Download Avogadro`s Number Determination

DETERMINATION OF AVOGADRO’S NUMBER
USING THE MICROLAB INTERFACE:
An Electrolysis Experiment for General Chemistry
Michael Collins, Professor of Chemistry Emeritus,
Viterbo University
MicroLab, Inc. April 8, 2011
Page 1
Avogadro’s Number Determination Using the MicroLab Interface:
An Electrolysis Experiment for General Chemistry
Pre-lab Assignment:
(1) Read the introduction.
(2) Before beginning of lab, answer and hand in questions 1 through 5 that appear in the "Introduction" section below.
Introduction
Historical Background
The mole concept is one of the most important concepts in chemistry. It derives from the atomic theory and the idea of relative atomic
masses. The original atomic theory of John Dalton was proposed to explain three very important laws of chemistry, laws that changed chemistry
from the arena of magic to the arena of understanding.
The first of these laws is the Law of Conservation of Mass: in any chemical reaction, there is no change in mass as the reaction proceeds,
provided all materials before and after the reaction are accounted for.
The second is the Law of Constant Composition: any pure compound always has the same elemental composition by mass. That is to say,
pure distilled water is always 11.2% hydrogen and 88.8% oxygen by weight; pure sodium chloride is always 39.3% Na and 60.7% Cl by weight; and
so on for any compound.
These mass percentages can also be expressed as mass ratios. Thus the mass ratio of O/H in water can be expressed as
O 88.8 7.93


H 11.2
1
Similarly in sodium chloride, the mass ratios are
Cl 60.7 1.54


Na 39.3
1
The converse of the Law of Constant Composition is that we can determine the purity of a compound from its elemental composition or elemental
mass ratios.
The third law is the Law of Multiple Proportions. This law acknowledges that certain elements may combine to form more than one
compound. For example, hydrogen and oxygen for two compounds with each other, water and hydrogen peroxide. These two compounds both
consist exclusively of hydrogen and oxygen, but have remarkably different properties. As you know, hydrogen peroxide is a strong oxidizing agent
used for bleaching hair and disinfecting wounds, among others. Hydrogen peroxide in its pure form is known to be 5.93% H and 94.07% O by
weight, with a weight ratio of 15.86/1.
There is an intriguing relationship between the mass ratios of water compared to hydrogen peroxide: if we divide the ratios we get
15.86
1 2
7.93
1
1
Similar relationships exist for other elements that form a family of compounds, and can be summarized in the Law of Multiple
Proportions: when two elements combine to form more than one compound, the ratio of those compounds’ mass ratios is a simple whole number.
-----------------------------------------------------------------------------------------------------------Q. 1 Copper forms two oxides. Red copper oxide is 88.8% Cu and 11.2% O. Black copper oxide is 79.9% Cu and 20.1% O. (a) Express these as mass
ratios and, for each compound, compare its mass ratio with the ratio of atomic masses. (b) Show that these compounds taken together obey the
Law of Multiple Proportions. (c) If black copper oxide is assumed to be CuO, what is the formula of red copper oxide? Use units and show work!
-----------------------------------------------------------------------------------------------------------John Dalton explained these laws in his original atomic theory. The reason mass is conserved is that atoms are conserved in chemical
reactions. The reason compounds always have the same composition by mass is that compounds always have the same composition by atoms. We
can describe the composition of compounds with formulas such as H2O1 or Na1Cl1 that specify the relative numbers of each type of atom that
combine to form the compound.
We can likewise understand how atoms can combine to form two compounds with mass ratios that are simple whole number ratios.
Stated atomically, using the water- hydrogen peroxide example, for a given number of hydrogen atoms in each compound (i.e. the denominators in
the ratios are the same, 1) there must be twice as many atoms of oxygen in hydrogen peroxide as in water. Thus if the formula of water is H 2O1,
then the formula of hydrogen peroxide must be H2O2!
2
A direct and crucial result of this thinking is that we can deduce the relative masses of the atoms of the elements if we know the formula
of the compound(s) they form with each other. Thus, for NaCl, if the mass ratio of Cl/Na is 1.54/1 and if one atom Cl combines with one atom Na,
then each Cl atom must have a mass 1.54 times that of each Na atom. Check it out using the table of atomic masses in your textbook: 35.45/23 =
1.54!
Similarly, for hydrogen peroxide, since there are equal numbers of hydrogen and oxygen atoms, and the mass ratio is 15.86/1, each
oxygen atom must have a mass that is 15.86 times that of each hydrogen atom.
Conversely, if we weigh out a mass of chlorine that is 1.54 times the mass of sodium taken, then we must have taken the same number of
atoms of chlorine as atoms of sodium! This idea was pivotal in elevating chemistry to a quantitative science and developing experiments to test the
atomic theory.
-----------------------------------------------------------------------------------------------------------Q. 2 What mass of chlorine has the same number of chlorine atoms as there are sodium atoms in a 4.6 gram sample of sodium metal? Use units
and show work!
th
The modern tables of atomic masses are the direct descendants of 19 Century tables that were the result of careful comparative work
on mass ratios in many simple compounds. The masses given are relative masses, based on mass ratios, and are independent of the system of
weights and measures used by society.
If we work in the metric system, we tend to measure masses in grams. If we take an amount of an element in grams equal to its relative
atomic mass, say 16 grams of oxygen, then we say that we have taken 1 mole of that material. A mole of oxygen atoms and a mole of sodium
atoms and a mole of chlorine atoms and a mole of hydrogen atoms all refer to the same number of atoms.
-----------------------------------------------------------------------------------------------------------Q. 3 How many moles of Cu atoms are in 63.54 g Cu? In 0.06354 g ? In 0.0100 g? Use units!
-----------------------------------------------------------------------------------------------------------How many atoms are in a mole? This is a very interesting and important question, but one that need not be answered operationally in
order to put a mole of copper atoms into a beaker. Chemists were working with moles for more than 75 years before they could measure the
number of atoms in a mole. But now that we can measure the number of atoms in a mole, it sure would be fun to try!
Measuring Avogadro’s Number
One type of chemical reaction is electrolysis, in which electricity is used to carry out a chemical reaction. Electrolysis may be used to
electroplate a metal that resists tarnish – gold, chromium, silver, copper – onto a metal that more readily tarnishes, such as iron or certain kinds of
non-stainless steels. Electrolysis can also be used to generate gases at an electrode: if an electric current is passed through water to which a small
amount of sulfuric acid has been added (to make the water an electrical conductor), hydrogen gas is generated at one electrode and oxygen gas is
generated at the other.
The electrode at which reduction (the lowering of the oxidation state of an element) occurs is called the cathode. The electrode where
oxidation (the increase of oxidation state) occurs is called the anode.
By convention, the wire connected to the cathode is color-coded red; that connected to the anode is black. The red wire is connected to
the electric source (e.g. battery) at the (+) terminal, the black wire is connected to the (-) terminal.
A simple electrolysis cell for the electroplating of copper can be schematically represented as follows:
Figure 1 A simple electrolytic cell
-
e (-)
Cathode
Battery
(+)
Cu
Cu
Anode
Aqueous CuSO4
In this cell, as electricity is passed through the battery, Cu ions enter the solution at the (+) electrode, leaving their electrons in the
electrode:
3
Cu

2+
Cu
-
+
2e
eq (1)
2+
as, simultaneously, Cu ions in solution plate onto the (-) electrode, gaining electrons that are “pumped” through the circuit from the anode by the
battery:
2+
Cu
+
-

2e
Cu
eq (2)
Ideally, the mass of copper lost from the anode should exactly equal the mass of copper plated at the cathode, and the CuSO 4 concentration should
remain unchanged.
th
In the late 19 Century, Michael Faraday discovered that the number of moles of material plated at or lost from an electrode was
proportional to the number of coulombs passed through the cell. We can measure the number of coulombs by measuring the current (I) in
amperes (1 ampere = 1 coulomb/second) and the time (t) over which the current is passed:
Just before the turn of the century, the charge of a single electron was determined to be
# coulombs  I (amps)  t (seconds)
1.60 x 10
-19
coulombs. Amazingly, this means that if we measure the number of coulombs, we can calculate the number of electrons
------------------------------------------------------------------------------------------------------------
# electrons 
I t
1.60 10  19 C/e
eq (3)
Q. 4 Suppose an electrolysis of copper/aqueous copper sulfate is carried out by passing a current (average) of 0.0200 amperes through a cell,
similar to that in Figure 1, for 600 seconds. (a) How many coulombs of electricity are passed? (b) how many electrons would be passed through the
cell? Use units!
-----------------------------------------------------------------------------------------------------------If one determines simultaneously the number of moles of electrode material involved in the electrolysis
Δ moles 
Δ mass
GAW
eq (4)
(GAW = gram atomic weight, also called the gram atomic mass) one has a relationship between the number of particles (electrons, and, through
the balanced half reactions, atoms) and the moles of material. This gives you the information needed to calculate Avogadro’s number, N A, by
combining equations 2, 3, and 4:
I t
N A  # atoms
1mole

19
1.6010  1atomCu
Δmass
GAW
2electrons
-----------------------------------------------------------------------------------------------------------Q. 5 Suppose in the experiment of Q. 4, you found that 0.0039 grams of copper were plated onto the cathode. Calculate Avogadro’s number. Use
units!
-----------------------------------------------------------------------------------------------------------Measuring current and time
Some modifications need to be made to Figure 1 in order to make the circuit suitable for measuring the current through the cell.
Specifically, a switch must be added so that the circuit can be turned on and off; and a means for measuring the current needs to be added, using
an ammeter connected in series with the cell, as in Fig. 2. The jagged line represents a resistor, R, which limits the amount of current through the
cell.
Figure 2. Circuit for Measuring Current in the Determination of Avogadro’s Number Using Cu Electrodes and CuSO 4 Electrolyte Using an
Ammeter
4
Battery
(-)Black
R
(+)Red
Ammeter
Cathode
Cu
Cu
Anode
Aqueous CuSO4
In order to measure Avogadro’s number with this circuit, one needs to measure the mass of the electrodes to at least +1 mg (+ 0.1 mg is
ideal), the time of the electrolysis, and the average current. Other electrode – electrolyte combinations may be used either in place of the Cu
system or in series with it, since the same current flows through all components in a series circuit. Another suitable electrode system is Ag/AgNO3.
The half reactions for the silver system are as follows:
+
-
Ag
+
e
Ag

Ag
+

Ag
+
e
-
The cell would be analogous to the Cu system:
Figure 3. Circuit for Measuring Current in the Determination of Avogadro’s Number Using Ag Electrodes and AgNO 3 Electrolyte
Battery
(-)Black
R
(+)Red
Ammeter
Cathode
Ag
Ag
Anode
Aqueous AgNO3
The calculation would differ only in that there is 1 mole of Ag involved at each electrode for each 1 mole of electrons.
Our actual circuit will be simplified considerably by the use of the MicroLab FS522 data acquisition unit in conjunction with the MicroLab
Model 270 electrolysis module. The Model 270 is a regulated power supply that substitutes for all but the chemical cell itself and a pair of
connecting wires. The FS522 unit and associated software on the PC serve to control the voltage output on the 270 as well as to acquire and display
the time/current data. The software also displays the number of coulombs of electrical charge passed through the circuit as well as the number of
moles of electrons.
Reproduce the appropriate portions of the following table in your laboratory notebook:
Electrolysis Data
5
Run 1
Run 2
Run 3
Mass of anode before electrolysis
_______________ g.
_______________ g.
_______________ g.
Mass of anode after electrolysis
_______________ g.
_______________ g.
_______________ g.
Mass change of anode
_______________ g.
_______________ g.
_______________ g.
Number of coulombs (anode) _______________ C
Atomic mass (g/mol)
-
Ratio (#atoms/ e )
NA
Average
_______________ C
_______________ g/mol
_______________ C
_______________ g/mol
_______________ g/mol
_______________/ mol
_______________/ mol
_______________
(anode)
_______________ /mol
(anode)
_______________ /mol
Mass of cathode before electrolysis
_______________ g.
_______________ g.
_______________ g.
Mass of cathode after electrolysis
_______________ g.
_______________ g.
_______________ g.
Mass change of cathode
_______________ g.
_______________ g.
_______________ g.
Number of coulombs
_______________ C
_______________ C
_______________ C
Atomic mass (g/mol)
_______________ g/mol
_______________ g/mol
_______________ g/mol
-
Ratio (#atoms/ e )
_______________
NA
_______________ /mol
(cathode)
Average (cathode)
_______________/ mol
_______________/ mol
_______________ /mol
6
Equipment needed:
PC with MicroLab software
MicroLab interface, power cord
MicroLab Model 270 Electrochemistry Module, leads, and power supply
One red and one black lead, each with a banana plug on one end and an alligator clip on the other
2 pieces Ag wire or 2 pieces Cu wire
Fine steel wool for cleaning electrodes
2 30 mL beakers and rubber stoppers to fit
1 10 mL graduated cylinder
30 mL of 0.1M CuSO4 or 0.1M AgNO3
Magnetic stirrer and stir bar to fit the beaker
Procedure:
CAUTION: Eye protection must be worn at all times.
CAUTION: Copper and silver salt solutions are considered heavy metals. They are toxic, and as such they must never be discarded down the
drain. When silver salts come in contact with skin, the silver is reduced and forms black stains on the skin that come off only when the skin
sloughs off over time. Use gloves and appropriate safety precautions when handling copper and silver salt solutions.
In this experiment, instead of a battery we will use a computer interface that can output a user-specified voltage and also measure current and
time. An accessory (the Model 270) also has an internal resistor. The software has on-off and start stop switches built in.

Set up a MicroLab interface connected to a laptop PC via USB.

Attach the Model 270 Electrochemistry Module to the Ethernet jack labeled “A” on the front panel of the MicroLAB unit. Plug all the
power cords into the same lab power strip assure a common ground.

Connect the MicroLAB unit to the correct USB port of the PC. Turn on power to the MicroLab unit and the electrolysis module. The green
pilot lights should come on.

Double click on the MicroLab icon to bring up the MicroLab software.

Click on the Electroplating experiment.
This software that controls the experiment is relatively simple to follow, and you should take the time now to examine the listing in the lower left
window of the PC screen to see what it is doing. Prepare a brief flow chart of the software in your lab notebook and include it in your report.
Notice that it is reading timer, current (I), and voltage (V) values, placing them in the spreadsheet in separate columns and repeating each half
second. The program specifies that I and V are from “CAT-5 A” (a CAT-5 connector is an Ethernet jack). Make sure you are plugged into CAT-5 A.
You will be assigned one of the systems shown in Figures 2 or 3. The electrolysis cell for Cu or Ag consists of a 30 mL beaker containing about 15
mL of the appropriate solution (CuSO4 or AgNO3), two electrodes, connecting wires, and a stirring magnet. The electrodes must not touch or a
“short circuit” will result. The best way to hold the Ag and Cu electrodes in place is to pin them near the walls of the beaker with a cork or rubber
stopper. The beaker should be clamped to prevent tipping. See the setup at the instructor’s desk for details. The “battery” will be the voltage
output of the interface unit. The resistor and the ammeter are internal to the Model 270. The on-off switch is also internal and is closed and
opened with the “Start” and “Stop” soft “buttons” at the bottom of the sensor window.

Clean the Cu or Ag electrodes with a fine gauge steel wool – but do it on a watch glass to avoid ruining the bench top. Rinse the
electrodes with distilled water, dry them with Kimwipes, and weigh them as accurately as possible, to at least + 0.001 gm (+0.0001 gm is
better).

When you are ready to begin, attach the leads to the Model 270 unit, red to red and black to black. Do not clip on to the metal electrodes
yet. Turn on the stirrer and stir the solution at a rate of about 2-4 rotations per second.

Open the electrolysis software and choose the proper voltage. This is not a trivial matter. You need to have sufficient voltage to initiate
the desired electrolysis, but too high a voltage will cause water electrolysis, which will lead to significant errors. We suggest a voltage less
than one volt sufficiently high to get about 30-50 mA current. Try 0.8-0.9 volts to start.

Click the start button, without connecting the leads, and follow the directions for obtaining the "open circuit offset current." This allows
the software to correct for any small stray current that may be part of the system. When the open circuit offset measurement is
complete, you will be prompted to connect the leads and start the experiment. When you do this, you should see a rapid rise in the
current to a plateau (30-50 mA is typical) and visible changes occurring at the electrodes.

Let the system electrolyze for at least 900 seconds (15 minutes). [You should be seeing about 30-50 mA of current.] When you are ready
to stop, click on the “Close” button.

You will be prompted to SAVE YOUR DATA! [Note 1] Do so!
7

Carefully rinse the electrodes with distilled water and set them in the oven for five minutes to dry thoroughly. When the electrodes are
cool, weigh them to obtain the mass changes. You may find that cathode forms needles on the surface that may fall off. You might think
about how you are going to get the mass change at the cathode, or whether it is all right to forget about the product that falls off.
Do triplicate runs on your system, re-using the solutions and electrodes. Return solutions, electrodes, wires, clips and other special equipment to
the proper locations.
DO NOT discard the solutions down the sink. Use the waste jars in the lab or return to the original solution as directed by your instructor.
Calculations:
For each run:
1. Obtain the number of coulombs of charge passed through the cell. See Note 2 below.
2. Calculate the number of electrons passed through the cell [Note 3]
3. Calculate the change in the number of moles of copper or silver at each electrode, according to which system you explored.
4. Use the balanced half reactions to convert number of electrons to number of atoms
5. Calculate Avogadro’s number from the number of atoms and the number of moles at each of your electrodes
Obtain the average value at each electrode for Avogadro’s Number.
Calculate the % error of your average value compared to the known value.
Calculate the % difference between your anode average and your cathode average.
Your Report: this will be a brief report rather than a formal report.
1. State the purpose of the study in a sentence.
2. Include the flow chart for logic of the data acquisition program.
3. Write the balanced “half-reaction” occurring at each electrode for the chemical system that you studied.
4. Show a sample calculation for one of your good runs.
5. Put all the results in tables. Generating separate tables for anode and cathode is a good idea. Your tables should include individual
cathode data and results and anode data and results and the average values of Avogadro’s Number at each electrode, using the
appropriate number of significant figures.
6. Report all individual anode values of Avogadro’s number and the average value, using the appropriate number of significant figures, for
all your good runs.
23
7. Report the % error at each electrode in each run and in the average, using 6.022 x 10 as the “true” value for Avogadro’s Number.
8. Include the carbon copy of your relevant lab notebook page(s) in your report.
9. Answer the three questions below.
Questions:
1. For the cell system studied, estimate the number of significant figures in each measurement. Which measurement limits the accuracy of your
results? Explain.
2. Which electrode should give a better value for Avogadro’s number? Explain.
3. Use the known chemistry of zinc, iron, and calcium to explain briefly why they are not suitable for determining Avogadro’s number by this
method.
Notes:
1. Of course the whole experiment could be done manually with a 9 Volt transistor battery, a Digital voltmeter operating as an ammeter, and
stopwatch in place of the computer. You would need to take a time and ampere reading every 30 seconds or so for about 20 minutes.
2.
There are two ways to go from current/time data to obtain the number of coulombs passed: (a) Get the average of all current measurements
over the course of the experiment. The average current multiplied by the total time should give the number of coulombs. (b) Integrate the
time/current curve. Integration, for those of you who have taken calculus, over a specified interval will give the area under the I-t curve

t
Charge(Cou lombs)  Idt
0
which is the number of coulombs. The software does the processing for you by the first method, updating the result after each measuement.
3.
The number of electrons is gotten from the number of Coulombs and equation 3.
8