TEMA PARA LA MATERIA DE QUÍMICA...t/t

1. Chemistry:
The Study of Change
TEACHER: QBA MIGUEL ANGEL CASTRO RAMÍREZ

2. Introduction to
Chemistry
and the
Scientific Method

3. ask a
draw a question
conclusion do research
analyze
data design an
experiment
make observations
and collect data

4. Chemistry: A Science for the 21st Century
Health and Medicine
•Sanitation systems
• Surgery with anesthesia
• Vaccines and antibiotics
Energy and the environment
•Fossil fuels
• Solar energy
• Nuclear energy
1.1

5. Chemistry: A Science for the 21st Century
Materials Technology
•Polymers, ceramics, liquid crystals
• Room-temperature superconductors?
• Molecular computing?
Food Technology
•Genetically modified crops
• “Natural” pesticides
• Specialized fertilizers
1.1

6. The Study of Chemistry
Macroscopic Microscopic
Chemists study the microscopic properties of matter, which in turn
produce matter’s observable macroscopic properties – thus, we
often switch back and forth between microscopic and macroscopic
views of matter in this course.
1.2

7. The scientific method is a systematic approach
to research. Although it is systematic, it is not a
rigid series of steps that must be done in a particular
order.
ask a
question
draw a
do research
conclusion
researcher’s form a
hidden bias hypothesis
analyze design an
data experiment
make observations
and collect data

8. A hypothesis is a tentative tested modified
explanation for a set of
observations that can be tested.
A theory is a unifying
principle that explains a body
of facts and/or those laws that
Atomic Theory are based on them.
A law is a concise statement of a
relationship between phenomena Force = mass x
that is always the same under the acceleration
same conditions.
1.3

9. Classification of
Matter

10. Substances
Matter is anything that has mass and occupies
space.
Matter that has a uniform and unchanging
composition is called a (pure) substance.
examples of pure substances include table salt, pure
water, oxygen, gold, etc.

11. States of Matter
Matter normally occupies one of three
phases, or states. These are:
P Solid
P Liquid
P Gas
* Plasma is a 4th state of matter in which the particles are
at extremely high temperatures (over 1,000,000 °C).

12. States of Matter
As we shall see in more detail later, the phase (or
state) of a substance is determined by the average
kinetic energy of the particles that make up the
substance, (i.e., temperature) and the strength of the
attractive forces holding the substance’s particles
together.
moderate
liquid
weak
gas strong
solid

13. States of Matter
Solids
 Solids have a definite shape and volume.
 The particles of a solid cannot exchange positions.
 Solids are incompressible.

14. States of Matter
Liquids
 Liquids have definite volumes
 Liquids do not have a fixed
shape
 Like solids, liquids are also
incompressible

15. States of Matter
Gases
 Gases take on the shape and
volume of their container
 Unlike solids and liquids,
gases are highly compressible

16. States of Matter
Technically, the word
“gas” refers to a
substance that is in the
gas phase at room
temperature.
The word “vapor” refers
to the gaseous state of a
substance that is normally
a solid or liquid at room
temperature.

17. Classification of Matter
Matter can be classified based on its characteristics
into the following categories and subcategories:
1. mixtures
• homogeneous (solution)
• heterogeneous
2. (pure) substances
• compounds
• elements

18. Classification of Matter
A pure substance is a form of matter that has a
definite composition and distinct properties.
examples: gold, salt, iron, pure water, sugar
A mixture is a combination of two or more
substances in which each substance retains its
own distinct identity.
examples: salt water, oil & vinegar dressing, granite, air

19. Classification Summary see page 13

20. Classification of Matter
Mixtures
Mixtures can be heterogeneous or homogeneous.
Heterogeneous mixture : the composition is not
uniform throughout. You can visibly see the
different components.
examples: cement, iron filings in
sand, granite, milk, oil and water,
etc.

21. Classification of Matter
Homogenous mixture (also called a solution): The
composition of the mixture is the same throughout.
Solutions are made up of two components:
(1) the solute which is dissolved in
(2) the solvent.
If the solvent is water, the solution
is called an aqueous solution
which is symbolized: (aq).

22. Classification of Matter
We often think of a solution as being a solid
dissolved in a liquid. However…
In a solution, both the solvent and solute can be in
any phase – solid, liquid or gas.
solvent solute example
liquid solid salt dissolved in water
liquid liquid gasoline (a mix of liquids)
gas gas air (O2 dissolved in nitrogen)
solid solid alloys (brass, bronze, etc.)

23. Classification of Matter
If a substance dissolves in another substance, we
say the first substance is soluble in the second. If
they do not dissolve, they are said to be insoluble.
example: carbon dioxide is soluble in air
gold is insoluble in water
In the case of liquids, we use a special term:
If two liquids completely dissolve in each other, they
are said to be miscible. If they do not, they are
immiscible.
example: alcohol and water are miscible
gasoline and water are immiscible

24. Classification of Matter
A mixture can be separated into its pure
components by simple physical methods.
Filtration is a means of separating a
solids from liquids. For example, we
can filter out the sand from a mix of
sand and water.
Magnetic substances can
be separated using a
magnet.

25. Separation of a Mixture
Fractional crystallization is a means of separating
two solids by adding a solvent that will dissolve one
of the solids but not the other; the mixture is then
filtered to separate out the insoluble solid. Finally,
the solvent is evaporated off to recover the
remaining solid.
For example, we can separate
salt from sand by adding hot
water to dissolve the salt, then
filter off the sand. The water is
then evaporated off, leaving the
salt behind.

26. Separation of a Mixture
Distillation is a means of separating two liquids
based on differences in their boiling points.
The substance with the lowest boiling point
“boils off” and is then cooled and condensed
back into a liquid. The liquid is collected in a
receiver flask.
This method is only effective for
substances that are liquids at
room temperature with
significant differences in their
boiling points.
distillation apparatus

27. Separation of a Mixture
Chromotography is the separation of a mixture
based on solubility in a “mobile” solvent coupled with
an adherence to a “stationary phase” medium, such
as paper or silica gel, etc.
column chromotography is a
common means of separating
components from a mixture
Thin Layer Chromotography
can be used to separate the
components of chlorophyll from
a crushed plant leaf.

28. Separation of a Mixture
Other means of separation:
Other techniques of separating a mixture include
sublimation, extraction, and leaching, etc.
If you had a jar containing both nails and marbles,
the only way to separate them would be by hand
speak to the
hand…

29. Example:
You are given a test tube which contains a mixture of water, methanol, aspirin,
acetanilide and aluminum oxide. (Acetanilide, aluminum oxide and aspirin are
all white, powdery solids at room temperature and are thus visibly
indistinguishable from each other. Water and methanol are both colorless
liquids at room temperature and are also visibly indistinguishable from each
other.) Assume your only source of heat is a Bunsen burner which can produce
a maximum temperature of 600C. Using the following information, devise a
method to separate this mixture. Be specific and complete in your answer.
substance melting point boiling point what it dissolves/does not dissolve in
water 0 C 100 C dissolves in cold or hot methanol
methanol – 97 C 65C dissolves in cold or hot water
aspirin 135 C decomposes at dissolves in methanol or water (if above
140C 10C)
aluminum 2072C 2980 C does not dissolve in either methanol or
oxide water at any temperature
acetanilide 114C 304C dissolves only in hot (50C) water or
warm (25°C) methanol

30. Pure Substances:
Elements and Compounds
 Mixtures are composed of two or more substances
physically combined.
 Recall that a (pure) substance is matter that has
a uniform, unchanging composition
 Pure substances may be elements or
compounds

31. Classification Summary see page 13

32. Classification of Matter
Elements
An element is a substance that cannot be
separated into simpler substances by chemical
means.
carbon sulfur
• 114 elements have been identified
mercury • 82 elements occur naturally on Earth
examples include carbon, sulfur,
copper iron
copper, iron, and mercury
• 32 elements have been synthesized by scientists.
examples: technetium, americium, and seaborgium

33. Symbols for Elements
Elements are identified by a one or two-letter symbol.
The first letter, which is ALWAYS capitalized, is
typically the first letter in the name of the element.
eg, C = carbon, H = hydrogen
The second letter (which is only used if other elements
have the same first letter) is NEVER capitalized.
eg, Cl = chlorine, He = helium.
Some symbols are based on the Latin name
eg, iron is Fe (for ferrum) and sodium is Na (for natrium)

34. Symbols for Elements
http://www.privatehand.com/flash/elements.html

35. Classification of Matter
Compounds
A compound is a substance composed of atoms
of two or more different elements chemically
bonded in fixed proportions.
As such, they can be chemically decomposed
into their component elements.
table salt (NaCl) sugar
Water (H2O)
Sucrose (C12H22O11)

36. Classification of Matter
The properties of a compound are different from
the properties of its component elements
For example, table salt is composed of sodium and chlorine.
Sodium is a soft, silver colored metal that reacts violently
with water, and chlorine is a pale-green poisonous gas – yet
when chemically combined, they form table salt, a white
crystalline solid you put on your eggs in the morning!
+ =

37. Compounds
Compounds can only be separated into their pure
components (elements) by chemical means.
For example:
Iron is separated from iron ore (Fe2O3)
by heating the ore in a blast furnace and
reacting it with carbon monoxide and
elemental carbon (in the form of “coke”).
Water can be separated into its
elements, hydrogen and oxygen, by
passing an electric current through it,
a process called electrolysis.

38. Compounds
There are TWO kinds of compounds, depending on
the nature of the chemical bond holding the atoms
together.
Molecules form when two or more neutral atoms
form bonds between them by sharing electrons
Note that some elements exist as molecules.
For example,the following elements occur in
nature as molecular diatomic elements:
H2 O2
H2, N2, O2, F2, Cl2, Br2 and I2
They are molecules, but they are NOT N2
compounds, because they have only Cl2
one kind of element present.

39. Compounds
Ionic compounds are composed of ions, which are
atoms that have a (+) or (-) charge.
+ ions are called cations and form +
─
+ ─ +
─ + +
when an atom loses ─ + ─
─ ─
electrons + ─ +
+ ─
ions are called anions and form
when an atom gains electrons
Ionic compounds form when cations and anions
form electrostatic attractions between them
(opposite charges attract)

40. Classification
MATTER Summary
can it be separated
YES NO
by physical means?
MIXTURE PURE
SUBSTANCE
is the mixture uniform can the substance be
throughout? chemically decomposed into
simpler substances?
YES NO YES NO
heterogeneous
solution compound element
mixture

41. Properties
of Matter

42. Physical & Chemical Properties
Physical Properties are measurable properties
• mass $ density
• boiling point $ solubility in water
Chemical Properties describe how a
substance reacts with other substances
• flammability $bonds with oxygen
• reacts with water $decomposes when
heated

43. Extensive and Intensive Properties
Physical properties can be classified as being either
extensive or intensive properties.
An extensive property of a material depends upon
how much matter is being considered. Extensive
properties are additive.
• mass
• length
• volume

44. Extensive and Intensive Properties
An intensive property of a material is independent
of the amount of matter is being considered, and is
not additive.
• density • melting point
• temperature •color
Note that ALL chemical properties
are intensive properties.

45. Physical & Chemical Changes
A physical change does not alter the composition
or identity of a substance.
sugar dissolving
ice melting
in water
A chemical change (reaction) alters the identity
or composition of the substance(s) involved.
hydrogen burns
in air to form
water

46. Physical & Chemical Changes
Evidence of a chemical reaction include:
1. Heat and light (both) produced
2. Gas produced (bubbles)
3. Solid precipitate forms
4. Color changes occur

47. Measurement

48. Measurement
The SI System of Measurement
Scientists around the world use a unified system of
measurement (Le Systeme Internationale d’Unites,
or SI for short).
There are seven fundamental “quantities” that
can be measured:
Length Temperature Luminous intensity
Mass Electric Current
Time Chemical quantity

49. International System of Units (SI)
Each base quantity is given a unit with a
specific name and symbol
page 16

50. International System of Units (SI)
The SI units are based on metrics. Each power
of ten change is given a special prefix used with
the base unit.
You must know these prefixes
see page 17

51. Measurements with SI Units
Length (SI unit = meter) The meter is often
divided into cm and mm. (10 mm = 1 cm ).
Your little finger is about 1 cm in width.
A dime is about 1 mm thick.
English/Metric equivalencies
1 inch = 2.54 cm
1 meter = 39.37 inches

52. Measurements with SI Units
Volume (SI unit = m3) Volume is the amount of
space occupied by something.
A more common unit is the dm3 =1 liter.
A smaller unit that we will use frequently is the cm 3.
1 cm3 = 1 ml
1000 ml = 1 liter
English/Metric equivalencies
1 liter = 1.057 quarts
1 ml ~ 15 drops

53. Measurements with SI Units
Measuring Volume
regular solids: volume = length x width x height
liquids—use a graduated cylinder. To read the
scale correctly, read the volume at the lowest part
of the meniscus - the curve of the liquid’s surface
in a container.
Your eye should be level with
the meniscus when reading the
volume
meniscus

54. Measurements with SI Units
Measuring Volume continued
irregular solids: volume is found by displacement.
Begin with a known volume of water. Add the solid.
The amount of water displaced is the volume of the
solid.
volume of solid =
6 6 volume displaced :
6.0 – 4.0 = 2.0 cm3
4 4
2 2

55. Measurements with SI Units
Mass (SI unit = kilogram): the amount of matter.
The mass of a given object is constant.
A kilogram is about 2.2 pounds -- this is too large a
unit for most chemistry labs, so we will use grams
instead.
Note that mass and weight are two different
things…

56. Measurements with SI Units
Weight is a measure of the force due to gravity
acting on a mass. The weight of an object
changes, depending on the gravitational force
acting on it.
For example, on the moon you would weigh only 1/6th what
you do on Earth, because the force of gravity on the moon
is only 1/6th that of Earth.
http://www.exploratorium.edu/ronh/weight/

57. The Importance of Units
On 9/23/99, the Mars Climate Orbiter entered Mar’s atmosphere 100 km (62
miles) lower than planned and was destroyed by heat because the
engineers that designed the rocket calculated the force provided by the
engines in pounds, but NASA engineers thought the force was given in the
units of Newtons (N) when they determined when to fire the rockets…
1 lb = 1 N
1 lb = 4.45 N
“This is going to be the
cautionary tale that will be
embedded into introduction
to the metric system in
elementary school, high
school, and college science
courses till the end of time.”

58. Measurements with SI Units
Measuring Mass
Triple beam balance
Electronic balance
We still use the term “weighing” even though we are finding
the mass of an object, not its weight…
English/Metric equivalencies
1 kg = 2.203 lbs 1 paperclip  1 gram
1 lb = 453.6 grams

59. Measurements with SI Units
Temperature (SI unit = kelvin) is a measure of the
average kinetic energy (energy due to motion) of
the atoms and molecules that make up a substance.
There are three common temperature scales
Fahrenheit (oF) – English system, based on the freezing
point of salt water.
Centigrade (oC) – metric system, based on the freezing and
boiling points of pure water
Kelvin (K) – SI unit, also called the “Absolute” scale; 0 K
(Absolute Zero) is defined as the temperature at which all
motion stops (kinetic energy = 0).

60. Temperature
Conversions:
K = oC + 273.15
273 K = 0 oC
373 K = 100 oC
o
C = 5 (oF – 32)
9
9
o
F= (oC) + 32
5
32 oF = 0 oC
212 oF = 100 oC

61. Temperature
Examples
A thermometer reads 12o F. What would this be in oC ?
The conversion formula from oF to oC is: oC = 5/9(oF – 32)
Inserting the values gives: : oC = 5/9(12oF – 32)
o
C = 5/9(-20) = -11.1oC
A thermometer reads 315.3 K. What would this be in oF ?
First convert K to oC: 315.3 K – 273.15 = 42.15oC
The conversion formula from oC to oF is: oF = 9/5(oC) +32.
Inserting the values gives: : oC = 9/5(42.15 oC) + 32
o
C = (75.9) + 32 = 107.9 oF

62. page 21

63. Measurements with SI Units
Time (SI unit = second). This is the only non-
metric SI unit. We still use 1 day = 24 hours,
1 hour = 60 minutes, 1 minute = 60 seconds
We do use metric fractions of time, however, such
as milliseconds (1/1000th of a second), etc.
Chemical Quantity ( SI unit = mole). Since atoms
are so tiny, it takes a LOT of them to make even
one gram. In fact, you would have to put
602,200,000,000,000,000,000,000 atoms of carbon
(that’s 6.022 X 1023) on a balance to get just 12
grams of carbon!

64. Measurements with SI Units
The Mole continued
That huge number (6.022 X 1023) is given a special name; it is
called “Avogadro’s Number,” symbolized NA, after the Italian
physicist, Lorenzo Romano Amedeo Avogadro who lived
between 1776-1856.
Just like 1 dozen = 12 things, we define:
1 mole = 6.022 X 1023 things
Avogadro

65. JUST HOW BIG IS AVOGADRO’S NUMBER??
 1 mole of oranges would cover the surface of the earth to a
depth of 9 miles!
 If you stacked 1 mole of notebook paper, it would take you
5,800 years, traveling at the speed of light (186,000,000
miles per second) to reach the top of the stack!
 If you were given 1 mole of dollar bills when the universe
began 13 billion years ago, and you immediately began
spending money at the rate of one million dollars per second,
you would still have about 190 billion trillion dollars left !
 but 1 mole of Hydrogen atoms would only mass about 1
gram!

66. Working with Numbers: Scientific Notation
The number of atoms in 12 g of carbon:
602,200,000,000,000,000,000,000
6.022 x 1023
The mass of a single carbon atom in grams:
0.0000000000000000000000199
1.99 x 10-23
N is a number n is a positive or
between 1 and 10 N x 10 n
negative integer

67. Measurements with SI Units
Derived Units
Although we need only seven fundamental SI units,
we can combine different units to obtain new units,
called derived units.
For example, speed is distance per unit time, so we
must combine the unit for distance (m) and time
(sec) to get the SI unit for speed:
speed = meters per second (m/s)
We will be working with many different derived
units in this course. It is important to pay attention
to the individual units that make up derived units!!

68. Derived Units
Density is the mass per unit volume of a substance. It is
calculated using the equation:
mass m
density = volume d= V
SI derived unit for density is kg/m 3 . This is not a convenient
unit in chemistry, so we usually use the units g/cm3 or g/mL
Every substance has a unique density. For example:
substance density You need to know the density of water.
gasoline 0.70 g/cm3
Any object that is more dense than water
water 1.00 g/cm 3
will sink in water; if it is less dense, it will
aluminum 2.70 g/cm3 float in water
lead 11.35 g/cm3

69. page 18

70. page 18

71. Dimensional Analysis:
A problem solving
technique
desired unit
given unit x = desired unit
given unit

72. The Mathematics of Units
In algebra, we learn that:
u x u = u2 and… (2u)3 = 8 u3
u
u = 1 (the u’s cancel!) and u x a = a
u
If we let “u” = units, then every measured quantity is
a number x a unit. We can solve problems by
setting them up so that the unit we do NOT want
gets cancelled out by dividing u/u in the problem.
Thus, if a/u is a conversion (say 100 cm/1 m) then
we can convert cm to meters etc. using this
conversion factor so that the cm cancel…

73. Dimensional Analysis
Method of Solving Problems
1. Determine which unit conversion factor(s) are needed
2. Carry units through calculation
3. If all units cancel except for the desired unit(s), then the
problem was solved correctly.
given quantity x conversion factor = desired quantity
given unit x desired unit = desired unit
given unit

74. Dimensional Analysis
Method of Solving Problems
Example: How many μm are in 0.0063 inches?
Begin with what units you have “in hand,” then make a list of
all the conversions you will need.
conversion factors needed:
0.0063 in = ? 1 inch = 2.54 cm
106 μm = 1 m 1 m = 100 cm
2.54 cm 1m 106 μm
0.0063 inch x x x = 160 μm
1 inch 10 cm
2 1m

75. Dimensional Analysis Method
of Solving Problems
Example: The speed of sound in air is about 343 m/s. What
is this speed in miles per hour? (1 mile = 1609 meters)
conversion units
meters to miles 1 mi = 1609
m
seconds to hours 1 min = 60 s 1 hour = 60 min
m 1 mi 60 s 60 min mi
343 x x x = 767
s 1609 m 1 min 1 hour hour

76. page 29

77. page 30

78. Uncertainty,
Precision & Accuracy
in Measurements

79. Measurements with SI Units
Uncertainty, Precision and Accuracy in
Measurements
When you measure length using a meterstick, you
often have to estimate to the nearest fraction of a
line.
The uncertainty in a measured value is partly due
to how well you can estimate such fractional units.
The uncertainty also depends on how accurate the
measuring device, itself, is.
http://
www.mhhe.com/physsci/chemistry/chang7/esp/folder_structure/ch/

80. Precision and Accuracy
Accuracy – how close a measurement is to the true or
accepted value
To determine if a measured value is accurate, you would
have to know what the true or accepted value for that
measurement is – this is rarely known!
Precision – how close a set of measurements are to
each other; the scatter of repeated measurements
about an average.
We may not be able to say if a measured value is accurate,
but we can make careful measurements and use good
equipment to obtain good precision, or reproducibility.

81. Precision and Accuracy
A target analogy is often used to compare accuracy and
precision.
accurate precise not accurate
& but &
precise not accurate not precise

82. Precision and Accuracy
example: which is more accurate: 0.0002 g or 2.0 g?
answer: you cannot tell, since you don’t know what the
accepted value is for the mass of whatever object this is that
you are weighing!
example: which is more precise: 0.0002 g or 2.0 g?
answer: surprisingly, the most precise value is 2.0 g, not the
0.0002 g. The number of places behind the decimal is not
what determines precision! If that were so, I could increase
my precision by simply converting to a different metric prefix
for the same measurement:
Which is more precise: 2 cm or 0.00002 km? They are, in
fact, identical!

83. Precision and Accuracy
Precision is a measure of the uncertainty in a measured value.
Any measured value is composed of those digits of which you
are certain, plus the first estimated digit.
1 2 3 4
1
The length of the object is at least 1.7 cm, and we might
estimate the last digit to be half a unit, and say it is 1.75 cm
long. Others might say 1.74 or possibly 1.76 – the last digit
is an estimate, and so is uncertain.

84. Precision and Accuracy
We always assume an uncertainty of ±1 in the last digit.
The percent error in a measured value is defined as:
% error = ± uncertainty x 100
measured value
The smaller the percent error, the greater the precision – the
smaller the % error, the more likely two measurements will be
close together using that particular measuring instrument.
Thus: 2.0 ± 0.1 has a % error of (0.1/2.0) x 100 = ±5%
but 0.0002 has a % error of (0.0001/0.0002) x 100 = ± 50%

85. Percent Difference
When determining the accuracy of an experimentally
determined value, it must be compared with the “accepted
value.” One common method of reporting accuracy is called
the percent difference ( %) – this gives how far off your
value is, as a percent, from the accepted value:
Percent difference:
experimental value – accepted value
% = x 100
accepted value

86. example: In an experiment, a student determines the
density of copper to be 8.74 g/cm 3. If the accepted value is
8.96 g/cm3, determine the student’s error as a percent
difference.
% = experimental value – accepted value x 100
accepted value
8.74 – 8.96 x 100 = − 2.46 %
% =
8.96
The (-) sign indicates the experimental value is 2.46%
smaller than the accepted value; a (+) % means the
experimental value is larger than the accepted value.

87. Precision and Accuracy
We will be doing math operations involving measurements
with uncertainties, so we need a method of tracking how the
uncertainty will affect calculated values – in other words, how
many places behind the decimal do we really get to keep the
answer?
The method requires us to keep track of significant digits.
Significant digits (or significant figures) are all of
the known digits, plus the first estimated or
uncertain digit in a measured value.

88. Significant Figures: Rules
• Any digit that is not zero is significant
1.234 kg 4 significant figures
• Zeros between nonzero digits are significant
606 m 3 significant figures
• Zeros to the left of the first nonzero digit are not significant
0.08 L 1 significant figure
• If a number is greater than 1, then all zeros to the right of
the decimal point are significant
2.0 mg 2 significant figures
• If a number is less than 1, then only the zeros that are at
the end and in the middle of the number are significant
0.00420 g 3 significant figures

89. Significant Figures: Rules
How many significant figures are in
each of the following measurements?
24 mL 2 significant figures
3001 g 4 significant figures
0.0320 m3 3 significant figures
6.4 x 104 molecules 2 significant figures
560 kg You cannot tell!!

90. Significant Figures: Rules
Suppose you wanted to estimate the number of jellybeans in a
jar, and your best guess is around 400.
Now – is the uncertainty in your estimate ±1 jellybean, or is it
±10 jellybeans, or maybe even ±100 jellybeans (if you weren’t
very good at estimating jellybeans…)
We need a way to write 400 and indicate in some way
whether that was 400 ±1 vs 400±10 vs 400±100. The plain
number “400” is ambiguous as to where the uncertain digit is.
Use scientific notation to remove the ambiguity:
400 ± 1 = 4.00 x 102 = 3 sig figs 400 ± 100 = 4 x 102 = 1 sig fig
400 ± 10 = 4.0 x 102 = 2 sig figs

91. Rounding Numbers
Given the number 6.82 and asked to round to 2 sig digits we
would write 6.8. We write 6.8 because 6.82 is closer to 6.8
than it is to 6.9
Given the number 6.88 and asked to round to 2 sig digits, we
would write 6.9. We write 6.9 because 6.88 is closer to 6.9
than it is to 6.8
You were taught this long ago.
You were also probably taught that, given the number 6.85,
and asked to round this to 2 sig digits, you would write 6.9.
My question is, WHY did you round UP? 6.85 is JUST as
close to 6.8 as it is to 6.9! Since it is in the middle, it could be
rounded either way! And we should round it “either way.”

92. Rounding Numbers
Since the rounding is “arbitrarily” up, this can introduce
some round-off errors in chain calculations involving this
number – the final value will be too large if you always
round up when the next digit is exactly 5.
Because rounding is “arbitrary” when the next digit is
exactly 5, we introduce the following “odd-even rounding
rule:
When the next digit is exactly 5, round up or
down to make the number an even number.
e.g. : 4.65 ≈ 4.6 but 4.75 ≈ 4.8
Note however, that 4.651 is closer to 4.7 than 4.6, so we
round it to 4.7: only invoke the “odd-even rule” when the
next digit is exactly 5.

93. Math Operations with Significant Digits
We need a set of rules to determine how the
uncertainty or error will “propagate” or move through a
series of calculations and affect the precision of our
final answer.
There is one rule for addition
and subtraction, and one rule
for multiplication and division.
Do not mix them and match
them and confuse them!

94. Significant Figures
Addition or Subtraction
The answer cannot have more digits to the right of the
decimal point than any of the original numbers.
89.332
+1.1 one digit after decimal point
90.432 round off to 90.4
3.70 two digits after decimal point
-2.9133
0.7867 round off to 0.79

95. Significant Figures
Addition or Subtraction
We often encounter two numbers that must be added that
are in scientific notation. We cannot add them and
determine the number of places “behind the decimal” unless
they have the same power of 10 – we may have to convert!
Example: What is the sum of 2.4 x 102 + 3.77 x 103 ?
3.77 x 103
Always convert the smaller power of 10 to the
0.24 x 103 larger power of 10
4.01 x 103
The answer is good to 2 behind the decimal
when written as x103 -- that is, the uncertain
digit is in the “tens” place (± 10)

96. Significant Figures
To determine the power of 10, visualize a see-saw when you
move the decimal point:
10n Increasing the power of 10
+ n means you must move the
decimal to the LEFT one
n
place for each power of 10
increase
10n + n Moving the decimal determines
both the magnitude and the +/-
n value of 10n
Decreasing the power of 10
+n means you must move the
10 n
decimal to the RIGHT by one
n
place for each power of 10
decrease.

97. Significant Figures
Example: What is the answer to the following, to the correct
number of significant digits?
3.0268 x 10-2
- - 0.012 x 10-2
0 1.2 x 10-4
3.0148 x 10-2 = 3.015 x 10-2

98. Significant Figures
Multiplication or Division
The number of significant figures in the result is set by the
original number that has the smallest number of significant
figures
4.51 x 3.6666 = 16.536366 = 16.5
3 sig figs round to
3 sig figs
6.8 ÷ 112.04 = 0.0606926 = 0.061
round to
2 sig figs 2 sig figs

99. page 25

100. Significant Figures
Exact Numbers
Numbers from definitions or numbers of objects are
considered to have an infinite number of significant figures
Example: Find the average of three measured lengths:
6.64, 6.68 and 6.70 cm.
These values each have 3 significant figures
6.64 + 6.68 + 6.70
= 6.67333 = 6.67 =7
3
Because 3 is an exact number the answer is not rounded to 7, but
rather reported to be 6.67 cm (three sig figures).
1.8

101. Atoms, Molecules and Ions
Chapter 2
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

102. Early Ideas
Our understanding of the structure of
matter has undergone profound changes
in the past century.
Nonetheless, what we know today did
not arrive on a sudden inspiration. We
can trace a fairly steady plodding
towards our current understanding,
starting as far back as 400 BCE…

103. Early Ideas
Democritus (c.a. 400 BCE)
All matter was composed of tiny, indivisible
particles called atoms (atomos = indivisible)
Each kind of matter had its own unique kind of
atom – ie., there were water atoms, air atoms, fire
atoms, bread atoms, etc.
The properties of matter could be explained by the
shape and size of its atoms.
Fire atoms water atoms
“ouch!” rolls & flows

104. Early Ideas
Most importantly, Democritus believed atoms
existed in a vacuum – that is, there was “nothing”
in the spaces between atoms…
Vacuum??
Aristotle, among others, refused
to believe in the existence of
“nothingness” that still occupied
space…
As a result, Democritus’ ideas were not very well
receieved. It would be some 1200 years before the
idea of atoms was revisited!

105. Early Ideas
Aristotle
Aristotle was the court philosopher to Alexander the
Great. Because of this, Aristotle’s ideas were given a
lot of weight .
Aristotle believed that all matter
was composed of four elements:
earth, air, fire and water.

106. Early Ideas
These elements could be “inter-converted” into each
other by exchanging the “properties” of hot, cold, dry
and wet. FIRE
hot dry
AIR EARTH
wet cold
WATER
example
Heating WATER exchanged “hot” for “cold” which created “AIR”
(which we see as steam…)
WATER (cold, wet) AIR (hot, wet)

107. Early Ideas
This idea that one kind of
element could be converted
into another eventually led
to the belief in Alchemy –
that one could turn lead into
gold by performing the right
chemical reaction!

108. Early Ideas
The “scientific method” of inquiry was developed
during the 17th and 18th centuries. The invention of
the balance and other instruments soon led to a new
understanding about the nature of matter.
The French chemist, Antoine-
Laurent Lavoisier (1743-1794),
presented two important ideas
which would later help lead to
a new, more developed atomic
theory of matter…

109. Lavoisier
1. The Law of Conservation of Matter: matter is not
created or destroyed in chemical reactions. Any
atomic theory would have to explain why matter is
not gained or lost in reactions.
2. Lavoisier defined element as any substance that
could not be chemically broken down into a simpler
substance.
Lavoisier was a meticulous experimenter. He also
helped develop the metric system of measurement.
He is often called the “Father of Modern Chemistry,”
in recognition of his pioneering works.

110. Lavoiser experimenting with respiration

111. Early Ideas
Joseph Proust, another 18th century French
scientist, proposed the Law of Definite
Proportion, which states that the mass ratios of
elements present in different samples of the same
compound do not vary.
For example, the percent by mass of the elements
present in sugar are always found to be:
53.3% oxygen, 40.0% carbon and 6.7% hydrogen.

112. John Dalton (1766-1824)
Dalton started out as an
apothecary's assistant (today,
we would call him a pharmacist).
He was also interested in both
meteorology and the study of
gases.
Dalton developed a new atomic theory of the nature
of matter based on several postulates. His theory
differed significantly from the early ideas of
Democritus, but they both agreed that the simplest
form of matter was the atom.

113. Dalton’s Atomic Theory (1808)
1. Elements are composed of extremely small particles
called atoms.
2. All atoms of a given element are identical, having the
same size, mass and chemical properties. The atoms of
a given element are different from the atoms of all other
elements.
3. Compounds are composed of atoms of more than one
element. In any compound, the ratio of the numbers of
atoms of any two of the elements present is either an
integer or a simple fraction.
4. A chemical reaction involves only the separation,
combination, or rearrangement of atoms; it does not
result in their creation or destruction.

114. Dalton’s Atomic Theory
Law of Conservation of Matter and
Definite Proportion Explained…
+ =
16 X + 8Y 8 X2Y

115. Law of Multiple Proportions
If Dalton’s ideas about atoms were correct, then he
proposed that the mass of a compound containing
different numbers of a given element (atom) would
vary by the mass of that one whole atom – that is:
If two elements can combine to form more than
one compound, then the masses of one element
that combine with a fixed mass of the other
element are in ratios of small, whole numbers.

116. Dalton’s Atomic Theory
Consider the mass ratio of oxygen to carbon in
the two compounds: CO and CO2
16
= 1.33
12
2.67 / 1.33 = 2
32 = 2.67
12
Note that the mass of oxygen that combines with 12 g of
carbon in carbon dioxide is 2 x greater than the mass of oxygen
that combines with 12 g of carbon in carbon monoxide.

117. Modern Ideas
In the late 19th and early 20th centuries, three
important experiments that shed light on the
nature of matter were conducted:
1. J.J. Thomson’s investigation of cathode rays that led
to the discovery of the electron.
2. Robert Millikan’s “Oil drop experiment” that
determined the charge and mass of the electron.
3. Ernest Rutherford’s “Gold foil experiment” that finally
gave us the current “nuclear” model of the atom.

118. Cathode rays, discovered by William Crookes,
are formed when a current is passed through an
evacuated glass tube. Cathode rays are
invisible, but a phosphor coating makes them
visible.

119. J.J. Thomson
The Electron is Discovered
J.J. Thomson helped show
that cathode rays are made up
of negatively charged particles
(based on their deflection by
magnetic and electric fields).
Sir Joseph John Thomson
1856-1940
N
S

120. J.J. Thomson
Thomson showed that all cathode rays are identical,
and are produced regardless of the type of metals
used for the cathode and anode in the cathode ray
tube.
Thomson was unable to determine either the actual
electric charge or the mass of these cathode ray
particles. He was, however, able to determine the
ratio of the electric charge to the mass of the
particles.

121. J.J. Thomson
To do this, he passed cathode rays simultaneously through
electric and magnetic fields in such a way that the forces
acting on the cathode ray particles (now called electrons)
due to the fields cancelled out. The ratio of the electric field
strength to the square of the magnetic field strength at this
point was proportional to the charge to mass ratio of the
electron.
Electric
field only +
Both
Magnetic
field only
_
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122. J.J. Thomson
The value he obtained, −1.76 x 108 C/g*, was always
the same, regardless of the source of the cathode
rays.
This value was nearly 2000 times larger than the
charge to mass ratio of a hydrogen ion (H+)!
This indicated that either the charge of the electron
was very large, or that the mass of the electron was
very, very small – much smaller than the mass of a
hydrogen atom, which was the lightest atom known.
*the SI unit of electric charge is the Coulomb (C)

123. J.J. Thomson
Thomson proposed that these electrons were
not just very small particles, but were actually a
sub-atomic particle present in all atoms.
We thus credit Thomson with the “discovery”
of the electron because of his work in determining
their physical characteristics, and his rather bold
hypothesis that they were present in all atoms
(which was later shown to be true).

124. The Plum Pudding Model
Since the atom is neutrally charged, if it has (-)
charged electrons, there must also be a (+) part to
the atom to cancel the negative electrons.
This showed that Dalton’s idea that atoms were
indivisible is NOT correct – instead, the atom is
composed of TWO oppositely charged parts.
Thomson thought the atom was a diffuse (+)
charged object, with electrons stuck in it, like
raisins in pudding (the plum pudding model).

125. Thomson’s Plum Pudding
Model of the Atom
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126. Millikan’s Oil Drop Experiment
Robert Millikan (1911) designed an experiment to
determine the actual charge of an electron.
He suspended charged oil
drops in an electric field.
The drops had become
charged by picking up free
electrons after passing
through ionized air.

127. Millikan’s Oil Drop Experiment
FELEC = E x q when the downward force of
gravity on the drop was
balanced by the upward force
FGRAVITY = m x g of the electric field, then:
E x q = m x g or q = mg/E
Knowing the mass (m) of the oil drop, and the
strength of the electric field (E), he was able to find
the charge (q) on the oil drop.

128. Millikan’s Oil Drop Experiment
To find the charge of the electron, he found the
smallest difference between the charges on any
two oil drops.
eg: Suppose you find three oil drops have the following
charges: 12.4, 7.6, 10.8. The differences between the
charges are:
12.4 – 10.8 = 1.6 10.8 – 7.6 = 3.2
12.4 – 7.6 = 4.8 4.8 – 3.2 = 1.6
You would conclude the charge of the electron was 1.6
charge units.

129. Millikan’s Oil Drop Experiment
Using this technique, Millikan was able to determine
the charge of an electron to be:
e = C 1.602 x 10C 19 C
Using Thomson’s charge to mass ratio and the
charge for the electron, Millikan determined the
mass of the electron to be 9.11 x 10-31 kilogram.
For his work, Millikan received the 1923 Nobel Prize in
Physics.
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130. Radioactivity was discovered in 1895
It was found that there are three distinct types of
radiation: (+) alpha particles, (-) beta particles, and
neutral gamma rays.
(Uranium compound)
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131. Rutherford’s
Gold Foil
Experiment
(1908 Nobel Prize in Chemistry)
Rutherford designed an experiment using these
newly discovered alpha-particles to test if Thomson’s
plum pudding model was correct.
He fired (+) alpha particles at the gold foil. If the
Thomson model was correct, most of the alpha
particles would pass through the foil with little
deflection.

132. Rutherford’s Experiment
Expected Results of Rutherford’s Experiment
The force of repulsion is directly proportional to the product of
the charges of the alpha particle and nucleus and inversely
proportional to the square of the distance between the center
of the two charges. F = kQ1Q2/R2
A large, diffuse positive charge is not able to repel a (+) alpha
particle very strongly, because the alpha particle cannot make
a close approach, so the angle of deflection,θ, would be fairly
small.
θ
(+) -particle R

133. Rutherford’s
Gold Foil
Experiment
When Rutherford performed the experiment,
nearly all the alpha particles passed through the
foil without deflection, as expected…
However, some particles were deflected
significantly, and perhaps one in 2000 were
actually deflected nearly 180 degrees!

134. Rutherford’s Experiment
Rutherford was stunned. This would be like firing a
machine gun at an apple, and having most of the
bullets pass through -- but every once in a while one
of the bullets would bounce off the apple and come
back and hit you! Why would this happen???
DUCK,
ERNIE!
?!?
something small and massive must be in there that ?
deflects only those bullets that directly hit it…

135. Rutherford’s Experiment
Only a positive charge with a very, very small
radius would allow the alpha particle to approach
close enough to experience a significant repulsion.
Strong repulsion!
-particle
θ
nucleus
R
By carefully measuring the angles of deflection, θ,
Rutherford was able to determine the approximate
size of this positive core to the atom.

136. Rutherford’s Experiment
Next, by measuring the kinetic energy of the alpha
particle before and after the collision, Rutherford
was able to apply conservation of momentum and
determine the mass of the atom’s positive core.
Putting it all together, he was able to conclude that
all the positive charge -- and about 99.9% of the
mass -- of an atom was concentrated in a very tiny
area in the middle of the atom, which he called the
nucleus.

137. Rutherford’s Experiment
Only the very few (+) α-particles that passed very
near this incredibly tiny (+) nucleus were strongly
deflected; most α-particles never came near the
nucleus and so were not deflected significantly.
*note carefully that the
(+) α-particles never
actually collide with
the (+) nucleus – the
repulsive force between
the like charges is too
great for that to occur!

138. Rutherford’s Model
of the Atom
The estimated size of this nucleus was such a tiny fraction of
the total volume of the atom, that at first Rutherford doubted
his own conclusion.
atomic radius ~ 100 pm = 1 x 10-10 m
nuclear radius ~ 5 x 10-3 pm = 5 x 10-15 m
“If the atom is the Houston Astrodome, then the
nucleus is a marble on the 50-yard line.”

139. As another size comparison, if the nucleus were the size of a
basketball, placed at PHS, the atom would be over 20 km in
diameter, reaching Martin to the North, and just missing the
US 131 Business Loop exit to the South!
The basketball-
sized nucleus would
also mass about
70,000,000,000
tons! This is
equivalent to about
100,000 cruise ship
ocean liners!

140. • Rutherford fired (+) charged alpha particles at thin sheets of gold foil and measured the angles at
which the alpha particles were deflected.
• Rutherford was testing the validity of Thomson’s plum pudding model. If this model were correct,
the (+) alpha particles would not be deflected by the diffuse (+) charge of Thomson’s atom.
• When Rutherford performed the experiment, he found that the majority of alpha particles did, in
fact, pass without significant deflection. However, a small number were significantly deflected, and
a very few were strongly deflected nearly 180 degrees.
• By measuring the angles of deflection, Rutherford was able to calculate the size and mass of the (+)
center that could produce the observed deflections. He found that all the (+) charge and about
99.9% of the atom’s mass was concentrated in a tiny region (about 1/100,000 the volume of the
atom).
• Only those alpha particles that passed very close to the nucleus experienced a strong enough
repulsion to produced significant deflections – most particles never came near the nucleus, and so
were not deflected.
• AP Extras:
• The repulsive force depends on 1/R 2 between the (+) alpha particle and the (+) charge of the
nucleus.
• He also relied on conservation of momentum to help him determine the mass of the nucleus
which was repelling the alpha particles.

141. Chadwick’s Experiment (1932)
(1935 Noble Prize in Physics)
Discovery of the Neutron
•H atoms have 1 p; He atoms have 2 p
?? •ratio of mass He/mass H should be 2/1 = 2
•measured ratio of mass He/mass H = 4 ???
James Chadwick discovered that when 9Be was bombarded
with alpha particles, a neutral particle was emitted, which was
named the neutron.
α + 9Be 1
n + 12C + energy
neutron (n) is neutral (charge = 0)
n mass ~ p mass = 1.67 x 10-24 g
Now the mass ratios can be explained if He has 2 neutrons
and 2 protons, and H has one proton with no neutrons

142. mass n  mass p  1,840 x mass e-

143. Atomic number, Mass number and Isotopes
Atomic number (Z) = number of protons in nucleus
Mass number (A) = number of protons + number of neutrons
also called the nucleon number = atomic number (Z) + number of neutrons
Isotopes are atoms of the same element (X) with different
numbers of neutrons in their nuclei
Mass Number A
ZX
Element Symbol nuclide
Atomic Number
1 2 3
examples
1H protium
1H (D) deuterium 1H (T) tritium
14 235
C Carbon-14 U Uranium-235
6 92

144. Atomic number, Mass number and Isotopes
Examples:
How many protons, neutrons, and electrons are in 14C ?
6
6 protons, 8 (14 - 6) neutrons, 6 electrons
How many protons, neutrons, and electrons are in 59 Fe ?
26
26 protons, 33 (59 - 26) neutrons, 26 electrons

145. see page 50

146. The Periodic Table
of the Elements
We now understand that the number of protons in
the nucleus of the atom is what “defines” the
element and gives each element its unique
properties.

147. The Periodic Table of the Elements
Transition metals
group
period

148. Elements
Properties of Metals
• malleable and ductile
• lustrous
• good conductors
• lose e- to form cations
Properties of Non-metals
• brittle
• dull
• poor conductors
• gain e- to form anions

149. Elements
Properties of Metalloids
• properties are intermediate between those of metals
and nonmetals
• semi-conductors
Names of Families or Groups
1A = alkali metals 5A = pnictides
2A = alkaline earths 6A = chalcogens
3A = boron family 7A = halogens
4A = carbon family 8A = noble gases
The chemical properties of elements within a
Family or Group are similar

150. Elements
Natural abundance
of elements in the
Earth’s crust
Natural abundance
of elements in the
human body

151. Molecules
and Ions

152. Molecules & Ions
A molecule is an aggregate of two or more neutral
atoms in a definite arrangement held together by
chemical forces
Note that some elements exist as molecules. For
example,the following elements occur in nature as molecular
diatomic elements:
H2, N2, O2, F2, Cl2, Br2 and I2
H2 F2
They are molecules, but they are NOT
compounds, because they have only
one kind of element present. O2 N2

153. Molecules & Ions
A polyatomic molecule contains more than two atoms
O3, H2O, NH3, C3H6O
An allotrope is one of two or more distinct molecular forms of
an element, each having unique properties. For example, O 2
and O3 are allotropes of oxygen; diamond, graphite and
buckminster fullerene (C60) are all different allotropes of
carbon.

154. Classification of Matter
Ionic compounds are composed of ions, which are
atoms that have a (+) or (-) charge.
+ ions are called cations and form when
C +
C + +
C +C + C
an atom loses electrons
C C
+ C
-ions are called anions andC + when +
form +
an atom gains electrons C
Ionic compounds form when cations and anions
form electrostatic attractions between them
(opposite charges attract)

155. Molecules and Ions
A monatomic ion contains only one atom
Examples: Na+, Cl-, Ca2+, O2-, Al3+, N3-
note that the convention is to indicate the magnitude of the
charge first, and then the sign: e.g., Ca2+, not Ca+2
A polyatomic ion contains more than one atom
Examples: ClO3-, NO2- , CN- , SO42-

156. Molecules and Ions
Examples
27 3+
How many electrons are in 13 Al ?
13 protons, so there are 13 – 3 =10 electrons
78
How many electrons are in 34 Se2- ?
34 protons, so there are 34 + 2 = 36 electrons

157. Charges of common monatomic ions
see page 54
Note that some atoms, especially transition metals, have multiple charge states
Note also that metals typically form (+) charged ions, nonmetals form (-)
charged ions.

158. Also note the relation between the magnitude of the charge
and the group number (1A, 5A, etc) for most elements.
The charge of representative metals (group 1A, 2A and 3A)
is equal to the group number
The charge of representative nonmetals (group 4A-7A) is
equal to: (the group number – 8)

159. Chemical
Nomenclature
Determining the names and
formulas of chemical compounds
IUPAC = International Union of Pure and Applied Chemists.
This is the group that determines the official rules of
nomenclature for all chemical elements and compounds

160. Chemical Formulas
A chemical formula is a combination of element
symbols and numbers that represents the
composition of the compound.
Subscripts following an element’s symbol
indicate how many of that particular atom are
present. If no subscripts are given, it is
assumed that only one of that atom is present in
the compound.
NH3 C3H6S P4O10
1 N + 3 H atoms 3 C + 6 H + 1 S atoms 4 P + 10 O atoms

161. Chemical Formulas
A molecular formula shows the exact number of
atoms of each element in the smallest unit of a
substance

162. Chemical Formulas
An empirical formula shows the simplest whole-
number ratio of the atoms in a substance
molecular empirical
H2O H2O
N2H4 NH2
C2H8O2 CH2O
C6H12O6 CH2O
note that different molecular compounds
may have the same empirical formula

163. Ionic Formulas
For ionic compounds the formula is always the
same as the empirical formula.
The sum of the charges of the cation(s) and anion(s) in each
formula unit must equal zero. Thus, the ratio
of cations to anions can always be reduced to simple, whole
number ratios. The ionic compound NaCl
Na+500Cl-500 = NaCl

164. Naming Binary
Molecular Compounds

165. Naming Molecular Compounds
We will only consider naming binary molecules.
Binary molecular compounds typically form
between two non-metals, or a non-metal and a
metalloid.
Naming Molecules:
1st element + root of 2nd element + “-ide”
e.g. : HCl = hydrogen chloride

166. Naming Molecular Compounds
See page 62
If there is more than one of
a given element, we use
prefixes to indicate the
number of each kind of
atom present.
The prefix mono is only
used for atoms that can
form more than one
compound with the second
element. For this class,
oxygen is the main
element that does this.

167. Naming Molecular Compounds
Examples of naming molecules
HI hydrogen iodide
NF3 nitrogen trifluoride
SO2 sulfur dioxide
N2Cl4 dinitrogen tetrachloride
NO2 nitrogen dioxide
N2O dinitrogen monoxide (laughing gas)

168. Naming Molecular Compounds
If the second element begins with a vowel, the
terminal vowel of the prefix is allowed to be
dropped.
For example
N2O4 could be called dinitrogen tetroxide, rather
than dinitrogen tetraoxide.
CO would be called carbon monoxide, not carbon
monooxide
Note, however, that the official IUPAC rule states
that the vowel is only dropped for “compelling
linguistic reasons.”

169. Naming Molecular Compounds
Naming Compounds containing Hydrogen
Compounds containing hydrogen can be named using the
Greek prefixes, but most have common names that are
accepted by IUPAC. The most common examples are:
B2H6 diboron hexahydride diborane
CH4 carbon tetrahydride methane
SiH4 silicon tetrahydride silane
NH3 nitrogen trihydride ammonia
phosphorus trihydride phosphine
PH3
dihydrogen monoxide water
H2O
dihydrogen sulfide hydrogen sulfide
H2S

170. Naming Molecular Compounds
Determining the formula of molecules from the
name
The subscripts tell you the number of each type
of element present, so naming molecules from
the formula is straightforward.
e.g. sulfur hexafluoride = SF6
dichlorine heptoxide = Cl2O7
The order in which the atoms are listed in molecules is based
on something called electronegativity. For now, we can predict
the order using the chart on the next slide…

171. Chemical Formulas
Order of Elements in Writing Molecular Formulas
H
B C N O F
Si P S Cl
Ge As Se Br
Sb Te I

172. Organic chemistry is the branch of chemistry that
deals with carbon compounds
Carbon is unique among all the elements in its
ability to catenate, or form long or branching
chains of carbon atoms.
We usually write these chains as “condensed formulas” that
assumes carbons are bonded to each other as follows:
H H H
= CH3CH2CH3
H C C C H
note that we could also
H
write this as: C3H8
H H

173. Organic molecules that contain only carbon and
hydrogen are called hydrocarbons.
The first 10 simple hydrocarbons
Hydrocarbon
compounds are
named based on
the number of
carbon atoms in
the “backbone” or
chain of carbon
atoms.

174. Naming Ionic Compounds

175. Naming Ionic Compounds
Ionic Compounds
Ionic compounds are typically composed
of a metal cation and a non-metal anion
$ name of cation = simply the name of the element
$ name of anion = root of element’s name + - “ide”

176. Naming Ionic Compounds
Binary ionic compounds are named:
name of metal ion + root of non-metal + “-ide”
e.g. BaCl2 barium chloride
K2O potassium oxide
Na2S sodium sulfide
Mg3N2 magnesium nitride
Al2O3 aluminum oxide

177. Formula of Ionic Compounds
Determining the formula of ionic compounds from
the name is a little more involved – unlike
molecular compounds, the name does not give us
the subscripts. These must be determined based
on the charges of each ion.
Remember that the total number of (+)
and (-) charges in any ionic compound
must sum to zero.

178. Formula of Ionic Compounds
2 x +3 = +6 3 x -2 = -6
aluminum oxide Al2O3
Al3+ O2-
1 x +2 = +2 2 x -1 = -2
calcium bromide CaBr2
Ca2+ Br-
1 x +2 = +2 1 x -2 = -2
magnesium sulfide MgS
Mg2+ S2-

179. Formula of Ionic Compounds
Note that if you take the magnitude of the charge of
the cation, and make it the subscript on the anion,
and take the magnitude of the anion’s charge and
make it the subscript of the cation, the compound
will always end up with a net neutral charge. Now, if
possible, reduce the subscripts to a simpler ratio,
and you have the correct formula for the compound!
+3 -2
Al O Al2O3
2 3
Al3+ O2-

180. see page 58

181. Pb
Cu
Multivalent ions:
W
The Non-Representative
Atoms
Fe
Mn
Co

182. Transition and other
multi-valent metal ions
Most elements form only ions with one charge.
However, most of the transition metals, as well as
Pb and Sn, have more than one possible charge
state. We say they are multi-valent.
e.g. : copper can exist in either a +1 or +2 charge
state: Cu+ or Cu2+
The formula or name of the compound must indicate
which charge state the metal cation is in.

183. Transition and other
multi-valent metal ions
Older method gives a common name for
each valence state
Cu+ cuprous Fe2+ ferrous
Cu2+ cupric Fe3+ ferric
Cr2+ chromous Hg22+ mercurous
Cr3+ chromic Hg2+ mercuric
e.g. CuCl = cuprous chloride Hg2I2 = mercurous iodide
Fe2O3 = ferric oxide PbO = plumbous oxide

184. Transition and other
multi-valent metal ions
To determine which charge state the cation is in, you
must look at the anion, and calculate the charge of
the cation…
CuS
S is always -2, and there is only one Cu to cancel this out,
so copper must be +2. Thus, this is cupric sulfide.
Fe2O3
Subscript on O is the charge of the iron! Thus, Fe is +3 and
this compound is ferric oxide.

185. Transition and other
multi-valent metal ions
Stock System:
We indicate charge on metal with Roman numerals
FeCl2 2 Cl- = -2 so Fe is 2+ iron(II) chloride
FeCl3 3 Cl- = -3 so Fe is 3+ iron(III) chloride
Cr2S3 3 S-2 = -6 so Cr is 3+ chromium(III) sulfide

186. 2-
NH4 +
C2O 4
Polyatomic
Ions
SO4 2-
2-
C2H3O 2

187. Naming Polyatomic Ions
There are certain groups of neutral atoms that bond
together, and then gain or lose one or more
electrons from the group to form what is called a
polyatomic ion. Most polyatomic ions are
negatively charged anions.
Examples:
OH- = hydroxide ion CN- = cyanide ion
NO3- = nitrate ion NH4+ = ammonium ion
SO42- = sulfate ion SO32- = sulfite ion

188. See page 60

189. Naming Polyatomic Ions
Naming ionic compounds containing polyatomic ions
is straightforward:
Name the cation + name the (polyatomic) anion
Examples:
NaOH = sodium hydroxide
K2SO4 = potassium sulfate
Fe(CN)2 = iron (II) cyanide
(NH4)2CO3 = ammonium carbonate

190. page 61

191. page 62

192. Compound
Summary
see page 64

193. NAMING ACIDS AND BASES
There are a different set of
rules for naming acids. Some
of the rules are based on a
much older system of
nomenclature, and so the
rules are not as simple as they
are for molecular and normal
ionic compounds.

194. Acids
An acid can be defined as a substance that yields
hydrogen ions (H+) when dissolved in water. These
H+ ions then bond to H2O molecules to form H3O+,
called the hydronium ion.
Many molecular gases, when
dissolved in water, become acids:
•HCl (g) = hydrogen chloride
•HCl (aq) = HCl dissolved in water
which forms (H3O+,Cl-) = hydrochloric acid

195. Acids
All acids have hydrogen as the first listed
element in the chemical formula.
For nomenclature purposes, there are two major
types of acids:
Oxoacids (also called oxyacids) = acids that
contain oxygen. eg: H2SO4, HC2H3O2
Non-oxo acids = acids that do not contain oxygen.
eg: HCl (aq), H2S (aq)

196. Acids
Rules for naming non-oxoacids
acid = “hydro-” + root of anion + “-ic acid”
see page 65
*
*note that we add an extra syllable for acids with sulfur and phosphorus:
it’s not hydrosulfic acid, but hydrosulfuric acid. Similarly, acids with
phosphorus will end in phosphoric, not phosphic acid.

197. Acids
An oxoacid is an acid that contains hydrogen,
oxygen, and another element –
That is, oxoacids are the protonated form of
those polyatomic ions that have oxygen in their
formulas.
examples:
HClO3 chloric acid
HNO2 nitrous acid
H2SO4 sulfuric acid

198. When naming oxoacids, NO “hydro” prefix is used.
Instead, the acid name is the root of the name of the
oxoanion + either “-ic” acid or “-ous” acid, as follows:
If the name of the polyatomic anion ends in
“ate,” drop the -ate and add “ic acid.”
eg: SO42- = sulfate anion H2SO4 = sulfuric acid
C2H3O2- = acetate anion HC2H3O2 = acetic acid
If the name of the polyatomic anion ends in
“ite,” drop the -ite and add “ous acid.”
eg: SO32- = sulfite anion H2SO3 = sulfurous acid
NO2- = nitrite anion HNO 2 = nitrous acid

199. Acids
Naming Oxoacids and Oxoanions
see page 66

200. Acids
As a mnemonic aid, I always use the following:
ic goes with ate because….”IC…I ATE it!
ite goes with ous like……tonsil-ITE-OUS, senior-ITE-OUS

201. Bases
A base can be defined as a substance
that yields hydroxide ions (OH-) when
dissolved in water.
NaOH sodium hydroxide
KOH potassium hydroxide
Ba(OH)2 barium hydroxide

202. Hydrates
Hydrates are compounds that have a specific
number of water molecules attached to them.
BaCl2•2H2O barium chloride dihydrate
LiCl•H2O lithium chloride monohydrate
MgSO4•7H2O magnesium sulfate heptahydrate
Sr(NO3)2 •4H2O strontium nitrate tetrahydrate
CuSO4•5H2O CuSO4
cupric sulfate anhydrous
pentahydrate cupric sulfate

203. Hydrates
Other terms associated with hydrates
Anhydrous: without water; this term describes
hydrated compounds after “drying.”
Hygroscopic: readily absorbs moisture directly
from the air.
Deliquescent: absorbs moisture from the air so
readily, that these compounds can take on enough
water to actually start to dissolve.
Water of hydration: the water absorbed and
incorporated into hygroscopic compounds

204. see page 68

205. Mass Relationships in Chemical
Reactions
Chapter 3
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

206. Relative Masses of the Elements
Micro World Macro World
atoms & molecules grams
Atomic mass is the mass of an atom in atomic
mass units (amu). This is a relative scale
based on the mass of a 12C atom.
By definition: 1 atom 12C “weighs” 12 amu
On this scale 1H = 1.008 amu and 16O = 16.00 amu

207. Relative Masses of the Elements
How do we find the relative masses of the
other elements?
Imagine we have 66.00 grams of CO2. The compound
is decomposed and yields 18.00 grams of C and 48
grams of O. Since there are two oxygen atoms for
every 1 carbon atom, we can say that
48 g O xygen 2 × 24 gra m s O xyg en 24 g O
= so = 1 .3 3 3
18 g C arbon 1 8 g C a rbon 18 g C
This means that the relative mass of each oxygen atom is
1.333 x the mass of a carbon atom (12.00 amu) , or…
mass of oxygen = 1.333 x 12.00 amu = 16.00 amu

208. Average Atomic Mass
The average atomic mass of an element is the
weighted average mass of that element, reflecting
the relative abundances of its isotopes.
example: consider lithium (Li), which has two
isotopes with the following relative percent
abundances: 7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
The Average atomic mass of lithium would be:
 7.42 + 
92.58
  6.015 amu   7.016 amu = 6.941 amu
 100   100 

209. Average Atomic Mass
IA The masses reported at the
1 bottom of the “box” for each
H
1. 0079 IIA element in the Periodic Table
3
Li
4
Be is the average atomic mass
6.941 9. 012 for that element, (in amu).
11 12
Na Mg IIIB IVB
22.99 24.305
19 20 21 22
K Ca Sc Ti
39.098 40.08 44.956 47.90

210. Average Atomic Mass
see page 79

211. The Mole & Avogadro’s Number
The mole (mol) is the SI unit for the amount of a
substance that contains as many “things” as there
are atoms in exactly 12.00 grams of 12C.
This number, called Avogadro’s number (NA),
has been experimentally determined to be
approximately 6.0221367 X 1023 things.
1 mol = NA = 6.022 x 1023 “things”
We can have 1 mole of atoms, or molecules, or even dump
trucks. The mole refers only to a number, like the term
“dozen” means 12.

212. The Mole & Avogadro’s Number
JUST HOW BIG IS AVOGADRO’S NUMBER??
• If you stacked 1 mole of notebook paper, it would take you
5,800 years, traveling at the speed of light (186,000,000 miles
per second) to reach the top of the stack!
• If you were given 1 mole of dollar bills when the universe
began 13 billion years ago, and you immediately began
spending money at the rate of one million dollars per second,
you would still have about 190 billion trillion dollars left !
• 1 mole of oranges would cover the surface of the earth to a
depth of 9 miles!
• but 1 mole of Hydrogen atoms would only mass about 1
gram!

213. The Mole & Molar Mass
Molar mass is the mass, in grams, of exactly
1 mole of any object (atoms, molecules, etc.)
Note that because of the way we defined the mole :
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
Thus, for any element
atomic mass (amu) = molar mass (grams)
For example: 1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li

214. The Mole & Molar Mass
One Mole of:
C = 12.01 g S = 32.06 g
Hg = 200.6 g
Cu = 63.55 g Fe = 55.85 g

215. The Mole & Molar Mass
Solving Mole Problems
We can now add the definitions of the mole, Avogadro’s
number, and molar mass to our repertoire of conversion
factors we can use in dimensional analysis problems.
Thus, given the mass, we can use the molar mass to
convert this to moles, and then use Avogadro’s number to
convert moles to particles, and vice versa…
M = molar mass in g/mol
NA = Avogadro’s number

216. Solving Mole Problems
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
conversion factors
1 mol K = 6.022 x 1023 atoms K
1 mol K 6.022 x 1023 atoms K
0.551 g K x x
39.10 g K 1 mol K
= 8.49 x 1021 atoms K

217. Solving Mole Problems
see page 81

218. Solving Mole Problems
see page 82

219. Solving Mole Problems
see page 82

220. Molecular Mass
Molecular mass (or molecular weight) is the sum of
the atomic masses of the atoms in a molecule.
Example: consider SO2
1S 32.07 amu
2O + 2 x 16.00 amu
SO2 64.07 amu
SO2 1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
As was the case for atoms, for any molecule
molecular mass (amu) = molar mass (grams)

221. Molecular Mass
see page 83

222. Formula Mass
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
1Na 22.99 amu
NaCl
1Cl + 35.45 amu
NaCl 58.44 amu
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
For any ionic compound
formula mass (amu) = molar mass (grams)

223. Formula Mass
What is the formula mass of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
3 Ca 3 x 40.08
2P 2 x 30.97
8O + 8 x 16.00
310.18 amu
Since the formula mass, in grams (per mole), is
numerically equal to the molar mass, in amu, we find
that the formula mass of Ca3(PO4)2 = 310.18 grams
per mole of Ca3(PO4)2.

224. Molecular/Formula Masses
Using Molecular/Formula Masses in Dimensional
Analysis Problems
We can now add molecular & formula masses to our list of
conversion factors. They are used similarly to the way we
used the molar mass of the elements as conversion factors.
Example: How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
conversion
factors
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms
72.5 g C3H8O x x x
60 g C3H8O 1 mol C3H8O 1 mol H atoms
= 5.82 x 1024 atoms H

225. Solving Mole Problems
see page 84

226. Solving Mole Problems
see page 85

227. The Mass Spectrometer
Atomic and molecular masses of unknown compounds are
determined using a mass spectrometer.
A gaseous sample of the unknown is bombarded with
electrons in an electron beam. This knocks electrons loose
from the unknown to produce cations. These cations are
then accelerated through perpendicular electric and
magnetic fields. The charge:mass ratio (e/m) of the
unknown ions determines the degree to which the particles
are deflected.
The greater the charge:mass ratio, the smaller the angle
through which the beam is deflected.

228. The Mass Spectrometer
We know the angle that a given e/m produces, so we can
identify the unknown ion when it registers on a special screen.
high e/m low e/m
Mass Spectrometer

229. Percent composition
Percent composition of an element in a compound is
the percent, by mass, of that element in the compound.
It can be calculated as follows:
n x molar mass of element
x 100%
molar mass of compound
where n is the number of moles of the
element in 1 mole of the compound
Knowing the percent composition, one can determine
the purity of a substance, (are there contaminants
present in the sample?) and you can even determine
the empirical formula of an unknown compound.

230. Percent composition
Example: What is the percent composition of
ethanol, which has the formula, C2H6O ?
First, we find the molecular mass of ethanol. This is found
to be: 2(12.01) + 6(1.008) + 1(16.00) = 46.07 grams/mole.
2 x (12.01 g)
% Composition: %C = x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H = x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O = x 100% = 34.73%
46.07 g
C2H6O
check: 52.14% + 13.13% + 34.73% = 100.0%

231. Percent composition
We can also determine the % by mass of groups
of atoms present in a compound in the same
manner.
Example: what is the percent water in epsom salts, which
has the formula: MgSO4 • 7 H2O ?
mass of water
% H2O = x 100
mass of compound
this is the molar
7(18.02) mass of water
= 24.31 + 32.07 + 4(16.00) + 7(18.02)
126.14 g H2O
= x 100 = 51.17% H2O
246.52 g cmpd

232. Percent composition
Example: How many grams of CaCl2 • 2 H2O must be weighed out
to obtain 12.20 grams of CaCl2?
There are two ways of solving this problem:
Method 1:
First determine the % CaCl2 in CaCl2 • 2 H2O:
110.98 g CaCl2
i. % CaCl2 = x 100 = 75.49%
147.02 g CaCl2 • 2 H2O
Then we note that the 12.20 g of CaCl2 desired must be
75.49% of the mass of the hydrate used:
ii. 75.49% of (X grams) of CaCl2•2 H2O = 12.20 g of CaCl2
 0.7549(X) = 12.20 or X = 12.20/0.7549 = 16.16 grams

233. Percent composition
Example: How many grams of CaCl2 • 2 H2O must be weighed out
to obtain 12.20 grams of CaCl2?
There are two ways of solving this problem:
Method 2:
Use dimensional analysis and molar masses:
1 mole CaCl2• 2 H2O
12.20 g CaCl2 x 1 mole CaCl2 x
110.98 g CaCl2 1 mole CaCl2
x 147.02 g CaCl2 • 2 H2O = 16.16 g
1 mole CaCl2• 2 H2O
note that, math-wise, both methods involve the exact same calculations
(i.e., the ratio of the molar mass of the hydrate to the molar mass of the
anhydrous form had to be determined). The only difference was the “logic”
you followed which led you to that calculation!

234. Percent Composition and
Empirical Formulas
Knowing the percent composition of a compound, one can
determine the empirical formula. It is essentially the same
process as finding the percent composition – only you work
backwards to find the molar mass of the compound…
1. First, you convert the % composition into grams. This is
easily done – suppose you had 100 grams of the
substance. Then, the mass, in grams, of each
component element is numerically the same as its
percent composition.
example: a sample of an iron ore is found to contain
69.94% Fe and 30.06% O. In 100 grams of the ore, there
would be 69.94 grams of Fe and 30.06 grams of oxygen.

235. Percent Composition and
Empirical Formulas
2. Next, knowing the mass of each element (in your 100
gram sample), determine the number of moles of that
element in your sample, by dividing the mass by the
molar mass of the element.
The number of moles of Fe and O in our sample of the iron
ore would be:
69.94 grams Fe x 1 mol Fe = 1.252 mol Fe
55.847 g
30.06 grams O x 1 mol O = 1.879 mol O
16.00 g

236. Percent Composition and
Empirical Formulas
3. To find the simplest mole ratio of the elements, divide
each by the smallest number:
in our iron ore sample, we would have:
1.879 mol O = 1.501 mol O per mole of Fe
1.252 mol Fe
4. If this ratio is a whole number, then you are done – if the
ratio is NOT a whole number, it must be converted to a
whole number ratio (we cannot have fractions of an
atom!)
Fe1.00O1.50 = Fe 2O 3 = Fe2O3
2 2

237. Percent Composition and
Empirical Formulas
The process is summarized in
Figure 3.5 on page 89 in your
textbook.