Problem Solving
• Chemistry  college  life is all
about solving problems
• Chemistry: it makes sense!
• Develop a logical plan (series of
steps) from your known to your
unknowns
http://www.geneseo.edu/~mcknight/
Problem Solving and
Dimensional Analysis
• Many problems in chemistry involve using
relationships to convert one unit of measurement
to another
• Conversion factors are relationships between two
units
– May be exact or measured
• Conversion factors generated from equivalence
statements
– e.g., 1 inch = 2.54 cm can give
2.54cm or 1in
1in
2.54cm
Problem Solving and
Dimensional Analysis
• Arrange conversion factors so given unit
cancels
– Arrange conversion factor so given unit is on
the bottom of the conversion factor
• May “string” conversion factors
– So we do not need to know direct relationships,
as long as we can find steps that leads to the
desired units (known  unknown)
desired unit
given unit 
 desired unit
given unit
“Must have a Plan”
• a conceptual plan is a visual outline that
shows the strategic route required to solve a
problem
• for unit conversion, the plan focuses on units
and how to convert one to another
• for problems that require equations, the
conceptual plan focuses on solving the
equation to find an unknown value
Concept Plans and
Conversion Factors
•
Convert inches into centimeters
1) Find relationship equivalence: 1 in = 2.54
cm
2) Write concept plan
in
cm
3) Change equivalence into conversion
factors with starting units on the bottom
2.54 cm
1 in
A Systematic Approach
• Sort the information from the problem
– identify the given quantity and unit, the quantity and unit
you want to find, any relationships implied in the
problem
• Design a strategy to solve the problem (roadmap)
– Concept plan
• sometimes may want to work backwards
• each step involves a conversion factor or equation
• Apply the steps in the concept plan
– check that units cancel properly
– multiply terms across the top and divide by each bottom
term
• Check the answer
– double check the set-up to ensure the unit at the end is the
one you wished to find
– check to see that the size of the number is reasonable
• since centimeters are smaller than inches, converting inches
to centimeters should result in a larger number
Example: Convert 1.76 yd. to centimeters
•
•
•
•
•
Sort
information
Strategize
Follow the
concept plan
to solve the
problem
Sig. figs. and
round
Check
Given:
Find:
Concept
Plan:
Relationship
Solution:
1.76 yd
length, cm
yd
m
cm
1.094 yd = 1 m
1 m = 100 cm
1m
100 cm
1.79 yd 

 160.8775 cm
1.094 yd
1m
Round:
160.8775 cm = 161 cm
Check: Units & magnitude are correct
Practice – Convert 30.0 mL to quarts
(1 L = 1.057 qt)
(Hint: 1000 mL makes 1 L)
Convert 30.0 mL to quarts
•
•
•
•
•
Sort
information
Strategize
Follow the
plan to
solve the
problem
Sig. figs.
and round
Check
Given:
Find:
Concept
Plan:
Relationship:
Solution:
30.0 mL
volume, qts
mL
L
qt
1 L = 1.057 qt
1 L = 1000 mL
1L
1.057 qt
30.0 mL 

 0.03171 qt
1000 mL
1L
Round:
0.03171 qt = 0.0317 qt
Check: Units & magnitude are correct
Problem Solving with Equations
• When solving a problem using an
equation, you are usually given all the
variables except the one you want to find
• Solve the equation for the variable you
wish to find, then substitute and compute
Using Density in Calculations
Concept Plans:
Mass
Density 
Volume
m, V
D
Mass
Volume 
Density
m, D
V
V, D
m
Mass  Density  Volume
Density Calculations
•
We can use density as a conversion
factor between mass and volume!!
–
–
density of H2O = 1.0 g/mL \ 1.0 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
= 45 g Pb
Question:
The mass of fuel in a jet must be
calculated before each flight to ensure
that the jet is not too heavy. A 747 jet is
fueled with 173,231 L of jet fuel. If the
density of the fuel is 0.738 g/mL, what is
the mass of the fuel in kilograms?
Example: What is the mass in kg of 173,231 L of
jet fuel whose density is 0.738 g/mL?
•
•
•
•
•
Sort
information
Strategize
Follow the
concept plan
to solve the
problem
Sig. figs.
and round
Check
Given:
Find:
Concept
Plan:
173,231 L
density = 0.738 g/mL
mass, kg
L
mL
g
kg
1 mL = 0.738 g, 1 mL = 10-3 L
Relationship
1 kg = 1000 g
Solution:
1 mL 0.738 g 1 kg
173,231 L 
-3
10 L

1 mL

1000 g
 1.33 105 kg
Round:
1.3 x 105 kg
Check: Units & magnitude are correct
Counting Atoms by Moles
•
If we can find the mass of a particular
number of atoms, we can use this
information to convert the mass of an
element sample into the number of atoms in
the sample.
The number of atoms we use is 6.022 x 1023 and
we call this a mole
•
–
1 mole = 6.022 x 1023 entities
•
Like 1 dozen = 12 entities
Avogadro’s Number
Example: Calculate the number of atoms
in 2.45 mol of copper
Given:
Find:
Concept
Plan:
2.45 mol Cu
atoms Cu
mol Cu
atoms Cu
6.0221023 atoms
1 mol
1 mol = 6.022 x 1023 atoms
Solution:
6.0221023 atoms
2.45 molCu 
1 mol
 1.481024 atomsCu
Check:
since the number is slightly greater than twice
Avogadro’s number, it make sense
Relationship Between
Moles and Mass
•
•
•
•
The mass of one mole of atoms is called
the molar mass
The molar mass of an element, in grams,
is numerically equal to the element’s
atomic mass, in amu
The lighter the atom, the less a mole
weighs
The lighter the atom, the more atoms
there are in 1 g
Mole and Mass Relationships
hydrogen
carbon
Weight of
Pieces in
1 atom
1 mole
1.008 amu 6.022 x 1023 atoms
12.01 amu 6.022 x 1023 atoms
Weight of
1 mole
1.008 g
12.01 g
oxygen
16.00 amu 6.022 x 1023 atoms
16.00 g
sulfur
32.06 amu 6.022 x 1023 atoms
32.06 g
calcium
40.08 amu 6.022 x 1023 atoms
40.08 g
chlorine
35.45 amu 6.022 x 1023 atoms
35.45 g
copper
63.55 amu 6.022 x 1023 atoms
63.55 g
Substance
1 mole
sulfur
32.06 g
1 mole
carbon
12.01 g
Example: Calculate the # moles of
carbon in 0.0265 g of pencil “lead”
Given:
Find:
Concept
Plan:
0.0265 g C
mol C
gC
mol C
1 mol
12.01g
1 mol C = 12.01 g
Solution:
1 mol
0.0265g C 
12.01g
 2.2110-3 molC
Check: since the given amount is much less than 1 mol
C, the number makes sense
Example: How many copper atoms are
in a copper penny weighing 3.10 g?
Given:
Find:
Concept
Plan:
Solution:
3.10 g Cu
atoms Cu
g Cu
mol Cu
1 mol
63.55g
1 mol Cu = 63.55 g,
1 mol = 6.022 x 1023
atoms Cu
6.0221023 atoms
1 mol
1 molCu 6.0221023 at oms
3.10 g Cu 

63.55g Cu
1 mol
 2.94 1022 at omsCu
Check: since the given amount is much less than 1 mol
Cu, the number makes sense