Transcript Dimensional Analysis - SandersScienceStuff

Dimensional
Analysis
DHS Chemistry
1
Note:
From this point on, unless told otherwise, it is
expected that all answers will be reported using
the sig fig rules. All numbers greater than 999
and less than 0.01 are to be reported in
scientific notation.
For multiplication and division, the measurement
with the lowest number of sig figs will determine
how many sig figs are in the answer. (Usually go
by the sig figs in the given)
2
Dimensional Analysis
Dimensional analysis is a powerful problemsolving method that is based on the fact
that any number or expression can be
multiplied by one without changing its
value. It is the most important skill you
will learn for the rest of your science
career.
3
Conversion factors =
relationships
To use dimensional analysis you will use
conversion factors and fraction
cancellation. Anytime you can say two
things are equal to each other (or the
same as each other), you can make 2
conversion factors out of it --- each
conversion factor is equal to 1.
4
Setting Up Ratios
100 centimeters = 1 meter can be written as
100 cm
OR
1 m___
1m
100 cm
8 slices = 1 pizza can be written as
1 pizza
OR
8 slices___
8 slices
1 pizza
5 g/mL can be written as 5 g OR 1 mL
1 mL
5g
5
$1.99 per pound can be written as $1.99 OR 1 lb
1 lb
$1.99
The Format
# x
1
#
#
x
#
#
x
#
#
=
OR
#
1
#
#
# =
#
Note: Unless you are using 1 as a spot filler,
all numbers should ALWAYS include units
6
Understanding order of operations &
typing in calculator keys correctly
Solve.
10
13
2
Method 1: 10 X 13 X 6 =
2 x4 =
6 =
4
780
8
then divide 97. 5
Method 2: 10 X 13 X 6 / 2 / 4 = 97.5
130 780 390 97.5
Method 3: 10 X 13 / 2 X 6 / 4 = 97.5
130 65 390 97.5
7
Try It
15 2
1 4
8
3
5
1.2
14
4=
5
0.30857
Canceling out units
Just as numbers are multiplied and divided,
units are too. Cancel out all units that are
located in both the numerator and
denominator to reveal the left over units.
cm
9
in
cm
feet
in
yard = yard
feet
Try It
week day
hr
1 week day
10
min sec = sec
hr min
Putting the numbers &
units together
3 weeks 7 day
24 hr 60 min 60 sec =
1
1 week 1 day
1 hr 1 min
=1814400 sec
Rules?
11
Using Dimensional Analysis
Using Dimensional Analysis
1) Start by writing GWR (given, want,
relationships)
2) Identify the given.
3) Determine the units you want.
4) Choose the relationships that will allow you
to convert from your given to the want.
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Using Dimensional Analysis
Set-up your problem:
5) Start with the given as a clean fraction.
6) Write your multiplication symbol and place
your relationships carefully so everything
above and below each other is equal to each
other, and make sure your units cancel out
diagonally.
7) Any units that did not cancel out are part of
the units in your answer.
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I. One-Step
Conversions
14
I. One- Step Conversions
EX 1: Determine the number of slices in 282.3 pizzas
Step 1) what are you given? 282.3 pizzas
Step 2) What unit do you want?
Number of slices
Step 3) List your relationships
1 pizza = 8 slices
(remember from pg 1 of the notes)
Step 4) Set up your problem
282.3 pizzas
1
15
8 slices =2258.4 slices
1 pizza =2258 slices (SF)
EX 2: Determine the number of dozen eggs in
3.8 x 103 eggs.
Step 1) what are you given? 3.8 X 103 eggs
Step 2) What unit(s) do you want? Dozen Eggs
Step 3) List your relationships
12 eggs = 1 dozen
Step 4) Set up your problem
3.8 X 103 eggs 1 dozen =316.67
1
12 eggs
Dozen of eggs
16
=320 dozen of eggs (SF)
Practice
1. Determine the number of feet in 2821 inches.
Relationship
1 foot = 12 inches
want
given
Start with what’s given
2821 in
1
17
1 foot
12 in
= 235.1 feet
Practice
2. Determine the number of g in 0.03455 kg
Relationship
1000g = 1kg
want
Start with what’s given
0.03455 kg
1
18
1000 g = 34.55 g
1 kg
given
given
want
Relationships
1 mi = 1609.3 meter
1 mi = 1.6093 km
Start with what’s given
37 mi
19
1
99 mi
1
1.6093 km = 59.5 km
1 mi
1.6093 km = 159 km
1 mi
Solving the same
problem, but in a
different way
20
given
want
Relationships
1 mi = 1609.3 meter
1 km = 1000 m
Start with what’s given
37 mi
1
21
1609.3 m
1 mi
1 km = 59.5 km
1000 m
For every problem, start
with:
Given:
Want:
Relationships:
22
II. Multi-Step
Conversions
23
II. Multi-step conversions
Of course, most problems are not simple 1step conversions. Most problems will
require multiple conversion factors. Note:
in this class you MUST go through the root
unit for any metric conversion. (root units:
meter, gram, liter …) Tip: Cancel out all
units until you get what you’re
looking for.
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Ex. 1 Convert 0.115 km to cm
Given: 0.115 km
Want: ____ cm
Relationships: 100 cm = 1 m
1 km = 1000 m
1000 m
100 cm
1m
1 km
Start with what’s given
0.115 km
1
= 11500 cm
same as 1.15 x 104 cm
25
Ex. 2 Convert 323 mL to cups
Given: 323 mL
Want: ____ cups
Relationships: 1000 mL = 1 L
1 L = 1.06 quarts
1 quart = 4 cups
Start with what’s given
323 mL
1
26
1L
1000 mL
1.06 qt.
1L
4 cups
1 qt.
1.36952= 1.37 cups
Practice 1: Convert 7.005 ft to mm
Given: 7.005 ft
Want: ____ mm
Relationships: 1 foot = 12 inches
1 inch = 2.54 cm
100cm = 1 m
1m = 1000mm
Start with what’s given
7.005 ft 12 in
1
1 ft
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2.54 cm 1 m 1000 mm
1 in
100 cm 1 m
2135 mm
Practice 2: calculate the number of seconds in 2 years
Given: 2 years
Want: ____ seconds
Relationships: 1 year = 365.25 days
24hrs = 1 day
60 min = 1 hr
1 min = 60 sec
Start with what’s given
2 years
1
28
365.25 days
1 year
24 hrs 60 min 60 sec
1 day
1 hr
1 min
63115200 = 6 X 10 7 seconds
What does that
price mean?
29
$1.99 = 1lb
Solving Word
Problems using
Dimensional
Analysis
30
If you are given multiple numbers in a
problem, only one number will be
your starting point. The other
numbers are relationships that you
will use in your problem. If you are
given multiple numbers and one of
them involves a “/” (e.g. m/s), then
always use the “/” as a relationship
and start with the other number.
31
If it helps, change
any combination
unit into a
relationship
ex. 0.05mL/s
0.05mL = 1 s
32
EX: A faucet is dripping at a rate of 0.05 mL/s.
How many liters of water will be lost in 1
day?
Given: 0.05mL/s
Want: ____ Liters
Relationships:
0.05mL = 1 sec
1 day = 24 hr
1000mL = 1 L
1 hr = 60 min
60 sec = 1 min
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1 day
Relationship
EX: A faucet is dripping at a rate of 0.05 mL/s.
How many liters of water will be lost in 1 day?
want
given
Relationships
0.05 mL = 1 s
60 min = 1 hr
1000mL = 1 L
24 hr = 1 day
60s = 1 min
Start with what’s given
1 day 24 hr 60 min 60 s 0.05 mL 1 L
1000 mL
1 1 day 1 hr 1 min 1 s
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= 4.32 L = 4 L (SF)
Finally, note dimensional
analysis can be used
anytime you can say
something is equal to
something else. It does not
have to involve standard
conversion factors.
35
EX 1: Motion pictures are shown at a speed of
24 frames, or individual pictures, each second. If
a standard frame is 1.9 cm long, how long will
the strand of film be (in ft) for a 2 ½ hr movie?
Given: 2.5 hours
Want: ____ feet
Relationships:
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24 frames = 1 sec
1 frame = 1.9cm
1 hr = 60 min
1 min = 60 sec
1 in = 2.54cm
1 ft = 12 in
EX 1: Motion pictures are shown at a speed of
24 frames, or individual pictures, each second. If a
Relationship
standard frame is 1.9 cm long, how long will the
relationship
want
strand of film be (in feet) for a 2 ½ hour movie?
want
Relationships
24 frames = 1 s
given
1 frame = 1.9cm
1 hr = 60 min
60 s = 1 min
1in = 2.54 cm
1 ft = 12 in
Start with what’s given
2.5 hr 60 min 60 s 24 frames 1.9 cm 1in 1 ft
1 1 hr 1 min 1 s
1 frame 2.54 12 in
= 13464.6 ft
cm
37
EX 2: The density of an unknown liquid is
0.5 g/mL. What is the mass (in kg) of
2.00 L?
Given: 2.00 L
Want: ____ kg
Relationships: 0.05 g = 1 mL
1000g = 1kg
1L = 1000mL
38
EX 2: The density of an unknown liquid
is 0.5 g/mL. What is the mass (in kg) of
Relationship
want
2.00
L?
Relationships
0.5 g = 1mL
given
1000g = 1 kg
1000mL = 1 L
2.00L 1000mL 0.5g
1
1L
1mL
1 kg
1000g
=1.00 kg
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EX 3: What is the volume of 5 kg of hydrogen
gas at room conditions? The density of
hydrogen at room conditions is 0.0899 g/L.
Given: 5 kg
Want: ____ Liters
Relationships: 0.0899 g = 1 L
1 kg = 1000 g
40
EX 3: What is the volume of 5 kg of hydrogen gas at
want
given
room conditions? The density of hydrogen at room
conditions is 0.0899 g/L.
Relationship
Relationships
0.0899 g = 1L
1000g = 1 kg
5 kg
1
1000g
1 kg
1L
0.0899g
=55600 L
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Practice Box Answers
1. The Toyota Prius gets 60 mi/gal of gas. If each
trip to school is 3.5 km, how many trips can I
make with 10 gallons of gas? 275.88 trips
2. Every second 2500 g of sulfuric acid flows out of
a pipe. How many kg of sulfuric acid will flow
in 1 day? 216000 kg
3. The density of aluminum is 2.7 g/cm3. What
volume would a 5.0 kg block of aluminum
have?
1851.85 cm3
4.Oxygen has a density of 1.43 g/L at 0oC and 1
atm. What would be the mass of 5.0 x 105 L of
oxygen at those conditions? 715000 g
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1. The Toyota Prius gets 60 mi/gal of gas. If each
trip to school is 3.5 km, how many trips can I
make with 10 gallons of gas?
Given: 10 gallons
Want: ____ trips
Relationships:
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60 mi = 1 gal
1000m = 1km
1 trip = 3.5km
1 mile = 1609.3 m
1. The Toyota Prius gets 60 mi/gal of gas. If
Relationship
each trip to school is 3.5 km, how many trips
relationship
can I make with 10 gallons of gas?
want
given
Relationships
60 miles = 1 gal
1 mi = 1609.3m
Start with what’s given
10gal 60 mi
1
1 gal
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1 trip = 3.5 km
1000m = 1km
1609.3 m 1 km
1 trip
1 mi
1000m 3.5 km
= 275.88 trips
= 300 trips (SF)
2. Every second 2500 g of sulfuric acid flows
out of a pipe. How many kg of sulfuric acid
will flow in 1 day?
Given: 1 day
Want: ____ kg
Relationships:
45
1 sec = 2500g
60 s = 1 min
1kg = 1000g
60 min = 1 hr
24 hrs = 1 day
2. Every second 2500 g of sulfuric acid flows out
Relationship
of a pipe. How many kg of sulfuric acid will
flow in 1 day?
want
given
Relationships
1 s = 2500g
1000g = 1kg
Start with what’s given
60s = 1 min
60 min = 1 hr
24 hr = 1 day
1 day 24 hr 60 min 60 s 2500g 1 kg
1 s 1000g
1
1 day 1 hr 1 min
= 216000 kg
= 2 x 105 kg
46
3. The density of aluminum is 2.7 g/cm3.
What volume would a 5.0 kg block of
aluminum have?
Given: 5.0kg
3
____
cm
Want:
Relationships: 2.7 g = 1 cm3
1kg = 1000g
47
3. The density of aluminum is 2.7 g/cm3. What
Relationship
volume would a 5.0 kg block of aluminum have?
want
given
Relationships
2.7 g = 1 cm3
1000g = 1 kg
5.0kg 1000g
1
1 kg
1 cm3
2.7g
=1851.85 cm3
48
=1900 cm3 (SF)
4. Oxygen has a density of 1.43 g/L at 0oC
and 1 atm. What would be the mass of
5.0 x 105 L of oxygen at those conditions?
Given: 1.43g/L
5.0 x 105L
Want: ____ g
Relationships:
49
4. Oxygen has a density of 1.43 g/L at 0oC and 1
Relationship
atm. What would be the mass of 5.0 x 105 L of
want
given
oxygen at those conditions?
Relationships
1.43 g = 1L
5.0 X 105L
1
1.43 g
1L
=715000 g
50
=72000 g (SF)= 7.2 x 104 g
Tips for studying
• Try doing the review without using the
conversion sheet
• Quiz yourself with the conversions
• Do extra practice problems (your textbook
may have some)
51
More Practice
52
Practice: Convert 5.2 L to cups
Given: 5.2 L
Want: ____ cups
Relationships: 1 L = 1.06 quarts
1 quart = 4 cups
Start with what’s given
5.2 L
1
53
1.06 qt.
1L
4 cups = 22.0 cups
1 qt.
A block of metal measuring 2.5 cm x 7.2 cm x 6.7
cm has a mass of 356 g. What volume of water will
Relationship
want
1500 grams of this metal displace?
given
Relationships
120.6 cm3 = 356g
1500g 120.6 cm3
1
356 g
54
=508 cm3
=510 cm3 (SF)