Download Chapter 1 Chemical Foundations 1.6 Dimensional Analysis

Chapter 1
Chemical Foundations
1.6
Dimensional Analysis
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
1
Equalities
Equalities
• use two different units to describe the same measured
amount.
• are written for relationships between units of the metric
system, U.S. units, or between metric and U.S. units.
For example,
1m
=
1 lb
=
2.205 lb =
1000 mm
16 oz
1 kg
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Exact and Measured Numbers in
Equalities
Equalities between units of
• the same system are definitions and use exact
numbers.
• different systems (metric and U.S.) use measured
numbers and count as significant figures.
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Some Common Equalities
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Equalities on Food Labels
The contents of packaged foods
• in the U.S. are listed as both metric and
U.S. units.
• indicate the same amount of a substance in
two different units.
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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Conversion Factors
A conversion factor
• is a fraction obtained from an equality.
Equality:
1 in. = 2.54 cm
• is written as a ratio with a numerator and denominator.
• can be inverted to give two conversion factors for
every equality.
1 in.
and 2.54 cm
2.54 cm
1 in.
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Learning Check
Write conversion factors for each pair of units.
A. liters and mL
B. hours and minutes
C. meters and kilometers
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Solution
Write conversion factors for each pair of units.
A. liters and mL
Equality: 1 L = 1000 mL
1L
and
1000 mL
1000 mL
1L
B. hours and minutes
1 hr
60 min
and
Equality: 1 hr = 60 min
60 min
1 hr
C. meters and kilometers
1 km
1000 m
and
Equality: 1 km = 1000 m
1000 m
1 km
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Conversion Factors in a Problem
A conversion factor
• may be obtained from information in a word problem.
• is written for that problem only.
Example 1:
The price of one pound (1 lb) of red peppers is $2.39.
1 lb red peppers and $2.39
$2.39
1 lb red peppers
Example 2:
The cost of one gallon (1 gal) of gas is $2.34.
1 gallon of gas and $2.34
$2.34
1 gallon of gas
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Percent as a Conversion Factor
A percent factor
• gives the ratio of the parts to the whole.
% =
Parts x 100
Whole
• uses the same unit to express the percent.
• uses the value 100 and a unit for the whole.
• can be written as two factors.
Example: A food contains 30% (by mass) fat.
30 g fat
100 g food
and
100 g food
30 g fat
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Percent Factor in a Problem
The thickness of the skin fold at
the waist indicates 11% body
fat. What percent factors can be
written for body fat in kg?
Percent factors using kg:
11 kg fat
and 100 kg mass
100 kg mass
11 kg fat
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
11
Learning Check
Write the equality and conversion factors for each of the
following.
A. square meters and square centimeters
B. jewelry that contains 18% gold
C. One gallon of gas is $2.27
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Solution
A. square meters and square centimeters
(1m)2 and (100 cm)2
(100 cm)2
(1m)2
B. jewelry that contains 18% gold
18 g gold and 100 g jewelry
100 g jewelry
18 g gold
C. One gallon of gas is $2.27
1 gal
and
$2.27
$2.27
1 gal
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Given and Needed Units
To solve a problem
• Identify the given unit
• Identify the needed unit.
Example:
A person has a height of 2.0 meters.
What is that height in inches?
The given unit is the initial unit of height.
given unit = meters (m)
The needed unit is the unit for the answer.
needed unit = inches (in.)
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Learning Check
An injured person loses 0.30 pints of blood. How
many milliliters of blood would that be?
Identify the given and needed units given in this
problem.
Given unit
= _______
Needed unit = _______
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Solution
An injured person loses 0.30 pints of blood. How
many milliliters of blood would that be?
Identify the given and needed units given in this
problem.
Given unit
=
Needed unit =
pints
milliliters
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Problem Setup
• Write the given and needed units.
• Write a unit plan to convert the given unit to the needed
•
•
unit.
Write equalities and conversion factors that connect the
units.
Use conversion factors to cancel the given unit and
provide the needed unit.
Unit 1
x
Given
unit
x
Unit 2
= Unit 2
Unit 1
Conversion = Needed
factor
unit
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Guide to Problem Solving (GPS)
The steps in the
Guide to Problem
Solving are useful
in setting up a
problem with
conversion
factors.
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Setting up a Problem
How many minutes are 2.5 hours?
Given unit =
2.5 hr
Needed unit =
min
Unit Plan
=
hr
min
Setup problem to cancel hours (hr).
Given Conversion
Needed
unit
factor
unit
2.5 hr x 60 min = 150 min (2 SF)
1 hr
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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Learning Check
A rattlesnake is 2.44 m long. How many centimeters
long is the snake?
1) 2440 cm
2) 244 cm
3) 24.4 cm
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Solution
A rattlesnake is 2.44 m long. How many
centimeters long is the snake?
2)
244 cm
Given
Conversion Needed
unit
factor
unit
2.44 m x 100 cm = 244 cm
1m
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Using Two or More Factors
• Often, two or more conversion factors are required
to obtain the unit needed for the answer.
Unit 1
Unit 2
Unit 3
• Additional conversion factors are placed in the
setup to cancel each preceding unit
Given unit x factor 1
Unit 1
x Unit 2
Unit 1
x factor 2
x Unit 3
Unit 2
= needed unit
= Unit 3
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Example: Problem Solving
How many minutes are in 1.4 days?
Given unit: 1.4 days
Factor 1
Plan:
days
Factor 2
hr
Set up problem:
1.4 days x 24 hr x 60 min
1 day
1 hr
2 SF
Exact
Exact
min
= 2.0 x 103 min
= 2 SF
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Check the Unit Cancellation
• Be sure to check your unit cancellation in the setup.
• The units in the conversion factors must cancel to
give the correct unit for the answer.
What is wrong with the following setup?
1.4 day
Units =
x 1 day
24 hr
x
1 hr
60 min
day2/min is not the unit needed
Units don’t cancel properly.
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Using the GPS
What is 165 lb in kg?
STEP 1 Given 165 lb Need kg
STEP 2 Plan
STEP 3 Equalities/Factors
1 kg = 2.20 lb
2.20 lb and
1 kg
1 kg
2.20 lb
STEP 4 Set Up Problem
165 lb x
1 kg = 75.0 kg
2.20 lb
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Learning Check
A bucket contains 4.65 L of water. How many
gallons of water is that?
Unit plan:
L
qt
gallon
Equalities: 1.06 qt = 1 L
1 gal = 4 qt
Set up Problem:
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Solution
Given: 4.65 L
Needed: gallons
Plan:
qt
L
gallon
Equalities: 1.06 qt = 1 L; 1 gal = 4 qt
Set Up Problem:
4.65 L x x 1.06 qt
1L
3 SF
3 SF
x
1 gal
4 qt
exact
= 1.23 gal
3 SF
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Learning Check
If a ski pole is 3.0 feet in length, how long is the
ski pole in mm?
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Solution
3.0 ft x 12 in x 2.54 cm x 10 mm =
1 ft
1 in.
1 cm
Calculator answer:
914.4 mm
Needed answer:
910 mm (2 SF rounded)
Check factor setup:
Check needed unit:
Units cancel properly
mm
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Learning Check
If your pace on a treadmill is 65 meters per minute,
how many minutes will it take for you to walk a
distance of 7500 feet?
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Solution
Given: 7500 ft
65 m/min
Need: min
Plan: ft
in.
cm
m
min
Equalities: 1 ft = 12 in. 1 in. = 2.54 cm 1 m = 100 cm
1 min = 65 m (walking pace)
Set Up Problem:
7500 ft x
12 in. x
1 ft
2.54 cm
1 in.
x 1m
x 1 min
100 cm 65 m
= 35 min final answer (2 SF)
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Percent Factor in a Problem
If the thickness of the skin fold
at the waist indicates an 11%
body fat, how much fat is in a
person with a mass of 86 kg?
percent factor
86 kg mass x
11 kg fat
100 kg mass
= 9.5 kg fat
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
32
Learning Check
How many lb of sugar are in 120 g of candy if the
candy is 25% (by mass) sugar?
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Solution
How many lb of sugar are in 120 g of candy if the
candy is 25%(by mass) sugar?
percent factor
120 g candy x 1 lb candy
454 g candy
x 25 lb sugar
100 lb candy
= 0.066 lb sugar
Ref: Timberlake, “Chemistry”, Pearson/Benjamin Cummings, 2006, 9th Ed.
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