
Protein, Triglycerides, and Ester Lab
Part 4: Models
Structures and names of some organic compounds
Examine the numbered models found in throughout the lab. Draw a condensed
structural
formula for
o o
o
each molecule represented. In the case of alcohols or amines, classify each as 1 , 2 or 3 . Name each
compound. If you wish to draw the condensed formulas on a separate page, that is fine.
Black spheres are carbon, white are hydrogen, blue are nitrogen, and red are oxygen.
Model
Condensed Formula
Classification
Name
1
2
3
4
5
6
7
8
4
Chemistry*108*
Protein,*Triglycerides,*and*Esters*Lab*
Lab #8 Prelab: Protein, Triglycerides, and Esters Lab
Purpose of the lab:
In Part 1, you will make cheese by separating the caseins from milk. In part 2, you will make soap
using the saponification of triglycerides reaction. In part 3, you will synthesize two esters that smell
very good!
Part 1: Proteins in milk: Separating the Caseins from Milk
The name protein is taken from the Greek protelos, which means first. This name is well chosen. Of all
chemical compounds, proteins could certainly be ranked first, for they are the substance of life. Proteins
make up a large part of the animal body, they hold it together, and they run it. They are found in all
living cells. They are the principal material of skin, muscle, tendons, nerves, and blood; of enzymes,
antibodies, and many hormones. (Only the nucleic acids, which control heredity, can challenge the
position of proteins; and the nucleic acids are important because they direct the synthesis of proteins.)
"Chemically, proteins are high polymers. They are polyamides, and the monomers from which they are
derived are the α-amino carboxylic acids. A single protein molecule contains hundreds or even
thousands of amino acids units; these units can be of twenty-odd different kinds. The number of
different protein molecules that are possible, is almost infinite. It is likely that tens of thousands of
different proteins are required to make up and run an animal body; and this set of proteins is not
identical with the set required by an animal of a different kind."
- Morrison & Boyd's Organic Chemistry
In this lab activity we will study proteins - described so eloquently in the above quote from Morrison &
Boyd's Organic Chemistry- from several different viewpoints. In Part 1, we will separate some amino
acids (including an unknown) by using paper chromatography. In Part 2 we will use a pH meter to
measure the acidity of solutions of three amino acids. In Part 3 we will separate
the caseins from milk. Caseins* (from Latin caseus "cheese") make up about 80%
of the proteins in milk. Caseins are phosphoproteins, that is, there are phosphate
groups attached to some of the amino acid side chains. It is interesting that proline
(see figure), an unsymmetrical amino acids makes up 20% of the amino acids
residues in some caseins. In proline, the side chain makes a ring structure with the
amino group. The result is a lack of ordered internal protein secondary structure
probably and facilitates attack by digestive enzymes in infants. The rest of the
milk protein is a mixture of soluble proteins that includes the albumins and
globulins; these are commonly called the whey proteins.
Casein consists of four different proteins, usually referred to as α, β. γ, and κ caseins. The α, β, and κ
forms are most abundant, making up roughly 50%, 33%, and 15% of the casein, respectively. Casein
exists in milk as the calcium salt, calcium caseinate. The protein has a negative overall charge and its
charge is balanced by the positive calcium ions. Surprisingly, though, it is found that calcium ions, at
the concentration of Ca2+ normally found in milk, cause α and β casein, singly or in combination to
precipitate. However, in milk, the κ casein is soluble and is thought to surround the α and β casein
forming a soluble micelle. Disruptions, such as denaturation or enzymatic bond cleavage, of the
micelles cause the casein proteins to coagulate; this is what the cheese-making process does. This can
be done with acids, bases, temperature, mechanical agitation, or with certain enzymes (as in the case of
yogurt); we will use acetic acid in the lab activity.
Chemistry*108*
Protein,*Triglycerides,*and*Esters*Lab*
Part 2: Saponification of Triglycerides: Soap Making
Soaps are amphipathic molecules used for cleaning and bathing. Soaps molecules are the base form,
the carboxylate ions, of fatty acids. Soaps are made using the saponification reaction that converts a
triglyceride molecule into 3 carboxylate ions (soap molecules) and a glycerol molecule. Since
carboxylate ions have a polar head and a nonpolar tail, they are able to emulsify oil and nonpolar
substances in the cleaning and bathing processes. An example of the saponification reaction is given
below.
Note that the fatty acid residues contained in the reactant triglyceride are arbitrary; any fatty acid
residues could have been used.
This reaction is simply three "hydrolysis of an ester" reactions. We first saw the hydrolysis of an ester
reaction in chapter 6 then once again in chapter 10. In the saponification of triglycerides, the reaction
is catalyzed by a strong base. Since the reaction occurs in a strong base (pH > pKa of fatty acids), it is
the base form of the fatty acids, the carboxylate ions, which are produced. The base used for catalysis
is usually sodium hydroxide. Sodium hydroxide is also called "lye". When one makes soap using
sodium hydroxide, the soap is called "lye soap".
Chemistry*108*
Protein,*Triglycerides,*and*Esters*Lab*
History of Soap Making (From Wikipedia):
(Note: I have removed the references from the literature cited, however, they can be found on the
Wikipedia site if you want more information or you wish to use this information in a report for another
class)
Early history. The earliest recorded evidence of the production of soap-like materials dates back to
around 2800 BC in ancient Babylon. A formula for soap consisting of water, alkali, and cassia oil was
written on a Babylonian clay tablet around 2200 BC.
The Ebers papyrus (Egypt, 1550 BC) indicates the ancient Egyptians bathed regularly and combined
animal and vegetable oils with alkaline salts to create a soap-like substance. Egyptian documents
mention a soap-like substance was used in the preparation of wool for weaving.
In the reign of Nabonidus (556–539 BC), a recipe for soap consisted of uhulu [ashes], cypress [oil] and
sesame [seed oil] "for washing the stones for the servant girls".
Ancient Rome. The word sapo, Latin for soap, first appears in Pliny the Elder's Historia Naturalis,
which discusses the manufacture of soap from tallow and ashes, but the only use he mentions for it is
as a pomade for hair; he mentions rather disapprovingly that the men of the Gauls and Germans were
more likely to use it than their female counterparts. Aretaeus of Cappadocia, writing in the first
century AD, observes among "Celts, which are men called Gauls, those alkaline substances that are
made into balls, called soap".
A popular belief claims soap takes its name from a supposed Mount Sapo, where animal sacrifices
were supposed to have taken place; tallow from these sacrifices would then have mixed with ashes
from fires associated with these sacrifices and with water to produce soap, but there is no evidence of a
Mount Sapo in the Roman world and no evidence for the apocryphal story. The Latin word sapo
simply means "soap"; it was likely borrowed from an early Germanic language and is cognate with
Latin sebum, "tallow", which appears in Pliny the Elder's account. Roman animal sacrifices usually
burned only the bones and inedible entrails of the sacrificed animals; edible meat and fat from the
sacrifices were taken by the humans rather than the gods.
Zosimos of Panopolis, circa 300 AD, describes soap and soapmaking. Galen describes soap-making
using lye and prescribes washing to carry away impurities from the body and clothes. According to
Galen, the best soaps were Germanic, and soaps from Gaul were second best. This is a reference to
true soap in antiquity.
Ancient China. Soap, or more accurately a detergent similar to soap, was manufactured in ancient
China from vegetation and herbs. True soap, made of animal fat, did not appear in China until the
modern era. Soap-like detergents were not as popular as ointments and creams.
Middle East. A 12th-century Islamic document describes the process of soap production. It mentions
the key ingredient, alkali, which later becomes crucial to modern chemistry, derived from al-qaly or
"ashes".
By the 13th century, the manufacture of soap in the Islamic world had become virtually industrialized,
with sources in Nablus, Fes, Damascus, and Aleppo.
Medieval Europe. Soapmakers in Naples were members of a guild in the late sixth century, and in
the eighth century, soap-making was well known in Italy and Spain. The Carolingian capitulary De
Chemistry*108*
Protein,*Triglycerides,*and*Esters*Lab*
Villis, dating to around 800, representing the royal will of Charlemagne, mentions soap as being one
of the products the stewards of royal estates are to tally. Soapmaking is mentioned both as "women's
work" and as the produce of "good workmen" alongside other necessities such as the produce of
carpenters, blacksmiths, and bakers.
15th–19th centuries. In France, by the second half of the 15th century, the semi-industrialized
professional manufacture of soap was concentrated in a few centers of Provence— Toulon, Hyères,
and Marseille — which supplied the rest of France. In Marseilles, by 1525, production was
concentrated in at least two factories, and soap production at Marseille tended to eclipse the other
Provençal centers. English manufacture tended to concentrate in London.
Finer soaps were later produced in Europe from the 16th century, using vegetable oils (such as olive
oil) as opposed to animal fats. Many of these soaps are still produced, both industrially and by smallscale artisans. Castile soap is a popular example of the vegetable-only soaps derived by the oldest
"white soap" of Italy.
Modern Times. In modern times, the use of soap has become universal in industrialized nations due
to a better understanding of the role of hygiene in reducing the population size of pathogenic
microorganisms. Industrially manufactured bar soaps first became available in the late 18th century, as
advertising campaigns in Europe and the United States promoted popular awareness of the relationship
between cleanliness and health.
Until the Industrial Revolution, soapmaking was conducted on a small scale and the product was
rough. Andrew Pears started making a high-quality, transparent soap in 1789 in London. His son-inlaw, Thomas J. Barratt, opened a factory in Isleworth in 1862. William Gossage produced low-priced,
good-quality soap from the 1850s. Robert Spear Hudson began manufacturing a soap powder in 1837,
initially by grinding the soap with a mortar and pestle. American manufacturer Benjamin T. Babbitt
introduced marketing innovations that included sale of bar soap and distribution of product samples.
William Hesketh Lever and his brother, James, bought a small soap works in Warrington in 1886 and
founded what is still one of the largest soap businesses, formerly called Lever Brothers and now called
Unilever. These soap businesses were among the first to employ large-scale advertising campaigns.
Liquid Soap (Detergent). Liquid soap was not invented until the 1800s. In 1865, William Shepphard
patented liquid soap. In 1898, B.J. Johnson developed a soap formula, and his company (the B.J.
Johnson Soap Company) introduced Palmolive soap the same year. This new soap was made of palm
and olive oils and became popular in a short amount of time; Palmolive became so popular that B.J.
Johnson Soap Company changed its name to Palmolive. At the turn of the century, Palmolive was the
world's best-selling soap.
In the early 1900s, other companies began to develop their own liquid soap. Products such as Pine-Sol
and Tide appeared on the market, making the process of cleaning clothing, counters and bathrooms
easier.
As a detergent, liquid soap tends to be more effective than flake soap, and there is a smaller chance of
residue being left on clothing with liquid soap. Liquid soap also works better for more traditional
washing methods, such as using a washboard.
Chemistry*108*
Protein,*Triglycerides,*and*Esters*Lab*
Part 3: Esterification
When an carboxylic acid (R-COOH) and an alcohol (R-OH) are mixed together and heated in the
presence of an acid catalyst (such as H2SO4), the two will react to form an ester (plus H2O). This
process is called esterification. Each ester has its own unique odor, and with a discriminating nose, one
can use this fact to help identify them. In this lab you will be reacting various organic acids (acetic acid
& salicylic acid) with various alcohols (1-pentanol & ethanol). You will make two different esters with
odors that should be familiar to you. You may have noticed that esterification is the reverse reaction
of the hydrolysis of esters seen in the saponification reaction.
Esterification: Addition of an alcohol with a strong acid catalyst creates an ester.
O
||
R–C–O–H
H+
+ H – O – R'
O
||
R – C – O – R' + H2O
PRELAB QUESTIONS:
1. Draw the structure of di-peptide Tyr-Phe as it would exist at pH = 1, 7, and 14.
pH = 1
pH = 7
pH = 14
Chemistry*108*
Protein*and*Esters*Lab*
2) Draw a triglyceride (do not draw the triglyceride used in the example on page 2 of this prelab, you
can draw any other triglyceride) :
3) Draw the products of the saponification of the triglyceride that you drew in question #2 above:
4. Draw and name the ester that would be formed in the reaction of propanoic acid with ethanol:
(see chapter 10 lecture notes if you need a reminder about naming esters)
6*
*