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Chapter 1
An Introduction to Chemistry
“Chemistry is the science of composition, structure,
Definition of chemistry properties and reactions of matter, especially atomic
and molecular systems.”
Broad field – includes everything in the whole universe,
animate and inanimate.
Agriculture – find a wheat kernel that makes best bread
Astronomy – what matter is there on Mars?
Animal science – find the best animal feed
Geology – composition of rocks, water inclusion in lava
Medicine – MRI
Material science – catalysts for fuel cells
Molecular biology – protein structure and folding
Despite the common belief that a chemist mixes chemicals
just for the love of it – and gets unimaginable results – the truth is different:
a chemist tries to understand the nature and natural phenomena,
composition and structure of complex systems (like biomolecules) or to synthesize new substances in order to advance
science in general and contribute to the well-being of humanity.
Scientific method
Collect data or facts relevant to the
problem.
Formulate prediction (hypothesis) to
explain the data
Plan and test the hypothesis with
additional experiments.
Modify the hypothesis.
laws
observation
hypothesis
experiment
theory
Hypothesis – tentative explanation
Theory – well-established hypothesis;
explanation of general principles of certain
phenomena with facts to prove it.
- Einstein’s theory of relativity
Scientific laws – simple statements of
natural phenomena to which no
exceptions are known.
- Law of buoyancy: any
object submerged in a fluid
is acted on by a force
with magnitude equal to
the weight of the fluid
displaced by the object.
Bias- a strong
preference that
inhibits impartial
judgment.
Aristotle believed that women
have less teeth than men.
Matter
Matter is composed of tiny particles called atoms. It can
be in any of the three physical states: solid, liquid or gas.
The physical state in which the matter exists depends on
the nature of the matter (i.e. attractive forces in the atoms
or molecules of the matter), and external factors
(pressure, temperature).
Physical transformations: (solid – liquid) freezing /
melting, liquid-gas: vaporization / condensation, and
(solid-gas) sublimation
Matter is divided into pure substances, and mixtures.
Pure substances are
particular form of matter
with definite and fixed
composition, e.g.
elements (Cu, Au, Al)
and
compounds
(salt, sugar, water).
Mixtures can be:
homogeneous - uniform in appearance and with
same properties throughout, also called: solution, or
heterogeneous (two distinct phases with boundary
between them).
Mixtures contain two or more substances
that can be separated physically (boiling,
magnet, filtration, etc.
Physical state and properties
State
Solid
Liquid
Gas
Shape
Definite
Indefinite
Indefinite
Volume
Definite
Definite
Indefinite
Compress.
V.Slight
Slight
High
State of common materials
Solids
Most metals
(gold, copper, zinc)
Salt
Sugar
Sand
0
100 Temperature, oC
Liquids
Water
Alcohol
Vinegar
Blood
Oil
Gases
Air (oxygen,
nitrogen,…)
Acetylene
Chlorine
Noble gases
(He, Ne …)
3
Properties and changes of matter
Set of properties gives unique identity to the matter.
Physical properties can be determined without destroying the substance (state,
color, density, taste/odor, melting/boiling point).
Chemical properties describe the ability of the substance to form new substances,
either by decomposition or reaction with other substances (corrosiveness,
flammability, acidity, toxicity, etc.).
No two substances have identical physical and chemical properties.
Physical changes are reversible changes (in size, shape, state of matter,
density) when no new substances are formed.
Heating of Pt wire changes its appearance from silvery metallic to glowing yellowred. Upon cooling, original appearance is restored and no new compounds are
formed. Shattered glass looks different, but its (chemical) composition is the same.
Chemical changes are irreversible as new compounds are formed. Upon
heating, Copper wire changes its appearance like Pt, but appears black after
it is cooled. Cu oxidizes to copper(II) oxide, CuO.
Physical change results in a
different form of the same
substance, while chemical
change results in a
completely different
substance, or substances.
Chapter 2
Measurements and significant figures
Measurement is a comparison of the quantity of the sample to a standard.
Measurement is important – a small variation of a drug quantity may kill you!
Measurement never gives an exact value. Its numerical value always contains
all certain figures plus one that is estimated.
Estimated figure you obtain
These are called significant figures.
The measuring device gives you the certain figures. by reading between the lines.
Measurement of a coin
Estimated
using a ruler marked in
figure: 5
centimeters gives the
diameter to be halfway between 2 and 3
cm, or about 2.5 cm;
uncertainty is in the
estimated decimal.
Certain figure: 2
This measurement gives the radius to
be 2.5 cm. That is, two significant figures.
Accuracy: closeness of the
measurement to the true value.
Using a ruler
marked in
millimeters
makes the 0.5
cm a certain
figure.
Estimated
figure: 0
Certain
figures:
2, 5
This measurement gives the radius to
be 2.50 cm: three significant figures.
Precision: closeness to one another of a series
of measurements made on the same object.
Using the first ruler
we got the
diameter of 2.5 cm.
What would be its
radius?
Radius = 2.5 cm / 2. Your calculator
would show 1.25 cm.
BUT: the second decimal place is
beyond your ability to estimate.
Thus, you must round the number to
a single decimal place, i.e. 1.3 cm.
You can keep only significant figures – the ones that are not beyond the
accuracy of the measuring device, plus your estimated figure.
Note that there are exact numbers: there is exactly 2 radii in a diameter!
Significant Figure Rules –
1. all non-zero digits are significant.
2. zero is NOT significant if:
– it is before the first nonzero digit or
- at the end of a number without decimal point.
Practice:
20.201
20.210
20.0002
120.
# of sig. figs.:
5
5
6
3
Leading (placeholder) zeros only
give the location of the decimal
point and are not significant.
0.06 inch has only one sig figs.
Decimal point tells you that the last zero is
an estimate (uncertainty +/- 1).
Without the decimal point, the number 120 can
be interpreted as 120 +/- 10, thus two sig. figs.
To avoid ambiguity, we use scientific notation.
Scientific notation
1.2
x=2
A = 1.2
1.23 x 102
1.23 x
10-1
1.23 x 100
4.6 x 104
4.6 x 10-4
x is an
exponent (or
power) of 10.
Exponent x is always
a whole number.
When x is positive, you move the
decimal point x places to the right.
Scientific notation
A x 10x
A is a number
> (or =) 1 and < 10
x 102
Decimal
point moves:
2 places right
1 place left
doesn’t move
4 places right
4 places left
When x is negative, you move the
decimal point x places to the left.
Standard notation
123.
0.123
1.23
46,000
0.00046
1.20 x 102
2 places left
120.
1.2 x 102
2 places left
120
no need for
decimal point
100 = 1 !
no decimal point!
Add 3 zeros after six
Add 3 zeros before four
Reverse movement of
decimal point when going
from standard to sci. notation
When doing math
Calculator often gives more digits than justified.
Rounding rule – use all digits you want to retain. Then check the digit
after the last one retained:
- If it is 4 or less, drop it and all the digits behind. Leave the retained number
unchanged.
- If it is 5 or more, drop it and all the digits behind. The last digit in the retained
number increases by one.
To avoid accumulating errors, always round the number at the end of calculation.
Sig. figs. rules for multiplication and division:
The result cannot have more significant figures than there are in the
measurement that has the smallest number of significant figures.
2.0 cm X 2 cm = 4 cm2
2.0 cm X 2.0 cm = 4.0 cm2
2 sig. figs. 1 sig. fig. 1 sig. fig.
2 sig. figs. 2 sig. fig. 2 sig. fig.
Sig. figs. rules for addition and subtraction:
The result can be no more certain than the least certain measurement in the
series.
123 cm
20.5 lb + 15 lb
35.5 lb
=
= 14.489…
+ 1.006 cm
2
2.45 ft x 1.00 ft
2.45 ft
+ 2.6 cm
= 14 lb/ft2
Calculator gives 126.606 cm
Uncertainty +/- 1 127 cm
Uncertainty +/- 1: Should be rounded to
36, or 2 significant figures
Number with a name - Unit of measure
A measurement makes no sense unless units are specified.
Say, you are to get to the airport in 3. Three what?
Minutes? Hours? Gallons of gas?
Remembering names of different units for the same
property, and the relationship between them can be
cumbersome.... 12 inches in a foot, 3 foot in a yard, 22
yards in a chain, 10 chains in a furlong, 8 furlongs in a mile …
8 fl. Oz in a cup, 4 cups in a quart, 4 quarts in a gallon,
31 gallons in a barrel…
… and if you make a mistake in conversion
from one unit into another, you are toast!
Not only in medieval times!
NASA wasted $125 million for a Martian Lander that
burned up in the red planet’s atmosphere because of
ambiguously expressed retrorocket force. The
company that built the rockets expressed the force in
US Customary units (in, ft, lbs), whereas NASA
assumed the data was given in metric units (meters,
kilograms, Newtons).
SI System
Whole world (except US!) now use standardized system of units, called SI system.
SI system uses an unique name for a physical property and prefixes to
describe bigger or smaller units. Make the relationship between bigger /
smaller units to the base unit in factor of 1000.
There is even relationship between units for different physical properties!
Base SI Units
Physical quantity
Length
Mass
Time
Temperature
Amount of subst.
Name
meter
kilogram
second
Kelvin
mole
Abbrev.
m
kg
s
K
mol
Derived SI Units
Physical quantity
Volume
Pressure
Energy
Electric charge
Name
Abbrev.
cubic meter m3
Pascal
Pa
joule
j
coulomb
C
A cube of 0.1 m length has a volume of 1 L; filled with water, its mass is 1 kg.
Because 1 m3 is a huge volume, a
commonly used, non-SI units for
volume are Liter (L) and milliliter
(mL). 1 m3 = 1000 L. 1 L = 1000 mL
1 L = (0.1 m)3 = 1 dm3 1 L = 1.057 qt.
1 m is 3.37 inches longer than a yard.
1 kg = 2.2046 lb
1 m = 1.0936 yd
SI System (continued)
Prefix - milli
Other commonly used, non-SI units are
Pressure in: millimeters of mercury (mm Hg),
atmospheres (atm)
Temperature in: degrees of Celsius (oC)
Physical quantity: meter
1mm = 10-3 m
1km = 103 m
1 MB = 106 B
1µm = 10-6 m
1mL = 10-3 L
1GB=109 B
Prefixes Used with SI Units
Greek Prefix
pico- (p)
nano- (n)
micro- (µ)
milli- (m)
centi- (c)
deci- (d)
Greek Prefix
kilo- (k)
mega- (M)
giga- (G)
tera- (T)
Meaning
one-trillionth (10-12)
one-billionth (10-9)
one-millionth (10-6)
one-thousandth (10-3)
one-hundredth (10-2)
one-tenth (10-1)
Meaning
one thousand (103)
one million (106)
one billion (109)
one trillion (1012)
Note: All prefixes for smaller units are lower
case, and those for larger are upper case!
Prefix kilo – is the only exception.
Note: Kelvin is
represented as
K, not oK!
K = oC + 273.15
oC = (5/9) (oF – 32)
oF = (9/5) oC + 32
Density
Density = mass of substance divided by the volume occupied by the mass. By
definition, the density of water is exactly 1 kg / L (hence the definition of L!)
Specific gravity is the ratio of densities of a substance to that of water. It is
dimensionless.
When immersed in water, substances with density greater than 1 kg/L (e.g. most
metals) sink, and those having density less 1 (ice, most liquids, wood, etc) float.
Density depends on the
way atoms are packed in
the crystal lattice.
Volume of an
irregularly shaped
object can be
measured by
displacement of water.
Density is an intensive property, i.e. does not depend on the amount of material.
Other intensive properties include: color, odor, taste, specific heat etc.
The opposite is extensive property, one that depends on amount present.
Mass and volume are both extensive properties, but their ratio (density) is intensive.
Unit Analysis
Recall – density is the ratio of mass and volume.
d=m/V
What to do if you are given the density and the mass of mercury, and asked
to calculate the volume? (Suppose m = 2.00 g and d = 13.6 g/mL)
One can do it in two ways.
Way #1
Use above eq. to find V as a function of the other two quantities.
Recall that equation stays unchanged if a mathematical
operation is applied to both sides of equal sign.
solved for m
m
x V
Multiply both sides with V:
Vx d=
Vxd=m
V
m
m
Vxd
V =
solved for V
Now divide both sides with d:
=
d
d
d
2.00 g
Plug in the numbers:
= 0.147 mL
V=
g
13.6
mL
1
g g x mL
g
x 4
1
1 =
Recall fractions: 2
Check the units:
=
=
g
g g x 1 = mL
x
3
2 3
mL
mL
4
HOW?
What to do if I cannot make one unit disappear immediately?
Try and try again!
Use multiple conversion factors; keep multiplying until you reach the desired
unit(s) and have all other “unwanted” units disappear.
Example 1:
Suppose you are given density of 2.50 g / mL, and asked to convert it to lb / gal.
2.50 g
1 lb
x
1 mL
1000 mL
x
436 g
3.7854 L
x
1L
1 gal
=
20.9 lb
1 lb = 436 g
1 gal = 3.7854 L
1 gal
Example 2:
How many seconds are there in (exactly) 3.5 weeks?
7 day
24 hour
3600. s
x
3.5 week
x
x
= 2,116,800. s
1 week
1 day
1 hour
Homework:
Chapter 2: 4 (a-d,f), 7, 9, 13, 18(b), 23, 41, 49, 55
Chapter 1: 1, 3, 8, 9, 12
1. Explain the difference between
(a) A hypothesis and a theory
(b) A theory and a scientific law
3. How many
phases are
present in the
graduated
cylinder?
12. Is the system
that contains
only one
substance
necessarily
homogeneous?
Explain.
Consider
Table 1.1
below and
answer the
following:
8. Which liquids are not mixtures?
9. Which of the gases are not pure substances?
4. State the abbreviation for each of the following
units: (a) milligram, (b) kilogram, (c) meter.
(d) nanometer, (f) microleter
7. How many significant figures are in each of
the following numbers? (a) 0.025, (b) 22.4,
(c) 0.0404, (d) 5.50 x 103
9. Round each of the following numbers to three
significant digits: (a) 93.246, (b) 0.02857, (c)
4.644, (d) 34.250
13. Solve the following problems:
(a) 12.62 + 1.5 + 0.25 = ?
(b) (2.25 x 103) (4.80 x 104) = ?
(c) (452) (6.2) / (14.3) = ?
(d) (0.0394) (12.8) = ?
(e) (0.4278) / (59.6) = ?
(f) 10.4 + (3.75) (1.5 x 104) = ?
18. Solve equation for the variable x:
(b) 8.9 g/mL = 40.90 g / x
23. After you have worked out at a gym on a
stationary bike for 45 min, the distance gauge
indicates you have traveled 15.2 miles. What was
your rate in km/hr?
41. A textbook is 27 cm long, 21 cm wide and 4.4
cm thick. What is the volume in (a) cubic
centimeters, (b) liters, (c) cubic inches?
49. The normal boiling point of ethanol is 173.3
oF. What is this temperature in oC?
55. Concentrated hydrochloric acid has a density
of 1.19 g/mL. Calculate the mass of 250 mL of it.
Chapter 3
Elements and Compounds
All known substances on Earth and probably the universe are formed
by combinations of more than 100 elements.
Most substances are compounds.
Compounds can be decomposed into two or more simpler substances.
Water can be decomposed into
hydrogen and oxygen.
Table salt can be decomposed
into sodium and chlorine.
An element is a fundamental or elementary substance that cannot
be broken down into simpler substances by chemical means.
Each element has a number.
Beginning with hydrogen as 1, the elements are numbered in order of
increasing complexity.
An atom is the smallest particle of an element that can exist.
It is the smallest unit of an element that can enter into a chemical reaction.
A molecule is a combination of two or more atoms.
If a molecule contains only atoms of one kind, it is an
element. Otherwise, it is a compound.
Carbon
Symbols of the elements:
Oxygen
One or two letters. First letter always capitalized.
Barium
Sodium
Elements are not distributed equally by nature.
C
O
Ba
Na
In the universe, the
most abundant
element is hydrogen
(91%) and the second
most abundant
element is helium
(8.75%).
Oxygen is the most
abundant element in
the crust of the earth
(49.2%), followed by
silicon.
Oxygen is the most
abundant element in
the human body
(65%), followed by
carbon.
A number of symbols appear to have no connection with the element.
Most symbols start with the same letter as the element.
Hydrogen is in a group of its own
leftmost
transition area
rightmost
nonmetals
Metals, Nonmetals and Metalloids
Metals Most elements are metals.
Physical properties
Good conductors of heat and electricity.
Have high luster.
High melting point and density.
Metals are solid at room temperature.
Mercury is the only exception.
Malleable (hammered into sheets);
Ductile (drawn into wires).
Mix with other metals to form alloys
Brass = copper + zinc
Bronze = copper + tin
Steel = Iron + carbon
Chemical properties
Little tendency to combine with
other metals
Readily combine with nonmetals
(especially oxygen) to form compounds.
In nature, minerals are formed by
combinations of the more reactive metals with
other elements.
Some less reactive metals are found in
nature in free state (elemental form).
only seven elements
Metalloids
Physical properties Intermediate
Be, Si, Ge, As, Sb, Te, and radioactive Po.
Nonmetals
Physical properties opposite of metals
Lack luster.
Poor conductors.
Low melting point.
Low density.
Can be found in all three phases (solid,
liquid, gas) at room temperature.
17 elements are nonmetals, including
rare and radioactive (At) and six noble
gases. Remaining ten nonmetals make
almost all compounds.
C, P, S, Se, I: solids
Br: liquid
H, O, N, F, Cl, noble gases: gases
Periodic Table
The periodic table was designed by Dimitri Mendelev in 1869
In the table each element’s symbol is placed inside of a box
Above the symbol of the element is its atomic number.
Elements are arranged in order of increasing atomic number
Elements with similar chemical properties are organized in
columns called families or groups.
7
N
He
These elements are known as the noble gases. They are
non-reactive. They are found in pure (elemental) form.
Ne
Noble metals can also be found in elemental form.
Most substances around us are mixtures or compounds.
Ar
Air is a mixture of nitrogen, oxygen, argon and traces
of other gases (e.g. water vapor and carbon dioxide).
Kr
H, O, N etc. cannot exist in
nature as single atoms. They
are always paired (H2, O2, N2).
These elements exist as
diatomic molecules.
Xe
Rn
Composition of Elements and Compounds
Hydrogen and oxygen are elements.
Elements cannot be decomposed into
simpler substances.
Free, or elemental hydrogen (H2) is
found in volcanoes; It can be prepared in
the laboratory.
Elemental oxygen (O2) is found in air (21
vol%); It can be prepared in the laboratory.
Water is a compound; its formula is H2O.
It is made of two
hydrogen atoms
and one oxygen
atom.
Water contains NO
free hydrogen or
free oxygen.
H2O
A compound is a distinct substance
that contains two or more elements
combined in a definite proportion by
weight.
H2
O2
Compounds can be decomposed
chemically into simpler substances – that
is, into simpler compounds or elements.
Atoms of the elements that constitute a
compound are always present in simple
whole number ratios. They are never
present as fractional parts.
Physical and chemical properties of
a compound differ from those of the
parent elements.
Types of Compounds
Compounds can be molecular or ionic.
A molecule is the smallest uncharged
individual unit of a compound formed by
the union of two or more atoms.
An ion is a positively or negatively
charged atom or group of atoms.
Cation is positively charged ion.
Anion is negatively charged ion.
Ionic compounds are
held together by
attractive forces
between positively and
negatively charged ions.
Table salt, sodium chloride is a colorless
crystalline ionic substance, 39.3% sodium
and 60.7% chlorine by mass.
The solid NaCl does not conduct electricity;
molten NaCl, and solution of salt in H2O do.
Compounds can be classified as molecular or ionic.
Molecular compounds are held together by covalent bonds.
Ionic compounds are held together by attractive forces between their
positive and negative charges.
Chemical Formulas
Serve as abbreviations of the names of compounds.
Tell which elements the compound is composed of
and how many atoms of each element are present in a
formula unit.
Show the symbols of the atoms of the elements
present in a compound.
CaCl
(1) 2
Show the ratio of the atoms of the elements present
in a compound.
calcium
chloride
calcium
chlorine
Ca calcium
Cl chlorine
2 Cl
1 Ca
Rules for Writing Chemical Formulas
Formulas do not necessarily represent the arrangement of atoms.
When a formula contains one atom of an element, the
symbol of that element represents the one atom. The
number one (1) is not used as a subscript.
When the formula contains more than one atom of an
element, the number of atoms is indicated by a
subscript written to the right of the symbol of that atom.
When the formula contains more than one of a
group of atoms that occurs as a unit, parentheses
are placed around the group, and the number of
units of the group is indicated by a subscript
placed to the right of the parentheses.
There is one phosphorus
atom in a phosphate group
Ba3(PO4)2
There are three
barium atoms
Indicates two
phosphate
(PO4)3- groups
There are four oxygen atoms
in a phosphate group
There are three barium, two phosphorus
and eight oxygen atoms in the compound!
H2O
NaCl
Indicates the
element
sodium
(one atom)
Indicates the
element
chlorine (one
atom)
H3PO4
Indicates the
element
hydrogen
Indicates 3
H atoms
Indicates
the element
oxygen
Indicates 4
O atoms
Indicates the
element phosphorus
(1 atom)
Chapter 4
Properties of Matter
Properties of a substance are characteristic of the substance.
Each substance has a set of properties that are characteristic
of that substance and give it a unique identity.
Color
Taste / Odor
Physical state
Melting point
Boiling point
Physical properties are the inherent characteristics of a substance
that are determined without changing its composition.
Chemical properties describe the ability of a substance to form new
substances, either by reaction with other substances or by decomposition.
Physical properties of chlorine gas (Cl2):
2.4 x heavier than air yellowish-green color
disagreeable odor
melting point: -101 oC
boiling point: -34.6 oC
Chemical properties of chlorine (Cl2):
Doesn’t burn
disinfectant
Bleaching agent supports combustion
Produces table salt with sodium
Physical and Chemical Changes
Physical changes are changes in physical
properties (such as size, shape, and density) or
changes in the state of matter without an
accompanying change in composition.
No new substances are formed.
tearing of paper
change of ice into
water
change of water into
steam
heating platinum wire
Chemical change is a change in which new substances are formed that have
different properties and composition from the original material.
The formation of copper(II)
oxide from copper and
oxygen is a chemical
change. The newly-formed
substance has set of
properties that are different
from those of copper.
Water decomposes when
electrical energy passes through
it. The products, elemental
hydrogen and oxygen, have
different properties from those of
water.
Chemical Reaction
Water decomposes into hydrogen and oxygen when electrolyzed.
Chemical symbols can be used to express chemical reactions
reactant
2H2O
yields
products
2H2 + O2
Conservation of Mass
No change is observed in the
total mass of the substances
involved in a chemical change.
sodium +
46.0 g
sulfur
→
32.1 g
78.1 g reactant
mass reactants
sodium sulfide
78.1 g
→
78.1 g product
=
mass products
Energy
Kinetic energy matter possesses due to its motion.
Mechanical Chemical
Electrical
Heat
Nuclear
Radiant
Potential energy could also thought as the
energy stored in an object. Chemical energy
is one form of potential energy.
The heat released when gasoline burns is
associated with a decrease in its chemical
potential energy.
The new substances formed by burning
have less chemical potential energy than
the gasoline and oxygen.
Increasing potential energy
Potential energy is the energy that an object
possesses due to its relative position.
Bouncing ball.
Running man.
Height
Energy is the capacity to do work.
Potential energy
rises with the
height increase
Energy in Chemical Reactions
In all chemical changes, matter either absorbs or releases energy.
Examples of chemical processes leading to energy release
Type of Energy
Energy Source
Electrical
Storage batteries
Light
A lightstick. Fuel combustion.
Heat and Light
Combustion of fuels.
Body
Chemical changes within body cells.
Examples of chemical processes leading to energy absorption
Type of Energy
Chemical Change
Electrical
Electroplating of metals. Decomposition
of water into hydrogen and oxygen
Light
Photosynthesis in green plants.
Conservation of Energy
An energy transformation occurs
whenever a chemical change occurs.
If energy is given off in a chemical
change, the products will have less
chemical potential energy than the
reactants.
If energy is absorbed during a chemical
change, the products will have more
chemical potential energy than the
reactants.
Energy can be neither created nor destroyed,
Law of Conservation of Energy though it can be transformed from one form of
energy to another form of energy.
higher potential energy
Electrolysis of Water
lower potential energy
Burning of hydrogen in air
Heat: Quantitative Measurement
Heat is a form of energy associated with small particles of matter.
Temperature is a measure of the intensity of heat, or of how hot or cold a system is.
The SI unit for heat energy is the joule (pronounced “jool”).
Another unit is the calorie.
1 calorie = 4.184 Joules (exactly)
This amount of heat energy
will raise the temperature of
1 gram of water by 1 oC.
Two calories is needed to
raise 1 gram of water by 2 oC.
Twice as much heat energy is
required to raise the temperature
of 200 g of water by 10 oC as
compared to 100 g of water.
specific
Heat =
heat
(
q = C
)(
x
mass
m
100
Ag
water
200
Bg
water
30oC
20
30oC
20
4184 J
8368 J
temperature
change
)(
x
)
∆t
Specific Heat
The specific heat of a substance is the quantity of heat required
to change the temperature of 1 g of that substance by 1 oC.
The units of specific heat in calories are:
 calories 
 gram oCelcius 


 cal 
 g oC 


The units of specific heat in joules are:
Joules


 gram oCelcius 


 J 
 g oC 


It takes about 5 times more energy to warm 1 g of water by 1 oC
as to warm the same amount of aluminum by 1oC.
Quantitative Calculation of Heat
The relation of mass, specific
heat, temperature change
I can be found if
(∆t), and quantity of heat lost
both sides were
or gained is expressed by the
divided by m x ∆t
general equation:
(
)(
specific heat
of substance
Problem 1:
Calculate the specific heat of a
solid in J/goC and if 1638 J raise
the temperature of 125 g of the
solid from 25.0oC to 52.6oC.
)
mass of
∆t = heat
substance
heat = 1638 J
mass = 125 g
∆t = 52.6 oC – 25.0 oC = 27.6 oC
spec.heat = ?
heat
specific heat =
mass x ∆t
=
1638 J
125 g x 27.6 oC
= 0.475
J
g oC
Heat = specific heat
Q
C
=
x
mass of water
m
x
x
x
change in temp.
∆t
Problem 2: How much heat (in J) it takes to warm a 2.000 lb block of iron from 50 oC to 75 oC?
J
453.6 g
m = 2000 lb
x 25 oC = 1.02 x 104 J
Q = m C ∆t = 2.000 lb x
x 0.449
g oC
1 lb
∆t = 25 oC
From Table 4.3:
C = 0.449 J/goC
Problem 3: What is the final temperature of a 20.0 g block of iron after it has absorbed 100.0 J
of heat, if its initial temperature was 25.0 oC? Hint: solve for ∆t, then add it to the initial temp!
∆t =
Q
mC
100.0 J
=
J
20.0 g x 0.449 g oC
Calorie Measurement
Place the food (e.g. Twinkie)
into a vessel (“bomb”),
surround it with water and
measure the water
temperature as you burn the
food.
= 11.1 oC
Final temp. = 25.0 oC + 11.1 oC = 36.1 oC
Advanced A sample of a metal with a mass of 212 g is heated to 125.0 oC and then
problem: dropped into 375 g of water at 24.0 oC. If the final temperature of the water
is 34.2 oC, what is the specific heat of the metal?
When the metal enters the water, it begins to cool, losing heat to the water. At the same time,
the temperature of the water rises. This process continues until the temperature of the metal
and the temperature of the water are equal (34.2 oC), at which point no net flow of heat occurs.
Heat gained by water = heat lost by metal
mW = 375 g
∆tW = 34.2 – 24.0 oC
Strategy:
∆tM = 125.0 – 34.2 oC
1. Calculate the heat (qW) gained by the water. mM = 212 g
2. Calculate the final temperature of the metal. spec.heat
o
3. Calculate the specific heat of the metal.
of water = 4.184 J/g C qW = qM
temperature gained by water
1.
∆tW = 34.2
oC
– 24.0
oC
= 10.2
3.
oC
heat gained by water
J
x 10.2 oC = 1.60 x 104 J
qW = 375 g x 4.184
g oC
qM = qW = 1.60 x 104 J
qM
spec.heat =
m x ∆t
1.60 x 104 J
spec.heat =
125 g x 90.8 oC
temperature lost by metal
2.
∆tM = 125.0
oC
– 34.2
oC
= 90.8
oC
spec.heat = 0.831
J
g oC
Homework, chapter 3 (paired exerc., p.62):
5. Write the formula for each compound:
a) Zinc oxide (1 atom Zn, 1 atom O)
b) Potassium chlorate (1 atom K, 1 atom
Cl, 3 atoms O)
c) Sodium hydroxide (1 atom Na, 1 atom
O, 1 atom H)
d) Ethyl alcohol (2 atoms C, 6 atoms H, 1
atom O)
9. How any atoms are represented in
each formula?
a) KF; b) CaCO3; c) K2Cr2O7; d)
NaC2H3O2; e) (NH4)2C2O4
19. Classify each material as an element,
compound or mixture: a) xenon; b) sugar;
c) crude oil; d) nitric acid
23. What percent of the first 36 elements
on the periodic table are metals?
Homework, chapter 4 (paired exerc., p.77):
7. State whether ech of the following
represents a chemical or physical
change: a) A steak is cooked until
well done; b) Students firepolish the end of a
glass rod to smoothen jagged edges; c)
Chlorine bleach removes coffee stains from
white lab coat; d) When two clear solutions
are mixed together, the solution becomes
yellow and cloudy;
13. Indicate with the plus sign (+) any of these
processes that requires energy, and a
negative sign (-) any that release energy: a)
melting ice; b) relaxing a taut rubber band; c) a
rocket launching; d) striking a match.
19. A 250.0 g metal bar requires 5.866 kJ to
change its temperature form 22 oC to 100.0
oC. Calculate the specific heat of the metal.
20. A 1.00 kg of antimony absorbed 30.7 kJ
thus raising its temperature from 20.0 oC to its
melting point (630.0 oC). Calculate the specific
heat of antimony.
31. One gram of anthracite coal gives off
7000. cal when burned. How many joules is
this? How many grams of anthracite are
required to raise the temperature of 4.0 L of
water from 20.0 oC to 100.0 oC?
Chapter 5
Early Atomic Theory
and Structure
Empedocle (440 BC): all matter consists
of 4 “elements” earth, fire, water and air.
Democritus (470-370 BC): matter is composed of
indivisible particles (Greek atomos - not cuttable).
Aristotle (384-322 BC): endorsed Empedoclean
theory, so that it dominated until 17th century.
John Dalton (1766-1844), 2000 years after
Democritus, revived concept of atoms:
Dalton’s rules:
1. elements are composed of indivisible particles-atoms.
2. atoms of the same element are alike in mass and size.
3. atoms of different elements are not alike – they have
different masses and sizes.
4. compounds are union of two or more atoms of
different elements.
5. compounds contain simple numerical ratios of atoms.
6. more than one compound can be formed of atoms of
two elements.
Dalton rules (cont.)
Water
While most of the Dalton’s theory still
holds, some modifications were necessary:
An atom is composed of subatomic
particles, it can be decomposed and may
have isotopes of different masses.
Hydrogen
peroxide
H2O and H2O2 have the same
atoms, but different number.
Law of definite composition – a compound contains two or more elements in
a definite proportion by mass.
Law of multiple proportions – atoms of two or more elements may combine
in different ratios to produce more than one compound.
compound
water
hydrogen peroxide
Copper (I) chloride
Copper(II) chloride
Methyl alcohol
Ethyl alcohol
formula
H2O
H2O2
CuCl
CuCl2
CH4O
C2H6O
percent composition
11.2 % H, 88.8% O
5.9% H, 94.1% O
64.2% Cu, 35.8% Cl
47.3% Cu, 52.7% Cl
37.5% C, 12.6% H, 49.9% O
52.1%C, 13.1% H, 34.7% O
Electric Charge
Matter may have electric charge
1. charge may be of two types, positive and negative
2. unlike charges attract, like repel
3. can be transferred from one object to the other by contact or induction
4. the smaller the distance, the stronger the force between two charges
x
(F – force, k - constant, q1, q2 - charges, r - distance)
F = k x q1 2q2
r
G.J. Stoney (1826-1911) named the unit of electricity electron.
Joseph Thompson (1856-1940) experimentally showed the existence of
electron, which is: negative in charge, deflected by magnetic and electric fields,
and capable of moving small paddle wheel.
Michael Faraday (1791-1867)
1. Some compounds dissolved in water conduct electricity, others decompose
2. Particles of some elements are attracted to the positive or negative electrode.
3. Concluded that they are charged and called them ions (Greek-wanderer).
Svante Arhenius (1859-1927)
1. Melted NaCl also conduct electricity. NaCl breaks up to Na+ and Cl- ions.
2. Ions move to oppositely charged electrode. Ion that goes towards positively
charged cathode is called cation. The other goes to positive anode is called anion.
Charge and Periodic Table
IA
H+
Li+
Na+
K+
Rb+
Cs+
IIA
Invented by Dmitri Mendeleev, the
Periodic Table shows recurring trends
in properties of elements and their
charges.
VIIIA
IIIA
Be2+
Mg2+
Ca2+
IIIB IVB
VB
Cr2+
Cr3+
Sr2+
VIB
VIIB
Fe2+
Fe3+
------ VIIIB -------
Al3+
IB
IIB
Cu2+
Cu+
Zn2+
Ag+
Cd2+
IVA
VA
VIA
N3-
O2-
P3-
S2-
VIIA
F-
ClBr-
I-
Ba2+
Group IA: Group IIA:
always 1+ always 2+
Charge = group number
Elements of B groups
(transition metals) form
ions of different charges
Group VA: Group VIA: Group VIIA:
often 3usually 2- always 1Charge = 8 - group number
Group VIIIA: no charge
-noble gases
Subatomic particles
Atom is extremely small, cannot be observed by optical microscope.
Atomic diameters range from 0.1 to 0.5 nm.
Consider this dot on the right.
If its diameter is 1 mm, 10 million H atoms could be arranged in it.
Yet, there are particles smaller than H atom. They are called subatomic particles.
Three subatomic particles are sufficiently long-lived to be used in experiments:
electron, proton and neutron.
Proton and electron have charges opposite in sign, identical in absolute value.
Proton and neutron have almost equal masses.
Number of other short-lived subatomic particles are detected in recent years.
quarks, neutrinos, antineutrinos, positrons and antiprotons ...
Their existence is still debated.
The Evolution of Atomic Theory
Becquerel in 1896 discovered radioactivity. Radioactive elements spontaneously
emit alpha particles, beta particles and gamma rays from their nuclei.
By 1907 Rutherford found that alpha particles emitted by
certain radioactive elements were helium nuclei.
J.J. Thomson discovered electron in 1897. Shortly after,
he proposed the first atomic ‘Plum-Pudding’ model where
negatively-charged electrons (‘plums’) are embedded into
a positively-charged sphere (‘pudding’).
+
Thomson and Goldstein discovered proton in 1907,
and Chadwick discovered neutron in 1932.
α particles should pass through the
Rutherford’s
experiment
showed
that the
plumpudding
model is
incorrect.
sphere because it is effectively neutral.
α particle velocity ~ 1.4 x 107 m/s
(~5% speed of light)
The Evolution of Atomic Theory (cont.) Indeed, most of them did, but a few (1 in
20,000) were deflected, some in great
If Thomson’s model was correct, αangles. It was “as if you were firing 15”
particles would go straight through the foil.
shell at a tissue paper, and it came back
and hit you”.
Rutherford’s Conclusions:
1. Atom consists of a nucleus and electrons.
2. Size of an atom is about 100,000 times
bigger than that of its nucleus.
Size of an atom: If the
nucleus is as big as a
marble, the atom is as
big as football field.
Atoms are mostly
empty space!
3. Atoms positive
charge is concentrated
in the nucleus.
4. Electron has negative
charge, and proton has
positive charge.
Mass of p is 1840 x mass of
e- (1.67 x 10-24 g).
In order to avoid falling onto the nucleus,
electron must be constantly rotating with
great speed, as if it was everywhere at once.
Problem: classical physics says that the
Nucleus has both protons and neutrons. moving charge (electron) radiates energy.
Creation of Ions
An neutral atom has equal
number of protons and
electrons.
Remember that the proton has
equal, but opposite, charge as
an electron
When one or more electrons
are lost from an atom, a
positive cation is formed.
When one or more electrons are
added to a neutral atom, an
anion is formed.
The word ‘cation’ comes from the
fact that positive ion (cation)
moves towards negativelycharged (cathode) in an
electrolyte solution. The negative
ion (anion) moves towards
positively-charged electrode
(anode).
Atomic Numbers of the Elements
The atomic number of an element is equal to the number of protons in the
nucleus of that element. It gives the identity of the element.
Atomic number (Z) = number of protons in nucleus
Mass number (A) = number of protons + number of neutrons
= atomic number (Z) + number of neutrons
Isotopes: atoms of the same element with different numbers of neutrons in their
nuclei. Neutrons can be envisioned as glue that holds nuclei together.
The mass of 12C is 12.000 g.
Mass Number
A
13
Element Symbol There exists 1.11 % C and
Atomic Number
Z
traces of 14C. The mass of
C atom (12.01 g) is weighted
average of all atoms.
Two isotopes of uranium
X
235
92
238
92
U
U
Protons: 92
Neutrons:
235-92=143
Protons: 92
Neutrons:
238-92=146
Three isotopes of hydrogen
Hydrogen
(Protium)
Deuterium
Tritium
(radioactive)
Masses of Elements and Isotopes
To overcome the problem in
measuring extremely small masses, a
system of relative atomic masses
using “atomic mass units” was devised
to express the masses of elements
using simple numbers.
A mass of exactly 12 atomic mass
units (amu) was assigned to 126 C.
Element masses are
average numbers,
because of isotopes.
Mass of H atom is
1.008 amu.
The mass of a single
atom is too small to
measure on a balance.
Using a mass spectrometer, the
mass of the hydrogen atom was
determined = 1.673 x 10-24 g.
The standard to which the masses
of all other atoms are compared to
was chosen to be the most abundant
isotope of carbon.
1 amu is defined
as exactly equal to
1/12 the mass of a
carbon - 12 atom:
1 amu = 1.6606 x 10-24 g
12
6
C
Masses of elements and Isotopes (cont.)
Isotopes of the same element have different masses.
The listed atomic mass of an element is the average relative mass of the isotopes
of that element compared to the mass of carbon-12 (exactly 12.0000…amu).
To calculate the atomic mass, multiply the atomic mass of each isotope by
its abundance and add the results.
Percent abundances are
Example: calculate average atomic mass of copper.
Isotope
multiplied by 100!
Average
Isotopic
Abundance
atomic
mass (amu)
(%)
mass (amu)
63
29
Cu
62.9298
69.09
65
29
Cu
64.9278
30.91
(62.9298 amu)
x
0.6909 =
(64.9278 amu)
x
0.3091 =
63.55
+
43.48 amu
20.07 amu
63.55 amu
Finding the Number of Protons, Neutrons and Electrons
Element
Chlorine
Gold
Bromine
Symbol
36Cl
Au
Br -
Z
Mass# #p
17
79
56
36
197
80
35
Fill in the table below.
Symbol Element
Z = #p.
2. Find mass number and #n
Mass# = #p + #n
3. Find charge or #electrons
(charge) = #p – #e
#e = #p – (charge)
35
#e
charge
19
118
79
16
17
79
18
0
0
+2
-2
45
36
-1
Charge = 0
Locate Cl in periodic table
Look up Atomic number (Z) for Cl
1. From Periodic table, find: Element,
symbol, atomic number and #protons.
At least one of those must be given, the
other three are read from Periodic Table.
17
79
#n
Mass # is 36 = #p + #n
#n = 36 – #p = 19
4. You can get #e and charge. Use step 3
to find #p and from that identify element.
See if you can solve the other
two examples. The solutions are:
135Ba2+, and 32S2-.
56
16
HW, Chapter 5 (p.100): 12, 15, 23, 27, 37
Chp. 10 (p.220): 13, 19, 33, 39
Chapter 10
The Modern Atomic Theory and Periodic Table
Rutherford model, based on classical (Newtonian) physics could not explain
periodicity of the elements, nor how electrons stay in orbit around nucleus. A new
model was needed.
Light moves with speed of 300,000 km/s.
It is characterized with wavelength λ.
When passed through a prism, a familiar
set of colors are seen. Spectrum is
continuous as colors blend.
Visible spectrum is just a tiny fraction of the electromagnetic radiation.
Bohr’s Theory
Niels Bohr in 1913 postulated that an
electron inside an atom can possess
only certain values of energy. Those
“packs” of energies are called quanta
(pl. of quantum).
Atom is in the lowest energy when its
electrons are closest to the nucleus.
When atom absorbs quantum of energy,
electrons are moved to higher orbits.
Electrons can only have allowed
energies, not the energies between the
allowed ones. It is like a person on a
ladder who can stand on a 1st, 2nd, 3rd
rung, but not between the rungs.
Allowed orbits are called shells. They are
characterized by the principal quantum
number, n that goes from n = 1 (shell
closest to the nucleus), to infinity.
The maximum number of electrons a
shell can hold increases with n as the
radius of the shell increases.
Li and Na have a single electron in
the valence shell.
Periodicity
Elements
have similar
chemical
properties
because their
valence shell
configurations F and Cl have 7 electrons
are similar.
in the valence shell.
The number of electrons in the
valence shell is equal to the
Roman-numeral group number
for the representative elements.
The dependence of chemical on
valence shell occupancy holds for
each group of elements.
Helium is an exception. It behaves as
a noble gas although it has only 2 ein the valence shell.
Group IA has one electron, Group IIA has two…
The lowest reactivity is found in
the group VIIIA (noble gases).
Elements of other groups can
attain the valence shell of group
VIIIA by producing compounds.
Line Spectra
Ground state
s: 2 e- p: 6 ed: 10 e- f: 14 e-
Excited state
New, refined model
includes subshells
that are very close in
energy and size.
Every subshell is
designated with a
letter (s, p, d, f).
s subhell can hold two
electrons. Each
subshell after it holds
four electrons more
than the previous.
Transition n=3 n=2 produces 13.1
eV – 11.2 eV = 1.9 eV, i.e. red color.
Similarly, n=4 n=2 produces 13.8 –
11.3 = 2.6, green color; other colors
in the visible spectrum are also found.
Bohr’s model correctly predicted all
transitions of H atom; transition into
n=1 produces a line in UV, into n=2
(visible) and into n≥3 (infrared). But
the model failed to predict spectrum
for any atom with more than one e-.
n
s
p
d
f
1
2
2
2 6
3
2 6 10
4
2 6 10 14
Each ‘box’ contains
2 electrons with
paired spins.
Filling of Subshells
Subshells exist for any n > 1.
They increase in energy:
s < p < d < f.
Subshells may cross each other;
the crossing is more pronounced
when spacing between shells
reduces as the principle quantum
number increases.
35Br
This crossing is the
reason for the socalled ‘potassium
problem’.
Present the electron
configuration of 19K.
1s2 2s2 2p6 3s2 3p6 4s1
Valence
electron
Orbital
diagram
is
slightly more
complicated.
1s2 2s2 2p6
3s2 3p6 4s2
3d10 4p5
Rearrange:
1s2 2s2 2p6
3s2 3p6 3d10
4s2 4p5
Valence electrons
Subshell fill up
S
No need to memorize crossings.
Just follow the periodic table.
Move He next to H to get four
blocks, each with different
width (s:2, p:6, d:10, f:14).
Electron configuration of 16S:
Try 26Fe
1s2 2s2 2p6
3s2 3p6 4s2 3d6
1st period: 1s2.
2nd period: 2s2 2p6.
3rd period: 3s2 3p4.
Valence electrons
2
8
+ 6
16
1s2 2s2 2p6 3s2 3p6 4s2 3d10
Gd:
64
4p6 5s2 4d10 5p6 6s2 5d1 4f 7.
When a transition metal produces cation, first to go are its
electrons in the s orbital of the valence shell !
Valence electron jumps from Na to Cl
The best way for Mg to
achieve octet is to lose two
electrons from its 3rd shell.
Theoretically, Mg could also gain 6 e-,
but the 6 e- excess is too much for Mg
nucleus to hold on to.
The best way for F to achieve
octet is to gain one e-.
Each e- from Mg 3rd shell jumps
into 2nd shell of an F atom.
Mg atom becomes Mg2+ cation.
Each F atom becomes F- anion.
All three ions have 8 valence e-.
Compound Formation
NaCl
MgF2
Neutral atoms:
1 valence e- in Na
7 valence e- in Cl
Note the size of atoms!
Ions: no electrons in
n = 3 shell of Na.
Na+ has 8e- in n=2 shell.
Cl- has 8e- in n=3 shell.
Both ions satisfy octet rule.
Note the size of ions!
Atomic Size
Understanding of atomic
size with Bohr’s model.
One consequence: shells
get larger as n increases.
A combination of same
valence shell and increased
nuclear charge shrinks the
atom in a group.
Quantum Mechanical Model
Electron has wavelike characteristics, it can tunnel
to appear in ‘forbidden’ places. Uncertainty principle
states that it is impossible to know at any given time
where the electron is, or where it is going. Instead, etravels around nucleus in an electron cloud or orbital.
The QM model also explains why e- stays in
orbit around the nucleus: An e- orbital is allowed
only if the electron wave closes in on itself.
Allowed
orbit must
close in on
itself.
Nonallowed
orbit does not.
Electron cloud exhibits wavelike motion, so that
different orbitals have different shapes. It has a
particular shape and energy determined by the
principle quantum number n.
Chapter 8
Chemical Reactions
In mid 1980s chemists combined Y2O3, CuO and BaCO3 and
obtained a new superconducting ceramic material, YBa2Cu3O7
that conducted electricity without resistance when cooled below
liquid nitrogen temperature. Magnetic lines cannot penetrate a
superconductor, so a magnet floats above it. Experimental trains
based on magnetic levitation (maglev) are in construction.
In a chemical reaction one or more substances (reactants)
are converted into new substances (products) with different
physical and chemical properties.
Reactants Products
Which reaction is chemical ?
H2O(l) H2O(g)
CO2 + O2 O2 + CO2
CO2 + H2O H2CO3
Gas molecules travel rapidly at room
temperature (hundreds of miles per
hour). Energy associated with
motion (kinetic energy) is absorbed
by colliding molecules.
H2
I2
H – H+ I – I 2 atoms H
2 HI
Energy is required to break
+ 2 atoms I
H–I
a chemical bond.
2 molecules HI
Balanced equation H – I
2 atoms H, 2 atoms I
Balanced equation must have smallest possible whole numbers as coefficients.
∆
2 Al(s) + Fe2O3(s)→ 2 Fe(s)+ Al2O3(s)
coefficient
coefficient
A chemical equation is a shorthand expression for a chemical change or reaction.
A chemical equation must be balanced.
Coefficients (whole numbers) are placed in front of substances to balance the
equation and to indicate the number of units (atoms, molecules, moles, or ions)
of each substance that are reacting.
A chemical equation uses the chemical symbols and formulas of the reactants
and products and other symbolic terms to represent a chemical reaction.
Symbol + means plus.
Symbol hν
ν means light.
Symbol means yields. Symbol means gas formation.
Symbol means precipitate.
Symbol ∆ means heat.
Conditions required to carry out the reaction may be placed above or below
the arrow.
The physical state of a substance is indicated by symbols such as (s) for
solid, (l) for liquid, (g) for gas and (aq) for aqueous solution.
Balancing chemical equations
H2 + I2 ?
HI
Not balanced
H2 + I2
2 HI
Balanced
H2 + I2
Impossible!
Never change subscripts in
the molecular formula!
Always adjust coefficients, i.e. numbers in
front of the molecules.
Practice
N2 + 3 H2 2 NH3
Balanced N: 1 2
N: 2
H: 2 6
H: 3 6
Try some more…
CH4 + 2 O2 CO2 + 2 H2O
C: 1
C: 1
H: 4
H: 2 4
O: 2 4
O: 3 4
And more…
C8H18 +
25 O
2
2
H2I2
H2I2 does not
exist. If it did, it
would be a
compound
different from HI.
Step1: Count atoms on each side.
Step 2: Start balancing metal first. Continue with
any nonmetal other than O or H.
Step 3: When all atoms other than O and H are
balanced, start balancing either O or H. if there is
either O2 or H2 on the left side, balance it last.
Step 1: count atoms.
Step 2: No metal atoms. Continue with C.
Step 3: Balance H next. Finish with O.
8 CO2 + 9 H2O
2 C8H18 + 25 O2 16 CO2 + 18 H2O
And more.
C6H12O6 + 6 O2 6 CO2 + 6 H2O
2 HCl + Na2CO3 2NaCl + CO2 + H2O
A Balanced Equation Shows:
Mole is a
number like a
dozen, only
much bigger. It
is used to
translate
number of
particles into
grams.
Formulas
Number of
molecules
Number
of atoms
Number
of moles
Molar
masses
Sum of masses
Must be balanced
44.09 g + 160.0 g
204.1 g
?
=
132.0 g + 72.08 g
204.1 g
Types of Reactions
Chemists try to classify a reaction by understanding what it is doing.
Ca2+(aq) + CO32-(aq) CaCO3(s)
Remember that aq, s, l, g indicate physical state.
The reaction is used to soften ‘hard’ water (i.e. water containing Ca2+ ions). There
are other reactions serving the same purpose.
Single-displacement reactions
One element replaces another
element in a compound.
A displaces B.
2 Al(s) + Fe2O3(s) Al2O3(s) + 2 Fe(l)
Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g)
A
+
A C
B C
+
B
Double-displacement reactions
PCl3(l) + 3 AgF(s) PF3(g) + 3 AgCl(s)
Two compounds exchanged partners.
A B + C D A D + C B
Decomposition reaction
One compound breaks into two
or more simpler substances.
2 HgO(s)
A B
Heat
2 Hg(l) + O2(g)
A +
B
H2(g) + I2(g) 2 HI(g) 2 Na(s) + Cl2(g) 2 NaCl(s)
Combination reaction
Two or more simple compounds
4 Ni(s) + 4 Al(s) + S8(s) 2 Ni2Al2S4(s)
combine to give one or more
A + B A B
complex substances.
Single displacement reactions
Occur if the reactant in elemental form is
more reactive than the element in the
compound.
Metal Activity Series
2 Al(s) + Fe2O3(s) Al2O3(s) + 2 Fe(l)
The reaction occurs
ONLY if the position
of the free element in
the activity series is
above the element in
compounds.
Mg(s) + PbS(s) → MgS(s) + Pb(s)
Ag(s) + CuCl2(s) → no reaction
Halogen Activity Series
Cl2(g) + CaBr2(s) → CaCl2(aq) + Br2(aq)
I2(g) + CaBr2(s) → no reaction
Double displacement reactions
Accompany the following processes:
release of heat
Reaction goes only
formation of water
when one of the
formation of a precipitate products leaves
release of gas bubbles
aqueous solution
Halogens
F2
Cl2
Br2
I2
increasing
activity
Metals
K
Ca
Na
Mg
Al
Zn
Fe
Ni
Sn
Pb
H
Cu
Ag
Hg
AB + CD
→ AD + CB
Double Displacement Reactions (cont.d)
1. Acid / base neutralization, or acid + metal oxide, produces water and heat
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
2HNO3(aq) + CuO(s) → Cu(NO3)2(aq) + H2O(l)
2. Formation of insoluble precipitate
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
3. Formation of a gas
Indirect gas formation
NH4Cl(aq) + NaOH(aq) → NaCl(aq) + NH4OH(aq)
Heat in Chemical
Reactions
NH4OH(aq) → NH3(g) + H2O(l)
Energy changes accompany
H2(g) + Cl2(g) → 2HCl(g) + 185 kJ (exothermic)
chemical reactions. Energy is
Energy released
2 mol
1 mol 1 mol
either released to, or
absorbed from surroundings.
per mole H2 or Cl2 NOT per mole HCl
Since the amounts of
N2(g) + O2(g) + 185 kJ → 2NO(g) (endothermic)
substances are expressed in
2 mol Energy absorbed
1 mol 1 mol
moles, the heat of the reaction
6CO2 + 6H2O + 2519 kJ → C6H12O6 + 6O2
is expressed per mole of
2519 kJ of energy is absorbed per 6 moles CO2, H2O, O2!
either reactant or product.
Energy of Activation
A certain amount of energy is
always required for a reaction
to occur. It is called the
energy of activation, Ea.
The reaction will occur only if the
activation energy is supplied.
The activation energy can take
.
the form of a spark or a flame.
CH4 + 2O2 → CO2 + 2H2O + 890 kJ
Although the reaction releases
energy, it needs Ea to start.
Exothermic Reaction:
Combined energy of products
is lower than that of reactants.
The difference is released.
Endothermic Reaction:
Combined energy of products
is larger than that of reactants.
The difference is absorbed.
N2(g) + O2(g) + 185 kJ → 2NO(g)
Chapter 6
Nomenclature of inorganic compounds
There are ~ 11 million chemical compounds. Some of the arbitrary (common) names
you may recognize (water = hydrogen oxide H2O, laughing gas = dinitrogen monoxide
N2O, quicksilver = mercury Hg). One cannot memorize arbitrary names of all of them.
Chemical nomenclature is the system of names for compounds.
Systematization of names was devised by IUPAC (International Union of Pure and
Applied Chemistry) in 1921 and is constantly updated
Elements:
Exceptions:
Chemical formula of an element is just the symbol of the element.
Some non-metals cannot exist as single atoms at normal
temperatures. They form polyatomic molecules.
Chemical formula of element hydrogen is not H, but H2.
Same is true for N2, O2, F2, Cl2, Br2, I2, S8 and P4.
Ions:
Charged particles produced by adding or
removing electrons from neutral atoms
are called cations and anions.
K K+ + eMg Mg2+ + 2eAl Al3+ + 3e-
F + e- FO + 2e- O2N + 3e- N3-
The bond between metals and non-metals is usually ionic. Metals give away their
electrons and become positively charged cations. Nonmetals accept them and
become anions. The ionic bond is formed as a result of attraction between oppositely
charged ions.
Charge and Periodic Table
IA
H+
Li+
Na+
K+
Rb+
Cs+
IIA
Invented by Dmitri Mendeleev, the
Periodic Table shows recurring trends
in properties of elements and their
charges.
VIIIA
IIIA
Be2+
Mg2+
Ca2+
IIIB IVB
VB
Cr2+
Cr3+
Sr2+
VIB
VIIB
Fe2+
Fe3+
------ VIIIB -------
Al3+
IB
IIB
Cu2+
Cu+
Zn2+
Ag+
Cd2+
IVA
VA
VIA
N3-
O2-
P3-
S2-
VIIA
F-
ClBr-
I-
Ba2+
Group IA: Group IIA:
always 1+ always 2+
Charge = group number
Elements of B groups
(transition metals) form
ions of different charges
Group VA: Group VIA: Group VIIA:
often 3usually 2- always 1Charge = 8 - group number
Group VIIIA: no charge
-noble gases
Ionic compound form three-dimensional ordered
network of cations and anions called ionic lattice.
There are no bonds.
Electronegativity and the
Polar Covalent Bond
Electrons are rarely shared equally
between atoms. Electronegativity
(EN) is numerical rating of an atoms
ability to attract to itself the shared
electrons in a covalent bond.
Generally, electronegativity of metals
is low, and that of nonmetals is high.
The difference in electronegativity
determines the nature of the bond.
∆EN = 0, covalent;
∆EN = 1.0, polar
covalent (23% ionic); ∆EN = 1.9, polar
covalent (60% ionic); ∆EN > 1.9, ionic.
Naming of a molecule follows the order
of increasing electronegativity.
Naming Chemical Compounds
While some compounds have common names, today millions of compounds are
known. Hence the need to name each compound by its chemical formula.
1. metal + nonmetal (ionic) and
2. nonmetal + nonmetal (covalent)
Ionic:
Less electronegative atom (usually metal) goes first; its name is unchanged;
nonmetal ion keeps the stem, adds suffix –ide. If the metal can form two
classes of compounds, the metal charge is given by the Roman numerals.
Covalent: Because of many possibilities when two nonmetals combine, prefixes are
used to tell how many atoms of each element are present in the molecular
formula. Lesser electronegative element’s name stays unchanged, the other
gets suffix –ide. If there is only one atom of lesser electronegative element,
prefix mono– is omitted.
Examples:
Mon(o)- 1
NaCl Sodium chloride
FeCl2
Iron(II) chloride
Di2
MgS Magnesium sulfide
FeCl3
Iron(III) chloride
Tri3
Li3N Lithium nitride
CuO
Copper(II) oxide
Tetr(a)- 4
Alluminum
oxide
Al2O3
Cu2O
Copper(I) oxide
Pent(a)- 5
SnO2
Hexa6
Tin(IV) oxide
CoCl3 Cobalt(III) chloride
Binary compounds (made of two elements):
PCl3 Phosphorus trichloride
P4O6 Tetraphosphorus hexaoxide
NO Nitrogen monoxide
NO2
N2O4
N2O5
Nitrogen dioxide
Dinitrogen tetroxide
Dinitrogen pentoxide
HeptaOctaNonaDeca-
7
8
9
10
Common names are used.
Polyatomic Ions
Ammonium is the only cation.
Mostly composed of a nonmetal
plus one or more oxygen atoms.
Exceptions: some transition
metal anions and CN-.
Most oxyanions exist in series
that differ only in the number of
oxygen atoms. If there are two
members in such a series, the
ion with fewer oxygens gets the
-ite ending, the one with more
oxygens gets the -ate ending.
If there are more oxyanions in a
series, the prefix hypo- is used
denotes the ion with fewest
oxygens, while the prefix perdenotes the oxyanions with the
most oxygens.
Naming: metal ion goes
first, using the rules for
binary compounds, followed
by the polyatomic anion.
ClOClO2ClO3ClO4-
hypochlorite
chlorite
chlorate
perchlorate
NaNO2
NaNO3
K2CO3
Sodium nitrite
Sodium nitrate
Potassium
carbonate
The molecule is neutral: the Cu(NO3)2 Copper(II) nitrate
sum of positive charges is Mg (PO ) Magnesium
3
4 2
equal to the sum of
phosphate
negative charges.
Nomenclature of Acids
A compound that dissociates in water to produce hydrogen ions (H+) is an acid.
If the acid does not contain oxygen, it
is named by adding prefix hydro- and
the suffix –ic acid.
The above names only apply for
aqueous solution, i.e. when acid can
dissociate. In gaseous phase
compounds are named according to
the rules for nonacidic compounds.
If the acid contains oxygen(s)
it is called oxyacid. The rules
follow the polyatomic naming.
If the anion name ends in
–ate, the acid is named by
changing the suffix to –ic.
If the anion name ends in
–ite, the acid is named by
changing the suffix from –ite
to –ous.
HF(aq) hydrofluoric acid
HCl(aq) hydrochloric acid
H2S(aq) hydrosulfuric acid
HCl(g)
hydrogen chloride
Note that the salt of these acids have
usual names of binary compounds.
NaCl
sodium chloride
Examples
no hydro- prefix
indicates oxyacid
containing sulfur
possible existence
of an –ous acid indicates
hydrogen
sulfuric acid
sulfite
sulfurous acid
nitrite
nitroous acid
2−
3
SO
H 2SO3
−
NO2
HNO 2
This is the way the
formula is written.
contains contains contains
hydrogen sulfur
oxygen
H 2SO 4
sulfate
sulfuric acid
nitrate
nitric acid
KMnO4
K
+
MnO
4
SO
2−
4
H 2SO4
−
NO3
HNO3
The ions are what is
actually present.
Polyatomic compound
Acid
(begins with H)
Other compounds
w/polyatomic ions
Anion containing
oxygen
1. Name cation(s)
2. Name anion
Check ending
on anion
-ite
Writing formula of a compound
-ate
Cation
Anion name ends
with –ous acid
Anion name ends
with –ic acid
K+
H+
Zn2+
Al3+
Homework:
Chpt. 8 (p.172): 3, 9, 13, 21
Chpt. 6 (p.122): 3, 6, 8, 15, 17, 25, 29
ClKCl
HCl
ZnCl2
AlCl3
Anions
SO42O2PO43K2SO4
K2O
K3PO4
H2SO4
HCl
H3PO4
ZnSO4
ZnO Zn3(PO4)2
Al2(SO4)3 Al2O3 AlPO4
Chapter 7
Quantitative Composition of Compounds
Making new chemicals is much like following a recipe from a cook book...
1 cup flour + 2 eggs
+ ½ tsp baking powder → 5 pancakes … except you
don’t get to lick
the spoon!
What if you want to make more (or less)?
Suppose you have plenty of flour and baking powder,
but only 8 eggs. How many pancakes can you make?
You can solve it using conversion factor:
8 eggs x
5 pancakes
2 eggs
= 20 pancakes
5 pancakes
2 eggs
Solve it in your head:
2 eggs makes 5
pancakes, so four
times more eggs
makes 20 (5x4)
pancakes.
Practice using the following mouthwashing, diet-buster recipe:
3 blocks cream cheese + 5 eggs + 1
cup sugar = 1 cheese cake.
How many cheese cakes can we make out of 15 eggs?
1 cake
15 eggs x 5 eggs = 3 cheese cakes
How much sugar do we need for 5 cheese cakes? (5)
Suppose you want to ‘whip’ a batch of hydrogen
iodide, following the balanced chemical equation:
H2 + I2 2 HI
How much H2 and I2 should you use to make 10 g of HI?
A common mistake is that H2 and I2 react in one-to-one mass ratio so:
5 g H2 + 5 g I2 10 g HI
The coefficients balancing the
equation refer to number of
atoms, not masses.
Introducing the mole. The mole is like a dozen, but much, much more.
The mole is Avogadro’s Number of items.
1 mole = 602,214,179,000,000,000,000,000 or 6.022 x 1023.
We need the mole because the mass of an atom is too small to
be measured on a balance. Remember: 1 amu = 1.6 x 10-24 g.
1 mole of
anything:
donuts, pancakes,
is always 6.022 x
atoms, molecules, ions… 1023 of that thing.
1 mole of soft drink cans is enough to cover the surface of
the earth to a depth of over 200 miles.
If we were able to count atoms at the rate of 10 million
per second, it would take about 2 billion years to count
the atoms in one mole.
The mole translates between the number of atoms (or
molecules, ions) and grams of atoms (molecules, ions).
It is defined as the mass of Avogadro’s number of
atoms 126C, which, in turn, weights exactly 12 g.
A mole of atoms weighs the same number of
grams as the atomic mass.
One mole of H atoms weighs 1.0079 g.
One mole of C atoms weighs 12.011 g.
H2
Amadeo Avogadro
The mole
Atomic mass refers to: the sum of protons and
neutrons in a single atom, weighted average mass
of all isotopes of an element and also to the
number of grams in one mole of atoms.
+
I2
=
2 HI
1 molecule
2 H atoms
1 molecule
2 I atoms
12 molecules
12 molecules
6.022 x 1023
molecules
1 mole
2.0158 g
6.022 x 1023
molecules
1 mole
253.81 g
2 molecule
2 x (1 atom H, 1 atom I)
24 molecules
1.204 x 1024
molecules
2 mole
255.8258 g
or any number of
molecules
1 mole of H2 weighs 2 x 1.0079 g = 2.0158 g
Conversion factors:
1 mole
6.022 x 1023 species
1 mole
molar mass
Mole - mass - atoms
conversions
Q1: How many atoms in 0.5 mole Au?
6.022 x 1023 atoms Au
0.5 mole Au x
1 mole Au
= 3.011 x 1023 atoms Au
Q1a: How many moles in 7.12 x 1024 atoms of
Cu?
1 mole Cu
7.12 x 1024 atoms Cu x
6.022 x 1023 atoms Cu
= 11.8 mol Cu
Q2: What is the mass of 0.5 mol Au?
196.967 g Au
0.5 mole Au x 1 mole Au
= 98.4835 g Au
Q3: How many atoms
in 15.00 g Au?
6.022 x 1023 atoms Au
1 mole Au
x
= 4.59 x 1022 atoms Au
15.00 g Au x
1 mole Au
196.987 g Au
Percent Composition
Percent composition is % mass that each
element in a molecule contributes to the
total molar mass of the compound.
Assume that you have one mole of the
compound.
What is the % composition of CH2O?
Total mass = 12.01 g + 2.016 g + 16.00 g
= 30.026 g
12.01 g
%C = 30.026 g x 100
%C = 40.00 %
%H = 6.71 %
Practice: What is the % composition of glucose?
Check your answer: it is the same as in CH2O!
Types of Formulas
+
%O = 53.29 %
100.00 %
CH2O is the empirical formula for glucose, C6H12O6
Empirical Formula: the formula of a
compound that expresses the
smallest whole number ratio of the
atoms present.
Formulas describe the relative
number of atoms (or moles) of each
element in a formula unit. It’s always
a whole number ratio.
If we can determine the relative
number of moles of each element in a
compound, we can determine a
formula for the compound.
Molecular Formula: the formula that states
the actual number of each kind of atom found
in one molecule of the compound.
1 molecule of C9H8O4
= 9 atoms of C, 8
atoms of H and 4
atoms of O.
1 mole of C9H8O4 = 9
mol of C, 8 mol of H
and 4 mol of O atoms.
Dr. Ent burned
0.5 g of the
sample and
obtained the
total of over 1
g of products.
How is that
possible?
Oxygen from
air is a
reactant!
From the mass
of the products
(water and
carbon dioxide)
we determine
the number of
moles of C, H,
and O, and from
them obtain the
empirical
formula of the
compound.
1. Determine the mass in grams of
each element present, if necessary.
Remember, % means “out of 100”.
2. Convert grams of CO2 and H2O
(or C and H) into moles of C and H
atoms.
3. Convert moles of C into grams
of C. Do the same for H.
4. Add masses for C and H and
subtract the sum from the mass of
the sample to obtain mass of O.
Convert the mass into moles of O.
5. Divide all number of moles with
the smallest to obtain the subscripts
of the empirical formula.
6. Divide the molar mass of the
compound by the molar mass of
the empirical formula. To find the
molecular formula, multiply all
subscripts in the empirical formula
by this product.
Note: steps 3, 4 apply only for finding
formulas from combustion analysis.
Combustion analysis shown 0.300 g H2O and
0.733 g CO2 from 0.500 g of sample. Find the
empirical and molecular formula if the molar mass
of the compound is 180.15 g/mol.
2 mol H atms
1 mol H2O
x
0.3 g H2O x 18.01 g H
1 mol H2O
2O
2.
= 0.0333 mol H at.
1 mol C atms
1 mol CO2
x
0.733 g CO2 x 44.01 g CO
1 mol CO2
2
= 0.0166 mol C at.
0.0333 mol H x
3.
1.008 g H
1 mol H = 0.0336 g H
12.01 g C
= 0.199 g C
1 mol C
g O = 0.5 – (0.0336 + 0.199) = 0.267 g O
1 mol O at.
0.267 g O x 16.00 g O = 0.0169 mol O at.
Empirical Molar mass
H: 0.0333 / 0.0166 = 2
formula
emp. form.
C: 0.0166 / 0.0166 = 1
CH2O
30.026
O: 0.0169 / 0.0166 ~ 1
0.0166 mol C x
4.
5.
6.
Molar mass sample
Molar mass emp. formula
There is 6 CH2O units
in the compound.
=
180.15
=6
30.03
Molecular formula: C6H12O6.
Find the empirical and molecular formulas if the % composition is 40.0% C,
6.70% H, 53.3% O, and the molar mass of the compound is 180.155 g/mol.
ank
h
T
,D
u
o
y
t!
n
r. E
1. Assume that you have 100.00 g sample; the mass of each
element is equal to the % composition. 40.0 g C, 6.70 g H, 53.3 g O.
2.
1 mol C
1 mol H
1 mol O
40.0 g C x 12.01 g C 6.70 g H x1.008 g H 53.3 g O x 16.00 g O
= 3.33 mol O
= 3.33 mol C
= 6.65 mol H
1. Determine the mass in grams of
each element present, if necessary.
Remember, % means “out of 100”.
2. Convert grams of CO2 and H2O (or
C and H) into moles of C and H atoms.
3. Convert moles of C into grams of
C. Do the same for H.
4. Add masses for C and H and
subtract the sum from the mass of
the sample to obtain mass of O.
Convert the mass into moles of O.
5. Divide all number of moles with the
smallest to obtain the subscripts of
the empirical formula.
6. Divide the molar mass of the
compound by the molar mass of the
empirical formula. To find the
molecular formula, multiply all
subscripts in the empirical formula
by this product.
Skip steps 3 and 4, they apply for combustion analysis only.
5. C: 3.33 / 3.33 = 1
H: 6.65 / 3.33 = 2
O: 3.33 / 3.33 = 1
Empirical formula CH2O.
Emp. Formula mass = 30.026
Molar mass sample
180.155
=
=6
Molar mass emp. formula
30.026
Thus, there are 6 (CH2O) units. Molecular formula: C6H12O6.
6.
Practice (answer in parenthesis):
1. A compound has an empirical formula of NO2. The colorless
liquid used in rocket engines has a molar mass of 92.0 g mole-1.
What is the molecular formula of this substance? (N2O4)
2. A sample of a brown gas, a major air pollutant, is found to contain
2.34 g N and 5.34 g O. Determine an empirical formula for this
substance. (NO2)
3. Percent composition of a compound is found to be 43.2% K,
39.1% Cl, and some O. Find the empirical formula. If the molar mass
of the compound is 90.550 g mol-1, find the molecular formula. (KClO)
Chapter 9
Calculations from Chemical Equations
The molar mass of an element is its atomic mass in grams.
It contains 6.022 x 1023 atoms (Avogadro’s number) of the element.
The molar mass of a compound is the sum of the atomic masses of
all its atoms. For instance: molar mass of NaCl is 22.99 + 35.45 = 65.44 g
For calculations of mole-mass-number_of_particle relationships:
Remember me?
Conversions go through moles.
1. Use balanced equation.
∆
2 Al + Fe2O3 Al2O3 + 2 Fe
2 mol
1 mol 1 mol
2 mol
2. The coefficient in front of a
formula represents the number of
moles of the reactant or product.
To quantitatively convert from one quantity
to another we introduce mole ratio:
1 mol Fe2O3
2 mol Al
1 mol Fe2O3
1 mol Al2O3
Mole ratio is found from the coefficients
of the balanced equation.
moles of desired substance
Mole ratio =
moles of starting substance
Which conversion factor will be used depends on starting and desired substance
A mole of a compound weighs the sum of all atoms in the compound.
Mole – Mole Conversions - Molecules
Example 1:How many moles of NaCl
result from the complete reaction of
3.4 mol of Cl2? Assume that there is
more than enough Na.
2 Na(s) + Cl2(g) → 2 NaCl(s)
desired substance
3.4 moles Cl2 x
1 mole
2 moles
2 moles NaCl
= 6.8 moles NaCl
1 mole Cl2
starting substance
The following examples refer to the equation:
Ca5(PO4)3F(s) + 5H2SO4(aq)→ 3H3PO4(aq) + HF(aq) + 5CaSO4(s)
1 mole
5 moles
3 moles
1 mole
5 moles
Example 2: Calculate the number of moles of phosphoric acid (H3PO4) formed
by the reaction of 10 moles of sulfuric acid (H2SO4) on phosphate rock:
10 moles H2SO4 x
3 moles H3PO4
5 moles H2SO4
= 6 moles H3PO4
Example 3: Calculate the number of moles of Ca5(PO4)3F needed to produce
6 moles of H3PO4.
6 moles H3PO4 x
1 mole Ca5(PO4)3F
3 moles H3PO4
= 2 moles Ca5(PO4)3F
Mass – Mole conversion
Example 4: Calculate the number
of moles of H2SO4 necessary to
yield 784 g of H3PO4.
g
Molar mass of H3PO4 = 97.994 mole
Ca5(PO4)3F(s) + 5H2SO4 → 3H3PO4 + HF + 5CaSO4
1 mole
5 moles 3 moles 1 mole 5 moles
1. Convert the starting
1 mole H3PO4
784 g H3PO4 x
substance into moles.
= 8.00 moles H3PO4.
97.994 g H3PO4
2. Convert moles of
starting substance into
5 moles H2SO4
8.00
moles
H
PO
x
= 13.3 moles H2SO4.
3
4
moles of desired
3 moles H3PO4
substance.
3. Convert moles of desired substance into the units specified in the problem.
done.
Ex. 5: Calculate the mass of phosphate rock, Ca5(PO4)3F needed to yield 200. g of HF.
Molar masses: Ca5(PO4)3F = 504.31 g/mol; HF = 20.008 g/mol
Step 1, 200. g HF x 1 mole HF = 10.0 moles HF x 1 mole Ca5(PO4)3F = 10.0 moles
1 mole HF
20.008 g HF
Ca5(PO4)3F
Step 2
504.3 g ph.r. = 5.00 kg Ca (PO ) F.
Step 3:
5
4 3
10.0 moles Ca5(PO4)3F x
1 mole ph.r
Step_by_step:
Mass – mass conversion
Ex. 6: Calculate the number of grams of H2SO4 necessary to yield 392 g of H3PO4.
Ca5(PO4)3F(s) + 5H2SO4 → 3H3PO4 + HF + 5CaSO4
1 mole
5 moles
3 moles
1 mole
5 moles
Molar mass H3PO4 = 97.994 g
mole
g
Molar mass H2SO4 = 98.086 mole
1. Convert the starting substance into moles.
1 mole H3PO4
392 g H3PO4 x
= 4.00 moles
97.994 g H3PO4
5 moles H2SO4
2. Convert moles of starting substance
4.00 moles H3PO4 x
3 moles H3PO4
into moles of desired substance.
= 6.67 moles
3. Convert moles of desired substance into the units specified in the problem.
6.67 moles H2SO4 x
392 g H3PO4 x
98.086 g
= 654 g H2SO4.
1 mole H2SO4
Combined steps:
1 mole H3PO4 x 5 moles H2SO4 x 98.086 g
= 654 g H2SO4.
97.994 g H3PO4 3 moles H3PO4 1 mole H2SO4
Example 7: Find the mass of glucose that can be synthesized from 58.5 g of CO2,
assuming that there is more than enough water to react with all the CO2. Molar
masses are 44.01 g (CO2) and 180.16 (glucose).
sunlight
6 CO 2(g) + 6 H 2O(l)  
→ 6 O 2(g) + C6 H12 O 6(aq)
58.5 g CO2 x
1 mole CO2
1 mole glucose 180.16 g glucose = 39.9 g glucose
x
x
44.01 g CO2
6 moles CO2
1 mole glucose
Conversion – General Case
Mass to moles
of starting compound
Step 1
Moles of starting compound
to moles of desired compound
Step 2
Moles of desired comp.
to units desired.
Step 3
Mass – mass: All 3 steps
Example 8: Calculate the mass of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2 → 2NH3
grams H2 → moles H2 → moles NH3 → grams NH3
Molar masses: H2: 2.016 g/mol; NH3: 17.034 g/mol
1 mole H2 2 moles NH3 17.034 g NH3
x
x
= 1420 g NH3 = 1.42 kg NH3.
2.016 g H2 3 moles H2
1 mole NH3
Starting
Step 1
compound
result
Step 2
Step 3
Moles – moles: Step 2 only
Example 9: Calculate the moles of NH3 formed by the reaction of 1.5 moles of H2.
2 moles NH
1.50 moles of H2 x 3 moles H 3 = 1.00 mole NH3.
2
112 g H2 x
Starting
compound
Step 2
result
Moles – mass: Step 2 and Step 3 only
Example 10: Calculate the mass of NH3 formed by the reaction of 1.50 moles of H2.
17.034 g NH
2 moles NH
1.50 moles of H2 x 3 moles H 3 x 1 mole NH 3 = 17.0 g NH3.
3
2
Starting
result
compound
Step 2
Step 3
Conversion – General Case (cont’d)
Mass – moles: Step 1 and Step 2 only
Example 11: Calculate the moles of NH3 formed by the reaction of 150. g H2.
1 mole H
2 moles NH
N2 + 3H2 → 2NH3
150. g H2 x 2.016 g H2 x 3 moles H 3 = 49.6 g NH3.
2
2
Starting
Step 1
Step 2
result
compound
Mass – particles: All 3 steps
Example 12: Calculate the # molecules of NH3 formed by the reaction of 150. g H2.
1 mole H2 2 moles NH3 6.022 x 1023 molecules NH3 = 2.23 x 1025
112 g H2 x
x
x
2.016 g H2 3 moles H2
1 mole NH3
molecules NH3.
Starting
Step 1
Step 3
result
compound
Step 2
Limiting Reactant and Yield Calculations
The amount of the product(s) depends on the reactant
that is used up during the reaction, i.e. limiting reactant.
One bicycle needs 1 frame, 1 seat and 2 wheels,
therefore not more than 3 bicycles can be made.
The number of seats is the limiting part (reactant);
one frame and two wheels are parts in excess; 3
bicycles is the yield.
Limiting Reactant and yield Calculations (cont’d)
Example 13: How many moles of Fe3O4 can be obtained by reacting 16.8 g Fe
with 10.0 g H2O? Which substance is the limiting reactant? Which substance is in
excess? How much of the reactant in excess remains unreacted?
Strategy:
1. Write and balance equation.
2. Calculate the number of moles of product for each reactant;
3. The reactant that gives the least moles of (the same!) product is the limiting reactant.
4. Find the amount of reactant in excess needed to react with the limiting reactant.
Subtract this amount from the starting quantity to obtain the amount in excess.
5. Find the yield from the limiting reactant.
Balanced equation: 3 Fe (s) + 4 H2O (g)
Fe3O4 (s) + 4 H2 (g)
Yield
1 mol Fe x 1 mol Fe3O4
yield
from Fe:
16.8 g Fe x
=
0.100
mol
Fe
O
.
3 4
55.85 g Fe
3 mol Fe
limiting reactant
Least moles Fe3O4?
1 mol H2O
1 mol Fe3O4
From H2O: 10.0 g H2O x 18.02 g H O x 4 mol H O = 0.139 mol Fe3O4.
2
2
Reacted
H2O
16.8 g Fe x
Excess:
10.0 g – 7.01 g = 2.99 g H2O.
1 mol Fe x
55.85 g Fe
18.02 g H2O
4 mol H2O
= 7.01 g H2O.
x
1 mol H2O
3 mol Fe
Answer: Yield is 0.100 mol Fe3O4, Fe is the limiting reactant, 2.99 g H2O is in excess.
Percent Yield
Calculations done so far assumed that the reaction gives maximum (100%) yield.
Many reactions (especially organic) do not give the 100% yield, due to:
side reactions, reversible reactions, product losses due to human factor.
Theoretical yield: Amount calculated from the chemical equation.
Actual yield: Amount obtained experimentally.
Actual yield
Percent yield:
x 100 %
Theor. yield
Strategy:
Find limiting reactant.
Calculate theoretical yield.
Calculate percent yield.
Example 14: If 65.0 g CCl4 was prepared by
CS2 + 3 Cl2
CCl4 + S2Cl2
reacting 100. g CS2 and 100. g of Cl2, calculate
the percent yield. Molar masses: CS2: 76.15; Cl2: 70.90; CCl4: 153.81 g/mol
1 mol CS2
1 mol CCl4
100. g CS2 x
x
= 1.31 mol CCl4.
76.15 g CS2 1 mol CS2
1 mol Cl2
1 mol CCl4
x
= 0.470 mol CCl4.
70.90 g Cl2
3 mol CS2
Limiting reactant
153.81 g CCl4
0.470 mol CCl4 x
= 72.3 g CCl4.
1 mol CCl4
Theoretical yield
65.0 g CCl4
65.0 g CCl4
HW, Chp. 7: 1, 5, 15, 26, 33
x 100 % = 89.9 %
Actual yield
72.3 g CCl4
Chp. 9: 3, 7, 13, 15, 23, 29
Percent yield
100. g Cl2 x
Chemical Bonds: Formation of Compounds
Chapter 11
1st shell
2nd shell
s
d subshell
p
d
f
This presentation has 8 groups.
Periods in the Periodic System of
elements are called shells;
subshells (s, p, d, f…) contain
orbitals (s:1, p: 3, d: 5, f: 7), and
each orbital can be occupied by 2
electrons with opposite spins.
Elements are classified in groups
having similar chemical
characteristics. Noble gases have
completely filled p orbital with 8 e-.
4s orbital is populated before 3d.
Before we continue with filling 4p,
10 e- must go into 3d. We put
them in a loop since we are
stalled filling 4th shell.
Atomic properties
Ionization energy
… is the energy
required to remove
an e- from the atom.
and increases
bottom to top,
up a group.
It increases left to
right across a
period…
That is the behavior
exactly opposite to
that of atomic radii!
Ionization Energy vs. Atomic Number
Atomic radii
increase right
to left across
the period,
and top to
bottom down
the group.
Chemical Bonding
A molecule is a collection of atoms bound together. It is considered as an element if all
atoms are of the same type (e.g. H2), or a compound if it is made of different atoms.
A bond between two atoms of hydrogen will occur spontaneously if the atoms are within
a close proximity, i.e. when attractive forces between the nucleus of one atom and the eof the other overcome repulsive forces among their e-, thus pulling the atoms together.
74 pm
In a covalent bond between two hydrogen
atoms, each atom apparently has 2 electrons in
its shell. That gives each H atom the electronic
configuration of the closest noble gas, He.
Electrons are shared between two atoms.
Covalent bond
H:H
can be
..H
H
presented as
H–H
any of the
following ways:
Remember that
the line stands
for a pair of e-!
1 pm = 10-12 m
The bond that is
formed is called
covalent bond.
(Co-means partner,
valent refers to
valence electrons).
It always releases
energy.
Filled-shell e- (or core e-) are almost
never involved in the bond as they
are too close to their own nucleus.
2 He atoms
repel each
other and
will never
form a bond.
Molecules and the Octet Rule
Elements ‘want’ to have the electronic configuration identical to that of noble gas (8e-).
When form a molecule, atoms achieve the octet (8e-) by sharing e- with other atoms.
Hydrogen is an exception, as it only needs
one more electron to fill its 1s orbital.
How many electrons an element needs
to satisfy the octet rule can be found by
observing the Roman numerals in the
periodic table.
C: 8eH: 2e-
All atoms in
the molecule
have the
electronic
configuration
identical to
that of the
closest noble
gas.
Core electrons
Valence
electrons
Nonmetals usually gain electrons, metals loose them.
O: 8eH: 2e-
Step 1: Find the total number of
Drawing Dot Diagrams
valence e- by adding up the group
numbers of all atoms. For ions,
CO2
CO
adjust the dot count accordingly
Group IVA (14)
Group IVA Group VIA
Group VIA (16)
(subtract e- for cation, add for anion).
4e- + 2 x 6e4 e- + 6 eStep 2: Unless told otherwise,
Total:
16e
assume that the first nonTotal: 10eO–C–O
hydrogen atom in the formula of
C–O
the group is the central atom.
16e – 4e = 12eConnect atoms with single bonds.
in lone pairs
The central atom must form multiple bonds,
hence H can never be the central atom.
Step 3: Subtract 2e- for each bond
from total #e- to get #e- in lone pairs.
Step 4: Put in the remaining
electrons, two at a time, as lone
pairs. Start with the terminal atoms,
and continue with the central atom if
there are any electron pairs left.
Step 5: Check that each atom has
octet satisfied (doublet for H). If not,
move electron pair(s) from the
adjacent atom to form multiple bonds.
Practice with CO32-, SO3, etc.
10e – 2e = 8ein lone pairs
..
: C – .O. :
..
..
:O
. . – C – .O. : or
..
..
:O
. . – C – .O. :
or
..
..
:O
. . – C – .O. :
..
..
O
..=C=O
..
..
: .O. – C Ξ O :
:CΞO:
..
:OΞC–O
. .:
Equivalent structures, or resonance forms.
..
: C – .O. :
C2H4
CO324e + (3 x 6e) + 2e = 24eO
O
C
total
O
24e – (3 x 2e) = 18ein lone pairs
..
:O
..
Which C is the central atom? 2x4e + 2x1e + 2x7e = 24etotal
BOTH ! 2x4e+4x1e=12e- total
H
H
C
cannot
H
H 12e – (5x2e) = 2e
pull
C
C
C
C
H
H
..
O:
..
H
H
:.O. :
Octet rule not
satisfied for C
C
C
H
H
..
:O
..
C
C
:.O. :
O:
..
C2H2Cl2
in lone pairs
H
H
:C
C
H
H
Octet rule not
satisfied for C
electrons
from Cl
: Cl
. . : : Cl
. .:
H
..
: Cl :
H
C
C
C
C
H
:Cl
..:
:Cl
. .:
H
..
:Cl :
H has no e- to give
in for double bonds
2Two more resonance
structures for CO32ion are possible.
SO3 is identical to this; try it yourself.
HW chapter 11: 1, 3, 13, 29, 37
The bond between metals and non-metals is usually ionic. Metals give away
their e- and become positively charged (cations). Nonmetals accept them
and become anions. The ionic bond is formed as a result of attraction
between oppositely charged ions. The compound is called ionic compound,
and the three-dimensional ordered network of the ions is called ionic lattice.
Electronegativity and the
Polar Covalent Bond
The difference in EN defines the bond.
∆EN = 0, covalent; ∆EN = 1.0, polar
covalent (23% ionic); ∆EN = 1.9, polar
covalent (60% ionic); ∆EN > 1.9, ionic.
Electrons are rarely
shared equally
between atoms.
Electronegativity (EN) is numerical rating
of an atoms ability to attract to itself the
shared electrons in a covalent bond.
Generally, electronegativity of metals is low,
and that of nonmetals is high.
The least electronegative atom (except
H!) is the central atom in dot structures.
Polar covalent bond is a covalent bond in
which e- are shared unequally (large ∆EN).
A partial negative charge (δ-)
occurs on the more EN atom.
A partial positive charge (δ+)
occurs on the less EN atom.
Ionic and covalent are two extremes at
the ends of a continuum bonding types.
The Shape of the Molecules
Valence Shell Electron Pair Repulsion (VSEPR) theory is the
model mostly used to predict molecular shape.
Electron pairs on the central atom repel one another.
The two dimensional dot structure of methane, CH4. gives the angles
between electron pairs of 90o. But the dot structure angles are arbitrary.
Molecules are three dimensional, and the electron pairs would be further
away if the third dimension is considered. In fact, the shape of methane
molecule is tetrahedral; the bond angles between electron pairs is 109.5o.
H
|
H–C–H
|
H
Four electron pairs around an atom assume tetrahedral arrangement.
When there are not enough electrons for single bonds the molecule
forms multiple bonds and the structure differs. VSEPR theory treats
each multiple bond as a single electron group, because it occupies
roughly the same region of space. The number of electron groups
around an atom is called the atom’s steric number (SN).
Dot structures of formaldehyde
and acetylene are arbitrarily
shown with angles of 90o.
Their true geometry has
bond angles of 120o and
180o, respectively.
:O:
||
H–C–H
H H
|
|
CΞC
formaldehyde
acetylene
Dot structures
:O:
||
C
H
H–CΞC–H
H
True geometry
If the central atom has a
lone pair of electrons, that
electron pair is included
in the molecular shape.
The dot diagram of
ammonia presents the
atom in a plane.
The steric number on
nitrogen is 4 (3 bonding
pairs and a lone pair).
The e- pairs on the N
assume tetrahedral
arrangement.
..
H–N–H
|
Electrons in the lone
H
pair occupy more
VSEPR arrangements of electron groups around an atom
having no lone pair electrons
O=C=O
We describe the shape
of the molecule, not that
of its electrons. Lone
pair(s) of electrons are
therefore ignored.
The molecule of NH3
has a pyramidal shape.
space than the
bonding pairs. They
squeeze H-N-H bond
angle to approx. 107o.
Rule of thumb: each lone pair of e- on a
period-2 atom compresses the remaining
bond angles around that atom by ~2o.
Using VSEPR
1. Draw a dot diagram.
2. Count the number of epairs around the central
atom, including lone pairs
(i.e. the steric number, SN).
A multiple bond counts as a
single e- group.
3. Find the best
arrangement of the
electrons using SN.
4. Pretending the lone epairs are invisible,
describe the resulting
shape of the molecule.
Practice on H2O and O3.
H-O-H
bond angle
~105o.
bent
H
|
: O:
O-O-O
bond angle
120o.
SN=4
bent
H
O
||
O:
O
SN=3
Chapter 13
Water and Properties of Liquids
Liquids have intermediate properties between solids
and gases. Liquids are almost incompressible, have
definite volume and assume the shape of the container.
Densities of
liquids are
usually lower
than that of their
solids. Water is
an exception.
Evaporation or vaporization is the escape of
molecules from liquid into gaseous state.
During evaporation, liquid that stays behind is
cooler. The opposite process is condensation.
Sublimation is the escape of molecules directly
from solid into gas, bypassing liquid state.
Vapor pressure is the pressure exerted by a gas at
evaporation
equilibrium with its liquid, so that:
liquid
gas
Vapor pressure depends only on
condensation
temperature, not on the amount of
liquid.
Open container
completely evaporates.
Closed container
reaches equilibrium
between liquid and gas.
Vapor Pressure Measurement
1 atm = 760 torr
20 oC
20
oC
a.
b.
a. The system is evacuated.
Manometer attached to the
flask shows equal pressure
in both legs.
b. Water is added.
Liquid evaporates.
Manometer shows
increase in pressure.
20 oC
30 oC
c.
d.
c. Equilibrium established.
Manometer shows constant
pressure difference, 17.5 torr.
d. Temperature raised to 30 oC.
Equilibrium reestablished.
Manometer shows constant
pressure difference of 31.8 torr.
Vapor pressure
and temperature
Vapor pressure of any
gas at the boiling point
is equal to the
atmospheric pressure.
Vapor pressure of
ethyl ether is the
highest at any temp.
TBP
TBP
Vapor pressure:
Ether > Alc. > Water.
TBP
Rate of evaporation:
Ether > Alc. > Water.
proportional to vapor
pressure.
Volatility
Boiling point:
Ether < Alc. < Water
Substances that readily evaporate are volatile.
Vapor pressure of ethyl ether at 20 oC: 442.2 torr
Volatile
Vapor pressure of water at 20 oC: 17.5 torr
Vapor pressure of mercury at 20 oC: 0.0012 torr
Moderately volatile
Nonvolatile
Boiling Point Curves
Normal Boiling Point
Boiling point at standard pressure
(1 atm, or 760 torr).
Each point on the curve represents a
vapor-liquid equilibrium at a
particular temperature and pressure.
At 500 torr, ethyl ether boils at
~22 oC, alcohol at ~68 oC, and
water at 89 oC.
Freezing or Melting Point
The temperature at which the solid
and liquid are in equilibrium.
Changes of State
Majority of substances change phases
upon heating: solid liquid gas.
1 atmosphere
pressure
TBP ethyl ether
TBP alcohol
TBP water
34.6oC
78.4oC
100.0oC
Heating curve for a pure
substance
CO2 is an exception (dry ice sublimes).
A – B: solid state
B – C: melting
C – D: liquid state D – E: evaporation
E – F: vapor state
Temperature is constant during melting
and boiling – all heat used to break
solid (at boiling point) or liquid forces.
liquid
solid
evaporation
condensation
melting
freezing
gas
liquid
Heat of Fusion and Heat of Vaporization
We learned before that amount of heat
Qheating = (mass) (spec.heat) (temp.change)
depends on mass and temp. change.
Energy (heat) needed to change 1 g
Energy (heat) needed to change 1 g
of a solid at its melting point into Constant
of a liquid at its boiling point into
liquid is heat of fusion.
temperature! vapor is heat of vaporization.
Qfusion = (mass) (spec.heat of fusion)
Qvaporization = (mass) (spec.heat of vaporization)
Example 1: How many joules is
needed to change 20.0 g of ice at
0 oC to steam at 100. oC?
Qheating = (mass) (spec.heat) (temp.change)
Qtot = Qfusion + Qheating + Qvaporization
Qfusion = (20.0 g) x (335 J/g)
Qheating = (20.0 g) x (4.184 J/goC) x (100. oC)
Qvaporization = (20.0 g) x (2260 J/g)
Hydrogen Bond
}
Qtot = 60.3 kJ
produces unusually high melting & boiling point
Hydrogen Bonding (cont.)
H bonding exists between H directly bonded
to one of the three most electronegative
elements (Fluorine, Oxygen, and Nitrogen),
and F, O or N of another molecule.
. . H bond . .
H – O :. . . H – O :
|
|
H
H
H bonded
to O
No H bond
H
H
| .. |
H–C–O–C-H
| .. |
H
H
Ethyl ether
Surface Tension and Capillary Action
A droplet of liquid
falling forms a
sphere due to
attractions to other
liquid molecules –
surface tension.
Cohesive forces within
Spontaneous rise mercury liquid (left) are
of liquid in a
stronger than adhesive
forces between Hg and
narrow tube –
walls of the container.
capillary action.
Opposite is true for H O.
2
No H bonded
to F, O, or N
H bonds are
intermolecular
forces.
Hydrates
Some ionic solutions retain water upon evaporation. It becomes
the part of the crystalline compound – water of crystallization.
The formula is written as: ionic compound, dot , # water molecules…
CuSO4 5 H2O and name them by adding # (Latin) hydrate.
.
Copper(II) sulfate pentahydrate.
Hydrates are true compounds and the water is an integral part of it.
Formula mass CuSO4 5 H2O: 63.55+32.07+64.00+5x18.02 = 249.7
Percent composition of water is (5x18.02 / 249.7) x 100 = 36.08%
dry CuSO4 – white
Hydrate = blue
.
. 5 H O(s) CuSO (s) + 5 H O(g)
2
4
2
..
O
δ-
..
Water can be removed by intense heat: CuSO4
The reaction is reversed when water is added.
Water, a Unique Liquid
Water indicator
H
H
Water covers ~75% of Earth. 97% of water is in the oceans. Only
3% is fresh water, of which 2/3 is locked up in ice polar caps.
δ+
Solid form (ice) has lower density than liquid water.
Water is very stable molecule, can stand temperatures up to 2000 oC. It does not
conduct electricity when pure, but decomposes into H2 and O2 in solutions of ions.
2 H2 + O2 --> 2 H2O + 484 kJ
Water can be formed by
2 C2H2(g) + 5 O2 4 CO2 + 2 H2O(l) + 1212 kJ
Combustion,
HCl(aq) + NaOH(aq) --> NaCl(aq) + 2 H2O
Neutralization,
Metabolic reaction
C6H12O6(aq) + 6 O2 6 CO2(g) + 6 H2O(l) + 2519 kJ
Water reactions with metals:
Cold water reacts with Na, K, Ca:
Steam reacts with Zn, Al and Fe:
Reactions of water
Na + H2O H2 + NaOH
Fe + H2O(g) --> H2 + Fe3O4
Remind yourself of the activity series: the above six metals are the most active. Another
three metals are more active than H: Pb, Sn, and Ni and react with acids only; Cu, Ag,
Hg and Au are below H in the series and do not react with acids or H2O.
Water also reacts with certain nonmetals.
Anhydride means:
without water.
Most reactive:
2 F2 + 2 H2O(l) --> 4 HF(aq) + O2
Less reactive:
Cl2(g) + H2O(l) HCl(aq) + HOCl(aq) To test whether a metal or
nonmetal is an anhydride, try
Least reactive: C(s) + H2O(g) CO(g) + H2(g)
Water reacts with metal and non-metal oxides:
Basic anhydride: CaO + H2O(l) Ca(OH)2(aq)
Acid anhydride:
CO2(g) + H2O(l) H2CO3(l)
Water Purification
Screening, flocculation and sedimentation,
sand filtration, aeration, disinfection.
Hard water contains Mg2+ and Ca2+ ions
Additional water purification is done by
distillation, Ca2+, Mg2+ precipitation, ion
exchange and demineralization.
HW (p.332): 7, 11, 17, 25
to remove H2O until all
hydrogen is removed.
Ca
OH ∆
OH
CaO + H2O
H2SO4 ∆ SO3 + H2O
Chapter 12
Compressibility
Shape
Volume
Gaseous State of Matter
Solid
N
Y
Y
Liquid
N
N
Y
Gas
Y
N
N
An Ideal Gas consists of point-like particles
that do not attract or repel one another at all.
Gas molecules move freely until they hit the wall.
Unlike liquids, gases expand to fill the container.
When air is removed from a can, there are no gas
molecules to oppose the atmospheric pressure
exerted from the outside, so the can collapses.
Air pump
Pressure
=
Force
Area
Pressure can be measured by the height
of mercury column. When air is removed
from the tube, there is no resistance to
the flow of mercury from the bottom
container. Mercury fills the tube until the
atmospheric pressure is compensated
by the weight of mercury column.
We do not feel atmospheric pressure due to
random motion of air. But you feel the wind!
Pressure imbalance in ear: A difference in
pressure across the eardrum membrane
causes the membrane to be pushed out –
we call it a “popped eardrum.”
Gas Pressure
Normal atmospheric pressure:
1 atm = 760 mm Hg = 760 torr
= 29.9 in Hg.
Behavior of an ideal gas can be totally described
with pressure (P), volume (V), temperature (T) and
the number of moles of gas (n).
Pressure (P) – atmospheres
Temperature (T) – Kelvin
Volume (V) – liters
Amount of Gas (n) – moles
If different units are
given, convert them
into atm, K, L, moles
Describing the Gas
Volume is directly proportional to
Pressure (P) proportional to temperature (T). temperature (Charle’s Law): V α T
Faster moving gas molecules exert higher
Pressure is inversely proportional to
pressure on the walls of the container (Guythe volume (or proportional to 1/V).
Lussac’s Law).
As the volume increases, pressure
PαT
decreases (Boyle’s Law).
T in K!
P α 1/V
α means proportional.
Pressure is proportional to the number of
moles of gas. More gas molecules results in
more collision with the walls of the container.
Pαn
P α nT/V or P = const. x
P=R
nT
V
R = 0.082 L atm
mol K
nT
V
PV = nRT
Ideal gas law
R is called ideal gas constant, and is
the same for any gas.
Examples
Example 1. What is the volume of 5.34 moles of an Ideal Gas with a pressure
of 0.692 atm and a temperature of 45°C?
PV = nRT
V = nRT
T = 45°C + 273.15K = 318K sig.figs?
P
(5.34 mol) x (0.082 L atm / mol K) x (318 K)
V=
= 201 L
0.692 atm
Example 2. What is the temperature (in °C) of 0.0520 g of H2 gas enclosed in
a 762 mL container, if its pressure is 24.6 inches Hg?
1 mole H2
1L
762 mL x
= 0.762 L
0.0520 g H2 x
= 0.0260 moles H2
2.016 g H2
1000 mL
PV
1 atm
T=
PV = nRT
24.6 in Hg x
= 0.822 atm
nR
29.9 in Hg
T=
0.822 atm x 0.762 L
= 294 K = 21°C
0.0260 mol x (0.082 L atm / mol K)
Example 3. What is the mass of O2 gas in a 3.4 L container at 25°C with a
pressure of 625 mm Hg?
1 atm
T = 25°C + 273K = 298K
P = 625 mm Hg x
= 0.822 atm
760. mm Hg
0.822 atm x 3.4 L
n=
= 0.11 moles O2 m = 0.11 mol x 32.0 g = 3.5 g O2
298 K x (0.082 L atm / mol K)
1 mol
Dalton’s Law of Partial Pressures
Ptotal = PA + PB + PC + …
The total pressure of a mixture of gases is
the sum of the partial pressures exerted
by each of the gases in the mixture.
When collecting oxygen over water (usual
way), water vapor contributes to the total
pressure. To determine the amount of O2, the
water vapor pressure must be subtracted form
the total pressure.
PO2 = Ptot – PH2O
Ptot = PO2 + PH2O
Example 4: What is the volume of the dry oxygen if it was
collected over water at 23oC and 760. torr in a 500 mL
container? Water vapor pressure at 23oC is 21.2 torr.
Step 2: Organize data:
Step 1: determine the pressure of dry O2:
P1 = 739 torr
P2 =760 torr
PO2 = Ptot – PH2O = 760 torr – 21.2 torr = 739 torr.
V1 = 500 mL
V2 = ?
Step 3: Solve as a Boyle’s law:
or P1V1 = P2V2
P α 1/V P = const. / V PV = const.
P1V1 = const.
P2V2 = const.
V2 =
P1V1
739 torr x 500 mL
= 486 mL dry O2.
=
P2
760. torr
Standard temperature and pressure (STP) is 1 atm and 0oC.
Avogadro’s Law
Equal volumes of different gases at the same
T, P contain the same number of molecules.
H2
+
1 molecule
1 mol
1 volume
Cl2 1 molecule
1 mol
1 volume
2 HCl
2 molecules
2 mol
2 volumes
Different gases at the same P, T have equal kinetic
energy, and since they occupy the same volume,
they must have the same number of molecules to
satisfy the relationship PV = nRT.
Example 5: Calculate the molar mass
Experimentally determined that 1 mol of
ANY gas occupies 22.4 L at STP.
If the mass and volume of the gas at
STP are known, one can calculate the
molar mass M.
Since M = mass / number of moles,
M=m/n n=m/M
Density of Gases
Density is: mass / volume and for a gas
is expressed in g/L.
Recall that densities of solids and liquids
are given in g/mL!
Gas densities are up to 1000 times lower than
the densities of their liquids.
of a gas that occupies 2.00 L at STP
and its mass is 3.23 g.
Remember units of molar mass: g/mol.
3.23 g 22.4 L
= 36.2 g/mol
M=
x
2.00 L 1 mol
Substituting in Ideal Gas Law:
m
mRT
PV = nRT PV = RT
M=
M
PV
Gas volume depends on the T and P.
Example 6: Calculate the density of
Cl2 at STP. Molar mass of Cl2 = 70.90 g/mol
70.90 g 1 mol
= 3.17 g/L
d = 1 mol x
22.4 L
Gas Stoichiometry
Volume - mol conversion using
Avogadro’s law if STP, or ideal
gas law if not STP conditions.
Mass - moles and atoms - moles
conversions as in chapter 9.
1 mol (@ STP)
22.4 L
Mass (g) x 1 mol
molar mass (g)
Volume (L) x
# atoms x
Volume B
Volume A
Mass A
Atoms A
Moles A
Moles B
Mass B
Atoms B
Mol A – Mol B from coeff. of balanced equation.
Mol B – Volume, Mass or atoms, use inverse
conversion factors.
Example 7: What volume of O2 at STP can you
make from 10. g KClO3 (p.c.)? 2 KClO3 2 KCl + 3 O2
1 mol
6.022 x 1023 atoms 10. g p.c.x1 mol KClO3 x 3 mol O2 x 22.4 L = 2.7 L
122.55 g p.c. 2 mol KClO3 1 mol O2
Example 8: What volume of H2
g - mol A mol A - mol B mol - Liters B
will be formed when 50.0 g of
Al reacts with HCl at 30.0 oC
2 Al(s) + 6 HCl (aq) 2 AlCl3(aq) + 3 H2(g)
and 700. torr?
3 mol H2
Find volume at given T,P: V = nRT/P
1 mol Al
Moles H2: 50.0 g Al x
x
26.98 g Al 2 mol Al
L atm
x
(0.082
2.78
mol
H
= 2.78 mol H2
mol K) x 303 K
2
700. torr x (1 atm/760. torr)
Not STP conditions, must use ideal gas law.
= 75.1 L H2
Real Gases
Most real gases behave nearly
as predicted by ideal gas law.
Deviations occur when molecules
are crowded (high P, low T).
Air Pollution
oxygen
Oxygen allotrope ozone is produced
and decomposed in stratosphere,
protecting us from damaging UV light.
Simultaneous processes occur:
Chlorofluorocarbons in
O2
O+O
UV light
aerosols, refrigerators
O3
O2 + O
decompose ozone
O2 + O O3
ozone
producing ozone hole.
formation of ozone decomposition of ozone
Now over Antarctica but
Concentration of CO2 in air has risen steadily since industrial may spread over
revolution (end of populated areas.
1800s),
producing green
house effect and
warming Earth:
global warming.
sunlight
Homework,
chp.12 (p. 301): 11, 15, 19, 27, 35, 43, 50
Solutions
Chapter 14
Homogeneous mixture
Solute
separation
A solution is a homogeneous mixture of
Heterogeneous mixture
two or more substances. Its composition is
the same throughout, i.e. concentration
of the substances is the same.
A solution consists of solute (or
solutes) present in lesser amount, and
solvent, present in greatest amount.
Solvent / solute separation
A solution can be made of
steps require energy.
Solvent separation
substances in different phases.
Solvation step
Solvent solute
example
must
Gas
gas
air
overcome
gas
solid
dusty air
separation
gas
liquid
humid air
steps for the
Liquid
solid
sugar solution
substance to
Solvation
liquid
liquid
vinegar
be
soluble.
liquid
gas
soda
Solid
solid
all alloys
Grinding and mixing two or more
solids will never yield a true solution. A
liquid (or gaseous) phase is needed.
Attractive forces release energy.
Solubility of ions in water
insoluble
soluble
all Na+, K+, NH4+ most CO32-, PO43-,
all NO3-, C2H3O2- OH-, S2-.
(except Na+,K+,NH4+)
most SO42-, Cl-,
Br-, I-
Solubility and effects of P,T
Solubility is the maximum mass of
solute that dissolves in 100 g of solvent.
Solubility of most solids increases with
an increase in temperature.
Solution containing the maximum mass
of solute in 100 g H2O is called
saturated solution; it can be
calculated from the graph.
Example: prepare saturated solution of
NaNO3 in 250 g of water at 20 oC.
250 g H2O x 90 g NaNO3 = 225 g NaNO3
100 g H2O
Soaps and detergents wash (non-polar)
grease by the action of their molecules,
consisting of polar head and non-polar
tail. “Like dissolves like”
Solubility of gases decreases with
temperature, but increases with pressure.
“Tadpole’s”
tail is
attracted to
non-polar
grease.
Water
makes
hydrogen
bonds with
polar head.
Saturated solution of NaNO3 is 90 g NaNO3 in 100 g H2O at 20 oC. If one
adds 100 g NaNO3 in 100 g H2O, 90 g dissolves, 10 g remain undissolved.
At a given temperature, the mass of solid that goes into solution
equals to the mass that precipitates from the saturated solution.
solute (undissolved)
solute (dissolved)
Unsaturated solution contains less
solute per 100 g H2O than
Supersaturated solution has more
the saturated solution.
Supersaturated solutions are unstable, a small disturbance will cause rapid precipitation of the excess of solute.
Cooling of a saturated solution makes the excess of solute precipitate.
Rate of dissolving depends on temperature, particle size, concentration and stirring.
In solid state no reaction will occur:
Solution as a Reaction Medium
NaCl(s) + AgNO3(s) no reaction
Water breaks the crystal lattice, reaction
Ions are locked within the crystal lattice.
proceeds. NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)
Na+(aq) + Cl-(aq) + Aq+(aq) + NO3-(aq) AgCl(s) + Na+(aq) + NO3-(aq)
Percent Composition
% composition by Grams of solute x 100
Volume of solution
mass / volume
Concentration can be expressed as
% composition by
Volume of solute x 100
percent composition. There are 3 types.
volume / volume
Volume of solution
Amount of solute
Example 1: What is the mass % composition
x 100
Concentration =
Amount of solution
of 273 g solution that contains 35.0 g NaCl?
Grams of solute
% composition by mass Grams of solute x 100
Mass % = Grams of solution x 100
Grams of solution
35.0 g NaCl
=
x 100 = 12.8 mass %
Units must be specified to prevent ambiguity!
273 g solution
Chemists usually work with moles.
Molarity
Molarity is defined as the number of moles
of solute dissolved in 1 L of solution…
Molarity (M) = Moles of solute
1 L of solution
Not # of moles of solute plus 1 L solvent!
To prepare 1 L of NaCl, measure molar
mass of NaCl, pour it to a 1L volumetric
flask with enough water to dissolve it,
and then add water to the mark.
Another way to do it is to use more
concentrated solution (called stock
solution), by dilution equation:
Mstock soln. x Vstock soln. = Mdiluted soln. x Vdiluted soln.
Moles before dilution = Moles after dilution
Find the number of milliliters of the stock
solution and dilute it with water.
Moles solute x Liters of solution = Moles solute
M
x Vdiluted soln.
Vstock soln. = diluted soln.
Liters solution
Mstock soln.
Dilution does not affect the
number of moles (M x V)!
M
V
Example 2a: Prepare 500. mL of 0.15 M
NaCl using solid NaCl.
1 L solution
0.15 mol NaCl
500. mL x
x
1000 mL
1 L solution
58.443 g NaCl
x
= 4.4 g NaCl
1.00 mol NaCl
Example 2b: Prepare 500. mL of 0.15 M
NaCl using stock solution of 0.50 M NaCl.
0.15 M NaCl x 500. mL soln..
Vstock soln. =
0.50 M NaCl
= 150 mL stock solution + 350 mL water
Molar mass of solute
dilute conc.
M1V1=M2V2
V2 =
Example 3: How many grams of NaCl is in
200. mL of 1.00 M NaCl?
1 L solution
1.00 mol NaCl
200. mL x
x
1000 mL
1 L solution
58.443 g NaCl
x
= 11.7 g NaCl
1.00 mol NaCl
M1V1
M2
Example 4: Prepare 400. mL of 2.00 M NaCl
using 3.00 M NaCl stock solution.
2.00 M NaCl x 400. mL soln..
Vstock soln. =
3.00 M NaCl
= 267 mL stock solution + 133 mL water.
Molality (m) is the number of moles of solute per kilogram of solvent. Note the
difference in writing molarity (M) and molality (m) and the difference in their
definitions. m = mol solute / kg solvent, therefore, molality is independent of volume.
Colligative Properties of Solutions
NaCl is often used to “melt” ice from streets; ethylene glycol/water mixture in car
radiators works both as antifreeze and boiling point elevation agent.
Freezing Point Depression, Boiling Point Elevation and vapor-pressure lowering
are known as colligative properties of solutions. They depend on the number of
solute particles in a solution and not on the nature of the particles.
Reasoning: Vapor pressure is the tendency of water molecules to escape from the liquid
surface into gas. If, say, 10 % of molecules on the surface molecules are nonvolatile, the
resulting effect is that the vapor pressure is lowered.
If the vapor pressure is lowered, the boiling point is elevated because liquid boils
once its vapor pressure equals that of the atmospheric pressure.
Also, if the vapor pressure is lowered,
so is the freezing point because liquid
vapor pressure curve of the solution
does not intercept the solid vaporpressure at the freezing point of the
pure solution, but rather below it.
∆tf = m x Kf
∆tb = m x Kb
oC
o
= mol solute x C kg solvent
kg solvent
mol solute
Example 5: Find molality of solution
prepared by dissolving 150.0 g C6H12O6
(molar mass 180.156) in 600.0 g H2O.
150.0 g x 1 mol
= 0.8326 mol
180.156 g
molality = 0.8326 mol / 0.6 kg = 1.388 m
Example 6: what is the freezing point of
solution made by dissolving 100.0 g
ethylene glycol in 200. g H2O?
molar mass C2H6O2 = 62.05 g
100.0 g (1mol / 62.05 g) = 1.61 mol
molality = 1.61 mol / 0.2 kg = 8.05 m
Kf = 1.86 oC kg solvent/mol solute
∆tf = m x Kf = 8.05 (mol solute/kg solvent) x
(1.86 oC kg solvent/mol solute) = 15.0 oC
The freezing point of solution is lowered
by 15 oC from that of solvent (water), so
The actual freezing point is:
0 oC – 15.0 oC = -15.0 oC (5.00 oF).
Osmosis
… is diffusion of a liquid through a semipermeable membrane. The membrane
allows solvent (usually water) to diffuse, but prevents diffusion of larger molecules.
Solvent diffuses from the place with more solvent (water) to the place with less solvent.
Blood cells: in blood plasma (0.15 M NaCl, 0.9% saline), they are normal.
in hypertonic solution (1.6% saline) the blood cells shrink because water leaves the
cell plasma. In hypotonic solution (0.2%
saline) blood cells swell because water
diffuses into them.
Osmosis is also the reason bacteria
cannot survive in sugar solutions.
Plants take H2O from ground by osmosis.
HW (p.365): 1, 3, 21, 27(a-c), 33
normal
hypertonic
hypertonicsoln.
soln. hypotonic
hypotonicsoln.
soln.
Chapter 15
Electrolytes, Acids and Bases
Acid (Latin acidus - sour): sour
taste; turns plant dye litmus red;
dissolves metals producing H2 gas.
Substances that are neither acids nor
bases were called neutral substances.
Electrolytes
Nonelectrolytes
Vinegar (HC2H3OH)
Acetone (C3H6O)
Hydrogen chloride (HCl)
Carbon monoxide
Baking soda (NaHCO3)
Ethanol (C2H5OH)
Table salt (NaCl)
Methane (CH4)
Milk of magnesia (Mg(OH)2) Sucrose (C12H22O11)
Sulfuric acid (H2SO4)
Turpentine (C10H22)
Base: bitter taste; turns plant dye
litmus blue; aqueous solutions feel
slippery to touch.
Electrolyte conducts electric current,
non-electrolyte does not.
Michael Faraday
(early 1800)
proposed that
some mobile
charged particles
must exist in
solution: ions.
Svante Arhenius hypothesized that ions ‘Radical, ridiculous’ Ph.D.
come from the dissociation of the solute. thesis, Nobel prize later.
An electrolyte is a solute that dissolves in A nonelectrolyte is a solute that dissolves
water and dissociates into ions, yielding in water without producing ions. The
a solution that conducts electricity.
solution consists of intact solute molecules.
Ionic compounds are usually metal plus
Water soluble molecular substances
nonmetal or group of nonmetals.
usually consist entirely of nonmetals.
Exceptions: HX (X-halide) - polar covalent,
Quick quiz: find electrolytes from the list:
produce acids in H2O; ammonium salts.
Al(NO3)3, (CH3)2O, (NH4)2SO4, CH3OH, HBr
Weak and Strong Electrolytes
HCl and HF are both hydrogen halides.
A strong electrolyte is one that
completely dissociates into ions
upon dissolving in water.
A weak electrolyte is one that only
partially dissociates into ions upon
dissolving in water.
HCl(g)
Dissolves in H2O
H+(aq) + Cl-(aq)
100% dissociation
All HCl molecules dissociate, no HCl
molecules – light bulb shines strongly.
HF(g)
Dissolves in H2O
H+(aq) + F-(aq)
Far more undissociated HF molecules
than ions – light bulb barely lights.
Arhenius defined acid as an
electrolyte that produces H+
ions when dissolved in water.
In fact, H+ are always hydrated with
a water molecule, it is more precise
to say that the ions are H3O+ ions.
A strong acid is a water-soluble
compound that dissociates
extensively to produce a large
Strong acids
#H3O+
Weak acids
Hydrochloric (HCl) 1 1
Hydrofluoric (HF)
number of H3O+ (hydronium) ions.
Hydrobromic (HBr) 1 1
Hypochlorous (HClO)
Depending on the maximum number
Hydroiodic (HI)
1 1
Acetic (HC2H3O2)
of H3O+ ions that can be produced
Nitric (HNO3)
1 2
Carbonic (H2CO3)
from 1 mol of acid, it is referred to as Sulfuric (H SO
2 3
Phosphoric (H3PO4)
2
4)
monoprotic, diprotic or triprotic
(1, 2 or 3 moles of H3O+). A di- or tri-protic acid may not be stronger than a monoprotic.
Weak acid
Dissociation constant, Keq,
shows how many protons can
be obtained from 1 mol of acid.
H3PO4 + H2O
H3O+ + H2PO4-
Keq = 7.5 x 10-3
H2PO4- + H2O
H3O+ + HPO42-
Keq = 6.2 x 10-8
H3O+ + PO43-
Keq = 4.2 x 10-13
Strong acid
H3O+ + HSO4-
Keq > 1.0 x 103
H3O+ + SO42-
Keq = 1.2 x 10-2
Phosphoric acid is considered a
HPO42- + H2O
weak acid because even the first
dissociation step is weak.
The 3rd dissociation step is the weakest. H2SO4 + H2O
Sulfuric acid is considered strong acid
HSO4- + H2O
because of the first dissociation step.
The second dissociation step is weak.
Bases are opposite of acids. When
added in a proper amount to an
acidic solution, the acidic
properties are destroyed and the
solution becomes neutral.
Arhenius’ definition: Any electrolyte that contains
a metal ion and hydroxide group and produces
hydroxide (OH-) ions when dissolved in water.
All bases given here are strong
and completely dissociated.
Bases can be monobasic,
Lithium hydroxide, LiOH
di- and tri- basic.
Potassium hydroxide, KOH
Calcium hydroxide, Ca(OH)2
Quick quiz:
What do you get when you add
HNO3(aq) to Ba(OH)2(aq)?
Sodium hydroxide, NaOH
Magnesium hydroxide, Mg(OH)2
Barium hydroxide, Ba(OH)2
HNO3(aq) + Ba(OH)2(aq) H2O(l) + Ba(NO3)2(aq)
Arhenius theory cannot explain
basicity of NH3 - it has no OH- ions.
Brønsted and Lowry independently
explained it, defining a base as a
substance that removes H3O+.
To accept a proton (H+) from the
hydronium ion, the base must have a
free electron pair. When a proton
leaves H3O+ ion, H2O stays behind.
Base - anything that accepts a proton.
Acid - anything that donates a proton.
This new theory can explain acetyleneamide reaction, and formation of
ammonium chloride in gas phase.
Note that acetate ion and carbonate ion
come from dissociation of weak acids!
When a solution is acidic neutral basic
Parenthesis
mean:
concentration
[H3O+]
>
=
<
Weak bases
Ammonia, NH3
Acetate ion, C2H3O2Carbonate ion, CO32-
[OH-]
H2O + H2O
Keq = 1.8 x 10-5
Keq = 5.6 x 10-10
Keq = 2.1 x 10-4
H3O+ + OH-
[H3O+] x [OH-] K = K x [H O]2 = [H O+] x [OH-]
w
eq
2
3
Water also dissociates by the Keq = [H O] x [H O]
2
2
constant
process called auto-ionization.
pH Scale
Auto-ionization of water produces equal
concentrations of H3O+ and OH- ions:
[H3O+] = [OH-] = 1 x 10-7 M
O+]
10-7
Acid solution: [H3
>1x
M.
Basic solution: [OH-] > 1 x 10-7 M.
When [H3O+] rises, [OH-] lowers to satisfy
the ionic product of water, Kw = 1 x 10-14.
Performing calculations with so small
concentrations is tedious. Instead,
scientists use logarithms.
pH = - log [H3O+]
Taking a logarithm is asking “to what power
must I raise 10 to get the displayed number?”
Example: log 558 = 2.75, because 102.75 = 558.
Note that each one-fold decrease in pH
represents tenfold increase in acidity.
To convert from pH to molar H3O+
concentration, use the formula:
[H3O+] = 10-(pH)
Kw = Keq x [H2O]2 = [H3O+] x [OH-]
Kw = [H3O] x [OH-] = [1 x 10-7] x [1 x 10-7]
Kw = 1 x 10-14
Reactions of Acids and Bases
acid + metal hydrogen + ionic salt
2 HCl(aq) + Ca(s) H2(g) + CaCl2(aq)
acid + base salt + water
HBr(aq) + KOH(aq) KBr(aq) + H2O(l)
acid + metal oxide salt + water
2 HCl(aq) + Na2O(aq) 2 NaCl(aq) + H2O(l)
acid + carbonate salt + H2O + CO2
2 HCl(aq) + Na2CO3(aq) 2 NaCl(aq)
+ H2O(l) + CO2(g)
2 KOH(aq) + 2 Al(s) + 6 H2O(l) 2 KAl(OH)4(aq)
amphoteric hydroxides are capable of + 3 H2(g)
reacting as a base or an acid
Reactions of amphoteric hydroxides
Zn(OH)2(s) + 2 HCl(aq) ZnCl2(aq) + 2 H2O(l)
Zn(OH)2(s) + 2 NaOH(aq) Na2Zn(OH)4(aq)
Reactions with certain metals
2 NaOH(aq) + Zn(s) + H2O(l)
Na2Zn(OH)4(aq) + H2(g)
Titrations
Stoichiometric
procedure to
determine solution
concentrations.
Done by acid / base
neutralization.
Needed:
1. buret with known
conc. of titrant
2. Known volume of
the titrate in the
Erlenmeyer flask
3. Indicator (usually
phenolphtalein).
Fill the buret with the titrant
to 0.00 mL mark, drain it by
opening the stopcock. Stop
when indicator changes
color and read the final
volume of the titrant.
Balanced eq.: NaOH(aq) + HCl(aq) H2O + NaCl(aq)
moles of (known) base = moles of (unknown) acid Mbase x Vbase = Macid x Vacid
0.100 mol NaOH 1 mol HCl
1
x1 mol NaOH x
MHCl = 0.02250 L NaOH x 1L NaOH
0.05000 L HCl
Writing Net Ionic Equations
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Na+(aq) + Cl-(aq) + H2O(l)
spectator ions cancel
H+(aq) + OH-(aq) H2O(l)
Formula eq.
Total ionic eq.
Net ionic eq.
All three equations must have balanced number of atoms and electrical charges.
Strong electrolytes are written in their ionic form.
Weak electrolytes, nonelectrolytes, precipitates and gases are written in their molecular form.
Net ionic equation includes only substances that have undergone a chemical change.
2 AgNO3(aq) + BaCl2(aq) BaNO3(aq) + 2 AgCl(s)
2 Ag+(aq) + 2 NO3-(aq) + Ba2+(aq) + 2 Cl-(aq) Ba2+(aq) + 2 NO3-(aq) + 2 AgCl(s)
AgCl precipitates, Ba2+ and NO3- are spectator ions.
Ag+(aq) + Cl-(aq) AgCl(s)
Try H2SO4 + Ba(OH)2; Mg + HCl, Na2CO3 + H2SO4.
Colloidal Dispersions
Colloids are particles that are intermediate between true
solution and suspension (e.g. fine sand in water, which settles
down once the shaking stops). Colloids (e.g. milk) neither settle
down, nor form a true solution; the colloidal particle sizes are
between true solute ions or molecules, and particles of
mechanical suspension (i.e. between 10-4 and 10-7 cm) with a
diameter of ~ 500 nm. Thus, colloidal particle is 1000 times
larger in diameter, and about 109 times larger in volume.
Ba2+ + SO42- BaSO4(s)
2 H+ H2(g)
2H+ + CO32- H2CO3(aq)
H2CO3(aq) H2O(l) + CO2(g)
Colloidal particles can be
removed by dialysis (as in
artificial kidneys). Colloidal
dispersion shows Tyndal effect
(light scatter).
Resisting pH Changes – Buffer Solutions
A buffer is a solution of a weak acid
and its conjugate base. Addition of
a strong acid or base to a buffer
changes pH only slightly.
Human blood (pH = 7.4) may not change
by more than 0.4 pH units. Gastric juice
has pH as low as 1.3. Only a fraction of a
drop of gastric juice would change the pH
of blood enough to kill you, if it were not
buffered by HCO3- / CO32- system.
Each acid has a conjugate base,
and each base has conjugate acid.
The difference is in one proton.
A buffer replaces added strong
acid with its weak acid, or added
strong base with its weak base.
OH- + HC2H3O2 C2H3O2- + H2O
Added
base
Weak acid
in buffer
Conjugate
weak base
Neutral
water
H3O+ + C2H3O2- HC2H3O2 + H2O
Added
acid
Weak base
in buffer
Conjugate
weak acid
Neutral
water
Weak acid conjugate has all properties of a
base. Strong acid conjugate is neutral.
In acetate buffer, the weak
acid, HC2H3O2 goes after
any added OH- ions.
Simultaneously, the acid
conjugate, C2H3O2- goes
after any added H3O+.
No buffer entirely cancels out
the effect of adding strong acid
or base. pH still changes, but a
lot less than in water.
Buffer is exhausted when either
weak acid or its conjugate gets used up.
HW (p.401-403) 1, 3, 5(a-c), 13, 15, 33
What to remember:
-Solubility, soluble
ions and effect of P,T
-Define saturated soln
-Define molarity and
molality of a solution
-Recognize electrolyte
-Define strong/weak
electrolyte
-Define hydronium
and hydroxide ions
-Define acid and base
-Concentration of
hydronium/hydroxide
ions in pure water
-Define pH
-Formula eq, total
ionic and net ionic eq.
-Define buffer
Note that you eat acidic, but not basic compounds!