Chapter 2 Quantitative measurements give results in a definite form, usually a number Example...

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Chapter 2Chapter 2

Quantitative measurementsQuantitative measurements

give results in a definite form, give results in a definite form, usually a numberusually a number

Example

ThermometerThermometer

Color of a reactionColor of a reaction

Qualitative measurementsQualitative measurements

gives results in a descriptive formgives results in a descriptive form

Example

non-numeric valuenon-numeric value

Reproducibility of the measurementReproducibility of the measurement

How close a measurement comes to How close a measurement comes to the true valuethe true value

AccuracyAccuracy

PrecisionPrecision

include all digits that can be known include all digits that can be known accurately plus a last digit that is accurately plus a last digit that is

estimatedestimated

Significant FiguresSignificant Figures

Rules for Significant FiguresRules for Significant Figures

(All have 3 Sig. Figs.)(All have 3 Sig. Figs.)

1.1. Every nonzero digit in a recorded Every nonzero digit in a recorded measurement is significantmeasurement is significant

Examples

65.65.22

0.0.262688

126126

2. Zeros between nonzero digits are 2. Zeros between nonzero digits are significantsignificant

Example

22000044

6600.39.39 8.68.60022

(All have 4 Sig. Figs.)(All have 4 Sig. Figs.)

3. Zeros in front of all nonzero digits 3. Zeros in front of all nonzero digits are are NOTNOT significant. They act as significant. They act as place-holders.place-holders.

Examples

00..00006622

00.58.58 00..0000002277

(2 Sig. Figs.) (2 Sig. Figs.)

4. Zeros at the end of a number and 4. Zeros at the end of a number and to the right of a decimal point are to the right of a decimal point are significantsignificant

Examples

61.61.0000 1.031.0300 8.8.000000

(4 Sig. Figs.)(4 Sig. Figs.)

Zeros at the end of a measurement Zeros at the end of a measurement and to the left of the decimal can and to the left of the decimal can be confusing. be confusing.

If they are place-holders to show If they are place-holders to show the magnitude of a number, they the magnitude of a number, they are not significantare not significant

Examples

300 700 300 700 2721027210

ambiguousambiguous

If the zeros were measured- If the zeros were measured- significantsignificant

700.

To avoid confusion use scientific To avoid confusion use scientific notationnotation

3.00 x 10 3.00 x 10 22

How Many Sig. Figs.?How Many Sig. Figs.?

123 123

0.123 0.123

40,506 40,506

9.8000 x 10 9.8000 x 10 44

98,000 98,000

0.07080 0.07080

0.078 0.078

33

55

33

55

22

22

44

Sig. Figs in calculationsSig. Figs in calculations

If digit following last sig fig is <5, all If digit following last sig fig is <5, all digits are droppeddigits are dropped

If digit following last sig fig is >5, digit in If digit following last sig fig is >5, digit in last sig fig place is increased by onelast sig fig place is increased by one

Round These…….

314.72314.72110.001770.0017755

64.32 x 10 64.32 x 10 -1-1

4 sig figs4 sig figs 314.7314.7

2 sig 2 sig figsfigs

0.00180.0018

1 1 sig sig figfig

60 x 10 60 x 10 --

1 1 or 6 or 6

Addition and SubtractionAddition and Subtraction

The answer cannot contain any The answer cannot contain any more digits to the right of the more digits to the right of the decimal point than are contained in decimal point than are contained in the measurement with the least the measurement with the least number of digits to the right of the number of digits to the right of the decimal pointdecimal point

12.52+

349.0

8.24

369.8 or 3.698 x 10 2

Sample Problem

+

Multiplication and DivisionMultiplication and Division

Answer must contain no more sig figs than Answer must contain no more sig figs than the measurement with the least number the measurement with the least number

of sig figs. of sig figs.

*Decimal point has nothing to do with *Decimal point has nothing to do with determining this!determining this!

Example Problems

755755 x .0 x .03434 = =2626

3 sig. figs. 2 sig. figs 2 sig. figs

2.4526 2.4526 8.4 8.4 ==

0.290.29

►Metric SystemMetric System

►SI system (International SI system (International System of Units)- System of Units)- Developed in 1790 in Developed in 1790 in FranceFrance

QuantityQuantity SI Unit Non-SI

Length m

Volume m3 Liters

Mass kg grams

Density g/cm³, g/ml

Temperature K oC

Time s

Pressure pa (pascal) atmospheres, mm Hg

Energy J (joules) calories

PrefixPrefix Symbol Symbol Magnitude Magnitude

kilokilo KK 1000x1000x

hecto hecto hh 100x100x

decadeca dada 10x10x

unitunit m, L, gm, L, g x x

decideci dd .1x.1x

centicenti cc .01x.01x

millimilli mm .001x.001x

micromicro uu 1 x 10 1 x 10 -6 -6 x x

nanonano nn 1 x 10 1 x 10 -9 -9 x x

picopico pp 1 x 101 x 10-12 -12 x x

Units of Measurement – Meter (m)Units of Measurement – Meter (m)

Units of Volume – Liter (L)Units of Volume – Liter (L)

Volume of a cube-10 cm x 10 cm Volume of a cube-10 cm x 10 cm x 10 cm= 1000 cm x 10 cm= 1000 cm 33

►1 L = 1 dm 1 L = 1 dm 33 = 1000 cm = 1000 cm 33

1 cm 1 cm 33 = 1 mL = 1 mL

►MassMass- quantity of matter- quantity of matter►WeightWeight- a force- a force►Unit of mass – gram (g) = Unit of mass – gram (g) =

mass of 1 cm mass of 1 cm 33 of H of H22O at 4 ºCO at 4 ºC

Units of Mass – (g)Units of Mass – (g)

►DensityDensity- ratio of mass of an - ratio of mass of an object to volumeobject to volume

D= M = g/mL or g/cm D= M = g/mL or g/cm 33

VV

►Specific GravitySpecific Gravity- - comparison of the density comparison of the density of a substance to the of a substance to the density of a reference density of a reference substancesubstance

Specific Gravity Specific Gravity

Specific Gravity Cont’dSpecific Gravity Cont’d

Sp. Gravity = density of substance g/cm Sp. Gravity = density of substance g/cm 33

density of Hdensity of H22O g/cm O g/cm 33

Sp. Gravity of a liquid is measured with Sp. Gravity of a liquid is measured with a hydrometera hydrometer

►TemperatureTemperature- Degree of hotness - Degree of hotness or coldness of an objector coldness of an object Determined direction of heat Determined direction of heat

transfertransfer►Heat transferHeat transfer- occurs when two - occurs when two

objects at different objects at different temperatures contact each other temperatures contact each other Heat goes from high Heat goes from high

temperature to low temptemperature to low temp

FahrenheitFahrenheit

32ºF - FP of H32ºF - FP of H22OO

212ºF - BP of H212ºF - BP of H22OO

Celsius (centigrade)Celsius (centigrade)

0ºC - 0ºC - FPFP of H of H22OO

100ºC - BP of H100ºC - BP of H22OO

KelvinKelvin

273 K - 273 K - FPFP of H of H22OO K = K = ooCC + 273 + 273

373 K - BP of H373 K - BP of H22OO

Temperature ScalesTemperature Scales

HeatHeat

►HeatHeat- measured in joule (J) or - measured in joule (J) or calories (cal)calories (cal) One calorie is the quantity of One calorie is the quantity of

heat that raises the heat that raises the temperature of 1 g of Htemperature of 1 g of H22O 1ºCO 1ºC►1 cal = 4.18 J1 cal = 4.18 J

Calorie in nutritional terms Calorie in nutritional terms means kilocaloriemeans kilocalorie

Heat Cont’dHeat Cont’d

►Heat CapacityHeat Capacity- quantity of heat - quantity of heat required to change an objects required to change an objects temperature by exactly 1ºCtemperature by exactly 1ºC Depends on mass and type of Depends on mass and type of

substancesubstance

Specific HeatSpecific Heat

►The quantity of heat required to The quantity of heat required to raise 1g of a substance 1raise 1g of a substance 1ºCºC Sp. Heat = heat = qSp. Heat = heat = q

mass x mass x T (m) (T) T (m) (T)

Units of Specific Heat:Units of Specific Heat:

J/gJ/g°C or cal/g°C°C or cal/g°C

Chapter Chapter 2B2B

Problem Solving in Problem Solving in ChemistryChemistry

► Three easy steps to problem Three easy steps to problem solving. . . solving. . .

Step 1 : AnalyzeStep 1 : Analyze

► Identify a knownIdentify a known Determine where you are starting Determine where you are starting

from.from. What information do you already What information do you already

have to work with?have to work with?

Step 1 cont’d…Step 1 cont’d…

►Identify an unknown Identify an unknown Where are you going?Where are you going? What are you looking for?What are you looking for?

►Plan a solutionPlan a solution How are you going to get there?How are you going to get there?

Step 2: CalculateStep 2: Calculate

► May involve substituting known May involve substituting known quantities and doing the arithmetic quantities and doing the arithmetic needed to solve for unknown.needed to solve for unknown.

► You may also need to convert.You may also need to convert.

Step 3: EvaluateStep 3: Evaluate

► Go over your answersGo over your answers Does the answer make sense?Does the answer make sense? Did you use correct units?Did you use correct units?

►Check your workCheck your work Make sure you copied down the given Make sure you copied down the given

information correctly.information correctly.

Sample ProblemSample Problem

►What is the mass, in What is the mass, in grams, of a piece of lead grams, of a piece of lead that has a volume of that has a volume of 19.84 cm19.84 cm³?³?

Step 1 : AnalyzeStep 1 : Analyze

►List Knowns and UnknownsList Knowns and Unknowns KnownsKnowns : :

►Volume of lead: 19.84 cmVolume of lead: 19.84 cm³³►Density of lead = 11.4 g/cm³ Density of lead = 11.4 g/cm³

(according to table 3.7)(according to table 3.7)►Density = massDensity = mass

volumevolume UnknownsUnknowns: mass = ?g: mass = ?g

► Density = mass -or- Mass = volume x Density = mass -or- Mass = volume x densitydensity

volumevolume

► Mass = 19.84 cm Mass = 19.84 cm ³ x 11.4 g/cm ³³ x 11.4 g/cm ³

= 226.176 g= 226.176 g

Mass = 226 gMass = 226 g

Step 2: CalculateStep 2: Calculate

Step 3: EvaluateStep 3: Evaluate

►Has the unknown been found? Has the unknown been found? Yes, problem asks for massYes, problem asks for mass

►Do you have the correct units? Do you have the correct units? Yes, units canceled correctly to yield grams Yes, units canceled correctly to yield grams

(g)(g)► Is the number of sig figs correct?Is the number of sig figs correct?

Yes, answer has 3 sig figsYes, answer has 3 sig figs

Mass = 226 gMass = 226 g

What is the volume, in What is the volume, in cubic centimeters of a cubic centimeters of a sample of cough syrup sample of cough syrup that has a mass of 50.0g? that has a mass of 50.0g? The density of cough The density of cough syrup is 0.950g/cmsyrup is 0.950g/cm³.³.

Volume = mass

density

Volume = 50.0g = 52.6316 cm³

0.950g/cm³

Volume = 52.6 cm³

Your school club has sold 600 Your school club has sold 600 tickets to a chili-supper tickets to a chili-supper

fundraising event, and you have fundraising event, and you have volunteered to make chili. You volunteered to make chili. You have a chili recipe that serves have a chili recipe that serves 10. The recipe calls for two 10. The recipe calls for two

teaspoons of chili powder. How teaspoons of chili powder. How much chili powder do you need much chili powder do you need

for 600 servings?for 600 servings?

Servings Needed = 600Servings Needed = 600

10 servings = 2 tsp chili powder10 servings = 2 tsp chili powder

Amount of chili powder = ? tspAmount of chili powder = ? tsp

600 servings x 2 tsp chili powder = 600 servings x 2 tsp chili powder =

10 servings10 servings

1200 tsp chili powder = 120 tsp chili 1200 tsp chili powder = 120 tsp chili powderpowder

1010

►How many cups are in 120 How many cups are in 120 teaspoons of chili powder?teaspoons of chili powder?

120 tsp x 1 tbs x 1 cup = 120 tsp x 1 tbs x 1 cup =

3 tsp 16 tbs3 tsp 16 tbs

120 cups = 120 cups = 2.5 cups2.5 cups

4848

►Express 750 dg in gramsExpress 750 dg in grams

Mass = 750 dgMass = 750 dg

1 g = 10 dg1 g = 10 dg

Mass = ?gMass = ?g

750 dg x 1g = 750 dg x 1g = 75 g75 g

10 dg 10 dg

►What is 0.073 km in cm?What is 0.073 km in cm?

0.073 km x 1000 m x 100 cm = 0.073 km x 1000 m x 100 cm =

1 km 1 m 1 km 1 m

7300 cm = 7300 cm = 7300 cm7300 cm

1 1

► How many seconds are in How many seconds are in one day?one day?

1 day x 24 h x 60 min x 60 s 1 day x 24 h x 60 min x 60 s = =

1 day 1 h 1 min1 day 1 h 1 min

86400 s = 8.64 x 1086400 s = 8.64 x 104 4 ss

11