1

Dimensional Analysis

DHS Chemistry

2

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)

3

Dimensional AnalysisDimensional analysis is a powerful

problem-solving 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.

4

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.

5

Setting Up Ratios100 centimeters = 1 meter can be written as

100 cm OR 1 m___ 1 m 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 5 g

$1.99 per pound can be written as $1.99 OR 1 lb

1 lb $1.99

6

The Format # x # x # x # = 1 # # #

OR# # # = 1 # #

Note: Unless you are using 1 as a spot filler, all numbers should ALWAYS include units

7

Understanding order of operations & typing in

calculator keys correctlySolve. 10 13 6 =

2 4

Method 1: 10 X 13 X 6 = 780 then divide 97. 5 2 x 4 = 8

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

8

Try It

15 2 3 1.2 4 = 1 4 5 14 5

0.30857

9

Canceling out unitsJust 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 in feet yard = cm in feet

yard

10

Try It

week day hr min sec = 1 week day hr min

sec

11

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?

12

Using Dimensional Analysis

Using Dimensional Analysis1) 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.

13

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.

14

I. One-Step Conversions

15

282.3 pizzas

I. One- Step ConversionsEX 1: Determine the number of slices in 282.3

pizzas Step 1) what are you given? Step 2) What unit do you want?

Step 3) List your relationships (remember from pg 1 of the

notes)

Step 4) Set up your problem8 slices

slices1 pizza

=2258.41

282.3 pizzas

Number of slices

1 pizza = 8 slices

=2258 slices (SF)

16

Dozen of eggs

1 dozen

EX 2: Determine the number of dozen eggs in 3.8 x 103 eggs.

Step 1) what are you given? Step 2) What unit(s) do you want? Step 3) List your relationships

Step 4) Set up your problem

3.8 X 103 eggs 12

eggs

=316.671

3.8 X 103 eggsDozen Eggs

12 eggs = 1 dozen

=320 dozen of eggs (SF)

17

Practice1. Determine the number of feet in 2821

inches. want given

2821 in

1 foot feet12 in

= 235.11

Relationship1 foot = 12 inches

Start with what’s given

18

Practice2. Determine the number of g in 0.03455

kgwant given

0.03455 kg

1000 g g1 kg

= 34.551

Relationship1000g = 1kg

Start with what’s given

19

wantgiven

37 mi 1.6093 km km1 mi

= 59.51

Relationships1 mi = 1609.3 meter

1 mi = 1.6093 km Start with what’s given

99 mi 1.6093 km km1 mi

= 1591

20

Solving the same problem, but in a

different way

21

1 km

wantgiven

37 mi 1609.3 m km1 mi

= 59.51

Relationships1 mi = 1609.3 meter

1 km = 1000 m Start with what’s given

1000 m

22

For every problem, start with:

Given:

Want:

Relationships:

23

II. Multi-Step Conversions

24

II. Multi-step conversions

Of course, most problems are not simple 1-step 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.

25

Ex. 1 Convert 0.115 km to cmGiven:

Want:

Relationships:

0.115 km

____ cm

100 cm = 1 m

1 km = 1000 m 0.115 km 1000

m

cm

1 km = 11500

same as 1.15 x 104 cm

1

Start with what’s given

1 m

100 cm

26

Ex. 2 Convert 323 mL to cupsGiven:

Want:

Relationships:

323 mL

____ cups

1000 mL = 1 L

1 L = 1.06 quarts

1 quart = 4 cups

1.06 qt.323 mL 1 L

cups

1000 mL

1.36952= 1.37

1

Start with what’s given

1 L 1 qt.

4 cups

27

Practice 1: Convert 7.005 ft to mmGiven:

Want:

Relationships:

7.005 ft____ mm

1 foot = 12 inches

1 inch = 2.54 cm

100cm = 1 m

1m = 1000mm

2.54 cm

7.005 ft12 in

mm

1 ft

2135

1

Start with what’s given

1 in 100 cm1 m 1000

mm1 m

28

Practice 2: calculate the number of seconds in 2 years

Given:

Want:

Relationships:

2 years____ seconds

1 year = 365.25 days

24hrs = 1 day

60 min = 1 hr

1 min = 60 sec

24 hrs

2 years 365.25 days

seconds

1 year

63115200 = 6 X 10 7

1

Start with what’s given

1 day 1 hr60 min

60 sec1 min

29

What does that price mean?

$1.99 = 1lb

30

Solving Word Problems using

Dimensional Analysis

31

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.

32

If it helps, change any

combination unit into a

relationship

ex. 0.05mL/s0.05mL = 1 s

33

Given:

Want:

Relationships:

0.05mL/s

____ Liters

1 day = 24 hr

1 hr = 60 min

EX: A faucet is dripping at a rate of 0.05 mL/s. How many liters of water will be

lost in 1 day?

1 day

0.05mL = 1 sec

1000mL = 1 L

60 sec = 1 min

34

1 min

Relationship

0.05 mL

60 min

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

1 day24 hr

L

1 day

= 4.32

1

Relationships0.05 mL = 1 s 60 min = 1

hr

1000mL = 1 L 24 hr = 1 day

60s = 1 minStart with what’s given

1 hr60 s

1 s

1 L1000 mL

= 4 L (SF)

35

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.

36

Given:

Want:

Relationships:

____ feet

24 frames = 1 sec

1 hr = 60 min

1 in = 2.54cm

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?

2.5 hours

1 frame = 1.9cm

1 min = 60 sec

1 ft = 12 in

37

12 in

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

feet) for a 2 ½ hour movie?

1 s

Relationship

1.9 cm60 s

wantrelationship

2.5 hr60 min

ft

1 hr= 13464.6

1

Relationships24 frames = 1 s 1 frame =

1.9cm

1 hr = 60 min 60 s = 1 min

1in = 2.54 cm 1 ft = 12 inStart with what’s given

1 min24 frames

1 frame

1in2.54 cm

want given

1 ft

38

Given:

Want:

Relationships:

____ kg

0.05 g = 1 mL

1000g = 1kg

1L = 1000mL

EX 2: The density of an unknown liquid is 0.5 g/mL. What is the mass (in kg) of

2.00 L?

2.00 L

39

1 L

EX 2: The density of an unknown liquid is 0.5 g/mL. What is the

mass (in kg) of 2.00 L?

12.00L

1000g

1 kg

=1.00

1000mL

Relationship

given

want

Relationships0.5 g = 1mL

1000g = 1 kg

1000mL = 1 L

0.5g1mL

kg

40

Given:

Want:

Relationships:0.0899 g = 1 L

____ Liters

1 kg = 1000 g

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.

5 kg

41

1 kg

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.

15 kg

=55600

1000g

Relationship

givenwant

Relationships0.0899 g = 1L

1000g = 1 kg

0.0899g

1L

L

42

Practice Box Answers1. 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

43

Given:

Want:

Relationships:

____ trips

60 mi = 1 gal

1 trip = 3.5km

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?

10 gallons

1000m = 1km

1 mile = 1609.3 m

44

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?

1000m

Relationship

1 trip1609.3 m

wantrelationship

10gal 60 mi

trips

1 gal

= 275.88

1

Relationships60 miles = 1 gal1 trip = 3.5 km

1 mi = 1609.3m 1000m = 1kmStart with what’s given

1 mi1 km

3.5 km

given

= 300 trips (SF)

45

Given:

Want:

Relationships:

____ kg

60 s = 1 min

60 min = 1 hr

24 hrs = 1 day

2. Every second 2500 g of sulfuric acid flows out of a

pipe. How many kg of sulfuric acid will flow in 1

day?

1 day

1 sec = 2500g

1kg = 1000g

46

1 kg1000g

2. Every second 2500 g of sulfuric acid flows

out of a pipe. How many kg of sulfuric acid

will flow in 1 day?

1 min

Relationship

2500g60 min

want

1 day 24 hr

kg

1 day

= 216000

1

Relationships1 s = 2500g 60s = 1 min

1000g = 1kg 60 min = 1 hr

24 hr = 1 day

Start with what’s given

1 hr60 s

1 s

given

kg= 2 x 105

47

Given:

Want:

Relationships:

____ cm3

2.7 g = 1 cm3

1kg = 1000g

3. The density of aluminum is 2.7 g/cm3. What volume would a 5.0 kg block of

aluminum have?

5.0kg

48

1 kg

3. The density of aluminum is 2.7 g/cm3. What

volume would a 5.0 kg block of aluminum

have?

15.0kg

=1851.85

1000g

Relationship

givenwant

Relationships2.7 g = 1

cm3

1000g = 1 kg

1 cm3

2.7gcm3

=1900 cm3 (SF)

49

Given:

Want:

Relationships:

1.43g/L

____ g

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?

5.0 x 105L

50

1 L

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?

15.0 X 105L

=715000

1.43 g

Relationship

givenwant

Relationships1.43 g = 1L

g=72000 g (SF)= 7.2 x 104 g

51

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)

52

More Practice

53

Practice: Convert 5.2 L to cupsGiven:

Want:

Relationships:

5.2 L

____ cups

1 L = 1.06 quarts

1 quart = 4 cups

1.06 qt.5.2 L cups

1= 22.0

Start with what’s given

1 L 1 qt.

4 cups

54

356 g

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 1500 grams of this metal displace?

11500g =508120.6 cm3

Relationship

given

want

Relationships120.6 cm3 = 356g

cm3

=510 cm3 (SF)