Matter and Its Classifications
Matter
o
Anything that has mass and occupies space
Mass
How much matter given object has
o Measure of object’s momentum, or resistance to change in motion
o
Weight
o
Force with which object is attracted by gravity
Example: Mass versus Weight
Astronaut on moon and on earth
o
Weight on moon = 1/ 6 weight on earth
o
Same mass regardless of location
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1
Matter and Elements
Chemical Reactions
o
Transformations that alter chemical compositions of
substances
Decomposition
o
A subset of chemical reactions
o
Chemical reaction where one substance is broken down into
two or more simpler substances
Example:
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2
Elements
• Substances that can’t be decomposed into simpler
materials by chemical reactions
• Substances composed of only one type of atom
• Simplest forms of matter that we can work with directly
• More complex substances composed of elements in
various combinations
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3
Chemical Symbols for Elements (1 of 2)
Chemical Symbol
o
One or two letter symbol for each element name
o
First letter capitalized, second letter lower case
Example: C = carbon
S = sulfur
Ca = calcium Ar = argon
Br = bromine H = hydrogen
Cl = chlorine O = oxygen
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Chemical Symbols for Elements (2 of 2)
•
Used to represent elements in chemical formulas
Example: Water = H2O
Carbon dioxide = CO2
o
Most based on English name
o
Some based on Latin or German names
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Chemical Symbols
English Name
Chemical Symbol
Latin Name
Sodium
Na
Natrium
Potassium
K
Kalium
Iron
Fe
Ferrum
Copper
Cu
Cuprum
Silver
Ag
Argentum
Gold
Au
Aurum
Mercury
Hg
Hydrargyrum
Antimony
Sb
Stibium
Tin
Sn
Stannum
Lead
Pb
Plumbum
Tungsten
W
Wolfram (German)
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Compound
• Formed from two or more atoms of different elements
• Always combined in same fixed ratios by mass
• Can be broken down into elements by some chemical
changes
Example: Water decomposed to elemental hydrogen and
oxygen
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Pure Substance versus Mixture
Pure substances
o
o
Elements and compounds
Composition always same regardless of source
Mixture
o
o
Can have variable compositions
Made up of two or more substances
Example: CO2 in water—varying amounts of “fizz” in soda
o
Two broad categories of mixtures:
•
Heterogeneous
•
Homogeneous
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Homogeneous Mixtures (1 of 2)
• Same properties throughout sample
• Solution
Example:
o
o
Liquid solution
•
Sugar in water
•
Coca-Cola (without ice)
Gas solution
•
Air
•
Contains nitrogen, oxygen, carbon dioxide and other gases
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Homogeneous Mixtures (2 of 2)
o
Solid solution
•
US 5¢ coin – Metal alloy
•
Contains copper and nickel metals
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Heterogeneous Mixtures
• Two or more regions of different properties
• Solution with multiple phases
• Separate layers
Example:
o
Salad dressing
•
o
Oil and vinegar
Ice and water
•
Same composition
•
Two different physical states
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11
Relationship of Elements, Compounds,
and Substances
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Learning Check: Classification
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13
Your Turn! 1 (1 of 2)
Coffee is a homogeneous mixture. When you add sugar to
it and the sugar dissolves…
A. the coffee becomes a heterogeneous mixture.
B. the coffee is still a homogeneous mixture.
C. the coffee, which used to be a pure compound, is no
longer a pure compound.
D. the coffee, which used to be an element, is no longer
an element.
E. the sugar reacts with water to form a new compound.
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Your Turn! 1 (2 of 2)
Coffee is a homogeneous mixture. When you add sugar to
it and the sugar dissolves…
A. Answer: the coffee becomes a heterogeneous
mixture.
B. the coffee is still a homogeneous mixture.
C. the coffee, which used to be a pure compound, is no
longer a pure compound.
D. the coffee, which used to be an element, is no longer
an element.
E. the sugar reacts with water to form a new compound.
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Physical Change
• No new substances formed
• Substance may change state or the proportions
Example: Ice melting
• Sugar or salt dissolving
• Stirring iron filings and sulfur together
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States of Matter (1 of 2)
Solids:
o
Fixed shape and volume
o
Particles are close together
o
Have restricted motion
Liquids:
o
Fixed volume, but take container
shape
o
Particles are close together
o
Are able to flow
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States of Matter (2 of 2)
Gases:
o
Expand to fill entire container
o
Particles separated by lots of space
Example: Ice, water, steam
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Chemical Change or Chemical Reaction
• Formation of new substance or compound
• Involves changing chemical makeup of substances
• New substance has different physical properties
• Can’t be separated by physical means
Example:
• Fool’s gold
• Compound containing sulfur and iron
o No longer has same physical properties of free elements
o Can’t be separated using magnet
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19
Learning Check 1:
• For each of the following, determine if it represents a
chemical or physical change:
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Your Turn! 2 (1 of 2)
Which one of the following represents a physical change?
A. when treated with bleach, some dyed fabrics change
color
B. grape juice left in an open unrefrigerated container
turns sour
C. when heated strongly, sugar turns dark brown
D. in cold weather, water condenses on the inside
surface of single pane windows
E. when ignited with a match in open air, paper burns
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Your Turn! 2 (2 of 2)
Which one of the following represents a physical change?
A. when treated with bleach, some dyed fabrics change
color
B. grape juice left in an open unrefrigerated container
turns sour
C. when heated strongly, sugar turns dark brown
D. Answer: in cold weather, water condenses on the
inside surface of single pane windows
E. when ignited with a match in open air, paper burns
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Physical Properties
Properties that can be observed without changing the
chemical makeup of a substance
Example:
o
Color
o
Electrical conductivity
o
Melting point and boiling point
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Chemical Properties
• Describes how a substance undergoes a chemical reaction –
its ‘reactivity’
o
o
o
Involves changing chemical makeup of substances
New substance has different physical properties
Can’t be separated by physical means
Example:
o
o
Fool’s gold
Compound containing sulfur and iron
• No longer has same physical properties of free elements
• Can’t be separated using magnet
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Intensive versus Extensive Properties
Intensive properties
o
o
Independent of sample size
Used to identify substances
Example: Color
Density
Boiling point
Melting point
Chemical reactivity
Extensive properties
o
Depend on sample size
Example: volume and mass
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Your Turn! 3 (1 of 2)
Which of the following is an extensive property?
A. Density
B. Melting point
C. Color
D. Temperature
E. Mass
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26
Your Turn! 3 (2 of 2)
Which of the following is an extensive property?
A. Density
B. Melting point
C. Color
D. Temperature
E. Answer: Mass
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Observations
• Fall into two categories
1. Quantitative observations
o
Numeric data
o
Measure with instrument
Example: Melting point, boiling point, volume, mass
2. Qualitative observations
o
Do not involve numerical information
Example: Color, rapid boiling, white solid forms
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Measurements Include Units
1. Measurements involve comparison
o
o
Always measure relative to reference
Example: Foot, meter, kilogram
Measurement = number + unit
Example: Distance between 2 points = 25
• What unit? inches, feet, yards, miles
• Meaningless without units
2. Measurements are inexact
o
Measuring involves estimation
Always have uncertainty
o
The observer and instrument have inherent physical limitations
o
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International System of Units (SI)
• Standard system of units used in scientific and engineering measurements
• Metric
• Seven Base Units
TABLE 1.2 The SI Base Units
Measurement
Unit
Abbreviation
Length
meter
m
Mass
kilogram
kg
Time
second
s
Electric current
ampere
A
Temperature
kelvin
K
Amount of substance
mole
mol
Luminous intensity
candela
cd
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SI Units
• Focus on first six in this book
• All physical quantities will have units derived from
these seven SI base units
Example: Area
o Derived from SI units based on definition of area
o length × width = area
o meter × meter = area
m  m = m2
o SI unit for area = square meters = m 2
Note: Units undergo same kinds of mathematical
operations that numbers do
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Learning Check 2
• What is the SI derived unit for velocity?
distance
Velocity (v) =
time
meters m
Velocity units =
=
seconds s
• What is the SI derived unit for volume of a cube?
Volume (V) = length × width × height
V = meter × meter × meter
V = m3
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Your Turn! 4 (1 of 2)
The SI unit of length is the
A. millimeter
B. meter
C. yard
D. centimeter
E. foot
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Your Turn! 4 (2 of 2)
The SI unit of length is the
A. millimeter
B. Answer: meter
C. yard
D. centimeter
E. foot
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Your Turn! 5 (1 of 2)
What is the SI derived unit for acceleration
( hint: acceleration = distance/time ) ?
2
A.
B.
C.
D.
E.
mm min
yd hr 2
m s2
ms
ft 3
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Your Turn! 5 (2 of 2)
What is the SI derived unit for acceleration
( hint: acceleration = distance/time ) ?
2
A. mm min
B. yd hr 2
C. Answer: m s 2
D. m s
E. ft 3
Copyright ©2014 John Wiley & Sons, Inc.
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Some Non-SI Metric Units Commonly
Used in Chemistry
TABLE 1.3 Some Non-SI Metric Units Commonly Used in Chemistry
Measurement
Unit
Abbreviation
Value in SI Units
Length
angstrom
A
1A = 0.1 nm = 10−10 m
Mass
atomic mass unit
u (amu)
metric ton
t
1t =103 kg
minute
min.
1 min. = 50 s
hour
h
1 h = 60 min. = 360 s
Temperature
degree Celsius
°C
TK = tC + 273.15
Volume
liter
L
1L = 1000 cm3
Time
Copyright ©2014 John Wiley & Sons, Inc.
1 u = 1.66054  10 −27 kg
( rounded to six digits )
37
Some Useful Conversions
TABLE 1.4 Some Useful Conversions
Measurement
English Unit
English/SI Equalitya
Length
inch
1 in. = 2.54 cm
yard
1 yd = 0.9144 m
mile
1 mi = 1.609 km
pound
1 lb = 453.6 g
ounce (mass)
1 oz = 28.35 g
gallon
1 gal = 3.785 L
quart
1 qt = 946.4 mL
ounce (fluid)
1 oz = 29.6 mL
Mass
Volume
aThese
equalities allow us to convert English to metric or metric to
English units.
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Using Decimal Multipliers
• Use prefixes on SI base units when number is too large
or too small for convenient usage
• Only commonly used are listed here
o
For more complete list see Table 1.5 in textbook
• Numerical values of multipliers can be interchanged
with prefixes
Example: 1mL = 10−3 L
o
1 km = 1000 m
−9
1ng
=
10
g
o
o 1,130, 000 m =1.13  106 m = 1.13 Mm
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Decimal Multipliers
TABLE 1.5 SI Prefixes—Their Meanings and Valuesa
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Laboratory Measurements (1 of 2)
• Four common
1. Length
2. Volume
3. Mass
4. Temperature
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Laboratory Measurements (2 of 2)
1. Length
• SI Unit is meter (m)
• Meter too large for most laboratory measurements
• Commonly use
o
Centimeter (cm)
• 1cm = 10−2 m = 0.01m
o
Millimeter (mm)
• 1mm = 10 −3 m = 0.001m
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2. Volume
( length )
• Dimensions of
• SI unit for Volume = m3
• Most laboratory measurements
use V in liters (L)
3
o 1L = 1 dm 3 ( exactly )
• Chemistry glassware marked in L or mL
o
1 L = 1000 mL
• What is a mL?
o 1mL = 1cm 3
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3. Mass
• SI unit is kilogram (kg)
Frequently use grams (g) in laboratory as more realistic size
1
1kg
=
1000g
1g
=
0.1000
kg
=
g
•
1000
o
• Mass is measured by comparing weight of sample with
weights of known standard masses
• Instrument used = balance
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4. Temperature (1 of 3)
• Measured with thermometer
• Three common scales
A. Fahrenheit scale
o
o
Common in US
Water freezes at 32 °F and boils at 212 °F
o
180 degree units between melting and boiling points of water
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4. Temperature (2 of 3)
B. Celsius scale
Most common for use in science
o Water freezes at 0 °C
o Water boils at 100 °C
o
o
100 degree units between melting and boiling points of water
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4. Temperature (3 of 3)
C. Kelvin scale
o
SI unit of temperature is kelvin (K)
•
o
Water freezes at 273.15 K and boils at 373.15 K
•
o
Note: No degree symbol in front of K
100 degree units between melting and boiling points
Only difference between Kelvin and Celsius scale is zero
point
Absolute Zero
o
o
Zero point on Kelvin scale
Corresponds to nature’s lowest possible temperature
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Temperature Conversions (1 of 2)
How to convert between °F and °C?
9 F
tF = 
 tC + 32 F
5 C
Example: 100 °C = ? °F
9 F
tF = 
100 C + 32 F
5 C
tF = 212 °F
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Temperature Conversions (2 of 2)
• Common laboratory thermometers are marked in
Celsius scale
• Must convert to Kelvin scale
1K
TK = (tC + 273.15 C)
1 C
Amounts to adding 273.15 to Celsius temperature
Example: What is the Kelvin temperature of a solution
at 25 °C?
1K
TK = (25 C + 273.15 C)
1 C
Copyright ©2014 John Wiley & Sons, Inc.
= 298 K
49
Learning Check: T Conversions (1 of 2)
1. Convert 121 °F to the Celsius scale.
 9 °F 
tF = 
 tC + 32 °F
 5 °C 
 5 °C 
tC = ( tF − 32 °F ) 

9
°F


 5 °C 
tC = (121 o F − 32 °F ) 
 = 49C
 9 °F 
2. Convert 121 °F to the Kelvin scale.
• We already have in °C so…
TK = (tC + 273.15 °C)
1K
1K
= (49 + 273.15 °C)
1 °C
1 °C
Copyright ©2014 John Wiley & Sons, Inc.
TK = 332 K
50
Learning Check: T Conversions (2 of 2)
3. Convert 77 K to the Celsius scale.
1K
TK = (tC + 273.15 °C)
1 °C
tC = (77 K − 273.15 K)
1 °C
1K
tC = (TK − 273.15 K)
1 °C
1K
= –196 °C
4. Convert 77 K to the Fahrenheit scale.
o
We already have in °C so
 9 °F 
tF = 
 ( − 196 °C) + 32 °F
 5 °C 
= –321 °F
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Your Turn! 6 (1 of 2)
In a recent accident some drums of uranium hexafluoride
were lost in the English Channel. The melting point of
uranium hexafluoride is 64.53 °C. What is the melting
point of uranium hexafluoride on the Fahrenheit scale?
A. 67.85 °F
B. 96.53 °F
C. 116.2 °F
D. 337.5 °F
E. 148.2 °F
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Your Turn! 6 (2 of 2)
In a recent accident some drums of uranium hexafluoride
were lost in the English Channel. The melting point of
uranium hexafluoride is 64.53 °C. What is the melting
point of uranium hexafluoride on the Fahrenheit scale?
A. 67.85 °F
B. 96.53 °F
C. 116.2 °F
D. 337.5 °F
E. Answer: 148.2 °F
 9 °F 
tF = 
 tC + 32 °F
 5 °C 
 9 °F 
tF = 
 64.53 °C + 32 °F
 5 °C 
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Your Turn! 7 (1 of 2)
On an absolute temperature scale, 100 °F is not double
50 °F (i.e., not ‘twice as hot’). What temperature, in °F,
would really be double 50 °F (hint: convert 50 °F to K,
double the Kelvin temperature, then convert back to °F)?
A. 560 °F
B. 25 °F
C. 200 °F
D. 283.15 °F
E. 566 °F
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Your Turn! 7 (2 of 2)
On an absolute temperature scale, 100 °F is not double
50 °F (i.e., not ‘twice as hot’). What temperature, in °F,
would really be double 50 °F (hint: convert 50 °F to K,
double the Kelvin temperature, then convert back to °F)?
A. Answer: 560 °F
B. 25 °F
C. 200 °F
D. 283.15 °F
E. 566 °F
 5 °C 
tC = ( 50 °F − 32 °F ) 
 = 10 °C
 9 °F 
10 o C + 273.15 K = 283.15 K
Doubled is 566.3 K
566.3 K − 273.15 K = 293.15 o C
 9 °F 
tF = 
 293.15 °C + 32 °F = 560 °F
 5 °C 
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Uncertainties in Measurements
• Measurements all inexact
o
Contain uncertainties or errors
• Sources of errors
o
Limitations of reading instrument
• Ways to minimize errors
o
Take series of measurements
Data clusters around central value
o
Calculate average or mean values
o
Report average value
o
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Limits in Reading Instruments (1 of 2)
• Consider two Celsius thermometers
• Left thermometer has markings every 1°C
o
o
o
o
T between 24 °C and 25 °C
About 3 10 of way between marks
Can estimate to 0.1 °C = uncertainty
T = 24.3  0.1 °C
• Right thermometer has markings every 0.1 °C
o
o
o
T reading between 24.3 °C and 24.4 °C
Can estimate 0.01 °C
T = 24.32  0.01 °C
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Limits in Reading Instruments (2 of 2)
• Finer graduations in markings
o
Means smaller uncertainties in measurements
• Reliability of data
o
Indicated by number of digits used to represent it
• What about digital displays?
Mass of beaker = 65.23 g on digital balance
o Still has uncertainty
o
o
Assume 1 2 in last readable digit
o
Record as 65.230  0.005 g
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Significant Figures
• Scientific convention:
o
All digits in measurement up to and including first
estimated digit are significant.
• Number of certain digits plus first uncertain digit
• Digits in measurement from first non-zero number on
left to first estimated digit on right
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Rules for Significant Figures (1 of 2)
1. All non-zero numbers are significant.
Example: 3.456 has 4 sig. figs.
3. Zeros between non-zero numbers are significant.
Example: 20, 089 or 2.0089  104 has 5 sig.figs
3. Trailing zeros always count as significant if number has
decimal point
Example: 500 or 5.00  102 has 3 sig.figs
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Rules for Significant Figures (2 of 2)
4. Final zeros on number without decimal point are NOT
significant
Example: 104,956, 000 or 1.04956  108 has 6 sig.figs
5. Final zeros to right of decimal point are significant
Example: 3.00 has 3 sig. figs.
6. Leading zeros, to left of first nonzero digit, are never
counted as significant
Example: 0.00012 or 1.2  10−4 has 2 sig.figs
Copyright ©2014 John Wiley & Sons, Inc.
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Learning Check 3
How many significant figures does each of the following
numbers have?
Scientific Notation
# of Sig. Figs.
1. 413.97
4.1397 10 2
5
2. 0.0006
6 10−4
1
3. 5.120063
5.120063
7
4. 161,000
1.61105
3
5. 3600.
3.600 103
4
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Your Turn! 8 (1 of 2)
How many significant figures are in 19.0000?
A. 2
B. 3
C. 4
D. 5
E. 6
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Your Turn! 8 (2 of 2)
How many significant figures are in 19.0000?
A. 2
B. 3
C. 4
D. 5
E. Answer: 6
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Your Turn! 9 (1 of 2)
How many significant figures are in 0.0005650850?
A. 7
B. 8
C. 9
D. 10
E. 11
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Your Turn! 9 (2 of 2)
How many significant figures are in 0.0005650850?
A. Answer: 7
B. 8
C. 9
D. 10
E. 11
• Could be rewritten as 5.650850 10−4
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Accuracy and Precision
Accuracy
o
o
How close measurement is to true or
accepted true value
Measuring device must be calibrated with
standard reference to give correct value
Precision
o
o
How well set of repeated
measurements of same quantity
agree with each other
More significant figures equals
more precise measurement
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Rounding to Correct Digit
1. If digit to be dropped is greater than 5, last remaining digit is
rounded up.
Example: 3.677 is rounded up to 3.68
2. If number to be dropped is less than 5, last remaining digit stays
the same.
Example: 6.632 is rounded to 6.63
3. If number to be dropped is exactly 5, then if digit to left of 5 is
a. Even, it remains the same.
Example: 6.65 is rounded to 6.6
b. Odd, it rounds up.
Example: 6.35 is rounded to 6.4
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Scientific Notation
• Clearest way to present number of significant figures
unambiguously
o
o
Report number between 1 and 10 followed by correct power of
10
Indicates only significant digits
Example: 75,000 people attend a concert
o
o
If a rough estimate
• Uncertainty  1000 people
• 7.50  10 4
If number estimated from aerial photograph
• Uncertainty 100 people
• 7.50  10 4
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69
Learning Check 4
Round each of the following to three significant figures.
Use scientific notation where needed.
1. 37.459
2. 5431978
3. 132.7789003
4. 0.00087564
5. 7.665
Copyright ©2014 John Wiley & Sons, Inc.
70
Learning Check 5
Round each of the following to three significant figures.
Use scientific notation where needed.
1. 37.459
2. 5431978
37.5 or 3.75  101
5.43  106
3. 132.7789003
4. 0.00087564
133 or 1.33  102
5. 7.665
7.66
8.77  10−4
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71
Your Turn! 10 (1 of 2)
Round 0.00564458 to four significant figures and express
using scientific notation.
A. 5.64 10−2
B. 5.000  10−3
C. 5.645 10−4
D. 0.56446
E. 5.645  10−3
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72
Your Turn! 10 (2 of 2)
Round 0.00564458 to four significant figures and express
using scientific notation.
A. Answer: 5.64 10−2
B. 5.000  10−3
C. 5.645 10−4
D. 0.56446
E. 5.645  10−3
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73
Significant Figures in Calculations (1 of 2)
Multiplication and Division
Number of significant figures in answer = number of
significant figures in least precise measurement
Example: 10.54 × 31.4 × 16.987 = 5620 = 5.62×10 3
4 sig. figs. × 3 sig. figs. × 5 sig. figs. = 3 sig. figs.
Example: 5.896 ÷ 0.008 = 700 = 7 ×102
4 sig. figs. ÷ 1 sig. fig. = 1 sig. fig.
Copyright ©2014 John Wiley & Sons, Inc.
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Your Turn! 11 (1 of 2)
Give the value of the following calculation to the correct
number of significant figures.
 635.4  0.0045 


2.3589


A. 1.21213
B. 1.212
C. 1.212132774
D. 1.2
E. 1
Copyright ©2014 John Wiley & Sons, Inc.
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Your Turn! 11 (2 of 2)
Give the value of the following calculation to the correct
number of significant figures.
 635.4  0.0045 


2.3589


A. 1.21213
B. 1.212
C. 1.212132774
D. Answer: 1.2
E. 1
Copyright ©2014 John Wiley & Sons, Inc.
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Significant Figures in Calculations (2 of 2)
Addition and Subtraction
• Answer has same number of decimal places as quantity with
fewest number of decimal places.
Example:
12.9753
Example:
4 decimal places
319.5
1 decimal place
+ 4.398
3 decimal places
336.9
1 decimal place
397
0 decimal places
– 273.15
2 decimal places
124
0 decimal place
Copyright ©2014 John Wiley & Sons, Inc.
77
Learning Check 6
For each calculation, give the answer to the correct
number of significant figures.
1. 10.0g +1.03 g + 0.243 g =
2. 19.556 °C – 19.552 °C =
3. 327.5 m × 4.52 m =
4. 15.985 g ÷ 24.12 mL =
Copyright ©2014 John Wiley & Sons, Inc.
78
Learning Check 7
For each calculation, give the answer to the correct
number of significant figures.
1. 10.0g +1.03 g + 0.243 g = 11.3 g or 1.13×101 g
2. 19.556 °C – 19.552 °C =
0.004 °C or 4×10 −3 °C
3. 327.5 m × 4.52 m = 1.48×103 m
4. 15.985 g ÷ 24.12 mL = 0.6627 g / mL or
6.627 × 10 –1 g / mL
Copyright ©2014 John Wiley & Sons, Inc.
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Your Turn! 12 (1 of 2)
When the expression,
412.272 + 0.00031 − 1.00797 + 0.000024 + 12.8
is evaluated, the result should be expressed as:
A. 424.06
B. 424.064364
C. 424.1
D. 424.064
E. 424
Copyright ©2014 John Wiley & Sons, Inc.
80
Your Turn! 12 (2 of 2)
When the expression,
412.272 + 0.00031 − 1.00797 + 0.000024 + 12.8
is evaluated, the result should be expressed as:
A. 424.06
B. 424.064364
C. Answer: 424.1
D. 424.064
E. 424
Copyright ©2014 John Wiley & Sons, Inc.
81
Your Turn! 13 (1 of 2)
When the expression,
152.6736 − 0.0028 − 125.00797 + 56.795 + 12.853
is evaluated, the result should be expressed to the:
A. tenths place
B. hundredths place
C. thousandths place
D. ten-thousands place
E. hundred-thousands place
Copyright ©2014 John Wiley & Sons, Inc.
82
Your Turn! 13 (2 of 2)
When the expression,
152.6736 − 0.0028 − 125.00797 + 56.795 + 12.853
is evaluated, the result should be expressed to the:
A. tenths place
B. hundredths place
C. Answer: thousandths place
D. ten-thousands place
E. hundred-thousands place
Copyright ©2014 John Wiley & Sons, Inc.
83
Learning Check 8
For the following calculation, give the answer to the
correct number of significant figures.
(0.002 m)
(71.359 m − 71.357 m)
1.
=
(3.2 s×3.67 s)
(11.744 s 2 )
= 2×10−4 m/s 2
2.
(13.674 cm× 4.35 cm× 0.35 cm)
(856 s +1531.1 s)
(20.818665 cm3 )
=
(2387.1 s)
= 0.0087cm 3 / s
Copyright ©2014 John Wiley & Sons, Inc.
84
Exact Numbers
• Numbers that come from definitions
o
12 in. = 1 ft
o
60 s = 1 min
• Numbers that come from direct count
o
Number of people in small room
• Have no uncertainty
• Assume they have infinite number of significant
figures
• Do not affect number of significant figures in
multiplication or division
Copyright ©2014 John Wiley & Sons, Inc.
85
Your Turn! 14 (1 of 2)
Express the answer to the following with the correct
number of significant figures:
 ft. 
36.00 in× 
=
 12 in. 
A. 3 ft.
B. 3.0 ft.
C. 3.00 ft.
D. 3.000 ft.
E. 3.00000 ft.
Copyright ©2014 John Wiley & Sons, Inc.
86
Your Turn! 14 (2 of 2)
Express the answer to the following with the correct
number of significant figures:
 ft. 
36.00 in× 
=
 12in. 
A. 3 ft.
B. 3.0 ft.
C. 3.00 ft.
D. Answer: 3.000 ft.
36.00 has four sig. figs.
The conversion is exact.
So the answer has four sig.
figs.
E. 3.00000 ft.
Copyright ©2014 John Wiley & Sons, Inc.
87
Your Turn! 15 (1 of 2)
For the following calculation, give the answer to the
correct number of significant figures.
(14.5 cm × 12.334 cm )
( 2.223 cm − 1.04 cm )
A. 179 cm 2
B. 1.18 cm
C.
D.
151.2 cm
151 cm
E.
178.843 cm 2
Copyright ©2014 John Wiley & Sons, Inc.
88
Your Turn! 15 (2 of 2)
For the following calculation, give the answer to the
correct number of significant figures.
(14.5 cm × 12.334 cm )
( 2.223 cm − 1.04 cm )
A. 179 cm 2
B. 1.18 cm
2
178.843
cm
(
)
C. 151.2 cm
D. Answer: 151 cm
E.
(1.183 cm )
178.843 cm 2
Copyright ©2014 John Wiley & Sons, Inc.
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Dimensional Analysis (1 of 2)
• Also called the Factor Label Method
• Not all calculations use specific equation
• Use units (dimensions) to analyze problem
Conversion Factor
• Fraction formed from valid equality or equivalence
between units
• Used to switch from one system of measurement
and units to another
Given Quantity × Conversion Factor = Desired Quantity
Copyright ©2014 John Wiley & Sons, Inc.
90
Conversion Factors
Example: How to convert a person’s height from
68.0 in to cm?
• Start with fact
2.54 cm = 1 in. (exact)
• Dividing both sides by 1 in. or 2.54 cm gives 1
2.54 cm 1 in.
×
=1
1 in.
1 in.
2.54 cm
1 in.
×
=1
2.54 cm 2.54 cm
• Cancel units
• Leave ratio that equals 1
• Use fact that units behave as numbers do in
mathematical operations
Copyright ©2014 John Wiley & Sons, Inc.
91
Dimensional Analysis (2 of 2)
• Now multiply original number by conversion factor
that cancels old units and leaves new
Given Quantity × Conversion Factor = Desired Quantity
2.54cm
68.0 in. ×
= 173 cm
1 in.
• Dimensional analysis can tell us when we have done
wrong arithmetic
68.0 in.×
o
1 in.
= 26.8 in 2 /cm
2.54 cm
Units not correct
Copyright ©2014 John Wiley & Sons, Inc.
92
Using Dimensional Analysis
Example: Convert 0.097 m to mm.
• Relationship is 1 mm = 1×10−3 m
• Can make two conversion factors
1mm
1×10−3 m
1×10−3 m
1 mm
• Since going from m to mm use one on left.
1mm
0.097 m ×
= 97 cm
−3
110 m
Copyright ©2014 John Wiley & Sons, Inc.
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Learning Check 9
Example: Convert 3.5 m3 to cm3 .
• Start with basic equality 1 cm = 0.01 m
• Now cube both sides
o Units and numbers
3
3
1cm
=
0.01m
) (
)
o (
o 1 cm3 =1×10−6 m3
• Can make two conversion factors
1cm3
1×10−6 m3
or
1×10−6 m3
1 cm3
3
3.5 m ×
1cm3
1  10−6 m3
= 3.5  106 cm 3
Copyright ©2014 John Wiley & Sons, Inc.
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Non-metric to Metric Units
Convert speed of light from 3.00  108 m s to mi hr
• Use dimensional analysis
• 1 min = 60 s
60 min = 1 hr
• 1 km = 1000 m
1 mi = 1.609 km
3.00 ×108 m 60 s 60 min
12
×
×
= 1.08×10 m/hr
1hr
s
1 min
12
1.08×10 m
1 km
1 mi
8
×
×
= 6.71× 10 mi/hr
hr
1000 m 1.609 km
Copyright ©2014 John Wiley & Sons, Inc.
95
Your Turn! 16 (1 of 2)
The Honda Insight hybrid electric car has a gas
mileage rating of 56 miles to the gallon. What is this
rating expressed in units of kilometers per liter?
1 gal = 3.784 L 1 mile = 1.609 km
A. 1.3×10 2 km L−1
B. 24 km L−1
C. 15 km L−1
D. 3.4 102 km L−1
E. 9.2 km L−1
Copyright ©2014 John Wiley & Sons, Inc.
96
Your Turn! 16 (2 of 2)
The Honda Insight hybrid electric car has a gas
mileage rating of 56 miles to the gallon. What is this
rating expressed in units of kilometers per liter?
1 gal = 3.784 L 1 mile = 1.609 km
A. 1.3×10 2 km L−1
−1
B. Answer: 24 km L
C. 15 km L−1
D. 3.4 102 km L−1
E. 9.2 km L−1
1gal
mi
1.609km
56
×
×
gal 3.784 L
1 mi
Copyright ©2014 John Wiley & Sons, Inc.
97
Your Turn! 17 (1 of 2)
The volume of a basketball is 433.5in 3 . Convert this to
mm3 .
-2
3
A. 1.101×10 mm
B. 7.104 ×106 mm3
4
3
7.104
×10
mm
C.
D. 1.101×104 mm 3
E. 1.101×106 mm 3
Copyright ©2014 John Wiley & Sons, Inc.
98
Your Turn! 17 (2 of 2)
The volume of a basketball is 433.5in 3 . Convert this to
mm3 .
-2
3
A. 1.101×10 mm
B. Answer: 7.104 ×106 mm3
C. 7.104 ×104 mm3
D. 1.101×104 mm 3
E. 1.101×106 mm3
 2.54 cm
433.5 in3  
in

3
 
m

 
  100 cm
3
  1000 mm 
  

m

 
Copyright ©2014 John Wiley & Sons, Inc.
3
99
Density (1 of 2)
• Ratio of object’s mass to its volume
mass
density=
volume
m
d=
V
• Intensive property (size independent)
o
Determined by taking ratio of two extensive properties
(size dependent)
o
Frequently ratio of two size dependent properties leads to
size independent property
Sample size cancels
o
• Units
3
o g/ml or g/cm
Copyright ©2014 John Wiley & Sons, Inc.
100
Learning Check 10
• A student weighs a piece of gold that has a volume
of 11.02 cm3 of gold. She finds the mass to be 212 g.
What is the density of gold?
m
d=
V
d=
212 g
3
=
19.3
g/cm
11.02 cm3
Another student has a piece of gold with a
volume of 1.00 cm3 . What does it weigh? 19.3 g
What if it were 2.00 cm3 in volume? 38.6 g
Copyright ©2014 John Wiley & Sons, Inc.
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Density (2 of 2)
• Most substances expand slightly when heated
o Same mass
o Larger volume
o Less dense
• Density decreases slightly as temperature increases
• Liquids and solids
o Change is very small
Can ignore except in very precise calculations
• Density useful to transfer between mass and volume of
substance
o
Copyright ©2014 John Wiley & Sons, Inc.
102
Learning Check 11
1. Glass has a density of 2.2 g/cm3 . What is the volume
occupied by 35 g of glass?
V =
m
35 g
=
= 16. g
3
d 2.2 g / cm
3
2. What is the mass of 4.00 cm of glass?
m = d  V = 2.2 g / cm3 × 400.cm3 = 8.8 × 102 cm 3
Copyright ©2014 John Wiley & Sons, Inc.
103
Your Turn! 18 (1 of 2)
Titanium is a metal used to make artificial joints. It
has a density of 4.54 g/cm3 . What volume will a
titanium hip joint occupy if its mass is 205 g?
A. 9.31×102 cm3
1
3
4.51×10
cm
B.
C. 2.21×10−2 cm3
D. 1.07×10−3 cm 3
E. 2.20×10−1 cm3
Copyright ©2014 John Wiley & Sons, Inc.
104
Your Turn! 18 (2 of 2)
Titanium is a metal used to make artificial joints. It
has a density of 4.54 g/cm3 . What volume will a
titanium hip joint occupy if its mass is 205 g?
A. 9.31×102 cm3
1
3
4.51×10
cm
B. Answer:
205g
V =
=
3
4.54 g / cm
C. 2.21×10−2 cm3
D. 1.07×10−3 cm 3
E. 2.20×10−1 cm3
Copyright ©2014 John Wiley & Sons, Inc.
105
Your Turn! 19 (1 of 2)
A sample of zinc metal ( density = 7.14 g cm ) was
-3
submerged in a graduated cylinder containing water.
The water level rose from 162.5 cm3 to 186.0 cm3 when the
sample was submerged. How many grams did the sample
weigh?
3
1.16×10
g
A.
3
1.33×10
g
B.
C. 23.5 g
2
1.68×10
g
D.
E.
3.29 g
Copyright ©2014 John Wiley & Sons, Inc.
106
Your Turn! 19 (2 of 2)
A sample of zinc metal ( density = 7.14 g cm −3 ) was
submerged in a graduated cylinder containing water.
The water level rose from 162.5 cm3 to 186.0 cm3 when the
sample was submerged. How many grams did the sample
weigh?
mass = density × volume
3
1.16×10
g
A.
volume = (186.0cm3 − 162.5cm3 )
3
1.33×10
g
B.
C. 23.5 g
= 23.5cm3
D. Answer: 1.68×102 g
E. 3.29 g
mass = 7.14 g cm −3  23.5 cm3
Copyright ©2014 John Wiley & Sons, Inc.
107
Specific Gravity
• Ratio of density of substance to density of water
density of substance
specific gravity =
density of water
• Unitless
• Way to avoid having to tabulate densities in all sorts
of different units
Copyright ©2014 John Wiley & Sons, Inc.
108
Learning Check 12
Concentrated sulfuric acid is sold in bottles with a label
that states that the specific gravity at 25 °C is 1.84. The
density of water at 25 °C is 0.995g / cm3 . How many
cubic centimeters of sulfuric acid will weigh 5.55
kilograms?
Analysis:
5.55 kg sulfuric acid = ? cm3 sulfuric acid
Solution:
density sulfuric acid = specific gravity × density water
dsulfuric acid = 1.84 × 0.995 g/cm3 = 1.83
Vsulfuric acid
m 5.55  103 g
3
= =
=
3.03
×
10
cm 3
3
d 1.83 g/cm
Copyright ©2014 John Wiley & Sons, Inc.
109
Your Turn! 20 (1 of 2)
Liquid hydrogen has a specific gravity of 7.08 10−2.
If the density of water is 1.05 g/cm3 at the same temperature,
what is the mass of hydrogen in a tank having a volume of
36.9 m3 ?
A.
B.
C.
D.
E.
7.43×10−2 g
2.74 g
274 g
2.74 ×106 g
2.61  106 g
Copyright ©2014 John Wiley & Sons, Inc.
110
Your Turn! 20 (2 of 2)
Liquid hydrogen has a specific gravity of 7.08 10−2.
If the density of water is 1.05 g/cm3 at the same temperature,
what is the mass of hydrogen in a tank having a volume of
36.9 m3 ?
A. 7.43×10−2 g
B. 2.74 g
C. 274 g
D. Answer: 2.74 ×106 g
E. 2.61  106 g
dsulfuric acid = specific gravitysulfuric acid × d H2O
dsulfuric acid = 7.08  10−2  1.05 g / cm3
= 7.43×10−2 g/cm3
g sulfuric acid = 7.43  10−2 g/cm3
Copyright ©2014 John Wiley & Sons, Inc.
3
100
cm


 36.9 m3  

 1m 
111
Importance of Reliable Measurements
To trust conclusions drawn from measurements
• Must know they are reliable
• Must be sure they are accurate
o
o
o
Measured values must be close to true values
Otherwise can’t trust results
Can’t make conclusions based on those results
• Must have sufficient precision to be meaningful
o So confident that two measurements are same
for two samples
o Difference in values must be close to uncertainty in
measurement
Copyright ©2014 John Wiley & Sons, Inc.
112
Learning Check 13 (1 of 2)
You have a ring, is it made of 24 K gold?
• Calculate density and compare to known
o
Density of 24 K gold = 19.3 g/mL
• Use inaccurate glassware
o
Volume of ring = 1.0 mL
• Use kitchen balance
o
Mass of ring = 18 ± 1 g
• Anywhere between 17 and 19 g
•
o
Density range is 17−19 g/mL
Could be 24 K gold or could be as low as 18 K gold
(density = 16.9 g/mL)
Copyright ©2014 John Wiley & Sons, Inc.
113
Learning Check 13 (2 of 2)
• Use more precise laboratory balance
o
Mass of ring = 18.153 ± 0.001 g
• Use more precise glassware
Volume of ring = 1.03 mL
o Density of ring = 18.153 g/1.03 mL = 17.6 g/mL
o
• Calculate difference between d24 K gold and dring
19.3 g/mL − 17.6 g/mL = 1.7 g/mL
o Larger than experimental error in density
o
~ ± 0.1 g/mL
o Conclusion: Ring is NOT 24 K gold!
o
Copyright ©2014 John Wiley & Sons, Inc.
114
Copyright
Copyright © 2014 John Wiley & Sons, Inc.
All rights reserved. Reproduction or translation of this work beyond that permitted in
Section 117 of the 1976 United States Act without the express written permission of the
copyright owner is unlawful. Request for further information should be addressed to the
Permissions Department, John Wiley & Sons, Inc. The purchaser may make back-up
copies for his/her own use only and not for distribution or resale. The Publisher assumes
no responsibility for errors, omissions, or damages, caused by the use of these programs or
from the use of the information contained herein.
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115