3. Significant Digits
• Any non-zero digit
• A zero between two nonzero digits
• Zeroes to the right of a nonzero digit and to
the right of the decimal point
10. A zero is significant at the end of a number that
includes a decimal point.
5 Significant Figures
000.55
Significant Figures
11. A zero is significant at the end of a number that
includes a decimal point.
5 Significant Figures
0391.2
Significant Figures
12. Non-Significant Digits
• Zeroes to the left of the leftmost nonzero digit
• Zeroes to the left of an implied decimal point
– 567 has an implied decimal point
– 450 has an implied decimal point we do not know
if the zero is significant therefore it is non-
significant
13. A zero is not significant when it is before the
first nonzero digit.
1 Significant Figure
600.0
Significant Figures
14. A zero is not significant when it is before the
first nonzero digit.
3 Significant Figures
907.0
Significant Figures
15. A zero is not significant when it is at the end of a
number without a decimal point.
1 Significant Figure
00005
Significant Figures
16. A zero is not significant when it is at the end of a
number without a decimal point.
4 Significant Figures
01786
Significant Figures
17. 12 inches = 1 foot100 centimeters = 1 meter
• Exact numbers have an infinite number of
significant figures.
• Exact numbers occur in simple counting
operations
Exact Numbers
• Defined numbers are exact.
12345
19. • Often when calculations are performed on a
calculator extra digits are present in the results.
• It is necessary to drop these extra digits so as to
express the answer to the correct number of
significant figures.
• When digits are dropped, the value of the last
digit retained is determined by a process known
as rounding off numbers.
20. 80.873
Rule 1. When the first digit after those you want to retain
is 4 or less, that digit and all others to its right are
dropped. The last digit retained is not changed.
4 or less
Rules for Rounding Off
21. 1.875377
Rule 1. When the first digit after those you want to retain
is 4 or less, that digit and all others to its right are
dropped. The last digit retained is not changed.
4 or less
Rounding Off Numbers
22. 5 or greater
5.459672
Rule 2. When the first digit after those you want to retain
is 5 or greater, that digit and all others to its right are
dropped. The last digit retained is increased by 1.
drop these figures
increase by 1
6
Rounding Off Numbers
24. The results of a calculation based on
measurements cannot be more
precise than the least precise
measurement.
It is important
to remember that:
26. • In multiplication or division, the answer must contain
the same number of significant figures as in the
measurement that has the least number of
significant figures.
27. (190.6)(2.3) = 438.38
438.38
Answer given by
calculator.
2.3 has two significant figures.
190.6 has four significant
figures.
The answer should have two significant
figures because 2.3 is the number with
the fewest significant figures.
Drop these three
digits.
Round off this digit to
four.
The correct answer is 440 or 4.4 x 102
29. The results of an addition or a
subtraction must be expressed to the
same precision as the least precise
measurement.
30. The result must be rounded to the same
number of decimal places as the value
with the fewest decimal places.
31. 1.039 - 1.020
Calculate
1.039
1.039 - 1.020
= 0.018286814
1.039
Answer given by
calculator.
1.039 - 1.020 = 0.019
0.019
= 0.018286814
1.039
The answer should have two significant
figures because 0.019 is the number
with the fewest significant figures.
280.018 6814
Two significant
figures.
Drop these
6 digits.
0.018286814
Correct answer.
32. Units of Measurement
• Recall that the unit part of a measurement
tells us the scale or standard used to
represent the results of a measurement
• The two most widely used systems of
measurement are the English System and the
Metric System
33. Measurement of mass, length and
volume
• In the United States, we use a fairly awkward system
of measurement for most things - the English system
• Scientists use the metric or SI system of units for the
measurement of physical quantities
• This system uses standard units based on very
precisely known properties of matter and light
• Metric Prefixes are used to change the size of the
fundamental or standard units.
35. The Metric or International System (SI,
Systeme International)
• Is a decimal system of units.
• It is built around standard units.
• It uses prefixes representing powers of 10
to express quantities that are larger or
smaller than the standard units.
36. International System’s
Standard Units of Measurement
Quantity Name of Unit Abbreviation
Length meter m
Mass kilogram kg
Temperature Kelvin K
Time second s
Amount of substance mole mol
Electric Current ampere A
Luminous Intensity candela cd
37. Because these fundamental units are not
always convenient size, the SI System uses
metric prefixes to change the size of the unit
38. Common Prefixes and Numerical Values for SI Units
Power of 10
Prefix Symbol Numerical Value Equivalent
giga G 1,000,000,000 109
mega M 1,000,000 106
kilo k 1,000 103
hecto h 100 102
deca da 10 101
— — 1 100
39. Prefixes and Numerical Values for SI Units
deci d 0.1 10-1
centi c 0.01 10-2
milli m 0.001 10-3
micro µ 0.000001 10-6
nano n 0.000000001 10-9
pico p 0.000000000001 10-12
femto f 0.00000000000001 10-15
Power of 10
Prefix Symbol Numerical Value Equivalent
41. The standard unit of length in the SI system is
the meter. 1 meter is the distance that light travels
in a vacuum during of a second.
1
299,792,458
42. The Meter
• 1 meter = 39.37 inches
• 1 meter is a little longer than a yard
43. Metric Units of Length
Exponential
Unit Abbreviation Metric Equivalent Equivalent
kilometer km 1,000 m 103
m
meter m 1 m 100
m
decimeter dm 0.1 m 10-1
m
centimeter cm 0.01 m 10-2
m
millimeter mm 0.001 m 10-3
m
micrometer µm 0.000001 m 10-6
m
nanometer nm 0.000000001 m 10-9
m
angstrom Å 0.0000000001 m 10-10
m
45. Converting Between Units
• The standard method to convert between two
different units is the factor-label or
dimensional analysis method
• Dimensional analysis converts a
measurement in one unit to another by the
use of a conversion factor
• Conversion factors are developed from
relationships between two units
46. Dimensional Analysis
• Dimensional analysis converts one unit to another
by using conversion factors.
• The conversion factor must accomplish two things:
1. It must cancel unit1
2. It must introduce unit2
unit1 x conversion factor = unit2
47. Conversion factors
• Unit factors - factors that relate a quantity in a
certain unit to one of another unit
e.g. 103
m = 1 km
• The conversion factor is created by dividing both
sides by the same quantity
103
m = 1 = 1 km
103
m 103
m or
103
m = 1 km = 1
1km 1 km
Each unit factor gives 2 possible conversion
factors.
48. Dimensional analysis
• Multiplying a quantity in one unit by an appropriate conversion
factor converts the number into the new unit
• Note that conversion factors are exact relationships
• Exact relationships have unlimited precision, so they can be
ignored for the purposes of deciding the number of significant
digits in a calculation
km1068.4
m10
km1
m468.0 4
3
−
×=×
49. Some Problems
• How many millimeters are there in 2.5 meters?
The conversion factor must
accomplish two things:
m x conversion factor = mm
It must cancel
meters.
It must introduce
millimeters
52. Use the conversion factor with millimeters in the
numerator and meters in the denominator.
1000 mm
x
1 m
2.5 m = 2500 mm
3
2.5 x 10 mm
How many millimeters are there in 2.5
meters?
1000 mm
1 m
53. How many millimeters are there in 2.5
meters?
Use the conversion factor with millimeters in
the numerator and meters in the denominator.
1000 mm
1 m
2.5 m
1000 mm
x
1 m
= 2500 mm
3
2.5 x 10 mm
54. 16.0 in
2.54 cm
x
1 in
= 40.6 cm
2.54 cm
1 in
Use this
conversion
factor
Convert 16.0 inches to centimeters.
55. Centimeters can be converted to micrometers by
a series of two conversion factors.
cm → m → µmeters
3
3.7 x 10 cm
1 m
x
100 cm
1
= 3.7 x 10 m
6
10 μm
x
1 m
7
= 3.7 x 10 μm
1
3.7 x 10 m
Convert 3.7 x 103
cm to micrometers.
56. Convert 3.7 x 103
cm to micrometers.
3
3.7 x 10 cm
1 m
x
100 cm
6
10 μm
x
1 m
7
= 3.7 x 10 μm
Centimeters can be converted to micrometers by
writing down conversion factors in succession.
cm → m → µmeters
59. The standard unit of mass in the SI system
is the kilogram. 1 kilogram is equal to the mass of
a platinum-iridium cylinder kept in a vault at
Sevres, France.
1 kg = 2.205 pounds
60. Metric Units of mass
Exponential
Unit Abbreviation Gram Equivalent Equivalent
kilogram kg 1,000 g 103
g
gram g 1 g 100
g
decigram dg 0.1 g 10-1
g
centigram cg 0.01 g 10-2
g
milligram mg 0.001 g 10-3
g
microgram µg 0.000001 g 10-6
g
62. An atom of hydrogen weighs 1.674 x 10-24
g. How
many ounces does the atom weigh?
1 lb
x
454 g
-24
1.674 x 10 g -27
3.69 x 10 lb=
16 oz
x
1 lb
-26
5.90 x 10 oz=-27
3.69 x 10 lb
1 lb = 454 g
16 oz = 1 lb
Grams can be converted to ounces using a series
of two conversion factors.
63. An atom of hydrogen weighs 1.674 x 10-24
g. How
many ounces does the atom weigh?
-24
1.674 x 10 g
1 lb
x
454 g
16 oz
x
1 lb
-26
5.90 x 10 oz=
Grams can be converted to ounces using a
single linear expression by writing down
conversion factors in succession.
65. Volume
• The amount of 3-dimensional space occupied
by a substance
• Fundamental unit of volume in the SI System
for volume is based on the volume of a cube
measuring
1m x 1m x 1m = (1m)3
= 1m3
= one cubic
meter
66. The liter (L) and milliliter (mL) are the
standard units of volume used in most
chemical laboratories.
• 1m3
is split into 1000 smaller
cubes ….each has a volume of
1dm3
• 1dm3
= 1L (liter)
• A cube with a volume of 1dm3
can be split into 1000 smaller
cubes each with a volume of
1cm3
= 1mL
(milliliter)
1L = 1000mL
67. Convert 4.61 x 102
microliters to
milliliters.
Microliters can be converted to milliliters
using a series of two conversion factors.
µL → L → mL
6
1 L
x
10 μL
2
4.61x10 μL
-4
4.61x10 L=
-1
= 4.61 x 10 mL
-4
4.61x10 L
1000 mL
x
1 L
68. Microliters can be converted to milliliters
using a linear expression by writing down
conversion factors in succession.
µL → L → mL
2
4.61x10 μL 6
1 L
x
10 μL
1000 mL
x
1 L
-1
= 4.61 x 10 mL
Convert 4.61 x 102
microliters to
milliliters.
69. An extensive property of a material depends upon
how much matter is is being considered.
An intensive property of a material does not
depend upon how much matter is is being
considered.
• mass
• length
• volume
• density
• temperature
• color
Extensive and Intensive Properties
71. Heat
• A form of energy that is associated with the
motion of small particles of matter.
• Heat refers to the quantity of this energy
associated with the system.
• The system is the entity that is being heated
or cooled.
72. Temperature
• A measure of the intensity of heat.
• It does not depend on the size of the system.
• Heat always flows from a region of higher
temperature to a region of lower
temperature.
73. Temperature Measurement
• The SI unit of temperature is the Kelvin.
• There are three temperature scales: Kelvin,
Celsius and Fahrenheit.
• In the laboratory, temperature is commonly
measured with a thermometer.
74. 3 Temperature Scales
1. Celsius Scale – used in Europe and in physical and life sciences
• Unit is o
C
• Freezing point of H2O is 0o
C
• Boiling point of H2O is 100o
C
2. Fahrenheit Scale – used in US and Britain
• Unit is o
F
• Freezing point of H2O is 32o
F
• Boiling point of H2O is 212o
F
3. Absolute or Kelvin Scale
Unit is K
• Freezing point of H2O is 273K
• Boiling point of H2O is 373K
75. Temperature Scales
• The size of each temperature unit (each
degree) is the same for the Celsius and
Kelvin Scale
- the difference between the freezing
point and boiling point of H2O is 100
units on both scales
• The Fahrenheit degree is smaller than
the C or K degree
- On the F scale there are 180 F degrees
between the boiling point and freezing
point of H2O compared to 100 in the
Celsius and Kelvin scales
• The zero points are different on all 3
scales
76. Converting Between Scales
• Conversions Between K and C Scales
• simple because the size of the units is the same
toC + 273 = tK
Convert 70o
C to K:
70 + 273 = tK
tK= 343K
Convert 77K to o
C:
toC= 77 -273
toC= -196 o
C
77. Conversions Between F and C Scales
• Requires 2 adjustments
1. Adjustment for different size of unit
2. Adjustment for different zero points
• To convert Co
to Fo
:
toF= 1.80(toC) + 32
Factor 1.8 is due to the fact that there are:
100C divisions/180F divisions =
1C division/1.8 F divisions or 1.8F div./1C div
• To convert Fo
to Co
:
Subtract 32 from the F temperature (so both scales start at
the same point) then multiply by the proper conversion unit
toC= (toF – 32) (1o
C/1.8o
F)
78. o o o
F - 32 = 1.8 x C
To convert between the scales, use the
following relationships:
o o o
F = 1.8 x C + 32
o
K = C + 273.15
o
o F - 32
C =
1.8
79. It is not uncommon for temperatures in the Canadian
plains to reach –60o
F and below during the winter.
What is this temperature in o
C and K?
o
o F - 32
C =
1.8
o o60. - 32
C = = -51 C
1.8
86. A 13.5 mL sample of an unknown liquid
has a mass of 12.4 g. What is the density of the liquid?
M
D
V
= 0.919 g/mL=
12.4g
13.5mL
=
87. 46.0 mL
98.1 g
A graduated cylinder is filled to the 35.0 mL mark with water.
A copper nugget weighing 98.1 grams is immersed into the
cylinder and the water level rises to the 46.0 mL. What is the
volume of the copper nugget? What is the density of copper?
35.0 mL
copper nugget final initialV = V -V = 46.0mL - 35.0mL = 11.0mL
g/mL8.92
mL11.0
g98.1
V
M
D ===
88. The density of ether is 0.714 g/mL. What is the
mass of 25.0 milliliters of ether?
Method 1
(a) Solve the density equation for mass.
mass
d =
volume
(b) Substitute the data and calculate.
mass = density x volume
0.714 g
25.0 mL x = 17.9 g
mL
89. The density of ether is 0.714 g/mL. What is the
mass of 25.0 milliliters of ether?
Method 2 Dimensional Analysis. Use density as a
conversion factor. Convert:
0.714 g
25.0 ml x = 17.9 g
mL
mL → g
g
mL x = g
mL
The conversion of units is
90. The density of oxygen at 0o
C is 1.429 g/L. What is
the volume of 32.00 grams of oxygen at this
temperature?
Method 1
(a) Solve the density equation for volume.
mass
d =
volume
(b) Substitute the data and calculate.
mass
volume =
density
2
2
32.00 g O
volume = = 22.40 L
1.429 g O /L
91. The density of oxygen at 0o
C is 1.429 g/L. What is the
volume of 32.00 grams of oxygen at this temperature?
Method 2 Dimensional Analysis. Use density as a
conversion factor. Convert:
2 2
2
1 L
32.00 g O x = 22.40 L O
1.429 g O
g → L
L
g x = L
g
The conversion of units is
92. Solving Problems
1. Read the problem carefully. Determine
what is known and what is to be solved for
and write it down.
– It is important to label all factors and units with
the proper labels.
93. 2. Determine which principles are involved and
which unit relationships are needed to solve
the problem.
– You may need to refer to tables for needed
data.
3. Set up the problem in a neat, organized and
logical fashion.
– Make sure unwanted units cancel.
– Use sample problems in the text as guides for
setting up the problem.
Basic Steps
94. 4. Proceed with the necessary mathematical
operations.
– Make certain that your answer contains the
proper number of significant figures.
5. Check the answer to make sure it is
reasonable.
Basic Steps
96. In studying the materials of the
Earth (and other parts of the Universe)
scientists have found that all matter can be
broken down chemically into about 118
different elements
It is quite amazing that millions of known
substances are composed of so few
fundamental elements ….
a phenomena not unlike the hundreds of
thousands of known words composed from
only 26 letters of the alphabet
97. Compounds
Are made by combining atoms of the various
elements…..
just as words are constructed from the 26 letters of
the alphabet
and you had to learn the letters of the alphabet
before you learned to read and write
You now must learn the names and
symbols of the Chemical Elements
before you can read and write chemistry!
98. The Elements
There are presently 118 known elements
88 occur naturally
the remainder were made in the lab and decompose
spontaneously into others by radioactive decay
the elements vary tremendously in abundance
-only 10 constitute over 99% of the Earth’s crust
-about 93% of the mass of our bodies is composed of only 3 elements: C, H, and O
-the list of elements found in living matter differs greatly from that of the earth’s
crust
*** Elements are fundamental to understanding
Chemistry
99. How Chemists use the word Element
Element
-single atom
of that element
-sample of
the element large
enough to weigh on
a balance
-some elements
that contain
molecules rather
than individual
atoms
100. Single Atoms of an Element – microscopic
form of an element
Sample of an Element Large Enough to
Weigh on a Balance –
Contains many, many atoms of the element and are
the macroscopic form of an element
Some Elements that Contain Molecules
Rather than Individual Atoms – macroscopic
form of elements like oxygen, hydrogen Nitrogen,
chlorine, bromine and iodine exist as O2 , H2 , N2 , Cl2 ,
Br2,F2 and I2
101. Names for Elements
The names of chemical elements come from a variety of
sources
- they are often derived from Greek, Latin or German words
that describe some property of the element
ie. Gold- originally called aurum (Latin for “shinning dawn”)
Lead- called plumbum (heavy)
Names of Cl2(chlorine) and I2(iodine) come
from Greek words describing their colors
Bromine- derived from a Greek word for “stench”
Some elements are named for the place they were
discovered: Francium, Germanium, Californium and
Americium
102. Element Symbols
Element symbols are abbreviations for
element names.
Consist of the first letter or first two letters
of the elements name
The first letter is always capitalized!
The second letter (if present) is never
capitalized.
103. Element Symbols
• Fluorine – F
• Oxygen – O
• Neon – Ne
• Silicon – Si
Some have two letters
which are not the first
and second letters in
the name
• Zinc – Zn
• Chlorine – Cl
• Cadmium – Cd
• Platinum - Pt
104. More Element Symbols
Some other elements
are based on the Latin
or Greek name.
• Gold (aurum) - Au
• Lead (plumbum)- Pb
• Sodium (natrium) - Na
• Iron (ferrum) - Fe
• All 118 elements and
their symbols are found
in the inside front cover
of your text….as well as
in the Periodic Table
• We will look at the
Periodic Table in detail
in chapters to come
105. Atomic Theory
As scientists of the 18th
Century studied the nature
of materials, several things became clear:
1. Most natural materials are mixtures of pure substances
2. Pure substances are either elements or combinations of elements
called compounds
3. A given compound always contains the same proportions by mass of its
elements
ie. H2O always contains 8g of oxygen for every 1g of hydrogen and
CO2always contains 2.7g of oxygen for every 1g of carbon
- this is known as the Law of Constant Composition
- this law says that a given compound has the same composition
regardless of where it comes from
106. John Dalton
• An English scientist and teacher
• He knew of these observations and offered an
explanation for them
• His explanation is known as:
107. DALTON’S ATOMIC THEORY
The main ideas of his theory include:
1. Elements are made of tiny particles called atoms.
2. All atoms of a given element are identical. Atoms of a given
element are different from those of any other element.
3. Atoms of one element can combine with atoms of other
elements to form compounds. A given compound always
has the same relative number and types of atoms
4. Atoms are indivisible in chemical processes.
** Atoms are not created or destroyed in chemical reactions.
A chemical reaction simply changes the way atoms are
grouped together.
108. * Dalton’s Theory offered simple explainations for some basic
laws of chemistry such as:
The Law of Conservation of Mass
Mass is neither created or destroyed.
If atoms are conserved in a reaction then mass must also be
conserved
The Law of Constant Composition
Tells us that a cmpd regardless of its origin or method of
preparation always contains the same elements in the same
proportions by weight
Law of Multiple Proportions
When 2 elements combine to form more than 1 cmpd the
masses of one element which combines with a fixed mass of
the other elelment are in a ratio of small whole numbers
such as 2:1
