1

INTRODUCTIONINTRODUCTION

andand

CHAPTER ONECHAPTER ONERead the Introduction and Chapter 1.

Chemistry is NOT a spectator sport. Work out complete solutions for all the bold numbered problems.

2

Welcome to the Welcome to the World of World of

ChemistryChemistry

3

Beginning Chemistry 200Beginning Chemistry 200

• Read the book before class!

• Study the examples and do the

practice exercises.

• Be prepared for each class

session, lecture and laboratory.

• Don’t get behind.

4

Outline for Chapter 1Outline for Chapter 1• Definitions

–Homogenous and Heterogeneous–Matter ?–3 states of matter–Chemical –vs.- Physical Change

• Calculations–Density–Temperature

»Kelvin Celsius Fahrenheit»Fahrenheit Celsius Kelvin

–Significant Figures (Sig Figs)

5

MatterMatter

6

MixturesMixtures

(a)HeterogeneousHeterogeneous

(b)SuspensionSuspension Blood

Salt water(c)HomogenousHomogenous

Cookie

7

The Language of ChemistryThe Language of Chemistry

• CHEMICAL ELEMENTSCHEMICAL ELEMENTS - pure - pure substances that cannot be decomposed by substances that cannot be decomposed by ordinary means to other substances.ordinary means to other substances.

SodiumBromineAluminum

8

The Language of ChemistryThe Language of Chemistry• The elements, their names, and The elements, their names, and

their symbols are given on the their symbols are given on the

PERIODIC TABLEPERIODIC TABLE• How many elements are there?How many elements are there?

9

The Periodic TableThe Periodic Table

Dmitri Mendeleev (1834 - 1907)Dmitri Mendeleev (1834 - 1907)

10

SodiumFind sodium, Na, on the chart.

11

Copper Copper atoms on atoms on silica silica surface.surface.

• An An atomatom is the smallest particle of an is the smallest particle of an element that has the chemical element that has the chemical properties of the element.properties of the element.

Find copper on the chart.

12

An An atomatom consists of a nucleus (of protons consists of a nucleus (of protons and neutrons) and electrons in space and neutrons) and electrons in space about the nucleus.about the nucleus.

The AtomThe Atom

Electron cloud

Nucleus

13

TheThe red compound is red compound is composed ofcomposed of• • Ni- NickelNi- Nickel• • C- CarbonC- Carbon• • O- OxygenO- Oxygen•• N- Nitrogen N- Nitrogen

CHEMICAL COMPOUNDSCHEMICAL COMPOUNDS are composed of atoms and so can be are composed of atoms and so can be

decomposed to those atoms.decomposed to those atoms.

FixedFixed compositioncomposition

14

MOLECULE MOLECULE The smallest unit of a compound that The smallest unit of a compound that retains the chemical characteristics of retains the chemical characteristics of the compound.the compound.

MOLECULAR FORMULAMOLECULAR FORMULAComposition of moleculesComposition of molecules

HH22OO CC88HH1010NN44OO22 - Caffeine

15

The Nature of MatterThe Nature of Matter

Chemists are interested in the nature of Chemists are interested in the nature of matter and how this is related to its atoms matter and how this is related to its atoms

and molecules.and molecules.

GoldAu

MercuryHg

16

Graphite Graphite layer layer structure of structure of carbon carbon atoms atoms reflects reflects physical physical properties.properties.

17

Chemistry & MatterChemistry & Matter• We can We can explore the MACROSCOPIC world explore the MACROSCOPIC world

— — what we can see — what we can see —

• Understand the PARTICULATE world Understand the PARTICULATE world

— — we cannot see —we cannot see —

We can write SYMBOLSWe can write SYMBOLS to describe these worlds. to describe these worlds.

18

A Chemist’s ViewA Chemist’s View

2 H2(g) + O2 (g) --> 2 H2O(g)

MacroscopicMacroscopic

Symbolic-Symbolic-

ParticulateParticulate

19

20

STATES OF MATTERSTATES OF MATTER• SOLIDS SOLIDS — have rigid shape, fixed — have rigid shape, fixed

volume. External shape can reflect the volume. External shape can reflect the atomic and molecular arrangement.atomic and molecular arrangement.

–Reasonably well understood.Reasonably well understood.

• LIQUIDS LIQUIDS — have no fixed shape and — have no fixed shape and may not fill a container completely. may not fill a container completely.

–Not well understood.Not well understood.

• GASES GASES — expand to fill their — expand to fill their container. container.

–Good theoretical understanding.Good theoretical understanding.

21

THE THREE STATES OF THE THREE STATES OF MATTERMATTER

Bromine (gas) Aluminum (solid) Water or H2O (liquid)

22

KINETIC NATURE OF MATTERKINETIC NATURE OF MATTERMatter consists of atoms and molecules Matter consists of atoms and molecules

in motion.in motion.

23

Physical Physical PropertiesProperties

What are some physical What are some physical propertiesproperties??

–ColorColor

–Melting and boiling Melting and boiling pointpoint

–OdorOdor

–ConductivityConductivity

–DensityDensity

24

Physical ChangesPhysical Changes

Some physical Some physical changeschanges would bewould be– boiling of a liquidboiling of a liquid

– melting of a solidmelting of a solid

– dissolving a solid in a dissolving a solid in a liquid to give a liquid to give a homogeneous mixture homogeneous mixture

25

DENSITYDENSITY - an important and useful - an important and useful

physical propertyphysical property

13.6 g/cm13.6 g/cm33

Gold

19.3 g/cm19.3 g/cm33

Mercury

26

Which is more dense?Which is more dense?

27

Relative Densities of the ElementsRelative Densities of the Elements

28

Sig Figs

Sig Fig PPT

29

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

30SOLUTIONSOLUTION

1. Get dimensions in common units.

2. Calculate volume in cubic centimeters.

3. Calculate the density.

. =0 95 mm •1cm

10 mm 0.095 cm3

(9.36 cm)(7.23 cm)(0.095 cm) = 6.4 cm3

6.4 cm3

57.54g= 9.0g/cm3

31

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? In pounds?of Hg? In grams? In pounds?

Solve the problem usingSolve the problem using

DIMENSIONAL DIMENSIONAL

ANALYSIS.ANALYSIS.

32

Then, use dimensional analysis to Then, use dimensional analysis to calculate mass.calculate mass.

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?of Hg?

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

See next slide

95 cm3 • 13.6 g

cm3 = 1.3 x 103 g

What is the mass in pounds?

33

The milliliter and the cubic centimeter are equivalent. Notice the units of 10’s.

back

34

What is the mass in pounds? (1 lb = 454 g)What is the mass in pounds? (1 lb = 454 g)

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?of Hg?

1.3 x 103 g • 1 lb

454 g = 2.8 lb

35DensityDensityPROBLEM: An object weighing 15.67 g is PROBLEM: An object weighing 15.67 g is placed in water starting at 6.8 mL, then placed in water starting at 6.8 mL, then displaces water to 20.2 mL. What is the displaces water to 20.2 mL. What is the density of the object?density of the object?

6.8

20.2

20.2mL - 6.8mL = 13.4mL

36

15.67g 1

(20.2 - 6.8)ml = 1.17g/cm3

PROBLEM: An object weighing 15.67 g is PROBLEM: An object weighing 15.67 g is placed in water starting at 6.8 mL, then placed in water starting at 6.8 mL, then displaces water to 20.2 mL. What is the displaces water to 20.2 mL. What is the density of the object?density of the object?

37

Chemical Properties and Chemical Properties and Chemical ChangeChemical Change

•Burning hydrogen (HBurning hydrogen (H22) in ) in

oxygen (O oxygen (O22) gives H) gives H22O.O.

38

Chemical PropertiesChemical Properties

• Similar to Physical Properties only with reference to a Chemical reactions

–Heat and or light produced

–Color

–Oder

39

Chemical Properties and Chemical Properties and Chemical ChangeChemical Change

• Chemical change Chemical change or or chemical reaction chemical reaction involves the involves the transformation of one or transformation of one or more atoms or molecules more atoms or molecules into one or more different into one or more different molecules.molecules.

• Burning hydrogen (HBurning hydrogen (H22) in ) in

oxygen (O oxygen (O22) gives H) gives H22O.O.

40

2 Al + 3 Br2 Al2Br6

Chemical ChangeChemical Change

41

Electrolyzing water

42

UNITS OF MEASUREMENTUNITS OF MEASUREMENT

• We make We make QUALITATIVEQUALITATIVE observations of reactions — observations of reactions — changes in color and physical changes in color and physical state.state.

• We also make We also make QUANTITATIVEQUANTITATIVE MEASUREMENTSMEASUREMENTS, which involve , which involve numbersnumbers..

• Use Use SI unitsSI units — based on the — based on the metric systemmetric system

43

UNITS OF MEASUREMENTUNITS OF MEASUREMENT Use Use SI unitsSI units — based on the — based on the

metric systemmetric system

length length (meter, m)(meter, m)

massmass (kilogram, kg, (kilogram, kg, and gram, g)and gram, g)

timetime (second)(second)

44

Units of LengthUnits of Length• 1 kilometer (km) = ? meters (m)1 kilometer (km) = ? meters (m)

• 1 meter (m) = ? centimeters (cm)1 meter (m) = ? centimeters (cm)

• 1 centimeter (cm) = ? millimeter (mm)1 centimeter (cm) = ? millimeter (mm)

• 1 nanometer (nm) = 1.0 x 101 nanometer (nm) = 1.0 x 10-9-9 meter meter

O—H distance =O—H distance =9.4 x 109.4 x 10-11 -11 mm9.4 x 109.4 x 10-9 -9 cmcm0.094 nm0.094 nm94 pm94 pm

45

MeasurementMeasurement• Learn the prefixes in Table 1.4

• Other Relationships1 cm3 = 1 mL = 0.001 L

1.00 lb = 454 g

1.00 in = 2.54 cm

1.06 qt = 1.00 L

Significant figures Page 47

Precision and accuracy Page 43

Examples

46

Temperature ScalesTemperature Scales• FahrenheitFahrenheit

• CelsiusCelsius

• KelvinKelvin

Anders Celsius1701-1744

Lord Kelvin(William Thomson)1824-1907

47

Temperature ScalesTemperature Scales

Notice that 1 Kelvin degree = 1 degree Celsius1 Kelvin degree = 1 degree Celsius

48

Calculations Using Calculations Using TemperatureTemperature

• Generally require temp’s in Generally require temp’s in KelvinKelvin

• T (K) = t (T (K) = t (°°C) + 273C) + 273

• Body temp = 37 Body temp = 37 ooC + 273 = 310. KC + 273 = 310. K

• Liquid nitrogen = -196 Liquid nitrogen = -196 ooC + 273 = 77 KC + 273 = 77 K

49

In Class Problems

• A rectangular box has dimensions of A rectangular box has dimensions of 20.0 cm 20.0 cm 15.0 cm 15.0 cm 8.00 mm. Calculate 8.00 mm. Calculate the volume of the box in liters. the volume of the box in liters.

• A standard sheet of paper has dimensions A standard sheet of paper has dimensions of 8.5 inch by 11 inch. A sheet of paper of 8.5 inch by 11 inch. A sheet of paper weighs on the average 0.150 g and has a weighs on the average 0.150 g and has a density of 0.710 g/cmdensity of 0.710 g/cm33. Calculate the . Calculate the thickness of the paper in cm. thickness of the paper in cm.

• A gallon (3.78 L) of latex paint can cover A gallon (3.78 L) of latex paint can cover 385 ft385 ft22 of the surface of a wall. What is of the surface of a wall. What is the average thickness of one coat of paint the average thickness of one coat of paint (in micrometers)?(in micrometers)?

50

Sample problems• Calculate the volume of 525 g of Calculate the volume of 525 g of

mercury, d=13.534g/cmmercury, d=13.534g/cm33..

• The melting point of tin is 505.5 K. The melting point of tin is 505.5 K. Calculate the Celsius temperature.Calculate the Celsius temperature.

• Find the symbol for gold and the Find the symbol for gold and the element name for K.element name for K.

• An iron sheet is 3.50 cm square and An iron sheet is 3.50 cm square and has a mass of 15.396 g. The density has a mass of 15.396 g. The density of iron is 7.87 g/cmof iron is 7.87 g/cm33. Calculate the . Calculate the thickness of the iron sheet in mm.thickness of the iron sheet in mm.

The End!The End!

51

DimensionalDimensional Analysis Analysis“English-English” (one conversion)

1) 6 in = ? ft

2) 3.5 gal = ? qt

6 in 1 ft 12 in

= 0.5 ft

3.5 gal 4 qt1 gal

= 14 qt

52

Dimensional AnalysisDimensional Analysis“Metric-Metric” (one conversion)

1) 5.0 cm = ? mm

4.0 dg 1 kg104 dg = 4.0 x 10-4 kg

5.0 cm 10 mm 1 cm

= 50. mm

2) 4.0 dg = ? kg

53

DimensionalDimensional Analysis Analysis

“Metric-English” (one conversion)

1) 200.0 cm = ? in

2) 34 qt = ? L

34 qt 1.00 L1.06 qt

= 32 L

200.0 cm 1.00 in2.54 cm

= 78.74 in

54

DimensionalDimensional Analysis Analysis

“English-English” (two or more conversions)

1) 6 in = ? mile

2) 3.5 gal = ? oz

= 450 oz

6 in 1 ft12 in

= 9 x 10-5 mile1 mile

5280 ft

3.5 gal 4 qt1 gal

32 oz

1 qt

55

DimensionalDimensional Analysis Analysis“Metric-Metric” (two or more conversions)

1) 5.0 cm = ? km

2) 4 kg = ? pg

= 4 x 1015 pg

5.0 cm 1 m100cm

= 5.0 x 10-5 km1 km

1000 m

4 kg 103 g1 kg

1012 pg

1 g

56

DimensionalDimensional Analysis Analysis

“Metric-English” (two or more conversions)

1) 200 m = ? in

2) 34 qt = ? mL

= 3.2 x 104 mL

200 m 100 cm1 m

= 8000 in1.00 in

2.54 cm

34 qt 1.00 L1.06 qt

103 mL

1 L

57

DimensionalDimensional Analysis Analysis

“Derived Unit Conversions” Area

1) 8.0 ft2 = ? cm2

2) 2.3 cm2 = ? nm2

= 2.3 x 1014 nm2

8.0 ft2 144 in2

1 ft2 = 7400 cm2(2.54)2 cm2

1.00 in2

2.3 cm2 1014 nm2

1 cm2

58

DimensionalDimensional Analysis Analysis

“Derived Unit Conversions” Volume

1) 445 dm3 = ? mL

2) 5 cm3 = ? mm3

= 5 x 103 mm3

445 dm3 103 cm3

1 dm3= 4.45 x 105mL 1 mL

1 cm3

5 cm3 103 mm3

1 cm3

59

DimensionalDimensional Analysis Analysis

“Other Conversions Problems”

1) 100. km/hr = ? mile/hr

2) 25 m/gal = ? nm/qt

= 6.2 x 109 nm/qt

100. km hr

105 cm 1 km

= 62.1 mile/hr

1.00 in

2.54 cm

25 m

gal 109 nm 1 m

1 ft

12 in

1 mile

5280 ft

1 gal

4 qt

60

DimensionalDimensional Analysis Analysis“Other Conversions Problems” (continue)

3) Calculate the density of a material if 45 mL of it has a mass of 128 g.

4) Calculate the volume in mL of 2.5 g of a material that has a density of 3.65 g/mL.

= 0.68 mL 2.5 g mL

3.65 g

= 2.8 g/mL 128 g 45 mL

61

DimensionalDimensional Analysis Analysis“Other Conversions Problems” (continue)

5) Calculate the mass of 20 L of a material that has a density of 8.54 g/mL.

6) How many grams of gold are in 48 g of an alloy that is 22.1% gold?

= 2 x 105 g 20 L 103 ml

1 L8.54 g

ml

= 11 g gold 48 g alloy 22.1 g gold

100.0 g alloy

62

Practice Problems1) 6.45 m = ? cm1) 6.45 m = ? cm

2) 12.4 kg = ? mg2) 12.4 kg = ? mg

3) 184 oz = ? g3) 184 oz = ? g

4) 24 oz/hr = ? L/day4) 24 oz/hr = ? L/day

5) Determine the volume (in L) of 2 kg of 5) Determine the volume (in L) of 2 kg of sodium chloride. (density = 2.17 g/mL)sodium chloride. (density = 2.17 g/mL)

6) How many grams of brass contain 50.0 g of 6) How many grams of brass contain 50.0 g of zinc? (This brass contains 15% Zn)zinc? (This brass contains 15% Zn)

1) 645 cm 2) 1.24 x 107 mg 3) 5220 g 4) 17 L/day

5) 0.9 L 6) 330 g

Return