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Chapter 1: Chemistry and Measurement
Problems:
2, 4-5, 7-10, 14, 16, 17, 24-26, 27a, 28-44, 47-52, 55-56, 59-77, 79-88, 93-96,
105-112, 115-132
science:
methodical study of nature followed by a logical explanation of the
observations
Chemistry:
1.1
study of matter and its properties, the changes that matter
undergoes, and the energy associated with those changes
MODERN CHEMISTRY
In what fields is chemistry used today? And how is it used?
1.2
EXPERIMENT AND EXPLANATION
Consider the following:
On the first day of school, you get in your car and turn the key on the ignition,
but nothing happens. What could be the problem?
The Scientific Method
1. Perform experiments and record observations on system
2. Analyze the data, and propose a hypothesis to explain observations
3. Conduct additional experiments to test hypothesis. If initial hypothesis holds
up to extensive testing, the hypothesis becomes a theory.
theory: a tested qualitative explanation of basic natural phenomena
If all or part of the hypothesis does not hold up to testing, then it is adjusted or
a new hypothesis is proposed to explain the observations.
law: a simple statement or mathematical equation about a fundamental
relationship
CHM 151: Chapter 1
page 1
1.5 MEASUREMENT AND SIGNIFICANT FIGURES
measurement: a number with attached units
When a measurement is recorded, all the numbers known with certainty are
given along with the last number, which is estimated. All the digits are significant
because removing any of the digits changes the measurement's uncertainty.
Ruler A
0
1
2
3
4
5
0
1
2
3
4
5
Ruler B
Ruler C
4.1
Ruler
4.2
4.3
Measurement/quantity
4.4
# of sig figs
A
____________
____
B
____________
____
C
____________
____
Which ruler gives the most exact measurement? _______________
Guidelines for Sig Figs (if measurement is given):
Count the number of digits in a measurement from left to right:
1. When there is a decimal point:
– For measurements greater than 1, count all the digits (even zeros).
– 62.4 cm has 3 sig figs, 5.0 m has 2 sig figs, 186.000 g has 6 s.f.
– For measurements less than 1, start with the first nonzero digit and
count all digits (even zeros) after it.
– 0.011 mL and 0.00022 kg each have 2 sig figs
2. When there is no decimal point:
– Count all non-zero digits and zeros between non-zero digits
– 125 g has 3 sig figs, 107 mL has 3 sig figs
CHM 151: Chapter 1
page 2
How many significant digits do the following numbers have?
# of sig figs
of sig figs
# of sig figs
a. 165.3
_____
c.
90.40
_____
e.
0.19600 _____
b. 105
_____
d.
100.00
_____
f.
0.0050 _____
SCIENTIFIC NOTATION
Some numbers are very large or very small  difficult to express.
For example,
Avogadro’s number = 602,000,000,000,000,000,000,000
an electron’s mass = 0.000 000 000 000 000 000 000 000 000 91 kg
Also, it's not clear how many sig figs there are in some measurements.
For example,
Express 100.0 g to 3 sig figs:
___________
Express 100.0 g to 2 sig figs:
___________
Express 100.0 g to 1 sig fig:
___________
To handle such numbers, we use a system called scientific notation.
All numbers can be expressed in the form:
where
N  10n
N =digit term= a number between 1 and 10, so there can only
be one number to the left of the decimal point: #.####
n = an exponent = a positive or a negative integer (whole #).
To express a number in scientific notation:
Count the number of places you must move the decimal point to get N
between 1 and 10.
Moving decimal point to the right (if # < 1) negative exponent.
Moving decimal point to the left (if # > 1)  positive exponent.
CHM 151: Chapter 1
page 3
Express the following numbers in scientific notation (each with 3 sig figs):
555,000

__________________
0.000888

__________________
602,000,000,000,000,000,000,000 ______________________
ROUNDING OFF NONSIGNIFICANT DIGITS
How do we eliminate nonsignificant digits?
• If first nonsignificant digit < 5, just drop ALL nonsignificant digits
• If first nonsignificant digit  5, raise the last sig digit by 1 then
drop ALL nonsignificant digits
For example, express 72.58643 with 3 sig figs:
last significant digit
72.58643 g
first nonsignificant digit
72.58643
to 3 sig figs
 
_______________________
Express each of the following with the number of sig figs indicated:
a.
376.276
 

to 3 sig figs
_______________________
b.
500.072
 

to 4 sig figs
_______________________
c.
0.00654321
d.
1,234,567
e.
2975
to 3 sig figs
 

to 5 sig figs
 

to 2 sig figs
 

_______________________
_______________________
_______________________
Express measurements in scientific notation when necessary to make it clear
how many sig figs there are in the measurement.
CHM 151: Chapter 1
page 4
ADDING/SUBTRACTING MEASUREMENTS
When adding and subtracting measurements, your final value is limited by
measurement with the largest uncertainty—i.e. the number with the fewest
decimal places.
Ex 1: 106.61 + 0.25 + 0.195 = 107.055
107.055 to the correct number of sig figs: _______________
Ex 2:
725.50 – 103 = 622.50
622.50 to the correct # of sig figs: ____________________
MULTIPLYING/DIVIDING MEASUREMENTS
When multiplying or dividing measurements, the final value is limited by the
measurement with the least number of significant figures.
Ex 1: 106.61  0.25  0.195 = 5.1972375
5.1972375 to the correct # of sig figs: ________________
Ex 2: 106.61  91.5 = 9754.815
w/ correct sig figs: _______________
MULTIPLYING/DIVIDING WITH EXPONENTIAL NUMBERS:
When multiplying or dividing measurements with exponents, use the digit term (N
in “N  10n”) to determine number of sig figs.
 1033
Ex. 1: (6.02  1023)(4.155  109) = 2.50131
How do you calculate this using your scientific calculator?
Step 1.
Enter “6.02  1023” by pressing:
6.02 then
EE or EXP (which corresponds to “ 10”)
then 23
Your calculator should read or something similar.
Step 2.
Multiply by pressing: 
CHM 151: Chapter 1
page 5
Step 3.
Enter “4.155  109” by pressing:
4.155 then
EE or EXP (which corresponds to “ 10”)
then 9
Your calculator should now read
Step 4.
Get the answer by pressing: 
Your calculator should now read:
or something similar indicating
2.50131 33 or 2.50131 E 33
10 33
2.50131
Thus, the answer with the correct sig figs = ________________________
Be sure you can do exponential calculations with your calculator. Many of
the calculations we do in chemistry involve very large and very small
numbers with exponential terms.
Ex. 2:(3.75  1015)(8.6  104) = 3.225  1020
with the correct sig figs: ___________________
Ex. 3: (3.75  1015)  (8.605  104) = 4.357931435  1010
with the correct sig figs: ___________________
EXACT NUMBERS
Although measurements can never be exact, we can count an exact number of
items. For example, we can count exactly how many students are present in a
classroom, how many M&Ms are in a bowl, how many apples in a barrel.
1.8 UNITS AND DIMENSIONAL ANALYSIS (FACTOR-LABEL METHOD)
UNIT EQUATIONS AND UNIT FACTORS
Unit equation: Simple statement of two equivalent values
Unit conversion factor = unit factor = equivalents:
- Ratio of two equivalent quantities
Unit equation
1 dollar = 10 dimes
CHM 151: Chapter 1
Unit factor
10 dimes
1 dollar
or
1 dollar
10 dimes
page 6
Equivalents are exact if we can count the number of units equal to another
or the units are in the same system (metric or English). For example, the
following unit factors and unit equations are exact:
365 days
1 year
7 days
1 week
12 inches
1 foot
and
1 yard 3 feet
Exact equivalents have an infinite number of sig figs
 never limit number of sig figs!
Note: When the relationship between two units or items is exact, the “”
(meaning “equals exactly”) is used instead of the basic “=” sign.
Other equivalents are inexact or approximate because they are
measurements or approximate relationships, such as
1.61 km
1 mile
65 mi
hour
speed of light =
3.00  108 m
s
Approximate equivalents do limit the sig figs for the final answer.
UNIT (DIMENSIONAL) ANALYSIS PROBLEM SOLVING
1. Write the units for the answer.
2. Determine what information to start with.
3. Arrange all other unit factors—showing them as fractions—with correct
units in the numerator and denominator, so all units cancel except for
the units needed for the final answer.
4. Check for correct units and number of sig figs in the final answer.
CHM 151: Chapter 1
page 7
Example 1:
If a marathon is 26.2 miles, how many inches are in the marathon?
(1 mile 5280 feet)
Example 2:
The speed of light is about 3.00 x 108 meters per second. Express
this speed in miles per hour. (1.609 km = 1 mile, 1000 m  1km)
1.6
SI Units (from French "le Système International d’Unités")
– standard units for measurement
Metric system: A unified decimal system of measurement with a basic unit for
each type of measurement
CHM 151: Chapter 1
quantity
basic unit
symbol
length
meter
m
mass
gram
g
volume
liter
L
time
second
s
page 8
Metric Prefixes
Multiples or fractions of a basic unit are expressed as a prefix
 Each prefix = power of 10
 The prefix increases or decreases the base unit by a power of 10.
Prefix
Symbol
Multiple/Fraction
mega
M
106
kilo
k
103
deci
d
0.1 = 10-1
centi
c
0.01 = 10-2
milli
m
0.001 = 10-3
micro
 (Greek “mu”)
10–6
nano
n
10–9
pico
p
10–12
KNOW these metric units above (highlighted in Table 1.3)!
Metric Conversion Factors
Ex. 1
Ex. 2
Complete the following unit equations:
a.
1 kg = __________ g
c.
1 cm = __________ m
b.
1 s = __________ ns
d.
1 L = __________ mL
Write 2 unit factors for each of the unit equations above.
CHM 151: Chapter 1
page 9
3.3
Metric-Metric Conversions: Solve the following using dimensional analysis.
Ex. 1 Convert 75 miles per hour into units of meters per second.
Ex. 2 Convert 12.0 kilograms into milligrams.
Metric-English Conversions
English system: Our general system of measurement.
Scientific measurements are exclusively metric. However, most Americans are
more familiar with inches, pounds, quarts, and other English units.
 A method of conversion between the two systems is necessary.
Know these conversions!
Quantity
English unit
Metric unit
English–Metric conversion
length
1 inch (in)
1 cm
1 in. = 2.54 cm (exact)
mass
1 pound (lb)
1g
1 lb = 453.6 g (approximate)
volume
1 quart (qt)
1 mL
1 qt = 946.4 mL (approximate)
All other metric-English conversions will be provided on quizzes and exams.
Ex. 1
What is the mass in kilograms of a person weighing 215 lbs?
CHM 151: Chapter 1
page 10
Ex. 2
What is the volume in cups for a 2.0-L bottle?
Ex. 3
A light-year (~5.88x1012 miles) is the distance light travels in one
year. Calculate the speed of light in meters per second.
(1 mile=1.61 km approx.)
Temperature:
— A measure of the average energy of a single particle in a system.
The instrument for measuring temperature is a thermometer.
Temperature is generally measured with these units:
Fahrenheit degree (°F) Celsius degree (°C)
References
English system
Metric system
freezing pt for water
32°F
0°C
boiling pt for water
212°F
100°C
Phoenix Summer day
113°F
45°C
Conversion between Fahrenheit and Celsius scales:
C =
CHM 151: Chapter 1
(F - 32)
1.8
F = (C  1.8) + 32
page 11
Ex. 1
In Europe, temperature is generally reported in the Celsius scale. A
European tourist asks you how hot it is. What is the equivalent Celsius
temperature if on that cool Summer day it’s only 95°F?
Kelvin Temperature Scale
There is a third scale for measuring temperature: the Kelvin scale.
The unit for temperature in the Kelvin scale is Kelvin (K, NOT °K!).
The Kelvin scale assigns a value of zero kelvins (0 K) to the lowest possible
temperature, which we call absolute zero and corresponds to –273.15°C.
(The term absolute zero is used because this is the theoretical lowest limit.)
Conversion between °C and K:
K = ˚C + 273.15
˚C = K – 273.15
Ex. 1
The average temperature in Reykjavik, Iceland during the summer is
about 15˚C. What is the equivalent Kelvin temperature? What is the
equivalent Fahrenheit temperature?
Ex. 2
In the movie Terminator 2, a tanker crashes and pours out liquid nitrogen.
This freezes the T-1000 because liquid nitrogen has a temperature of 77 K.
What is the equivalent temperature in degrees Celsius?
CHM 151: Chapter 1
page 12
1.7
DERIVED UNITS
Volume: Amount of space occupied by a solid, gas, or liquid.
– measured using graduated cylinder, a buret, a pipet, a volumetric flask, etc.
– generally in units of liters (L), milliliters (mL), or cubic centimeters (cm3)
– Know the following:
1 L  1 dm3
1 mL  1 cm3
These are both exact!
DETERMINING VOLUME
Volume is determined in three principal ways:
1. Volume of any liquid can be measured directly using calibrated glassware
(graduated cylinder, pipets, burets, etc.)
2. Volume of a solid with a regular shape (rectangular, cylindrical,
uniformly spherical or cubic, etc.) can be determined by calculation.
– e.g. volume of rectangular solid = length x width x thickness
3. Volume of an irregular solid is found indirectly by the amount of liquid it
displaces. This technique is called volume by displacement.
Volume By Displacement
a. Fill a graduated cylinder halfway with water, and record the initial volume.
b. Carefully place the object into the graduated cylinder so as not to splash
or lose water.
c. Record the final volume.
d. Volume of object = final volume – initial volume
CHM 151: Chapter 1
page 13
density: The amount of mass in a unit volume of matter
density =
mass
volume
or
d
=
m
V
generally in units of g/cm3 or g/mL
For water: 1.00 g of water occupies a volume of 1.00 cm3
d=
m
1.00 g
=
= 1 . 00 g /cm 3
3
V 1.00 cm
Applying Density as a Unit Factor
Given the density for any matter, you can always write two unit factors. For
example, the density of ice is 0.917 g/cm3.
Two unit factors would be:
0.917 g
cm 3
or
cm 3
0.917 g
Ex. 1 Aluminum has a density of 2.70 g/cm3. What is the mass of a piece of
aluminum with a volume of 0.525 cm3?
Ex. 2 Ethanol is used in alcoholic beverages and has a density of 0.789 g/mL. What
volume of ethanol (in liters) would have a mass of 500.0 mg?
CHM 151: Chapter 1
page 14
Ex. 3 A piece of silver metal weighing 194.3 g is placed in a graduated cylinder
containing 242.0 mL of water. The volume of water now reads 260.5 mL.
Calculate the density of silver.
1.3
CONSERVATION OF MASS
matter: anything that has mass and volume
mass: A measure of the amount of matter an object possesses.
– measured with a balance and NOT AFFECTED by gravity
MASS  WEIGHT = MASS x Acceleration due to gravity
Mass is not affected by gravity!
EARTH
mass = 68 kg
weight = 150 lbs
MOON
mass = 68 kg
weight = 25 lbs
SPACE
mass = 68 kg
weight = 0 lbs
law of conservation of mass: Matter is neither created nor destroyed
reaction (rxn):
REACTANTS (starting materials)
(substances before rxn)
 PRODUCTS
(substances after rxn)
Mass of the product(s) in a reaction must = mass of the reactant(s)!
For example:
Example:
CHM 151: Chapter 1
11 g hydrogen + 89 g oxygen = 100 g water
A 95-g sample of sodium chloride, NaCl, contains 58 g
of chlorine. Determine the mass of sodium in the sample.
page 15
1.4
MATTER: PHYSICAL STATE AND CHEMICAL COMPOSITION
Matter exists in one of three physical states: solid, liquid, gas (See Fig. 1.11)
solid:
Has definite shape, fixed volume
– At molecular level, particles are packed closely, in a fixed position
liquid: Fixed volume, but shape can change—takes shape of container
– Particles are packed closely together but can move past each other
gas (or vapor):
Volume is variable, particles are widely spaced
 If volume expands, particles move apart
If volume decreases, particles, move closer together
 Takes shape of container
ELEMENTS, COMPOUNDS, AND MIXTURES
PHYSICAL & CHEMICAL PROPERTIES
Physical Properties:
physical state (solid, liquid, gas)
color
density
melting and boiling points
electrical & heat conductivity
solubility
hardness
odor
Chemical Properties: how a substance reacts with other substances
PHYSICAL & CHEMICAL CHANGES
physical change:
– a process that does not alter the chemical composition of a substance
– eg. changing shape, changing physical state, dissolving
– eg. boiling water, melting ice, hammering gold into foil
chemical change or reaction:
– a process that changes the chemical composition (and thus the chemical
formula) of starting materials (reactants)
— eg. oxidation of matter (burning or rusting)
release of gas bubbles (fizzing)
formation of insoluble solid (precipitation)
release of heat or light
CHM 151: Chapter 1
page 16
pure substances:
– Matter having constant composition, definite and consistent properties
Two types of pure substances:
elements:
— consist of only one type of atom
— cannot be broken down by chemical reaction
— eg. carbon (C), hydrogen (H2), sulfur (S8), copper wire (Cu)
compounds:
— consist of more than one type of atom and has a specific formula
— can be broken down by chemical reaction
— eg. ethanol (C2H5OH)can be broken down to C, H, & O
Two or more pure substances can combine to form mixtures:
mixtures:
— consist of various compounds and/or elements, with no specific formula
— Matter having variable composition with either definite or varying
properties depending on the sample
— can be broken down into individual components
— eg. Any alloy like brass, steel, 10-K to 18-K gold;
course mixtures like sea water, carbonated soda, salt and iron
fillings; air consists mostly of nitrogen and oxygen gases
Also, be able to identify elements, compounds, and mixtures given images at the
atomic/molecular level. (See Concept Check 1.1 on p. 14)
CHM 151: Chapter 1
page 17
Chapter 2: Atoms, Molecules, and Ions
Problems:
2.1
2.1, 2.5-2.6, 2.8,2.11-2.14, 2.17-2.19, 2.21, 2.26, 2.29-2.32, 2.35-2.40, 2.43-2.54,
2.57-2.94, 2.99-2.118
ATOMIC THEORY OF MATTER
Postulate from John Dalton’s Model of the Atom
1. All matter is composed of indivisible atoms.
2. An element is composed of only one type of atom
– All atoms of one type of element always behave the same way.
– Atoms of different elements do not behave the same way.
3. Two or more elements combine to form compounds.
law of constant composition: A compound always has the same
elements in the same proportion by mass
– i.e. a compound always has the same formula
law of multiple proportions: Two elements can combine to form different
compounds (e.g. C and O can combine to form CO and CO2), and the
masses of each element for each compound are always in a fixed ratio
(because always whole #’s of atoms of each element)
4. A chemical reaction involves only the separation, combination, or
rearrangement of atoms—NEVER their creation or destruction.
– This is also called the law of conservation of mass.
Ex. 1:
If 2.50 g of iron powder react with 1.44 g of yellow sulfur powder,
what is the mass of the iron sulfide product?
Ex. 2:
If 10.11 g of limestone (calcium carbonate) decomposes with heat to
8.51 g of calcium oxide and CO2 gas, what mass of CO2 is produced?
CHM 151: Ebbing Chapter 2 Notes
page 1
2.2
THE STRUCTURE OF THE ATOM
Discovery of the Electron
J. J. Thomson (1897)
– demonstrated that cathode rays are composed of tiny, negatively
charged subatomic particles  electrons (e–)
Eugen Goldstein (late 1880s)
– canal rays are composed of positively charged subatomic particle
 protons (p+)
And decades later:
– James Chadwick (1935), Nobel Prize winner for his discovery
 neutron (n) = neutral subatomic particle
Nuclear Model of the Atom
Ernest Rutherford's Alpha-Scattering Experiment (Figure 2.8)
– Alpha () particles shot at a thin gold foil
– A detector set up around the foil to determine what happens to the  particles
– Most of a particles went straight through, some deflected, a few bounced back
Rutherford’s interpretation of the scattering experiment results
–Most alpha () particles pass through foil
 Atom is mostly empty space with electrons moving around the space
–Some  particles deflected or bounce back
 Atom must also contain a dense region, and particles colliding with this
region are deflected or bounce back towards source
 region = atomic nucleus (contains atom’s protons and neutrons)
 Why this is called the “Nuclear Model of the Atom”
Rutherford also estimated the size of the atom and its nucleus:
nucleus (d~10-15 m)
atom (diameter ~10-10 m)
If nucleus = size of a small marble, then atom BankOne Ballpark!
CHM 151: Ebbing Chapter 2 Notes
page 2
Properties of Protons, Neutrons, and Electrons
Subatomic Particle
Charge
Location
Mass (amu)
proton
+1
inside nucleus
1.00728
neutron
0
inside nucleus
1.00866
electron
-1
outside nucleus
0.00055
2.3
NUCLEAR STRUCTURE; ISOTOPES
Nuclide Symbol (also called Atomic Notation):
– shorthand for keeping track of protons and neutrons in the nucleus
atomic number (Z): whole number of p+ = number of e– in neutral atom
mass number (A): whole number sum of protons and neutrons in an atom
– Note electrons contribute almost no mass to an atom
mass number=A
atomic number=Z
E  element symbol
Ex. 1:
Write the atomic notation for carbon-14. How many neutrons are in
each neutral carbon-14 atom?
Ex. 2:
Write the atomic notation for uranium-235 (Z=92). How many neutrons
are in each neutral uranium-235 atom?
Isotopes
Each element always has the same number of protons, but the number of
neutrons may vary. Atoms with different numbers of neutrons are called
isotopes.
– e.g. Carbon exists as carbon-12, carbon-13, and carbon-14 where each
carbon atom has 6 protons but 6, 7, or 8 neutrons
– Isotopes are identified with an element name followed by the mass number
– eg. uranium-235 (U-235), carbon-12 (C-12), cobalt-60 (Co-60), etc.
CHM 151: Ebbing Chapter 2 Notes
page 3
2.4
ATOMIC WEIGHTS
Atoms are too small to weigh directly
– eg. one carbon atom has a mass of 1.99 x 10-23 g–too inconvenient!
 need more convenient unit for mass
 atomic mass unit (amu)
Carbon-12 was chosen and given a mass value of 12 amu
 1 amu = 1/12 the mass of carbon-12
 Mass of all other atoms measured relative to mass of carbon-12
Average Atomic Mass of an Element
– The mass of carbon on the Per. Table is 12.01 amu, NOT 12.00 amu–WHY?!
– Atomic masses reported on the Periodic Table are weighted averages of all
naturally occurring isotopes for each element.
Ex. 1 If 98.89% of carbon exists as carbon-12, which has a mass of
12.00000, while 1.11% exists as carbon-13, which has a mass of
13.00335, calculate the average atomic mass for carbon.
average atomic mass = (0.9889)(12.00000 amu) + (0.0111)(13.00335 amu)
= ____________________
Ex. 2 The atomic masses of the two stable isotopes for boron, B-10
(19.78%) and B-11 (80.22%), are 10.0129 amu and 11.0093 amu,
respectively. Calculate the average atomic mass for boron.
2.5
PERIODIC TABLE OF THE ELEMENTS
A vertical column is called a group or family.
– Elements belonging to the same group exhibit similar chemical properties
– e.g. Li, Na, and K all react vigorously with water; He, Ne, and Ar do not
react with any other substances
A horizontal row is called a period or series.
CHM 151: Ebbing Chapter 2 Notes
page 4
Main-Group (Representative or A Group) Elements
Those elements in groups 1, 2, 13, 14, 15, 16, 17, 18 (or IA to VIIIA)
– Group 1 or IA: alkali metals
– Group 2 or IIA: alkaline earth metals
– Group 17 or VIIA: halogens
– Group 18 or VIIIA: noble gases (because they are all unreactive gases)
Transition Metals (or B Group Elements)
– Elements in Groups 3 to 12 (middle of the Periodic Table)
Dimitri Mendeleev proposed that elements display recurring properties
according to increasing atomic mass
 first Periodic Table arranged elements according to increasing atomic mass
H.G.J. Moseley’s high-energy X-ray radiation experiments of atomic nuclei
 Repeating properties of elements more clearly reflected by the
arrangement of elements according to increasing atomic number
 Periodic Table’s arrangement today
– Trends for increasing atomic mass are identical with those for increasing
atomic number, except for Ni & Co, Ar & K, Te & I.
NAMES & SYMBOLS OF THE ELEMENTS
– Every element has an individual name, symbol, and number.
Convention for writing chemical symbols
– Use first letter (capitalized) of element name: hydrogen  H, carbon  C
– If symbol already used, include second letter (in lower case) of name:
helium  He, calcium Ca, cobalt  Co
– Some symbols come from Latin names
lead=plumbum  Pb
gold= “shining dawn”=aurum  Au
Know the first 56 elements of the periodic table, as well as lead (Pb) and
uranium (U) for Exam #1. You will be given a periodic table with only the
symbols written. Spelling counts!
CHM 151: Ebbing Chapter 2 Notes
page 5
SOLIDS, LIQUIDS, AND GASES
KNOW the physical state of each element!
In their natural state under room temperature conditions:
– Only mercury (Hg) and bromine (Br) are liquid
– H, N, O, F, Cl, and all Noble gases (group VIIIA) are gases
– All other elements are solids
METALS, NONMETALS, & METALLOIDS (or SEMIMETALS)
metals
shiny appearance
malleable, ductile
conduct heat & electricity
react with nonmetals
e.g. aluminum, copper, gold
nonmetals
dull appearance
brittle
nonconductor
react with metals and nonmetals
e.g. carbon, oxygen, sulfur
metalloids (or semimetals): Have properties of metals and nonmetals
Organization of the nonmetals, semimetals, metals on the Periodic Table
– Nonmetals (except H) are concentrated on the top-right of the Periodic Table
– Semimetals are along the stair-step
– All remaining elements are metals
Diatomic Molecules: consist of two atoms
– Recognize that some elements exist as diatomic molecules
– H2, O2, N2, Cl2, F2, I2, Br2
– These are often called “the magic seven” since there are seven of them, and
six of them form a 7 on the periodic table
2.6
CHEMICAL FORMULAS: MOLECULAR AND IONIC SUBSTANCES
molecule: a compound of two or more nonmetal atoms, which are held together
by covalent bond
A molecular formula indicates how many atoms are actually present
– e.g. in water, H2O, there are 2 H atoms and 1 O atom
CHM 151: Ebbing Chapter 2 Notes
page 6
chemical formulas:
– Symbolically express the number of atoms of each element in a compound
– Number of atoms is indicated by a subscript following the element’s symbol
(If there is no subscript, only one atom of that element is in the compound.)
– Some chemical formulas use parentheses
 more than one of that subunit present in the compound
–e.g. C2H4(OH)2 has 2 C, 6 H, and 2 O
Example: How many atoms of each element are in TNT: C7H5(NO2)3?
___ C atoms, ___ H atoms, ___ N atom, ___ O atoms
IONIC SUBSTANCES
– When atoms lose or gain electrons, they form charged particles called ions.
Metals lose e–s
 positively charged ions = cations
Nonmetals gain e–s,
 negatively charged ions = anions
Main-group elements generally form ions--i.e. gain or lose electrons--to get the
same number of electrons as a noble gas
 Ions from representative metals are usually isoelectronic with–i.e. have the
same number of electrons as–one of the noble gases!
Charges shown as superscripts
– Group IA elements  +1 charge: Li+, Na+, K+, etc. (“+” = “+1”)
– Group IIA elements  +2 charge: Mg+2, Ca+2, Sr+2, Ba+2,etc.
– Group IIIA elements  +3 charge: Al+3
– Group IVA elements  +4 charge: Sn+4, Pb+4
– Group VA elements  –3 charge: N–3, P–3
– Group VIA elements  –2 charge: O–2, S–2, Se–2
– Group VIIA elements  –1 charge: F–, Cl–, Br–, I–,etc.
monoatomic ions: from a single atom (eg Na+, Cl–, O2–)
polyatomic ions: from 2 or more atoms (eg. OH–, MnO4–, SO42–)
CHM 151: Ebbing Chapter 2 Notes
page 7
ionic compound: a compound composed of a metal + nonmetal(s), which
are held together by an ionic bond (electrostatic attraction)
An ionic compound is actually a network of ions, with each cation surrounded
by anions, and vice versa. A 2-dimensional slice would look like this:
 Why ionic compounds have the highest melting points compared to
other substances
– to melt the crystal solid, all the bonds between ions have to be broken
formula unit: simplest unit of an ionic compound--e.g. NaCl, Fe2O3, Fe2(SO4)3
– represent the ratio of ions present in the compound, not the
actual number of ions
NAMING IONIC COMPOUNDS
CATIONS: positively charged ions
– Metal atoms lose valence electrons to form cations.
I. Groups IA, IIA, IIIA elements, silver (Ag), zinc (Zn) and cadmium (Cd) form
only one type of ion each:
– Group IA elements  +1 charge always (e.g. Li+=lithium ion)
– Group IIA elements  +2 charge always (e.g. Mg+2=magnesium ion) –
– Group IIIA elements  +3 charge always (e.g. Al+3=aluminum ion)
+
2+
2+
– silver ion = Ag ; zinc ion = Zn ; cadmium ion = Cd
CHM 151: Ebbing Chapter 2 Notes
page 8
II. The Stock system is used to name most transition metals, Sn, Pb, and other
metals that form more than one ion:
+2
+3
– e.g. iron (Fe), a transition metal, forms 2 different ions: Fe and Fe ,
+2
+4
– e.g. lead (Pb), in Group IVA, forms 2 different ions: Pb and Pb
When a metal can form more than one ion, each ion is named:
element name (charge in Roman numerals) + ion
Cu + = copper (I) ion
Cu 2+= copper (II) ion
Pb 2+= lead (II) ion
Pb 4+= lead (IV) ion
+
Na = ________________________
3+
Co = ________________________
2+
= ________________________
Ba = ________________________
3+
= ________________________
Cd = ________________________
Ni
Al
Fe 2+= iron (II) ion
Fe 3+= iron (III) ion
2+
2+
ANIONS: formed only by nonmetals
When a nonmetal forms an ion, it is named:
element stem name + -ide suffix + ion
e.g.
S
2–
 O2– = oxide ion
N = nitrogen atom  N3– = nitride ion
O = oxygen atom
= ________________________
–
F = ________________________
–
Br = ________________________
–
Cl = ________________________
POLYATOMIC IONS
– Know the formulas and names of the following polyatomic ions:
NH4+ = ammonium ion
Hg22+ = mercury (I) ion
MnO4– = permanganate ion
C2H3O2– = acetate ion
PO43– = phosphate ion
CN– = cyanide ion
CHM 151: Ebbing Chapter 2 Notes
CrO42– = chromate ion
Cr2O72– = dichromate ion
SO42– = sulfate ion
SO32– = sulfite ion
NO3– = nitrate ion
NO2– = nitrite ion
OH– = hydroxide ion
CO32– = carbonate ion
HCO3– = hydrogen
carbonate ion
–
ClO = hypochlorite ion
ClO2– = chlorite ion
ClO3– = chlorate ion
ClO4– = perchlorate ion
page 9
WRITING CHEMICAL FORMULAS
formulas of compounds:
Cation + anion symbols and number of each
Compounds should be neutral  +ve charges = -ve charge
Simple techniques for writing chemical formulas:
1. If both ions have charges that are exactly opposite (+1 & -1, 2+ & -2, etc.),
 formula contains one of each (This also applies for polyatomic ions.)
Na+ + Cl–  NaCl
and
Na+ + CN–  NaCN
K+ + Cl–  ___________
K+ + NO3–  ___________
Mg2+ + O2–  ___________
Mg2+ + SO42–  ___________
Al3+ + N3–  ___________
Al3+ + PO43–  ___________
2. For monatomic ions with different charges, use the crossover rule:
 Make the negative charge the subscript of cation, and make positive charge
Ba
Al
2+
3+
–
+ Cl 
2–
+ O

3. For polyatomic ions, where ions have different charges, also use the crossover rule
— Express more than one polyatomic ion with subscripts and parentheses.
Ba
Al
2+
3+
–
+ NO3 
+ CO3
2–

EXCEPTION FOR CROSSOVER RULE: Ions with +4 and 2– charges!
Pb
+4
2–
+ O

Ionic compound formulas must reflect lowest ratio of elements
 Instead of Pb2O4, it should be PbO2 !
CHM 151: Ebbing Chapter 2 Notes
page 10
Also applies to polyatomic ions!
Pb
+4
+ SO4
2–

NAMING COMPOUNDS
Given known charges for some elements  get charge on transition metals!
I. If one of each (cation & anion) present  same but opposite signs
(a)CuCl: Cl ion has what charge? ______
 Cu ion must have what charge? ______ and what formula? ______
 What is the name for the Cu ion in this compound? _______________
 What is the name of this compound? _______________
(b)FeS: S ion has what charge? ______
 Fe ion must have what charge? ______ and what formula? ______
 What is the name for the Fe ion in this compound? _______________
 What is the name of this compound? _______________
Also applies to polyatomic ions:
(a)CuSO4: SO4 ion has what charge? ______
 Cu ion must have what charge? ______ and what formula? ______
 What is the name for the Cu ion in this compound? _______________
 What is the name of this compound? _______________
(b)FePO4: PO4 ion has what charge? ______
 Fe ion must have what charge? ______ and what formula? ______
 What is the name for the Fe ion in this compound? _______________
 What is the name of this compound? _______________
CHM 151: Ebbing Chapter 2 Notes
page 11
II. If ions have different charges  reverse crossover, making subscripts the
charges with positive sign on cation and negative sign on anion.
(Also applies to polyatomic ions.)
Get the individual ions for each compound below:
Co2O3 
Fe2(CO3)3 
EXCEPTION FOR CROSSOVER RULE: Ions with +4 and 2– charges
PbS2 
Sulfide ion must have -2 charge, so 2 sulfide ions  total charge = –4
To get overall charge of zero on compound, Pb must have +4 charge!
Naming Ionic Compounds:
1. Get the individual ions for each compound
2. CATION NAME + ANION NAME, minus “ion”  Name of compound
Ex.
CaCl2 =
______________  ______________________
individual ions
name of compound
Fe2S3 =
______________  ______________________
BaSO4 =
______________  ______________________
Co(NO3)3 =
______________  ______________________
Na3P =
______________  ______________________
Cu2CO3 =
______________  ______________________
TiO2 =
______________  ______________________
CHM 151: Ebbing Chapter 2 Notes
page 12
Given the name of a compound, predict the formula:
— KNOWING charges on ions formed by representative elements!
— KNOWING how to use polyatomic ions and their charges when given to you!
cesium bromide:
 ____________________
_____________
individual ions
formula of compound
copper (I) sulfide: _____________ 
_______________________
barium chloride:
_______________________
_____________ 
lead (IV) phosphide:
_____________  _______________________
iron (III) nitrite:
_____________ 
_______________________
silver hydroxide:
_____________ 
_______________________
tin (II) permanganate: _____________  _______________________
calcium phosphate:
_____________  _______________________
Binary Molecular Compounds:
composed of 2 or more nonmetals
NAMING:
# of atoms of element indicated by Greek prefix before element name
1. For first element, Greek prefix + element name
2. For second element, Greek prefix + element name stem + "ide"
— If only one atom present, “mono-” is generally omitted,
except in a few cases (eg. CO=carbon monoxide)
# of atoms
Greek prefix
# of atoms
Greek prefix
1
mono
6
hexa
2
di
7
hepta
3
tri
8
octa
4
tetra
9
nona
5
penta
10
deca
CHM 151: Ebbing Chapter 2 Notes
page 13
Examples: CO2= carbon dioxide
P4O10=
_________________________________
CCl4=
_________________________________
SF6=
_________________________________
Cl2O5=
_________________________________
Some binary molecular compounds also have common names
–e.g. everyone knows (or should know) H2O is water
Common compounds and names you need to know:
NH3 = ammonia
CH4 = methane
H2O2 = hydrogen peroxide
ACIDS: Aqueous solutions of a compound that releases H+ ions
– usually have H in front, in water indicated by aqueous (aq)
– naming depends on the ion from which the acid forms
add # of H's equal
to negative charge
 HF (aq) = hydrofluoric acid
F– = fluoride ion  
add # of H's equal
to negative charge
–
 HNO2 (aq) = nitrous acid
NO2 = nitrite ion  
–
add # of H's equal
to negative charge
 HNO3 (aq) = nitric acid
NO3 = nitrate ion  
For some acids, the stem name changes:
2–
add # of H's equal
to negative charge
2–
add # of H's equal
to negative charge
 H2SO3 (aq) = sulfurous acid
SO3 = sulfite ion  
 H2SO4 (aq) = sulfuric acid
SO4 = sulfate ion  
Exercises:
add # of H's equal
to negative charge
–
 ________ (aq) = __________________
Cl = ___________  
add # of H's equal
to negative charge
2–
 ________ (aq) = _________________
CrO4 = ___________  
-
add # of H's equal
to negative charge
 _______ (aq) = _______________
C2H3O2 = ___________  
3–
add # of H's equal
to negative charge
 ________ (aq) = ________________
PO4 = ___________  
CHM 151: Ebbing Chapter 2 Notes
page 14
Hydrates: Has a specific # of water molecules bonded per formula unit.



CuSO4·5 H2O (s)
hydrate
(contains 5 "waters of hydration")
Naming a hydrate:
name of anhydrous salt
CuSO4 (s)
+
5 H2O (l)
anhydrous salt
(has no waters)
+ Greek prefix indicating
number of waters
+
"hydrate"
CuSO4·5 H2O: ____________________________
CaCl2·6 H2O: ____________________________
2.9
WRITING CHEMICAL EQUATIONS (EQNS)
chemical equation: formulas and symbols describing a chemical rxn
A + B
reactants

starting
materials
C + D
products
substance(s) resulting
from chemical rxn
Physical state of all reactants and products are indicated using subscripts:
(s) = solid
(l) = liquid
(g)= gas
(aq) = aqueous (ions or compounds in solution)
Example:
HCl (aq) + NaHCO3 (s)

NaCl (aq) + H2O (l) + CO2 (g)
2.10 BALANCING CHEMICAL EQUATIONS
coefficient: Whole #s in front of each reactant or product, indicating how
many of each is present
subscript: Whole # following each element in a compound, indicating the # of each
element present
CHM 151: Ebbing Chapter 2 Notes
page 15
Balancing by Inspection GUIDELINES
1. Count the # of elements on both sides of the equation
2. Change the coefficients (NEVER the subscripts) to get the same # of elements on
both sides of the equation
– Balance the equation using the following order:
– Metals
– Polyatomic ions – Balance as a whole!
– Hydrogen
– Carbon
– Oxygen
– All other atoms
Examples:
______ H2 (g)
+
______ Cl2 (g)
______ Al (s)
+
______ O2 (g)
______ C5H12 (l)
+
______ O2 (g)

______ HCl (g)
______ Al2O3 (s)

______ H2O (g) + ______ CO2 (g)

Treat polyatomic ions as ONE UNIT—Do not break them up into atoms!
______ Ca(C2H3O2) 2 (aq)
+
______ K3PO4 (aq)
______ Ca3(PO4) 2 (s) +
______ Al2(SO4) 3 (aq)
+
CHM 151: Ebbing Chapter 2 Notes
______ KC2H3O2 (aq)
______ Ba(NO3) 2 (aq)
______ BaSO4 (s) +


______ Al(NO3) 3 (aq)
page 16
Chapter 3: Calculations with Chemical Formulas and Equations
Problems: 3.2-3.3, 3.6-3.7, 3.10, 3.13, 3.17-3.94, 3.97-3.106
3.2 The Mole Concept
Avogadro’s Number (NA) = 6.02 x 1023 (to 3 sig figs)
1 mole (abbreviated mol) = 6.02 x 1023 entities
Similar to: 1 dozen = 12 entities:
1 dozen doughnuts = 12 doughnuts
1 mole of doughnuts = 6.02 x 1023 doughnuts
How many eggs are in 3 dozen eggs? ___________________
How many eggs are in 3 moles of eggs?
How many C atoms are in 3 mol of C atoms?
Atomic weights and molar masses:
— The mass of 1 C atom (on average) is 12.01 amu
— The mass of 1 mole of C atoms is 12.01 g (or 12.01 g/mol)
1 mole (6.02 x 1023) is the amount of atoms of any element
that has a mass in grams equal to the mass of ONE atom in amu.
The atomic masses reported for each element in the Periodic Table gives
the atomic weight (or molecular/formula weight for compounds)
in amu and the molar mass in g/mol.
Ex.
What is the molar mass for each of the following?
(Use the atomic masses reported for each in the Periodic Table.)
a. Mg
c. Ar
b. Si
d. Sn
CHEM 151 Chapter 3
page 1 of 14
Molar mass (MM): Mass in grams of 1 mole of any element/compound
– To obtain, multiply the molar mass of each element by the number of each
present, then add up all the constituent parts.
Ex. Determine the molar mass of each of the following compounds:
a. O 2:
2 (molar mass of O) = 2 (16.00 g/mol) = 32.00 g/mol
b. H3PO4:
c. Al2(SO4)3:
Mole Calculations
Ex. 1 How many moles of Ne are in 0.500 g Ne?
Ex. 2 How many Ne atoms are in 0.500 g of Ne?
Ex. 3 How many moles of CO2 are in 0.500 g of CO2?
CHEM 151 Chapter 3
page 2 of 14
Ex. 4 How many CO2 molecules are in 0.500 g of CO2?
Ex. 5 How many oxygen atoms are in 0.500 g of CO2?
Molar Volume: Volume occupied by 1 mole of any gas
Avogadro's Law:
At the same temperature and pressure, equal volumes
of gases contain the same number of molecules
Standard temperature and pressure (STP): T=0˚C and P=1.00 atm
At STP, 1 mole of gas occupies 22.4L!
(3 sig figs)
Molar Volume Calculations
Ex. 1 How many moles of He occupy a volume of 3.36L at STP?
Ex.
2 What mass of SO3 occupies a volume of 2.05 L at STP?
CHEM 151 Chapter 3
page 3 of 14
Ex. 3
What is the volume occupied by 5.000 g of NH3 at STP?
Ex. 4
How many Ne atoms are present in 0.124 L of Ne gas at STP?
3.3
Mass Percentages from the Formula
mass percentage:
The mass of one element in a compound divided by
the mass of the entire compound
Steps to determine percentage composition:
1. Calculate the mass of each individual element in the compound
2. Add up all the masses of each element to get the total mass of compound
3. Divide the mass of each individual element with the total mass of
compound
Ex. 1 What is the percent composition of C and O in CO2?
Ex. 2 What is the percent composition of Fe and O in rust, Fe2O3?
Ex. 3 What is the mass of iron in 5.00 g of rust?
CHEM 151 Chapter 3
page 4 of 14
3.4
Elemental Analysis; Percentages of Carbon, Hydrogen, &
Oxygen
When a hydrocarbon (i.e. a compound containing only carbon and hydrogen)
or a hydrocarbon derivative (containing carbon, hydrogen, and oxygen) is
burned, the products are carbon dioxide and steam:
CxHyOz (g) +
O2 (g)

CO2 (g) + H2O (g)
Thus, if you have an unknown hydrocarbon is burned, we can use the amount
of carbon dioxide and steam formed to determine the mass percentage of
carbon and hydrogen in the original hydrocarbon.
What to keep in mind:
1. All the C in the CO2 came from original compound
 mass of C in the CO2 equals mass of C present in the original compound
2. All the H in the H2O came from original compound
 mass of H in the H2O equals mass of H present in the original compound
3. If there is O in the original compound:
 mass of O in original = mass of original – mass of C – mass of H
4. Divide the mass of each component by the mass of the compound to get the
mass percentage
Ex. 1 Combustion of an 11.5-mg sample of ethanol produces 22.0 mg of CO2
and 13.5 mg of H2O. What is the mass percentage of carbon, hydrogen,
and oxygen in ethanol?
CHEM 151 Chapter 3
page 5 of 14
Ex. 2 Cumene is a hydrocarbon—i.e., it contains only carbon and hydrogen,
and it is used in the production of acetone and phenol. Combustion of
47.6 mg of cumene produces 156.8 mg of CO2 and 42.8 mg of water.
Calculate the mass percentage of carbon and hydrogen in cumene.
3.5
Determining Formulas
Empirical Formula: Simplest whole-number ratio of atoms in a compound
Molecular Formula: Chemical formula of a compound that expresses the
actual number of atoms present in one molecule.
– The molecular formula will either be exactly the
same as or some multiple of the empirical formula!
What is the empirical formula of benzene, C6H6? __________
What is the molecular formula of benzene, C6H6? __________
What is the empirical formula of ascorbic acid, C6H8O6? _____________
What is the molecular formula of ascorbic acid, C6H8O6? ____________
Guidelines for Determining the Empirical Formula of a Compound
1. Find the # mols of each element in the compound
2. Divide each # mols of each element by smallest # mols to get ratio of
atoms
3. Get a whole number ratio for all atoms in the compound:
– If within 0.1 of a whole number, round to that whole number
– If any ratio ends close to .5  multiply ALL subscripts by 2
– If any ratio ends close to .33 or .66  multiply ALL subscripts by 3
CHEM 151 Chapter 3
page 6 of 14
Empirical Formula from Composition
Ex. 1: Determine the empirical formula for iron oxide if a sample of iron oxide
consists of 3.497 g of Fe and 1.503 g of O?
Determining Molecular Formula
1. Follow the same steps for determining empirical formula.
2. For the Molecular Formula, you need to be given the molar mass of the
compound. Find the molar mass of the empirical formula.
3. Divide the molar mass of the compound by the molar mass of the
empirical formula to get the factor with which to multiply each subscript in
the empirical formula.
Ex. 2: Acetylene is a hydrocarbon, a compound that consists only of carbon
and hydrogen. A 4.500-g sample of acetylene was analzyed and found
to contain 4.151 g of carbon. If acetylene has a molar mass of
26.0 g/mol, determine both the empirical formula and the molecular
formula for acetylene.
CHEM 151 Chapter 3
page 7 of 14
Empirical and Molecular Formulas from Percent Composition:
1. Assume 100.0 g of the compound change percentages to grams!
2. Follow the same steps for determining empirical formula.
3. Divide the molar mass of the compound by the molar mass of the
empirical formula to get the factor with which to multiply each subscript in
the empirical formula.
Ex. 3: Quinine is used as an antimalarial drug. An analysis of quinine
Indicates the compound consists of 74.03% carbon, 7.47% hydrogen,
8.64% nitrogen, and 9.86% oxygen. If quinine's molar mass is
325 g/mol, determine the empirical and molecular formulas for quinine.
3.4
Molar Interpretation of a Chemical Equation
A + B

products
starting
materials
substance(s) resulting
from chemical reaction
Interpreting a Chemical Equation
H2 (g) +
Cl2 (g)    
1 molecule
1 molecule
2 H2 (g)
+
____ molecule(s)
CHEM 151 Chapter 3
C + D
reactants
2 HCl (g)
2 molecules
O2 (g)

____ molecule(s)
page 8 of 14
2 H2O (g)
____ molecules
It follows that any multiples of these coefficients will be in same ratio!

2 H2 (g)
+
O2 (g)
2 H2O(g)
1000
_____ molecule(s)
_____ molecule(s)
_____ molecule(s)
N
_____ molecule(s)
_____ molecule(s)
_____ molecule(s)
Since N = Avogadro’s # = 6.021023 molecules = 1 mole
2 H2 (g)
___ mole(s)
+
O2 (g)

___ mole(s)
2 H2O(g)
___ mole(s)
Thus, the coefficients in a chemical equation give the mole ratios of reactants
and products in a chemical equation.
Stoichiometry (STOY-key-OM-e-tree): calculation of the quantities of
reactants and products involved in a chemical reaction
3.7
Amounts of Substances in a Chemical Reaction
Consider the following:
CH4 (g) + 2 O2 (g)  CO2 (g)
+
2 H2O (g)
1. Use unit factors to determine how many moles of O2 are needed to
completely react with 2.25 moles of CH4.
2. How many moles of CO2 form when 5.25 moles of O2 completely react?
3. How many moles of H2O form when 7.50 moles of CO2 form?
CHEM 151 Chapter 3
page 9 of 14
Mass-Mass Stoichiometry Problems
MASS OF
KNOWN
Ex. 1:
Molar
Mass
MOLES OF MOLE-MOLE
Ratio
KNOWN
MOLES OF
UNKNOWN
Molar
Mass
MASS OF
UNKNOWN
Chlorofluorocarbons (CFCs) have a negative impact on the
environment. One of the most common CFCs is CFC-12 or freon-12,
which has the formula CCl2F2 and can be prepared from the reaction
between hydrogen fluoride gas, HF, and carbon tetrachloride, CCl4.
2 HF (g) + CCl4 (l)  CCl2F2 (g) + 2 HCl (g)
1. Calculate the mass of CCl4 necessary to react completely with 50.0 g of HF.
2. Calculate the mass of CCl2F2 produced when 50.0 g of HF reacts completely.
Mass-Volume Stoichiometry Problems
MASS OF
KNOWN
Molar
Mass
MOLES OF MOLE-MOLE
Ratio
KNOWN
Molar Volume
MOLES OF
of a GAS =
UNKNOWN 22.4 L/mol
VOLUME OF
UNKNOWN
MOLES OF
UNKNOWN
MASS OF
UNKNOWN
OR
Molar Volume
VOLUME OF
of a GAS =
22.4 L/mol
KNOWN
CHEM 151 Chapter 3
MOLES OF
KNOWN
MOLE-MOLE
Ratio
page 10 of 14
Molar
Mass
Ex. 1:
Calculate the volume (in liters) of hydrogen gas produced when
0.165 g of aluminum metal reacts completely with dilute HCl at STP,
2 Al (s) + 6 HCl (aq) 
2 AlCl3 (aq) + 3 H2 (g)
Ex. 2:
An automobile airbag inflates when N2 gas results from the explosive
decomposition of sodium azide (NaN3),
2 NaN3 (s)
spark
2 Na (s) + 3 N2 (g)
Calculate the mass of NaN3 needed to produce 50.0 L of N2 gas at STP.
3.8 LIMITING REACTANT (LIMITING REAGENT)
In practice, reactants will not always be present in the exact amounts necessary
for all reactants to be converted completely into products.
Only in a limited supply of the some reactants (usually the more expensive) are
present, so these are completely used up
 “limiting reagent(s)” since they limit the amount of product that can be
made
Some reactants (usually the least expensive) are present in larger amounts and
are never complete used up
 “reactant(s) in excess”
GUIDELINES FOR SOLVING LIMITING REAGENT PROBLEMS:
1. Calculate the mass of any product that can be made from each reactant
– Use mass-mass or mass-volume conversions
2. Whichever reactant produces the smaller amount of product
 limiting reagent
3. All other reactant(s)  reactant in excess
CHEM 151 Chapter 3
page 11 of 14
 PANCAKES
MAKING BISQUICK
2 cups Bisqui ck 
1 cup milk
2 eggs
14 pancakes
2 cups of Bisquick + 1 cup milk + 2 eggs  14 pancakes
Ex. 1 If you have 10 cups of Bisquick, 10 cups of milk, and 14 eggs, how many
pancakes can you make? Indicate the limiting reagent(s) and the
reagent(s) in excess.
Consider:
1 lb Bisquick= 6 cups of Bisquick
1 gallon of milk = 16 cups milk
If you have 3.0 lbs of Bisquick, 2.0 gallon of milk, and a dozen eggs,
a. How many dozen pancakes can you make?
b. What is the limiting reagent?
c. What are the reagents in excess?
CHEM 151 Chapter 3
page 12 of 14
Consider the reaction to produce ammonia:
N2 (g)
+
3 H2 (g) 
2 NH3 (g)
Ex. 1
How many moles of ammonia form if 10.0 moles of N2 reacts with
20.0 moles of H2? (Indicate the limiting reagent and reagent in excess.)
Ex. 2.
If 50.0 g of N2 reacts with 10.0 g of H2,
a. What mass of ammonia is produced?
b. Indicate the limiting reagent and the reactant in excess.
c. What mass of the reactant in excess remains after the reaction?
CHEM 151 Chapter 3
page 13 of 14
Experimental yield; Percent Yield
Percent yield =
actual yield
 100%
theoretical yield
theoretical yield: Amount of product one should get based on the chemical
equation and the amount of reactants present
– One generally calculates this in grams from info given
actual yield: Amount of product one actually obtains
– Generally smaller than the theoretical yield because of
impurities and other adverse conditions in the lab
– This value is generally given for a problem
Example: For the reaction of 50.0 g of N2 with 10.0 g of H2, we determined the
theoretical yield of anmmonia to be what? (See Ex. 2 on p. 13)
theoretical yield = ________________________
If 49.6 g of ammonia was actually produced, calculate the percent
yield for the reaction.
percent yield =
CHEM 151 Chapter 3
page 14 of 14
Chapter 4: Chemical Reactions
Problems: 1-3, 5-6, 11, 18, 20-40, 43b, 44a, 46, 47-56, 59-72, 79-92, 95-98, 101-104
solution: composed of a solute dissolved in a solvent
solute: component present in smaller amount
solvent: component present in greater amount
aqueous solution: solution where water is the dissolving medium (the solvent)
Evidence for Chemical Reactions
1. A gas is produced—indicated by bubbles
2. A precipitate (ppt) is formed when 2 solutions are combined
3. Heat energy change is noted
– exothermic reaction: releases heat (reaction vessel feels hot)
– endothermic reaction: absorbs heat (reaction vessel feels cold)
Types of Chemical Reactions
 Precipitation Reactions (also called double-replacement reactions)
 Acid-Base Neutralization Reaction
 Oxidation-Reduction (Redox) Reactions (also called combination,
decomposition, and single-replacement reactions)
4.3
PRECIPITATION REACTIONS
soluble
= ions stay in solution (no solid/precipitate)
insoluble = solid/precipitate forms
Solubility Rules: Indicate if an ionic compound is soluble/insoluble in water.
Solubility Rules for Ionic Compounds in Water
Soluble if the ionic compound contains:
Insoluble if the ionic compound contains:
1. Li+, Na+, K+, NH4+ (ALWAYS!)
6. Carbonate ion (CO32-), but Li+, Na+, K+, NH4+
2. Nitrate ion (NO3–)
7. Chromate ion (CrO42-), but Li+, Na+, K+, NH4+
3. Acetate ion (C2H3O2–)
8. Phosphate ion (PO43-), but Li+, Na+, K+, NH4+
4. Halide ions (X–): chloride (Cl–), bromide
(Br–), or iodide ion (I–), but AgX, PbX2,
and Hg2X2 are insoluble
9. Sulfide ion (S2–), but compounds with Li+, Na+,
K+, NH4+, and CaS, SrS, and BaS are soluble.
(SO42-),
but CaSO4, BaSO4,
5. Sulfate ion
and PbSO4 are insoluble.
CHM 151: Chapter 4 Notes
10. Hydroxide ion (OH–), but compounds with
Li+, Na+, K+, NH4+, Ca(OH)2, Sr(OH)2, and
Ba(OH)2 are soluble.
page 1 of 18
Ex. 1 Use the Solubility Rules to predict whether the following ionic compounds
are soluble or insoluble. Indicate the physical state as (aq) for soluble
compounds and (s) for insoluble compounds.
a. NaCl
e. CaS
i. K2CO3
b. CuS
f. Li2CrO4
j. Ag3PO4
c. Na2SO4
g. Mg(OH)2
k. Li2S
d. KOH
h. BaSO 4
l. (NH4)2S
In a precipitation reaction, two solutions react to form a precipitate:
AX (aq) + BZ (aq)
  AZ (s) + BX (aq)
To balance and complete the following reactions:
1. Exchange the anions, writing the formulas for the products based on
the charges of the ions!
2. Use the Solubility Rules to determine if each product is soluble or insoluble
– If one (or more) is/are insoluble, a precipitate reaction has occurred,
so write the formulas for the products, indicating the precipitate as (s), then
balance the equation.
– If both products are soluble—both (aq)—then write NR=no reaction.
1.
MgSO4 (aq)
2.
CaCl2 (aq)
3.
KC2H3O2 (aq)
CHM 151: Chapter 4 Notes
NaOH (aq) 
+
AgNO3 (aq) 
+
+
LiNO3 (aq) 
page 2 of 18
4.4
ACID-BASE (NEUTRALIZATION) REACTIONS:
Arrhenius Definitions:
acid: A substance that releases hydrogen ions (H+) when dissolved in water
– Some acids are monoprotic (release only H+ per molecule)
– e.g. HCl, HBr, HI, HNO3, HClO4
– Some acids are polyprotic (release more than on H+ per molecule)
– e.g. H2SO4 and H2CO3 are both diprotic, H3PO4 is triprotic
base: A substance that releases hydroxide ions (OH–) when dissolved in water
In an acid-base reaction,
– Hydrogen ions (H+) from acid react with the hydroxide ions (OH–) from base
 water, H2O
– The cation (M+) from the base combines with the anion from the acid (X–)
 the salt.
A general equation for an acid-base neutralization reaction is shown below:
HX (aq) + MOH (aq)
acid
 H2O (l) + MX (aq)
base
water
salt
Because water is always produced, an acid always reacts with a base!
Ex. 1 Complete and balance each of the equations below:
NaOH (aq) 
a.
HCl (aq) +
b.
H2SO4 (aq) +
KOH (aq) 
c.
H3PO4 (aq) +
Ca(OH)2 (aq) 
CHM 151: Chapter 4 Notes
page 3 of 18
Brønsted-Lowry Definitions:
acid: A substance that donates a proton (H+)
base: A substance that accepts a proton (H+)
– It need not contain hydroxide ion (OH–).
According to Brønsted-Lowry, an acid-base reaction simply involves a
proton transfer, not necessarily the formation of water and a salt.
Consider the following acid-base reaction:
HCl (aq)
+
NH3 (aq) NH4+ (aq)
B-L acid
+
Cl– (aq)
B-L base
Ex. 1 Indicate the Brønsted-Lowry acid and base in each of the following:
a.
NH3 (aq)
b.
NH3 (aq)
+
H2O (l) 
+
NH4+ (aq)
HNO3 (l)  NH4+ (aq)
OH– (aq)
+
+
NO3– (aq)
Acid-Base Reactions with Gas Formation
– Some acid-base reactions involve the formation of carbon dioxide gas, CO2
(g), in addition to water and a salt.
– When the base contain carbonate ion (CO32–) or hydrogen carbonate ion
(HCO3–), then the products of the acid-base reaction are water, carbon
dioxide gas, and a salt.
The general equations for the unbalanced acid-base reactions are below:
HX (aq) + MCO3 (aq)
acid
base
HX (aq) + MHCO3 (aq)
acid
base
 H2O (l) + CO2 (g) + MX (aq)
water
carbon dioxide
salt
 H2O (l) + CO2 (g) + MX (aq)
water
carbon dioxide
salt
Because water is always produced, an acid always reacts with a base!
CHM 151: Chapter 4 Notes
page 4 of 18
Ex. 1 Complete and balance each of the equations below:
Na2CO3 (aq) 
a.
HCl (aq) +
b.
HNO3 (aq) +
CaCO3 (aq) 
c.
H2SO4 (aq) +
KHCO3 (aq) 
4.1 Ionic Theory of Solutions and Solubility Rules
Strong Acids
Strong Bases
HCl, HBr , HI,
HNO3, HClO4, H2SO4
LiOH, NaOH, KOH,
Ca(OH)2, Sr(OH)2, Ba(OH)2
strong electrolytes: substances that are strong/good conductors of
electricity
– strong acids, strong bases, all soluble salts
– These substances dissociate to produce many ions in water
 many ions present to move electrons/conduct electricity
 strong electrolyte
weak electrolytes: substances that are weak/poor conductors of electricity
– weak acids, weak bases, insoluble salts
– These substances barely dissociate to produce only a few ions in water
 few ions to move electrons/conduct electricity
 weak electrolyte
nonelectrolytes: substances that cannot conduct electricity
– sugar (e.g. sucrose), ethanol (C2H5OH), and other molecules that are not
acids
– These molecules do not break down into ions.
– these compounds remain intact as neutral molecules that have no charge
 no ions to move electrons/conduct electricity.
CHM 151: Chapter 4 Notes
page 5 of 18
4.2 MOLECULAR AND IONIC EQUATIONS
molecular equation:
chemical equation showing reactants and products
are compounds
total/complete ionic equation:
– shows strong electrolytes as individual ions while all solids, liquids, gases,
and weak electrolytes remain intact as compounds
spectator ions: ions that do not form solids, liquids, gases, or weak
electrolytes
– appear on both sides of total ionic equation as ions
net ionic equations: shows only solids, liquids, gases, weak electrolytes
(weak acids and weak bases), and ions undergoing
reaction
– excludes spectator ions
Guidelines for Writing Net Ionic Equations
1. Balance the chemical equation.
2. Convert the molecular equation to total ionic equation
– Leave solids, liquids, gases, and weak acids and bases as compounds
– Break down strong acids, strong bases, all aqueous salts—show as (aq)
3. Cancel spectator ions to get net ionic equation
– If canceling spectator ions eliminates all ions  NO REACTION (NR)
– If coefficients can be simplified, do so to get the lowest ratio.
4. Make sure total charges (+ve and –ve) are equal on both sides of equation.
Write the net ionic equation for each of the following:
a.
H2SO4 (aq) +
CHM 151: Chapter 4 Notes
NaOH (aq)

H2O (l) +
Na2SO4 (aq)
page 6 of 18
b.
H3PO4 (aq) +
c.
Na2SO4 (aq) +
d.
KBr (aq) +

KOH (aq)
H2O (l) +
K3PO4 (aq)
 NaNO3 (aq) +
Ba(NO3)2 (aq)

Cu(C2H3O2)2 (aq)
CuBr2 (aq) +
BaSO4 (s)
KC2H3O2 (aq)
4.10 OXIDATION-REDUCTION (REDOX) REACTIONS
Combination Reactions: A + Z  AZ
– Usually, a metal and a nonmetal react to form a solid ionic compound:
metal + nonmetal
where




ionic compound (s)
indicates the reactants are also heated.
Ex. 1 Complete and balance each of the equations below:

a.
Na (s) +
Cl2 (g)

b.
Al (s) +
O2 (g)

c.
Zn (s) +
S8 (s)

CHM 151: Chapter 4 Notes


page 7 of 18
Decomposition Reactions:
AZ  A + Z
Balance the following decomposition reactions:
a. _____ Ca(HCO3)2 (s)  _____ CaCO3 (s) + _____ H2O (l) + _____ CO2 (g)
b. _____ KClO3 (s)
MnO 2
 ,

 _____ KCl (s) + _____ O2 (g)
ACTIVITY SERIES: Relative order of elements arranged by their
Li > K > Ba > Sr > Ca > Na > Mg > Al > Mn > Zn >
Fe > Cd > Co > Ni > Sn > Pb > (H) > Cu > Ag > Au
Note: The Activity Series will be given to you on quizzes and exams.
Displacement Reactions: A + BZ
metal A + aqueous soln B
 AZ + B
 aqueous soln A + metal B
To balance and complete the following rxns:
– Check the Activity Series to see which metal is more active.
  The more active metal will prefer to be in solution (aq).
1.
Mg (s) +
CdSO4 (aq)  
2.
Cd (s) +
CuSO4 (aq)  
3.
Cu (s) +
ZnSO4 (aq)  
CHM 151: Chapter 4 Notes
page 8 of 18
To balance and complete the following rxns:
– Check the Activity Series to see which metal is more active, the metal or H.
  The more active metal will prefer to be in solution (aq).
metal A + aqueous acid solution
1.
Zn (s) +
2.
Cu (s) +
 aqueous solution A + H2 (g)
HCl (aq)  
HCl (aq)  
ACTIVE METALS: Li > K > Ba > Sr > Ca > Na
— React directly with water
active metal + H2O (l)  metal hydroxide + H2 (g)
1.
Na +
H2O (l) 
2.
Fe +
H2O (l) 
COMBUSTION RXNS:
1.
C3H8 (g) +
2.
C6H6O (l) +
3.
C2H2 (g) +
CHM 151: Chapter 4 Notes

CxHy + O2 (g) 
 H2O (g) + CO2 (g)

CxHyOz + O2 (g) 
 H2O (g) + CO2 (g)
O2 (g)
O2 (g)
O2 (g)





page 9 of 18
OXIDATION NUMBERS: actual or hypothetical charge of an atom in a
compound if it existed as a monatomic ion
Guidelines for Assigning Oxidation Numbers
1. The oxidation number of an element in its natural form is 0.
– e.g. the oxidation number is zero for each element in H2, O2, Cl2, P4, Na, etc.
2. The oxidation number of a monatomic ion is the charge on the ion.
– e.g. the oxidation number of Na in Na+ is +1; the oxidation number of N in
N3– is -3; the oxidation numbers for Al2O3 are +3 for Al and –2 in O.
3. In a compound or polyatomic ion,
– Group I elements are always +1.
– Group II elements are always +2.
– Fluorine is always –1.
– Oxygen is usually –2 (except in the peroxide ion, O22–, when O is –1)
– Hydrogen is usually +1
(except when it is with a metal, like NaH or CaH2, then it is –1)
4. In a compound, the sum of all oxidation numbers must equal 0.
In a polyatomic ion, the sum of all oxidation numbers must equal charge.
Example: Determine the oxidation number for each element in the following:
a. H2SO4: H: _____, S: _____, O: _____
b. KClO3: K: _____, Cl: _____, O: _____
c. CaCr2O7: Ca: _____, Cr: _____, O: _____
d. C2O42–: C: _____, O: _____
e. Ni(OH)2: Ni: _____, O: _____, H: _____
Oxidation: lose electrons (oxidation number goes up)
Reduction: gain electrons (oxidation number goes down)
In a redox reaction
– One reactant Loses Electrons/is Oxidized (LEO)
– Another reactant Gains Electrons/is Reduced (GER)
An easy way to remember is “LEO the lion goes GER!”
– The element or reactant that is oxidized is the reducing agent.
– The element or reactant that is reduced is the oxidizing agent.
CHM 151: Chapter 4 Notes
page 10 of 18
For each of the following reactions,
1. Balance the equation.
2. Identify the reactant that is oxidized and the reactant that is reduced.
3. Identify the oxidizing agent and the reducing agent.
a.
Zn (s) +
b.
Mn (s) +
O2 (g)
c.
Ca (s) +
H2O (l)
d.
C2H2 (g) +
e.
KClO3 (s) 
f.
H2O2 (aq)

AgNO3 (aq)
O2 (g)

Zn(NO3)2 (aq) +
MnO2 (s)


KCl (s) +

Ag (s)
H2O (l) +
Ca(OH)2 (aq) +
CO2 (g) +
H2 (g)
H2O (g)
O2 (g)
O2 (g)
Example f is a disproportionation reaction, where an element in one
oxidation state is simultaneously oxidized and reduced.
CHM 151: Chapter 4 Notes
page 11 of 18
WORKING WITH SOLUTIONS
4.7
Molar Concentration = Molarity
Molarity =
moles of solute
(reported in units of M=molar)
liters (L) of solution
Ex. 1
Find the molarity of a solution prepared by dissolving 0.100 mol of
NaCl in 250.0 mL of solution:
Ex. 2
Find the molarity of a solution prepared by dissolving 1.25 g of KOH
in 250.0 mL of solution:
Ex. 3
Indicate the molarity of each ion in the solutions indicated below:
Ex. 4
a.
[Cl–] = _____________ in 0.500 M CaCl2 (aq)
b.
[Na+] = _____________ in 0.125 M Na3PO4 (aq)
c.
[NO3–] = _____________ in 1.500 M Mg(NO3)2 (aq)
d.
[SO42–] = _____________ in 1.250 M Al2(SO4)3 (aq)
Circle the solution in each set with the highest [H+]:
a.
0.100 M HF (aq)
b.
0.100 M H2CO3 (aq)
Ex. 5
0.100 M HCl (aq)
0.100 M H2SO3 (aq)
0.100 M HNO2
0.100 M H2SO4(aq)
Explain why the hydroxide ion concentration, [OH–], in a 1.00 M
NH4OH solution is not 1.00 M.
CHM 151: Chapter 4 Notes
page 12 of 18
4.8
Diluting Solutions
Dilution Equation:
M1 V1 = M2 V2
where
M1=initial molarity, V1=initial volume, M2=final molarity, V2=final volume
Ex. 1:
What is the molarity of a HCl solution prepared by diluting 15.0 mL of
6.00 M HCl to give a total volume of 100.0 mL?
Ex. 2:
What is the molarity of a NaOH solution prepared by diluting 12.5 mL
of 0.500 M NaOH to give a total volume of 50.0 mL?
Writing Molar Concentration Unit Factors and Molarity Calculations:
Ex. 1 Write 2 unit factors for each of the following:
a. 6.00 M HCl solution
b. 0.125 M NaCl solution
Ex. 2 Calculate the number of moles of HCl present in 50.0 mL of 6.00 M HCl.
Ex. 3 Calculate the mass of NaOH in 25.0 mL of a 0.500 M NaOH solution.
CHM 151: Chapter 4 Notes
page 13 of 18
Ex. 4 What volume (in L) of a 0.250 M NaCl solution contains 5.00 g of
NaCl?
Solution Stoichiometry
Ex1. One important property of oxalic acid (H2C2O4) is its ability to remove rust
(Fe2O3), as shown in the following equation:
Fe2O3 (s) + 6 H2C2O4 (aq)  2 Fe(C2O4)3–3(aq) + 3 H2O (l) + 6 H+(aq)
Calculate the mass of rust in grams that can be removed with 175 mL of a
0.250 M oxalic acid solution.
Ex2. Barium hydroxide and sodium sulfate react to form barium sulfate
precipitate.
2 AgNO3 (aq) + CaCl2 (aq)  2 AgCl (s) + Ca(NO3)2 (aq)
Calculate the amount of precipitate formed when 22.75 mL of 0.820 M
silver nitrate reacts with excess calcium chloride.
CHM 151: Chapter 4 Notes
page 14 of 18
4.6 Volumetric Analysis
standard solution: an acid or base solution where the concentration is
known, generally to 3 sig figs
— used to analyze properties of substances, such as neutralizing power of
commercial antacids, tartness of wine, etc.
acid-base indicators:
– Solutions that are pH sensitive & change color
– Generally have their color change occurring for pH7 since reactions
monitored are neutralization reactions, which are complete at pH=7
titration: The gradual addition of standard solution to another solution of
unknown concentration until the reaction between the two is
complete, as signaled by an indicator changing color
endpoint:
When one reactant has completely reacted with the other
reactant, as evidenced by an indicator changing color
Ex1. Find the molarity of a HCl solution if 25.50 mL of HCl is required to
neutralize 0.375 g of Na2CO3 as shown in the following equation:
2 HCl (aq) + Na2CO3 (aq)

2 NaCl (aq) + H2O (l) + CO2 (l)
Ex2. Find the molarity of a NaOH solution if 42.15 mL of NaOH is required to
neutralize 0.424 g of oxalic acid, H2C2O4, as shown in the following
balanced equation:
2 NaOH (aq) + H2C2O4 (aq)
CHM 151: Chapter 4 Notes

Na2C2O4 (aq) + 2 H2O (l)
page 15 of 18
Ex 3. A 10.0 mL sample of vinegar (or acetic acid, HC2H3O2) requires 37.55 mL
of a 0.255 M NaOH solution for complete neutralization. Calculate the
molarity of the acetic acid solution if the balanced equation for the reaction
is:
HC2H3O2 (aq) + NaOH (aq)
 H2O (l) + NaC2H3O2 (aq)
Ex 4. A 10.0 mL sample of battery acid (H2SO4) is titrated with 0.275 M NaOH. If
the acid concentration is 0.555 M, what volume of NaOH is required for
the titration?
H2SO4 (aq) + 2 NaOH (aq)  2 H2O (l) + Na2SO4 (aq)
Ex 5. Citric acid (abbreviated H3Cit) is a triprotic acid—ie. it has three H+ ions
that can react to produce water. If 36.10 mL of 0.223 M NaOH is used to
neutralize a 0.515 g sample of citric acid, what is the molar mass of the
acid?
H3Cit (aq) + 3 NaOH (aq)
CHM 151: Chapter 4 Notes
 Na3Cit (aq) + 3 H2O (l)
page 16 of 18
solution: composed of a solute dissolved in a solvent
solute: component present in smaller amount
solvent: component present in greater amount
MASS PERCENT CONCENTRATION (M/M%)
M/M% =
mass of solute
mass of solute
 100% 
 100%
mass of solution
mass of solute + mass of solvent
Ex1. What is the mass percent concentration of a solution made by dissolving
25.0 g of HCl in 65.0 g of water? (What is the solute, and what is the
solvent?)
Ex2. A person accused of a DUI violation submitted a 5.00 g sample of blood
for alcohol content analysis. The analysis determined the presence of
4.59 mg of alcohol in the blood. If a person with a blood alcohol content
(mass percent of alcohol in the blood) of 0.08% is considered legally
impaired, was this person driving while impaired—i.e., was the blood
alcohol content greater than or equal to 0.08%?
Ex3. Intravenous injections of glucose are sometimes administered to patients
with low blood sugar. If a normal glucose solution is 5.00%, what is the
mass of solution that contains 12.7 g of glucose?
CHM 151: Chapter 4 Notes
page 17 of 18
Ex4. Intravenous saline injections are sometimes administered to restore
electrolyte balance in trauma patients. What is the mass of water required
to dissolve 2.00 g of NaCl for a 0.90% saline solution?
CHM 151: Chapter 4 Notes
page 18 of 18
CHAPTER 5: GASES AND THE KINETIC-MOLECULAR THEORY
Problems: 1, 10, 20, 23-46, 51-81, 83-94, 101-103, 105-107, 109, 113-114
AN OVERVIEW OF THE PHYSICAL STATES OF MATTER
At "normal atmospheric conditions" (25°C and 1 atm)
– Elements that are gases: H2, N2, O2, F2, Cl2, ozone (O3), all noble gases
– No ionic compounds exist as gases
– Some molecules are gases (CO, CO2, HCl, NH3, CH4); most are solids or liquids
Physical characteristics of gases
– Gases assume the volume and shape of their containers
– Gases are the most compressible of the states of matter
– Gases will mix evenly and completely when confined to the same container
– Gases have much lower densities than liquids and solids; in units of g/L
5.1
GAS PRESSURE AND ITS MEASUREMENT
– Because gas molecules are in constant motion, gases exert pressure on
any surface they encounter
Atmospheric Pressure:
– pressure exerted by column of air on an area exposed to Earth's
atmosphere
– depends on location, temperature, and weather conditions
– decreases as altitude increases because air becomes thinner
! About 760 mmHg at sea level
barometer: instrument that measures atmospheric pressure
Units of Pressure
Standard Atmospheric Pressure (1 atm): 760 mmHg at 0°C at sea level
1 atm ! " 760 mmHg " 760 torr = 101.325 kPa
Ex. 1: If the atmospheric pressure is measured to be 725 mmHg on a given
day in Phoenix, express this atmospheric pressure in torr, atm, and
kPa.
CHM 151: Chapter 5 Notes
Page 1 of 10
5.2 EMPIRICAL GAS LAWS
Boyle’s Law: Pressure-Volume Changes
– When T and n are constant, V is indirectly proportional to pressure of the gas
Charles’ Law: Volume-Temperature Changes
– When P and n are constant, V is directly proportional to the absolute T
Gay-Lussac’s Law: Pressure-Temperature Changes
– When V and the n are constant, P is directly proportional to the absolute T
Combined Gas Law
– Consider when there are changes in P,V, and T for a gas, but n and R
remain constant
Exercises: If a value (P,V,n, or T) is not given, that value is constant.
Ex. 1.
If 25.0 mL of hydrogen gas are heated from 225K to 675K, calculate the
new volume.
CHM 151: Chapter 5 Notes
Page 2 of 10
Ex. 2.
A 250.0-mL sample at 1.20 atm is compressed to 125.0 mL. Calculate
the new pressure.
Ex. 3.
A sample of CO2 gas at 795 torr is cooled from 25°C until the new
pressure is 125 torr. Calculate the new temperature of the gas.
Ex. 4.
A sample of krypton gas at –80.0°C and 1245 torr occupies 50.5 mL.
What is the volume at STP (1.00 atm and 0.00°C)?
5.3 The Ideal Gas Law:
PV=nRT
where P=pressure (in atm), V=volume (in L), n=# of moles of gas,
L ! atm
T=temperature (in K), and the ideal gas constant, R = 0.0821
mol ! K
Standard temperature and pressure (STP): 0°C and 1 atm
Ex. 1:
Calculate the volume for 1.00 mole of gas at STP.
CHM 151: Chapter 5 Notes
Page 3 of 10
Ideal Gas Calculations NOT at STP
Ex. 2:
How many moles of NO2 gas occupy a volume of 5.00 L at 50.00°C and
735 torr?
Ex. 3.
Calculate the mass of nitrogen gas that occupies a volume of 75.0 L at
35.00°C and 2.50 atm. (Remember that nitrogen exists as a diatomic
molecule.)
Further Applications of the Ideal Gas Law: Density and Molar Mass
Ex. 1:
Calculate the density (in g/L) of H2S (g) at STP.
Ex. 2:
An unknown gas having a mass of 2.041g occupies a volume of 1.15 L at
740 torr and 20.0°C. Calculate the molar mass of the unknown gas.
Ex. 3:
Calculate the density of ammonia (NH3) in g/L at 645 mmHg and 65°C.
– Use R and solve using unit analysis.
CHM 151: Chapter 5 Notes
Page 4 of 10
5.4
STOICHIOMETRY PROBLEMS INVOLVING GAS VOLUMES
– The first step in every stoichiometry problem is getting moles!
At STP, use
PV
22.4L
, and for other conditions, start with: n =
mol
RT
Ex. 1 Magnesium nitride reacts with water to give ammonia gas:
Mg3N2 (s) + 6 H2O (l) " 3 Mg(OH)2 (s) + 2 NH3 (g)
What volume of ammonia would be produced at STP given 5.00 g of
magnesium nitride and excess water?
Ex. 2 When heated, NH4NO3 decomposes to give off steam and nitrogen gas:
NH4NO3 (s) " 2 H2O (g) + N2 (g)
How many grams of ammonium nitrate are required to produce 6.37 L of
steam at 25.00°C and 720.0 torr?
Ex. 3 The active agent in many hair bleaches is hydrogen peroxide, H2O2. The
amount of hydrogen peroxide present can be determined by titration with a
standard permanganate, MnO4–, solution:
2 MnO4– (aq) + 5 H2O2 (aq) + 6 H+ (aq) " 5 O2 (g) + 2 Mn+2 (aq) + 8 H2O (l)
Calculate the molarity of hydrogen peroxide if 28.75 mL of hydrogen
peroxide produced 695 mL of oxygen gas at 0.950 atm and 315 K?
CHM 151: Chapter 5 Notes
Page 5 of 10
The Volume-Amount Relationship: Avogadro's Law
Avogadro's Law:
At constant temperature and pressure, the volume of a gas is directly proportional
to the number of moles of gas present.
" When the reactants and products in a chemical equation are gases, we can
relate the amounts of each gas to each other in terms of a volume-to-volume
ratio, just like the mole-to-mole ratio.
Example:
2 SO2 (g)
+
O2 (g)
" 2 SO3 (g)
2 mol
1 mol
2 mol
2L
1L
2L
All Gaseous Species in a Reaction
– If all the reactants and products involved in a problem are gases, use volumevolume ratios since Avogradro’s Law applies.
Ex. 1:
Consider the reaction to produce sulfur trioxide gas:
3 H2 (g) + N2 (g) " 2 NH3 (g)
Calculate the volume of ammonia formed if 5.00 L of hydrogen gas
reacts with 3.00 L of nitrogen gas.
5.5 GAS MIXTURES; LAW OF PARTIAL PRESSURES
partial pressure: pressures of individual gas components in a mixture
Dalton's Law of Partial Pressure:
– total pressure of a mixture of gases is the sum of the partial pressures of the
gases present
PTotal = P1 + P2 + P3 + ...
Example: A mixture of gases contains nitrogen (N2), oxygen (O2), and trace
gases. Given the following partial pressures for the gases:
P N2 = 0 .78 atm
P O2 = 0 .19 atm
P trace = 0 .05 atm
Determine the total pressure for the mixture.
CHM 151: Chapter 5 Notes
Page 6 of 10
mole fraction: ratio of the number of moles of one component to sum total of all the
moles of all components
X A !"
# of moles of A
total # of moles of all gases in mixture
We can use mole fraction to calculate the partial pressure of a gas in a mixture
– For system with more than many gases, the partial pressure of nth component:
Pn = Xn PTotal
Ex. 1 A mixture of gases contains 4.46 mol of neon, 0.74 mol of argon, and
2.15 mol of xenon. Calculate the partial pressures of the gases if the total
pressure is 2.00 atm at a given temperature.
5.6
KINETIC THEORY OF AN IDEAL GAS
1. Particle volume: Gas molecules are so small they occupy no volume
– i.e. "points" that have mass but negligible volume
2. Particle motion: Gas molecules move in constant, random, straight-line
motion.
3. Particle attraction: Gas molecules do not attract or repel one another
– i.e. they have no effect on the motion of other particles except upon collision
4. Particle collisions: Collisions are perfectly elastic
– i.e. energy totally transferred from one molecule to another, and overall
energy of the system remains the same; no energy lost to friction
5. The average kinetic energy (KE or energy of motion) of the molecule is
proportional to the temperature of the gas in Kelvins
"
at higher temperatures, gas molecules move more quickly
"
at lower temperatures, gas molecules move more slowly
CHM 151: Chapter 5 Notes
Page 7 of 10
5.7
MOLECULAR SPEEDS; DIFFUSION AND EFFFUSION
kinetic energy: energy associated with the motion of an object
KE =
1
(mass)(speed)2
2
Molecular Speeds
– In 1860, Maxwell formulated an equation to determine the speed of molecules
at a given instant
Combining ideal gas equation and KMT, we can get root-mean-square speed,
urms:
urms =
where R=0.0821
3RT
MM
L ! atm
, T in Kelvins, MM=molar mass
mol ! K
This equation indicates that the higher the molar mass of a particle, the
slower the particle moves.
Same Gas at Two Different Temperatures
Example: Given two distribution curves corresponding to N2 at 0°C and 500°C,
match each curve with the corresponding temperature.
# of
molecules
molecular speeds
Note:
At higher temperatures, the curve for any gas flattens out because more
molecules have higher molecular speeds.
CHM 151: Chapter 5 Notes
Page 8 of 10
Different Gases at the Same Temperature: Relationship Between Molar
Mass and Molecular Speed
Example: Given the distribution curves of equal samples of H2 and N2 at STP,
match each curve with the corresponding gas, H2 or N2.
# of
molecules
molecular speeds
Note: Heavier molecules move slower than lighter molecules
Diffusion and Effusion
Diffusion: gradual mixing of molecules of one gas with molecules of another
by virtue of their kinetic properties
Example: Consider a tube where gas molecules can be injected into both
ends. If a sample of neon gas and a sample of fluorine gas are
injected into opposite ends of the tube at the same time, predict
where along the tube the two gases would meet. Explain why.
Ne
F2
Effusion: process of a gas under pressure escaping from a container via a
small opening.
Example: Consider a container sealed with a
small opening. A mixture of several
gases is added to the container.
Assuming the gases do not react,
indicate the order that the gases
escape out of the container,
starting with the gas that escapes
the fastest. Explain why.
CHM 151: Chapter 5 Notes
Ar
Ne
Kr
N2
O2
Page 9 of 10
Graham’s Law of Effusion
– rate of effusion of molecules " #
For 2 gases:
1
MMg as
(same temp. and pressure)
MMGas2
rate of effusion of Gas1
!
rate of effusion of Gas2
MMGas1
Ex. 1 Calculate the ratio of effusion rates for molecules of methane and
carbon dioxide.
Ex. 2 If it takes 10.0 s for a sample of neon to escape from a hole in a
container, how long would it take for an equal amount of nitrogen
gas, N2, to escape from the hole in the same container?
CHM 151: Chapter 5 Notes
Page 10 of 10
Chapter 6: Thermochemistry
Problems: 1, 6, 8-11, 13, 18-22, 25-28, 33-60, 63-78, 85-94, 97-102
heat: form of energy that is transferred from a body at a higher temperature
to one at a lower temperature
– "heat flow" means heat transfer
thermodynamics: study of heat and its transformations
thermochemistry: study of heat flow that accompanies chemical reactions
6.1 ENERGY AND ITS UNITS
energy: potential or capacity to move matter
kinetic energy (KE): energy associated with an object’s motion
– e.g. a car moving at 75 mph has much greater KE than the same car
moving at 15 mph
! Greater damage if the car crashes at 75 mph than at 15 mph
KE =
1
mv2
2
potential energy (PE): energy associated with an object’s position or its
chemical bonds
– A 10-lb bowling ball has higher PE when it’s 10 feet off the ground as
opposed to 10 inches off the ground
! Greater damage on your foot if it hit your foot after falling 10 feet than
if it hit after falling only 10 inches
– In terms of chemical bonds, the stronger the bond
!
more energy required to break the bond
! higher the potential energy of the bond
joule (J):
1J=
1 kg " m2
s2
– SI (i.e. standard) unit of energy
– To appreciate the size of a joule, note that
1 watt = 1
! So a 100-watt bulb uses 100 J every second.
J
s
– Heat can also be reported in kilojoules (kJ), where 1 kJ = 1000 J
CHM 151: Chapter 6 Thermochemistry
page 1 of 12
Example: Calculate the kinetic energy (in joules) Randy Johnson fastball
if the baseball has a mass of 143 g and travels at 95 miles per
hour.
(1 mph = 0.4469 m/s approximately)
calorie (cal): unit of energy used most often in the U.S.
– amount of energy required to raise the temperature of 1 g of water by 1˚C
1 cal ! 4.184 J
(Note: This is EXACT!)
Example: What is the kinetic energy in calories of the Randy Johnson
fastball described above?
law of conservation of energy:
6.2
energy is neither created nor destroyed
but converted from one form to another
HEAT OF REACTION
system: that part of the universe being studied
surroundings: the rest of the universe outside the system
Direction and Sign of Heat Flow
Let q = heat flow,
q is + when heat flows into the system from the surroundings
q is – when heat flows out of the system into the surroundings
CHM 151: Chapter 6 Thermochemistry
page 2 of 12
heat of reaction (qreaction):
heat associated with a chemical reaction
Where Does the Heat of Reaction Come From?
Bond Energy
– energy required to break a particular bond in 1 mol of gaseous molecules
– always positive since breaking a bond always requires energy
– a quantitative measure of the strength of a bond (i.e. stability of molecule)
Breaking and Forming Bonds
– energy is absorbed by reactants when their bonds are broken, and
energy is released by products when their bonds are formed
If
! energy released $
! energy required $
&
#
&
#
& > # when products' &
#
to break
&&
##
&&
##
" bonds are formed%
"reactants' bonds %
'
endothermic reaction
If
! energy released $
! energy required $
&
#
&
#
& < # when products' &
#
to break
&&
##
&&
##
" bonds are formed%
"reactants' bonds %
'
exothermic reaction
Endothermic reaction: qreaction = +
– Heat flows from surroundings to reaction system
' surroundings feel cooler
– e.g. water evaporating is endothermic
H2O (l)
' H2O (g)
qreaction = +ve
Exothermic reaction: qreaction = –
– Heat flows from reaction system to surroundings:
' surroundings feel hotter
– e.g. propane burning is exothermic
C2H5OH (l) + 3 O2 (g) ' 3 H2O (g) + 2 CO2 (g)
qreaction = –ve
6.6 MEASURING HEATS OF REACTION
Heat Capacity and Specific Heat
specific heat (s): amount of heat necessary to raise the temperature of 1
gram of any substance by 1°C; has units of J/g°C
– water has relatively high specific heat (4.184 J/g°·C)
– because the earth is covered by so much water
' why temperatures on earth do not vary by large degree
CHM 151: Chapter 6 Thermochemistry
page 3 of 12
heat capacity (C):
amount of heat necessary to raise the temperature of a
given amount of any substance by 1°C; in units of J/°C
molar heat capacity: heat capacity per mole of a substance (in J/mol·°C)
Use the following equations to solve problems:
q = heat capacity ! ! T
or
q = (specific heat) ! (mass) ! ! T
where !T=change in temperature
Ex. 1.
How much heat is released by a 150.0-g sample of copper that cools
from 100.0°C to 25.0°C? (The molar heat capacity for copper is
24.6 J/mol·°C)
Ex. 2
To raise the temperature from 23.5°C to 100.0 °C for a 15.5-g sample of
silver, 279.9 J is required. What is the specific heat of silver?
Ex. 3
A beaker with 250.1 g of water is heated from 25.0°C to its boiling
point. If the specific heat of water is 4.184 J/g·°C, how much heat is
required to heat the water?
CHM 151: Chapter 6 Thermochemistry
page 4 of 12
Ex. 4 When drinking an ice-cold beverage, a person must raise the
temperature of the beverage to 37.0°C (normal body temperature). One
argument for losing weight is to drink ice-cold beverages since the
body must expend about 1 calorie per gram of water per degree
Celsius—i.e. the specific heat of water = 1.00 cal/g·°C —to consume the
drink.
a. Calculate the amount of energy expended (in cal) to consume a 12oz beer (about 355 mL) if the beer is originally at 4.0°C. Assume the
drink is mostly water and its density is 1.01 g/mL.
b. If the label indicates 103 Calories (where 1 nutritional Calorie (Cal)
equal 1 kcal), what is the net calorie gain or loss when a person
consumes this beer? Is this a viable weight loss alternative?
Measurement of Heat of Reaction
calorimeter:
– an instrument that measures heat changes for physical and chemical processes
– insulated, so the only heat flow is between reaction system and calorimeter
Coffee-Cup Calorimeter
– also called constant-pressure calorimeter since under atmospheric pressure
– polystyrene cup partially filled with water
– since polystyrene is good insulator, very little heat lost through cup walls
– heat evolved by a reaction is absorbed by water, and the heat capacity of
calorimeter is the heat capacity of the water
heat of reaction:
CHM 151: Chapter 6 Thermochemistry
qreaction = – mwater !
4.184 J
!""T
g !˚C
page 5 of 12
Ex. 1 A 28.2 g sample of nickel is heated to 99.8°C and placed in a coffee cup
calorimeter containing 150.0 g of water at 23.5°C. After the metal cools,
the final temperature of metal and water is 25.0°C. Calculate the specific
heat of the metal assuming no heat is lost to the surroundings or the
calorimeter.
6.2
Enthalpy and Enthalpy Change
Enthalpy (H): refers to heat flow into and out of a system under constant pressure
Under constant pressure (usually the case since under atmospheric pressure),
qreaction = !H = Hproducts – Hreactants
6.4
For an endothermic reaction:
! H = +ve
For an exothermic reaction:
! H = –ve
Thermochemical Equations:
thermochemical equation: shows both mass and enthalpy relationships
Consider: Ice melting to form water at 0°C and 1 atm
— requires energy (6.01 kJ/mol)
We can represent the melting of 1 mol of ice as an equation:
H2O (s) " H2O (l)
!H = +6.01 kJ
Note: ! H is +ve since ice must absorb heat to form water
CHM 151: Chapter 6 Thermochemistry
page 6 of 12
Consider: The formation of water releases heat:
2 H2 (g) + O2 (g) ! 2 H2O (g)
"H = –571.6 kJ
Note: " H is –ve since heat is lost to the surroundings
Guidelines for Thermochemical Equations
1. Sign of "H indicates if reaction is exothermic ("H<0) or endothermic ("H>0).
2. The coefficients in the chemical equation represent the numbers of moles
of reactants and products for the "H given.
– e.g. " H is –185 kJ for 1 mol H2 + 1 mol Cl2 to form 2 mols of HCl
3. The physical states must be indicated for each reactant and product.
– e.g. For water, the solid and the liquid states vary by 6.01 kJ
4. "H generally reported for reactants and products at 25°C.
6.5 APPLYING STOICHIOMETRY TO HEATS OF REACTION
Rule 1.
The magnitude of " H is directly proportional to the amount of
reactant or product.
! Consider heat as you would a reactant or product in a mole-tomole ratio, where "H is the heat released or absorbed for the
moles of reactants and products indicated in the equation
Example: Consider
H2 (g) + Cl2 (g) ! 2 HCl (g)
"H = –185 kJ
a. What is "H for when 5.00 mol of hydrogen gas reacts completely?
b. What is "H for the formation of 1.00 g of hydrogen chloride gas?
CHM 151: Chapter 6 Thermochemistry
page 7 of 12
Rule 2. For a reverse rxn, ! H is equal in magnitude but opposite in sign.
If
H2O (s)
" H2O (l)
then
H2O (l)
"
! H fusion):
heat of fusion (!
H2O (s)
!H = +6.01 kJ
!H = –6.01 kJ
heat associated with a solid melting, in kJ/mol;
i.e. solid " liquid
!H!"!!Hfusion
! H vapor):
heat of vaporization (!
heat associated with a liquid vaporizing,
in kJ/mol; i.e. liquid " gas
!H!"!!Hvapor
Example: If a 24.6-g ice cube requires 8.19 kJ of heat to melt completely,
calculate !Hfusion for ice (in kJ/mol)?
Rule 3: If all the coefficients in a chemical equation are multiplied by a factor n
" !H is multiplied by factor n:
If
H2O (s)
" H2O (l)
!H = +6.01 kJ
"
2 [H2O (s) " H2O (l)] = 2 H2O (s) "2 H2O (l)
!H = 2(+6.01 kJ) = 12.0 kJ
"
1
1
1
[H2O (s) " H2O (s)] = H2O (s) " H2O (l)
2
2
2
!H =
6.7
1
(+6.01 kJ) =3.01kJ
2
Hess’s Law
Hess’s Law of heat summation: The value of !H for a reaction is the same
whether it occurs in one step or in a series of steps
! H = ! H1 + ! H2 + ! H3 + ...
CHM 151: Chapter 6 Thermochemistry
page 8 of 12
Example 1:
Cgraphite (s) + 2 H2 (g)
! CH4 (g)
Here are the reactions involved in the formation of CH4:
(a) Cgraphite (s) + O2 (g) !
CO2 (g)
"H = –393.5 kJ
(b) 2 H2 (g) + O2 (g) ! 2 H2O (l)
"H = –571.6 kJ
(c) CH4 (g) + 2 O2 (g) ! CO2 (g) + 2 H2O (l)
"H = –890.4 kJ
Rearranging the data allows us to calculate "H for the reaction:
(a)Cgraphite (s) + O2 (g)
(b)2 H2 (g) + O2(g)
(c)
!
!
CO2 (g)
"H = –393.5 kJ
2 H2O (l)
"H = –571.6 kJ
C O2 (g) + 2 H2O (l)
!
Cgraphite (s) + 2 H2 (g)
!
CH4 (g) + 2 O2 (g)
CH4 (g)
"H =
890.4 kJ
"H = –74.7
kJ
If the coefficients in an equation can be simplified, multiply by the
necessary factors to cancel all intermediate compounds and get the
correct coefficients for your final equation.
Ex. 2
Rearrange the following data:
(a) Cgraphite (s) + O2 (g) ! CO2 (g)
"H = –393.5 kJ
(b) 2 CO (g) + O2 (g)
"H = 566.0 kJ
! 2 CO2 (g)
to calculate the enthalpy change for the reaction:
2 Cgraphite (s) + O2 (g) ! 2 CO (g)
CHM 151: Chapter 6 Thermochemistry
page 9 of 12
Ex. 3
From the following data:
(a)N2 (g) + 3 H2 (g) ! 2 NH3 (g)
"H = –92.6 kJ
(b)N2 (g) + 2 O2 (g) ! 2 NO2 (g)
"H =
(c)
"H = –571.6 kJ
2 H2 (g) +
O2 (g) ! 2 H2O (l)
67.70 kJ
Calculate the enthalpy change for the reaction:
4 NH3 (g) + 7 O2 (g) ! 4 NO2 (g) + 6 H2O (l)
6.6
Standard Enthalpies of Formation ( "Hf! )
Definitions of standard state (or reference form)
1. A gas at 1 atm
2. An aqueous solution with a concentration of 1 M at a pressure of 1 atm
3. Pure liquids and solids
4. The most stable form of elements at 1 atm and 25°C
allotrope:
one or 2 or more forms of an element in the same physical state
– e.g. diamond and graphite are allotropes of carbon;
O 2 (g) and ozone, O3 (g) are allotropes of oxygen
standard enthalpy of formation ( "H!f )
– enthalpy change for formation of 1 mole of an element or compound from its
elements in their standard states (i.e. naturally occurring form at 1 atm and 25˚C)
Note: "H!f =0 for any element in its naturally occurring (most stable) form
CHM 151: Chapter 6 Thermochemistry
page 10 of 12
Examples of !H!f are as follows:
1
Ag (s) +
Cl2 (g) " AgCl (s)
2
""
1
N2 (g) + O2 (g) " NO2 (g)
2
!H = –127.1 kJ
"
!H!f (AgCl, s) = -127.1 kJ
!H = +33.2 kJ
""" !H!f (NO2, g) = +33.2 kJ
Example: Explain for each of the following if !H˚ is a heat of formation.
a.
N2 (g) + 3 H2 (g) " 2 NH3 (g)
b.
H2O (l) " H2O (g)
c.
Cgraphite (s) + 2 H2 (g)
d.
2 CO (g) + O2 (g)
e.
CO2 (s) " CO2 (g)
CHM 151: Chapter 6 Thermochemistry
" CH4 (g)
" 2 CO2 (g)
page 11 of 12
Calculation of ! H°
– superscript ° denotes taken under 1 atm and 25°C
For the reaction:
aA + bB "
cC + dD
where a,b,c,d = stoichiometric
coefficients
!
= # n !H!f (products) – # m !H!f (reactants)
!Hrxn
= [c !H!f (C) + d !H!f (D)] – [a !H!f (A) + b !H!f (B)]
!
Use Table 6.2 to solve for !Hrxn
for each of the following:
Ex. 1: C graphite (s) + O2 (g) "
CO2 (g)
!
= (1mol) !H!f (CO2, g) – [(1mol) !H!f (Cgraphite, s) + (1mol) !H!f (O2, g)]
!Hrxn
=
Ex. 2:
2 CH3OH (l) + 5 O2 (g) " 2 CO2 (g) + 4 H2O (g)
!
=
!Hrxn
Ex. 3
Consider the reaction for the combustion of glucose:
C6H12O 6 (s) + 6 O2 (g) " 6 CO2 (g) + 6 H2O (g)
!
!Hrxn
= –2803 kJ
Calculate the enthalpy of formation for glucose using Table 6.2.
CHM 151: Chapter 6 Thermochemistry
page 12 of 12
Chapter 7: Quantum Theory of the Atom
Problems: 1, 8-11, 15, 18, 25, 27-40, 49-54, 63-66, 71-72, 83-84
7.1 The Wave Nature of Light
– Light travels through space as a wave
– Waves have the following characteristics (see Fig. 7.4)
! =Greek “lambda”): distance between successive peaks
wavelength (!
!=
distance
; generally in units of m, cm, nm
wave
" =Greek “nu”): number of waves passing by a given point in 1 s
frequency ("
"=
wave
1
cycle
; generally in hertz (Hz) =
or
time
s
s
Electromagnetic (EM) spectrum:!! continuum of radiant energy (Fig. 7.5)
visible region:!! ! portion of the EM spectrum that we can perceive as color
For example, a "red-hot" or "white-hot" iron bar freshly removed from a hightemperature source has forms of energy in different parts of the EM spectrum
– red or white glow = radiation within the visible region
– warmth = radiation within the infrared region
speed of light, c=3.00 x 108 m/s, depends on frequency and wavelength
c
"
!
#
"
distance distance wave
"
#
time
wave
time
Know how to convert between wavelength and frequency using the speed of light!
Ex. 1
The wavelength for the electromagnetic radiation responsible for a blue
sky is about 473 nm. What is the frequency of this radiation in Hz?
CHM 151: Chapter 7 Notes
page 1 of 6
7.2
Quantum Effects and Photons
Classical Descriptions of Matter
John Dalton (1803)
– atoms are hard, indivisible, billiard-like particles
– atoms have distinct masses (what distinguishes on type of atom from another)
– all atoms of same element are the same
JJ Thomson (1890s)
– discovered charge-to-mass ratios of electrons!
! atoms are divisible because the electrons are one part of atom
Ernest Rutherford (1910)
– shot positive alpha particles at a thin foil of gold!
! discovery of the atomic nucleus
James Maxwell (1873)
– visible light consists of electromagnetic waves
Transition between Classical and Quantum Theory
Max Planck (1900); Blackbody Radiation
– heated solids to red or white heat
– noted matter did not emit energy in continuous bursts, but in whole-number
multiples of certain well-defined quantities
! matter absorbs/emits energy in bundles = "quanta"
(single bundle of energy= "quantum")
Albert Einstein (1905); Photoelectric Effect
– Photoelectric Effect: Light shining on a clean metal ! ! emission of electrons
– Einstein applied Planck's quantum theory to light
! !!light exists as a stream of "particles" called photons
Energy is proportional to the frequency (") and wavelength (#) of radiation, and the
proportionality constant (h) is now called Planck's constant
E " h" =
CHM 151: Chapter 7 Notes
hc
#
where
h = 6.626#10–34 J·s
page 2 of 6
Ex. 1.
Excited mercury atoms emit light strongly at a wavelength of 436 nm.
a. What is the energy (in J) for one photon of this light?
b. What is the energy (in kJ/mol) for a mole of photons of this light?
Ex. 2.
Certain elements emit light of a specific color when they are burned.
When an potassium solution is burned in a flame test, the energy of
the light emitted is 4.909!10-19 J. Calculate the wavelength for this
light, and use the visible spectrum (Fig. 7.5) to determine the color of the
light.
7.4 Quantum Mechanics
Louis de Broglie (1924); Dual Nature of the Electron
– If light can behave like a wave and a particle
! " matter (like an electron) can behave like waves
– if electron behaves like a standing wave
! an electron can only have specific wavelengths
"
! an electron can only have specific frequencies and thus, energies
"
for wave: E = mc2
for matter: E = h"
Combining the two equations, we can solve for the wavelength for any matter.
! de Broglie relation#
h
#"$"
mv
CHM 151: Chapter 7 Notes
page 3 of 6
Ex. 1 a. A baseball with mass 0.143 kg is thrown towards a batter at a velocity of
42.5 m/s, calculate the wavelength (in m) associated with the baseball’s
motion.
b. How does the !! compare in size to the baseball (diameter"0.08 m)?
The baseball’s !!"""""""! baseball’s size. (Circle one)
"
>>
<<
Ex. 2 a. Calculate the wavelength (in m) associated with an electron traveling at the
same velocity, 42.5 m/s. (The mass of the electron is 9.1095 #10–31 kg.)
b. How does the !! compare in size to the electron (diameter"1#10–10 m)?
The electron ‘s !!"""""""! electron’s size. (Circle one)
"
>>
<<
Thus, although all matter can have wave properties, such properties are
only significant for submicroscopic particles.
7.3 Bohr Theory of the Hydrogen Atom
Bohr Postulates: Bohr Model of the Atom
1. Energy-level Postulate
– Electrons move in discrete (quantized), circular orbits around the nucleus
– "tennis ball and stairs" analogy for electrons and energy levels
– a ball can bounce up to or drop from one stair to another, but it can
never be halfway between two levels
– Each orbit has a specific energy associated with it, indicated as n=1, 2,...
– ground state or ground level (n = 1): lowest energy state for atom
–!when the electron is in most stable orbit
– excited state: when the electron is in a higher energy orbit (n = 2,3,4,...)
CHM 151: Chapter 7 Notes
page 4 of 6
2. Transitions Between Energy Levels
– When the atom absorbs energy, an electron can jump from a lower energy
level to a higher energy level.
– When an electron drops from a higher energy level to a lower energy level,
the atom releases energy, sometimes in the form of visible light.
Atomic Line Spectra
Emission Spectra: continuous or line spectra of radiation emitted by substances
– a heated solid (e.g. the filament in an incandescent light bulb) emits light that can be
spread out to give a continuous spectrum = spectrum containing all
wavelengths of light, like a rainbow
– an atom in the gas phase emits light only at specific wavelengths = line spectrum
– each element has a unique line spectrum ! !!can be used to identify
unknown atoms in chemical analysis
– why line spectrum for each element called "atomic fingerprint"
Limitations of the Bohr Model ! Quantum Mechanical Model
Unfortunately, the Bohr Model failed for all other elements that had more than one
proton and one electron. (The multiple electron-nuclear attractions, electronelectron repulsions, and nuclear repulsions make other atoms much more
complicated than hydrogen.)
In 1920s, a new discipline, quantum mechanics, was developed to describe the
motion of submicroscopic particles confined to tiny regions of space.
– Quantum mechanics makes no attempt to specify the position of a small particle
at a given instant or how the electron got there
– It only gives the probability of finding small particles
– Just like taking snapshot of a location and estimating where greatest
number of people are likely to be
! Instead we take a snapshot of the atom at different times and “see”
where the electrons are usually found (See. Figs. 7.24, 7.26)
Erwin Schrödinger (1926)
– developed a differential equation that allows us to find the electron's wave
" ), which ultimately allows us to determine the probability of finding
function ("
the electron in a given place
– probability density for an electron is called the "electron cloud"
! !“shape”
Quantum Numbers, Energy Levels, and Orbitals
– 4 quantum numbers describe distribution and behavior of electrons in atoms
– Each wave function corresponds to a set of 3 quantum numbers and is referred
to as an atomic orbital
CHM 151: Chapter 7 Notes
page 5 of 6
We will only discuss the first 2 Quantum Numbers:
First (or Principal) Quantum Number (n): n=1,2,3,...
– relates the average distance of electron from nucleus
– higher n means electron is further from nucleus and higher energy (less
stable) orbital
Second (or Angular Momentum) Quantum Number (l);
! Sublevels (s, p, d, f)
– gives the "shape" of the electron clouds associated with each orbital
– The limitations on n and l
! ! for n=1, only 1s sublevel
! ! for n=2, only 2s and 2p sublevels
! ! for n=3, there are 3s, 3p, and 3d sublevels
! ! for n=4, there are 4s, 4p, 4d, and 4f sublevels
Atomic Orbital Shapes:
s orbital: spherical (see Fig. 7.25)
p orbitals:
dumbbell-shaped (see Fig. 7.26)
– only for n=2 or greater
– 3 types: px, py, pz (where x, y, and z give axis on which orbital aligns)
d orbitals:
various shapes (see Fig. 7.27)
– only for n=3 or greater
– 5 types: dxy, dxz, dyz, d x 2 " y 2 , dz2
f orbitals: Don’t worry about these.
The interesting part...
– Consider Fig. 7.26 showing the electron distribution for the 2px orbital.
– Note that the electron is never along the y-axis. There is a “node” in that
region, indicating the electron is never found there.
! !How can the electron be on either side of the y-axis without going through it?
CHM 151: Chapter 7 Notes
page 6 of 6
Chapter 8: Electron Configurations and Periodicity
Problems: 7-8, 12, 15, 19, 23, 33-56, 61-66, 71-72
8.2
Building-Up Principle and the Periodic Table
Electrons are distributed in orbitals of increasing energy levels, where the
lowest energy orbitals are filled first.
Once an orbital has the maximum number of electrons it can hold, it is
considered “filled.” Remaining electrons must then be placed into the next
highest energy orbital, and so on.
– Parking garage analogy
Orbitals in order of increasing energy:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 5d < 6p
electron configuration:
– Shorthand description of the arrangement of electrons by sublevel
according to
increasing energy
REMEMBER!
– s orbitals can hold 2 electrons
– a set of p orbitals can hold 6 electrons
– a set of d orbitals can hold 10 electrons
– a set of f orbitals can hold 14 electrons
Ex. 1
Li!! atomic number=3! ! 3 eelectron configuration for Li:
Ex. 2
F ! ! _____ e–
electron configuration for F:
Ex. 3
Fe ! _____ e–
electron configuration for Fe:
CHEM 151: Chapter 8 Notes
page 1 of 8
Exceptions to the Building-Up Principle
Atoms gain extra stability when their d subshells are half-filled or completely
filled.
! If we can fill or half-fill a d subshell by promoting an electron from an s orbital
to a d orbital, we do so to gain the extra stability.
Ex. 1
Cr!! _____ e–
electron configuration for Cr:
actual electron configuration for Cr:
Ex. 2
Ag!!_____ e–
electron configuration for Ag:
actual electron configuration for Ag:
8.3 Writing Electron Configurations Using the Periodic Table
Electron Configuration from the Periodic Table
– The Periodic Table's shape actually corresponds to the filling of energy
sublevels.
– See Fig. 8.12 (p. 325), to see how electrons for each element are distributed
into the energy sublevels.
Electron configurations of atoms with many electrons can become cumbersome.
! Abbreviated electron configurations (“noble-gas core” notation):
– Since noble gases are at the end of each row in the Periodic Table, all of
their electrons are in filled orbitals.
– Such electrons are called “core” electrons since they are more stable
(less reactive) when they belong to completely filled orbitals.
valence electrons: electrons that are in the outermost shell (unfilled orbitals)
Noble gas electron configurations can be used to abbreviate the “core”
electrons of all elements.
[He]
[Ne]
[Ar]
[Kr]
[Xe]
= 1s2
= 1s2 2s2 2p6
= 1s2 2s2 2p6 3s2 3p6
= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6
CHEM 151: Chapter 8 Notes
page 2 of 8
[Fe] = 1s2 2s2 2p6 3s2 3p6 4s2 3d6
=
[Cd] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10
=
[Ni] =
[Y]
=
[I]
=
Remember again that for some transition metals, extra stability with filled
and half-filled d orbitals, so electrons are excited from the s orbitals to d
orbitals:
[Cu] =
[Mo] =
8.1
Electron Spin and the Pauli Exclusion Principle
Each electron in an atom can have one of two possible orientations, spin ! or spin ".
electron configuration: arrangement of electrons in an atom among sublevels
CHEM 151: Chapter 8 Notes
page 3 of 8
orbital diagram: diagram showing how electrons exist within an atom’s orbitals
Pauli Exclusion Principle: no 2 e-s in an atom can have same four quantum #s
! Two electrons in the same orbital must have opposite spins
– For example, with the helium atom, there are three ways to represent two
electrons in 1s orbital (where spin is represented with the electron pointing
up or down):
for He:
""
##
"#
(a)
(b)
(c)
– but the Pauli exclusion principle rules out (a) and (b) since these show
two electrons in the same orbital with the same spin.
8.4 Orbital Diagrams of Atoms; Hund’s Rule
Hund's Rule: the most stable arrangement of electrons in subshells has the
greatest number of parallel spins
– i.e. distribute electrons with same spin (up or down) and do
not pair electrons until all subshells have an electron
For example, if carbon’s electron configuration is: 1s2 2s2 2p2
! carbon’s orbital diagram can be shown in the following ways:
(a)
"#
"#
"#
"#
"#
"
#
"#
"#
"
"
1s
(b)
1s
(c)
1s
2s
2s
2s
2p
2p
2p
– but using Hund's rule, we know (c) would be the most stable.
General Rules for Assigning Electrons in Atomic Orbital Diagrams
1. First, determine the electron configuration.
2. There is only one s orbital for each level: one 1s, one 2s, one 3s, etc.
– There are 3 p orbitals for each p sublevel.
– There are 5 d orbitals for each d sublevel.
3. Each orbital orbital can only hold 2 electrons
! Each s orbital can hold 2 electrons, each p sublevel can hold 6, d’s can hold 10.
4. Electrons in the same orbital must have opposite spins.
5. To fill sublevels, put one electron in each orbital (with same spin) before pairing.
CHEM 151: Chapter 8 Notes
page 4 of 8
Ex. 1
Draw the atomic orbital diagram for oxygen.
Ex. 2
Draw the atomic orbital diagram for phosphorus. (Use full notation)
Ex. 3
Draw the atomic orbital diagram for the valence (outermost shell)
electrons in cobalt. (Use core notation.)
Magnetic Properties of Atoms
paramagnetic: substance that contains unpaired electrons
! weakly attracted to magnetic field
diamagnetic:
substance that contains only paired electrons
! slightly repelled by magnetic field
To determine if a substance is paramagnetic or diamagnetic, you must draw
its orbital diagram and determine if it has an unpaired electrons.
Ex.
Is cobalt paramagnetic or diamagnetic?
CHEM 151: Chapter 8 Notes
____________________
page 5 of 8
8.3 Mendeleev’s Predictions from the Periodic Table
Dimitri Mendeleev proposed that elements display recurring properties
according to increasing atomic mass
! Periodic Table originally arranged with elements in order of increasing atomic
mass
Henry G.J. Moseley’s high-energy x-ray radiation experiments of atomic nuclei
! Repeating properties of elements more clearly reflected by the arrangement
of elements according to increasing atomic number
! Periodic Table’s arrangement today
Trends for increasing atomic mass are identical with those forincreasing
atomic number, except for Ni & Co, Ar & K, Te & I.
Neils Bohr’s introduction of electron energy levels
! Periodic Table’s shape
– Indicates filling of electron orbitals and element’s electron configuration
periodic law: When elements are arranged in terms of increasing atomic
number, the trends within a row or column form patterns that
allow us to predict the physical and chemical properties of
elements
8.6 Some Periodic Properties
Atomic radius: distance from nucleus to outermost electrons
– Increases down a group: More p+, n, and e– ! bigger radius
– Decreases from left to right along a period.
effective nuclear charge:
– positive charge an electron experiences because of charge from protons
in the nucleus minus any shielding due to electrons closer to the nucleus
Consider the atoms of Ti and Ni:
Thus, the higher the effective nuclear charge, the smaller the atomic radius!
CHEM 151: Chapter 8 Notes
page 6 of 8
Atomic radius trend:
– Trend from top to bottom ! like a snowman
– Trend from left to right ! like a snowman that fell to the right
First Ionization Energy:
Energy necessary to remove first electron from a
neutral atom in gaseous state to form negatively
charged ion.
First Ionization Energy TRENDS
– Decreases down a group:
Bigger the atom, the further away e–s are from +vely charged nucleus
! e–s held less tightly and are more easily removed
– Increases from left to right along a period:
– Elements with fewer (1–3) valence e–s can more easily give up e–s to
gain noble gas configuration (stability)
– Elements with more (4–7) valence e–s can more easily gain e–s togain
noble gas configuration (stability)
Trend from top to bottom ! like an upside-down snowman
Trend from left to right ! like a upside-down snowman that fell to the right
Variations in Successive Ionization Energies
– Recognize that it becomes more difficult to remove electrons from stable ions.
– See Table 8.3 on p. 337.
– For example, the ionization energy to remove the first electron from Li is
much smaller than the any successive ionization energy since electrons
are now removed from a stable ion (with a positive charge AND
sometimes a noble gas electron configuration).
CHEM 151: Chapter 8 Notes
page 7 of 8
We can indicate first and successive ionization energies in the following way:
First ionization energy = IE1
Second ionization energy = IE2
Third ionization energy = IE3
Ex. 1:
Between which two ionization energies (IE1 and IE2, IE2 and IE3, etc.)
would you expect there to be the largest jump for the following
elements? Explain.
a. Mg: Between _____ and _____
Explain why:
b. K: Between _____ and _____
Explain why:
c. Al: Between _____ and _____
Explain why:
Ex. 2:
8.7
This 3rd period element has a large jump between IE5 and IE6.
Explain how you can identify the element.
Periodicity in the Main-Group Elements
Metallic character:
– Decreases from left to right along a period:
Metals concentrated on left-hand side of P.T., nonmetals on right-hand side
– Increases down a group: Looking at groups IVA and VA, go from
nonmetals (C & N) to semimetals (Si & As) to metals (Sn & Bi)
! Same snowman trends as for atomic radius!
CHEM 151: Chapter 8 Notes
page 8 of 8
Chapter 6: Thermochemistry
Problems: 1, 6, 8-11, 13, 18-22, 25-28, 33-60, 63-78, 85-94, 97-102
heat: form of energy that is transferred from a body at a higher temperature
to one at a lower temperature
– "heat flow" means heat transfer
thermodynamics: study of heat and its transformations
thermochemistry: study of heat flow that accompanies chemical reactions
6.1 ENERGY AND ITS UNITS
energy: potential or capacity to move matter
kinetic energy (KE): energy associated with an object’s motion
– e.g. a car moving at 75 mph has much greater KE than the same car
moving at 15 mph
 Greater damage if the car crashes at 75 mph than at 15 mph
KE =
1
mv2
2
potential energy (PE): energy associated with an object’s position or its
chemical bonds
– A 10-lb bowling ball has higher PE when it’s 10 feet off the ground as
opposed to 10 inches off the ground
 Greater damage on your foot if it hit your foot after falling 10 feet than
if it hit after falling only 10 inches
– In terms of chemical bonds, the stronger the bond

more energy required to break the bond
 higher the potential energy of the bond
joule (J):
1J=
1 kg  m2
s2
– SI (i.e. standard) unit of energy
– To appreciate the size of a joule, note that
1 watt = 1
 So a 100-watt bulb uses 100 J every second.
J
s
– Heat can also be reported in kilojoules (kJ), where 1 kJ = 1000 J
CHM 151: Chapter 6 Thermochemistry
page 1 of 12
Example: Calculate the kinetic energy (in joules) Randy Johnson fastball
if the baseball has a mass of 143 g and travels at 95 miles per
hour.
(1 mph = 0.4469 m/s approximately)
calorie (cal): unit of energy used most often in the U.S.
– amount of energy required to raise the temperature of 1 g of water by 1˚C
1 cal  4.184 J
(Note: This is EXACT!)
Example: What is the kinetic energy in calories of the Randy Johnson
fastball described above?
law of conservation of energy:
6.2
energy is neither created nor destroyed
but converted from one form to another
HEAT OF REACTION
system: that part of the universe being studied
surroundings: the rest of the universe outside the system
Direction and Sign of Heat Flow
Let q = heat flow,
q is + when heat flows into the system from the surroundings
q is – when heat flows out of the system into the surroundings
CHM 151: Chapter 6 Thermochemistry
page 2 of 12
heat of reaction (qreaction):
heat associated with a chemical reaction
Where Does the Heat of Reaction Come From?
Bond Energy
– energy required to break a particular bond in 1 mol of gaseous molecules
– always positive since breaking a bond always requires energy
– a quantitative measure of the strength of a bond (i.e. stability of molecule)
Breaking and Forming Bonds
– energy is absorbed by reactants when their bonds are broken, and
energy is released by products when their bonds are formed
If
 energy released 
 energy required 




 >  when products' 

to break




 bonds are formed
reactants' bonds 

endothermic reaction
If
 energy released 
 energy required 




 <  when products' 

to break




 bonds are formed
reactants' bonds 

exothermic reaction
Endothermic reaction: qreaction = +
– Heat flows from surroundings to reaction system
 surroundings feel cooler
– e.g. water evaporating is endothermic
H2O (l)
 H2O (g)
qreaction = +ve
Exothermic reaction: qreaction = –
– Heat flows from reaction system to surroundings:
 surroundings feel hotter
– e.g. propane burning is exothermic
C2H5OH (l) + 3 O2 (g)  3 H2O (g) + 2 CO2 (g)
qreaction = –ve
6.6 MEASURING HEATS OF REACTION
Heat Capacity and Specific Heat
specific heat (s): amount of heat necessary to raise the temperature of 1
gram of any substance by 1°C; has units of J/g°C
– water has relatively high specific heat (4.184 J/g°·C)
– because the earth is covered by so much water
 why temperatures on earth do not vary by large degree
CHM 151: Chapter 6 Thermochemistry
page 3 of 12
heat capacity (C):
amount of heat necessary to raise the temperature of a
given amount of any substance by 1°C; in units of J/°C
molar heat capacity: heat capacity per mole of a substance (in J/mol·°C)
Use the following equations to solve problems:
q = heat capacity   T
or
q = (specific heat)  (mass)   T
where T=change in temperature
Ex. 1.
How much heat is released by a 150.0-g sample of copper that cools
from 100.0°C to 25.0°C? (The molar heat capacity for copper is
24.6 J/mol·°C)
Ex. 2
To raise the temperature from 23.5°C to 100.0 °C for a 15.5-g sample of
silver, 279.9 J is required. What is the specific heat of silver?
Ex. 3
A beaker with 250.1 g of water is heated from 25.0°C to its boiling
point. If the specific heat of water is 4.184 J/g·°C, how much heat is
required to heat the water?
CHM 151: Chapter 6 Thermochemistry
page 4 of 12
Ex. 4 When drinking an ice-cold beverage, a person must raise the
temperature of the beverage to 37.0°C (normal body temperature). One
argument for losing weight is to drink ice-cold beverages since the
body must expend about 1 calorie per gram of water per degree
Celsius—i.e. the specific heat of water = 1.00 cal/g·°C —to consume the
drink.
a. Calculate the amount of energy expended (in cal) to consume a 12oz beer (about 355 mL) if the beer is originally at 4.0°C. Assume the
drink is mostly water and its density is 1.01 g/mL.
b. If the label indicates 103 Calories (where 1 nutritional Calorie (Cal)
equal 1 kcal), what is the net calorie gain or loss when a person
consumes this beer? Is this a viable weight loss alternative?
Measurement of Heat of Reaction
calorimeter:
– an instrument that measures heat changes for physical and chemical processes
– insulated, so the only heat flow is between reaction system and calorimeter
Coffee-Cup Calorimeter
– also called constant-pressure calorimeter since under atmospheric pressure
– polystyrene cup partially filled with water
– since polystyrene is good insulator, very little heat lost through cup walls
– heat evolved by a reaction is absorbed by water, and the heat capacity of
calorimeter is the heat capacity of the water
heat of reaction:
CHM 151: Chapter 6 Thermochemistry
qreaction = – mwater 
4.184 J
T
g ˚C
page 5 of 12
Ex. 1 A 28.2 g sample of nickel is heated to 99.8°C and placed in a coffee cup
calorimeter containing 150.0 g of water at 23.5°C. After the metal cools,
the final temperature of metal and water is 25.0°C. Calculate the specific
heat of the metal assuming no heat is lost to the surroundings or the
calorimeter.
6.2
Enthalpy and Enthalpy Change
Enthalpy (H): refers to heat flow into and out of a system under constant pressure
Under constant pressure (usually the case since under atmospheric pressure),
qreaction = H = Hproducts – Hreactants
6.4
For an endothermic reaction:
 H = +ve
For an exothermic reaction:
 H = –ve
Thermochemical Equations:
thermochemical equation: shows both mass and enthalpy relationships
Consider: Ice melting to form water at 0°C and 1 atm
— requires energy (6.01 kJ/mol)
We can represent the melting of 1 mol of ice as an equation:
H2O (s)  H2O (l)
H = +6.01 kJ
Note:  H is +ve since ice must absorb heat to form water
CHM 151: Chapter 6 Thermochemistry
page 6 of 12
Consider: The formation of water releases heat:
2 H2 (g) + O2 (g)  2 H2O (g)
H = –571.6 kJ
Note:  H is –ve since heat is lost to the surroundings
Guidelines for Thermochemical Equations
1. Sign of H indicates if reaction is exothermic (H<0) or endothermic (H>0).
2. The coefficients in the chemical equation represent the numbers of moles
of reactants and products for the H given.
– e.g.  H is –185 kJ for 1 mol H2 + 1 mol Cl2 to form 2 mols of HCl
3. The physical states must be indicated for each reactant and product.
– e.g. For water, the solid and the liquid states vary by 6.01 kJ
4. H generally reported for reactants and products at 25°C.
6.5 APPLYING STOICHIOMETRY TO HEATS OF REACTION
Rule 1.
The magnitude of  H is directly proportional to the amount of
reactant or product.
 Consider heat as you would a reactant or product in a mole-tomole ratio, where H is the heat released or absorbed for the
moles of reactants and products indicated in the equation
Example: Consider
H2 (g) + Cl2 (g)  2 HCl (g)
H = –185 kJ
a. What is H for when 5.00 mol of hydrogen gas reacts completely?
b. What is H for the formation of 1.00 g of hydrogen chloride gas?
CHM 151: Chapter 6 Thermochemistry
page 7 of 12
Rule 2. For a reverse rxn,  H is equal in magnitude but opposite in sign.
If
H2O (s)
 H2O (l)
then
H2O (l)

 H fusion):
heat of fusion (
H2O (s)
H = +6.01 kJ
H = –6.01 kJ
heat associated with a solid melting, in kJ/mol;
i.e. solid  liquid
HHfusion
 H vapor):
heat of vaporization (
heat associated with a liquid vaporizing,
in kJ/mol; i.e. liquid  gas
HHvapor
Example: If a 24.6-g ice cube requires 8.19 kJ of heat to melt completely,
calculate Hfusion for ice (in kJ/mol)?
Rule 3: If all the coefficients in a chemical equation are multiplied by a factor n
 H is multiplied by factor n:
If
H2O (s)
 H2O (l)
H = +6.01 kJ

2 [H2O (s)  H2O (l)] = 2 H2O (s) 2 H2O (l)
H = 2(+6.01 kJ) = 12.0 kJ

1
1
1
[H2O (s)  H2O (s)] = H2O (s)  H2O (l)
2
2
2
H =
6.7
1
(+6.01 kJ) =3.01kJ
2
Hess’s Law
Hess’s Law of heat summation: The value of H for a reaction is the same
whether it occurs in one step or in a series of steps
 H =  H1 +  H2 +  H3 + ...
CHM 151: Chapter 6 Thermochemistry
page 8 of 12
Example 1:
Cgraphite (s) + 2 H2 (g)
 CH4 (g)
Here are the reactions involved in the formation of CH4:
(a) Cgraphite (s) + O2 (g) 
CO2 (g)
H = –393.5 kJ
(b) 2 H2 (g) + O2 (g)  2 H2O (l)
H = –571.6 kJ
(c) CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
H = –890.4 kJ
Rearranging the data allows us to calculate H for the reaction:
(a)Cgraphite (s) + O2 (g)
(b)2 H2 (g) + O2(g)
(c)


CO2 (g)
H = –393.5 kJ
2 H2O (l)
H = –571.6 kJ
C O2 (g) + 2 H2O (l)

Cgraphite (s) + 2 H2 (g)

CH4 (g) + 2 O2 (g)
CH4 (g)
H =
890.4 kJ
H = –74.7
kJ
If the coefficients in an equation can be simplified, multiply by the
necessary factors to cancel all intermediate compounds and get the
correct coefficients for your final equation.
Ex. 2
Rearrange the following data:
(a) Cgraphite (s) + O2 (g)  CO2 (g)
H = –393.5 kJ
(b) 2 CO (g) + O2 (g)
H = 566.0 kJ
 2 CO2 (g)
to calculate the enthalpy change for the reaction:
2 Cgraphite (s) + O2 (g)  2 CO (g)
CHM 151: Chapter 6 Thermochemistry
page 9 of 12
Ex. 3
From the following data:
(a)N2 (g) + 3 H2 (g)  2 NH3 (g)
H = –92.6 kJ
(b)N2 (g) + 2 O2 (g)  2 NO2 (g)
H =
(c)
H = –571.6 kJ
2 H2 (g) +
O2 (g)  2 H2O (l)
67.70 kJ
Calculate the enthalpy change for the reaction:
4 NH3 (g) + 7 O2 (g)  4 NO2 (g) + 6 H2O (l)
6.6
Standard Enthalpies of Formation ( Hf )
Definitions of standard state (or reference form)
1. A gas at 1 atm
2. An aqueous solution with a concentration of 1 M at a pressure of 1 atm
3. Pure liquids and solids
4. The most stable form of elements at 1 atm and 25°C
allotrope:
one or 2 or more forms of an element in the same physical state
– e.g. diamond and graphite are allotropes of carbon;
O 2 (g) and ozone, O3 (g) are allotropes of oxygen
standard enthalpy of formation ( Hf )
– enthalpy change for formation of 1 mole of an element or compound from its
elements in their standard states (i.e. naturally occurring form at 1 atm and 25˚C)
Note: Hf =0 for any element in its naturally occurring (most stable) form
CHM 151: Chapter 6 Thermochemistry
page 10 of 12
Examples of Hf are as follows:
1
Ag (s) +
Cl2 (g)  AgCl (s)
2

1
N2 (g) + O2 (g)  NO2 (g)
2
H = –127.1 kJ

Hf (AgCl, s) = -127.1 kJ
H = +33.2 kJ
 Hf (NO2, g) = +33.2 kJ
Example: Explain for each of the following if H˚ is a heat of formation.
a.
N2 (g) + 3 H2 (g)  2 NH3 (g)
b.
H2O (l)  H2O (g)
c.
Cgraphite (s) + 2 H2 (g)
d.
2 CO (g) + O2 (g)
e.
CO2 (s)  CO2 (g)
CHM 151: Chapter 6 Thermochemistry
 CH4 (g)
 2 CO2 (g)
page 11 of 12
Calculation of  H°
– superscript ° denotes taken under 1 atm and 25°C
For the reaction:
aA + bB 
cC + dD
where a,b,c,d = stoichiometric
coefficients

=  n Hf (products) –  m Hf (reactants)
Hrxn
= [c Hf (C) + d Hf (D)] – [a Hf (A) + b Hf (B)]

Use Table 6.2 to solve for Hrxn
for each of the following:
Ex. 1: C graphite (s) + O2 (g) 
CO2 (g)

= (1mol) Hf (CO2, g) – [(1mol) Hf (Cgraphite, s) + (1mol) Hf (O2, g)]
Hrxn
=
Ex. 2:
2 CH3OH (l) + 5 O2 (g)  2 CO2 (g) + 4 H2O (g)

=
Hrxn
Ex. 3
Consider the reaction for the combustion of glucose:
C6H12O 6 (s) + 6 O2 (g)  6 CO2 (g) + 6 H2O (g)

Hrxn
= –2803 kJ
Calculate the enthalpy of formation for glucose using Table 6.2.
CHM 151: Chapter 6 Thermochemistry
page 12 of 12
Chapter 7: Quantum Theory of the Atom
Problems: 1, 8-11, 15, 18, 25, 27-40, 49-54, 63-66, 71-72, 83-84
7.1 The Wave Nature of Light
– Light travels through space as a wave
– Waves have the following characteristics (see Fig. 7.4)
 =Greek “lambda”): distance between successive peaks
wavelength (
=
distance
; generally in units of m, cm, nm
wave
 =Greek “nu”): number of waves passing by a given point in 1 s
frequency (
=
wave
1
cycle
; generally in hertz (Hz) =
or
time
s
s
Electromagnetic (EM) spectrum: continuum of radiant energy (Fig. 7.5)
visible region:  portion of the EM spectrum that we can perceive as color
For example, a "red-hot" or "white-hot" iron bar freshly removed from a hightemperature source has forms of energy in different parts of the EM spectrum
– red or white glow = radiation within the visible region
– warmth = radiation within the infrared region
speed of light, c=3.00 x 108 m/s, depends on frequency and wavelength
c




distance distance wave


time
wave
time
Know how to convert between wavelength and frequency using the speed of light!
Ex. 1
The wavelength for the electromagnetic radiation responsible for a blue
sky is about 473 nm. What is the frequency of this radiation in Hz?
CHM 151: Chapter 7 Notes
page 1 of 6
7.2
Quantum Effects and Photons
Classical Descriptions of Matter
John Dalton (1803)
– atoms are hard, indivisible, billiard-like particles
– atoms have distinct masses (what distinguishes on type of atom from another)
– all atoms of same element are the same
JJ Thomson (1890s)
– discovered charge-to-mass ratios of electrons
 atoms are divisible because the electrons are one part of atom
Ernest Rutherford (1910)
– shot positive alpha particles at a thin foil of gold
 discovery of the atomic nucleus
James Maxwell (1873)
– visible light consists of electromagnetic waves
Transition between Classical and Quantum Theory
Max Planck (1900); Blackbody Radiation
– heated solids to red or white heat
– noted matter did not emit energy in continuous bursts, but in whole-number
multiples of certain well-defined quantities
 matter absorbs/emits energy in bundles = "quanta"
(single bundle of energy= "quantum")
Albert Einstein (1905); Photoelectric Effect
– Photoelectric Effect: Light shining on a clean metal   emission of electrons
– Einstein applied Planck's quantum theory to light
 light exists as a stream of "particles" called photons
Energy is proportional to the frequency () and wavelength () of radiation, and the
proportionality constant (h) is now called Planck's constant
E  h =
CHM 151: Chapter 7 Notes
hc

where
h = 6.62610–34 J·s
page 2 of 6
Ex. 1.
Excited mercury atoms emit light strongly at a wavelength of 436 nm.
a. What is the energy (in J) for one photon of this light?
b. What is the energy (in kJ/mol) for a mole of photons of this light?
Ex. 2.
Certain elements emit light of a specific color when they are burned.
When an potassium solution is burned in a flame test, the energy of
the light emitted is 4.90910-19 J. Calculate the wavelength for this
light, and use the visible spectrum (Fig. 7.5) to determine the color of the
light.
7.4 Quantum Mechanics
Louis de Broglie (1924); Dual Nature of the Electron
– If light can behave like a wave and a particle
  matter (like an electron) can behave like waves
– if electron behaves like a standing wave
 an electron can only have specific wavelengths

 an electron can only have specific frequencies and thus, energies

for wave: E = mc2
for matter: E = h
Combining the two equations, we can solve for the wavelength for any matter.
 de Broglie relation
h

mv
CHM 151: Chapter 7 Notes
page 3 of 6
Ex. 1 a. A baseball with mass 0.143 kg is thrown towards a batter at a velocity of
42.5 m/s, calculate the wavelength (in m) associated with the baseball’s
motion.
b. How does the  compare in size to the baseball (diameter0.08 m)?
The baseball’s  baseball’s size. (Circle one)

>>
<<
Ex. 2 a. Calculate the wavelength (in m) associated with an electron traveling at the
same velocity, 42.5 m/s. (The mass of the electron is 9.1095 10–31 kg.)
b. How does the  compare in size to the electron (diameter110–10 m)?
The electron ‘s  electron’s size. (Circle one)

>>
<<
Thus, although all matter can have wave properties, such properties are
only significant for submicroscopic particles.
7.3 Bohr Theory of the Hydrogen Atom
Bohr Postulates: Bohr Model of the Atom
1. Energy-level Postulate
– Electrons move in discrete (quantized), circular orbits around the nucleus
– "tennis ball and stairs" analogy for electrons and energy levels
– a ball can bounce up to or drop from one stair to another, but it can
never be halfway between two levels
– Each orbit has a specific energy associated with it, indicated as n=1, 2,...
– ground state or ground level (n = 1): lowest energy state for atom
–when the electron is in most stable orbit
– excited state: when the electron is in a higher energy orbit (n = 2,3,4,...)
CHM 151: Chapter 7 Notes
page 4 of 6
2. Transitions Between Energy Levels
– When the atom absorbs energy, an electron can jump from a lower energy
level to a higher energy level.
– When an electron drops from a higher energy level to a lower energy level,
the atom releases energy, sometimes in the form of visible light.
Atomic Line Spectra
Emission Spectra: continuous or line spectra of radiation emitted by substances
– a heated solid (e.g. the filament in an incandescent light bulb) emits light that can be
spread out to give a continuous spectrum = spectrum containing all
wavelengths of light, like a rainbow
– an atom in the gas phase emits light only at specific wavelengths = line spectrum
– each element has a unique line spectrum  can be used to identify
unknown atoms in chemical analysis
– why line spectrum for each element called "atomic fingerprint"
Limitations of the Bohr Model  Quantum Mechanical Model
Unfortunately, the Bohr Model failed for all other elements that had more than one
proton and one electron. (The multiple electron-nuclear attractions, electronelectron repulsions, and nuclear repulsions make other atoms much more
complicated than hydrogen.)
In 1920s, a new discipline, quantum mechanics, was developed to describe the
motion of submicroscopic particles confined to tiny regions of space.
– Quantum mechanics makes no attempt to specify the position of a small particle
at a given instant or how the electron got there
– It only gives the probability of finding small particles
– Just like taking snapshot of a location and estimating where greatest
number of people are likely to be
 Instead we take a snapshot of the atom at different times and “see”
where the electrons are usually found (See. Figs. 7.24, 7.26)
Erwin Schrödinger (1926)
– developed a differential equation that allows us to find the electron's wave
 ), which ultimately allows us to determine the probability of finding
function (
the electron in a given place
– probability density for an electron is called the "electron cloud"
 “shape”
Quantum Numbers, Energy Levels, and Orbitals
– 4 quantum numbers describe distribution and behavior of electrons in atoms
– Each wave function corresponds to a set of 3 quantum numbers and is referred
to as an atomic orbital
CHM 151: Chapter 7 Notes
page 5 of 6
We will only discuss the first 2 Quantum Numbers:
First (or Principal) Quantum Number (n): n=1,2,3,...
– relates the average distance of electron from nucleus
– higher n means electron is further from nucleus and higher energy (less
stable) orbital
Second (or Angular Momentum) Quantum Number (l);
 Sublevels (s, p, d, f)
– gives the "shape" of the electron clouds associated with each orbital
– The limitations on n and l
  for n=1, only 1s sublevel
  for n=2, only 2s and 2p sublevels
  for n=3, there are 3s, 3p, and 3d sublevels
  for n=4, there are 4s, 4p, 4d, and 4f sublevels
Atomic Orbital Shapes:
s orbital: spherical (see Fig. 7.25)
p orbitals:
dumbbell-shaped (see Fig. 7.26)
– only for n=2 or greater
– 3 types: px, py, pz (where x, y, and z give axis on which orbital aligns)
d orbitals:
various shapes (see Fig. 7.27)
– only for n=3 or greater
– 5 types: dxy, dxz, dyz, d x 2  y 2 , dz2
f orbitals: Don’t worry about these.
The interesting part...
– Consider Fig. 7.26 showing the electron distribution for the 2px orbital.
– Note that the electron is never along the y-axis. There is a “node” in that
region, indicating the electron is never found there.
 How can the electron be on either side of the y-axis without going through it?
CHM 151: Chapter 7 Notes
page 6 of 6
Chapter 8: Electron Configurations and Periodicity
Problems: 7-8, 12, 15, 19, 23, 33-56, 61-66, 71-72
8.2
Building-Up Principle and the Periodic Table
Electrons are distributed in orbitals of increasing energy levels, where the
lowest energy orbitals are filled first.
Once an orbital has the maximum number of electrons it can hold, it is
considered “filled.” Remaining electrons must then be placed into the next
highest energy orbital, and so on.
– Parking garage analogy
Orbitals in order of increasing energy:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 5d < 6p
electron configuration:
– Shorthand description of the arrangement of electrons by sublevel
according to
increasing energy
REMEMBER!
– s orbitals can hold 2 electrons
– a set of p orbitals can hold 6 electrons
– a set of d orbitals can hold 10 electrons
– a set of f orbitals can hold 14 electrons
Ex. 1
Li atomic number=3  3 eelectron configuration for Li:
Ex. 2
F   _____ e–
electron configuration for F:
Ex. 3
Fe  _____ e–
electron configuration for Fe:
CHEM 151: Chapter 8 Notes
page 1 of 8
Exceptions to the Building-Up Principle
Atoms gain extra stability when their d subshells are half-filled or completely
filled.
 If we can fill or half-fill a d subshell by promoting an electron from an s orbital
to a d orbital, we do so to gain the extra stability.
Ex. 1
Cr _____ e–
electron configuration for Cr:
actual electron configuration for Cr:
Ex. 2
Ag_____ e–
electron configuration for Ag:
actual electron configuration for Ag:
8.3 Writing Electron Configurations Using the Periodic Table
Electron Configuration from the Periodic Table
– The Periodic Table's shape actually corresponds to the filling of energy
sublevels.
– See Fig. 8.12 (p. 325), to see how electrons for each element are distributed
into the energy sublevels.
Electron configurations of atoms with many electrons can become cumbersome.
 Abbreviated electron configurations (“noble-gas core” notation):
– Since noble gases are at the end of each row in the Periodic Table, all of
their electrons are in filled orbitals.
– Such electrons are called “core” electrons since they are more stable
(less reactive) when they belong to completely filled orbitals.
valence electrons: electrons that are in the outermost shell (unfilled orbitals)
Noble gas electron configurations can be used to abbreviate the “core”
electrons of all elements.
[He]
[Ne]
[Ar]
[Kr]
[Xe]
= 1s2
= 1s2 2s2 2p6
= 1s2 2s2 2p6 3s2 3p6
= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6
CHEM 151: Chapter 8 Notes
page 2 of 8
[Fe] = 1s2 2s2 2p6 3s2 3p6 4s2 3d6
=
[Cd] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10
=
[Ni] =
[Y]
=
[I]
=
Remember again that for some transition metals, extra stability with filled
and half-filled d orbitals, so electrons are excited from the s orbitals to d
orbitals:
[Cu] =
[Mo] =
8.1
Electron Spin and the Pauli Exclusion Principle
Each electron in an atom can have one of two possible orientations, spin  or spin .
electron configuration: arrangement of electrons in an atom among sublevels
CHEM 151: Chapter 8 Notes
page 3 of 8
orbital diagram: diagram showing how electrons exist within an atom’s orbitals
Pauli Exclusion Principle: no 2 e-s in an atom can have same four quantum #s
 Two electrons in the same orbital must have opposite spins
– For example, with the helium atom, there are three ways to represent two
electrons in 1s orbital (where spin is represented with the electron pointing
up or down):
for He:



(a)
(b)
(c)
– but the Pauli exclusion principle rules out (a) and (b) since these show
two electrons in the same orbital with the same spin.
8.4 Orbital Diagrams of Atoms; Hund’s Rule
Hund's Rule: the most stable arrangement of electrons in subshells has the
greatest number of parallel spins
– i.e. distribute electrons with same spin (up or down) and do
not pair electrons until all subshells have an electron
For example, if carbon’s electron configuration is: 1s2 2s2 2p2
 carbon’s orbital diagram can be shown in the following ways:
(a)











1s
(b)
1s
(c)
1s
2s
2s
2s
2p
2p
2p
– but using Hund's rule, we know (c) would be the most stable.
General Rules for Assigning Electrons in Atomic Orbital Diagrams
1. First, determine the electron configuration.
2. There is only one s orbital for each level: one 1s, one 2s, one 3s, etc.
– There are 3 p orbitals for each p sublevel.
– There are 5 d orbitals for each d sublevel.
3. Each orbital orbital can only hold 2 electrons
 Each s orbital can hold 2 electrons, each p sublevel can hold 6, d’s can hold 10.
4. Electrons in the same orbital must have opposite spins.
5. To fill sublevels, put one electron in each orbital (with same spin) before pairing.
CHEM 151: Chapter 8 Notes
page 4 of 8
Ex. 1
Draw the atomic orbital diagram for oxygen.
Ex. 2
Draw the atomic orbital diagram for phosphorus. (Use full notation)
Ex. 3
Draw the atomic orbital diagram for the valence (outermost shell)
electrons in cobalt. (Use core notation.)
Magnetic Properties of Atoms
paramagnetic: substance that contains unpaired electrons
 weakly attracted to magnetic field
diamagnetic:
substance that contains only paired electrons
 slightly repelled by magnetic field
To determine if a substance is paramagnetic or diamagnetic, you must draw
its orbital diagram and determine if it has an unpaired electrons.
Ex.
Is cobalt paramagnetic or diamagnetic?
CHEM 151: Chapter 8 Notes
____________________
page 5 of 8
8.3 Mendeleev’s Predictions from the Periodic Table
Dimitri Mendeleev proposed that elements display recurring properties
according to increasing atomic mass
 Periodic Table originally arranged with elements in order of increasing atomic
mass
Henry G.J. Moseley’s high-energy x-ray radiation experiments of atomic nuclei
 Repeating properties of elements more clearly reflected by the arrangement
of elements according to increasing atomic number
 Periodic Table’s arrangement today
Trends for increasing atomic mass are identical with those forincreasing
atomic number, except for Ni & Co, Ar & K, Te & I.
Neils Bohr’s introduction of electron energy levels
 Periodic Table’s shape
– Indicates filling of electron orbitals and element’s electron configuration
periodic law: When elements are arranged in terms of increasing atomic
number, the trends within a row or column form patterns that
allow us to predict the physical and chemical properties of
elements
8.6 Some Periodic Properties
Atomic radius: distance from nucleus to outermost electrons
– Increases down a group: More p+, n, and e–  bigger radius
– Decreases from left to right along a period.
effective nuclear charge:
– positive charge an electron experiences because of charge from protons
in the nucleus minus any shielding due to electrons closer to the nucleus
Consider the atoms of Ti and Ni:
Thus, the higher the effective nuclear charge, the smaller the atomic radius!
CHEM 151: Chapter 8 Notes
page 6 of 8
Atomic radius trend:
– Trend from top to bottom  like a snowman
– Trend from left to right  like a snowman that fell to the right
First Ionization Energy:
Energy necessary to remove first electron from a
neutral atom in gaseous state to form negatively
charged ion.
First Ionization Energy TRENDS
– Decreases down a group:
Bigger the atom, the further away e–s are from +vely charged nucleus
 e–s held less tightly and are more easily removed
– Increases from left to right along a period:
– Elements with fewer (1–3) valence e–s can more easily give up e–s to
gain noble gas configuration (stability)
– Elements with more (4–7) valence e–s can more easily gain e–s togain
noble gas configuration (stability)
Trend from top to bottom  like an upside-down snowman
Trend from left to right  like a upside-down snowman that fell to the right
Variations in Successive Ionization Energies
– Recognize that it becomes more difficult to remove electrons from stable ions.
– See Table 8.3 on p. 337.
– For example, the ionization energy to remove the first electron from Li is
much smaller than the any successive ionization energy since electrons
are now removed from a stable ion (with a positive charge AND
sometimes a noble gas electron configuration).
CHEM 151: Chapter 8 Notes
page 7 of 8
We can indicate first and successive ionization energies in the following way:
First ionization energy = IE1
Second ionization energy = IE2
Third ionization energy = IE3
Ex. 1:
Between which two ionization energies (IE1 and IE2, IE2 and IE3, etc.)
would you expect there to be the largest jump for the following
elements? Explain.
a. Mg: Between _____ and _____
Explain why:
b. K: Between _____ and _____
Explain why:
c. Al: Between _____ and _____
Explain why:
Ex. 2:
8.7
This 3rd period element has a large jump between IE5 and IE6.
Explain how you can identify the element.
Periodicity in the Main-Group Elements
Metallic character:
– Decreases from left to right along a period:
Metals concentrated on left-hand side of P.T., nonmetals on right-hand side
– Increases down a group: Looking at groups IVA and VA, go from
nonmetals (C & N) to semimetals (Si & As) to metals (Sn & Bi)
 Same snowman trends as for atomic radius!
CHEM 151: Chapter 8 Notes
page 8 of 8
Chapter 9: Ionic and Covalent Bonding
Problems: 13, 15-16, 18, 22-24, 26-28, 31-38, 47-70, 75-78, 81-84, 91
9.1
Describing Ionic Bonds
A metal transfers valence electrons to a nonmetal
 metal cations and nonmetal anions
– the charged ions are attracted to each other
 ionic bond: electrostatic attraction between positively charged cation
and negatively charged anion
core electrons: innermost electrons belonging to filled electron shells
valence electrons: Electrons in the outermost shell
– Since atoms want filled electron shells to be most stable, they’ll combine with
other atoms with unfilled shells (gaining or losing e–s) to get stability.

Valence e–s lead to chemical bonds and reactions between atoms
For Main Group (A) elements, Group #   # of valence e–s
– Elements in Group IA: each have 1 valence e–
– Elements in Group IIA: each have 2 valence e–
– Elements in Group IIIA: each have 3 valence e–
– Elements in Group IVA: each have 4 valence e–
– Elements in Group VA: each have 5 valence e–
– Elements in Group VIA: each have 6 valence e–
– Elements in Group VIIA: each have 7 valence e–
– Elements in Group VIIIA: each have 8 valence e–
How many valence electrons do each of the following have?
Mg: ____ val. e–
N: ____ val. e–
Br: ____ val. e–
Al: ____ val. e–
Rb: ____ val. e–
Si: ____ val. e–
Se: ____ val. e–
Xe: ____ val. e–
Lewis Electron-Dot Symbols
– Shows the atom or ion of an element with its valence electrons
1. Element symbol representing the nucleus and core electrons
2. Dots representing the valence e–
CHM 151: Chapter 9 Notes
page 1
Rules for writing Electron Dot Symbol
1. Write down the element symbol
2. Determine the number of valence electrons using the group number
3. For ions, account for electrons gained or lost leading to the charge.
4. Assume the atom has four sides, and distribute electrons with one electron
per side before pairing electrons.
Ex. 1 Give the Lewis electron-dot symbol for the following elements:
Si
K
Cl
P
Al
Ex. 2 Give the Lewis electron-dot symbol for the following ions:
calcium ion:
potassium ion:
sulfide ion:
Why do ions form if formation requires energy (i.e. ionization energy)?
– When ionic compounds form, energy is released due to so many bonds formed
 amount of energy released more than makes up for the ionization energy
Properties of Ionic Substances:
– An ionic compound is actually a network of ions, with each cation surrounded
by anions, and vice versa. A 2-dimensional slice would look like this:
Most ionic compounds are solid at room temperature and have very high
melting points, since every bond between each ion must be broken to melt
the substance.
CHM 151: Chapter 9 Notes
page 2
Coulomb's law: Relative strength of ionic bond given by the following:
1. Charges of ions: Higher the charge  the stronger the bond
– The bond in CaO (+2 and –2) is stronger than that in NaCl (+1 and –1)
 why melting point of CaO (2927°C) much higher than NaCl's (801°C)
2. Distance between two ions: Shorter distance  stronger the bond
– internuclear distance for NaCl (0.276 nm) is shorter than KBr's (0.328 nm).
 why NaCl melting point (801°C) is higher than KBr's (734°C).
9.2
Electron Configuration of Ions
Ions of the Main-Group Elements
– Representative elements generally form ions—ie. gain or lose electrons—to
achieve a noble gas electron configuration
 Ions from representative metals are usually isoelectronic with—i.e. have
the same electron configuration as—one of the noble gases!
For IONS, one must account for the loss or gain of electrons:
# electrons = atomic # – (charge = change in # of valence electrons)
Or you can simply use the Periodic Table
– Find out with which element the ion is isoelectronic
— Move to the left for electrons lost
— Move to the right for electrons gained
 write the electron configuration for that element
Fill in the table below:
Ion
Isoelectronic
with this element
Electron Configuration
(using core notation)
P–3
Na+
Al+3
CHM 151: Chapter 9 Notes
page 3
Transition-Metal Ions
– Transition metals lose their s electrons first when forming ions
Ion
Electron Configuration for
element (use core notation)
Electron Configuration for ion
(use core notation)
Zn2+
Ag+
Fe+3
9.3
IONIC RADII:
cation radius < neutral atom radius < anion radius
(loses e-s)
(gains e-s)
Cl atom
Na+ ion
Cl – ion
Na atom
9.4 Describing Covalent Bonds
covalent bond:
sharing of a pair of electrons between 2 nonmetal atoms
– the type of bond that holds atoms together in a molecule
H. +
•
H
H—H
bonding electrons: electrons involved in covalent bond formation
nonbonding electrons (or lone pairs): belong solely to an individual atom
bonding electron pair
: :
: :
+
: :
: :
:F .
:F :
nonbonding electrons
or lone pairs
: F—F :
Coordinate covalent bond:
– When one atom donates both electrons to make the bond
:F :
–
: :
CHM 151: Chapter 9 Notes
: :
H+ +
H—F :
page 4
9.5 Polar Covalent Bonds; Electronegativity
Electronegativity (EN): Ability of an atom in a bond to draw e–s to itself
– F is the most electronegative (EN) element
 Elements are less electronegative the further away from F
EN increases
Periodic Table
EN
increases
Except for H, which has EN between B and C.
In some covalent bonds, one of the two atoms holds bonding e–s more tightly
 polar covalent bond results between two atoms
Because the electrons spend more time around F, they must spend less time
around H  F gets a partial –ve charge (–) and H gets partial +ve charge (+):

: :

H—F :
In other covalent bonds, both atoms have equal EN values
 share the bonding electrons equally  nonpolar covalent bond
9.6 Writing Lewis Electron-Dot Formulas
1. Count the total number of valence electrons present
– from atoms
– from charge if polyatomic ion (add or subtract e–s for charge)
2. Write the skeleton structure of the compound
– Put least EN atom as central atom and surround with other atoms
– Note: H and F atoms will always be outer atoms
3. Connect all atoms by drawing single bonds between all atoms, then distribute the
remaining valence electrons as lone pairs around outer atoms so each has 8 e–s
octet rule: atoms form bonds such that all atoms get eight electrons
– BUT hydrogen (H) can only have 2 electrons (to fill 1s orbital)
CHM 151: Chapter 9 Notes
page 5
4. If there are not enough electrons for each atom to have an octet, make double
and/or triple bonds between central atom and surrounding atoms
– BUT fluorine can only form a single bond
– Note that double bonds are shorter than single bonds, and
triple bonds are shorter than double bonds
5. For polyatomic ions, square brackets are drawn around the Lewis structure, and
the charge is put in the upper right-hand corner
Give the Lewis formula for the following:
a. H2O
b. O 2
c. NF3
d. HCN
e. NH4+
f. NO2–
g. SO4–2
h. CO32–
CHM 151: Chapter 9 Notes
page 6
Lewis Electron-Dot Formulas for Ternary Oxyacids (e.g. HNO3, H2SO4, etc.)
– Ternary oxyacids contain hydrogen, oxygen, and one other element.
– In ternary oxyacids, the central atom is the “other element” and hydrogen atoms
are bonded directly to oxygen atoms.
 more than one central atom in the molecule
Example:
Draw the electron-dot formula for each of the following:
a. HNO3 (aq)
9.9
b.
H2SO4 (aq)
Formal Charge and Lewis Formulas
Example:
Give two different Lewis formulas for CO2 that satisfy the octet rule.
Because there are many ways to draw Lewis structures, we need a way to
determine the most plausible structure
 formal charge: hypothetical charge an atom would have if bonding electrons are
shared equally and lone pairs belong solely to a single atom
 total number of 
 total number of
formal charge =  valence electrons —  nonbonding  —  total number 
 of bonds 
 in free atom 
 electrons 




1. For neutral molecules, sum of formal charges must equal zero.
2. For ions, the sum of the formal charges must equal charge.
CHM 151: Chapter 9 Notes
page 7
Guidelines for formal charges and plausible structures:
– Lewis formulas with no formal charges is preferable to one with formal charges.
– Lewis formula with large formal charges (+2, +3, and/or –2, –3, etc.) is less
plausible than one with lower formal charges (+1 and –1, etc.)
– When Lewis formulas have similar formal charge distributions, most plausible
formula has negative formal charge on more electronegative atom
Ex. 1 Determine the most plausible formula for CO2 above using formal charges.
Ex. 2 Determine the most likely Lewis structure for formaldehyde, CH2O, starting the
with following skeletal structures and by determining formal charges:
H
H
C
O
H
C
O
H
9.7
Delocalized Bonding; Resonance
Given the Lewis structure for ozone, we expect either of the following structures:
:
:
O
:
:
:
:
:
O
:
O
:
O
:
:
O
:
O
so one bond (O—O) bond should be longer than the other (O=O).
BUT experimental evidence indicates that both oxygen-oxygen bonds in ozone
are identical, so neither of the structures accurately represents the molecule.
page 8
:
O
:
O
:
CHM 151: Chapter 9 Notes
O
:
These types of electrons are called delocalized electrons
because they are spread between more than two atoms.
:
The actual structure is a cross between the two structures,
where the electron pair is actually spread over all three atoms:
To correctly represent such delocalized electrons using Lewis formulas we show
all the Lewis formulas with a double-arrow between each:
:
:
O
:
:
:
O
:
:
:
:
O
:
O
:
O
:
O
where each of these structures is called a resonance structure.
resonance structure:one of two or more Lewis structures representing a single
molecule that cannot be described fully with only one Lewis structure
resonance: the use of two or more Lewis structures to represent a molecule
– a real ozone molecule doesn't oscillate between the resonance
structures above but is a unique, stable structure we cannot
adequately represent with one Lewis structure
Ex. 1
Give the resonance structures for NO2–:
O
Ex. 2
9.8
N
O
O
N
O
Give the resonance structures for the carbonate ion, CO32–:
Exceptions to the Octet Rule
The Incomplete Octet:
– Some atoms are satisfied with less than an octet
– Be (1s22s2) is stable with only four valence electrons
:F
B
: :
: :
– Boron (1s2 2s2 2p1) also tends to form compounds
with less than eight electrons
H—Be—H
F:
:F :
:
CHM 151: Chapter 9 Notes
page 9
: : : :
The Expanded Octet:
– Atoms in and beyond the 3rd period can
have more than eight electrons when in a
compound
:F
:F
: :
: : : :
. : :
:
Odd-Electron Molecules:
– Some molecules have an odd number of electrons
can't satisfy octet rule; usually N has the odd number
N=O
:F :
F:
S
:F :
F:
Examples: Draw the Lewis structure for the following molecules:
BeCl2:
BH3:
PF5:
XeF4:
NO2:
triiodide ion, I3–:
CHM 151: Chapter 9 Notes
page 10
9.10 Bond Length and Bond Order
Bond Length: distance between 2 nuclei of two bonded atoms
– shorter the bond length, the stronger the bond
– triple bonds are the shortest, so they are the strongest bonds
– double bonds are the next shortest, so they are next in strength
– single bonds are longest, so they are weaker than double and triple bonds
Bond order: number of pairs of electrons in a bond
– single bond: bond order=1
– double bond: bond order=2
– triple bond: bond order=3
9.11 Bond Energy
bond energy: energy required to break a particular bond in the gas phase
– always positive since it takes energy to break a bond
– a quantitative measure of a bond’s strength (i.e. stability of molecule)
– higher the bond energy  stronger the bond
Use of Bond Energies in Thermochemistry
– We can estimate the enthalpy change (H˚) for any reaction using bond energies:
H° = BE (reactants) – BE (products)
= total energy input – total energy released
Note:
if
if
BE (reactants)
BE (reactants)
>
<
BE (products)
BE (products)
 endothermic reaction
 exothermic reaction
Ex. 1 Use Table 9.5 to calculate the enthalpy of reaction for the following:
H2 (g) + F2 (g)  2 HF (g)
CHM 151: Chapter 9 Notes
page 11
Ex. 2 Use Table 9.5 to calculate the enthalpy of reaction for the following:
H
:NN: (g)
CHM 151: Chapter 9 Notes
+
2 H2 (g) 
H
H—N—N—H
(g)
page 12
Chapter 10: Molecular Geometry and
Chemical Bonding Theory
Problems: 2-4, 7-9, 17-18, 23-32, 35-40, 43-44, 47-48, 55-60
10.1 The Valence-Shell Electron-Pair Repulsion (VSEPR) model
– refers to three-dimensional arrangement of atoms in molecule
– responsible for many physical and chemical properties of molecules
Repulsion between electrons causes them to be as far apart as possible
 valence-shell electron-pair repulsion (VSEPR) model
– accounts for geometric arrangements of electron pairs around central atom in
terms of repulsion between electron pairs
General Rules
1. Consider double and triple bonds like single bonds
(an approximation since extra electrons occupy space but sufficient for
qualitative purposes)
2. If two or more resonance structures can be drawn for a molecule, VSEPR
model can be applied to any one of them
3. Don't show formal charges
If there are only two atoms, the molecule must be linear.
Ideal Geometries with Two to Six Electron Pairs on the Central Atom
(where central atom has no lone pairs)
— consider a molecule composed of only two types of atoms, A and X
A=central atom
X=outer atoms
For three or more atoms in a molecule, general formula: AX# (where #=2 to 6)
CHM 151: Chapter 10 Notes
page 1 of 13
Molecules Where Central Atom Has No Lone Pairs
# of
BONDS
General
Formula
2
AX 2
3
AX 3
4
AX 4
MOLECULAR
GEOMETRY
180˚
120˚
109.5˚
NAME
Examples
linear
BeCl2
trigonal
planar
BF3
tetrahedral
CH4, NH4+
trigonal
bipyramidal
PCl5
octahedral
SF6
90˚
5
AX 5
6
AX 6
120˚
all angles are 90˚
CHM 151: Chapter 10 Notes
page 2 of 16
AX 2: linear
– the two outer atoms are 180° from each other
– e.g. HCN
AX 3: trigonal planar
– three outer atoms at the corners of an equilateral triangle
– each outer atom is 120° from the other two outer atoms
– e.g. CH2O
AX 4: tetrahedral (tetra = four) since four-sided, or four faces
– maximum distance between electrons requires 3D structure with
– each outer atom is 109.5° from the other outer atoms
– e.g. CH4
AX 5: trigonal bipyramidal
– trigonal = three outer atoms form planar triangle around central atom
– bipyramidal = two outer atom directly above and below central atom,
connecting outer atom forms two 3-sided pyramids
– equatorial positions: ends of planar triangle
– 3 of outer atoms are at equatorial positions, 120° from each other
– axial positions: above and below central atom
– 2 atoms are at axial positions, 90° from equatorial atoms
– e.g. PF5
AX 6: octahedral (octa=eight) connecting six atoms eight faces
– all outer atoms are 90° away from each other
– the terms "axial" and "equatorial" do not apply because all six positions
are identical
– e.g. SF6
Molecules Where Central Atom Has One Or More Lone Pairs
A central atom with lone pairs has three types of repulsive forces
lone - pair vs.
lone - pair
repulsion
>
lone - pair vs.
bonding - pair
repulsion
>
bonding - pair vs.
bonding - pair
repulsion
– bonding-pair: takes up less space than lone pairs since held by attractive
– lone pairs: take up more space than bonding-pair
A=central atom
CHM 151: Chapter 10 Notes
X=outer atoms
E=lone pairs
page 3 of 13
Molecules Where Central Atom Has Lone Pairs
Original Shape
General
Formula
Molecular Geometry
Name
trigonal planar (AX3)
<120˚
120˚
tetrahedral (AX4)
AX2E
bent or
angular
AX3E
trigonal
pyramidal
< 109.5˚
109.5˚
trigonal
pyramidal
AX3E
< 109.5˚
AX2E: bent
– start with AX3 molecule (trigonal planar) and replace an X atom w/ lone pair
– bond angles are now less than 120°
– e.g. SO2
AX3E: trigonal pyramidal (central atom + 3 outer atoms make a pyramid)
– start with AX4 molecule (tetrahedral) and replace a X atom w/ lone pair
– bond angles are now less than 109.5°
– e.g. NH3
AX2E2: bent
– start with AX4 molecule (tetrahedral) and replace 2 X atoms with 2 lone pairs
– bond angles are now less than 109.5°
–eg. H2O
CHM 151: Chapter 10 Notes
page 4 of 16
Molecules Where Central Atom Has Lone Pairs (Cont’d)
Original Shape
General
Formula
Molecular Geometry
Name
< 180˚
AX4E
See-saw
< 90˚
trigonal bipyramidal
(AX5)
90˚
< 120˚
< 180˚
AX3E2
T-shaped
< 90˚
120˚
AX2E3
linear
180˚
AX4E: seesaw
– start with AX5 molecule and replace one X atom with one lone pair
– X atom can be taken from an axial or an equatorial position
–from axial: lone pair is 90° from three equatorial X atoms and
– from equatorial: lone pair is 90° from two axial and 120° from
two other equatorial X atoms
– taking X atom from equatorial position maximizes space between lone
pair and X atoms
– bond angles are now less than 90° and less than 120°
– eg. SF4
CHM 151: Chapter 10 Notes
page 5 of 13
AX3E2:T-shaped
– start with AX5 molecule and replace two X atoms with two lone pairs
–both X atoms taken from equatorial positions to maximize distance between
the lone pairs  bond angles for remaining atoms are now less than 90°
–eg. ClF3
AB2E3:linear
– start with AX5 molecule and replace three X atoms with three lone pairs
–first two X atoms taken from equatorial positions; third X atom taken:
– from axial: third lone pair would be 90° from other lone pairs
– from equatorial: third lone pair would be ~120° from other lone pairs
taking third X atom from equatorial position maximizes space
– bond angle is now 180°
– eg. XeF2
Molecules Where Central Atom Has Lone Pairs (Cont’d)
Original Molecular
Geometry
General
Formula
Molecular
Geometry
Name
octahedral (AX6)
square
pyramidal
AX5E
all angles are < 90˚
square
planar
AX4E2
all angles are 90˚
all angles are 90˚
CHM 151: Chapter 10 Notes
page 6 of 16
AX 5E: square pyramidal (central atom + 5 X atoms make 4-faced pyramid)
– start with AX6 (octahedral) and replace one X atom with one lone pair
– since all six outer positions are identical, doesn't matter which one you take 
– bond angles are now less than 90°
– e.g. BrF5
AX4E2: square planar (central atom + 4 X atoms form square all in 1 plane)
– start with AX6 (octahedral) and replace 2 X atoms with 2 lone pairs
– doesn't matter which X atom taken first, second X atom taken 180° away from it
to maximize space between lone pairs  square planar shape
– bond angles are now exactly 90° since lone pairs balance each other
– e.g. XeF4
Guidelines for Applying the VSEPR Model
1. Draw Lewis formula
2. Count number of atoms and lone pairs around central atom, treating
double and triple bonds as single bonds
– Consider molecule's shape if lone pairs around central atom were
actually bonded pairs, then remove number of outer atoms equal to
number of lone pairs.
– Remember that lone pair(s) occupy more space than bonded pairs of
electrons, so the bond angles are compressed when the central atom has
lone pairs.
Example: For each of the following,
i. Draw the Lewis formula
ii. Indicate the molecular geometry (or shape).
iii. Indicate the bond angles.
PCl3:
CHM 151: Chapter 10 Notes
IBr4–:
page 7 of 13
NH2–:
SCN–:
SeF4:
SnCl5–:
Geometry of Molecules with More than One Central Atom
– Overall geometry of entire molecule cannot be described
 instead describe shape around each central atom in molecule
– e.g. CH3OH
Lewis structure
H
H
: :
C
H—C—O—H
H
H
H
O
tetrahedral around C and
H
bent around O
What is the geometry around each central atom in CH3COOH?
O
: :
H
: :
H—C—C—O—H
H
CHM 151: Chapter 10 Notes
page 8 of 16
10.2
Dipole Moment and Molecular Geometry
Polar Covalent Bonds
– Electrons concentrate around the more electronegative atom in a molecule
 Atom gains a partial –ve charge
– Since atoms spend less time around the other atom
 Other atom gains a partial +ve charge
– molecules that have separated +ve and –ve ends are polar molecules
– overall dipole indicated with an arrow pointing to more electronegative end

–
:
H—F
Nonpolar covalent bond: Covalent bond between 2 atoms w/ same EN
– Best example is found between two identical atoms:
– H2, O2, N2, Cl2, F2, I2, Br2
– Nonpolar covalent bonds can also occur between different atoms which
have identical EN values: (See table of EN values in Table 9.15 on p. 367)
– eg. N—Cl, C–S, S—I, etc.
Example: Indicate the type of bond (ionic, polar covalent, nonpolar covalent)
for each of the following bonds:
a. S—O in SO2
b. C—C in C2H4
c. Cl—Ba in BaCl2
ionic
ionic
ionic
polar covalent
polar covalent
polar covalent
nonpolar covalent
nonpolar covalent
nonpolar covalent
For diatomic molecules:
– nonpolar molecules: when the 2 atoms have equal EN values
– polar molecules: when the 2 atoms have different EN values
–have dipole (+ve and –ve ends)
For molecules of three of more atoms:
— polarity depend on individual bonds and geometry around central atom
Example: Determine whether the following have dipole moments:
BeCl2:
H2O:
Cl —Be—Cl
CHM 151: Chapter 10 Notes
O
H
H
page 9 of 13
CCl4 versus CHCl3:
H
CCl4
CHCl3
10.3 Valence Bond (VB) Theory
Basic Theory
– a covalent bond consists of a pair of electrons in atomic orbitals
H atom has the atomic orbital diagram:

1s
Each H in a H2 (H—H) molecule

1s
Hybridization of Atomic Orbitals (explaining polyatomic molecules)
hybridization: mixing of atomic orbitals in an atom (usually central atom) to
generate a new set of atomic orbitals, called hybrid orbitals
sp Hybridization: mixing one s orbital and one p orbital
– Consider BeCl2:
– the ground state orbital diagram for Be should be

2s
2p
– BUT this indicates that Be does not form covalent bonds with Cl since Be's
electrons are already paired
one of Be's electrons must be promoted to 2p:

2s
CHM 151: Chapter 10 Notes

2p
page 10 of 16

2s

2p
– Now there are two Be orbitals for bonding, but this indicates that the two Be–Cl
bonds are different since one forms from a 2s orbital and other from a 2p
– BUT VSEPR predicts BeCl2 is linear and experiment indicates that the two
Be–Cl bonds are equivalent
 hybridize a 2s and a 2p orbital to get 2 equivalent sp hybrid orbitals:


2s


unhybridized p orbitals
sp
2p
2p
linear shape (180˚ bond angle) corresponds to sp orbitals
sp2 Hybridization: mixing one s orbital and two p orbitals
– Consider BF3:

– ground state orbital diagram for B should be

2s
2p

and promoting one of Be's 2s electrons to 2p gives:

2s

2p
– Now there are three B orbitals for bonding, but this indicates that two of the
B–F bonds (from 2s) should be the same but one (from 2s) should be different.
But VSEPR predicts BF3 is trigonal planar and experiment indicates that
the three B–F bonds are equivalent
hybridize a 2s and two 2p orbitals to get 3 equivalent sp2 hybrid orbitals:


2s

2p


sp 2

last unhybridized
p orbital
2p
2
trigonal planar shape (120˚ bond angle) corresponds to sp orbitals
CHM 151: Chapter 10 Notes
page 11 of 13
sp3 Hybridization: mixing one s orbital and three p orbitals
— Consider CH4:

— ground state orbital diagram for C should be

2s
2p


— form 4 bonds, C’s orbital diagram should be:


2s

2p
But VSEPR predicts CH4 is tetrahedral and experiment indicates that
the four C–H bonds are equivalent
hybridize a 2s and three 2p orbitals to get 4 equivalent sp3 orbitals:



2s




sp 3
2p
tetrahedral shape (109.5˚ bond angle) corresponds to sp 3 orbitals
sp3 hybridization also applies to ammonia (NH3), where N's e–s are arranged as




sp 3
– The lone pair on the N takes up more space than the e–s in N–H bonds, so the
H–N–H angles are all less than 109.5°.
sp3d Hybridization: mixing one s orbital, three p orbitals, and one d orbital
— Consider PCl5:




3p
— ground state orbital diagram for P should be 3s
— the necessary configuration for P to form 5 bonds should be:


3s


,

3d
3p
– But VSEPR predicts PCl5 is trigonal bipyramid and experiment indicates that
the five P–Cl bonds are equivalent
hybridize the 3s, all three 3p, and one 3d orbital to get 5 equal sp3d orbitals:



sp 3d


3d
3
trigonal bipyramidal shape corresponds to sp d orbitals
CHM 151: Chapter 10 Notes
page 12 of 16
sp3d2 Hybridization: mixing one s orbital, three p orbitals, two d orbitals
– Consider SF6:

 

3p
– ground state orbital diagram for S should be 3s
– the necessary configuration for S to form 6 bonds should be:



3s


,

3d
3p
– Now VSEPR predicts SF6 is octahedron and experiment indicates that the
six S–F bonds are equivalent
hybridize the 3s, all three 3p, and two 3d’s to get 6 equal sp3d2 orbitals:






sp 3d 2
3d
octahedral shape corresponds to sp 3d 2 orbitals
Predicting what hybrid orbitals form:
1. Draw Lewis structure to determine total number of bonds on central atom
2. Given the shape of molecule or bond angle from the VSEPR model, we can
determine what hybrid orbitals must be involved:
Molecular Geometry (Shape)
Hybrid Orbitals
linear (e.g. HCN, BeCl2, CO2)
sp
trigonal planar (120˚), bent (<120˚)
sp 2
tetrahedral (109.5˚) or shapes with <109.5˚ angles
sp 3
trigonal bipyramidal, see-saw, T-shape, linear with
expanded octet on central atom
sp 3d
octahedral (90˚) or shapes with all 90˚ or <90˚ angles
sp3d2
Example: What are the hybrid orbitals for the central atom in each of the
following molecules from pages 7-8 of your notes?
PCl3:
IBr4–:
NH2–:
SCN–:
SeF4:
SnCl5–:
CHM 151: Chapter 10 Notes
page 13 of 13
Chapter 11: States of Matter; Liquids and Solids
Problems: 1, 5-6, 9, 10, 12-13, 27, 29-32, 35-36, 45-46, 49-50, 53-64
11.1 Comparison of Gases, Liquids, and Solids
phase (=physical state): solid, liquid, or gas
Solids have the lowest kinetic energy (KE)—i.e. do not move very much
– Highest attraction between particles
 particles are stuck in specific sites = very confined
Liquids have slightly higher KE—i.e. particles moving more than in solid
– Particles are still attracted and maintain contact with one another
but can move past one another
 particles are less confined
Gases have greatest KE—i.e. particles move quickly and randomly
– Attractive forces almost (if not) completely overcome, so particles can fly
 particles are far away from each other = unrestricted
11.5 Intermolecular Forces; Explaining Liquid Properties
intermolecular forces: Attraction between 2 or more molecules
– e.g. between 2 water molecules or between a water and a CO2 molecule
London (Dispersion) Forces
– Attraction between temporary or induced dipoles in adjacent molecules
– Electrons are constantly shifting and can sometimes concentrate in one
region
 instantaneous (temporary) dipole that goes away once electrons shift
– Chocolate-chip cookie dough analogy
– Most common type and weakest of intermolecular forces, found in all
molecules
– Only type of intermolecular force between nonpolar molecules
– The strength of theses forces depends on
– Number of electrons in a molecule
– How easily the electrons are dispersed or (polarized) to form
temporary dipoles
– In general, larger molecules have more electrons and are more
“polarizable”
 as molar mass increases, dispersion/London forces
become stronger
 boiling point of nonpolar molecules increases with molar
mass
CHEM 151: Chapter 11
page 1 of 8
Dipole-Dipole Forces: Attraction between polar molecules
– strong because attraction is due to permanent electrostatic charges
Note: Van der Waals Forces refer to intermolecular forces due to either
London forces or dipole-dipole forces.
Hydrogen Bonds:
– Exist between molecules with following bonds: H–F, H–O, H–N
– Special type of dipole-dipole force caused by small radii and large
electronegativity differences between H and O, N, and F atoms
– strongest type of intermolecular bond
– Responsible for relatively high boiling points of molecules like H2O,
NH3, and HF
– Responsible for many of water's unusual properties, including why
the density of ice (d=0.917 g/cm3) is less than the density of liquid
water (d=1.000 g/cm3)
– Arrangement of water molecules in ice crystal results in "holes" or
empty space, but in water, the molecules arrange to fill in the
holes
Note:
Although hydrogen bonds are the strongest type of intermolecular
bonds, ionic and covalent bonds will always be stronger than
hydrogen bonds because ionic and covalent bonds hold the ions or
atoms together in a compound.
How to determine type of intermolecular forces involved:
yes
polar
Is the molecule
polar or nonpolar
hydrogen
bonding
Are there H–F,
H–O, or H–N bonds
no
nonpolar
CHEM 151: Chapter 11
dispersion
(London) forces
dipole-dipole
forces
page 2 of 8
Example
Identify the type of bond holding atoms together in the molecules
and the intermolecular forces between the molecules for the two
examples below:
O
O
H
H
H
H
H
H
H
H
H
H
Cl
Cl
H
H
O
H
H
O
H
H
O
H
H
O
H
H
O
H
H
H
Cl
H
H
H
H
H
H
H
Cl
H
H
H
Cl
H
H
Cl
Cl
Ex. 2: Circle the type of bond or attraction described for each below:
a. The C–C bond in C2H6.
dispersion forces
ionic bond
dipole-dipole forces
polar covalent bond
hydrogen bond
nonpolar covalent bond
b. The Na–Cl bond in NaCl.
dispersion forces
ionic bond
CHEM 151: Chapter 11
dipole-dipole forces
polar covalent bond
hydrogen bond
nonpolar covalent bond
page 3 of 8
c.
What attracts HF molecules to each other in a sample.
dispersion forces
ionic bond
d.
dipole-dipole forces
polar covalent bond
hydrogen bond
nonpolar covalent bond
The H–O bond in a H2O molecule.
dispersion forces
ionic bond
e.
dipole-dipole forces
polar covalent bond
hydrogen bond
nonpolar covalent bond
What holds atoms together in a molecule of NH3.
dispersion forces
ionic bond
Ex. 2:
dipole-dipole forces
polar covalent bond
hydrogen bond
nonpolar covalent bond
The stronger the intermolecular forces, the more energy is required to
break the attraction between molecules in liquids and solids
 higher melting and boiling points
Explain each the following:
a. Why O2's boiling point is -183°C while NO's boiling point is -151°C.
b. Why N2's boiling point is -196°C while Br2's boiling point is 59°C.
c. Why H2S’s boiling point is –61°C and H2O’s boiling point is 100˚C.
CHEM 151: Chapter 11
page 4 of 8
Physical Properties and Intermolecular Forces:
Vapor Pressure: partial pressure exerted by gas molecules above
the liquid in equilibrium with the liquid
– varies for different liquids, varies for different T’s
– more gas molecules  higher vapor pressure
– weaker the intermolecular forces
more molecules can go from liquid to vapor
higher vapor pressure
 H vap):
molar heat of vaporization (
– energy required to vaporize one mole of a liquid
– stronger intermolecular forces  higher molar heat of vaporization
Practical example of molar heat of vaporization:
– rubbing alcohol on your hands
– heat from your hands increases the KE of alcohol molecules
evaporates
– this is an endothermic process
– perspiration: body heat vaporizes water of perspiration from
body's surface
Boiling Point: temperature where vapor pressure of liquid is equal to
external pressure (usually atmospheric pressure)
– normal boiling point is boiling point at 1 atm
– e.g. water boils at 100°C at 1 atm, but it boils at ~95°C in
Albuquerque where the atmospheric pressure is ~0.85 atm
– both Hvap and boiling point depend on intermolecular forces
Surface Tension: attraction between surface molecules in a liquid
– stronger intermolecular forces = stronger attraction
 surface molecules are held together more strongly
 higher surface tension
11.2 Phase Transition: change from one physical state to another
sublimation
SOLID
fusion
freezing
LIQUID
vaporization
condensation
GAS
deposition
CHEM 151: Chapter 11
page 5 of 8
The Equilibrium Nature of Phase Changes
dynamic equilibrium: rate of forward process is exactly equal to the rate
of reverse process
Liquid-Gas Equilibrium:
liquid
vapor
evaporation (or vaporization): liquid  vapor
condensation: vapor  liquid
– the process of a gas liquefying
– can result from two techniques:
1. cooling sample of gas  lower KE, and molecules start to
aggregate to form small drops of liquid
2. applying pressure to gas  minimize space between
molecules, so molecules are attracted to each other
Liquid-Solid Equilibrium
freezing: liquid  solid
melting (or fusion): solid  liquid
melting point: temperature at which solid and liquid phases coexist in
equilibrium
– normal melting point is melting point at 1 atm
Consider solid-liquid equilibrium of water and ice (at 0°C and 1 atm):
ice
water
When ice cubes are placed into a glass of water, the ice cubes begin to
melt, but some water between the ice cubes freezes, causing ice cubes to
fuse
 H fus): energy required to melt one mole of solid
molar heat of fusion (
supercooling: a liquid can be temporarily cooled to below its freezing
point
– results when a liquid is cooled so rapidly that molecules don't have time
to arrange themselves properly
– unstable condition; gentle stirring or adding "seed" crystal will cause
solidification
– Practical example: Sometimes happens in your refrigerator if a bottle
of carbonated soda supercools.
– You notice ice crystals upon opening.
CHEM 151: Chapter 11
page 6 of 8
Solid-Gas Equilibrium:
Consider the dynamic equilibrium:
solid
vapor
Dry ice, CO2 (s), sublimes at room temperature, completely skipping a
liquid phase.
sublimation:solid  gas (with no liquid phase)
deposition: gas  solid (with no liquid phase)
 H sub): Hsub = Hfus + Hvap
molar heat of sublimation (
– energy required to sublime one mole of solid
– given by sum of heats of fusion and vaporization
– another illustration of Hess's Law
Heat of Phase Transition
heating-cooling curve:
Shows the phase changes that occur when heat
is added or removed from a sample
Heating Curve
Temperature (˚C)
Heat added
Draw a heating curve indicating the following:
1. Regions for solid only, liquid only, gas only, solid-liquid, liquid-gas
2. The relationship between melting point and phases present
3. The relationship between boiling point and phases present
4. Where the curve is flat, where the slope is positive
CHEM 151: Chapter 11
page 7 of 8
11.3 Phase Diagrams
– summarize the conditions at which a substance exists as solid, liquid, or
gas
– allow us to determine melting and boiling points at different external
pressures
Water: phase diagram shown in Fig. 11.11
– graph divided into three regions corresponding to each phase
– lines separating two regions indicate conditions when both phases exist
– triple point: point at which all three curves meet
– when all three phases can be in equilibrium with each other
– for water, at 0.01°C and about 4.56 mm Hg
Phase Diagram for Water versus that for CO2:
– Note that line separating solid and liquid phases has positive slope for
CO2 but negative slope for H2O
– for CO2, the solid is more dense than the liquid, so increasing pressure
converts the liquid to a solid
– for H2O, the solid is less dense than the liquid, so increasing pressure
converts the solid to a liquid
Critical Temperature and Pressure:
critical temperature (Tc): above which its gas form cannot be made to
liquefy, no matter how great the applied pressure
– highest temperature at which a substance can exist as a liquid
– intermolecular attraction is a finite quantity for given substance
– below Tc, molecules are moving slowly enough to maintain contact
– above Tc, molecular motion so energetic that molecules will always
break away from attraction
critical pressure (Pc):
minimum pressure that must be applied to
liquefy sample at the critical temperature
Given a Phase Diagram, be able to do the following:
– Determine what phase(s) is/are present at a given temperature and
pressure
– Indicate the boiling point or melting point at a given pressure
– Describe what phase change occurs when temperature is changed at a
constant pressure
– Describe what phase change occurs when pressure is changed at a
constant temperature.
CHEM 151: Chapter 11
page 8 of 8