Download Lab 1

IVC Chemistry 100
Laboratory Manual
Name__________________________________
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Table of Contents
Lab 1: DIMENSIONAL ANALYSIS .................................................................................................................. 4
Data for Lab 1: Dimensional Analysis.................................................................................................... 11
ADDITION AND SUBTRACTION...................................................................................................... 12
MULTIPLICATION & DIVISION OF SIGNIFICANT DIGITS ................................................................ 12
CALCULATORS AND SIGNIFICANT DIGIETS ................................................................................... 13
MATH CONCEPTS IN CHEMISTRY .......................................................................................................... 11
SIGNIFICANT DIGITS ...................................................................................................................... 11
ROUNDING OFF NONSIGNIECANT DIGITS .................................................................................... 11
SCIENTIFIC NOTATION ...................................................................Error! Bookmark not defined.
TEMPERATURE .............................................................................................................................. 15
METRIC - METRIC CONVERSIONS .................................................................................................. 16
METRIC - ENGLISH CONVERSIONS ................................................................................................ 16
GENERAL EXERCISES...................................................................................................................... 16
Lab 2: Atomic Structure ............................................................................................................................ 19
Discussion.......................................................................................................................................... 19
Experimental Procedures .................................................................................................................. 20
Data for Lab 2: Atomic Structure .......................................................................................................... 21
Lab 3: Periodic Properties ......................................................................................................................... 27
Data for Lab 3: Electronic Configurations and Periodic Trends ............................................................ 31
Lab 4 Nomenclature.................................................................................................................................. 35
Group 1 Ionic Compounds .................................................................................................................... 35
Class 1. Fixed Oxidation State (O.S.) ................................................................................................. 35
Class 2. Fixed Oxidation States ......................................................................................................... 36
Group 2: Covalent ................................................................................................................................. 38
Class 3. Covalent Non-acid ................................................................................................................ 38
Class 4. Binary Acids .......................................................................................................................... 38
Oxyoacids .......................................................................................................................................... 38
Data for Lab 4: Nomenclature .............................................................................................................. 41
Lab 5 Lewis Dot ......................................................................................................................................... 43
VSEPR: Valence shell electron-pair repulsion theory ....................................................................... 46
Data for Lab 5: Lewis Structure ............................................................................................................. 49
Lab 6 Heat Capacity of Metals .................................................................................................................. 51
Heat and Enthalpy............................................................................................................................. 52
Experimental Procedure ................................................................................................................... 53
Calculations ....................................................................................................................................... 53
Data for Lab 6: Heat Capacity ............................................................................................................... 55
Lab 7: Chemical EQUATION: WRITING BALANCED EQUATIONS ............................................................... 57
INSTRUCTIONS: ................................................................................................................................. 57
Lab Procedure ....................................................................................................................................... 58
Data for Lab 7: Writing Chemical Equations ......................................................................................... 59
Part 2. Rewrite the following word equations into balanced chemical equations. ..................... 61
PART 4: .......................................................................................................................................... 61
Lab 8 Formula of a Hydrate....................................................................................................................... 63
Data for Lab 8: Formula of a Hydrate ................................................................................................... 65
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Lab 9 Equilibrium ...................................................................................................................................... 69
Data Sheet for Lab 9 Equilbrium ........................................................................................................... 73
Lab 10 Standard Molar Volume of a Gas .................................................................................................. 75
Procedure .............................................................................................................................................. 76
Data for Lab 10: Electronic Configurations and Periodic Trends .......................................................... 77
Lab 11 Net Ionic Equations (NIE) .............................................................................................................. 79
Data Sheet Lab 11 NIE ........................................................................................................................... 83
Lab 12 Titration ......................................................................................................................................... 87
Titration General Information ...................................................................................................... 88
Data Sheet Lab 12 Titration .................................................................................................................. 91
30 points Extra Credit: Nomenclature (due when your instructor tells you :-) ........................................ 93
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Lab 1: DIMENSIONAL ANALYSIS
Rules for Addition: An answer obtained by adding or subtracting has the same number of decimal
places as the measurement with the fewest decimal places (look to the right of the decimal).
25.2
one decimal place
+ 1.34
two decimal places
26.54 calculated answer
Final Answer
26.5 answer with one decimal place
Another problem:
235.05
a
+19.6
+ 2.
256.65 rounds to 257
14400000
+ 72000000
86000000
Your final answer goes to the furthes right
as your least significant number
86000000
Rules for Multiplication: An answer obtained by multiplying or dividing has the same number of
significant figures as the measurement with the fewest significant figures (look for the fewest
significant figures).
Use rounding to limit the number of digits in the answer.
110.5 x 0.048 = 5.304 (calculator)
4 SF
2 SF
The final answer is rounded off to give
2 significant figures = 5.3 (2 SF).
Another problem:
40.311 ÷ 0.007 = 60
Mixed Mutliplication and Addition: Do each operation seperately.
Dimensional analysis
This is a very important lab. You will use what you learn in this lab EVERYDAY during lecture and lab.
Dimensional analysis is a problem-solving technique.
What will you learn: A Mathematical Technique, to solve EVERY problem we encounter.
Every Problem You Solve Will Look Like:
numerator
denomeator

numerator
denomeator
 ... 
4
The Math Principle is a Beginning Algebra concept of cross cancellation:
A
B

B
C
C
1

A
You are expected to solve ALL of your problems using this method, hence the reason for spending 2
days on this lab.
What Problems Can You Solve Using This Method?
You'll learn to solve many types of problems. For example how many mL’s are in 2.83qts, or how many
inches are in 259 cm? You'll learn to use the steps of DIMENSIONAL ANALYSIS as a model to solve
these types of problems.
To solve a dimensional analysis problem we must learn what a conversion factor is first.
Here are 2 examples of conversion factors.
1 ft = 12in 
1 ft
12 in
2.54 cm = 1 in 
2.54 cm
1 in
or
1 in
2.54 cm
or
12 in
1 ft
There are millions of conversion factors!
Note: these are exact relationships between two totally different units i.e. ft  in or cm  in
A conversion factor is important in math because they are equal to 1 and are therefore useful in
chemistry to help us do calculations.
For example, we know that 1 ft = 12 in, how is this equal to 1.
We know 1 ft = 12 in
1 ft
12 in
or
we had to divide by 12 in. and 1 ft respective
12 in
1 ft
1 ft 12 in
1 ft = 12 in, now divide both sides by 12 in or

1
12 in 12 in
To obtain the conversion factors
1 ft = 12 in now divide both sides by 1ft or
12 in 1 ft

1
1 ft 1 ft
We use these conversion factors when solving Dimensional Analysis Problems.
Dimensional Analysis
5
Let’s examine the problem already asked: How many inches are in 259 cm?
Use the conversion factor found on the first page that shows the relationship between inches and
centimeters. The answer would look like the following:
? inches =
259 cm
1 inch

= 102 inches
1
2.54 cm
Use unit analysis to solve the following problems.
1. 4 dollars =
Conversion factor
dimes
Hint: there are 10 dimes = 1 dollar
Solution :
4 dollars
10 dimes
×
= 40 dimes
1
1 dollar
Notice how the dollar is canceled in the numerator and dominator!
Try the following: SHOW ALL OF YOUR WORK!
2. 5 ft =
inches
Conversion Factor Hint: there are 12 inches = 1 ft. You fill in the numerator and denominator with the
listed conversion factor, and then do the multiplication.
inches? = 5 ft 
3. 24 hrs =
= _______ inches
minutes
What is the Conversion Factor? How many minutes in an hour?
What is Dimensional Analysis? This is what it looks like.
old unit 
new unit
= answer in new units
old unit
Notice how the “old unit” cancels in the numerator and denominator. This is why Dimensional Analysis
is so easy, you do NOT have to think in advance of all the steps. Dimensional Analysis is a VISUAL
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MATEMATICAL METHOD; you have a visual way of looking at your solution by looking at what units
cancel.
If you are still trying to solve these problems in your head I hope the next problem convinces you to
give up your old ways and try Dimensional Analysis!
In the fictitious country of Nolenville, 1.00 nolenzels equals 0.50 US dollars. How many nolenzels are
equal to 700 US dollars?
Solution using Dimensional Analysis:
?nolenzles  $700 
1.00 nolenzels
 1400nolenzels
$0.50
“$”
Notice that the
cancel. Only “nolenzels” remains in the numerator and since it does not cancel
so that, is your final unit.
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Calculate the standard deviation for the following set of coins, and express to the appropriate number
of significant figures:
7.691 g, 7.753 g, 7.658 g, 7.664 g, 7.553 g, 7.672 g, 7.561g
Solution:
First we must find the mean or average.
x
i
xi
n
Σ = is a math symbol for adding. The xi is a set of numbers, in this case the set of number in our
problem.
Σ: Note when adding line numbers over the decimal
7.961
7.753
7.658
7.664
7.553
7.672
7.561
53.552
Above, when adding we look for the least number of accurate places to the right. So above, all of the
numbers are 3 places to the right, so our answer, 53.552, must have 3 places to the right; which makes
it a 5 sig fig number.
mean  x 
53.822
 7.6889 g
7
Our final answer, 7.6889 has 5 sig figs, the same number of sig figs as 53.822 because division looks for
the number with the fewest sig figs (not like addition) so 53.822 has 5 sig figs and “7” is just a counting
number (it is not a 1 sig fig number, but an infinite number of sig figs).
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Next we work on the following function.

 

xi  x 
 
i
 
s

n

1








2
1
2
It is best to set up a table. Work in the inner most bracket first, always pay attention to powers.
Remember, when we do subtraction,
look to the right, so the first column
only has 3 places to the right, so our
final answer can only have 3 places to
the right.
Taking the square of a number is
multiplying 2 numbers together, so,
the number of sig figs in our final
answer can only have as many sig figs
as our worst measurement
xi  x (sig figs)
1
2
3
4
5
6
7
Σ
7.691 – 7.6889 =
7.753 – 7.6889 =
7.658 – 7.6889 =
7.664 – 7.6889 =
7.553 – 7.6889 =
7.672 – 7.6889 =
7.561 – 7.6889 =
53.552


   x  x  
 i i
 
s

n

1








2
1
2
1
2
 .0409 
s
  .0826
 7 1 
.002
.064
-.031
-.025
-.136
-.017
-.128
(1)
(2)
(2)
(2)
(3)
(2)
(3)
x  x
2
i
(sig figs)
.000004 (1)
.0041 (2)
.00096 (2)
.00063 (2)
.0185 (3)
.00029 (2)
.0164 (3)
.0409
When adding numbers alien the
numbers over the decimal, and look
for the number with the least
amount of accuracy to the right. In
this example 2nd, 5th, and 7th numbers
only have accuracy 4 places to the
right, so the final number can only
have 4 places past the decimal of
accuracy.
.0409 is a 3 sig fig so in the final
multiplication sequence, our answer
can only have 3 sig figs.
9
10
Data for Lab 1: Dimensional Analysis
Name ________________________________
Date _________________________________
Labmates ______________________________
MATH CONCEPTS IN CHEMISTRY
State the basic unit and symbol for the following quantizes.
a. length =
b: volume =
c, mass =
d. temperature =
e. time =
f. density =
SIGNIFICANT DIGITS
State the number of significant digits in the following measurements.
a. 0.707 g =
b. 0.10000 cm =
c. 2500 mL =
d. 0.0000110 g =
e. 100,000. cm =
f. 100,000 mL =
ROUNDING OFF NONSIGNIECANT DIGITS
Round off the following measurements to three significant digitsa. 12.59 mL =
b. 0.03662 g =
c. 31559745 cm =
d. 202577.94 mL =
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ADDITION AND SUBTRACTION
1. Perform the indicated mathematical operation. Express the answer with the proper units and
significant digits.
5000.
+
cm
1.0254 mL
.0256 cm
172.0
mL
0.00050 cm
+ 15.99
mL
60200
km
19.0 km
+
9.100 km
MULTIPLICATION & DIVISION OF SIGNIFICANT DIGITS
2. Perform the indicated mathematical operation. Express the answer with the proper units and
significant digits.
a. 21.1 cm × 20 cm
b. 5.15 cm × 2.55 cm × 1.00 cm
131.78 cm3
c.
19.26 cm
d.
131.78g + 45.43g − 17.9g
(20.117 × 29.0909)g + 432.449g
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Standard Deviation
Calculate the standard deviation for the following numbers:
9.845, 9.653, 8.998
Show all of your work.
CALCULATORS AND SIGNIFICANT DIGIETS
Using a Scientific Calculator
Students often purchase calculators that cost upwards of $200.00 and yet they do not know how to
use them. In this section, you should learn to
MASTER your scientific calculator.
All Calculators have different key setups ….so you will have to figure out where you SCIENTIFIC
NOTATION button is on your calculator. Locate ONE of the following scientific notation buttons on
your calculator (it may not even be listed here):
1. EE (Some Sharp)
2. EXP (Casio, TI, Shape, HP)
3. 2nd Function EE (many TI’s)
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At this point you will want to practice with you calculator.
First, enter the following
1  102
Times
5  103
=
You should get either 500,000 or 5  105 and either is correct. IF YOU DID NOT GET THAT ANSWER
THEN…….read on……….
DO NOT’s of Scientific calculators.
1. DO NOT USE the 10X key. DO NOT enter the above numbers as 1 times 102 times 5 times 103
even though you get the same answer….IT IS A WASTE OF KEY STROKES…you do not have time
to waste typing extra keystrokes. USE THE EE OR EXP BUTTON---THAT’S WHAT YOUR VERY
EXPENSIVE CALCULATOR IS MADE FOR.
2. DO NOT use the X2 or X3 buttons, again a waste of time.
Try this one
1  10-12
Times
5  10-5
=
You can only get ONE ANSWER 5  10-17…… IF YOU DID NOT GET THAT ANSWER THEN…….read on……….
You MUST ALSO learn to switch your calculator into SCIENTIFIC MODE.
1. Get your manual out and figure out how to get your calculator into SCI Mode.
2. You MUST see ME if you do not get this answer because your Chemistry 100 life will be a living
hell and it is very SIMPLE for me to show you how to work your calculator.
I hope this Helps!? If Not…SEE Me or ask someone for help!
Perform the indicated mathematical operation. Express the answer in decimal or Ordinary form (not
scientific notation) with the correct number of significant digits.
a. (4.65  105) (9.5  102)
96.77 10  0.0015110 
c.
3.49 10 
7
 4.97 10  3.24 10 
b.
56.7110 
8
4
2
8
10
14
TEMPERATURE
Learn the following 3 formulas for temperature formulas for your next exam
F =
9
C + 32
5
°C =
5
(°F - 32)
9
K = C + 273
Express the following temperatures in degrees Celsius.
(1) 1000. F
(2) -40. F
Convert the following temperatures in degrees Fahrenheit.
(1) 420.0 C
(2) -200. C
Express the following in Kelvin units.
(1) 190.0 F
(2) 980 F
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METRIC - METRIC CONVERSIONS
Convert the following metric system measurements using two conversion factors.
a. 120 mm to cm
a. 1.55cm to km
METRIC - ENGLISH CONVERSIONS
Convert the given measurements into the units indicated.
a. 1.01 lb to g
b. 360 cm to ft
GENERAL EXERCISES
Solve the following: SHOW ALL OF YOUR WORK FOR FULL CREDIT!
1. How many feet in 365.0 yards?
2. You threw the Frisbee 190. ft. How many yards did you throw the Frisbee?
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3. What is the cost of 1.5 pound of sugar if sugar costs $1.37 per 5.0 pounds?
4. The following relationships describe an antiquated set of British liquid units.
1 hogshead = 7 firkin
18 pottle = 1 firkin
140 pottle = 1 puncheon
504 pottle = 1 tun
Use dimensional analysis to determine the number of tuns in 144 hogshead. NOTE: It HELPS to write all
of your units out!!!
5. How many aspirin tablets can be made from 100. g of aspirin if each tablet contains 5.00 grains
of aspirin? (7.00 × 103 grains is equal to one pound) (454g = 1 lb)
6. What is the cost of 300.0 grams of aspirin if each tablet contains 5.00 grains of aspirin and 100
tablets cost $2.35? (7.00 × 103 grains is equal to one pound, 454 g is equal one pound.)
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7. Find the number of cm3 in 2.60 cubic foot.
8. How many µL are in 1 gallon
9. If one atom has a diameter of 0.372 nm and you have 2.62  1022 atoms (typical number of
atoms in a backpack), and you alien those atoms, like in a chain, one atom after another, how
many miles would this atom thick string be? The distance to the sun is 93,000,000 miles, how
many times can you go to the sun and back (round trips).
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Lab 2: Atomic Structure
Goals
•Write the correct symbols or names of some elements.
•Describe some physical properties of the elements you observe.
•Categorize an element as a metal or nonmetal from its physical properties.
•Given the complete symbol of an atom, determine its mass number, atomic number, and the number
of protons, neutrons, and electrons.
Discussion
Primary substances, called elements, build all the materials about you. Some look similar, but others
look unlike anything else. In this experiment, you will describe the physical properties of elements in a
laboratory display and determine the location of
elements on a blank periodic table.
A. Physical Properties of Elements
Metals are elements that are usually shiny or have a metallic luster. They are usually good conductors
of heat and electricity, ductile (can be drawn into a wire), and malleable (can be molded into a shape).
Some metals such as sodium or calcium may have a white coating of oxide formed by reacting with
oxygen in the air. If these are cut, you can see the fresh shiny metal underneath. In contrast, nonmetals are not good conductors of heat and electricity, are brittle (not ductile), and appear dull, not shiny.
B. Periodic Table
The periodic table, shown on the inside front cover of this lab manual and your textbook, contains
information about each of the elements. On the table, the horizontal rows are periods, and the vertical
columns are groups. Each group contains elements that have similar physical and chemical properties.
The groups are numbered across the top of the chart. Elements in Group I are the alkali metals,
elements in Group 2 are the alkaline earths, and Group 7 contains the halogens. Group 8 contains the
noble gases, which are elements that are not very reactive compared to other elements. A dark zigzag
line that looks like a staircase separates the metals on the left side from the nonmetals on the right
side.
C. Subatomic Particles
There are different kinds of atoms for each of the elements. Atoms are made up of smaller bits of
matter called subatomic particles. Protons are positively charged particles, electrons are negatively
charged, and neutrons are neutral (no charge). In an atom, the protons and neutrons are tightly packed
in the tiny center called the nucleus. Most of the atom is empty space, which contains fast-moving
electrons. Electrons are so small that their mass is considered to be negligible compared to the mass of
the proton or neutron. The atomic number is equal to the number of protons. The mass number of an
atom is the number of protons and neutrons.
atomic number = number of protons (p+)
mass number = number of protons plus number of neutrons (p+ + n°)
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D. Isotopes
Isotopes are atoms of the same element that differ in the number of neutrons. This means that
isotopes of an element have the same number of protons, but different mass numbers. The following
example represents the symbol of a sulfur isotope that has 16 protons and 18 neutrons.
Complete Symbol of an Isotope
mass number (p+ and n°)  34
symbol of element  S
atomic number (p+= e-)  16p+  16e-
Meaning
34
16
S
This atom has 16 protons and 18 neutrons.
The element is sulfur.
The atom has 16 protons and 16 electrons
Experimental Procedures
A. Physical Properties of Elements
Identify each element as a metal (M) or a nonmetal (NM).
B. Periodic Table
Materials: Periodic table, pencil, display of elements
B.1 On the incomplete periodic table provided in the report sheet, write the atomic numbers and symbols of the elements you observed in part A. Write the group number at the top of each column of the
elements (Groups 1-18). Write the period numbers for each of the horizontal rows shown (1-7). Shade
(with a pencil) & lable the alkali metals, alkaline earths, halogens, and noble gases, and the transition
elements. Draw a heavy line to separate the metals and nonmetals.
B.2 Use the periodic table to decide whether the elements listed on the report sheet would be metals
or nonmetals.
C. Subatomic Particles
For each of the neutral atoms described in the table, write the atomic number, mass number, and
number of protons, neutrons, and electrons.
D. Isotopes
Complete the information for each of the isotopes of calcium: the complete nuclear symbol and the
number of protons, neutrons, and electrons.
E. Isotopic mass calculations.
20
Data for Lab 2: Atomic Structure
Name ________________________________
Date _________________________________
Labmates ______________________________
1. On the following list of elements, circle the symbols of the transition elements and underline the
symbols of the halogens:
Mg
Cu
Br
Ag
Ni
Cl
Fe
2.Complete the list of names of elements and symbols:
Name of Element
Symbol
Name of Element
Symbol
Potassium
Na
Sulfur
P
Nitrogen
Fe
Magnesium
Cl
Copper
Ag
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A. Properties of Elements
Element
Symbol
Number
Metal/Nonmetal
Aluminum
______
______
_______
Carbon
______
______
_______
Copper
______
______
_______
Iron
______
______
_______
Magnesium
______
______
_______
Nickel
______
______
_______
Nitrogen
______
______
_______
Oxygen
______
______
_______
Phosphorus
______
______
_______
Silicon
______
______
_______
Silver
______
______
_______
Sulfur
______
______
_______
Tin
______
______
_______
Zinc
______
______
_______
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B. Periodic Table B.1 (Note follow B.1 instructions in the Experimental Procedure above)
Questions and Problems
Q.1 From their positions on the periodic table, categorize the following elements as metals (M) or
nonmetals (NM).
Na
S
Cu
F
Fe
C
Ca
_____
_____
_____
_____
_____
_____
_____
Q.2 Give the name of each of the following elements:
a. Noble gas in Period 2 _________________
b. Halogen in Period 2 _________________
c. Alkali metal in Period 3 _________________
d. Halogen in Period 3 _________________
e. Alkali metal in Period 4 _________________
f. The first metal in Group 14 _________________
g. The second nonmetal in Group 16 _________________
23
B.2
Element
Identify as either a Metal/Nonmetal
Chromium
Gold
Carbon
Cadmium
Sulfur
C. Subatomic Particles
Element
Atomic
Number
Mass
Number
Protons
Neutrons
Iron
Electrons
30
27
13
Bromine
80
Gold
197
19
20
53
74
D. Isotopes
Nuclear Symbol
40
20
43
20
Protons
Neutrons
20
22
Ca
Ca
24
46
20
Electrons
20
Ca
24
E. Isotopic Mass Calculations
Isotopes are the same element…with different numbers of neutrons (see above).
If you look at the periodic table you notice that the atomic mass of lithium is 6.94 amu.
Mass  % Abundance
= Mass  (% Abundance/100)
6
Li
6.015
7.42
0.4463
7
Li
7.016
92.58
6.4954
100.00 %
6.94
Total
Li
To calculate the Atomic Mass or Atomic Weight of Li, as seen on a Periodic Table we must find the total
mass of all the Isotopes of Li as we find them in nature. The general formula is as follows:
mass of isotope #1  % abundence #1 mass of isotope #2  % abundence #2

+ ...
100
100
= Ave AMU
Calculating Atomic Weight for Lithium
Li = Mass 6Li  (% Abundance 6Li /100) + Mass 7Li  (% Abundance 7Li /100)
total
or
Li = 6.015  0.0742 + 7.016  0.9258 = 6.94
total
1. A sample of chlorine has two naturally occurring isotopes. The isotope Cl-35 (mass 35.11 amu)
makes up 75.83% of the sample, and the isotope Cl-37 (mass = 37.69 amu) makes up 24.17% of the
sample. What is the average atomic mass for chlorine?
2. Copper has two naturally occurring isotopes. A typical sample consists of 69.17%
amu) and 30.83%
65
29
63
29
Cu (62.9395
Cu (64.9273 amu). Calculate the atomic mass of copper.
25
26
Lab 3: Periodic Properties
Goals
Draw a graph of atomic diameter against atomic number. Interpret the trends in atomic radii within a
family and a period.
Electron Configuration
In an electron configuration, electrons are arranged by subshells starting with the lowest energy. The
number of electrons in each subshell is written as a superscript. The electron arrangement of an
element is related to its position in the periodic table. The electron configuration can be written by
following the subshell blocks across the periodic table starting with period 1. The s- block is formed by
Groups 1 and 2. The p-block includes the elements in Groups 13 to 18. The d- block includes the
elements in Groups 3 to 12.
Examples of Groups 1, 16 and 18 with Periods 2, 3 and 4 are as follows:
Li 1s22s1
O 1s22s22p4
Ne ls22s22p6
Na ls22s22p63s1
S ls22s22p63s23p4
Ar ls22s22p63s23p6
K ls22s22p63s23p64s1
Se ls22s22p63s23p64s23d104p4
Kr ls22s22p63s23p64s23d104p6
A. Graphing a Periodic Property: Atomic Radius
Since the 1800s scientists have recognized that chemical and physical properties of certain groups of
elements tend to be similar. A Russian scientist, Dmitri Mendeleev, found that the chemical properties
of elements tended to recur when the elements were arranged in order of increasing atomic mass. This
repetition of similar characteristics is called periodic behavior. He used this periodic pattern to predict
the characteristics of elements that were not yet discovered. Later, H. G. Moseley established that the
similarities in properties were associated with the atomic number.
In the electron arrangement of an element, the electrons in the highest or outermost energy level are
called the valence electrons; note in the above electron configurations the bolded shells, subshells and
electrons. The valence electrons determine the chemical properties of the elements. If the elements
are grouped according to the number of valence electrons, their chemical and physical properties are
similar. The similarities of behavior occur periodically as the number of valence electrons is repeated.
In this exercise, you will graph the relationship between the atomic radius of an atom and its atomic
number. Such a graph will show a repeating or periodic trend. Observe the graph in Figure 1, which
was obtained by plotting the average temperature of the seasons. The graph shows that a cycle of high
and low temperatures repeats each year. Such a tendency is known as a periodic property. There are
three cycles on this particular graph, one full cycle occurring every year. When such cycles are known,
the average temperatures for the next year could be predicted.
27
Figure 1 A graph of average seasonal temperatures.
Lab Information
Time: 1 1/2 hr
Comments: Obtain a periodic table or use the inside cover of your textbook.
Tear out the Lab 3 report sheets and place them beside the procedures.
In neutral atoms, the number of electrons is equal to the number of protons.
Related Topics: Electrons and protons, energy levels, and electron arrangement
Experimental Procedures:
A. Electron Configuration
Write the electron configuration of each atom listed on the laboratory report. Indicate the number of
valence electrons and the group number for the element.
B. Graphing a Periodic Property: Atomic Radius
The atomic radii for elements with atomic numbers 1-25 are listed in Table 1. On the graph, plot the
atomic radius of each element against the atomic number of the element on the graph. Be sure to
connect the points. Use the completed graph to answer questions in the report sheet about valence
electrons and group number.
28
Table 1 Atomic Radii for the Elements with Atomic Numbers 1-25
Element Symbol
first period
hydrogen H
helium He
Atomic Number
Atomic Radius (pm*)
1
2
37
50
second period
lithium Li
beryllium Be
boron B
carbon C
nitrogen N
oxygen O
fluorine F
neon Ne
3
4
5
6
7
8
9
10
152
111
88
77
70
66
64
70
third period
sodium Na
magnesium Mg
aluminum Al
silicon Si
phosphorus P
sulfur S
chlorine Cl
argon Ar
11
12
13
14
15
16
17
18
186
160
143
117
110
104
99
94
19
20
21
22
23
24
25
231
197
160
150
135
125
125
fourth period
potassium K
calcium Ca
scandium Sc
titanium Ti
vanadium V
chromium Cr
manganese Mn
*(picometer = 10-12 m)
Electron Configuration and Periodic Properties
29
30
Data for Lab 3: Electronic Configurations and Periodic Trends
Name ________________________________
Date _________________________________
Labmates ______________________________
B. Electron Configuration
Atom
Electron Configuration
Number of Valence
Electrons
Group
4
4
O
Na
Ca
Ti
V
5
Br
Sr
Cs
1
Xe
8
Bi
5
Pb
4
15
Ra
31
Questions and Problems
# orbitals
Subshells
Max # electrons
1
s
2
3
p
6
5
d
10
7
f
14
Q.2 Complete the following electron shells, subshells, and configurations:
Number of subshells in 2nd energy level
Number of orbitals in the 2p subshell
Number of total electrons in only a 3d subshell
Number of total electrons in only a single 3p orbital
Period and Group number of carbon
First element that that begins to fill after the 4s is filled
Number of valence electrons in As
Q.3 Give the symbol of the element that meets the following information:
Element that fills the 3s sublevel
Period 4 element in the same group as F (fluorine)
Element with 3d6 electron configuration
Element with a filled 5p level
Element with five 3p electrons
The element with the electron configuration: 1s22s22p63s23p64s23d104p3
Element that completes 2nd energy level
Period 6 element in the same group as Mg
32
Graphing a Periodic Property: Atomic Radius
Atomic Radius vs. Atomic Number
Describe the change in the atomic radii for the elements in Period 2 from lithium to neon.
Why are the atomic radii of of elements in Period 3 larger then in Period 2?
33
34
Lab 4 Nomenclature
There are 5 (main) types/classes of Molecules in Inorganic Nomenclature…See Flow-Chart Handout
There are 2 Groups of Inorganic Compounds
1. Ionic
2. Covalent
Group 1: Ionic
There are 2 types of compounds in this group:
1. Fixed Oxidation States
2. Variable Oxidation States
Group 1 Ionic Compounds
Class 1. Fixed Oxidation State (O.S.)
Certain Metals have Fixed O.S., like Na, is always +1, and Sulfur is Always –2, so the names are written
using their metal and non-metal-ide name.
Full metal name + Nonmetal with an –ide ending.
NaCl = sodium chloride
Bromine
Carbon
Chlorine
Fluorine
Hydrogen
bromide
carbide
chloride
Fluoride
Hydride
BrC4ClFH-
Same rules applies to other main group ions Iodine, Nitrogen, Oxygen, Phosphorus, Sulfur
Examples:
Na2S
CaBr2
Ca4C
sodium sulfide
calcium bromide
calcium carbide
not disodium sulfide
not Calcium dibromide
not tetracalcium carbide
35
Class 2. Fixed Oxidation States
Metals with Fixed Oxidation States do not need any special Prefix because the O.S. IMPLIES how many
non-metals are around it.
HOWEVER------Transition Metals….need Special endings because they have Variable O.S.
Variable Oxidation states for Transition Metals
Chromium Cr (II) Cr2+, Cr (III) Cr3+
Iron Fe (II) Fe2+, Fe (III) Fe3+
Copper Cu+, Cu2+
Zinc Zn2+
Tin Sn2+, Sn4+
Lead Pb (II) Pb2+, Pb (IV) Pb4+
We will not use the –ous (lowest oxidation state) and –ic (highest oxidation state)
[Fe (II) Fe2+ ferrous, Fe (III) Fe3+ ferric or Cu (I) Cu+ cuprous, Cu2+ cupric]
Examples
PbS =
PbO2 =
Titanium (II) Chloride =
=
TiCl3
Titanium (IV) Chloride =
Metals with Polyatomic Ions
(Hint---1st determine if the metal has a fixed or variable OS before naming the compound)
The Anion is going to have a negative charge (except ammonium & Hydronium).
Nitride
N3Nitrite
NO2Nitrate
NO3-ide = no oxygen atoms
-ite = least # of oxygen atoms (helpful hint: little)
-ate = maximum # of oxygen atoms (helpful hint: lots )
Sulfide
Sulfite
Sulfate
S2SO32SO42-
Hydrogen Sulfate
Hydrogen Sulfite
HSO4HSO336
Phosphide
Phosphite
Phosphate
P3PO33PO43-
Hydrogen Phosphate HPO42Dihydrogen Phosphate
H2PO4Carbonate
CO32Cyanide
CNAmmonium NH4+
Hydronium H3O+
Permanganate MnO4Chromate
CrO42Dichromate Cr2O72Writing Polyatomic Ions
{cation}
{transition metal (oxidation state)}
NH4+ ClCa2+ OHCa2+ SO4 2-
NH4Cl
+
+
{anion}
{suffix of anion}
ammonium chloride
written CaOH2 or Ca(OH)2 calcium hydroxide
not Ca2(SO4)2 but CaSO4
calcium sulfate
Examples:
Fe (II) with sulfite, Fe (III) with sulfite
Cr(IV) with sulfate, Cr (III) with sulfite
Hydrate
“ “ = hydrate
Greek Prefix
1-mono” “
2-di
3-tri
4-tetra
5-penta
6-hexa
7-hepta
8-octa
9-nona
10-deca
Ba3(PO3)26 H2O =
calcium sulfate trihydrate =
37
Group 2: Covalent
There are 3 types of inorganic compounds in this group
1. Covalent Non-acid
2. Binary non-oxyo-acid and Oxyo-acids
Class 3. Covalent Non-acid
Use Greek Prefixes except….Never use mono if there is only one 1st element
Name the following:
CH4
BF3
P 2 O5
SO3 (note…no charge…this is NOT an ion)
Class 4. Binary Acids
When a molecular formula starts with an H then this is an acid. The H is called hydro- followed by the
anion-ic acid
Note: Never use the Greek Prefixes to describe the number of Hydrogen’s…the number of hydrogen’s
is assumed based on the OS of the anion
Name the following:
HF (aq)
HCl (aq)
H2S (aq)
Oxyoacids
Oxyoacids are formed from F, Cl, Br, and I. Do not confuse these with the Phosphates and Sulfates
already mentioned.
Polyatomic names ending with -ate  -ic
And names ending with –ite  ous
38
An example of Chlorine
Polyatomic
ClO4-
Polyatomic
Name
perchlorate
Potassium
salt
KClO4
ClO3-
chlorate
KClO3
ClO2-
chlorite
KClO2
ClO-
hypochlorite
KClO
Name
Acid
Acid Name
potassium
perchlorate
potassium
chlorate
potassium
chlorite
potassium
hypochlorite
HClO4
HClO3
perchlorate
acid
chlorate acid
HClO2
chlorite acid
HClO
hypochlorite
acid
So…it is possible to have—only the oxygenated halogen do this!
Polyatomic
Polyatomic
FO4-
Polyatomic
Name
Perfluoate
Polyatomic
BrO4-
Polyatomic
Name
Perbromate
IO4-
Polyatomic
Name
periodate
FO3-
Fluorate
BrO3-
Bromate
IO3-
iodate
FO2-
Fluorite
BrO2-
Bromite
IO2-
iodite
FO-
Hypofluorite
BrO-
hypobromite
IO-
hypoiodite
End Discussion
How do I look at nomenclature? This is how my mind works out nomenclature.
1st. Always Look at the First Element
1. Is it a metal
2. Is it a non-metal
2nd. If it is a metal
1. Does it have fixed OS
2. Does it have a variable OS
3. If it is a variable OS I figure out the OS
4. Figure out the anion
3rd. At this point it must be a covalent compound
1. Look for oxygen
2. If no oxygen is it
a. If a Hydrogen comes first it must be an acid
b. Otherwise it is Covalent
c. Figure out the anion
3. If it has oxygen it is an oxyo-acid
The following Chart covers the rules just covered. Put the appropriate “Class Number” next to the
molecule. This is always the first step in determining the name of your molecule.
39
40
Data for Lab 4: Nomenclature
Name ________________________________
Date _________________________________
Labmates ______________________________
1. SF2
14. Co2(C2O4)3
2. HgC2H3O2
15. Cu(CN)2
3. H2
16. CS2
4. KClO3
17. Co(NO3)3
5. H3PO4 (aq)
18. Al2(HPO4)3
6. KCN
19. Na2S
7. H2S (aq)
20. HFO4
8. LiBrO4
21. HFO2
9. P2O5
22. H2SO4
10. Ni(OH)2
23. CS2
11. ZnCO3
24. HBr(aq
12. HF(aq)
25. Lithium nitrite
13. HFO4
26. Dinitrogen heptaphosphide
41
27. Ammonium chloride
36. Titanium (IV) iodate
28. Chlorine
37. Titanium (III) iodate barium sulfate
29. Iron (IV) carbonate
38. calcium permanganate
30. dicarbon tetrasulfide
39. Iron (III) oxalate
31. hypoiodous acid
40. manganese (II) nitride
32. Sulfurous acid
41. Lithium bromide
33. Sodium chloride
42. Disulfur pentafluoride
34. Boron trifluoride
43. Fluoric acid
35. Aluminum chloride
44. Hypofluorous acid
42
Lab 5 Lewis Dot
G. N. Lewis, 1916 published a paper in JACS introducing the theory that a bond between two atoms
resulted from the sharing of 2 electrons. (Lewis was from Berkeley and is also known for his theory of
acid-base chemistry.) Lewis introduced this topic to his students to help them visualize how bonding
produced molecular shapes.
Lewis Structures are based on the idea that:
1. A bond consist of sharing 2 electrons
2. Atoms with partially filled valance strive to fill their valance shell
A simple example of the first idea “A bond consist of sharing 2 electrons” is illustrated below between two
p-orbitals atoms.
p-orbital
p-orbital
To describe the second idea “Atoms with partially filled valance strive to fill their valance shell” a new term
Octet-Rule is introduced. This rule applies to nonmetals usually in the first two periods, and roughly
applies to the remaining 4 periods as well.
1. Atoms continue to form bonds until all their vacant orbitals are filled. Octet refers to the number
8, which is the maximum number of electrons it takes to fill ns2 np6 period.
2. Once the ns2 np6 orbitals are filled the atoms has the same number of electrons as the Noble or
Inert Gasses. The word Inert doesn’t necessarily imply unreactive, more then it implies containing
the maximum number of electrons.
43
An example of fluorine: 19F 2s2 2p5
Fluorine is one electron from having a full octet. Adding a second electron fills the octet for Fluoride. No
more electrons can be added to fluorine, fluorine will NOT continue to form bonds.
+e2p5
2p6
2s2
2s2
F-
F
RULES FOR WRITING SIMPLE LEWIS STRUCTURES
Step 1: Draw a basic skeleton keeping the North, South, East, and West positions of the atomic symbol
in mind.
Step 2: Count the number of electron charge clouds surrounding the atom of interest.
Step 3: Place the electrons on the OUTSIDE first.
Step 4: Any remaining electrons go on the central atom.
Step 5: Check for the Octet
Step 6: Check for Resonances Structures
Step 7: Predict electron pair shape name, the molecular shape name and bond angles.
Examples of these rules with PF3, PO4-3 and NO3l. Write the chemical formula for the molecule (or ion) and determine the total number of valence
electrons in the molecule.
Valance Electrons for P Valance Electrons for F
Total Valance Electrons
PF3:
1 5
37
26
PO4-3: (1 5) + (4 6) + (3  1) = 32
NO3-: (1 5) + (3 6) + (1  1) = 24
44
2. Draw the skeletal arrangement of the molecule showing single bonds connecting the atoms.
F
F
O
P
O
O
O
N
P
O
F
O
O
3. Assume that each bond in the skeleton requires two valence electrons (an electron-pair bond). After
subtracting two electrons for each bond from the total number of valence electrons, assign the remaining
electrons to give each atom an octet, or share of eight electrons.
F
F
O
P
O
O
O
N
P
O
F
O
O
4. If after each atom has been given a share of eight electrons, additional electrons remain, assign the
extra electrons to the central atom of the molecule.
F
F
O
P
O
O
N
P
O
F
24 e-
O
O
O
24 e-
32 e-
5. If there are not enough electrons to give each atom a share of eight electrons, then form multiple bonds
between atoms by moving electron pairs to form double (or triple) bonds.
F
F
O
P
F
26 e-
O
O
32 e-
O
N
P
O
O
O
24 e-
45
Resonance Structures.
1
2
O
1
O
2
O
O
1
2
O
O
N
N
N
O
O
O
3
3
3
Possible Resonance Structures for NO3-
1
2
O
1
O
2
O
O
1
2
O
O
N
N
N
O
O
O
3
3
3
VSEPR: Valence shell electron-pair repulsion theory
To derive the VSEPR Electron Pair Name and VSEPR Molecular Shape Name you must first find the Lewis
Structure.
Lewis Structure  VSEPR Electron Pair Shape Name  VSEPR Molecular Shape Name
Electron pair is 1. Bonding Pair or 2. Lone Pair
Bond Polarity: based on Electronegativity
Pauling Electronegativity scale: Linus Pauling devised a scale of Electronegativity based on the relative
ability of a bonded atom to attract the shared electrons.
Pauling gave Fluorine a EN value of 4.0 and Lithium 1.0. All other atoms fall between 0.8 and 4.0. The
element with the smallest EN value is Cesium and Francium = 0.8.
46
H 2.1
Li 1.0 Be 1.5 …. B 2.0 C 2.5 N 3.0 O 3.5 F 4.0
Na 0.9 Mg 1.2 …. Al 1.5 Si 1.8 P 2.1 S 2.5 Cl. 3.0
K 0.8 Ca 1.0 ….
Br 2.8
For reference: Period 2 elements Fluorine, 4.0, has the GREATEST ability to Attract Electrons. Li has the
LEAST ability to Attract Electrons.
i.e. C-F, C-Cl, C-Br, C-I….rank from most to least EN.
Two types of bonds
a. Ionic---metal & nonmetal
b. Covalent---nonmetal & nonmetal
Bonds are further classified based on electronegativity.
1. Ionic (transfer of electrons)
2. Polar covalent (unequal sharing of electrons)
3. Nonpolar covalent (equal sharing of electrons)
Examples of Polar Covalent using Partial Positive + and Partial Negative -
C

C
F

F
Which is a Polar/Nonpolar: SO3 -vs- SO2
O
O
S
O
O
S
O
Polarity is determined by:
1. Electronegativity charge calculation
2. Presence of Lone Pairs (except in Triganol bipyramidal –Linear and Octahedral-square planer)
3. Symmetry of molecule
47
***Polarity is draw Polarity based on the Molecular Shape
VSEPR provides 3D arrangement of Groups around the central atom…directly related to bond angle
1st draw the Lewis dot structure
2nd determine the molecular 3D shape of the electron-pair name (Linear, Trigonal Planar, Tetrahedral)
3rd determine the VSEPR molecular shape name (Linear, Trigonal Planar, Angular (V-Bent), Trigonal
pyramidal, Tetrahedral)
Repulsion Ordering (deviations from expected bond angle)
Lone pair-Lone pair >> Lone pair-Bonding pair >> Bonding pair-Bonding pair
VESPR electron-pair name/molecular shape name
Trigonal
Angular
Linear
Trigonal Planar
pyramidal
Tetrahedral
(V-Bent)
molecular
molecular
molecular
molecular
molecular shape,
shape
shape
shape, Always
shape
Always Polar
Polar
Linear
electron-pair
Trigonal Planar
electron-pair
Tetrahedral
electron-pair
X
CS2, HCN
180
X
SO3, BF3
120
X
SO2, O3
<120
X
H2O, OF2
<<<109.5
X
NH3, PF3
<109.5
X
CH4, SiCl4
109.5
NOTE: The VSEPR GEOMETRY/Molecular shape NAME only takes into account the Shape between the
Atoms (bonding pair) attached to the central atom and disregards the lone pair of electrons:
REMEMBER THIS CHART!
48
Data for Lab 5: Lewis Structure
Name ________________________________
Date _________________________________
Labmates ______________________________
Molecule
# V.E.
Lewis Dot Diagram
Electron pair
shape name
Molecular shape Bond
name
Angles
Indicate with
polarity arrows
if polar or
nonpolar
SF2
COCl2
H2O
NH3
NH4+
49
N2O (the
central
atom is
the first
N)
O3
PCl3
CH2Cl2
BrO3-
SCN-
CS2
CH2O
50
Lab 6 Heat Capacity of Metals
Equipment:
1. 2 Styrofoam cups
2. thermometer
3. 100 mL graduated cylinder
4. Ring stand, iron ring, Bunsen burner
5. 600 mL beaker
6. 2 large test-tubes with fitted rubber stoppers
7. unknown metal
Today you will determine the identify of an unknown metal by calculating the heat capacity of your metal;
all substances have a different heat capacity.
Background: Today’s lab may seem little like a physics lab. However, you will be calculating a physical
characteristic of a metal hence, the connection to chemistry. The physical characteristic you will be
measuring is called heat capacity. Again, all substances have a different heat capacity. This lab causes
problems for all beginning chemistry students because we measure heat capacity indirectly. That is we
measure everything but heat capacity to make this determination. Do not be discouraged, all formulas are
carefully laid out for your convenience. If you can walk away from this lab understanding how heat works
you will be well on your way to understanding the principle of how heat works.
Often students don’t see all of the interconnectivity of what the lab is composed of. So lets take a look at
what heat is. Heat simply is how fast a molecule is moving. The faster a molecule moves the “hotter” or
more heat is posses. Think of a pool table setup before the queue ball breaks the triangle of balls. The
triangle of billiard balls are stationary. You aim the queue ball at the billiard balls, use the queue stick to
send the queue ball down the table to “break” apart the billiard balls. The force of the queue ball sets all
of the stationary billiard balls into motion. That is exactly how heat works. Fast moving molecules come
into contact with slower molecules and the faster moving molecules sets in motion the slower moving
molecules. This is why heat “flows” into cold. If you grab a glass of water with ice, the heat flows out of
your hand into the cool glass of water. In today’s lab we use the exact same principle. We put a hot object
into a cold object and measure the temperature change of the cold object; pretty simple.
What is Heat Capacity?
Cp 
energy
temperatur e change
Since energy is measured in Joules (J) the typical units for Cp are J/K or J/C. Another unit of energy is the
Calorie (cal).
51
Two types of Cp:
1. Molar Heat Capacity:
2. Specific Heat:
Cp 
C pm 
J
gK
J
mol  K
, we’ll be measuring this today in lab.
In a typical experiment to measure Cp, energy is measured using a water bath and temperature change is
measured with a thermometer. The reason for the water bath is joules and calories, units of energy, are
defined as the quantity of energy to increase the temperature of 1 g of water 1C. Therefore, if the
appropriate experimental conditions are chosen carefully, measuring Cp experiments can be preformed
using simple equipment….a Styrofoam cup for example.
Heat and Enthalpy
Heat is defined as:
Heat = q = Enthalpy + work
In chemistry, we often ignore the work term. Imagine a car motor where the pistons are allowed to move
up and down. The pistons turn a crankshaft hence the car moves. We know pistons move up and down
because gas gets in the cylinder, is ignited, and expands to move the pistons….the pistons must move
against the pressure of the atmosphere. Under these conditions, the volume of the system is constantly
changing and pressure is constant. However, in chemistry, we like to keep the volume constant and let
pressure change because we don’t “like” our experiments to do we work, we just want to observe the
reaction take place…we want all the heat to go into the molecules themselves. So chemist often define
Heat as:
Heat = q = Enthalpy (H)
We can now write our Heat or Enthalpy equation as we see it in the chemistry laboratory:
H = q = mCp T
For any lab experiment it is easy to identify the Cp of any substance if we know it’s mass and it’s change in
temperature (T). T is change in temperature and always written as T = Tf – Ti.
Our final statement about Heat is the principle called “Conservation of Energy.” This says, in the simplest
terms, energy in is equal to energy out, or you have to account for all the energy in the system and it must
equal 0. Today’s lab we will measure the
Heat in (cold object) + Heat out (hot object) = 0
Heat in (cold object) = -Heat out (hot object)
52
Experimental Procedure
1. Obtain a 600 mL beaker (hot water bath) and fill it ¾ full with deionized water (DI) and begin
heating the water to boiling
2. Obtain an unknown metal, record its number in your lab report. Using a scale weigh, recording the
EXACT weight, approximately 80 g of metal by weighing it into a beaker. Transfer the metal to a
large test tubes and repeat for the second test tube. Place the two test tubes with metal into the
400 mL beaker. LOOSLEY fit a rubber stopper on top of each test tube.
3. Allow the test tubes of metal to heat in the hot water bath (boiling water) for at least 40 minutes.
4. Dry the inside and outside of each Styrofoam cup, placing one cup inside the other, forming a
double-walled insulated cup; both cups are nestled inside each other.
5. Measure Exactly 50.0 mL of water, USING a 50.0 mL Pipet with Pipet Bulb into the Styrofoam cup
setup (2 Styrofoam Cups/dry/nestled inside each other).
6. Carefully pour one test tube of hot metal into the cold water, and while constantly stirring record
the Highest temperature in the Styrofoam cups…it will take between 30 seconds to 2 minutes for
this to happen. After recording your highest temperature achieved carefully pour out the water
into the sink, then dump the metal back on a piece of paper.
7. Repeat step 4-6 for the second hot metal USING fresh water…you do not need to record the
temperature of the cold water again…use the same temperature you recorded for the first
experiment.
8. Dry metal before putting back into it’s jar
Calculations
You must first measure the Heat capacity of the Styrofoam cups. Remember we are adding hot water to
cold water. While most of the heat from the hot water is transferred to the cold water, some of the heat
goes into the Styrofoam cups; it only makes sense. To account for all the heat we write the following.
Hcal + HCW + HHW = 0
Or
qcal + qCW + qHW = 0
Where cal = Styrofoam cups, cw = cold water, and hw = hot water
Since q = mCp T, we can be rewrite the above as:
mCp T +
Styrofoam
Cup
mCp T + mCp T = 0
cool
water
hot
water
So referring back to the previous equations we see that heat is being absorbed by the Styrofoam cup and
according to the Law of Conservation of Energy, we need to keep track of heat lost to the Styrofoam cup.
Of course any heat added to a Styrofoam cup with cold water, heat is going to be lost to the glass
thermometer and air of the surroundings, however these are negligible for our purposes.
53
Also note what T is. This is the change in temperature of the cold and hot water. In other words the
cold water starts cold but warms up once the hot water is put into it. The hot water starts hot but cools
down when put in the cold water. Since T
= Tf – Ti and since we mix BOTH the cold and hot water the
FINAL TEMPURATURE or Tf are BOTH THE SAME! Keep this in mind.
Careful inspection shows that the only constant we don’t know is C p for the Styrofoam cups; Cp for cool
and hot water (4.184 J/gK).
We change mCp T for the Styrofoam cup to BTCW to account for mass of Styrofoam and glass
thermometer, this is just an effective value.
From before: Hcal + HCW + HHW = 0
Final form for the calorimeter is:
Hcal = -(HCW + HHW)
or
BTCW +
mCWCp TCW + mHWCp THW = 0
Styrofoam
Cup
cool
water
Or
B=
hot
water
H cal
Tcw
Here, you do not calculate “B.” For two 16 oz Styrofoam cups you can use the value of “25 J/g.”
NOTE: Measure using your Graduated Cylinders the 40.0 mL remember that the density of water is as
follows: 1 gH2O = 1mLH2O, so 40.0 mL of H2O = 40.0 g H2O.
Your job is to calculate the Cp of your metal. In the following formula you know every variable expect
Cpmetal
BTCW + mCWCp TCW + mmetalCpmetal Tmetal = 0
Styrofoam
Cup
cool
water
hot
water
Compare your heat capacity (Cp) with the following to identify your unknown metal; all of units of J/gK
Know Heat Capacities of metals
H2O = 4.18
Mg = 1.04
Co = 0.456
Sn = 0.217
Al = 0.904
Fe = 0.473
Ni = 0.444
Cu = 0.387
Zn = 0.386
Pb = 0.128
Au = 0.129
54
Data for Lab 6: Heat Capacity
Name ________________________________
Date _________________________________
Labmates ______________________________
Data: Mass RECORD
ALL DIGITS FROM THE DIGITAL DEVICE
1. Unknown Metal number ________________________
2. Mass of metal in test tube #1 _____________________________
3. Mass of metal in test tube #2 _____________________________
Data: Temperatures
4. Temperature of Cold water (in cup) _____________________
5. Temperature of hot water (boiling water in beaker) _____________________
6. Final Temperature (cold water and metal) of Metal experiment #1 _____________________
7. Final Temperature (cold water and metal) of Metal experiment #2 _____________________
Calculations
Part 1.
1. Referring to your data tables above
8. Experiment #1 TCW = Tf – Ti = #6 - #4 = __________________
9. Experiment #1 THM = Tf – Ti = #6 - #5 = __________________ (this is a negative #)
10. Experiment #2 TCW = Tf – Ti = #7 - #4 = __________________
11. Experiment #2 THM = Tf – Ti = #7 - #5 = __________________ (this is a negative #)
*Since you used exactly 50.0 mL of cold water = 50.0 g
55
WATCH YOUR SIG FIGS!!!!! **Note: an average “B” is = 25 J/(C)
mCWCp TCW + mmetalCpmetal Tmetal = 0
BTCW +
Part 2.
12: Exp #1: BTCW = 25 J/(C)  _________ (8 above) = ________________ Note units
13: Exp #1: mCWCp CWTCW = 50.0 g  4.186 J/(gC)  _________ (8 above) = ________________
14: Exp #2: BTCW = 25 J/(gC)  _________ (10 above) = ________________
15: Exp #2: mCWCp CWTCW = 50.0 g  4.186 J/(gC)  _________ (10 above) = ________________
Exp #3 Cpmetal =
-
BT  H CW

mmetal THM





#3 & #7 found on the first data page.
Exp #4 Cpmetal =
-
BT  H CW

mmetal THM
Average your two values of Cpmetal. Show work.

(units 3 pages above)
Cpmetal =
UNKNOWN Metal = _______________________________ identify with metal in above table.
Now, take your Cpmetal and divide it into 24.9, this is known as the Delong Petit formula. This will find the
molar mass of your metal.
Also try: (24.9/Cp)
24.9 (
𝐽
𝑚𝑜𝑙𝐾
𝐶𝑝 𝑦𝑜𝑢𝑟𝑠
)
𝐽
( )
𝑔𝐾
=
𝑔
𝑚𝑜𝑙
;
where Cp is your metals calculated heat capcity.
56
Lab 7: Chemical EQUATION: WRITING BALANCED EQUATIONS
OBJECTIVES: Given the names of the reactants and products, you should be able to write the correct
formulas for these in equation form.
You should also be able to write the balanced equations given the above information, and to identify the
following types of reactions.
1. Combination or Synthesis
2. Decomposition
3. Complete oxidation or combustion (burning of organic compounds)
4. Single Replacement (a-type of Redox or oxidation / reduction reaction).
5. Double Rep1acement (ion combination) of the following types
a, Precipitation Rxn.
b. Neutralization Rxn.
INSTRUCTIONS:
Part 1. Perform the experiments for Combination, Decomposition, Combustion, Single and Double
replacements.
Parts 2-4. Write the balanced chemical equation for each reaction described below. Remember that the
process of writing equations involves two steps:
1. Write the correct formula for each reactant and product. The subscripts must be correct. If
necessary, consult your ion chart and periodic table.
2. Ba1ance the number of atoms of each element on each side of the equation only by using
coefficients (DO NOT CHANGE SUBSCRIPTS) If the coefficient is one, no coefficient is necessary.
3. For these reactions assume the oxidation number of the metal does not change upon reactions to
form products
Examples
1. Combination or Synthesis Rxn.
Ammonia (g) and su1furic acid combine to form ammonium sulfate.
2NH3 (g) + H2SO4  (NH4)2SO4
2. Decomposition Rxns.
When heated, potassium chlorate(s) decomposes into oxygen gas and potassium chloride solid.
KClO4  2O2 + KCl
57
3. Combustion or oxidation of organic compounds.
Propane(g), C3H8 burns in air (with O2).
C3H8 + 5O2  3CO2 + 4H2O
4. Single Replacement (Oxidation-Reduction)
Hydrogen(g) is released when aluminum metal reacts with hydrochloric acid.
Al + HCl  AlCl3 + H2
5. Double Replacement (neutralization or precipitate formation)
Barium carbonate precipitates from the reaction of barium chloride and sodium carbonate solutions.
BaCl2 + Na2CO3  BaCO3 + 2NaCl
Lab Procedure
DO EXPERIMENT 4 1st! It takes 30 Minutes.
Part 1.
1. Combination Reaction
a. Obtain 4 mL of Ni(H2O)62+ in a 50mL beaker (look for NiCl2, same thing: NiCl2 + H2O  Ni(H2O)62+)
b. To this solution add about 20 mL of water
c. Next add about 2 g NH4NO3
d. Then add 5 mL of con. NH3 (look for NH4OH, same thing: NH3 + H2O  NH4OH )
2. Decomposition
a. In a small test tube add about 1 gram of copper(II) sulfate pentahydrate (CuSO4•5H2O)
b. Holding the test tube with test tube tongs gently warm the test tube until you begin to see water
droplets form at the top of the test tube
3. Complete oxidation or combustion (burning of organic compounds)
Use extreme caution, open flames are hazardous.
a.
b.
c.
d.
Obtain a watch glass and a dropper of ethanol
Place about 3 drops of ethanol on the watch glass
Ignite a wood splint using Bunsen burner
Using the burning wood split ignite the ethanol on the watch glass.
4. Single Replacement (a type of Redox or oxidation / reduction reaction).
a. Obtain 20mL of 1.0M CuSO4 solution in a 50mL beaker, And 2 mL of 1.0M CuSO4 in a test tube;
keep the test tube for the end of the 30 min time period to compare the color of the original color
of the solution (in your test tube) the color of the beaker at the end of the experiment.
b. To the beaker with solution add a single piece of zinc metal
c. At the end of the experiment, recover the zinc metal, wash and dry it.
5. Double Rep1acement (ion combination) of the following
a. Obtain about 5mL of 1.0M iron(III) chloride (FeCl3) in a small beaker
b. Obtain about 5mL of 1.0M potassium thiocyanate (KSCN) in a small beaker
c. Pour the two reagents (reactants) together
58
Data for Lab 7: Writing Chemical Equations
Name ________________________________
Date _________________________________
Labmates ______________________________
Observation:
Part 1. Combination Reaction
Ni2+(aq) + NH3(ag)  Ni(NH3)62+(aq)
What is the color of the original Ni2+ solution _____________
What is the color of the final Ni(NH3)62+ solution _____________
Part 2. Decomposition
CuSO4•5H2O(s)  CuSO4(s) + 5H2O(g)
What is the color of the original CuSO4•5H2O solid _____________
What is the color of the final CuSO4 solid _____________
Part 3. Complete oxidation or combustion (burning of organic compounds)
Use extreme caution when handling ethanol (C2H5OH), do allow it to become wet (contact with wet fingers
or wet test tube), and do NOT dispose of remaining CaC2 in the sink.
The combustion reaction of acetylene is as is as follows: Balance the equation.
_____ C2H5OH(g) + _____O2(g)  _____CO2(g) + ____H2O(l)
59
What observation did you notice (smell, sound, color, light etc…) when you ignited the acetylene gas (be
brief).
Part 4. Single Replacement (a type of Redox or oxidation / reduction reaction).
Make observations as follows:
1. Initial color of solution and metal:
2. Color of solution and metal after 15 minutes:
3. Color of solution compared to your small test tube and appearance of new metal after 30 minutes:
Write a Single Replacement reaction between zinc metal and CuSO4 below
Part 5. Double Replacement (ion combination) of the following
Observations:
1. Color of each original solution FeCl3 _______________ & KSCN ___________________
2. Final color of mixed solution ______________________________________
Write the double replacement reaction below
60
Part 2. Rewrite the following word equations into balanced chemical equations.
1. Sulfur trioxide combines with water to form sulfuric acid.
2. Magnesium metal and a solution of silver nitrate will react to form silver metal and a blue solution.
3. Sodium metal reacts violently with water to form hydrogen and sodium hydroxide
PART 3: Beneath each word equation write the formula equation and balance it
4. Zinc metal + sulfuric acid  Zinc sulfate + hydrogen (g)
5. Aluminum metal + hydrochloric acid  aluminum chloride + hydrogen
PART 4: Complete and balance the following double replacement reaction equation. Assume all reactions
go to completion
7. Na2CO3 + HCl 
8. NH4Cl + NaOH 
9. H2SO3 + NH4OH 
61
62
Lab 8 Formula of a Hydrate
A. Introduction
Many compounds are formed in reactions that take place in water solutions. The water is then evaporated
to obtain the crystalline compound. In some cases water molecules are weakly attracted to the ions or
molecules that make up the compound and are retained within the crystal structure. Crystalline
compounds that retain water during evaporation are referred to as being hydrated or are said to contain
water of hydration. The ratio of moles of water to moles of compound is a small whole number. The
formula for the hydrated compound copper (II) sulfate is
CuSO45H2O
The dot shows that for every mole of CuSO4 in the crystal, there are 5 moles of H2O. The ratio of moles of
H2O to moles of compound can be determined experimentally in most cases by heating to remove the
water. The compound with the water removed is anhydrous. In this experiment you will determine the
formula for a hydrated unknown compound. The formula is determined by comparing the mass of the
hydrated and anhydrous forms of the compound.
Definitions:
∆
CuSO45H2O
Hydrate
CuSO4 + 5H2O
anhydrous water
The “” means to heat.
B. Procedure and Data
1. Clean the crucible first, remove excess residue 1st before heating. After this point NEVER touch your
crucible and lid with your fingers, the oils on your fingers will actually affect your mass.
2. Obtain a clean crucible and lid, support there on a clay triangle, and heat with an intense flame for 5
minutes. Allow cooling. Weigh the crucible and lid together. Handle the crucible and lid with the crucible
tongs (or a test tube holder) for the rest of the experiment; do not use your fingers.
Note: Use a ring stand, and Bunsen burner. The Blue Inner-Cone is the hottest part of the Flame. Position
your crucible a few cm above the Inner-Cone.
63
3. Add a maximum of 1.5 (1-1.5g) g of the CuSO4 hydrate (CuSO45H2O) to the crucible and weigh the
crucible, lid, and sample together. Record.
Note: Use the Analytical Scales for the entire lab.
4. Place the crucible with the cover on a clay triangle over a laboratory burner.
5. At first, heat the sample slowly and then gradually intensify the heat. Do not allow the crucible to
become red-hot. This can cause the anhydrous salt to decompose. Heat the sample for 15 minutes. Cool to
room temperature, and weigh the crucible, lid, and sample together.
6. Repeat with a second sample---2 Trial
7. Collect ALL Ba solid in waste container
64
Data for Lab 8: Formula of a Hydrate
Name ________________________________
Date _________________________________
Labmates ______________________________
RECORD ALL DIGITS FROM THE DIGITAL DEVICE
Trial 1
1. Empty “CLEAN” Crucible & Cover mass after heating 2 minutes __________________
2. Crucible & Cover with sample (before heating = hydrate) __________________
3. Crucible & Cover with sample after 20 min heating __________________
4. Mass of hydrated compound used (2-1) __________________
5. Mass of anhydrous compound produced (3- 1 ) __________________
6. Mass of water lost (4 - 5 ) _________________
Trial 2
7. Empty Crucible & Cover mass after heating __________________
8. Crucible & Cover with sample (before heating = hydrate) __________________
9. Crucible & Cover with sample after 20 min heating __________________
10. Mass of hydrated compound used (8-7) __________________
11. Mass of anhydrous compound produced (9 - 7 ) __________________
12. Mass of water lost (10 - 11 ) _______________
65
C. Calculations and Results
1. Determine the percentage of water in the hydrated compound for each trial and the average
percentage of water.
%H 2O 
mass H 2O
 100
mass of hydrate
Trial #1:
%H 2O 
mass H 2O
 100
mass of hydrate
Trial #2:
%H 2O 
mass H 2O
 100
mass of hydrate
Average %H2O: Show Work.
%H2O =
2. Calculate the formula based on the number of moles of anhydrous compound produced and the
number of moles of water lost.
moles CuSO 4 
moles H 2 O 
___g CuSO 4 1 mole CuSO 4

# moles CuSO 4
1
159.5 g CuSO 4
___g H 2 O 1 mole H 2 O

# moles H 2 O
1
18.0 g H 2 O
Show your work for Trial #1 & Trial #2
66
3. Determine the formula for the hydrate
For example if you find you get 0.0010 moles of CuSO4 and 0.0052 moles of H2O do the following by taking
the smaller number into the larger number:
0.0052/0.0010 = 5.2 moles of H2O
0.0010/0.0010 = 1.0 mole CuSO4
Your molecular formula would be CuSO45.2H2O. You do NOT need to get whole numbers---for This
Experiment ONLY---so do not round.
Show your formulas from each trial below
4. Calculate the theoretical % error.
What is the % error? As in 4 above the example showed the formula to be CuSO45.2H2O. The
experimental
formal is supposed to be: CuSO45H2O:
Base on the value obtained in 4 (for illustrational purposes only) my error would be
% Error 
% Error 
Exp-Theory
 100
Theory
5.20-5.00
100  4.0% error
5.00
Do this for each trial, then find the average error.
WRITE YOUR FINAL FORMULA OUT HERE:
67
68
Lab 9 Equilibrium
PURPOSE
To observe a number of interesting and colorful chemical reactions that are examples of chemical systems
at equilibrium. To see how these systems respond to changes in the concentrations of reactants or
products or to changes in temperature. To see that the direction of the shift in the equilibrium tends to at
least partially offset the change in conditions, a principle first dearly stated by Le Châtelier.
PRE-LAB PREPARATION
Any chemical equation that describes an actual chemical reaction sums up the result of an experiment
(more often, several experiments). Someone had to measure out the reactants, carry out the reaction,
identify the products, and measure quantitatively the number of moles of products produced per mole of
reactant consumed. The last step is the writing of a balanced chemical equation that concisely summarizes
the experimental observations.
Many chemical reactions go essentially to completion, but some do not, stopping at some point of
equilibrium that lies between no reaction and an essentially complete reaction. A state of equilibrium is
the point of balance where the rate of the forward reaction equals the rate of the backward reaction. For
these reactions, the same point of equilibrium can be reached from either end of the reaction, by mixing
together either the reactants or the products, dearly indicating that chemical reactions can go either
forward or backward.
In principle, every chemical reaction is a two-way reaction, but if the point of equilibrium greatly favors the
reactants, we say that there is no reaction. If the point of equilibrium greatly favors the products, we say
the reaction is complete. For either of these two extremes it is difficult to measure experimentally the
concentrations of all of the reactants and products at equilibrium. In the first case (reactants greatly
favored), the concentration of products is practically zero. In the second case (products greatly favored),
no significant quantity of the reactants will be produced when the products are mixed together. Thus,
when the equilibrium greatly favors the products, the reaction appears to go in only one direction:
reactants  products.
Nearly every chemical reaction consumes or releases energy. (We call these, respectively, endothermic or
exothermic reactions.) For these reactions, we can regard energy as if it were a reactant or product, and by
adding energy to a system (by heating it) or removing energy from a system (by cooling it), we can produce
a shift in the point of chemical equilibrium. The concentrations of reactants and products change to reflect
the new equilibrium.
There is an added complication that must be considered. Not all chemical reactions are rapid In fact, some
are so slow that you might be fooled into thinking that no reaction occurs, whereas the equilibrium point,
when it is finally reached, might greatly favor the products.'rhe rate at which a chemical reaction reaches
equilibrium will be taken up when you study the kinetics of chemical reactions.
For reactions that rapidly reach equilibrium, the point of equilibrium may be approached from either the
side of the reactants or the side of the products, which emphasizes the dynamic nature of chemical
69
reactions. The reactions you will study in this experiment are all rapid reactions, so you will be able to
observe almost immediately the effects of changing the concentration of either reactants or products. In
particular you will observe that when you change the concentration of a reactant or product, the point of
equilibrium shifts in a
direction that tends to offset the change. This behavior can he summarized in a general principle that was
first htuy stated in 1884 by the French chemist Henri Louis Le Chateher: A chemical reaction thin is
displaced from equilibrium by a change ht conditions (concewratiorts, temperature, pressure, volume)
proceeds toward a new equilibrium state in the direction that cat least partially o/]sets the change in
conditions.
This introductory experiment is designed to show qualitatively several important features of chemical
equilibria-ln subsequent experiments you will learn about the quantitative aspects of chemical equilibria,
measuring the equilibrium constants for a variety of reactions, and making calculations of equilibrium
concentrations from previously measured equilibrium constants.
COMPLEX ION EOUILIBRIA. It is common for cations (especially those with +2 or +3 charge) to attract
negatively charged ions (or neutral molecules with lone pairs of electrons) to form aggregates called
complexes. If the resulting aggregate has a net charge, it is called a complex ion. The composition of these
complexes may vary with the proportion and concentration of reactants.
Part 1: The Thiocyanatoiron (III) Complex Ion. This ion, sometimes called the ferric thiocyanate complex
ion, is formed as a blood-red substance described by the following equilibrium equation.
Fe3+ + SCN- ⇄ Fe(SCN)2+
(1)
In a 100-ml, beaker add 2 ml of 0.1 M Fe(NO3)3, to 2 mL of 0.1 M KSCN. Dilute by adding 60-70 mL of water
until the deep-red color is reduced in intensity, making further changes (either increases or decreases in
color) easy to see. Split this solution into 4 test tubes, so each test tube is about 2/3 full. To test tube #1
add 1 mL (about 25 drops) of 0.1 M Fe(NO3)3 (Equation #1 in Data Section). To the second test tube, add I
mL of 0.1 M KSCN (Equation #2 in Data Section). To a third add 5 to 6 drops of 6 M NaOH (Equation #3 in
Data Section). NOTE: Fe(OH)3, is very insoluble. Use the fourth portion as a control (use this test tube to
compare color with 1-3). Compare the relative intensity of the red color of the thiocyanato complex in
each of the first three test tubes to that of the original solution in the fourth tube. Interpret your
observations, using Le Châtelier’s principle and considering the equilibrium shown in Equation (1) above.
Part 2: The temperature-Dependent Equilibrium of Co(II) Complex Ions. The chloro complex that consists
of cobalt(II), CoCl42- is tetrahedral and has a blue color. The aquo complex of cobalt(II) Co(H2O)62-, is
octahedral and has a pink color. (Figure 1 shows the geometry of the tetrahedral and octahedral
complexes.) There is an equilibrium between the two forms in aqueous solution, and because the
conversion of one form to another involves a considerable energy change, the equilibrium is temperature
dependent:
CoCl42- + 6H2O ⇄ 4Cl- + Co(H2O)62- + energy
(2)
70
Le Chatelier's principle applied to Equation (2) predicts that if energy is removed (by cooling the system),
the equilibrium tends to shift toward the aquo complex, because a shift in this direction produces some
energy, thus partly offsetting the change.
Put 2 mL of 0.10 M CoCl2 (in methanol) into a 13 x 100 mm test tube (medium sized test tube); methanol is
used as a solvent so that you can observe the effects of adding water. Using a dropper, add just enough
water to the purple (it may be light pink to begin with) methanol solution to change the color to that of the
pink aquo complex. Next, add concentrated (12 M) HCl (Equation #4 in Data Section) one drop at a time
until you observe a color change to a blue color. Record your observations. Cool this test tube (Equation #5
in Data Section) in an ice bath and record your observations. Heat the test tube (Equation #6 in Data
Section) in a beaker of hot water (65 to 70°C). (Caution: Methanol vapors are toxic and flammable.) Use
EXTREME CAUTION to not let the methanol solution come into contact with your open flame. You
should note a color change. (If you do not, you probably did not add enough water to the original
methanol solution. Try again.) The color change is reversible. Cooling the solution in an ice bath will
restore the original pink color. Repeat the cycle of heating and cooling to verify this. Record your
observations, and interpret them by applying Le Châtelier’s principle described below.
H2O
OH2
Co
2+
OH2
ClH2O
-
Cl
H2O
H2O
-
Co2+
Cl-
Cl
Blue Tetrahedral
Co(II)
(hot)
Pink Octahedral
Co(II)
(cold)
FIGURE 1
The cobalt(II) complexes with chloride ion and water have different molecular geometries and different colors.
Procedures:
1. EXTREME CAUTION to not let the methanol solution come into contact with your open flame
2. Addition of reagents is approximate, no need to measure reagents i.e. use about a mL of reagent
3. Add enough reagent until you see a chemical reaction take place
4. All solutions are in the hood
5. Dispose of the FeSCN and any Co+2 solutions in appropriate, separate waste containers in the fume
hood
71
Applying Le Châtelier’s Principle
1. Write the equilibrium equation for the system.
2. Write the equation for the added substance
3. Use arrow(s) to show the stress. Up arrows,  , for increased concentrations and down arrows,  ,
for decreased concentrations.
4. Use an arrow to show the shift in position of equilibrium.  or 
5. Use arrows to show the resulting change in concentrations due to the shift described in step
3
6. Similar to 4, though use an arrow  or  to show the change in color or other noted change
Example of how to write Reaction #1. Fe(SCN)+2 plus Fe +3
#1 Fe(SCN)+2 plus Fe+3
Fe+3 + SCN-
1.
2.
Fe(SCN)+2
Fe+3 + 3NO3-
Fe(NO3)3
3–6
Fe+3 + SCN3
Fe(SCN)+2
5
5
4
6
~colorless
dark red
72
Data Sheet for Lab 9 Equilbrium
Name ________________________________
Date _________________________________
Labmates ______________________________
Observations based on color:
#1 Fe(SCN)+2 plus Fe+3 _____________
#2 Fe(SCN)+2 plus SCN- _____________
#3 Fe(SCN)+2 plus OH- ______________
#4 Co+2 plus HCl ________________
#5 Co+2 minus heat (cool) _____________
#6 Co+2 plus heat _______________
#2 Fe(SCN)+2 plus SCN+3
1. Fe + SCN
2. KSCN
common Ion
3–6
Fe+3 + SCN-
Fe(SCN)+2
K+ + SCNFe(SCN)+2
#3 Fe(SCN)+2 plus OH1. Fe+3 + SCN-
Fe(SCN)+2
2. Fe+3 + 3OHprecipitation reaction.
Fe(OH)3 (s)
3–6
Fe+3 + SCN-
notice, Fe3+ is removed through a
Fe(SCN)+2
73
#4 Co+2 plus HCl
1.
CoCl4-2 + 6H2O
2.
HCl  H+ + Cl-
3–6
CoCl4-2 + 6H2O
4Cl- + Co(H2O)6+2 +
energy
4Cl- + Co(H2O)6+2 +
energy
#5 Co+2 minus heat (cool)
1.
CoCl4-2 + 6H2O
4Cl- + Co(H2O)6+2 +
energy
4Cl- + Co(H2O)6+2 +
energy
2. N/A
3–6
#6 Co+2 plus heat
-2
1. CoCl4 + 6H2O
2. N/A
3–6
74
Lab 10 Standard Molar Volume of a Gas
Purpose and Objectives
Determine the volume and moles of H2 (g) collected at room temperature and pressure. To apply Dalton's
Partial pressure Law, the combined gas law, and the Ideal Gas Law. To determine the molar mass of the
unknown metal.
Discussion
The Ideal Gas Law relates the variables associated with gases through the equation:
PV = nRT
where P is measured in mmHg, but is converted to atmospheres by converting 760 mmHg/1 atm before
using the equation. Volume is measured in mL in the laboratory, but is converted to liters before using in
the equation. Moles, n, is calculated as n = mass/ MM (MM is the molar mass and defined as grams/mole).
T is measured in °C in the lab, but must be converted to Kelvin:
C + 273.15 = K
before using in the equation. R is the gas law constant,
R=
0.0821 L  atm
mol  K
The Combined Gas Law relates the P, V, and T for a gas in which one or more of these three parameters
(measurable quantities) have changed, but in which the moles, n, have remained the same. The law is
really a simplified version of the ideal gas law:
P1 V1
PV
= 2 2
T1
T2
Note that “1’s” and “2’s” represent initial (1) and final (2) conditions.
Dalton's Law of Partial Pressures, PT = P1 + P2 + P3 + ...Pn, relates the moles of a mixture of gases to the
pressure of the total amount of gases. If a gas mixture was 50% O 2(g) and 50% H2(g), the total pressure
would be the sum of the two pressures. H2O(g), usually called vapor pressure of water is the gas we usually
make corrections for in laboratory work. In dealing with gases, it is convenient to establish a standard
reference point, called Standard Temperature and Pressure, (STP). At STP, any gas occupies exactly 22.4 L
for exactly 1 mole. Since this is exactly one mole, it contains one Avogadro number of molecules or atoms
of gas. Therefore, the Molecular Volume given contains exactly one gram-molecular mass (or one gramatomic mass) of the gas.
In this experiment we will collect H2(g) from the reaction of a metal which reacts according to the
following equation:
M (s) + 2 HCl (aq)  H2(g) + MCl2(aq)
75
Note: this is one of the 4 special “Gas Forming” reactions in chemistry; metals react with strong acids to
produce hydrogen gas (H2 (g)) and a metal salt.
The metal we will use will be an unknown metal that will ionize to the +2 state. Thus one mole of the metal
will form from one mole of HZ(e). The molar mass of the metal and the identity of the metal will be
determined.
Procedure
1. Fill a large beaker to about 3/4 full with water.
2. Weigh no more then 0.03 g of magnesium ribbon and record to three decimal places, then roll it and tie
a copper wire through it and leave a two inch tail of copper. Wrap the tail side of the copper through a one
hole rubber stopper. The metal ribbon should be suspended about 1 cm from the small end of the stopper.
3. Fill to about the 20mL mark on the gas buret with 3M hydrochloric acid (HCI). Carefully fill the buret
with water using your wash bottle to the brim. Any air left will be mistaken for the H 2 gas.
4. Carefully fit the rubber stopper into the top, place your finger over the hole in the stopper, invert the
stopper end of the buret into the large beaker containing water, release your finger, and hold the buret
upright. The metal will then react with the acid. (The acid is denser than the pure water and slowly moves
to the bottom of the tube where it can react with the metal). The reaction is complete when there are no
more bubbles coming from the bottom, and the metal is completely gone. If some metal breaks off and
floats to the top, wait until it completely reacts.
5. Accurately read and record to one decimal place the volume of gas collected, holding the buret
vertically, and reading the bottom of the meniscus.
6. Read the barometer, and record the pressure in mmHg. Read the thermometer on the barometer and
record the room temperature to the tenths place. This temperature is taken as the gas temperature T 1.
Record the vapor pressure of water at room temperature.
Vapor Pressure of Water
Temp mm g
Temp
19
16.5
24
20
7.5
25
21
18.6
26
22
19.8
27
23
21.2
28
mmHg
22.4
23.8
25.2
26.7
28.3
7. Repeat steps 2-5 for second piece of metal for a second trial.
76
Data for Lab 10: Electronic Configurations and Periodic Trends
Name ________________________________
Date _________________________________
Labmates ______________________________
RECORD ALL DIGITS FROM THE DIGITAL DEVICE
Trial 1
Trial 2
Metal Mass (g)
V1 = Volume Hydrogen collected
in L
T1 = Temperature of air (close
enough) in K
Atmospheric Pressure (PT) i.e.
barometric pressure in mmHg
Vapor Pressure H2O (PH2O) found
in table above in mmHg
Partial pressure of H2 is: PT = PH2O + PH2, therefore PH2 = PT –PH2O
P1 = Partial Pressure H2 (PH2) =
77
Calculations: WATCH YOUR SIG FIGS
1. Calculate the volume (V2) of hydrogen gas produced, at STP conditions for each run. Use the Combined
Gas Law where P2 =760 mmHg and T2 = 273 K. Calculate the average volume.
P1 V1
PV
= 2 2
T1
T2
Trial 1
Trial 2
V2 = Volume H2 (STP) in L
Average Volume (V2) in L =
2. Calculate the moles of H2 produced used:
Mg + 2HCl  MgCl2 + H2
____ g Mg 1 mole Mg 1 mol H 2


 ____ mol H 2
1
24.3g Mg 1 mol Mg
Trial 1
Trial 2
Moles H2
Average mol =
3. Calculate the molar volume at STP from your averaged numbers. Use
V2
.
mol
78
Lab 11 Net Ionic Equations (NIE)
There are two Instructor demonstrations.
1. The Instructor will demonstrate, with a conductivity apparatus, the following:
1.
2.
3.
4.
Di-ionized water
Tap water
Di-ionized water with NaCl(s) added
Di-ionized water with C6H12O6 (sugar) added
Your instructor will explain the results of these experiments.
I. OBJECTIVES:
1. To become familiar with the use, of a Table of Solubility Rules and the application of a Table of
Electrolytes.
2. To practice identifying and drawing the form that a chemical species is present in aqueous solution
in a beaker.
3. To become familiar with writing chemical equations as a molecular, ionic, and net ionic equations.
II. THEORY:
Read in the text sections discussing electrolytes, net ionic & ionic equations.
REMEMBER!
1. Molecular Equation: Shows all substances present before and after the reaction- Write these in
"molecular form"2. Ionic Equation: Shows what actual species (ions or molecules) are present before and after the
reaction, including all participating and "spectator" ions3. Net Ionic Equation Shows which species; took part in the solution reaction (eliminating all
"spectator ions").
Here are some HINTS, for writing correct "ionic" and "net ionic equations".
A. Molecular equations
1. First, write a "normal" molecular equation (as we have been doing) and BALANCE it!!
2. Check the solubility chart and identify each species as (s), (aq), (g), or (l).
B- Ionic equations
1. For each soluble substance (aq), check the electrolyte chart to see which should be written as ions:
a. If it is a strong electrolyte write the cation and anion as separate ions
example: Na2SO4 (aq) : 2Na+(aq) + SO42- (aq)
79
b. If it is a weak electrolyte write it as a "molecule"
example: NH4OH (aq) : NH4OH (aq)
HC2H3O2 (aq) : HC2H3O2 (aq)
c. If it is a nonelectrolyte write it as a molecule.
example: methanol (aq) : CH3OH (aq)
water (l)
: H2O (l)
2. For insoluble substances ; precipitates, and gases and pure liquids write these as molecules
example: BaCrO4(s), AgCl (s), HCl(g), H2O(1), CO2(g)
3. Write the full ionic equation, being sure to show ALL CHARGES for ions and correct coefficients
C. Net ionic Equations:
1. Cross out all "spectator ions" (those which do not participate and do not change the reaction).
2. Show only the species which actually participated in the reaction. These are the species which
underwent a chemical change and are not identical on both sides of the equation.
*****All equations must be balanced, both atoms and charges!!
Example:
Molecular:
NaBr(aq) + AgNO3(aq) 
Ionic:
Na+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq)  AgBr(s) + Na+(aq) + NO3-(aq)
Net ionic:
Ag+(aq) + Br-(aq)  AgBr (s)
AgBr(s) + NaNO3(aq)
2. The Instructor will demonstrate the NaBr and AgNO3 reaction for the
students after writing the equation on the board to demonstrate the reliability of the solubility rules.
So how do I write net ionic equations???
I ask - What am I starting with? To answer this, I consider the reactants: Is each one a solid (s), liquid (1), or
a gas (g) ? Or is it in solution (aq) ?
A. If it is a (s), (1), or a (g), leave it in molecular form. Be sure the formula is correct.
80
B. If it is an (aq), then consider what type of electrolyte it is: STRONG - Write it as separated ions consult
the ion chart if necessary weak - most are molecules, so I write them as molecules in my equation. non leave it as molecules because all are in molecular (aq) form.
I ask - What am I forming? To answer this, I must consider each product:
A. If it is a precipitate, it must NOT be soluble in water.
1. Consult the solubility rules to check if this is so.
2. Write the formula correctly, including (s).
3. Leave the formula in MOLECULAR form - all ions stay together.
B. If it is soluble, I ask - is it a strong, weak or non-electrolyte? HC1, HBr, HI, H2SO4, HNO3, and HCIO4 are all
strong acids.
1. All others are WEAK
2. Almost all soluble salts are strong electrolytes, separated into ions with an (aq).
3. If the species is a weak electrolyte or a non-electrolyte, leave as moleculesREMEMBER: H+(aq) + OH-(aq)  H2O(l)
C. If it is a gas, what is it?
4 common gas forming reactions are:
If you see the following 3 as products, replace those molecules with the appropriate gas and H 2O (l).
1. H2CO3  H2O(l) + CO2(g)
2. H2SO3  H2O(l) + SO2(g)
3. NH4OH  NH3(g) + H2O(l)
This is a common method for making hydrogen gas
4. H+ + some metals  metal salt (aq) + H2(g)
Spectator ions undergo ABSOLLUTLY NO CHANGE - include in ( ) after the balanced net ionic equation is
written.
An equation is balanced only if both atoms and charges are conserved (the same number on both sides total).
81
82
Data Sheet Lab 11 NIE
Name ________________________________
Date _________________________________
Labmates ______________________________
This is a Paper and Pencil Lab. Students predict the results of these reactions using solulbility rules.
1. sodium carbonate reactions with phosphoric acid.
Molecular
Ionic
Net Ionic
2. Potassium iodide reacts with lead (II) nitrate to form a precipitate.
Molecular
Ionic
Net Ionic
83
3. Copper (II) sulfate reacts with silver nitrate.
Molecular
Ionic
Net Ionic
4. Lithium sulfite reacts with a solution of hydrochloric acid.
Molecular
Ionic
Net Ionic
5. Solutions of aluminum sulfate and calcium iodide reacts.
Molecular
Ionic
Net Ionic
84
6. Solutions of potassium acetate and lithium bromide are mixed
Molecular
Ionic
Net Ionic
7. Sulfuric acid solution neutralizes a solution of potassium hydroxide.
Molecular
Ionic
Net Ionic
8. When baking a cake we sometimes combine baking soda (NaHCO3) and vinegar (HC2H3O2) to create
bubbles that cause the cake to rise. Write an equation for this reaction.
Molecular
Ionic
Net Ionic
85
9. calcium hydrogen sulfite + sulfuric acid
Molecular
Ionic
Net Ionic
10. Silver hydrogen carbonate + sulfuric acid
Molecular
Ionic
Net Ionic
11. Aluminum metal reactions with phosphouric acid to make aluminum phosphate and hydrogen gas
Molecular
Ionic
Net Ionic
86
Lab 12 Titration
Demo by Instructor
1. Buret cleaning, usage
2. adding solids to flask, when the flask is on a scale
This is a Quantitative Analysis Lab, meaning the aim of the lab is to measure precision and accuracy.
The reaction for this lab is an acid base reaction. All acid base reactions produce a salt and water
Acid + Base
Salt + H2O
For example:
H3PO4 + NaOH
Na3PO4 + H2O
Looking at the above example, we can write the NIE
H3PO4(aq) + NaOH(aq)
Na3PO4(aq) + H2O(l)
which leads to:
H+ + OH-
H2O(l)
This is referred to as a neutralization reaction.
The point at which mol H+ = mol OH- is referred to as the Equivalency Point. In today’s lab we us an
indicator to let us know when the EQ point has been reached. Phenolphthalein, our indicator, is clear
under acidic conditions and turns a light pink under basic conditions which is at the point after the H + is
consumed.
H+
+
H
H+
OHOHOH-
H+
H+
H2O
OHOH-
87
I. Purpose:
The purpose of this experiment is to:
1. Prepare a 0.1 M sodium hydroxide solution
2. Standardize it with a primary standard acid
II. Theory:
In the neutralization of an acid with a base, the end point of the titration is reached when the number of
moles of hydrogen ion exactly equals the number of moles of hydroxide ion. Calculations can be make
using balanced chemical equations and standard stoichiometric methods.
III. Procedure:
Titration of H2C2O4 with NaOH Solution. (*DEMO)
Set up and clean a 50 mL buret as demonstrated by your instructor. Be sure that the buret is clean,
then rinse three times with 5 ml, portions of the NaOH solution you will be doing the titrations with;
approximate 0.1 M NaOH solution. Fill the buret, and be sure all air bubbles are eliminated from the tip.
Start by transfering approximately 0.01 gram (record accurately) of oxalic acid dehydrate (your
sample), H2C2O42H2O, to a clean, dry 125 ml, or 250 ml, Erlenmeyer flasks. Add about 30 mL of distilled
water and two-four drops of phenolphthalein indictor to each sample. Titrate the samples with the
approximately 0.1M NaOH solution to the first appearance of faint pink end point that persists for at least
30 seconds. NOTE: You should use approximately 18 mL of NaOH for each titration…in
other words, adjust your sample size so you USE between 18-25 mL of NaOH. The 0.01 g
is JUST a starting point, you WILL need to add more sample.
Initial and final buret readings are to made to the nearest 0.01 mL. Do a minimum of 4 titrations
with "good" end points. Calculate the molarity of the NaOH solution for each titration.
Note: H2C2O42H2O is a diprotic acid, it takes 2 moles of NaOH for every one mole of oxalic acid.
Dispose of all solutions into the sink, run water as you do this.
Titration General Information
1. This experiment is to be done in pairs.
2. The grade for this experiment is based entirely on the results.
3. All measurements should be made to the appropriate number of figures to be acceptable. Mass
measurements should be made to the nearest 0.001 g and volume measurements should be made to the
nearest 0.01 mL with the exception of the volumetric flask, which should be measured to the nearest 0.1
mL.
4. The data section must contain only data! Note the special way of setting up the data and the calculation
sections for this experiment.
Buret readings 3-4 sig figs with 2 decimal places
Mass readings 4 sig figs with 3 decimal places
88
***Critical
In order to be more accurate and to avoid losing precious time during the lab periods, beware of these
common mistakes:
Buret:
1. not rinsing the buret one time with deionized water three times with your base solution
2. not checking to see if the buret is clean
3. not watching for leaks around the stopcock
4. not removing all the air from the dropper tip
Making Solutions:
1. not weighting solids accurately (i.e. dropping crystals, not weighing to the nearest 0.0001 g)
2. trying to get all the solid samples the same size
3. Using dirty equipment
Titrating:
1. trying to make the initial volume read exactly 0.00 mL
2. not reading the initial and final volumes carefully and consistently to the nearest 0.01 mL
3. forgetting the indicator
4. not washing down tip of the buret and the neck of the flask with deionized water
5. not swirling flasks over white paper while titrating
6. not running enough "good" trials
7. misreading labels on chemicals used
8. misrecording data
9. making math errors
89
90
Data Sheet Lab 12 Titration
Name ________________________________
Date _________________________________
Labmates ______________________________
#1
mass acid
volume reading
RECORD THE MASS
CORRECLTY
READ THE BURRET CORRECTLY
________
final______
start______
Total Volume _____
#2
mass acid
volume reading
________
final______
start______
Total Volume ______
#3
mass acid
volume reading
________
final______
start______
Total Volume ______
#4
mass acid
volume reading
________
final______
start______
Total Volume ______
91
Molarity Calculations:
Molarity is defined as Mol/L. Here you will be determining the molarity of your NaOH solution.
____ g oxylic acid 
1 mole oxylic acid
____ mol NaOH
1


 ______ M NaOH
______ g oxylic acid _____ mol oxylic acid ____ L NaOH
Show your work for ALL 4 titrations.
What is your Average Molarity of NaOH =
92
30 points Extra Credit: Nomenclature (due when your instructor tells you :-)
Give an appropriate name:
Name__________________________________
1. NaIO4
15. HIO3
2. NaIO3
16. HIO2
3. NaIO2
17. H2O
4. NaIO
18. FeCl3
5. KBrO
19. Fe2(SO4)3
6. KBrO2
20. FeS
7. KBrO3
21. HCl (aq)
8. KBrO4
22. HClO3
9. HBrO4
23. H2CO3 (aq)
10. HBrO3
24. CaCO3
11. HBrO2
25. Be2C
12. HBrO
26. SnSO3
13. HIO
27. (NH4)2S
14. HIO4
28. N2O4
93
29. K2HPO4
43. Pb(Cr2O7)2
30. Fe(NO3)3
44. Ca(OH)2
31. P4O6
45. AlP
32. NH4Cl
46. Ni2Cr2O7
33. K2CrO4
47. Be(NO2)2
34. Na2SO4
48. CuSO3
35. KMnO4
49. Fe(MnO4)3
36. WO3
50. CuCN
37. Ca(ClO2)2
51. N2O
38. FeSO4
52. Al2S3
39. P4O10
53. NaH2PO4
40. Sn(ClO3)4
54. Ti2O
41. K2Cr2O7
55. TiO
42. PbCr2O7
56. TiO2
94
57. Sodium iodite
73. Aluminum carbide
58. Chromium (VI) oxide
74. Magnesium phosphate
59. Potassium bromate
75. Nitrogen dioxide
60. Potassium perbromate
76. Ammonium sulfate
61. Sodium hypofluorite
77. Barium carbonate
62. Potassium fluorate
78. Tungsten (IV) carbonate
63. Dinitrogen tetrasulfide
79. Sodium hydrogen carbonate
64. Fluorous acid
80. Calcium dihydrogen phosphate
65. Hypobromous acid
81. Disulfur dichloride
66. Fluoric acid
82. Sodium iodite
67. Perfluoric acid
83. Sodium hypoiodite
68. Iodic acid
84. Sodium iodate
69. Peroidic acid
85. Sodium periodate
70. Barium chloride
86. Potassium bromate,
71. Tin (II) nitrate
87. Potassium perbromate
72. Tin (IV) nitrate
88. Sodium hypofluorite
95
89. Potassium fluorate
105.
Iron (III) hydrogen phosphate
90. Potassium fluoride
106.
Iron (III) dihydrogen phosphate
91. Hydrofluoric acid
107.
Magnesium iodide
92. Lithium fluoride
108.
Hypoiodous acid
93. Hypobromous acid
109.
Periodic acid
94. Perfluoric acid
110.
Hypobromous acid
95. Iodic acid
111.
Magnesium periodate
96. hydroiodic acid
112.
Zinc iodate
97. Sulfuric acid
113.
Copper (I) iodite
98. Hydrosulfuric acid
114.
Copper (II) iodite
99. Lithium sulfide
115.
Hypofluorous acid
100.
Sulfourous acid
116.
Periodic Acid
101.
Iron (II) phosphate
117.
Tin (IV) hypoiodite
102.
Iron (II) hydrogen phosphate
118.
Tin (II) hypoiodite
103.
Iron (II) dihydrogen phosphate
119.
Calcium hypoiodite
104.
Iron (III) phosphate
120.
hydrocyanic acid
96