Dimensional Dimensional Analysis in Analysis in ChemistryChemistry

UNITS OF UNITS OF MEASUREMENTMEASUREMENT

Use Use SI unitsSI units — based on the metric — based on the metric systemsystem

Length Length

MassMass

VolumeVolume

TimeTime

TemperatureTemperature

Meter, mMeter, m

Kilogram, kgKilogram, kg

Seconds, sSeconds, s

Celsius degrees, ˚CCelsius degrees, ˚Ckelvins, Kkelvins, K

Liter, LLiter, L

Metric PrefixesMetric Prefixes

Conversion FactorsConversion Factors

Fractions in which the numerator and Fractions in which the numerator and denominator are EQUAL quantities expressed denominator are EQUAL quantities expressed in different unitsin different units

Example: 1 in. = 2.54 cm

Factors: 1 in. and 2.54 cm

2.54 cm 1 in.

Learning Check

Write conversion factors that relate each of Write conversion factors that relate each of the following pairs of units:the following pairs of units:

1. Liters and mL1. Liters and mL

2. Hours and minutes2. Hours and minutes

3. Meters and kilometers3. Meters and kilometers

How many minutes are in 2.5 hours?

Conversion factor

2.5 hr x 2.5 hr x 60 min 60 min = 150 min = 150 min

1 hr1 hr

cancel

By using dimensional analysis / factor-label method, By using dimensional analysis / factor-label method, the UNITS ensure that you have the conversion right the UNITS ensure that you have the conversion right side up, and the UNITS are calculated as well as the side up, and the UNITS are calculated as well as the

numbers!numbers!

Sample Problem

• You have $7.25 in your pocket in You have $7.25 in your pocket in quarters. How many quarters do you quarters. How many quarters do you have?have?

7.25 dollars 4 quarters7.25 dollars 4 quarters

1 dollar1 dollar X = 29 quarters= 29 quarters

Learning Check

A rattlesnake is 2.44 m long. How A rattlesnake is 2.44 m long. How long is the snake in cm?long is the snake in cm?

a) a) 2440 cm2440 cm

b)b) 244 cm244 cm

c)c) 24.4 cm24.4 cm

Learning Check

How many seconds are in 1.4 days?

Unit plan: days hr min seconds

1.4 days x 24 hr x ?? 1 day

Solution

Unit plan: days hr min seconds

1.4 day x 24 hr x 60 min x 60 sec

1 day 1 hr 1 min

= 1.2 x 105 sec

Wait a minute!

What is What is wrongwrong with the following setup? with the following setup?

1.4 day x 1.4 day x 1 day 1 day x x 60 min 60 min x x 60 sec 60 sec

24 hr 1 hr 1 min24 hr 1 hr 1 min

English and Metric English and Metric ConversionsConversions

• If you know ONE conversion for If you know ONE conversion for each type of measurement, you each type of measurement, you can convert anything!can convert anything!

• You will need to know and use You will need to know and use these conversions:these conversions:

–Mass: 454 grams = 1 poundMass: 454 grams = 1 pound

–Length: 2.54 cm = 1 inchLength: 2.54 cm = 1 inch

–Volume: 0.946 L = 1 quartVolume: 0.946 L = 1 quart

Learning CheckLearning Check

An adult human has 4.65 L of blood. How An adult human has 4.65 L of blood. How many gallons of blood is that?many gallons of blood is that?

Unit plan: L qt gallon

Equalities: 1 quart = 0.946 L

1 gallon = 4 quarts

Your Setup:

Steps to Problem SolvingSteps to Problem Solving

Read problemRead problem Identify data Identify data Make a unit plan from the initial unit to the Make a unit plan from the initial unit to the

desired unitdesired unit Select conversion factorsSelect conversion factors Change initial unit to desired unitChange initial unit to desired unit Cancel units and checkCancel units and check Do math on calculator Do math on calculator Give an answer using significant figuresGive an answer using significant figures

Dealing with Two Units – Honors Dealing with Two Units – Honors OnlyOnly

If your pace on a treadmill is 65 meters If your pace on a treadmill is 65 meters per minute, how many seconds will it per minute, how many seconds will it take for you to walk a distance of 8450 take for you to walk a distance of 8450 feet?feet?

InitialInitial

8450 ft x 8450 ft x 12 in. 12 in. x x 2.54 cm 2.54 cm x x 1 m 1 m

1 ft1 ft 1 in. 1 in. 100 cm 100 cm

x x 1 min 1 min x x 60 sec 60 sec = 2400 sec = 2400 sec 65 m65 m 1 min 1 min

Solution

Temperature ScalesTemperature Scales• FahrenheitFahrenheit

• CelsiusCelsius

• KelvinKelvin

Anders Celsius1701-1744

Lord Kelvin(William Thomson)1824-1907

Temperature ScalesTemperature Scales

Notice that 1 kelvin = 1 degree Celsius1 kelvin = 1 degree Celsius

Boiling point Boiling point of waterof water

Freezing point Freezing point of waterof water

CelsiusCelsius

100 ˚C100 ˚C

0 ˚C0 ˚C

100˚C100˚C

KelvinKelvin

373 K373 K

273 K273 K

100 K100 K

FahrenheitFahrenheit

32 ˚F32 ˚F

212 ˚F212 ˚F

180˚F180˚F

Calculations Calculations Using Using TemperatureTemperature

• Generally require temp’s in Generally require temp’s in kelvinskelvins

• T (K) = t (˚C) + 273.15T (K) = t (˚C) + 273.15

• Body temp = 37 ˚C + 273 = 310 KBody temp = 37 ˚C + 273 = 310 K

• Liquid nitrogen = -196 ˚C + 273 = 77 KLiquid nitrogen = -196 ˚C + 273 = 77 K

• Generally require temp’s in Generally require temp’s in kelvinskelvins

• T (K) = t (˚C) + 273.15T (K) = t (˚C) + 273.15

• Body temp = 37 ˚C + 273 = 310 KBody temp = 37 ˚C + 273 = 310 K

• Liquid nitrogen = -196 ˚C + 273 = 77 KLiquid nitrogen = -196 ˚C + 273 = 77 K

Fahrenheit Formula – Honors OnlyFahrenheit Formula – Honors Only

180°F180°F = = 9°F 9°F == 1.8°F 1.8°F 100°C 100°C 5°C 5°C 1°C1°C

Zero point: 0°C = 32°FZero point: 0°C = 32°F

°F = 9/5 °C + 32°F = 9/5 °C + 32

Celsius Formula Celsius Formula

Rearrange to find T°CRearrange to find T°C

°F °F = = 9/5 °C + 329/5 °C + 32

°F - 32 = °F - 32 = 9/5 °C ( +32 - 32)9/5 °C ( +32 - 32)

°F - 32°F - 32 = = 9/5 °C9/5 °C

9/5 9/5 9/5 9/5

(°F - 32) * 5/9 = (°F - 32) * 5/9 = °C°C

Temperature Conversions – Honors OnlyTemperature Conversions – Honors Only

A person with hypothermia has a body A person with hypothermia has a body temperature of 29.1°C. What is the body temperature of 29.1°C. What is the body temperature in °F? temperature in °F?

°F °F = = 9/5 (29.1°C) + 329/5 (29.1°C) + 32

= = 52.4 + 3252.4 + 32

= = 84.4°F84.4°F

Learning Check – Honors OnlyLearning Check – Honors Only

The normal temperature of a chickadee is The normal temperature of a chickadee is 105.8°F. What is that temperature in °C?105.8°F. What is that temperature in °C?

1) 73.8 °C 1) 73.8 °C

2) 58.8 °C2) 58.8 °C

3) 41.0 °C3) 41.0 °C

Learning Check – Honors Only

Pizza is baked at 455°F. What is that in °C?

1) 437 °C1) 437 °C

2) 235°C2) 235°C

3) 221°C3) 221°C

What is Scientific Notation?What is Scientific Notation?

• Scientific notation is a way of Scientific notation is a way of expressing really big numbers or expressing really big numbers or really small numbers.really small numbers.

• It is most often used in It is most often used in “scientific” calculations where the “scientific” calculations where the analysis must be very precise.analysis must be very precise.

• For very large and very small For very large and very small numbers, scientific notation is numbers, scientific notation is more concise.more concise.

Scientific notation consists Scientific notation consists of two parts:of two parts:

• A number between 1 and 10A number between 1 and 10

• A power of 10A power of 10

N x 10N x 10xx

Are the following in scientific notation?

To change standard form To change standard form to scientific notation…to scientific notation…

• Place the decimal point so that there is Place the decimal point so that there is one non-zero digit to the left of the one non-zero digit to the left of the decimal point.decimal point.

• Count the number of decimal places Count the number of decimal places the decimal point has “moved” from the decimal point has “moved” from the original number. This will be the the original number. This will be the exponent on the 10.exponent on the 10.

• If the original number was less than 1, If the original number was less than 1, then the exponent is negative. If the then the exponent is negative. If the original number was greater than 1, original number was greater than 1, then the exponent is positive.then the exponent is positive.

ExamplesExamples

• Given: 289,800,000Given: 289,800,000

• Use: 2.898 (moved 8 places)Use: 2.898 (moved 8 places)

• Answer:Answer: 2.898 x 102.898 x 1088

• Given: 0.000567Given: 0.000567

• Use: 5.67 (moved 4 places)Use: 5.67 (moved 4 places)

• Answer:Answer: 5.67 x 105.67 x 10-4-4

To change scientific To change scientific notation to standard notation to standard

form…form…• Simply move the decimal point to Simply move the decimal point to

the right for positive exponent 10. the right for positive exponent 10.

• Move the decimal point to the left Move the decimal point to the left for negative exponent 10.for negative exponent 10.

(Use zeros to fill in places.)(Use zeros to fill in places.)

ExampleExample

• Given: 5.093 x 10Given: 5.093 x 1066

• Answer: Answer: 5,093,0005,093,000 (moved 6 places (moved 6 places to the right)to the right)

• Given: 1.976 x 10Given: 1.976 x 10-4-4

• Answer: Answer: 0.00019760.0001976 (moved 4 places (moved 4 places to the left)to the left)

Learning Check

• Express these numbers in Express these numbers in Scientific Notation:Scientific Notation:

1)1) 405789405789

2)2) 0.0038720.003872

3)3) 30000000003000000000

4)4) 22

5)5) 0.4782600.478260

DENSITYDENSITY - an important - an important and useful physical propertyand useful physical property

Density mass (g)volume (cm3)

Density mass (g)volume (cm3)

Mercury

13.6 g/cm13.6 g/cm33 21.5 g/cm21.5 g/cm33

Aluminum

2.7 g/cm2.7 g/cm33

Platinum

ProblemProblem A piece of copper has a A piece of copper has a mass of 57.54 g. It is 9.36 cm long, mass of 57.54 g. It is 9.36 cm long, 7.23 cm wide, and 0.95 mm thick. 7.23 cm wide, and 0.95 mm thick. Calculate density (g/cmCalculate density (g/cm33).).

Density mass (g)volume (cm3)

Density mass (g)volume (cm3)

StrategyStrategy1. Get dimensions in common units.1. Get dimensions in common units.

2.2. Calculate volume in cubic centimeters. Calculate volume in cubic centimeters.

3. Calculate the density.3. Calculate the density.

SOLUTIONSOLUTION

1. Get dimensions in common units.1. Get dimensions in common units.

2.2. Calculate volume in cubic centimeters. Calculate volume in cubic centimeters.

3. Calculate the density.3. Calculate the density.

0.95 mm • 1cm

10 mm = 0.095 cm

57.54 g

6.4 cm3 = 9.0 g / cm3

(9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm(9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm33

Note only 2 significant figures in the answer!Note only 2 significant figures in the answer!

PROBLEM: Mercury (Hg) has a density PROBLEM: Mercury (Hg) has a density of 13.6 g/cmof 13.6 g/cm33. What is the mass of 95 mL . What is the mass of 95 mL of Hg in grams? of Hg in grams?

PROBLEM: Mercury (Hg) has a density PROBLEM: Mercury (Hg) has a density of 13.6 g/cmof 13.6 g/cm33. What is the mass of 95 mL . What is the mass of 95 mL of Hg in grams? of Hg in grams?

Solve the problem usingSolve the problem using DIMENSIONAL DIMENSIONAL ANALYSIS.ANALYSIS.

StrategyStrategy

1.1. Use density to calc. mass (g) Use density to calc. mass (g) from from volume.volume.

PROBLEM: Mercury (Hg) has a density of PROBLEM: Mercury (Hg) has a density of 13.6 g/cm13.6 g/cm33. What is the mass of 95 mL of Hg?. What is the mass of 95 mL of Hg?PROBLEM: Mercury (Hg) has a density of PROBLEM: Mercury (Hg) has a density of 13.6 g/cm13.6 g/cm33. What is the mass of 95 mL of Hg?. What is the mass of 95 mL of Hg?

First, note thatFirst, note that 1 cm1 cm33 = 1 mL = 1 mL

1.1. Convert volume to massConvert volume to mass

PROBLEM: Mercury (Hg) has a density of 13.6 PROBLEM: Mercury (Hg) has a density of 13.6 g/cmg/cm33. What is the mass of 95 mL of Hg?. What is the mass of 95 mL of Hg?PROBLEM: Mercury (Hg) has a density of 13.6 PROBLEM: Mercury (Hg) has a density of 13.6 g/cmg/cm33. What is the mass of 95 mL of Hg?. What is the mass of 95 mL of Hg?

95 cm3 • 13.6 g

cm3 = 1.3 x 103 g

Learning CheckLearning Check

Osmium is a very dense metal. What is its Osmium is a very dense metal. What is its

density in g/cmdensity in g/cm3 3 if 50.00 g of the metal occupiesif 50.00 g of the metal occupies

a volume of 2.22 cma volume of 2.22 cm33??

1) 2.25 g/cm1) 2.25 g/cm33

2)2) 22.5 g/cm22.5 g/cm33

3)3) 111 g/cm111 g/cm33

Solution

2) Placing the mass and volume of the osmium 2) Placing the mass and volume of the osmium metal into the density setup, we obtainmetal into the density setup, we obtain

D = D = massmass = = 50.00 g 50.00 g = = volumevolume2.22 cm2.22 cm33

= 22.522522 g/cm= 22.522522 g/cm3 3 == 22.5 g/cm22.5 g/cm33

Volume DisplacementVolume Displacement

A solid displaces a matching volume of A solid displaces a matching volume of water when the solid is placed in water.water when the solid is placed in water.

33 mL33 mL25 mL 25 mL

Learning CheckLearning Check

What is the density (g/cmWhat is the density (g/cm33) of 48 g of a metal if ) of 48 g of a metal if the metal raises the level of water in a graduated the metal raises the level of water in a graduated cylinder from 25 mL to 33 mL? cylinder from 25 mL to 33 mL?

1) 0.2 g/ cm1) 0.2 g/ cm33 2) 6 g/cm 2) 6 g/cm33 3) 252 g/cm3) 252 g/cm33

33 mL33 mL

25 mL25 mL

Learning CheckLearning Check

Which diagram represents the liquid layers in the Which diagram represents the liquid layers in the cylinder?cylinder?

(K) Karo syrup (1.4 g/mL), (V) vegetable oil (0.91 (K) Karo syrup (1.4 g/mL), (V) vegetable oil (0.91 g/mL,) (W) water (1.0 g/mL)g/mL,) (W) water (1.0 g/mL)

1)1) 2) 2) 3) 3)

K

K

W

W

W

V

V

V

K

SolutionSolution

(K) Karo syrup (1.4 g/mL), (V) vegetable oil (K) Karo syrup (1.4 g/mL), (V) vegetable oil (0.91 g/mL,) (W) water (1.0 g/mL)(0.91 g/mL,) (W) water (1.0 g/mL)

1)1)

K

W

V

Learning CheckLearning Check

The density of octane, a component of The density of octane, a component of gasoline, is 0.702 g/mL. What is the gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane?mass, in kg, of 875 mL of octane?

1) 0.614 kg1) 0.614 kg

2) 614 kg2) 614 kg

3) 1.25 kg3) 1.25 kg

Learning CheckLearning Check

If blood has a density of 1.05 g/mL, how If blood has a density of 1.05 g/mL, how many liters of blood are donated if 575 g many liters of blood are donated if 575 g of blood are given?of blood are given?

1) 1) 0.548 L0.548 L

2) 2) 1.25 L1.25 L

3) 3) 1.83 L1.83 L

Learning CheckLearning Check

A group of students collected 125 empty A group of students collected 125 empty aluminum cans to take to the recycling center. aluminum cans to take to the recycling center. If 21 cans make 1.0 pound of aluminum, how If 21 cans make 1.0 pound of aluminum, how many liters of aluminum (D=2.70 g/cmmany liters of aluminum (D=2.70 g/cm33) are ) are obtained from the cans?obtained from the cans?

1) 1.0 L1) 1.0 L 2) 2.0 L2) 2.0 L 3) 4.0 L3) 4.0 L