Scientific NotationUnits of Measurement

Significant FiguresConversion Calculations

Density

Chapter 2: Measurement

CH104 1

Scientific Notation

CH104 2

If a number is larger than 1

• Move decimal point X places left to get a number between 1 and 10.

Scientific notation

1 2 3 , 0 0 0 , 0 0 0.

• The resulting number is multiplied by 10X.

= 1.23 x 108

CH104 3

If a number is smaller than 1•Move decimal point X places right to get a number between 1 and 10.

Scientific notation

0. 0 0 0 0 0 0 1 2 3 = 1.23 x 10-7

• The resulting number is multiplied by 10-X.

CH104 4

Write in Scientific Notation:

25 =

8931.5 =

0.000593 =

0.0000004 =

3,210. =

Examples

2.5 x 10 1

8.9315 x 10 3

5.93 x 10 - 4

4 x 10 - 7

3.210 x 103

CH104 5

x 10

1.44939 x 10-2 =

Scientific notation

+

-1

/

x

0

2 3

4 5 6

7 8 9

.

CE

EE

log

ln

1/x

x2

cos tan

1.44939 E-2

0.0144939

On Calculator 1.44939 (-) 2EE

Means x 10 Change

Sign

CH104 6

Chapter 2 Measurements

2.1Units of Measurement

CH104 14

Units are important

Measurements in chemistry

has little meaning, just a number

has some meaning - money

more meaning - person’s salary

CH104 15

Metric SI CommonConversions

Length

Units of Measurement

meter (m) meter (m)

1 m = 1.09 yd2.54 cm = 1 in

1 m = 100 cm = 1000 mm

CH104 17

Metric SI CommonConversions

Length

Volume

Units of Measurement

meter (m) meter (m)

1 m = 1.09 yd2.54 cm = 1 in

1 m = 100 cm = 1000 mm

liter (L) cubic meter (m3)

1 L = 1.06 qt

946 mL = 1 qt

1 L = 1000 mL

CH104 18

Metric SI CommonConversions

Length

Volume

Mass

Units of Measurement

meter (m) meter (m)

1 m = 1.09 yd2.54 cm = 1 in

1 m = 100 cm = 1000 mm

liter (L) cubic meter (m3)

1 L = 1.06 qt

946 mL = 1 qt

1 L = 1000 mL

gram (g) Kilogram (kg)1 kg = 2.20 lb

1 kg = 1000 g

454 g = 1 lbCH104 19

The amount of material in an object Mass:Mass (in g’s) of

a 1L Bowling Ball > a 1 L Balloon

Weight:

Mass Vs. Weight

Pull of Gravity on an object.

Weight of Person on Earth > Person on Moon

CH104 20

Mass Vs. Weight

How much would you weigh on another planet?

http://www.exploratorium.edu/ronh/weight/

CH104 21

Metric SI CommonConversions

Length

Volume

Mass

Time

Temp

Units of Measurement

meter (m) meter (m) 1 m = 1.09 yd 2.54 cm = 1 in

liter (L) cubic meter (m3)

gram (g) Kilogram (kg) 1 kg = 2.20 lb

1 L = 1.06 qt

946 mL = 1 qt

Celsius (oC) Kelvin (K) oC = (oF-32)/1.8 K = oC + 273

second (s) second (s) 60 s = 1 min

CH104 22

For each of the following, indicate whether the unit describes

1) length, 2) mass, or 3) volume.

____ A. A bag of tomatoes is 4.5 kg.

____ B. A person is 2.0 m tall.

____ C. A medication contains 0.50 g of aspirin.

____ D. A bottle contains 1.5 L of water.

23

Learning Check

CH104

For each of the following, indicate whether the unit describes

1) length, 2) mass, or 3) volume.

____ A. A bag of tomatoes is 4.5 kg.

____ B. A person is 2.0 m tall.

____ C. A medication contains 0.50 g of

aspirin.

____ D. A bottle contains 1.5 L of water. 24

Solution

1

2

2

3

CH104

Identify the measurement that has an SI unit.

A. John’s height is _____.1) 1.5 yd 2) 6 ft 3) 2.1

m

B. The race was won in _____. 1) 19.6 s 2) 14.2 min 3)

3.5 h

C. The mass of a lemon is _____.1) 12 oz 2) 0.145 kg 3)

0.62 lb

D. The temperature is _____.1) 85 °C 2) 255 K 3) 45

°F25

Learning Check

CH104

26

Solution

CH104

Identify the measurement that has an SI unit.

A. John’s height is _____.1) 1.5 yd 2) 6 ft 3) 2.1

m

B. The race was won in _____. 1) 19.6 s 2) 14.2 min 3)

3.5 h

C. The mass of a lemon is _____.1) 12 oz 2) 0.145 kg 3)

0.62 lb

D. The temperature is _____.1) 85 °C 2) 255 K 3) 45

°F

2.2Measured Numbers

and Significant Figures

27

Chapter 2 Measurements

CH104

Measured & Exact Numbers

Exact Numbers = from counting or by definition

12 coins per package12 coins per package

12 coins 1 package

12 coins 1 package

1 package12 coins

1 package12 coins=

12 coins 1 dozen coins 12 coins

1 dozen coins 1 dozen coins

12 coins 1 dozen coins

12 coins=

CH104 28

29

Examples of Exact Numbers

from numbers in a defined relationship

when objects are counted

CH104

Measured & Exact Numbers

Measured Numbers = estimated using a tool

All measurements contain some uncertainty.

•We make errors

•Tools have limits

CH104 30

Length of object is between 6.7 and 6.8The next digit would be a guess.            

Significant figures

If use 6.76 then have error of + 0.01cm and have 3 significant figures.

CH104 33

Expresses accuracy & precision.

You can’t report values more accurate than the methods of measurement used .

6.76 units = 3 significant figures

Significant figures

Certain Digits

UncertainDigit

CH104 34

We can estimate the value to be 8.45 mL but cannot be more accurate than that.

8.45 has 3 sig figs.

Significant figuresMeniscus is between

8.4 and 8.5The next digit would

be a guess.

Sig Figs don’t depend on the decimal point.

255 millimeters 25.5 centimeters 2.55 decimeters 0.255 meters 0.0255 decameters

Significant figures

CH104 38

Significant figures: Rules for zeros

0.00421 Leading zeroLeading zero

Captive zeros are significant. 4012

Trailing zeros behind decimal are significant.114.20

Captive zeroCaptive zero

Trailing zeroTrailing zero

Leading zeros are not significant.3 sig figs

4 sig figs

5 sig figs

CH104 39

32,000Are the 0’s significant?

2 sig figs =

3 sig figs =

4 sig figs =

5 sig figs =

Significant figures: Rules for zeros

3.2 x 104

3.20 x 104

3.200 x 104

3.2000 x 104

32,000.CH104 40

1025 km

2.00 mg

0.00570

520

Significant figures: Rules for zeros

Three (only trailing zero behind decimal

is significant, leading zeros are not)

Four (Captive zeros are significant)

Three (trailing zeros behind decimal

are significant)

Two (No decimal, zero assumed insignif)

CH104 41

In scientific notation:All digits, including zeros in the coefficient, are

significant.

Scientific Notation Number of Significant

Figures___________8 x 104 m8.0 x 104 m8.00 x 104 m

42

Significant Figures inScientific Notation

123

CH104

State the number of significant figures in each of the following measurements:

A. 0.030 m

B. 4.050 L

C. 0.0008 g

D. 2.80 m43

Learning Check

CH104

State the number of significant figures in each of the following measurements:

A. 0.030 m

B. 4.050 L

C. 0.0008 g

D. 2.80 m

44

Solution

2

4

1

3

CH104

45

Learning Check

CH104

A. Which answer(s) contains 3 significant figures?1) 0.4760 2) 0.00476 3) 4.76 x 103

B. All the zeros are significant in

1) 0.00307 2) 25.300 3) 2.050 x 103

C. The number of significant figures in 5.80 x 102 is1) one 3) two 3) three

46

Solution

CH104

A. Which answer(s) contains 3 significant figures?1) 0.4760 2) 0.00476 3) 4.76 x 103

B. All the zeros are significant in

1) 0.00307 2) 25.300 3) 2.050 x 103

C. The number of significant figures in 5.80 x 102 is1) one 3) two 3) three

Classify each of the following as exact (E) or measured (M) numbers. Explain your answer.

A. __ Gold melts at 1064 °C.

B. __ 1 yard = 3 feet

C. __ The diameter of a red blood cell is 6 x 10-4 cm.

D. __ There are 6 hats on the shelf.

E. __ A can of soda contains 355 mL of soda.

51

Learning Check

CH104

52

Solution

M

M

M

E

E

CH104

Classify each of the following as exact (E) or measured (M) numbers. Explain your answer.

A. __ Gold melts at 1064 °C.

B. __ 1 yard = 3 feet

C. __ The diameter of a red blood cell is 6 x 10-4 cm.

D. __ There are 6 hats on the shelf.

E. __ A can of soda contains 355 mL of soda.

Chapter 2 Measurements

2.3Significant Figures in

Calculations

53CH104

Write with 4 Significant Figures:

2.579 035

Rounding

1st insignificant digit1st insignificant digit

34.20 221 becomes 34.20

becomes 2.5805

4

> 5 round up

< 5 round down.

> 5 round up

< 5 round down.

Sometimes a calculated answer shows too many significant digits so we need to round.

CH104 54

41.50.212

Adding Significant ZerosSometimes a calculated answer requires more significant digits so we need to add zeros.

Calculated answer

Zeros added to give 3 significant figures

4.00

1.50

0.200

12.0

CH104 55

Adjust the following calculated answers to giveanswers with three significant figures.

A. 824.75 cm

B. 0.112486 g

C. 8.2 L

56

Learning Check

CH104

Adjust the following calculated answers to giveanswers with three significant figures.

A. 824.75 cm

B. 0.112486 g

C. 8.2 L

Solution

First digit dropped is greater than 4.

825 cm

First digit dropped is 4.0.112 g

Significant zero is added.8.20 L

CH104 57

An answer can’t have greater significance than the quantities used to produce it.

Significant figuresand calculations

speed = 1.00 km 3.0 min

+

-1

/

x

0

2 3

4 5 6

7 8 9

.

CE

EE

log

ln

1/x

x2

cos tan

0.3333333333

= ?

Example How fast did you run if youwent 1.00 km in 3.0 minutes?

CH104 58

Multiplication & Division Problems:Do calculations.

Simplified rules for significant figures

•Look at sig figs for each value in calculation. (Constants don’t count.)

•Report answer with same sig figs as least significant value.

•Round off as needed.

speed = 1.00 km 3.0 min

= 0.333333333 km min

= 0.33 km min

2 sig figs

3 sig figs

CH104 59

Simplified rules for significant figuresAddition & Subtraction Problems:• Do calculations.

•Look at least significant place for each value in calculation.

•Report answer to least significant place.

•Round off as needed.

1.9+ 18.65 20.55

= 20.6

Significant to .1

Significant to .01

Significant to .1

CH10461

(1.9 + 18.65 ) =2.153

Add & Sub mixed w/ Mult & Div Problems:

( 20.55 ) = 2.153

• Do Addition & Subtraction calculations 1st.

• Make note of the least significant place.

3 sig figs (after addition)

4 sig figs

CH10462

• Do Multiplication & Division calculations.

9.54482118

• Round to least # sig fig.

9.549.54

(1.9 + 18.65 ) =2.153

( 20.55 ) = 2.153

3 sig figs (after addition)

Add & Sub mixed w/ Mult & Div Problems:

4 sig figs

CH10463

Chapter 2 Measurements

2.4Prefixes and Equalities

CH104 68

A prefix in front of a unit increases or decreases the size of that unit by one or more factors of 10

indicates a numerical value

Prefix Value1 kilometer = 1000 meters1 kilogram = 1000 grams

69

Prefixes

CH104

Metric prefixesPrefix

(Symbol)Factor

(multiple)Common

Conversion

mega (M)

kilo (k)

deci (d)

centi (c)

milli (m)

micro (m)nano (n)

1,000,000 = (106) 1Mm = 1,000,000 m

1,000 = (103) 1km = 1,000 m

0.1 = (10-1) 1m = 10 dm

0.01 = (10-2) 1m = 100 cm

0.001 = (10-3) 1m = 1,000 mm0.000001 = (10-6) 1m = 1,000,000 mm

0.000,000,001 = (10-9) 1m = 1,000,000,000 nm

Indicate the unit that matches the description:1. a mass that is 1000 times greater than 1 gram

1) kilogram 2) milligram 3) megagram

2. a length that is 1/100 of 1 meter 1) decimeter 2) centimeter 3) millimeter

3. a unit of time that is 1/1000 of a second 1) nanosecond 2) microsecond 3) millisecond

Learning Check

CH104 73

Solution

= 0.01 of 1 meter

= 0.001 of a sec

CH104 74

Indicate the unit that matches the description:1. a mass that is 1000 times greater than 1 gram

1) kilogram 2) milligram 3) megagram

2. a length that is 1/100 of 1 meter 1) decimeter 2) centimeter 3) millimeter

3. a unit of time that is 1/1000 of a second 1) nanosecond 2) microsecond 3) millisecond

Select the unit you would use to measure A. your height

1) millimeters 2) meters 3) kilometers

B. your mass 1) milligrams 2) grams 3) kilograms

C. the distance between two cities 1) millimeters 2) meters 3) kilometers

D. the width of an artery 1) millimeters 2) meters 3) kilometers

Learning Check

CH104 75

Solution

CH104 76

Select the unit you would use to measure A. your height

1) millimeters 2) meters 3) kilometers

B. your mass 1) milligrams 2) grams 3) kilograms

C. the distance between two cities 1) millimeters 2) meters 3) kilometers

D. the width of an artery 1) millimeters 2) meters 3) kilometers

States the same measurement in two different units

Metric Equalities

Length: 1 meter is the same as 100 cm or 1000 mm.1 m = 100 cm

1 m = 1000 mm

Volume: 1 L is the same as 1000 cm3.1 L = 10 cm X 10cm X 10 cm

1 L = 1000 mLMass: 1 kg = 1000 g

1 g = 1000 mg1 mg = 0.001 g1 mg = 1000 µg

CH104 77

Indicate the unit that completes each of the followingequalities:

A. 1000 m = 1) 1 mm 2) 1 km 3) 1dm

B. 0.001 g = 1) 1 mg2) 1 kg 3) 1dg

C. 0.1 s = 1) 1 ms 2) 1 cs 3) 1ds

D. 0.01 m = 1) 1 mm 2) 1 cm 3) 1dm

Learning Check

CH104 78

Indicate the unit that completes each of the followingequalities:

A. 1000 m = 1) 1 mm 2) 1 km 3) 1dm

B. 0.001 g = 1) 1 mg2) 1 kg 3) 1dg

C. 0.1 s = 1) 1 ms 2) 1 cs 3) 1ds

D. 0.01 m = 1) 1 mm 2) 1 cm 3) 1dm

Solution

CH104 79

Chapter 2 Measurements

2.5Writing Conversion Factors

82CH104

See Handout Sheet of Common conversion factors

&Handout of

Conversion Problems

Measurements in chemistry

CH104 83

Some Common Equalities

84CH104

Write equalities and conversion factors for each pair of units:A. liters and mL

B. hours and minutes

C. meters and kilometers

85

Learning Check

CH104

Write equalities and conversion factors for each pair of units:A. liters and mL

B. hours and minutes

C. meters and kilometers

86

Solution

Equality: 1 L = 1000 mL

1 L and 1000 mL 1000 mL 1 L

Equality: 1 hr = 60 min 1 hr and 60 min

60 min 1 hr

Equality: 1 km = 1000 m

1 km and 1000 m

1000 m 1 kmCH104

Write the equality and conversion factors for each of the following:A. meters and centimeters

B. jewelry that contains 18% gold

C. one liter of gas is $ 0.95

87

Learning Check

CH104

Write the equality and conversion factors for each of the following:A. meters and centimeters

B. jewelry that contains 18% gold

C. one liter of gas is $ 0.95

88CH104

Solution

1 m and 100 cm 100 cm 1 m

18 g gold and 100 g jewelry100 g jewelry 18 g gold

1 L and $0.95 $0.95 1 L

Chapter 2 Measurements

2.6Problem Solving

89CH104

Example: Metric Conversion How many milligrams (mg) are in 5 kilograms (kg)?

Factor label methodIdentify your conversions factors.

Conversion of units

1 kg = 11000 g

1000 g = 11 kg

1 g = 11000 mg

1000 mg = 11 g

CH104 90

• Identify what is to the problem.

• Identify how you want the answer to look.

5 kg = mg

Example: Metric Conversion How many milligrams are in 5 kilograms?

CH104 91

• Multiply by conversion factors until units cancel.

• If the words work, the numbers will work.

5 kg

1

= mg1000 g

1 kg

1000 mg

1 g

5,000,000

Example: Metric Conversion How many milligrams are in 5 kilograms?

CH104 92

Example: English-Metric Conversion

You have a pen of rats each with an average weight of 0.75 lb. How many mg rubbing alcohol will it take to kill ½ of the population if theLD50 is 5000. mg/kg ?

• Identify your conversions factors.

1 kg Bw = 15000 mg Alc

5000 mg Alc = 11 kg Bw

1.0 kg Bw = 12.2 lb Bw

2.2 lb Bw = 11.0 kg Bw

CH104 97

0.75 lbBW

Example: English-Metric Conversion

You have a pen of rats each with an average weight of 0.75 lb. How many mg rubbing alcohol will it take to kill ½ of the population if theLD50 is 5000. mg/kg ?

1.0 kgBW

2.2 lbBW

5000. mgAlc

1 kg BW= mgAlc 1704.545

1700 mg = 1.7 x 103

Identify what is unique to the problem.

Identify how you want the answer to look.

CH104 98

Given (unique) = 1.60 days

Needed unit = ? min

Plan = days hours min

Example:How many minutes are 1.60 days?

Set up problem to cancel hours (h). 1.60 days x 24 hrs x 60 min = 1 day 1 hr

3 SigFigs Exact Exact = 3 SigFigs

2300 min = 2.3 x 103

2304 min

CH104 102

13 males x 100 = 37.1429 %

35 Students

PercentagesPart x 100 =Whole

%

___100

Secret code for

37% male

CH104110

Percentages as Conversion FactorsExample: The population of the automotive repair course is 37% male. Of the 75 students in the class how many are men?

Secret code for

37% male = 37 male 100 students

100 students 37 male

CH104 111

• Identify your conversions factors.

= males

Percentages as Conversion FactorsExample: The population of the automotive repair course is 37% male. Of the 75 students in the class how many are men?

• Identify what is to the problem.

• Identify how you want the answer to look.

75 students

1

37 males

100 students27.75

28 males

CH104 112

=

100 mL

10 % Alcohol

PercentagesPart x 100 =Whole

%

___100

Secret code for

10 mL Alcohol

Solution

CH104 115

Percentages as Conversion FactorsExample: An athlete normally has 15 % body fat. How many lbs of fat does a 74 kg athlete have?

Secret code for

15% Body Fat = 15 lb Fat 100 lb BW

100 lb BW 15 lb Fat

CH104 116

• Identify your conversions factors.

2.2 lbBw

1.0 KbBw= lb fat

Percentages as Conversion Factors

• Identify what is to the problem.

• Identify how you want the answer to look.

74 KgBw

24.42

24 lb fat

Example: An athlete normally has 15 % body fat. How many lbs of fat does a 74 kg athlete have?

15 lb Fat 100 lb BW

CH104 117

Learning Check:If the thickness of the skin fold at the waist indicates an 11% body fat, how much fat is in a person with a mass of 86 kg?

CH104 118

Solution:

11% body fat means 11 kg fat 100 kg

86 kg x 11 kg fat = 9.5 kg of fat 100 kg

If the thickness of the skin fold at the waist indicates an 11% body fat, how much fat is in a person with a mass of 86 kg?

CH104 119

2.7Density

120

Chapter 2 Measurements

CH104

Water 1.0 Urine 1.01 - 1.03Air 0.0013 Bone 1.7 - 2.0Gold 19.3 Gasoline 0.66 - 0.69

DensityDensity =

Mass

Volume

1cc = 1 cm3 = 1 ml = 1 g water 1cc = 1 cm3 = 1 ml = 1 g water

g cm3

g mlor At 4 o C

CH104 121

Example.Density calculation

What is the density of 5.00 ml of serum if it has a mass of 5.230 grams?

d = m V

d = 5.230 g 5.00 ml

= 1.05 g ml

CH104 123

Osmium is a very dense metal. What is its density in g/cm3 if 50.0 g of osmium has a volume of 2.22 cm3?

1) 2.25 g/cm3

2) 22.5 g/cm3

3) 111 g/cm3

124

Learning Check

CH104

Given: mass = 50.0 g ,volume = 22.2 cm3

Need: Density

D = mass = 50.0 g volume 2.22 cm3

Calculator = 22.522522 g/cm3

Final answer (2 SF) = 22.5 g/cm3

Solution

CH104 125

Osmium is a very dense metal. What is its density in g/cm3 if 50.0 g of osmium has a volume of 2.22 cm3?

1) 2.25 g/cm3

2) 22.5 g/cm3

3) 111 g/cm3

A solid completely submerged in water displaces its own volume of water.The volume of the solid is calculated from the volume difference.Volume of solid = 45.0 mL – 35.5 mL = 9.5 mL = 9.5 cm3

128

CH104

The density of the zinc object is calculated from its mass and volume. mass = 68.60 g = 7.2 g/cm3 volume 9.5 cm3

Density Using Volume Displacement

What is the density (g/cm3) of 48.0 g of a metal if the level of water in a graduated cylinder rises from 25.0 mL to 33.0 mL after the metal is added?

1) 0.17 g/cm3 2) 6.0 g/cm3 3) 380 g/cm3

Learning Check

CH104 129

33.0 mL

25.0 mL object

25.0 mL

Given: 48.0 g Volume of water = 25.0 mL

Volume of water + metal = 33.0 mL

Need: Density (g/cm3 )

Volume of metal = 33.0 mL – 25.0 mL = 8.0 mL

8.0 mL x 1 cm3 = 8.0 cm3

1 mLSet up problem:

Density = 48.0 g = 6.0 g = 6.0 g/cm3

8.0 cm3 1 cm3 (2 SF)

Solution

CH104 130

What is the density (g/cm3) of 48.0 g of a metal if the level of water in a graduated cylinder rises from 25.0 mL to 33.0 mL after the metal is added?

1) 0.17 g/cm3 2) 6.0 g/cm3 3) 380 g/cm3

Sink or FloatIce floats in water because the density of ice is less than the density of water. Aluminum sinks in water because its density is greater than the density of water.

131CH104

Which diagram correctly represents the liquid layers in the cylinder? Karo (K) syrup (1.4 g/mL), vegetable (V) oil (0.91 g/mL), water (W) (1.0 g/mL)

1 2 3

Learning Check

K

K

W

W

W

V

V

V

K

CH104 132

1)

vegetable oil (0.91 g/mL)

water (1.0 g/mL)

Karo syrup (1.4 g/mL)

133

Solution

K

W

V

CH104

Density as a ConversionA liquid sample with a density of 1.09 g/mL is

found to weigh 7.453 grams. What is the volume of the liquid in mLs?

A liquid sample with a density of 1.09 g/mL is found to weigh 7.453 grams. What is the volume of the liquid in mLs?

1.09 g1 ml

1 ml1.09 g

• Identify any conversion factors.

•How should the answer look?

7.453 g = ml

• What is unique to the problem?

1 ml1.09 g

6.837614 = 6.84 ml

CH104 134

If olive oil has a density of 0.92 g/mL, how many liters of olive oil are in 285 g of olive oil?

1) 0.26 L 2) 0.31 L 3) 310 L

Learning Check

CH104 137

Given: D = 0.92 g/mL mass = 285 g

Need: volume in L

Plan: g mL L

Equalities: 1 mL = 0.92 g 1 L = 1000 mL

Set up: 285 g x 1 mL x 1 L = 0.31 L

0.92 g 1000 mL density metric

factor factor

SolutionIf olive oil has a density of 0.92 g/mL, how many liters of olive oil are in 285 g of olive oil?

1) 0.26 L 2) 0.31 L 3) 310 L

CH104 138

Specific Gravity =

Specific gravitydensity of substance g

mldensity of reference g

ml

Referencecommonly

water at 4oC

•Specific Gravity is unitless.

•density = specific gravity (if at 4oC)

CH104 143

Specific gravity

•Commonly used to test sugar in urine.

Hydrometer

• Float height will be based on Specific Gravity.