Transcript Introductory Chemistry, 2nd Edition Nivaldo Tro

Introductory Chemistry, 2nd Edition
Nivaldo Tro
Chapter 2
Part 2: Problem
Solving &
Dimensional Analysis
1
Units
Always write every number with its
associated unit
Always include units in your calculations
– you can do the same kind of operations on
units as you can with numbers
cm × cm = cm2
cm + cm = cm
cm ÷ cm = 1
– using units as a guide to problem solving is
called dimensional analysis
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2
Problem Solving and Dimensional
Analysis
Many problems in Chemistry use
relationships to convert one unit to
another
Conversion Factors are relationships
between two units which
Conversion factors are generated from
equivalence statements:
– e.g. 1 inch = 2.54 cm can give
Tro's Introductory Chemistry, Chapter 2
1in
2.54cm
or 2.54cm
1in
3
Problem Solving and
Dimensional Analysis
Arrange conversion factors so starting unit
cancels
– Arrange conversion factor so starting unit is on
the bottom of the conversion factor
May string conversion factors
– So we do not need to know every relationship,
as long as we can find something else the
beginning and ending units are related to
unit 1 x
unit 2
unit 1
= unit 2
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4
Solution Maps
Solution map = a visual outline
showing strategic route required to
solve a problem
For unit conversion, the solution map
focuses on units and how to convert
one to another
For problems that require equations,
the solution map focuses on solving
the equation to find an unknown value
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Systematic Approach
1) Write down given amount and unit
2) Write down what you want to find and unit
3) Write down needed conversion factors or
equations
a) Write down equivalence statements for each
relationship
b) Change equivalence statements to
conversion factors with starting unit on the
bottom
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Systematic Approach
4)
Design a solution map for the problem
–
–
5)
Apply the steps in the solution map
–
–
6)
order conversions to cancel previous units
or
arrange Equation so Find amount is isolated
check that units cancel properly
multiply terms across the top and divide by
each bottom term
Check the answer to see if its reasonable
–
correct size and unit
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7
Solution Maps and Conversion
Factors
Convert Inches into Centimeters
1) Find Relationship Equivalence: 1 in = 2.54
cm
2) Write Solution
Map
in
cm
3) Change Equivalence into Conversion
Factors with Starting Units on the Bottom
2.54cm
1in
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Convert 7.8 km to miles
1.
Write down the Given
quantity and its unit
2.
Write down the quantity
you want to Find and unit
3.
Write down the appropriate
Conversion Factors
4.
Write a Solution Map
Given:
7.8 km
Find:
? miles
Conversion
Factors:
Solution
Map:
Solution:
1 km = 0.6214 mi
km
mi
0.6214 mi
1 km
5.
Follow the Solution Map to
Solve the problem
6.
Sig. Figs. and Round
Round:
7.
Check
Check: Units & Magnitude are
correct
7.8 km 
0.6214 mi
 4.84692 mi
1 km
4.84692 mi = 4.8 mi
9
Solution Maps and
Conversion Factors
Convert Cups into Liters
1) Find Relationship Equivalence: 1 L = 1.057 qt,
1 qt = 4 c
2) Write Solution Map
c
qt
L
3) Change Equivalence into Conversion Factors
with Starting Units on the Bottom
1 qt
4c
1L
1.057 qt
Tro's Introductory Chemistry, Chapter 2
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How many cups of cream is 0.75 L?
1.
Write down the Given
quantity and its unit
2.
Write down the quantity
you want to Find and unit
3.
Write down the appropriate
Conversion Factors
4.
Write a Solution Map
Given:
0.75 L
Find:
? cu
Conversion
Factors:
Solution
Map:
Solution:
1 L = 1.057 qt
1 qt = 4 cu
L
qt
1.057 qt
1L
cu
4 cu
1 qt
5.
Follow the Solution Map to
Solve the problem
6.
Sig. Figs. and Round
Round:
7.
Check
Check: Units & Magnitude are
correct
0.75 L 
1.057 qt 4 cu

 3.171 cu
1L
1 qt
3.171 cu = 3.2 cu
Solving Multistep Unit
Conversion Problems
Example:
An Italian recipe for making creamy pasta sauce
calls for 0.75 L of cream. Your measuring cup
measures only in cups. How many cups should
you use?
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An Italian recipe for making
creamy pasta sauce calls for
0.75 L of cream. Your
measuring cup measures only
in cups. How many cups
should you use?
Write down the given quantity and its units.
Given:
0.75 L
Tro's Introductory Chemistry, Chapter 2
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An Italian recipe for making
creamy pasta sauce calls for
0.75 L of cream. Your
measuring cup measures only
in cups. How many cups
should you use?
Information
Given:
0.75 L
Write down the quantity to find and/or its units.
Find:
? cups
Tro's Introductory Chemistry, Chapter 2
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An Italian recipe for making
creamy pasta sauce calls for
0.75 L of cream. Your
measuring cup measures only
in cups. How many cups
should you use?
Information
Given:
0.75 L
Find: ? cu
Collect Needed Conversion Factors:
4 cu = 1 qt
1.057 qt = 1 L
Tro's Introductory Chemistry, Chapter 2
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An Italian recipe for making
creamy pasta sauce calls for
0.75 L of cream. Your
measuring cup measures only
in cups. How many cups
should you use?
Information
Given:
0.75 L
Find: ? cu
Conv. Fact. 4 cu = 1 qt;
1.057 qt = 1 L
Write a Solution Map for converting the units :
L
qt
1.057 qt
1L
cu
4 cu
1 qt
Tro's Introductory Chemistry, Chapter 2
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An Italian recipe for
making creamy pasta
sauce calls for 0.75 L of
cream. Your measuring
cup measures only in
cups. How many cups
should you use?
Information
Given:
0.75 L
Find: ? cu
Conv. Fact. 4 cu = 1 qt;
1.057 qt = 1 L
Sol’n Map: L  qt  cu
1.057 qt
1L
4 cu
1 qt
Apply the Solution Map:
1.057 qt 4 cu
0.75 L 

1L
1 qt
• Sig. Figs. & Round:
= 3.171 cu
= 3.2 cu
Tro's Introductory Chemistry, Chapter 2
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An Italian recipe for
making creamy pasta
sauce calls for 0.75 L of
cream. Your measuring
cup measures only in
cups. How many cups
should you use?
Information
Given:
0.75 L
Find: ? cu
Conv. Fact. 4 cu = 1 qt;
1.057 qt = 1 L
Sol’n Map: L  qt  cu
1.057 qt
1L
4 cu
1 qt
Check the Solution:
0.75 L = 3.2 cu
The units of the answer, cu, are correct.
The magnitude of the answer makes sense
since cups are smaller than liters.
Tro's Introductory Chemistry, Chapter 2
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Solution Maps and Conversion
Factors
Convert Cubic Inches into Cubic Centimeters
1) Find Relationship Equivalence: 1 in = 2.54 cm
2) Write Solution Map
in3
cm3
3) Change Equivalence into Conversion Factors
with Starting Units on the Bottom
3
2.543 cm3 16.4 cm3
 2.54 cm 


 
3
3
1 in
1 in 3
 1 in 
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Convert 2,659 cm2 into square meters
1.
Write down the Given
quantity and its unit
2.
Write down the quantity
you want to Find and unit
3.
Write down the
appropriate Conversion
Factors
Given:
2,659 cm2
? m2
Find:
Conversion
Factors:
1 cm = 0.01 m
cm2
m2
4.
Write a Solution Map
Solution
Map:
5.
Follow the Solution Map
to Solve the problem
Solution:
6.
Sig. Figs. and Round
Round:
0.2659 m2
7.
Check
Check:
Units & Magnitude
are correct
 0.01 m 


 1 cm 
2,659 cm 
2
2
1104 m 2
1 cm
2
 0.2659 m 2
Example 2.12:
Converting Quantities Involving
Units Raised
to a Power
Example:
A circle has an area of 2,659 cm2. What is the
area in square meters?
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Example:
A circle has an area of
2,659 cm2. What is the
area in square meters?
Write down the given quantity and its units.
Given:
2,659 cm2
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Example:
A circle has an area of
2,659 cm2. What is the
area in square meters?
Information
Given:
2,659 cm2
Write down the quantity to find and/or its units.
Find:
? m2
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Example:
A circle has an area of
2,659 cm2. What is the
area in square meters?
Information
Given:
2,659 cm2
Find: ? m2
Collect Needed Conversion Factors:
1 cm = 0.01m
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Example:
A circle has an area of
2,659 cm2. What is the
area in square meters?
Information
Given:
2,659 cm2
Find: ? m2
Conv. Fact.: 1 cm = 0.01 m
Write a Solution Map for converting the units :
cm2
m2
 0.01 m 


 1 cm 
2
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Information
Given:
2,659 cm2
Find: ? m2
Conv. Fact. 1 cm = 0.01 m
2
2
2
 0.01 m 
Sol’n Map: cm  m  1 cm 
Example:
A circle has an area of
2,659 cm2. What is the
area in square meters?


Apply the Solution Map:
2,659cm2 
110-4 m2
1 cm2
 m2
= 0.265900 m2
• Sig. Figs. & Round:
= 0.2659 m2
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Example:
A circle has an area of
2,659 cm2. What is the
area in square meters?
Information
Given:
2,659 cm2
Find: ? m2
Conv. Fact. 1 cm = 0.01 m
2
2
2
 0.01 m 
Sol’n Map: cm  m  1 cm 


Check the Solution:
2,659 cm2 = 0.2659 m2
The units of the answer, m2, are correct.
The magnitude of the answer makes sense
since square centimeters are smaller
than square meters.
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Density
Mass & Volume
Two main characteristics of matter
Cannot be used to identify what type of
matter something is
– if you are given a large glass containing 100 g
of a clear, colorless liquid and a small glass
containing 25 g of a clear, colorless liquid - are
both liquids the same stuff?
even though mass and volume are
individual properties - for a given type of
matter they are related to each other!
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Mass vs Volume of Brass
Mass
grams
Volume
cm3
20
2.4
32
3.8
40
4.8
50
6.0
100
11.9
150
17.9
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Volume vs Mass of Brass
y = 8.38x
160
140
120
Mass, g
100
80
60
40
20
0
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
16.0
18.0
Volume, cm3
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Density
Ratio of mass:volume
Solids = g/cm3
Mass
– 1 cm3 = 1 mL
Density 
Liquids = g/mL
Volume
Gases = g/L
Volume of a solid can be determined by
water displacement – Archimedes Principle
Density : solids > liquids >>> gases
– except ice is less dense than liquid water!
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Mass
Density
Density 
Volume
For equal volumes, denser object has larger
mass
For equal masses, denser object has smaller
volume
Heating objects causes objects to expand
– does not affect their mass!!
– How would heating an object affect its
density?
In a heterogeneous mixture, the denser object
sinks
– Why do hot air balloons rise?
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Using Density in Calculations
Solution Maps:
Mass
Density 
Volume
m, V
D
Mass
Volume 
Density
m, D
V
V, D
m
Mass  Density  Volume
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Applying Density in Problem Solving
Platinum has become a popular metal for fine
jewelry. A man gives a woman an engagement
ring and tells her that it is made of platinum.
Noting that the ring felt a little light, the woman
decides to perform a test to determine the ring’s
density before giving him an answer about
marriage. She places the ring on a balance and
finds it has a mass of 5.84 grams. She then
finds that the ring displaces 0.556 cm3 of water.
Is the ring made of platinum? (Density Pt = 21.4
g/cm3)
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She places the ring on a balance and finds it has a
mass of 5.84 grams. She then finds that the ring
displaces 0.556 cm3 of water. Is the ring made of
platinum? (Density Pt = 21.4 g/cm3)
Given: Mass = 5.84 grams
Volume = 0.556 cm3
Find: Density in grams/cm3
Equation: m
V
D
Solution Map:
m and V  d
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She places the ring on a balance and finds it has a mass of
5.84 grams. She then finds that the ring displaces 0.556 cm3
of water. Is the ring made of platinum? (Density Pt = 21.4
g/cm3)
Apply the Solution Map:
m
D
V
5.84 g
0.556 cm
3
g
 10.5
cm 3
Since 10.5 g/cm3  21.4 g/cm3 the ring cannot be
platinum. Should she marry him? ☺
Tro's Introductory Chemistry, Chapter 2
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Density as a Conversion
Factor
• Can use density as a conversion factor between
mass and volume!!
– density of H2O = 1 g/mL \ 1 g H2O = 1 mL H2O
– density of Pb = 11.3 g/cm3 \ 11.3 g Pb = 1 cm3 Pb
• How much does 4.0 cm3 of Lead weigh?
4.0 cm3 Pb x
11.3 g Pb
1 cm3 Pb
Tro's Introductory Chemistry, Chapter 2
= 45 g Pb
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Measurement and Problem Solving
Density as a Conversion Factor
The gasoline in an automobile gas tank has a mass
of 60.0 kg and a density of 0.752 g/cm3. What is the
volume?
Given: 60.0 kg
Find: Volume in L
Conversion Factors:
– 0.752 grams/cm3
– 1000 grams = 1 kg
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Measurement and Problem Solving
Density as a Conversion Factor
Solution Map:
kg  g  cm3
3
1000 g 1 cm
4
3
60.0 kg 

 7.98 10 cm
1 kg 0.752 g
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Density as a Conversion
Factor: Another Example
Example:
A 55.9 kg person displaces 57.2 L of water when
submerged in a water tank. What is the density
of the person in g/cm3?
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Example:
A 55.9 kg person displaces
57.2 L of water when
submerged in a water tank.
What is the density of the
person in g/cm3?
Write down the given quantity and its units.
Given:
m = 55.9 kg
V = 57.2 L
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Example:
A 55.9 kg person displaces
57.2 L of water when
submerged in a water tank.
What is the density of the
person in g/cm3?
Information
Given: m = 55.9 kg
V = 57.2 L
Write down the quantity to find and/or its units.
Find: density, g/cm3
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Example:
A 55.9 kg person displaces
57.2 L of water when
submerged in a water tank.
What is the density of the
person in g/cm3?
Information:
Given: m = 55.9 kg
V = 57.2 L
Find: density, g/cm3
Design a Solution Map:
m, V
D
m
D
V
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Example:
A 55.9 kg person displaces
57.2 L of water when
submerged in a water tank.
What is the density of the
person in g/cm3?
Information:
Given: m = 55.9 kg
V = 57.2 L
Find: density, g/cm3
Equation: D  m
V
Collect Needed Conversion Factors:
 Mass:
 Volume:
1 kg = 1000 g
1 mL = 0.001 L; 1 mL = 1 cm3
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Example:
A 55.9 kg person displaces
57.2 L of water when
submerged in a water tank.
What is the density of the
person in g/cm3?
Information:
Given: m = 55.9 kg
V = 57.2 L
Find: density, g/cm3
Solution Map: m,VD
m
D
Equation:
V
Conversion Factors: 1 kg = 1000 g
1 mL = 0.001 L
1 mL = 1 cm3
Write a Solution Map for converting the Mass units
kg
g
1000g
1 kg
Write a Solution Map for converting the Volume units
L
cm3
mL
1 mL
0.001 L
1 cm3
1 mL
49
Example:
A 55.9 kg person displaces
57.2 L of water when
submerged in a water tank.
What is the density of the
person in g/cm3?
Information:
Given: m = 55.9 kg
V = 57.2 L
Find: density, g/cm3
Solution Map:
m,VD
Equation: D  m
V
Apply the Solution Maps
1000g
55.9kg 
g
1 kg
= 5.59 x 104 g
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Example:
A 55.9 kg person displaces
57.2 L of water when
submerged in a water tank.
What is the density of the
person in g/cm3?
Information:
Given:
m = 5.59 x 104 g
V = 57.2 L
Find: density, g/cm3
Solution Map:
m,VD
Equation: D  m
V
Apply the Solution Maps
3
1 mL 1 cm
3
57.2L 

 cm
0.001L 1 mL
= 5.72 x 104 cm3
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Example:
A 55.9 kg person displaces
57.2 L of water when
submerged in a water tank.
What is the density of the
person in g/cm3?
Information:
Given:
m = 5.59 x 104 g
V = 5.72 x 104 cm3
Find: density, g/cm3
Solution Map:
m,VD
Equation:
m
D
V
Apply the Solution Maps - Equation
m
5.59 x 104 g
D

4
3
V 5.72 x 10 cm
= 0.9772727 g/cm3
= 0.977 g/cm3
52
Example:
A 55.9 kg person displaces
57.2 L of water when
submerged in a water tank.
What is the density of the
person in g/cm3?
Information:
Given:
m = 5.59 x 104 g
V = 5.72 x 104 cm3
Find: density, g/cm3
Solution Map:
m,VD
Equation: D  m
V
Check the Solution
D = 0.977 g/cm3
The units of the answer, g/cm3, are correct.
The magnitude of the answer makes sense.
Since the mass in kg and volume in L are
very close in magnitude, the answer’s
magnitude should be close to 1.
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