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chapt05.ppt
Gases
Chapter 5
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.
Elements that exist as gases at 250C and 1 atmosphere
5.1
5.1
		Gases assume the volume and shape of their containers.
		Gases are the most compressible state of matter.
		Gases will mix evenly and completely when confined to the same container.
		Gases have much lower densities than liquids and solids.
5.1
Physical Characteristics of Gases
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
5.2
Pressure = 
(force = mass x acceleration)
Barometer
Force
Area
Sea level
1 atm
4 miles
0.5 atm
10 miles
0.2 atm
5.2
5.2
Figure 5.4
5.3
As P (h) increases
V decreases
P a 1/V
P x V = constant
P1 x V1 = P2 x V2
5.3
Boyle’s Law
Constant temperature
Constant amount of gas
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
P1 x V1 = P2 x V2
P1 = 726 mmHg
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 = 
= 4460 mmHg
5.3
P1 x V1
V2
726 mmHg x 946 mL
154 mL
=
As T increases
V increases
5.3
Variation of gas volume with temperature
at constant pressure.
5.3
V a T
V = constant x T
V1/T1 = V2/T2
T (K) = t (0C) + 273.15 
Charles’ & Gay-Lussac’s Law
Temperature must be
in Kelvin
A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
V1 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 = 
= 192 K
5.3
V1/T1 = V2/T2
T1 = 125 (0C) + 273.15 (K) = 398.15 K
V2 x T1
V1
1.54 L x 398.15 K
3.20 L
=
Avogadro’s Law
V a number of moles (n)
V = constant x n
V1/n1 = V2/n2
5.3
Constant temperature
Constant pressure
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?
At constant T and P
5.3
4NH3 + 5O2 4NO + 6H2O
1 mole NH3 1 mole NO
1 volume NH3 1 volume NO
5.3
5.3
5.3
Ideal Gas Equation
5.4
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
R is the gas constant
PV = nRT
1
P
Boyle’s law: V a (at constant n and T)
nT
P
V a 
nT
P
nT
P
V = constant x = R
PV = nRT
R = 0.082057 L • atm / (mol • K)
5.4
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
PV
nT
R = 
(1 atm)(22.414L)
(1 mol)(273.15 K)
=
Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
PV = nRT
T = 0 0C = 273.15 K
P = 1 atm
V = 30.6 L
5.4
nRT
P
V = 
1 mol HCl
36.45 g HCl
n = 49.8 g x 
= 1.37 mol
L•atm
mol•K
1.37 mol x 0.0821 x 273.15 K
1 atm
V =
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant
= constant
= 1.48 atm
5.4
nR
V
P
T
=
P1
T1
P2
T2
=
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
T2
T1
P2 = P1 x 
358 K
291 K
= 1.20 atm x 
Density (d) Calculations
d = 
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
M =
d is the density of the gas in g/L
5.4
m
V
PM
RT
=
dRT
P
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.00C. What is the molar mass of the gas?
5.3
M =
dRT
P
M =
m
V
d = 
4.65 g
2.10 L
=
= 2.21 
g
L
2.21 
g
L
L•atm
mol•K
1 atm
 x 0.0821 x 300.15 K
M =
54.6 g/mol
Gas Stoichiometry
5.60 g C6H12O6
= 0.187 mol CO2
V = 
= 4.76 L
5.5
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
g C6H12O6 mol C6H12O6 mol CO2 V CO2
1 mol C6H12O6
180 g C6H12O6
x
6 mol CO2
1 mol C6H12O6
x
nRT
P
L•atm
mol•K
0.187 mol x 0.0821 x 310.15 K
1.00 atm
=
Dalton’s Law of Partial Pressures
V and T are constant
P1
P2
Ptotal = P1 + P2
5.6
Consider a case in which two gases, A and B, are in a container of volume V.
nA is the number of moles of A
nB is the number of moles of B
PT = PA + PB
PA = XA PT
PB = XB PT
Pi = Xi PT
5.6
nART
V
PA = 
nBRT
V
PB = 
nA
nA + nB
XA = 
nB
nA + nB
XB = 
ni
nT
mole fraction (Xi) = 
A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
Xpropane = 
PT = 1.37 atm
= 0.0132
Ppropane = 0.0132 x 1.37 atm
= 0.0181 atm
5.6
0.116
8.24 + 0.421 + 0.116
Bottle full of oxygen gas and water vapor
5.6
2KClO3 (s) 2KCl (s) + 3O2 (g)
PT = PO + PH O
2
2
5.6
Chemistry in Action:
Scuba Diving and the Gas Laws
5.6
		Depth (ft)		Pressure (atm)
		0		1
		33		2
		66		3
P
V
Kinetic Molecular Theory of Gases
		A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
		Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.
		Gas molecules exert neither attractive nor repulsive forces on one another.
		The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
5.7
Kinetic theory of gases and …
		Compressibility of Gases
		Boyle’s Law
P a collision rate with wall
Collision rate a number density
Number density a 1/V
P a 1/V
		Charles’ Law
P a collision rate with wall
Collision rate a average kinetic energy of gas molecules
Average kinetic energy a T
P a T
5.7
Kinetic theory of gases and …
		Avogadro’s Law
P a collision rate with wall
Collision rate a number density
Number density a n
P a n
		Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the presence of another gas
Ptotal = SPi
5.7
Apparatus for studying molecular speed distribution
5.7
The distribution of speeds
for nitrogen gas molecules
at three different temperatures
5.7
The distribution of speeds
of three different gases
at the same temperature

3RT
M
urms = 
Chemistry in Action: Super Cold Atoms
Gaseous Rb Atoms
1.7 x 10-7 K
Bose-Einstein Condensate
Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.
5.7
NH3
17 g/mol
HCl
36 g/mol
NH4Cl
Deviations from Ideal Behavior
1 mole of ideal gas
PV = nRT
5.8
Repulsive Forces
Attractive Forces
PV
RT
n = 
= 1.0
Effect of intermolecular forces on the pressure exerted by a gas.
5.8
5.8
Van der Waals equation
nonideal gas
an2
V2
P + (V – nb) = nRT
(
)
}
corrected
pressure
}
corrected
volume
chapt01.ppt
Chemistry: 
The Study of Change
Chapter 1
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.
 
Chemistry: A Science for the 21st Century
		 Health and Medicine
		 Sanitation systems
		 Surgery with
anesthesia
		 Vaccines and antibiotics
		Energy and the Environment
		 Fossil fuels
		 Solar energy
		 Nuclear energy
1.1
Chemistry: A Science for the 21st Century
		 Materials and Technology
		 Polymers, ceramics, liquid crystals
		 Room-temperature superconductors?
		 Molecular computing?
		 Food and Agriculture
		 Genetically modified crops
		 “Natural” pesticides
		 Specialized fertilizers
1.1
The scientific method is a systematic approach to research
1.3
A law is a concise statement of a relationship between phenomena that is always the same under the same conditions.
A hypothesis is a tentative explanation for a set of observations
tested modified
Chemistry In Action:
In 1940 George Gamow hypothesized that the universe began with a gigantic explosion or big bang.
Experimental Support
		 expanding universe
		 cosmic background radiation
		 primordial helium
1.3
Primordial Helim and the Big Bang Theory
		 Matter is anything that occupies space and has mass.
		 A substance is a form of matter that has a definite composition and distinct properties.
Chemistry is the study of matter and the
changes it undergoes
water, ammonia, sucrose, gold, oxygen
1.4
A mixture is a combination of two or more substances in which the substances retain their distinct identities.
		 Homogenous mixture – composition of the mixture is the same throughout.
		 Heterogeneous mixture – composition is not uniform throughout.
1.4
soft drink, milk, solder
cement, 
iron filings in sand
Physical means can be used to separate a mixture into its pure components.
1.4
magnet
distillation
An element is a substance that cannot be separated into simpler substances by chemical means.
		 113 elements have been identified
		 82 elements occur naturally on Earth
gold, aluminum, lead, oxygen, carbon
		 31 elements have been created by scientists
technetium, americium, seaborgium
1.4
A compound is a substance composed of atoms of two or more elements chemically united in fixed proportions.
Compounds can only be separated into their pure components (elements) by chemical means.
1.4
Water (H2O)
Glucose (C6H12O6)
Ammonia (NH3)
1.4
Three States of Matter
1.5
solid
liquid
gas
Physical or Chemical?
A physical change does not alter the composition or identity of a substance.
A chemical change alters the composition or identity of the substance(s) involved.
1.6
ice melting
sugar dissolving 
in water
hydrogen burns in air to form water
An extensive property of a material depends upon how much matter is is being considered.
An intensive property of a material does not depend upon how much matter is is being considered.
		 mass
		 length
		 volume
		 density
		 temperature
		 color
Extensive and Intensive Properties
1.6
Matter - anything that occupies space and has mass.
mass – measure of the quantity of matter
SI unit of mass is the kilogram (kg)
1 kg = 1000 g = 1 x 103 g
weight – force that gravity exerts on an object
1.7
weight = c x mass
on earth, c = 1.0
on moon, c ~ 0.1
A 1 kg bar will weigh
1 kg on earth
0.1 kg on moon
1.7
1.7
Volume – SI derived unit for volume is cubic meter (m3)
1 cm3 = (1 x 10-2 m)3 = 1 x 10-6 m3
1 dm3 = (1 x 10-1 m)3 = 1 x 10-3 m3
1 L = 1000 mL = 1000 cm3 = 1 dm3
1 mL = 1 cm3
1.7
Density – SI derived unit for density is kg/m3 
1 g/cm3 = 1 g/mL = 1000 kg/m3
1.7
m = d x V
= 21.5 g/cm3 x 4.49 cm3 = 96.5 g
density = 
mass
volume
d =
m
V
A piece of platinum metal with a density of 21.5 g/cm3 has a volume of 4.49 cm3. What is its mass?
d =
m
V
K = 0C + 273.15
1.7
273 K = 0 0C 
373 K = 100 0C
32 0F = 0 0C 
212 0F = 100 0C 
9
5
0F = x 0C + 32
Convert 172.9 0F to degrees Celsius.
1.7
9
5
0F = x 0C + 32
9
5
0F – 32 = x 0C
9
5
x (0F – 32) = 0C
9
5
0C = x (0F – 32)
9
5
0C = x (172.9 – 32) = 78.3
Chemistry In Action
On 9/23/99, $125,000,000 Mars Climate Orbiter entered Mar’s atmosphere 100 km lower than planned and was destroyed by heat.
1.7
1 lb = 1 N
1 lb = 4.45 N
“This is going to be the cautionary tale that will be embedded into introduction to the metric system in elementary school, high school, and college science courses till the end of time.”
1.8
Scientific Notation
6.022 x 1023
1.99 x 10-23
N x 10n
N is a number 
between 1 and 10
n is a positive or 
negative integer
The number of atoms in 12 g of carbon:
602,200,000,000,000,000,000,000
The mass of a single carbon atom in grams:
0.0000000000000000000000199
Scientific Notation
1.8
568.762
n > 0
568.762 = 5.68762 x 102
0.00000772
n < 0
0.00000772 = 7.72 x 10-6
Addition or Subtraction
		Write each quantity with the same exponent n
		Combine N1 and N2 
		The exponent, n, remains the same
4.31 x 104 + 3.9 x 103 =
4.31 x 104 + 0.39 x 104 =
4.70 x 104
move decimal left
move decimal right
Scientific Notation
1.8
Multiplication
		Multiply N1 and N2
		Add exponents n1 and n2
(4.0 x 10-5) x (7.0 x 103) =
(4.0 x 7.0) x (10-5+3) =
28 x 10-2 =
2.8 x 10-1 
Division
		Divide N1 and N2
		Subtract exponents n1 and n2
8.5 x 104 ÷ 5.0 x 109 =
(8.5 ÷ 5.0) x 104-9 =
1.7 x 10-5 
Significant Figures
1.8
		Any digit that is not zero is significant
1.234 kg 4 significant figures
		Zeros between nonzero digits are significant
606 m 3 significant figures
		Zeros to the left of the first nonzero digit are not significant
0.08 L 1 significant figure
		If a number is greater than 1, then all zeros to the right of the decimal point are significant
2.0 mg 2 significant figures
		If a number is less than 1, then only the zeros that are at the end and in the middle of the number are significant
0.00420 g 3 significant figures
How many significant figures are in each of the following measurements?
24 mL
2 significant figures
3001 g
4 significant figures
0.0320 m3
3 significant figures
6.4 x 104 molecules
2 significant figures
560 kg
2 significant figures
1.8
Significant Figures
1.8
Addition or Subtraction
The answer cannot have more digits to the right of the decimal
point than any of the original numbers.
89.332
1.1
+
90.432
round off to 90.4
one significant figure after decimal point
3.70
-2.9133
0.7867
two significant figures after decimal point
round off to 0.79
Significant Figures
1.8
Multiplication or Division
The number of significant figures in the result is set by the original number that has the smallest number of significant figures
4.51 x 3.6666 = 16.536366
= 16.5
6.8 ÷ 112.04 = 0.0606926 
= 0.061
3 sig figs
round to
3 sig figs
2 sig figs
round to
2 sig figs
Significant Figures
1.8
Exact Numbers
Numbers from definitions or numbers of objects are considered
to have an infinite number of significant figures
The average of three measured lengths; 6.64, 6.68 and 6.70?
Because 3 is an exact number
6.64 + 6.68 + 6.70
3
= 6.67333 = 6.67 
= 7
Accuracy – how close a measurement is to the true value
Precision – how close a set of measurements are to each other
accurate
&
precise
precise
but
not accurate
not accurate
&
not precise
1.8
1.9
Dimensional Analysis Method of Solving Problems
		Determine which unit conversion factor(s) are needed
		Carry units through calculation
		If all units cancel except for the desired unit(s), then the problem was solved correctly.
1 L = 1000 mL
How many mL are in 1.63 L?
1L
1000 mL
1.63 L x
= 1630 mL
L2
mL
1L
1000 mL
1.63 L x
= 0.001630
The speed of sound in air is about 343 m/s. What is this speed in miles per hour?
1 mi = 1609 m
1 min = 60 s
1 hour = 60 min
meters to miles
seconds to hours
1.9
m
s
343
1 mi
1609 m
x
 60 s
 1 min
x
60 min
 1 hour
x
mi
hour
= 767
chapt02.ppt
Atoms, Molecules and Ions
Chapter 2
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.
Dalton’s Atomic Theory (1808)
		 Elements are composed of extremely small particles called atoms. All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of one element are different from the atoms of all other elements.
		 Compounds are composed of atoms of more than one element. The relative number of atoms of each element in a given compound is always the same.
		 Chemical reactions only involve the rearrangement of atoms. Atoms are not created or destroyed in chemical reactions. 
2.1
2
2.1
8 X2Y
2.1
16 X
8 Y
+
J.J. Thomson, measured mass/charge of e- 
(1906 Nobel Prize in Physics)
2.2
Cathode Ray Tube
2.2
 e- charge = -1.60 x 10-19 C
Thomson’s charge/mass of e- = -1.76 x 108 C/g
 e- mass = 9.10 x 10-28 g
Measured mass of e- 
(1923 Nobel Prize in Physics)
2.2
(Uranium compound)
2.2
2.2
		atoms positive charge is concentrated in the nucleus
		proton (p) has opposite (+) charge of electron (-)
		mass of p is 1840 x mass of e- (1.67 x 10-24 g)
		 particle velocity ~ 1.4 x 107 m/s
(~5% speed of light)
(1908 Nobel Prize in Chemistry)
2.2
 atomic radius ~ 100 pm = 1 x 10-10 m
nuclear radius ~ 5 x 10-3 pm = 5 x 10-15 m
Rutherford’s Model of 
the Atom
2.2
“If the atom is the Houston Astrodome, then the nucleus is a marble on the 50-yard line.”
Chadwick’s Experiment (1932)
H atoms - 1 p; He atoms - 2 p
mass He/mass H should = 2
measured mass He/mass H = 4
neutron (n) is neutral (charge = 0)
n mass ~ p mass = 1.67 x 10-24 g
2.2
a + 9Be
1n + 12C + energy
mass p = mass n = 1840 x mass e-
2.2
Atomic number (Z) = number of protons in nucleus
 Mass number (A) = number of protons + number of neutrons 
 = atomic number (Z) + number of neutrons
Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei
2.3
X
A
Z
H
1
1
H (D)
2
1
H (T)
3
1
U
235
92
U
238
92
Mass Number 
Atomic Number
Element Symbol
2.3
6 protons, 8 (14 - 6) neutrons, 6 electrons
6 protons, 5 (11 - 6) neutrons, 6 electrons
Do You Understand Isotopes?
2.3
C
14
6
How many protons, neutrons, and electrons are in
?
C
11
6
How many protons, neutrons, and electrons are in
?
2.4
Period
Group
Alkali Metal
Noble Gas
Halogen
Alkali Earth Metal
Chemistry In Action
Natural abundance of elements in Earth’s crust
Natural abundance of elements in human body
2.4
A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical bonds
A diatomic molecule contains only two atoms
H2, N2, O2, Br2, HCl, CO
A polyatomic molecule contains more than two atoms
O3, H2O, NH3, CH4
2.5
H2
H2O
NH3
CH4
An ion is an atom, or group of atoms, that has a net positive or negative charge.
cation – ion with a positive charge
	If a neutral atom loses one or more electrons
	it becomes a cation.
anion – ion with a negative charge
	If a neutral atom gains one or more electrons
	it becomes an anion.
2.5
Na
11 protons
11 electrons
Na+
11 protons
10 electrons
Cl
17 protons
17 electrons
Cl-
17 protons
18 electrons
A monatomic ion contains only one atom
A polyatomic ion contains more than one atom
2.5
Na+, Cl-, Ca2+, O2-, Al3+, N3-
OH-, CN-, NH4+, NO3-
13 protons, 10 (13 – 3) electrons
34 protons, 36 (34 + 2) electrons
Do You Understand Ions?
2.5
Al
27
13
3+
How many protons and electrons are in ?
Al
78
34
2-
How many protons and electrons are in ?
2.5
2.6
A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance
An empirical formula shows the simplest 
whole-number ratio of the atoms in a substance
H2O
C6H12O6
CH2O
O3
O
N2H4
NH2
2.6
H2O
molecular
empirical
ionic compounds consist of a combination of cations and an anions
		 the formula is always the same as the empirical formula
		 the sum of the charges on the cation(s) and anion(s) in each 	formula unit must equal zero
The ionic compound NaCl
2.6
Formula of Ionic Compounds
Al2O3
2.6
Al3+
O2-
CaBr2
Ca2+
Br-
Na2CO3
Na+
CO32-
2 x +3 = +6
3 x -2 = -6
1 x +2 = +2
2 x -1 = -2
1 x +2 = +2
1 x -2 = -2
2.6
2.7
Chemical Nomenclature
		Ionic Compounds
		often a metal + nonmetal
		anion (nonmetal), add “ide” to element name
BaCl2
barium chloride
K2O
potassium oxide
Mg(OH)2
magnesium hydroxide
KNO3
potassium nitrate
2.7
		Transition metal ionic compounds
		indicate charge on metal with Roman numerals
FeCl2
2 Cl- -2 so Fe is +2
iron(II) chloride 
FeCl3
3 Cl- -3 so Fe is +3
iron(III) chloride 
Cr2S3
3 S-2 -6 so Cr is +3 (6/2)
chromium(III) sulfide
2.7
		Molecular compounds
nonmetals or nonmetals + metalloids
common names
		H2O, NH3, CH4, C60
element further left in periodic table is 1st
element closest to bottom of group is 1st
if more than one compound can be formed from the same elements, use prefixes to indicate number of each kind of atom
last element ends in ide
2.7
HI
hydrogen iodide
NF3
nitrogen trifluoride
SO2
sulfur dioxide
N2Cl4
dinitrogen tetrachloride
NO2
nitrogen dioxide
N2O
dinitrogen monoxide
Molecular Compounds
2.7
TOXIC!
Laughing Gas
2.7
An acid can be defined as a substance that yields 
hydrogen ions (H+) when dissolved in water.
HCl
		Pure substance, hydrogen chloride
		Dissolved in water (H+ Cl-), hydrochloric acid
An oxoacid is an acid that contains hydrogen, oxygen, and another element.
2.7
HNO3
HNO3
nitric acid
H2CO3
carbonic acid
H2SO4
sulfuric acid
2.7
2.7
2.7
A base can be defined as a substance that yields 
hydroxide ions (OH-) when dissolved in water.
2.7
NaOH
sodium hydroxide
KOH
potassium hydroxide
Ba(OH)2
barium hydroxide
2.7
Particle
 
Mass
 
(g)
 
Charge 
(Coulombs)
 
Charge 
(units)
 
Electron (e
-
)
 
9.1 x 10
-
28
 
-
1.6 x 10
-
19
 
-
1
 
Proton (p
+
)
 
1.67 x 10
-
24
 
+1.6 x 10
-
19
 
+1
 
Neutron (n)
 
1.67 x 10
-
24
 
0
 
0
 
 
 
NH
4
+
 
ammonium
 
SO
4
2
-
 
sulfate
 
CO
3
2
-
 
carbonate
 
SO
3
2
-
 
sulfite
 
HCO
3
-
 
bicarbonate
 
NO
3
-
 
nitrate
 
ClO
3
-
 
chlorate
 
NO
2
-
 
nitrite
 
Cr
2
O
7
2
-
 
dichromate
 
SCN
-
 
thiocyanate
 
CrO
4
2
-
 
chromate
 
OH
-
 
hydroxide
 
 
 
chapt03.ppt
Mass Relationships in Chemical Reactions
Chapter 3
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.
By definition:
 1 atom 12C “weighs” 12 amu
On this scale
1H = 1.008 amu
 16O = 16.00 amu
Atomic mass is the mass of an atom in atomic mass units (amu)
Micro World
atoms & molecules
Macro World
grams
3.1
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
= 6.941 amu
3.1
Average atomic mass of lithium:
7.42 x 6.015 + 92.58 x 7.016
100
Average atomic mass (6.941)
The mole (mol) is the amount of a substance that 
contains as many elementary entities as there 
are atoms in exactly 12.00 grams of 12C
3.2
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA) 
Molar mass is the mass
of 1 mole of in grams
eggs
shoes
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
 atomic mass (amu) = molar mass (grams)
3.2
One Mole of:
C
S
Cu
Fe
Hg
3.2
1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu
x
3.2
NA = Avogadro’s number
1 12C atom
12.00 amu
12.00 g
6.022 x 1023 12C atoms
=
1.66 x 10-24 g
1 amu
M
= molar mass in g/mol
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K 
1 mol K = 6.022 x 1023 atoms K
0.551 g K
8.49 x 1021 atoms K
3.2
Do You Understand Molar Mass?
1 mol K
39.10 g K
x
x
6.022 x 1023 atoms K
1 mol K
=
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
For any molecule
 molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2 
3.3
SO2
1S
32.07 amu
2O
+ 2 x 16.00 amu 
SO2
64.07 amu
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol H = 6.022 x 1023 atoms H
5.82 x 1024 atoms H
3.3
1 mol C3H8O molecules = 8 mol H atoms
72.5 g C3H8O
Do You Understand Molecular Mass?
1 mol C3H8O
60 g C3H8O
x
8 mol H atoms
1 mol C3H8O
x
6.022 x 1023 H atoms
1 mol H atoms
x
=
KE = 1/2 x m x v2
v = (2 x KE/m)1/2
F = q x v x B
3.4
Light
Light
Heavy
Heavy
Percent composition of an element in a compound =
n is the number of moles of the element in 1 mole of the compound
52.14% + 13.13% + 34.73% = 100.0%
3.5
n x molar mass of element
molar mass of compound
x 100%
C2H6O
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
3.6
g of O = g of sample – (g of C + g of H)
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
6.0 g C = 0.5 mol C
1.5 g H = 1.5 mol H
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
g CO2
mol CO2
mol C
g C
g H2O
mol H2O
mol H
g H
3.7
A process in which one or more substances is changed into one or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O 
reactants
products
How to “Read” Chemical Equations
2 Mg + O2 2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7
IS NOT
Balancing Chemical Equations
		Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. 
Ethane reacts with oxygen to form carbon dioxide and water
		Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 
3.7
C2H6 + O2
CO2 + H2O
2C2H6
NOT
C4H12
Balancing Chemical Equations
		Start by balancing those elements that appear in only one reactant and one product. 
3.7
start with C or H but not O
multiply CO2 by 2
multiply H2O by 3
C2H6 + O2
CO2 + H2O
2 carbon
on left
1 carbon
on right
C2H6 + O2
2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
C2H6 + O2
2CO2 + 3H2O
Balancing Chemical Equations
		Balance those elements that appear in two or more reactants or products. 
3.7
= 7 oxygen
on right
remove fraction
multiply both sides by 2
2 oxygen
on left
4 oxygen
(2x2)
C2H6 + O2
2CO2 + 3H2O
+ 3 oxygen
(3x1)
7
2
multiply O2 by 
7
2
C2H6 + O2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
Balancing Chemical Equations
		Check to make sure that you have the same number of each type of atom on both sides of the equation. 
3.7
2C2H6 + 7O2
4CO2 + 6H2O
Reactants
Products
4 C
12 H
14 O
4 C
12 H
14 O
4 C (2 x 2)
4 C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
		Write balanced chemical equation
		Convert quantities of known substances into moles
		Use coefficients in balanced equation to calculate the number of moles of the sought quantity
		Convert moles of sought quantity into desired units
Mass Changes in Chemical Reactions
3.8
Methanol burns in air according to the equation
If 209 g of methanol are used up in the combustion, 
what mass of water is produced?
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
209 g CH3OH
235 g H2O
3.8
2CH3OH + 3O2 2CO2 + 4H2O
grams CH3OH
moles CH3OH
moles H2O
grams H2O
1 mol CH3OH
32.0 g CH3OH
x
4 mol H2O
2 mol CH3OH
x
18.0 g H2O
1 mol H2O
x
=
Limiting Reagents
3.9
6 green used up
6 red left over
In one process, 124 g of Al are reacted with 601 g of Fe2O3
Calculate the mass of Al2O3 formed.
124 g Al
367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
3.9
Do You Understand Limiting Reagents?
2Al + Fe2O3 Al2O3 + 2Fe
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed
g Fe2O3
mol Fe2O3
mol Al needed
g Al needed
OR
1 mol Al
27.0 g Al
x
1 mol Fe2O3
2 mol Al
x
160. g Fe2O3
1 mol Fe2O3
x
=
Start with 124 g Al
need 367 g Fe2O3
Use limiting reagent (Al) to calculate amount of product that
can be formed.
124 g Al
234 g Al2O3
3.9
g Al
mol Al
mol Al2O3 
g Al2O3
1 mol Al
27.0 g Al
x
1 mol Al2O3
2 mol Al
x
102. g Al2O3
1 mol Al2O3
x
=
2Al + Fe2O3 Al2O3 + 2Fe
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
3.10
Actual Yield
Theoretical Yield
% Yield = 
x 100
Chemistry In Action: Chemical Fertilizers
Plants need: N, P, K, Ca, S, & Mg
3H2 (g) + N2 (g) 2NH3 (g)
NH3 (aq) + HNO3 (aq) NH4NO3 (aq)
2Ca5(PO4)3F (s) + 7H2SO4 (aq)
3Ca(H2PO4)2 (aq) + 7CaSO4 (aq) + 2HF (g)
fluorapatite
chapt04.ppt
Reactions in Aqueous Solution
Chapter 4
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.
4.1
A solution is a homogenous mixture of 2 or more substances
The solute is(are) the substance(s) present in the smaller amount(s)
The solvent is the substance present in the larger amount
Soft drink (l)
Air (g)
Soft Solder (s)
H2O
N2
Pb
Sugar, CO2
O2, Ar, CH4
Sn
Solution
Solvent
Solute
An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.
4.1
nonelectrolyte
weak electrolyte
strong electrolyte
Strong Electrolyte – 100% dissociation
Weak Electrolyte – not completely dissociated
Conduct electricity in solution?
Cations (+) and Anions (-)
4.1
NaCl (s) Na+ (aq) + Cl- (aq)
H2O
CH3COOH CH3COO- (aq) + H+ (aq)
Ionization of acetic acid
CH3COOH CH3COO- (aq) + H+ (aq)
4.1
Acetic acid is a weak electrolyte because its ionization in water is incomplete.
A reversible reaction. The reaction can occur in both directions.
Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner.
4.1
d+
d-
H2O
Nonelectrolyte does not conduct electricity?
No cations (+) and anions (-) in solution
4.1
C6H12O6 (s) C6H12O6 (aq)
H2O
Precipitation Reactions
Precipitate – insoluble solid that separates from solution
molecular equation
ionic equation
net ionic equation
Na+ and NO3- are spectator ions
4.2
Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3-
PbI2
Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq)
precipitate
Pb2+ + 2I- PbI2 (s)
4.2
Writing Net Ionic Equations
		Write the balanced molecular equation.
		Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions.
		Cancel the spectator ions on both sides of the ionic equation
4.2
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
Ag+ + NO3- + Na+ + Cl- AgCl (s) + Na+ + NO3- 
Ag+ + Cl- AgCl (s)
Write the net ionic equation for the reaction of silver 
nitrate with sodium chloride.
Chemistry In Action:
An Undesirable Precipitation Reaction
4.2
CO2 (aq) CO2 (g)
Ca2+ (aq) + 2HCO3 (aq) CaCO3 (s) + CO2 (aq) + H2O (l)
-
Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrus
fruits contain citric acid.
React with certain metals to produce hydrogen gas.
React with carbonates and bicarbonates to produce carbon 
dioxide gas
4.3
Cause color changes in plant dyes.
Aqueous acid solutions conduct electricity.
2HCl (aq) + Mg (s) MgCl2 (aq) + H2 (g)
2HCl (aq) + CaCO3 (s) CaCl2 (aq) + CO2 (g) + H2O (l)
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Bases
4.3
Cause color changes in plant dyes.
Aqueous base solutions conduct electricity.
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
4.3
Hydronium ion, hydrated proton, H3O+
4.3
A Brønsted acid is a proton donor
A Brønsted base is a proton acceptor
acid
base
acid
base
4.3
A Brønsted acid must contain at least one 
ionizable proton!
Monoprotic acids
Strong electrolyte, strong acid
Strong electrolyte, strong acid
Weak electrolyte, weak acid
Diprotic acids
Strong electrolyte, strong acid
Weak electrolyte, weak acid
Triprotic acids
Weak electrolyte, weak acid
Weak electrolyte, weak acid
Weak electrolyte, weak acid
4.3
HCl H+ + Cl-
HNO3 H+ + NO3-
CH3COOH H+ + CH3COO-
H2SO4 H+ + HSO4-
HSO4- H+ + SO42-
H3PO4 H+ + H2PO4-
H2PO4- H+ + HPO42-
HPO42- H+ + PO43-
Identify each of the following species as a Brønsted acid, base, or both. (a) HI, (b) CH3COO-, (c) H2PO4-
Brønsted acid
Brønsted base
Brønsted acid
Brønsted base
4.3
HI (aq) H+ (aq) + Br- (aq)
CH3COO- (aq) + H+ (aq) CH3COOH (aq)
H2PO4- (aq) H+ (aq) + HPO42- (aq)
H2PO4- (aq) + H+ (aq) H3PO4 (aq)
Neutralization Reaction
4.3
acid + base salt + water
HCl (aq) + NaOH (aq) NaCl (aq) + H2O
H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O
H+ + OH- H2O
Oxidation-Reduction Reactions
(electron transfer reactions)
Oxidation half-reaction (lose e-)
Reduction half-reaction (gain e-)
4.4
2Mg (s) + O2 (g) 2MgO (s)
2Mg 2Mg2+ + 4e-
O2 + 4e- 2O2-
2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-
2Mg + O2 2MgO 
4.4
Zn is oxidized
Cu2+ is reduced
Zn is the reducing agent
Cu2+ is the oxidizing agent
4.4
Ag+ is reduced
Ag+ is the oxidizing agent
Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)
Zn Zn2+ + 2e-
Cu2+ + 2e- Cu
Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s)
Cu Cu2+ + 2e-
Ag+ + 1e- Ag
Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
		Free elements (uncombined state) have an oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
		In monatomic ions, the oxidation number is equal to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
		The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1. 
4.4
		The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.
6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.
		Group IA metals are +1, IIA metals are +2 and fluorine is always –1.
HCO3-
O = -2
H = +1
3x(-2) + 1 + ? = -1
C = +4
4.4
Oxidation numbers of all the elements in HCO3- ?
Figure 4.10 The oxidation numbers of elements in their compounds
4.4
NaIO3
Na = +1
O = -2
3x(-2) + 1 + ? = 0
I = +5
IF7
F = -1
7x(-1) + ? = 0
I = +7
K2Cr2O7
O = -2
K = +1
7x(-2) + 2x(+1) + 2x(?) = 0
Cr = +6
4.4
Oxidation numbers of all the elements in the following ?
Types of Oxidation-Reduction Reactions
Combination Reaction
Decomposition Reaction
0
0
+4
-2
+1
+5
-2
+1
-1
0
4.4
A + B C
S + O2 SO2
2KClO3 2KCl + 3O2
C A + B
Types of Oxidation-Reduction Reactions
Displacement Reaction
Hydrogen Displacement
Metal Displacement
Halogen Displacement
4.4
0
+1
+2
0
0
+4
0
+2
0
-1
-1
0
A + BC AC + B
Sr + 2H2O Sr(OH)2 + H2
TiCl4 + 2Mg Ti + 2MgCl2
Cl2 + 2KBr 2KCl + Br2
The Activity Series for Metals
Hydrogen Displacement Reaction
M is metal
BC is acid or H2O
B is H2
4.4
Figure 4.15
M + BC AC + B
Ca + 2H2O Ca(OH)2 + H2
Pb + 2H2O Pb(OH)2 + H2
Types of Oxidation-Reduction Reactions
Disproportionation Reaction
Element is simultaneously oxidized and reduced.
0
+1
-1
4.4
Cl2 + 2OH- ClO- + Cl- + H2O
Chlorine Chemistry
Precipitation
Acid-Base
Redox (H2 Displacement)
Redox (Combination)
4.4
Ca2+ + CO32- CaCO3
NH3 + H+ NH4+
Zn + 2HCl ZnCl2 + H2
Ca + F2 CaF2
Classify the following reactions.
Chemistry in Action: Breath Analyzer
4.4
3CH3COOH + 2Cr2(SO4)3 + 2K2SO4 + 11H2O
+6
+3
3CH3CH2OH + 2K2Cr2O7 + 8H2SO4 
Solution Stoichiometry
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
M KI
M KI
500. mL
= 232 g KI
4.5
moles of solute
liters of solution
M = molarity =
What mass of KI is required to make 500. mL of
a 2.80 M KI solution?
volume KI
moles KI
grams KI
166 g KI
1 mol KI
x
2.80 mol KI
1 L soln
x
1 L
1000 mL
x
4.5
Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.
4.5
Dilution
Add Solvent
Moles of solute
before dilution (i)
Moles of solute
after dilution (f)
=
MiVi
MfVf
=
MiVi = MfVf
Mi = 4.00
Mf = 0.200
Vf = 0.06 L
Vi = ? L
4.5
= 0.003 L = 3 mL
3 mL of acid
+ 57 mL of water
= 60 mL of solution
How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
Vi =
MfVf
Mi
=
0.200 x 0.06
4.00
Gravimetric Analysis
4.6
		Dissolve unknown substance in water
		React unknown with known substance to form a precipitate
		Filter and dry precipitate
		Weigh precipitate
		Use chemical formula and mass of precipitate to determine amount of unknown ion
Titrations
In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
 equivalence point 
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
4.7
4.7
25.00 mL
= 158 mL
What volume of a 1.420 M NaOH solution is
Required to titrate 25.00 mL of a 4.50 M H2SO4 
solution?
WRITE
THE CHEMICAL EQUATION!
volume acid
moles acid
moles base
volume base
H2SO4 + 2NaOH 2H2O + Na2SO4
4.50 mol H2SO4
1000 mL soln
x
2 mol NaOH
1 mol H2SO4
x
1000 ml soln
1.420 mol NaOH
x
M
acid
rx
coef.
M
base
Chemistry in Action: Metals from the Sea
CaCO3 (s) CaO (s) + CO2 (g)
Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l)
CaO (s) + H2O (l) Ca2+ (aq) + 2OH (aq)
-
Mg2+ (aq) + 2OH (aq) Mg(OH)2 (s)
-
Mg2+ + 2e- Mg
2Cl- Cl2 + 2e-
MgCl2 (l) Mg (l) + Cl2 (g)