PACING • 45 min pp. 67 – 71 TR 17 Law of Conservation of Mass TR 18 Law of Multiple ProportionsEXT Historical Chemistry Modern Alchemy, p. 67

TE Reading Skill Builder Interpreting Vocabulary, p. 66 b

TE Lesson Starter, p. 67 g TE Visual Strategy, p. 68 g TE Visual Strategy, p. 69 g SE Careers in Chemistry, Physical Chemist, p.70 g SE QuickLab Constructing a Model, p. 71 ◆ g ANC Datasheets for In-Text Labs

Section 1 The Atom: From Philosophical Idea to Scientific Theory

PACING • 90 min pp. 72 – 76Section 2 The Structure of the Atom

TR 19 Gold Foil Experiment TR 20 Gold Foil Experiment on the Atomic Level TR 17A Properties of Subatomic Particles

TE Lesson Starter, p. 72 g TE Demonstration, p. 72 ◆ TE Visual Strategy, p. 68 g TE Demonstration, p. 73 ◆ TE Reading Skill Builder Prediction Guides, p. 74 b TE Visual Strategy, p. 74 g TE Visual Strategy, p. 75 g TE Reading Skill Builder Prediction Guides, p. 75 b

66A Chapter 3 Atoms: The Building Blocks of Matter

SECTIONS LABS, DEMONSTRATIONS, AND ACTIVITIES TECHNOLOGY RESOURCES

PACING • 135 min pp. 77 – 87Section 3 Counting Atoms

TR 21 Solving Mole Problems TR 22 Determining the Mass from the Amount in

Moles EXT Graphing Calculator Calculating Numbers of

Protons, Electrons, and Neutrons, p. 91

TE Lesson Starter, p. 77 g TE Visual Strategy, p. 77 g TE Visual Strategy, p. 78 g TE Table Strategy, p. 78 g TE Table Strategy, p. 79 g SE Historical Chemistry Discovery of Element 43, p.81

g

TE Table Strategy, p. 82 g TE Reading Skill Builder Sequencing, p. 82 b TE Class Discussion, p. 83 b SE Chapter Lab Conservation of Mass, pp. 94 – 95 ◆

MICRO INQUIRY

ANC Datasheets for In-Text Labs

Atoms: The Building Blocks of Matter

To shorten your instruction because of time limitations, omit Section 3.

Compression Guide

CHAPTER 3

SE Chapter Highlights, p. 88 SE Chapter Review, pp. 89 – 91 SE Math Tutor, p. 92 SE Standardized Test Prep, p. 93 ANC Chapter Test A gANC Chapter Test B a OSP Test GeneratorOSP Scoring Rubrics and Classroom Management

Checklists

CHAPTER REVIEW, ASSESSMENT, AND STANDARDIZED TEST PREPARATION

PACING • 90 minOnline and Technology ResourcesOnline and Technology Resources

Visit go.hrw.com to find a variety of online resources. To access this chapter’s extensions, enter the keyword HC6ATMXT and click the “go” button. Click Holt Online Learning for an online edition of this textbook, and other interactive resources.

• Holt Calendar Planner• Customizable Lesson Plans• Editable Worksheets• ExamView® Version 6

Assessment Suite

• Interactive Teacher’s Edition• Holt PuzzlePro® • Holt PowerNotes®

Presentations• MindPoint® Quiz Show

This DVD package includes:

SE Section Review, p. 71ANC Section Review *ANC Quiz *

National ScienceEducation StandardsPS 1a, 2bUCP 2, 4SAI 1– 2HNS 1– 3

SE Sample Problems and Practice Problems, pp. 79, 84, 85, 86 g

TE Additional Sample Problems, pp. 80, 84, 85, 86 gEXT Additional Practice Problems gANC Problem-Solving Workbook Chapter 5 * a SE Math Tutor Conversion Factors, p. 92 g

TE Alternative Assessment, p. 75 g SE Section Review, p. 76ANC Section Review *ANC Quiz *

PS 1a, 1b, 1cUCP 1– 2SAI 2HNS 1– 3

TE Alternative Assessment, p. 82 g TE Alternative Assessment, p. 83 b SE Section Review, p. 87ANC Section Review *ANC Mixed Review *ANC Quiz *

PS 1bUCP 1– 3SAI 2

Chapter 3 Planning Guide 66B

KEY SE Student Edition TE Teacher Edition ANC Ancillary Worksheet

OSP One-Stop Planner CD CD or CD-ROM TR Teaching Transparencies

EXT Online Extension * Also on One-Stop Planner ◆ Requires advance prep

SKILLS DEVELOPMENT RESOURCES REVIEW AND ASSESSMENT CORRELATIONS

www.scilinks.orgMaintained by the National Science Teachers Association.

Topic: Atomic TheorySciLinks Code: HC60120Topic: CarbonSciLinks Code: HC60214

Topic: Subatomic ParticlesSciLinks Code: HC61473Topic: IsotopesSciLinks Code: HC60820

• Chapter Summaries Audio Program• Student One Stop• Virtual Investigations• Visual Concepts

Search for any lab by topic, standard, difficulty level, or time. Edit any lab to fit your needs, or create your own labs. Use the Lab Materials QuickList software to customize your lab materials list.

ClassroomTechnology

Atoms:The Building

Blocks of Matter

Chapter OverviewSection 1 covers the historyand development of atomictheory, from Democritus toDalton to the modern era.

Section 2 covers the experi-ments that led to the discoveryof the electron and the nucleusas well as the principal proper-ties of these subatomic particles.

Section 3 outlines the man-ner in which the number ofatoms of an element and thenumber of an atom’s subatomicparticles can be expressed andmeasured.

Concept BaseStudents may need a review ofthe following concepts:

• characteristics of matter,Chapter 1

• calculations and units,Chapter 2

CHAPTER 3

66

Atoms: The BuildingBlocks of Matter

An atom is the smallest particle of an element that retains the chemical properties of that element.

C H A P T E R 3

Reading Skill Builder

INTERPRETINGVOCABULARY Show studentsthe derivation of the word atomfrom its Greek roots.

a-: not; tomos: cuttingAsk them what idea about matteris conveyed by that meaning.Explain that the idea, which alsooriginated in ancient Greece, isthat matter can be subdividedonly as small as an elemental particle: the atom.

STM Image of Impure Gold Surface

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 67

Lesson StarterWrite the following two statementson the board: “Young people shouldnot smoke,” and “Smoking at anearly age may make it more difficultto quit smoking later.” Have studentsdetermine which statement is anopinion and which is a theory. Whichis similar to Aristotle’s statements?

An opinion is an unsupported idea; atheory uses reason to explain obser-vations and experimental data. Havestudents bring in articles from maga-zines or newspapers. Then have themhighlight opinions and theories.

ApplicationHave students think of examples ofevidence that invisible particles exist.Possible examples include perfumediffusing across a room or the sensa-tion caused by wind.

Teaching Tip

GENERAL

SECTION 1

67

SECTION 1

OBJECTIVES

Explain the law of conservation of mass, the law of definite proportions,and the law of multiple proportions.

Summarize the five essentialpoints of Dalton’s atomictheory.

Explain the relationshipbetween Dalton’s atomictheory and the law of conser-vation of mass, the law ofdefinite proportions, and thelaw of multiple proportions.

The Atom: FromPhilosophical Idea toScientific Theory

When you crush a lump of sugar, you can see that it is made up ofmany smaller particles of sugar.You may grind these particles into a veryfine powder, but each tiny piece is still sugar. Now suppose you dissolvethe sugar in water.The tiny particles seem to disappear completely. Evenif you look at the sugar-water solution through a powerful microscope,you cannot see any sugar particles. Yet if you were to taste the solution,you’d know that the sugar is still there. Observations like these led earlyphilosophers to ponder the fundamental nature of matter. Is it continu-ous and infinitely divisible, or is it divisible only until a basic, invisibleparticle that cannot be divided further is reached?

The particle theory of matter was supported as early as 400 B.C. bycertain Greek thinkers, such as Democritus. He called nature’s basic par-ticle an atom, based on the Greek word meaning “indivisible.” Aristotlewas part of the generation that succeeded Democritus. His ideas had alasting impact on Western civilization, and he did not believe in atoms.He thought that all matter was continuous, and his opinion was accept-ed for nearly 2000 years. Neither the view of Aristotle nor that ofDemocritus was supported by experimental evidence, so each remainedspeculation until the eighteenth century. Then scientists began to gatherevidence favoring the atomic theory of matter.

Foundations of Atomic Theory

Virtually all chemists in the late 1700s accepted the modern definitionof an element as a substance that cannot be further broken down byordinary chemical means. It was also clear that elements combine toform compounds that have different physical and chemical propertiesthan those of the elements that form them.There was great controversy,however, as to whether elements always combine in the same ratiowhen forming a particular compound.

The transformation of a substance or substances into one or morenew substances is known as a chemical reaction. In the 1790s, the studyof matter was revolutionized by a new emphasis on the quantitative

Historical ChemistryGo to go.hrw.com for a full-lengtharticle on the history of atomic theory and transmutation.

Keyword: HC6ATMX

✔

analysis of chemical reactions.Aided by improved balances, investigatorsbegan to accurately measure the masses of the elements and compoundsthey were studying. This led to the discovery of several basic laws. Oneof these laws was the law of conservation of mass, which states that massis neither created nor destroyed during ordinary chemical reactions orphysical changes. This discovery was soon followed by the assertionthat, regardless of where or how a pure chemical compound is prepared,it is composed of a fixed proportion of elements. For example, sodiumchloride, also known as ordinary table salt, always consists of 39.34% bymass of the element sodium, Na, and 60.66% by mass of the elementchlorine, Cl. The fact that a chemical compound contains the sameelements in exactly the same proportions by mass regardless of the size ofthe sample or source of the compound is known as the law of definiteproportions.

It was also known that two elements sometimes combine to formmore than one compound. For example, the elements carbon and oxygenform two compounds, carbon dioxide and carbon monoxide. Considersamples of each of these compounds, each containing 1.00 g of carbon. Incarbon dioxide, 2.66 g of oxygen combine with 1.00 g of carbon. Incarbon monoxide, 1.33 g of oxygen combine with 1.00 g of carbon. Theratio of the masses of oxygen in these two compounds is 2.66 to 1.33, or 2to 1. This illustrates the law of multiple proportions: If two or moredifferent compounds are composed of the same two elements, then theratio of the masses of the second element combined with a certain mass ofthe first element is always a ratio of small whole numbers.

Dalton’s Atomic Theory

In 1808, an English schoolteacher named John Dalton proposed anexplanation for the law of conservation of mass, the law of definite pro-portions, and the law of multiple proportions. He reasoned that el-ements were composed of atoms and that only whole numbers of atomscan combine to form compounds. His theory can be summed up by thefollowing statements.1. All matter is composed of extremely small particles called atoms.2. Atoms of a given element are identical in size, mass, and other prop-

erties; atoms of different elements differ in size, mass, and otherproperties.

3. Atoms cannot be subdivided, created, or destroyed.4. Atoms of different elements combine in simple whole-number ratios

to form chemical compounds.5. In chemical reactions, atoms are combined, separated, or rearranged.

According to Dalton’s atomic theory, the law of conservation of massis explained by the fact that chemical reactions involve merely the combination, separation, or rearrangement of atoms and that duringthese processes atoms are not subdivided, created, or destroyed. This

C H A P T E R 368

SECTION 1

68

SECTION 1

Visual StrategyStress that no matter

how few sodium chloride crystals one inspects, the mass percentage of sodium and chloride remainsunchanged. This is an example of the law of definite proportions.

FIGURE 1

GENERAL

The ability of scientists to divideatoms into protons, neutrons, andelectrons and to then further divideprotons and neutrons into quarks isan example of why atoms are nolonger thought of as indivisible.

Teaching Tip

FIGURE 1 Each of the salt crys-tals shown here contains exactly39.34% sodium and 60.66% chlorineby mass.

CHAPTER CONNECTION

Students will use the law of conser-vation of mass to balance chemicalequations in Chapter 8.

• Learning Disabled

Ask students to write the five princi-ples of Dalton’s Theory on five individ-ual index cards. For each principle,have the students draw a model ordescribe an example that explains ordefines each principle. The cards maybe used for individual or small groupstudy or displayed on a classroom bulletin board. Students can showtheir understanding of the concept bypresenting their examples to the classor to a small group of other students.

StrategiesStrategiesINCLUSIONINCLUSION

✔

www.scilinks.orgTopic: Atomic TheoryCode: HC60120

idea is illustrated in Figure 2 for the formation of carbon monoxidefrom carbon and oxygen.

The law of definite proportions, on the other hand, results from thefact that a given chemical compound is always composed of the samecombination of atoms (see Figure 3). As for the law of multipleproportions, in the case of the carbon oxides, the 2-to-1 ratio of oxygenmasses results because carbon dioxide always contains twice as manyatoms of oxygen (per atom of carbon) as does carbon monoxide. Thiscan also be seen in Figure 3.

Modern Atomic Theory

By relating atoms to the measurable property of mass, Dalton turnedDemocritus’s idea into a scientific theory that could be tested by experi-ment. But not all aspects of Dalton’s atomic theory have proven to becorrect. For example, today we know that atoms are divisible into evensmaller particles (although the law of conservation of mass still holdstrue for chemical reactions). And, as you will see in Section 3, we knowthat a given element can have atoms with different masses. Atomic the-ory has not been discarded, however. Instead, it has been modified toexplain the new observations. The important concepts that (1) all mat-ter is composed of atoms and that (2) atoms of any one element differin properties from atoms of another element remain unchanged.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 69 69

SECTION 1

Visual StrategyPoint out to students that

the number of oxygen atoms and thenumber of carbon atoms are thesame before and after each reactionshown. This is an illustration of thelaw of conservation of mass.

The mass of oxygen in amolecule of carbon dioxide is twicethe mass of oxygen in a molecule ofcarbon monoxide. This is an illustra-tion of the law of multiple propor-tions. Have students draw analogiesfrom the world around them. Forexample, a bicycle has a frame andtwo wheels, and a tricycle has aframe and three wheels. The ratio ofwheels of a bicycle to those of a tri-cycle is always 2:3, a ratio of smallwhole numbers.

Figures 2 and 3 also illustrate thelaw of definite proportions. (Eachtype of molecule is always composedof the same ratio of atoms.)

Scientists now know that all atoms ofa given element do not necessarilyhave the same mass. In nature, mostelements consist of various isotopes,atoms with the same number of protons but a different number ofneutrons.

Teaching Tip

FIGURE 3

FIGURE 2

GENERAL

FIGURE 2 (a) An atom of carbon, C, and an atom of oxygen,O, can combine chemically to form a molecule of carbon monoxide, CO.The mass of the CO molecule isequal to the mass of the C atom plus the mass of the O atom.(b) The reverse holds true in a reaction in which a CO molecule is broken down into its elements.

FIGURE 3 (a) CO molecules arealways composed of one C atom andone O atom. (b) CO2 molecules arealways composed of one C atom andtwo O atoms. Note that a moleculeof carbon dioxide contains twice as many oxygen atoms as does a molecule of carbon monoxide.

Carbon, CMass x

Carbon, CMass x

(a)

(b)

Oxygen, OMass y

Oxygen, OMass y

Carbon monoxide, COMass x + Mass y

Carbon monoxide, COMass x + Mass y

+

+

=

=

Carbon, C

(a)

+

+ +

=

=

(b)Oxygen, O Carbon monoxide, CO

Carbon, C Oxygen, O Oxygen, O Carbon dioxide, CO2

✔

www.scilinks.orgTopic: CarbonCode: HC60214

C H A P T E R 370

SECTION 1

70

SECTION 1

Careers in ChemistryAnswers1. physics and mathematics

2. The electrons of surface atoms aredetected by STM. Physical Chemist

Physical chemists focus on under-standing the physical properties ofatoms and molecules. They are dri-ven by a curiosity of what makesthings work at the level of atoms,and they enjoy being challenged. Inaddition to chemistry, they studymathematics and physics extensively.Laboratory courses involving experi-ence with electronics and optics aretypically part of their training. Often,they enjoy working with instrumentsand computers. Physical chemistscan be experimentalists or theoreti-cians. They use sophisticated instru-ments to make measurements, orhigh-powered computers to performintensive calculations. The instru-ments used include lasers, electronmicroscopes, nuclear magnetic reso-nance spectrometers, mass spec-trometers, and particle accelerators.Physical chemists work in industry,government laboratories, researchinstitutes, and academic institutions.Because physical chemists work on a wide range of problems, takingcourses in other science disciplines is important.

Scanning TunnelingMicroscopyFor years, scientists have yearned forthe ability to “see” individualatoms. Because atoms are so small,this had been nothing more than adream. Now, the scanning tunnelingmicroscope, STM, gives scientists the ability to look at individualatoms. It was invented in 1981 by

Gerd Binnig andHeinrich Rohrer,scientists working for IBM in Zurich,Switzerland. Theyshared the 1986Nobel Prize inphysics for their discovery.

The basic principleof STM is based onthe current thatexists between ametallic needle that is sharpened toa single atom, the probe, and a con-ducting sample. As the probe passesabove the surface of the sample at adistance of one or two atoms, elec-trons can “tunnel” from the needletip to the sample’s surface. Theprobe moves across, or “scans,” thesurface of the sample. When theprobe comes close to the electronsof an individual atom, a signal isproduced. A weaker signal is pro-duced between atoms. These signalsbuild a topographical (hill and val-ley) “map” of conducting and non-conducting regions. The resultingmap shows the position and spacingof atoms.

Surface chemistry is a developingsubdiscipline in physical chemistry,and STM is an important tool in thefield. Scientists use STM to study sur-face reactions, such as those thattake place in catalytic converters.Other areas of research in which STMis useful include semiconductors and

microelectronics. Usually, STM is usedwith materials that conduct, but ithas also been used to study biologi-cal molecules, such as DNA.

One innovative application of STMis the ability to position individualatoms. The figure shows the result ofmoving individual atoms. First, ironatoms were placed on a copper sur-face. Then, individual iron atomswere picked up by the probe andplaced in position. The result is a“quantum corral” of 48 iron atomson the surface of copper. The diame-ter of the corral is about 14 nm.

1. In addition to chemistry, whatkinds of courses are importantfor a student interested in aphysical chemistry career?

2. What part of an atom is detect-ed by STM?

Questions

This STM image shows a “corral” of iron atoms on acopper surface.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 71 71

SECTION 1

Materials• can covered by a sock sealed

with tape• one or more objects that fit

in the container• metric ruler• balance

Wear safety goggles and an apron.

Students will make inferences aboutwhat is inside the can based on bothindirect and direct observations.Remind students to not look insidethe container because scientists mustdraw conclusions without being ableto “see” the answer. Do not placeanything in the can that can spill ordrip or harm students.

Discussion1. Scientists use both indirect anddirect observations to gather data.In part 1, observations were madewithout contact with the objectsthemselves. These observations wereindirect. In part 2, students madedirect observations by touching theobjects. Between the two types ofobservations, students could get agood idea of the identity of theobjects.

2. Determinations of the mass, size,and number of objects are quantita-tive. Shape, material, and texturedeterminations are qualitative.

3. Answers will vary.

1. The five main points of Dalton’stheory are listed on page 68.

2. The laws of conservation of mass,definite proportions, and multipleproportions can be explained byDalton’s theory.

3. Comparing the K/O mass ratiosgives 2.44/1.22 = 2; 4.89/2.44 = 2, and4.89/1.22 = 4. These data support thelaw of multiple proportions becausethe three compounds have mass ratiosconsisting of the small, whole numbers2, 2, and 4, respectively.

SECTION REVIEW

1. List the five main points of Dalton’s atomic theory.

2. What chemical laws can be explained by Dalton’stheory?

Critical Thinking

3. ANALYZING INFORMATION Three compounds containing potassium and oxygen are compared.Analysis shows that for each 1.00 g of O, the com-pounds have 1.22 g, 2.44 g, and 4.89 g of K,respectively. Show how these data support the law of multiple proportions.

SECTION REVIEW

QuestionHow can you construct amodel of an unknown objectby (1) making inferencesabout an object that is in aclosed container and (2)touching the object withoutseeing it?

ProcedureRecord all of your results in a data table.

1. Your teacher will provide youwith a can that is covered by asock sealed with tape. Withoutunsealing the container, try todetermine the number ofobjects inside the can as well asthe mass, shape, size, composi-tion, and texture of each. Todo this, you may carefully tiltor shake the can. Record yourobservations in a data table.

2. Remove the tape from the topof the sock. Do not look insidethe can. Put one hand throughthe opening, and make thesame observations as in step 1by handling the objects. Tomake more-accurate estima-tions, practice estimating thesizes and masses of someknown objects outside the can.

Constructing a Model

Then compare your estimatesof these objects with actualmeasurements using a metricruler and a balance.

Discussion

1. Scientists often use more thanone method to gather data.How was this illustrated in theinvestigation?

2. Of the observations you made,which were qualitative andwhich were quantitative?

3. Using the data you gathered,draw a model of the unknownobject(s) and write a brief sum-mary of your conclusions.

C H A P T E R 372

SECTION 1

72

SECTION 2

Lesson StarterShow students two different shapesmade from identical pieces of a toyconstruction set. Demonstrate howeven though the two shapes look dif-ferent, the characteristics of the vari-ous parts that compose them are thesame. The same is true with theatom. Though atoms of different ele-ments display different properties,isolated subatomic particles have thesame properties, regardless of theirsource. As technology improved, sci-entists learned to isolate the parts ofatoms that display similar properties.For example, the cathode-ray tubeled to the isolation of a subatomicparticle, the electron.

Did You Know?In a television tube, a cathode rayconstantly undergoes deflections thatresult from the varying magnetic fieldwithin the tube. The ray strikes acoated screen, where a luminescentimage is created.

DEMONSTRATIONBring in an oscilloscope or an oldcomputer monitor. Show how theimage on the screen seems to be distorted by a magnetic field, whichalters the path of electrons as theyapproach the screen.

Visual StrategyAlert students to the dif-

ference between the flow of electrons, or the cathode ray, and the conventional direction of electriccurrent. The figure depicts a cathoderay, created by electrons, flowingfrom the negatively charged cathodeto the positively charged anode. Thedirection of the electric current isfrom the anode to the cathode.

FIGURE 4

GENERALSECTION 2 The Structure of

the Atom

A lthough John Dalton thought atoms were indivisible, investigatorsin the late 1800s proved otherwise. As scientific advances allowed adeeper exploration of matter, it became clear that atoms are actuallycomposed of several basic types of smaller particles and that the num-ber and arrangement of these particles within an atom determine thatatom’s chemical properties.Today we define an atom as the smallest par-ticle of an element that retains the chemical properties of that element.

All atoms consist of two regions. The nucleus is a very small regionlocated at the center of an atom. In every atom, the nucleus is made up ofat least one positively charged particle called a proton and usually one ormore neutral particles called neutrons. Surrounding the nucleus is a regionoccupied by negatively charged particles called electrons. This region isvery large compared with the size of the nucleus. Protons, neutrons, andelectrons are often referred to as subatomic particles.

Discovery of the Electron

The first discovery of a subatomic particle resulted from investigationsinto the relationship between electricity and matter. In the late 1800s,many experiments were performed in which electric current was passedthrough various gases at low pressures. (Gases at atmospheric pressure

don’t conduct electricity well.) These experi-ments were carried out in glass tubes likethe one shown in Figure 4. Such tubes areknown as cathode-ray tubes.

Cathode Rays and ElectronsInvestigators noticed that when currentwas passed through a cathode-ray tube,the surface of the tube directly oppositethe cathode glowed. They hypothesizedthat the glow was caused by a stream ofparticles, which they called a cathode ray.The ray traveled from the cathode to theanode when current was passed throughthe tube. Experiments devised to test this

OBJECTIVES

Summarize the observedproperties of cathode raysthat led to the discovery ofthe electron.

Summarize the experimentcarried out by Rutherford andhis co-workers that led to thediscovery of the nucleus.

List the properties of protons,neutrons, and electrons.

Define atom.

Voltage source

Cathode rayGas at low pressure

+–

Cathode (metal disk)

Anode (metal disk)

FIGURE 4 A simple cathode-raytube. Particles pass through the tubefrom the cathode, the metal diskconnected to the negative terminalof the voltage source, to the anode,the metal disk connected to the posi-tive terminal.

hypothesis revealed the followingobservations.1. Cathode rays were deflected by a

magnetic field in the same manneras a wire carrying electric current,which was known to have a nega-tive charge (see Figure 5).

2. The rays were deflected away froma negatively charged object.

These observations led to thehypothesis that the particles thatcompose cathode rays are negativelycharged. This hypothesis was stronglysupported by a series of experimentscarried out in 1897 by the English physicist Joseph John Thomson. Inone investigation, he was able to measure the ratio of the charge ofcathode-ray particles to their mass. He found that this ratio was alwaysthe same, regardless of the metal used to make the cathode or thenature of the gas inside the cathode-ray tube. Thomson concluded thatall cathode rays are composed of identical negatively charged particles,which were named electrons.

Charge and Mass of the ElectronCathode rays have identical properties regardless of the element used toproduce them. Therefore it was concluded that electrons are present inatoms of all elements. Thus, cathode-ray experiments provided evidencethat atoms are divisible and that one of the atom’s basic constituents isthe negatively charged electron.Thomson’s experiment also revealed thatthe electron has a very large charge-to-mass ratio. In 1909, experimentsconducted by the American physicist Robert A. Millikan measured thecharge of the electron. Scientists used this information and the charge-to-mass ratio of the electron to determine that the mass of the electron isabout one two-thousandth the mass of the simplest type of hydrogenatom, which is the smallest atom known. More-accurate experiments con-ducted since then indicate that the electron has a mass of 9.109 × 10−31 kg,or 1/1837 the mass of the simplest type of hydrogen atom.

Based on what was learned about electrons, two other inferenceswere made about atomic structure.1. Because atoms are electrically neutral, they must contain a positive

charge to balance the negative electrons.2. Because electrons have so much less mass than atoms, atoms must

contain other particles that account for most of their mass.Thomson proposed a model for the atom that is called the plum puddingmodel (after the English dessert). He believed that the negative electronswere spread evenly throughout the positive charge of the rest of the atom.This arrangement is similar to that of seeds in a watermelon: the seeds arespread throughout the fruit but do not contribute much to the overallmass. However, shortly thereafter, new experiments disproved this model.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 73 73

SECTION 2

DEMONSTRATIONIf you have a cathode-ray tube suchas the one pictured in Figure 5, useit to demonstrate the rays and theirdeflection by a magnet. Follow theinstructions that accompany the tubeand the power supply. Keep students3 m away from the apparatus, weargoggles, and have students weargoggles. Do not touch the terminalsor the tube while the power supply ison. Keep magnets away from theends of the tube.

FIGURE 5 A magnet near thecathode-ray tube causes the beam tobe deflected. The deflection indi-cates that the particles in the beamhave a negative charge.

CathodeAnode

Millikan’s oil-drop experiment did notmeasure the mass of an electrondirectly. Millikan’s experiment meas-ured the charge of an electron. Usingthe charge determined by Millikan’sexperiment and the charge-to-massratio of an electron determined byThomson’s experiment, scientistswere able to calculate the mass of an electron.

Teaching Tip✔

www.scilinks.orgTopic: Subatomic ParticlesCode: HC61473

Discovery of the Atomic Nucleus

More detail of the atom’s structure was provided in 1911 by NewZealander Ernest Rutherford and his associates Hans Geiger andErnest Marsden. The scientists bombarded a thin piece of gold foil withfast-moving alpha particles, which are positively charged particles withabout four times the mass of a hydrogen atom. Geiger and Marsdenassumed that mass and charge were uniformly distributed throughoutthe atoms of the gold foil. They expected the alpha particles to passthrough with only a slight deflection, and for the vast majority of theparticles, this was the case. However, when the scientists checked for thepossibility of wide-angle deflections, they were shocked to find thatroughly 1 in 8000 of the alpha particles had actually been deflected backtoward the source (see Figure 6). As Rutherford later exclaimed, it was“as if you had fired a 15-inch [artillery] shell at a piece of tissue paperand it came back and hit you.”

After thinking about the startling result for a few months,Rutherford finally came up with an explanation. He reasoned that thedeflected alpha particles must have experienced some powerful forcewithin the atom. And he figured that the source of this force must occu-py a very small amount of space because so few of the total number ofalpha particles had been affected by it. He concluded that the forcemust be caused by a very densely packed bundle of matter with a pos-itive electric charge. Rutherford called this positive bundle of matterthe nucleus (see Figure 7).

Rutherford had discovered that the volume of a nucleus was verysmall compared with the total volume of an atom. In fact, if the nucleuswere the size of a marble, then the size of the atom would be about thesize of a football field. But where were the electrons? This question wasnot answered until Rutherford’s student, Niels Bohr, proposed a modelin which electrons surrounded the positively charged nucleus as theplanets surround the sun. Bohr’s model will be discussed in Chapter 4.

C H A P T E R 374

SECTION 1

74

SECTION 2

Visual StrategyHave students explain

what is happening in the figure.Students should realize thatFigure 6a shows merely the setupof the Rutherford foil experiment andthat Figure 6b shows the results.

If an atom were the size of a largefootball stadium, the nucleus wouldbe about the size of a marble. Thismodel suggests that the atom ismostly empty space. Ask students toconsider how many alpha particlespassed through the gold foil in theRutherford experiment compared withthe number that were deflected.

Teaching Tip

FIGURE 6

GENERAL

(a) (b)

Screen to detect deflected particles

Thin gold foil

Lead box containing radioactive source of fast-moving particles

Particlesdeflected by foil

FIGURE 6 (a) Geiger andMarsden bombarded a thin piece of gold foil with a narrow beam of alpha particles. (b) Some of the particles were deflected by the gold foil back toward their source.

✔

Reading Skill Builder

PREDICTION GUIDES Writethe following statements on thechalkboard:• An atom cannot be broken

down into smaller parts.• An atom is the same throughout.• An atom is made up of several

different, smaller parts.Ask students their opinions of thestatements. Have them discusstheir opinions and try to justifythem. Save a list of the opinionsfor discussion after completingSection 2.

Composition of the Atomic Nucleus

Except for the nucleus of the simplest type of hydrogen atom (discussedin the next section), all atomic nuclei are made of two kinds of particles,protons and neutrons. A proton has a positive charge equal in magni-tude to the negative charge of an electron. Atoms are electrically neu-tral because they contain equal numbers of protons and electrons. Aneutron is electrically neutral.

The simplest hydrogen atom consists of a single-proton nucleus with asingle electron moving about it. A proton has a mass of 1.673 × 10−27 kg,which is 1836 times greater than the mass of an electron and 1836/1837,or virtually all, of the mass of the simplest hydrogen atom. All atomsbesides the simplest hydrogen atom also have neutrons. The mass of aneutron is 1.675 × 10−27 kg—slightly larger than that of a proton.

The nuclei of atoms of different elements differ in their number ofprotons and therefore in the amount of positive charge they possess.Thus, the number of protons determines that atom’s identity. Physicistshave identified other subatomic particles, but particles other than elec-trons, protons, and neutrons have little effect on the chemical propertiesof matter. Table 1 on the next page summarizes the properties of elec-trons, protons, and neutrons.

Forces in the NucleusGenerally, particles that have the same electric charge repel one another.Therefore, we would expect a nucleus with more than one proton to be unstable. However, when two protons are extremely close to each other, there is a strong attraction between them. In fact, as many as 83

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 75 75

SECTION 2

Visual StrategyExplain to students that

the size of the nucleus and the num-ber of deflected alpha particles havebeen exaggerated for emphasis. Inreality, the nucleus occupies much lessspace in an atom than indicated here.In Rutherford’s experiment, almost allof the alpha particles passed undis-turbed through the atoms composingthe foil.

Did You Know?The forces holding the nucleus togeth-er, when unleashed, provide the ener-gy associated with nuclear power orthe detonation of an atomic weapon.

Alternative AssessmentHave students build a timeline for themajor events in the development ofthe modern atomic theory. Start withevents in the text, and then awardextra credit for events not mentionedin the text that contribute to a morecomplete chronology.

GENERAL

FIGURE 7

Beam of positive particles

Large deflection

Small deflection

Nucleus

Electrons surroundnucleus

FIGURE 7 Rutherford reasonedthat each atom in the gold foil contained a small, dense, posi-tively charged nucleus surroundedby electrons. A small number of thealpha particles directed toward thefoil were deflected by the tiny nu-cleus (red arrows). Most of the parti-cles passed through undisturbed(black arrows).

Reading Skill Builder

PREDICTION GUIDES Havestudents revisit their opinions fromthe beginning of the section. Havethem discuss whether their opinionshave changed or remain the same.Have them cite passages in the textthat account for their decisions.

protons can exist close together to help form a stable nucleus. A similarattraction exists when neutrons are very close to each other or whenprotons and neutrons are very close together. These short-range proton-neutron, proton-proton, and neutron-neutron forces hold the nuclearparticles together and are referred to as nuclear forces.

The Sizes of Atoms

It is convenient to think of the region occupied by the electrons as anelectron cloud—a cloud of negative charge. The radius of an atom is thedistance from the center of the nucleus to the outer portion of this elec-tron cloud. Because atomic radii are so small, they are expressed usinga unit that is more convenient for the sizes of atoms. This unit is thepicometer. The abbreviation for the picometer is pm (1 pm = 10−12 m =10−10 cm). To get an idea of how small a picometer is, consider that 1 cmis the same fractional part of 103 km (about 600 mi) as 100 pm is of 1 cm.Atomic radii range from about 40 to 270 pm. By contrast, the nuclei ofatoms have much smaller radii, about 0.001 pm. Nuclei also have incred-ibly high densities, about 2 × 108 metric tons/cm3.

C H A P T E R 376

SECTION 1

76

SECTION 2

1. a. the smallest particle of an ele-ment that retains the chemical prop-erties of that elementb. a negatively charged subatomicparticle located outside the nucleusc. the positively charged, dense cen-tral portion of an atom that containsnearly all the atom’s mass but takesup a very small fraction of its volumed. the subatomic particle of thenucleus that has a positive chargeequal in magnitude to the negativecharge of the electrone. the electrically neutral subatomicparticle found in atomic nuclei

2. a. concluded that electrons werefound in all atomsb. confirmed the negative charge ofthe electron and helped determine apossible massc. determined that most of the massof an atom is found in the nucleusand that the nucleus occupies verylittle space within an atom

3. Electrons surround the nucleusand have a negative charge andsmall mass relative to protons andneutrons. Most of an atom’s mass isin its nucleus, which is in the centerand is composed of protons and neu-trons. Protons have a positive charge;neutrons have no charge.

4. because charges will flow throughgases only at very low pressures

5. The protons and neutrons arepacked very close together, as shownby Rutherford. But like-charged parti-cles repel each other, so the protonswould not be expected to be close toother protons. The forces present thatprevent protons from repelling eachother are the nuclear forces.

SECTION REVIEW

Relativeelectric Mass Relative mass Actual mass

Particle Symbols charge number (amu*) (kg)

Electron e−, −10e −1 0 0.000 5486 9.109 × 10−31

Proton p+, 11H +1 1 1.007 276 1.673 × 10−27

Neutron n°, 10n 0 1 1.008 665 1.675 × 10−27

*1 amu (atomic mass unit) = 1.660 540 × 10−27 kg

TABLE 1 Properties of Subatomic Particles

1. Define each of the following:

a. atom c. nucleus e. neutron

b. electron d. proton

2. Describe one conclusion made by each of the fol-lowing scientists that led to the development ofthe current atomic theory:

a. Thomson b. Millikan c. Rutherford

3. Compare the three subatomic particles in terms oflocation in the atom, mass, and relative charge.

4. Why is the cathode-ray tube in Figure 4 connect-ed to a vacuum pump?

Critical Thinking

5. EVALUATING IDEAS Nuclear forces are said tohold protons and neutrons together. What is itabout the composition of the nucleus that requiresthe concept of nuclear forces?

SECTION REVIEW

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 77 77

SECTION 3

Lesson StarterHave students consider a familiarexample of a weighted average.For instance, if a student’s semestergrade depends 60% on exam scoresand 40% on laboratory explorations,then the student’s exam scores countmore heavily toward his or her finalgrade. Hand out sample exam scoresand laboratory scores, and have students determine the weightedaverage of the two. In this section,students will learn that the atomicmass of an element is a weightedaverage of the masses of the naturallyoccurring isotopes of that element.

CommonMisconceptionMake it clear to students that theidentity of the atom is determined by the number of protons, not thenumber of electrons or neutrons. Thenumber of electrons and the numberof neutrons can each vary and theatom will still be of the same ele-ment. But if the number of protonschanges, then the atom becomes an atom of a different element.

Visual StrategyHave students identify

the atomic number, atomic mass, andchemical symbol for different ele-ments on the classroom’s periodictable. Have them deduce the numberof electrons from the atomic numberfor neutral atoms.

FIGURE 8

GENERAL

GENERALSECTION 3

OBJECTIVES

Explain what isotopes are.

Define atomic number andmass number, and describehow they apply to isotopes.

Given the identity of a nuclide,determine its number of pro-tons, neutrons, and electrons.

Define mole, Avogadro’s num-ber, and molar mass, and statehow all three are related.

Solve problems involvingmass in grams, amount inmoles, and number of atomsof an element.

Counting Atoms

C onsider neon, Ne, the gas used in many illuminated signs. Neon is aminor component of the atmosphere. In fact, dry air contains only about0.002% neon.And yet there are about 5 × 1017 atoms of neon present ineach breath you inhale. In most experiments, atoms are much too smallto be measured individually. Chemists can analyze atoms quantitatively,however, by knowing fundamental properties of the atoms of each ele-ment. In this section, you will be introduced to some of the basic prop-erties of atoms. You will then discover how to use this information tocount the number of atoms of an element in a sample with a knownmass.You will also become familiar with the mole, a special unit used bychemists to express amounts of particles, such as atoms and molecules.

Atomic Number

All atoms are composed of the same basic particles. Yet all atoms arenot the same. Atoms of different elements have different numbers ofprotons. Atoms of the same element all have the same number of pro-tons. The atomic number (Z) of an element is the number of protons ofeach atom of that element.

Turn to the inside back cover of this textbook. In the periodic tableshown, an element’s atomic number is indicated above its symbol. Noticethat the elements are placed in order of increasing atomic number. Atthe top left of the table is hydrogen, H, which has atomic number 1. Allatoms of the element hydrogen have one proton. Next in order is heli-um, He, which has two protons. Lithium, Li, has three protons (seeFigure 8); beryllium, Be, has four protons; and so on.

The atomic number identifies an element. If you want to know whichelement has atomic number 47, for example, look at the periodic table.You can see that the element is silver, Ag. All silver atoms have 47 pro-tons. Because atoms are neutral, we know from the atomic number thatall silver atoms must also have 47 electrons.

Isotopes

The simplest atoms are those of hydrogen. All hydrogen atoms haveonly one proton. However, like many naturally occurring elements,hydrogen atoms can have different numbers of neutrons.

3Li

Lithium6.941

[He]2s1

FIGURE 8 The atomic number inthis periodic table entry reveals thatan atom of lithium has three protonsin its nucleus.

Three types of hydrogen atoms are known. The most common typeof hydrogen is sometimes called protium. It accounts for 99.9885% ofthe hydrogen atoms found on Earth. The nucleus of a protium atomconsists of one proton only, and it has one electron moving about it.There are two other known forms of hydrogen. One is called deuterium,which accounts for 0.0115% of Earth’s hydrogen atoms. Each deuteri-um atom has a nucleus with one proton and one neutron.The third formof hydrogen is known as tritium, which is radioactive. It exists in verysmall amounts in nature, but it can be prepared artificially. Each tritiumatom has one proton, two neutrons, and one electron.

Protium, deuterium, and tritium are isotopes of hydrogen. Isotopesare atoms of the same element that have different masses. The isotopes ofa particular element all have the same number of protons and electronsbut different numbers of neutrons. In all three isotopes of hydrogen, thepositive charge of the single proton is balanced by the negative chargeof the electron. Most of the elements consist of mixtures of isotopes.Tinhas 10 stable isotopes, for example, the most of any element.

Mass Number

Identifying an isotope requires knowing both the name or atomic num-ber of the element and the mass of the isotope. The mass number is thetotal number of protons and neutrons that make up the nucleus of an iso-tope. The three isotopes of hydrogen described earlier have mass num-bers 1, 2, and 3, as shown in Table 2.

C H A P T E R 378

SECTION 1

78

SECTION 3

Visual StrategyNote that the only differ-

ence between the hydrogen isotopesis in their number of neutrons.

Table 2 How are hydrogen’s isotopes different from one another?(They have different numbers of neutrons.) How does this informationalter Dalton’s atomic theory? (Atomsof a single element do not alwayshave the same mass, as Dalton suggested.) Hydrogen is the only element whose isotopes have names.Isotopes of other elements are identified by their mass number.

The mass numbers of the stable tenisotopes of tin are 112, 114, 115,116, 117, 118, 119, 120, 122, and124.

Teaching Tip

GENERALTABLE STRATEGY

FIGURE 9

GENERAL1 Neutron

1 Proton 1 Proton

2 Neutrons

1 Proton

Protium Deuterium Tritium

FIGURE 9 The nuclei of differentisotopes of the same element havethe same number of protons but dif-ferent numbers of neutrons. This isillustrated above by the three isotopes of hydrogen.

Atomic number(number of Number of Mass numberprotons) neutrons (protons + neutrons)

Protium 1 0 1 + 0 = 1

Deuterium 1 1 1 + 1 = 2

Tritium 1 2 1 + 2 = 3

TABLE 2 Mass Numbers of Hydrogen Isotopes

✔

Designating Isotopes

The isotopes of hydrogen are unusual in that they have distinct names.Isotopes are usually identified by specifying their mass number. Thereare two methods for specifying isotopes. In the first method, the massnumber is written with a hyphen after the name of the element.Tritium,for example, is written as hydrogen-3. We will refer to this method ashyphen notation. The uranium isotope used as fuel for nuclear powerplants has a mass number of 235 and is therefore known as uranium-235. The second method shows the composition of a nucleus using theisotope’s nuclear symbol. For example, uranium-235 is written as 235

92U.The superscript indicates the mass number (protons + neutrons) andthe subscript indicates the atomic number (number of protons). Thenumber of neutrons is found by subtracting the atomic number from themass number.

mass number − atomic number = number of neutrons235 (protons + neutrons) − 92 protons = 143 neutrons

Thus, a uranium-235 nucleus is made up of 92 protons and 143 neutrons.Table 3 gives the names, symbols, and compositions of the isotopes of

hydrogen and helium. Nuclide is a general term for a specific isotope ofan element. We could say that Table 3 lists the compositions of five dif-ferent nuclides, three hydrogen nuclides and two helium nuclides.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 79 79

SECTION 3

Table 3 Draw students’ attention tothe meaning of the two numbersassociated with an isotope’s nuclearsymbol. The number on the bottom is the number of protons (the atomicnumber). For a neutral atom, thenumber of protons equals the num-ber of electrons. The number on thetop is the sum of the protons and the neutrons (the mass number).Have students determine the numberof protons, neutrons, and electronsfor some other elements.

GENERALTABLE STRATEGY

Nuclear Number of Number of Number ofIsotope symbol protons electrons neutrons

Hydrogen-1 (protium) 11H 1 1 0

Hydrogen-2 (deuterium) 21H 1 1 1

Hydrogen-3 (tritium) 31H 1 1 2

Helium-3 32He 2 2 1

Helium-4 42He 2 2 2

TABLE 3 Isotopes of Hydrogen and Helium

Given: name and mass number of chlorine-37Unknown: numbers of protons, electrons, and neutrons

atomic number = number of protons = number of electronsmass number = number of neutrons + number of protons

SOLUTION

1 ANALYZE

2 PLAN

SAMPLE PROBLEM A

How many protons, electrons, and neutrons are there in an atom of chlorine-37?

www.scilinks.orgTopic: IsotopesCode: HC60820

Relative Atomic Masses

Masses of atoms expressed in grams are very small. As we shall see,an atom of oxygen-16, for example, has a mass of 2.656 × 10–23 g. Formost chemical calculations it is more convenient to use relative atomicmasses. As you read in Chapter 2, scientists use standards of measure-ment that are constant and are the same everywhere. In order to set upa relative scale of atomic mass, one atom has been arbitrarily chosen asthe standard and assigned a mass value. The masses of all other atomsare expressed in relation to this defined standard.

The standard used by scientists to compare units of atomic mass is thecarbon-12 atom. It has been arbitrarily assigned a mass of exactly 12 atom-ic mass units, or 12 amu. One atomic mass unit, or 1 amu, is exactly 1/12 the mass of a carbon-12 atom. The atomic mass of any other atom is determined by comparing it with the mass of the carbon-12 atom.The hydrogen-1 atom has an atomic mass of about 1/12 that of the carbon-12 atom, or about 1 amu. The precise value of the atomic mass of a hydrogen-1 atom is 1.007 825 amu. An oxygen-16 atom has about16/12 (or 4/3) the mass of a carbon-12 atom. Careful measurements show the atomic mass of oxygen-16 to be 15.994 915 amu. The mass of a magnesium-24 atom is found to be slightly less than twice that of a carbon-12 atom. Its atomic mass is 23.985 042 amu.

C H A P T E R 380

SECTION 1

80

SECTION 3

Go to go.hrw.com formore practice problemsthat ask you to workwith numbers ofsubatomic particles.

Keyword: HC6ATMX

The mass number of chlorine-37 is 37. Consulting the periodic table reveals that chlorine’satomic number is 17. The number of neutrons can be found by subtracting the atomic number from the mass number.

mass number of chlorine-37 − atomic number of chlorine =number of neutrons in chlorine-37

mass number − atomic number = 37 (protons plus neutrons) − 17 protons= 20 neutrons

An atom of chlorine-37 is made up of 17 electrons, 17 protons, and 20 neutrons.

The number of protons in a neutral atom equals the number of electrons. And the sum of the protons and neutrons equals the given mass number.

3 COMPUTE

4 EVALUATE

1. How many protons, electrons, and neutrons make up an atom ofbromine-80?

2. Write the nuclear symbol for carbon-13.

3. Write the hyphen notation for the isotope with 15 electrons and15 neutrons.

Answers in Appendix EPRACTICE

A-1 How many protons, electrons,and neutrons make up an atom of carbon-13?

Ans. 6 protons, 6 electrons,7 neutrons

A-2 Write the nuclear symbol for oxygen-16.

Ans. 168O

A-3 Write the hyphen notation forthe element whose atoms have7 electrons and 9 neutrons.

Ans. nitrogen-16

Practice Answers1. 35 protons, 35 electrons,45 neutrons

2. 1316C

3. phosphorus-30

You may wish to point out that 1 amuis equal to 1.660 540 × 10−27 kg or1.660 540 × 10−24 g.

Teaching Tip

ADDITIONALSAMPLEPROBLEMS GENERAL

✔

Some additional examples of the atomic masses of the naturallyoccurring isotopes of several elements are given in Table 4 on the nextpage. Isotopes of an element may occur naturally, or they may be madein the laboratory (artificial isotopes). Although isotopes have differentmasses, they do not differ significantly in their chemical behavior.

The masses of subatomic particles can also be expressed on the atom-ic mass scale (see Table 1). The mass of the electron is 0.000 5486 amu,that of the proton is 1.007 276 amu, and that of the neutron is1.008 665 amu. Note that the proton and neutron masses are close to butnot equal to 1 amu. You have learned that the mass number is the totalnumber of protons and neutrons that make up the nucleus of an atom.You can now see that the mass number and relative atomic mass of agiven nuclide are quite close to each other. They are not identicalbecause the proton and neutron masses deviate slightly from 1 amu andthe atomic masses include electrons.Also, as you will read in Chapter 21,a small amount of mass is changed to energy in the creation of a nucleusfrom its protons and neutrons.

Average Atomic Masses of Elements

Most elements occur naturally as mixtures of isotopes, as indicated inTable 4. The percentage of each isotope in the naturally occurring ele-ment on Earth is nearly always the same, no matter where the elementis found. The percentage at which each of an element’s isotopes occursin nature is taken into account when calculating the element’s averageatomic mass. Average atomic mass is the weighted average of the atom-ic masses of the naturally occurring isotopes of an element.

The following is a simple example of how to calculate a weightedaverage. Suppose you have a box containing two sizes of marbles. If25% of the marbles have masses of 2.00 g each and 75% have masses of3.00 g each, how is the weighted average calculated? You could countthe number of each type of marble, calculate the total mass of the mix-ture, and divide by the total number of marbles. If you had 100 marbles,the calculations would be as follows.

25 marbles × 2.00 g = 50 g75 marbles × 3.00 g = 225 g

Adding these masses gives the total mass of the marbles.

50 g + 225 g = 275 g

Dividing the total mass by 100 gives an average marble mass of 2.75 g.A simpler method is to multiply the mass of each marble by the dec-

imal fraction representing its percentage in the mixture. Then add theproducts.

25% = 0.25 75% = 0.75(2.00 g × 0.25) + (3.00 g × 0.75) = 2.75 g

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 81 81

SECTION 3

CommonMisconceptionMany students think that protons andneutrons have the same mass as indi-vidual particles as they do as part ofthe nucleus of an atom. However, themass of a nucleus is less than the sumof the masses of the protons and neu-trons making up the nucleus. The dif-ference is due to the binding energythat holds the nucleus together.

Discovery of Element 43The discovery of element 43, tech-netium, is credited to Carlo Perrier andEmilio Segrè, who artificially producedit in 1937. However, in 1925, a Germanchemist named Ida Tacke reported thediscovery of element 43, which shecalled masurium, in niobium ores. Atthe time, her discovery was not accept-ed because it was thought technetiumcould not occur naturally. Recent stud-ies confirm that Tacke and coworkersprobably did discover element 43.

CHAPTER CONNECTION

Chapter 21 outlines how the difference between the mass of anatom and the sum of the masses of the same number of individualprotons, neutrons, and electrons isconverted into the energy that holdsthe nucleus together. This energy iscalled nuclear binding energy. This is why the actual mass of a nuclide isslightly different from the mass cal-culated from its constituent particles.

Did You Know?Isotope abundances are measuredvery precisely using an instrumentcalled a mass spectrometer. High-energy electrons collide with anatom, knocking away one of its elec-trons and giving the particle a posi-tive charge. The charged particle isdirected through a magnetic field. Theisotopes of an element are separatedfrom each other because they havethe same charge but different masses.The particles with greater mass aredeflected less than those of less mass.The data collected gives both themass of the particles and their per-centage abundance in the sample.

Calculating Average Atomic MassThe average atomic mass of an element depends on both the mass andthe relative abundance of each of the element’s isotopes. For example,naturally occurring copper consists of 69.15% copper-63, which has anatomic mass of 62.929 601 amu, and 30.85% copper-65, which has anatomic mass of 64.927 794 amu. The average atomic mass of copper canbe calculated by multiplying the atomic mass of each isotope by its rel-ative abundance (expressed in decimal form) and adding the results.

0.6915 × 62.929 601 amu + 0.3085 × 64.927 794 amu = 63.55 amu

The calculated average atomic mass of naturally occurring copper is63.55 amu.

The average atomic mass is included for the elements listed inTable 4. As illustrated in the table, most atomic masses are known tofour or more significant figures. In this book, an element’s atomic mass isusually rounded to two decimal places before it is used in a calculation.

Relating Mass to Numbers of Atoms

The relative atomic mass scale makes it possible to know how manyatoms of an element are present in a sample of the element with a mea-surable mass. Three very important concepts—the mole, Avogadro’snumber, and molar mass—provide the basis for relating masses ingrams to numbers of atoms.

C H A P T E R 382

SECTION 1

82

SECTION 3

Table 4 Point out to students that the average atomic mass of anelement is often closest to the atomicmass of the most abundant isotope.

Alternative AssessmentFollow the example on this page to confirm the average atomic mass of hydrogen, carbon, oxygen, copper,and uranium using the percentageabundances and isotope masses listed in the table on this page.

See the Math Tutor at the end ofChapter 4 for more information onweighted averages and atomic mass.

Teaching Tip

GENERAL

GENERALTABLE STRATEGY AverageMass Percentage natural Atomic mass atomic mass

Isotope number abundance (amu) of element (amu)

Hydrogen-1 1 99.9885 1.007 8251.007 94

Hydrogen-2 2 0.0115 2.014 102

Carbon-12 12 98.93 12 (by definition)Carbon-13 13 1.07 13.003 355 12.0107

Oxygen-16 16 99.757 15.994 915Oxygen-17 17 0.038 16.999 132 15.9994Oxygen-18 18 0.205 17.999 160

Copper-63 63 69.15 62.929 60163.546

Copper-65 65 30.85 64.927 794

Cesium-133 133 100 132.905 447 132.905

Uranium-234 234 0.0054 234.040 945Uranium-235 235 0.7204 235.043 922 238.029Uranium-238 238 99.2742 238.050 784

TABLE 4 Atomic Masses and Abundances of Several Naturally Occurring Isotopes

Reading Skill Builder

SEQUENCING After studentshave read the text on relatingmass and numbers of atoms tothemselves, ask them to discussthe relationships illustrated inFigure 11. Have them work inpairs to write in their own wordsthe steps they would follow tochange from one expression to theother. Ask for volunteers to list thesteps on the board.

✔

The MoleThe mole is the SI unit for amount of substance. A mole (abbreviatedmol) is the amount of a substance that contains as many particles asthere are atoms in exactly 12 g of carbon-12. The mole is a countingunit, just like a dozen is. We don’t usually order 12 or 24 ears of corn;we order one dozen or two dozen. Similarly, a chemist may want 1 molof carbon, or 2 mol of iron, or 2.567 mol of calcium. In the sections thatfollow, you will see how the mole relates to masses of atoms andcompounds.

Avogadro’s NumberThe number of particles in a mole has been experimentally deter-mined in a number of ways. The best modern value is 6.022 141 79 ×1023. This means that exactly 12 g of carbon-12 contains 6.022 141 79 ×1023 carbon-12 atoms. The number of particles in a mole is known asAvogadro’s number, named for the nineteenth-century Italian scien-tist Amedeo Avogadro, whose ideas were crucial in explaining therelationship between mass and numbers of atoms. Avogadro’s num-ber—6.022 141 79 × 1023—is the number of particles in exactly onemole of a pure substance. For most purposes, Avogadro’s number isrounded to 6.022 × 1023.

To get a sense of how large Avogadro’s number is, consider the fol-lowing: If every person living on Earth (6 billion people) worked tocount the atoms in one mole of an element, and if each person countedcontinuously at a rate of one atom per second, it would take about3 million years for all the atoms to be counted.

Molar MassAn alternative definition of mole is the amount of a substance that con-tains Avogadro’s number of particles. Can you figure out the approxi-mate mass of one mole of helium atoms? You know that a mole ofcarbon-12 atoms has a mass of exactly 12 g and that a carbon-12 atomhas an atomic mass of 12 amu. The atomic mass of a helium atom is4.00 amu, which is about one-third the mass of a carbon-12 atom. It fol-lows that a mole of helium atoms will have about one-third the mass ofa mole of carbon-12 atoms. Thus, one mole of helium has a mass ofabout 4.00 g.

The mass of one mole of a pure substance is called the molar mass ofthat substance. Molar mass is usually written in units of g/mol.The molarmass of an element is numerically equal to the atomic mass of the ele-ment in atomic mass units (which can be found in the periodic table). Forexample, the molar mass of lithium, Li, is 6.94 g/mol, while the molarmass of mercury, Hg, is 200.59 g/mol (rounding each value to two deci-mal places).

The molar mass of an element contains one mole of atoms. For exam-ple, 4.00 g of helium, 6.94 g of lithium, and 200.59 g of mercury all con-tain a mole of atoms. Figure 10 shows molar masses of three commonelements.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 83 83

SECTION 3

Class DiscussionHave students decide whether singlemoles of all materials have the samemass. Divide students into groups offour to develop an answer and anexample. Then have one person fromeach group report the answer andexample to the class.

Alternative AssessmentHave students determine the molarmasses of different elements on theperiodic table. The answer will beequal to the average atomic mass,expressed in g/mol. One may alsocalculate the molar mass of a com-pound; the molar mass of water is18.02 g/mol, for example.

CommonMisconceptionIt is easy to confuse the terms atomicmass and molar mass. Although theyare often the same number, atomicmass is the mass of one atom,expressed in amu, and molar mass isthe mass of one mole of particles,expressed in g/mol.

(a)

(b)

(c)

FIGURE 10 Shown is approxi-mately one molar mass of eachof three elements: (a) carbon (graphite), (b) iron (nails), and(c) copper (wire).

Chapter 5 of the Problem-SolvingWorkbook (also found on the OneStop Planner CD-ROM) includesmore worked-out samples andadditional practice problemsinvolving gram/mole calculations.

Problem-Solving Practice ChemFile

HOLT

Gram/Mole ConversionsChemists use molar mass as a conversion factor in chemical calcula-tions. For example, the molar mass of helium is 4.00 g He/mol He. Tofind how many grams of helium there are in two moles of helium, mul-tiply by the molar mass.

2.00 mol He × = 8.00 g He

Figure 11 shows how to use molar mass, moles, and Avogadro’s numberto relate mass in grams, amount in moles, and number of atoms of anelement.

4.00 g He

1 mol He

C H A P T E R 384

SECTION 1

84

SECTION 3

The technique of using conversionfactors (see the Math Tutor at theend of this chapter) can be extendedto a more general case, called unitanalysis. Using the example in thetext, we see that 4.00 g He is equiva-lent to 1 mol He. The conversion fac-tors possible are ⎯41

.0m0ogl H

Hee⎯ or ⎯4

1.0m0ogl H

Hee

⎯. If we choose the first conversion factorand treat the units as algebraic quan-tities, the units “mol He” will canceland the unit “g He” remains, whichtells us we have correctly solved theproblem.

Teaching TipMass of element

in gramsNumber of atoms

of element

Amount of element

in moles

� �

��1 mol

molar massof element

molar massof element

1 mol

6.022 � 1023 atoms��

� �1 mol6.022 � 1023 atoms

1 mol

FIGURE 11 The diagram showsthe relationship between mass ingrams, amount in moles, and numberof atoms of an element in a sample.

Given: 3.50 mol CuUnknown: mass of Cu in grams

amount of Cu in moles ⎯→ mass of Cu in grams

According to Figure 11, the mass of an element in grams can be calculated by multiplyingthe amount of the element in moles by the element’s molar mass.

moles Cu × = grams Cu

The molar mass of copper from the periodic table is rounded to 63.55 g/mol.

3.50 mol Cu × = 222 g Cu

Because the amount of copper in moles was given to three significant figures, the answerwas rounded to three significant figures. The size of the answer is reasonable because it issomewhat more than 3.5 times 60.

63.55 g Cu

1 mol Cu

grams Cu

moles Cu

SOLUTION

1 ANALYZE

2 PLAN

3 COMPUTE

4 EVALUATE

SAMPLE PROBLEM B

What is the mass in grams of 3.50 mol of the element copper, Cu?

For more help, go to the Math Tutor at the end of this chapter.

✔

B-1 What is the mass in grams of 3.6 mol of the element carbon, C?

Ans. 43 g

B-2 What is the mass in grams of0.733 mol of the element sulfur, S?

Ans. 23.5 g

ADDITIONALSAMPLEPROBLEMS GENERAL

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 85 85

SECTION 3

Practice Answers1. 126 g Fe

2. 14.7 g K

3. 0.310 g Na

4. 957 g Ni

Go to go.hrw.com formore practice problemsthat ask you to convertfrom amount in molesto mass.

Keyword: HC6ATMX

1. What is the mass in grams of 2.25 mol of the element iron, Fe?

2. What is the mass in grams of 0.375 mol of the element potassium, K?

3. What is the mass in grams of 0.0135 mol of the element sodium, Na?

4. What is the mass in grams of 16.3 mol of the element nickel, Ni?

Answers in Appendix EPRACTICE

Go to go.hrw.com formore practice problemsthat ask you to convertfrom mass to amount inmoles.

Keyword: HC6ATMX

Given: 11.9 g AlUnknown: amount of Al in moles

mass of Al in grams ⎯→ amount of Al in moles

As shown in Figure 11, amount in moles can be obtained by dividing mass in grams bymolar mass, which is mathematically the same as multiplying mass in grams by the reciprocalof molar mass.

grams Al × = moles Al

The molar mass of aluminum from the periodic table is rounded to 26.98 g/mol.

11.9 g Al × = 0.441 mol Al

The answer is correctly given to three significant figures. The answer is reasonable because11.9 g is somewhat less than half of 26.98 g.

1 mol Al

26.98 g Al

moles Al

grams Al

SOLUTION

1 ANALYZE

2 PLAN

3 COMPUTE

4 EVALUATE

SAMPLE PROBLEM C

1. How many moles of calcium, Ca, are in 5.00 g of calcium?

2. How many moles of gold, Au, are in 3.60 × 10−5 g of gold?

3. How many moles of zinc, Zn, are in 0.535 g of zinc?

Answers in Appendix E

A chemist produced 11.9 g of aluminum, Al. How many moles of aluminum were produced?

PRACTICE

For more help, go to the Math Tutor at the end of this chapter.

C-1 How many moles of copper,Cu, are in 3.22 g of copper?

Ans. 0.0507 mol

C-2 How many moles of lithium, Li,are in 2.72 × 10–4 g of lithium?

Ans. 3.92 × 10–5 mol

Practice Answers1. 0.125 mol Ca

2. 1.83 × 10−7 mol Au

3. 8.18 × 10−3 mol Zn

Avogadro’s number is a countingunit, just as a pair of gloves is equalto two gloves or a ream of paper is equal to 500 sheets. We typicallyuse Avogadro’s number as the“counting unit” for atoms and mol-ecules. By definition, 1 mole of any-thing has 6.022 × 1023 items; 1 moleof C atoms, 1 mole of water mol-ecules, or even 1 mole of studentswould all have 6.022 × 1023 atoms,molecules, or students, respectively.The atomic masses listed in the peri-odic table represent the averagemass in amu for one atom or themass in grams of 1 mole of atoms.

Teaching Tip

ADDITIONALSAMPLEPROBLEMS GENERAL

✔

Conversions with Avogadro’s NumberFigure 11 shows that Avogadro’s number can be used to find the num-ber of atoms of an element from the amount in moles or to find theamount of an element in moles from the number of atoms. While thesetypes of problems are less common in chemistry than convertingbetween amount in moles and mass in grams, they are useful in demon-strating the meaning of Avogadro’s number. Note that in these calcula-tions, Avogadro’s number is expressed in units of atoms per mole.

C H A P T E R 386

SECTION 1

86

SECTION 3

Go to go.hrw.com formore practice problemsthat ask you to convertbetween atoms andmoles.

Keyword: HC6ATMX

Given: 3.01 × 1023 atoms of AgUnknown: amount of Ag in moles

number of atoms of Ag ⎯→ amount of Ag in moles

From Figure 11, we know that number of atoms is converted to amount in moles by dividing by Avogadro’s number. This is equivalent to multiplying numbers of atoms by the reciprocal of Avogadro’s number.

Ag atoms × = moles Ag

3.01 × 1023 Ag atoms × = 0.500 mol Ag

The answer is correct—units cancel correctly and the number of atoms is one-half ofAvogadro’s number.

1 mol Ag

6.022 × 1023 Ag atoms

moles Ag

Avogadro’s number of Ag atoms

SOLUTION

1 ANALYZE

2 PLAN

3 COMPUTE

4 EVALUATE

SAMPLE PROBLEM D

1. How many moles of lead, Pb, are in 1.50 × 1012 atoms of lead?

2. How many moles of tin, Sn, are in 2500 atoms of tin?

3. How many atoms of aluminum, Al, are in 2.75 mol of aluminum?

Answers in Appendix E

How many moles of silver, Ag, are in 3.01 � 1023 atoms of silver?

PRACTICE

For more help, go to the Math Tutor at the end of this chapter.

D-1 How many moles of carbon, C,are in 2.25 × 1022 atoms of carbon?

Ans. 0.0374 mol

D-2 How many moles of oxygen,O, are in 2,000,000 atoms of oxygen?

Ans. 3 × 10–18 mol

D-3 How many atoms of sodium,Na, are in 3.80 mol of sodium?

Ans. 2.29 × 1024 atoms

E-1 What is the mass in grams of 5.0 × 109 atoms of neon, Ne?

Ans. 1.7 × 10–13 g

E-2 How many atoms of carbon, C,are in 0.020 g of carbon?

Ans. 1.0 × 1021 atoms

E-3 What mass of silver, Ag, con-tains the same number of atoms as10.0 g of boron, B?

Ans. 99.8 g

Practice Answers1. 2.49 × 10−12 mol Pb

2. 4.2 × 10−21 mol Sn

3. 1.66 × 1024 atoms Al

ADDITIONALSAMPLEPROBLEMS GENERAL

Given: 1.20 × 108 atoms of CuUnknown: mass of Cu in grams

SOLUTION

1 ANALYZE

SAMPLE PROBLEM E

What is the mass in grams of 1.20 � 108 atoms of copper, Cu?

For more help, go to the Math Tutor at the end of this chapter.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 87 87

SECTION 3

Practice Answers1. 7. 3 × 10−7 g Ni

2. 7.51 × 1022 atoms S

3. 66 g Au

1. a. the number of protons in thenucleus of an atom of that elementb. the total number of protons andneutrons in a nucleusc. the mass of an atom given inatomic mass unitsd. the weighted average of the masses of the naturally occurring isotopes of an elemente. the amount of a substance thatcontains a number of particles equalto the number of atoms in exactly12 g of carbon-12f. the number of particles in exactlyone mole of a pure substanceg. the mass of one mole of a pure sub-stance, usually given in units of g/molh. an atom that has the same num-ber of protons as another atom but a different number of neutrons

2. a. 11 protons, 11 electrons,12 neutronsb. 20 protons, 20 electrons,20 neutronsc. 29 protons, 29 electrons,35 neutronsd. 47 protons, 47 electrons,61 neutrons

3. a. 2814Si, silicon-28

b. 5626Fe, iron-56

4. 39.10 amu, 39.10 g/mol

5. a. 28.0 g Nb. 17.7 g Cl

6. a. 0.5000 mol Mgb. 0.249 mol F

7. Beaker B; Both beakers contain thesame number of atoms.

SECTION REVIEW

Go to go.hrw.com formore practice problemsthat ask you to convertamong atoms, grams,and moles.

Keyword: HC6ATMX

number of atoms of Cu ⎯→ amount of Cu in moles ⎯→ mass of Cu in grams

As indicated in Figure 11, the given number of atoms must first be converted to amount inmoles by dividing by Avogadro’s number. Amount in moles is then multiplied by molarmass to yield mass in grams.

Cu atoms × × = grams Cu

The molar mass of copper from the periodic table is rounded to 63.55 g/mol.

1.20 × 108 Cu atoms × × = 1.27 × 10−14 g Cu

Units cancel correctly to give the answer in grams. The size of the answer is reasonable—108 has been divided by about 1024 and multiplied by about 102.

63.55 g Cu

1 mol Cu

1 mol Cu

6.022 × 1023 Cu atoms

grams Cu

moles Cu

moles Cu

Avogadro’s number of Cu atoms

2 PLAN

3 COMPUTE

4 EVALUATE

1. What is the mass in grams of 7.5 × 1015 atoms of nickel, Ni?

2. How many atoms of sulfur, S, are in 4.00 g of sulfur?

3. What mass of gold, Au, contains the same number of atoms as9.0 g of aluminum, Al?

Answers in Appendix EPRACTICE

1. Define each of the following:

a. atomic number e. mole

b. mass number f. Avogadro’s number

c. relative atomic mass g. molar mass

d. average atomic mass h. isotope

2. Determine the number of protons, electrons, andneutrons in each of the following isotopes:

a. sodium-23 c. 6429Cu

b. calcium-40 d. 10847Ag

3. Write the nuclear symbol and hyphen notation foreach of the following isotopes:

a. mass number of 28 and atomic number of 14

b. 26 protons and 30 neutrons

4. To two decimal places, what is the relative atomicmass and the molar mass of the element potas-sium, K?

5. Determine the mass in grams of the following:

a. 2.00 mol N

b. 3.01 × 1023 atoms Cl

6. Determine the amount in moles of the following:

a. 12.15 g Mg

b. 1.50 × 1023 atoms F

Critical Thinking

7. ANALYZING DATA Beaker A contains 2.06 mol ofcopper, and Beaker B contains 222 grams of silver.Which beaker contains the larger mass? Whichbeaker has the larger number of atoms?

SECTION REVIEW

C H A P T E R H I G H L I G H T S

C H A P T E R 38888

1. a. Because all chemical reactionsare only the rearrangements ofatoms, mass is neither creatednor destroyed in such changes.b. Atoms of each element havetheir own characteristic mass, socompounds consisting of theseatoms always have the samecomposition by mass.c. Only whole atoms combine inchemical compounds, so differentcompounds between the sametwo elements must result fromthe combination of differentwhole numbers of atoms.

2. 5 mass units; 13 mass units

3. a. the smallest particle of an el-ement that retains the chemicalproperties of that elementb. the nucleus and the surround-ing electrons

4. An electron is negativelycharged, has a mass approxi-mately 1/2000 that of a hydro-gen atom, has a fixedcharge-to-mass ratio, and is pres-ent in atoms of all elements.

5. His model had most of the massin the nucleus. He bombardedgold atoms with positivelycharged particles; most wentstraight through the atom, butsome were deflected and a fewbounced back.

6. the atomic number

7. a. atoms of an element thathave the same number of pro-tons but a different number ofneutronsb. They have the same numberof protons and electrons.c. They have different numbersof neutrons.

REVIEW ANSWERS

CHAPTER REVIEW

• Cathode-ray tubes supplied evidence of the existence of elec-trons, which are negatively charged subatomic particles thathave relatively little mass.

• Rutherford found evidence for the existence of the atomicnucleus by bombarding gold foil with a beam of positivelycharged particles.

• Atomic nuclei are composed of protons, which have an electriccharge of +1, and (in all but one case) neutrons, which have noelectric charge.

• Atomic nuclei have radii of about 0.001 pm (pm = picometers;1 pm × 10−12 m), and atoms have radii of about 40–270 pm.

• The idea of atoms has been around since the time of theancient Greeks. In the nineteenth century, John Dalton pro-posed a scientific theory of atoms that can still be used toexplain properties of most chemicals today.

• Matter and its mass cannot be created or destroyed in chemi-cal reactions.

• The mass ratios of the elements that make up a given com-pound are always the same, regardless of how much of thecompound there is or how it was formed.

• If two or more different compounds are composed of the sametwo elements, then the ratio of the masses of the second el-ement combined with a certain mass of the first element canbe expressed as a ratio of small whole numbers.

law of conservation of masslaw of definite proportionslaw of multiple proportions

atomnuclear forces

Vocabulary

Vocabulary

The Atom: From Philosophical Idea to Scientific Theory

The Structure of the Atom

• The atomic number of an element is equal to the number ofprotons of an atom of that element.

• The mass number is equal to the total number of protons andneutrons that make up the nucleus of an atom of that element.

• The relative atomic mass unit (amu) is based on the carbon-12atom and is a convenient unit for measuring the mass of atoms.It equals 1.660 540 × 10−24 g.

• The average atomic mass of an element is found by calculatingthe weighted average of the atomic masses of the naturallyoccurring isotopes of the element.

• Avogadro’s number is equal to approximately 6.022 × 1023. Asample that contains a number of particles equal to Avogadro’snumber contains a mole of those particles.

atomic numberisotopemass numbernuclideatomic mass unitaverage atomic massmoleAvogadro’s numbermolar mass

Vocabulary

Counting Atoms

C H A P T E R R E V I E WFor more practice, go to the Problem Bank in Appendix D.

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 89 89

CHAPTER REVIEW

8. Isotope Protons Electrons Neutrons

Si-28 14 14 14Si-29 14 14 15Si-30 14 14 16

9. a. its number of protonsb. the total number of protonsand neutrons in an isotopec. atomic = 1, mass = 2

10. any isotope of any element

11. a. helium-4 c. potassium-39b. oxygen-16

12. a. carbon-12 b. exactly 12 amu

13. a. 4 amu b. 54 amu

14. a. the number of particles equalto the number of atoms in exact-ly 12 g of carbon-12b. molc. 6.022 × 1023

d. Avogadro’s number

15. a. the mass of one mole ofatoms of that elementb. 12.01 g/mol, 20.18 g/mol,55.85 g/mol, 238.03 g/mol

16. a. divide by molar massb. divide by molar mass, thenmultiply by Avogadro’s number

17. a. 6.94 g d. 55.8 gb. 27.0 g e. 12.01 gc. 40.1 g f. 107.9 g

18. a. 1.000 mol c. 1.57 × 103 molb. 0.5000 mol d. 2.81 × 10−13 mol

19. 39.95 amu

20. 10.00 amu

21. a. 9.03 × 1023 atomsb. 4.068 × 1024 atomsc. 1.50 × 1023 atoms

22. a. 9.500 g d. 1.12 × 10−3 gb. 6.05 g e. 7.6 × 10−21 gc. 2.65 × 10−10 g f. 3 × 10−22 g

23. a. 3.01 × 1023 atomsb. 1.51 × 1023 atomsc. 2.31 × 1022 atomsd. 7.872 × 1019 atomse. 3.06 × 1011 atoms

The Atom: From PhilosophicalIdea to Scientific TheorySECTION 1 REVIEW

1. Explain each of the following in terms ofDalton’s atomic theory:a. the law of conservation of massb. the law of definite proportionsc. the law of multiple proportions

2. According to the law of conservation of mass, ifelement A has an atomic mass of 2 mass unitsand element B has an atomic mass of 3 massunits, what mass would be expected for com-pound AB? for compound A2B3?

The Structure of the AtomSECTION 2 REVIEW

3. a. What is an atom?b. What two regions make up all atoms?

4. Describe at least four properties of electronsthat were determined based on the experimentsof Thomson and Millikan.

5. Summarize Rutherford’s model of the atom, andexplain how he developed this model based onthe results of his famous gold-foil experiment.

6. What number uniquely identifies an element?

Counting AtomsSECTION 3 REVIEW

7. a. What are isotopes?b. How are the isotopes of a particular element

alike?c. How are they different?

8. Copy and complete the following table concern-ing the three isotopes of silicon, Si.(Hint: See Sample Problem A.)

9. a. What is the atomic number of an element?b. What is the mass number of an isotope?c. In the nuclear symbol for deuterium, 2

1H,identify the atomic number and the massnumber.

10. What is a nuclide?11. Use the periodic table and the information that

follows to write the hyphen notation for eachisotope described.a. atomic number = 2, mass number = 4b. atomic number = 8, mass number = 16c. atomic number = 19, mass number = 39

12. a. What nuclide is used as the standard in therelative scale for atomic masses?

b. What is its assigned atomic mass?13. What is the atomic mass of an atom if its mass is

approximately equal to the following?a. ⎯1

3⎯ that of carbon-12

b. 4.5 times as much as carbon-1214. a. What is the definition of a mole?

b. What is the abbreviation for mole?c. How many particles are in one mole?d. What name is given to the number of parti-

cles in a mole?15. a. What is the molar mass of an element?

b. To two decimal places, write the molar mass-es of carbon, neon, iron, and uranium.

16. Suppose you have a sample of an element.a. How is the mass in grams of the element

converted to amount in moles?b. How is the mass in grams of the element

converted to number of atoms?

PRACTICE PROBLEMS

17. What is the mass in grams of each of the follow-ing? (Hint: See Sample Problems B and E.)a. 1.00 mol Li d. 1.00 molar mass Feb. 1.00 mol Al e. 6.022 × 1023 atoms Cc. 1.00 molar mass Ca f. 6.022 × 1023 atoms Ag

18. How many moles of atoms are there in each ofthe following? (Hint: See Sample Problems Cand D.)a. 6.022 × 1023 atoms Ne c. 3.25 × 105 g Pbb. 3.011 × 1023 atoms Mg d. 4.50 × 10−12 g O

Number Number of Number ofIsotope of protons electrons neutrons

Si-28Si-29Si-30

90

24. a. 80.9 g e. 413 gb. 29.5 g f. 8.24 × 109 gc. 19.3 g g. 3.50 × 10−4 gd. 1.59 × 103 g

25. Particle Symbol Mass no.

Electron 0–1e, e– 0

Proton 11H, p+ 1

Neutron 10n, n0 1

Actual Relative mass charge

9.109 × 10–31 kg −11.673 × 10–27 kg +11.675 × 10–27 kg 0

26. a. 1 amu = ⎯112⎯ of the mass of one

carbon-12 atomb. the mass of an atom as itcompares with the mass of a carbon-12 atom

27. a. central portion of the atomcontaining most of the massb. Ernest Rutherfordc. protons and neutrons

28. a. 1.00 mol e. 0.0474 molb. 0.500 mol f. 6.95 × 10−5 molc. 0.100 mol g. 37.4 mold. 4.7 mol h. 8.30 × 10−23 mol

29. If two or more compounds arecomposed of the same two el-ements, then the ratio of themasses of the second elementcombined with a certain mass ofthe first element is always a ratioof small whole numbers; CO andCO2 are examples.

30. a. 144 amu b. 6 amu

31. a small, negatively charged parti-cle of small mass found outsidethe nucleus

32. Using carbon and oxygen: The lawof definite proportions states thatany amount of CO will contain aconstant percentage by mass of Cand O. The law of multiple propor-tions accounts for the differencebetween the oxygen:carbon massratio in the two compounds. Theoxygen:carbon mass ratio in car-bon dioxide is twice the ratio incarbon monoxide.

CHAPTER REVIEW CHAPTER REVIEW

19. Three isotopes of argon occur in nature—3618Ar,

3818Ar, and 40

18Ar. Calculate the average atomicmass of argon to two decimal places, given thefollowing relative atomic masses and abun-dances of each of the isotopes: argon-36(35.97 amu; 0.337%), argon-38 (37.96 amu;0.063%), and argon-40 (39.96 amu; 99.600%).

20. Naturally occurring boron is 80.20% boron-11(atomic mass = 11.01 amu) and 19.80% of someother isotopic form of boron. What must theatomic mass of this second isotope be in orderto account for the 10.81 amu average atomicmass of boron? (Write the answer to two deci-mal places.)

21. How many atoms are there in each of the following?a. 1.50 mol Na c. 7.02 g Sib. 6.755 mol Pb

22. What is the mass in grams of each of the following?a. 3.011 × 1023 atoms F e. 25 atoms Wb. 1.50 × 1023 atoms Mg f. 1 atom Auc. 4.50 × 1012 atoms Cld. 8.42 × 1018 atoms Br

23. Determine the number of atoms in each of thefollowing:a. 5.40 g B d. 0.025 50 g Ptb. 0.250 mol S e. 1.00 × 10−10 g Auc. 0.0384 mol K

24. Determine the mass in grams of each of the following:a. 3.00 mol Alb. 2.56 × 1024 atoms Lic. 1.38 mol Nd. 4.86 × 1024 atoms Aue. 6.50 mol Cuf. 2.57 × 108 mol Sg. 1.05 × 1018 atoms Hg

MIXED REVIEW

25. Copy and complete the following table concern-ing the properties of subatomic particles.

26. a. How is an atomic mass unit (amu) related tothe mass of one carbon-12 atom?

b. What is the relative atomic mass of an atom?27. a. What is the nucleus of an atom?

b. Who is credited with the discovery of theatomic nucleus?

c. Identify the two kinds of particles that makeup the nucleus.

28. How many moles of atoms are there in each ofthe following?a. 40.1 g Ca e. 2.65 g Feb. 11.5 g Na f. 0.007 50 g Agc. 5.87 g Ni g. 2.25 × 1025 atoms Znd. 150 g S h. 50 atoms Ba

29. State the law of multiple proportions, and givean example of two compounds that illustratethe law.

30. What is the approximate atomic mass of anatom if its mass isa. 12 times that of carbon-12?b. ⎯1

2⎯ that of carbon-12?

31. What is an electron?

32. Organizing Ideas Using two chemical com-pounds as an example, describe the differencebetween the law of definite proportions and thelaw of multiple proportions.

33. Constructing Models As described inSection 2, the structure of the atom was deter-mined from observations made in painstakingexperimental research. Suppose a series ofexperiments revealed that when an electric cur-rent is passed through gas at low pressure, thesurface of the cathode-ray tube opposite the

CRITICAL THINKING

C H A P T E R 390

Mass Actual RelativeParticle Symbol number mass charge

ElectronProtonNeutron

CHAPTER REVIEW

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 91 91

33. a. from the anode to the cathodeb. positive

34. The density of the nucleus is 9 ×1012 times the density of osmi-um, or 9 trillion times as dense.

35. a. amount of O2 presentb. CO poisoning, because CObinds to hemoglobin morestrongly than O2 doesc. CO produced by a heating system cannot be adequatelyventilated.

Refer to the One-Stop PlannerCD-ROM for appropriate scoringrubrics for items 36–39. Look forthese points in each report:

36. Chadwick bombarded beryllium-9 with alpha particles to obtaincarbon-12 and a single neutron.

37. Avogadro’s work with gases

38. the wave nature of electrons;quantum theory; an essentialtool in the study of science

39. Students’ reports should mentionelements and compounds com-monly used in nuclear medicine.

40. Answers will vary.

41. 12 in each isotope

CHAPTER REVIEW

anode glows. In addition, a paddle wheel placedin the tube rolls from the anode toward thecathode when the current is on.a. In which direction do particles pass through

the gas?b. What charge do the particles possess?

34. Analyzing Data Osmium is the element withthe greatest density, 22.58 g/cm3. How does thedensity of osmium compare to the density of atypical nucleus of 2 × 108 metric tons/cm3?(1 metric ton = 1000 kg)

35. Group 14 of the Elements Handbook describesthe reactions that produce CO and CO2.Review this section to answer the following:a. When a fuel burns, what determines whether

CO or CO2 will be produced?b. What happens in the body if hemoglobin

picks up CO?c. Why is CO poisoning most likely to occur in

homes that are well sealed during cold wintermonths?

36. Prepare a report on the series of experimentsconducted by Sir James Chadwick that led tothe discovery of the neutron.

37. Write a report on the contributions of AmedeoAvogadro that led to the determination of thevalue of Avogadro’s number.

38. Trace the development of the electron micro-scope, and cite some of its many uses.

39. The study of atomic structure and the nucleusproduced a new field of medicine called nuclearmedicine. Describe the use of radioactive tracers to detect and treat diseases.

RESEARCH & WRITING

USING THE HANDBOOK

40. Observe a cathode-ray tube in operation, andwrite a description of your observations.

41. Performance Assessment Using colored clay,build a model of the nucleus of each of carbon’sthree naturally occurring isotopes: carbon-12,carbon-13, and carbon-14. Specify the numberof electrons that would surround each nucleus.

ALTERNATIVE ASSESSMENT

Graphing Calculator Calculating Numbersof Protons, Electrons, and Neutrons

Go to go.hrw.com for a graphing calculatorexercise that asks you to calculate numbersof protons, electrons, and neutrons.

Keyword: HC6ATMX

Math Tutor CONVERSION FACTORS

92

1. a. 2.25 gb. 59 300 L

2. a. 7.2 × 101 mgb. 3.98 × 103 km

ANSWERS

CHAPTER REVIEW

SAMPLE 1 SAMPLE 2

Most calculations in chemistry require that all measurements of the same quantity(mass, length, volume, temperature, and so on) be expressed in the same unit. Tochange the units of a quantity, you can multiply the quantity by a conversion factor.With SI units, such conversions are easy because units of the same quantity are relatedby multiples of 10, 100, 1000, or 1 million. Suppose you want to convert a givenamount in milliliters to liters. You can use the relationship 1 L = 1000 mL. From thisrelationship, you can derive the following conversion factors.

and

The correct strategy is to multiply the given amount (in mL) by the conversion factorthat allows milliliter units to cancel out and liter units to remain. Using the secondconversion factor will give you the units you want.

These conversion factors are based on an exact definition (1000 mL = 1 L exactly),so significant figures do not apply to these factors. The number of significant figures ina converted measurement depends on the certainty of the measurement you start with.

1 L⎯1000 mL

1000 mL⎯

1 L

A sample of aluminum has a mass of 0.087 g.What is the sample’s mass in milligrams?

Based on SI prefixes, you know that 1 g =1000 mg. Therefore, the possible conversion fac-tors are

and

The first conversion factor cancels grams, leav-ing milligrams.

0.087 g × = 87 mg

Notice that the values 0.087 g and 87 mg eachhave two significant figures.

1000 mg⎯

1 g

1 g⎯1000 mg

1000 mg⎯

1 g

A sample of a mineral has 4.08 � 10�5 mol ofvanadium per kilogram of mass. How many micro-moles of vanadium per kilogram does the mineralcontain?

The prefix micro- specifies ⎯1 0010 000⎯ or 1 × 10−6 of

the base unit.So, 1 mmol = 1 × 10−6 mol. The possible conver-

sion factors are

and

The first conversion factor will allow moles tocancel and micromoles to remain.

4.08 × 10−5 mol × = 40.8 mmol

Notice that the values 4.08 × 10−5 mol and40.8 mmol each have three significant figures.

1 mmol⎯⎯1 × 10−6 mol

1 × 10−6 mol⎯⎯

1 mmol

1 mmol⎯⎯1 × 10−6 mol

C H A P T E R 392

1. Express each of the following measurementsin the units indicated.

a. 2250 mg in gramsb. 59.3 kL in liters

2. Use scientific notation to express each of thefollowing measurements in the units indicated.

a. 0.000 072 g in microgramsb. 3.98 × 106 m in kilometers

PRACTICE PROBLEMS

Standardized Test Prep

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 93 93

To give students practice under morerealistic testing conditions, give them60 minutes to answer all of the ques-tions in this Standardized TestPreparation.

1. C

2. D

3. C

4. A

5. B

6. A

7. B

8. D

9. C

10. Argon-40 has 22 neutrons (40 −18 = 22), and potassium-40 has21 neutrons (40 − 19 = 21).

11. 6.17 g

12. All cathode rays are the same,regardless of their source.Therefore, the particles responsi-ble for the cathode rays must bepresent in all atoms. The particlesare electrons.

13. When the average atomic mass iscalculated, it is 10.811. Becausethe atomic mass is the same asthe atomic mass of boron, mythi-um was not a new element.

TEST ANSWERS

CHAPTER REVIEW

Answer the following items on a separate piece of paper.

MULTIPLE CHOICE

1.A chemical compound always has the same el-ements in the same proportions by mass regard-less of the source of the compound. This is astatement ofA. the law of multiple proportions.B. the law of isotopes.C. the law of definite proportions.D. the law of conservation of mass.

2.An important result of Rutherford’s experi-ments with gold foil was to establish thatA. atoms have mass.B. electrons have a negative charge.C. neutrons are uncharged particles.D. the atom is mostly empty space.

3.Which subatomic particle has a charge of +1?A. electronB. neutronC. protonD. meson

4.Which particle has the least mass?A. electronB. neutronC. protonD. All have the same mass.

5.Cathode rays are composed ofA. alpha particles.B. electrons.C. protons.D. neutrons.

6. The atomic number of an element is the sameas the number ofA. protons.B. neutrons.C. protons + electrons.D. protons + neutrons.

7.How many neutrons are present in an atom oftin that has an atomic number of 50 and a massnumber of 119?A. 50B. 69C. 119D. 169

8.What is the mass of 1.50 mol of sodium, Na?A. 0.652 gB. 0.478 gC. 11.0 gD. 34.5 g

9.How many moles of carbon are in a 28.0 g sample?A. 336 molB. 72.0 molC. 2.33 molD. 0.500 mol

SHORT ANSWER

10.Which atom has more neutrons, potassium-40 orargon-40?

11.What is the mass of 1.20 × 1023 atoms of phosphorus?

EXTENDED RESPONSE

12.Cathode rays emitted by a piece of silver and apiece of copper illustrate identical properties.What is the significance of this observation?

13.A student believed that she had discovered anew element and named it mythium. Analysisfound it contained two isotopes. The composi-tion of the isotopes was 19.9% of atomic mass10.013 and 80.1% of atomic mass 11.009. Whatis the average atomic mass, and do you thinkmythium was a new element?

Choose the best possible answerfor each question, even if you think there is anotherpossible answer that is not given.

13-1 TEACHER‘S NOTES

RECOMMENDED TIME30 min

RATINGSTEACHER PREPARATION 1STUDENT SETUP 2CONCEPT LEVEL 2CLEANUP 1

MATERIALS(for each lab group)• 2 L plastic soda bottle• 17 g baking soda (sodium hydro-

gen carbonate)• 100 mL graduated cylinder• 270 mL vinegar (5% acetic acid

solution)• balance, triple beam or electronic• clear plastic cups, 2• hook-insert cap for bottle• microplunger

1. Use vinegar and baking sodainstead of diluted glacial acetic acid.2. To make the hook-insert caps forthe bottles, use hot-melt glue to attacha wire hook to the inside center of thebottle cap, as shown in Figure 1. Thehook may be handmade or purchasedfrom a hardware store.

CHAPTER LAB

94

Conservation of Mass

CHAPTER LAB MICRO-L A B

BACKGROUNDThe law of conservation of mass states that matter is neither created nor destroyed during a chemicalreaction. Therefore, the mass of a system shouldremain constant during any chemical process. In thisexperiment, you will determine whether mass is con-served by examining a simple chemical reaction andcomparing the mass of the system before the reac-tion with its mass after the reaction.

SAFETY

For review of safety, please see Safety in theChemistry Laboratory in the front of your book.

PREPARATION1. Make two data tables in your lab notebook, one

for Part I and another for Part II. In each table,create three columns labeled “Initial mass (g),”“Final mass (g),” and “Change in mass (g).”Each table should also have space for observa-tions of the reaction.

PROCEDURE—PART I1. Obtain a microplunger, and tap it down into

a sample of baking soda until the bulb end ispacked with a plug of the powder (4–5 mL ofbaking soda should be enough to pack the bulb).

2. Hold the microplunger over a plastic cup, andsqueeze the sides of the microplunger to loosenthe plug of baking soda so that it falls into the cup.

3. Use a graduated cylinder to measure 100 mL of vinegar, and pour it into a second plastic cup.

4. Place the two cups side by side on the balancepan, and measure the total mass of the system

OBJECTIVES

• Observe the signs of a chemical reaction.

• Compare masses of reactants and products.

• Design experiments.

• Relate observations to the law of conservation of mass.

MATERIALS

• 2 L plastic soda bottle

• 5% acetic acid solution (vinegar)

• balance

• clear plastic cups, 2

• graduated cylinder

• hook-insert cap for bottle

• microplunger

• sodium hydrogen carbonate (baking soda)

TEACHER‘S NOTES

EASY HARD

1 2 3 4

C H A P T E R 394

FIGURE A Slowly add the vinegar to prevent the reaction from getting out of control.

Wire hook

Hot-melt glueattaching hook toinside of bottle cap

2 L plasticsoda bottle cap

FIGURE 1

INQUIRYL A B?

13-1 TEACHER‘S NOTES

95

3. To make the microplungers, cutgraduated pipets, and thread a loopof string or thread through the tip asshown in Figure 2.

REQUIRED PRECAUTIONS• Safety goggles and a lab apron

must be worn at all times.• Read all safety precautions, and

discuss them with your students.• If, instead of using vinegar, glacial

acetic acid is used to prepare a 5%acetic acid solution, wear safetygoggles, a face shield, impermeablegloves, and a lab apron whilepreparing the solution. Work ina fume hood that is known to be in good working order, and haveanother person stand by to call for help in case of an emergency.

• In case of a spill, use a dampenedcloth or paper towel (or more thanone towel if necessary) to mop upthe spill. Then, rinse the cloth inrunning water at the sink, wring it out thoroughly, and put it in the trash.

Continued on page 95A

ANALYSIS AND INTERPRETATION—PART I

1. Drawing Conclusions: What evidence was therethat a chemical reaction occurred?

2. Organizing Data: How did the final mass of thesystem compare with the initial mass of thesystem?

3. Resolving Discrepancies: Does your answer to the previous question show that the law of conservation of mass was violated? (Hint:Another way to express the law of conservationof mass is to say that the mass of all of the products equals the mass of all of the reactants.)What do you think might cause the mass difference?

ANALYSIS AND INTERPRETATION—PART II

1. Drawing Conclusions: Was there any new evidence in Part II indicating that a chemicalreaction occurred?

2. Organizing Ideas: Identify the state of matter for each reactant in Part II. Identify the state of matter for each product.

CONCLUSIONS1. Relating Ideas: What is the difference between

the system in Part I and the system in Part II?What change led to the improved results in Part II?

2. Evaluating Methods: Why did the procedure forPart II work better than the procedure for Part I?

EXTENSIONS1. Applying Models: When a log burns, the result-

ing ash obviously has less mass than theunburned log did. Explain whether this loss ofmass violates the law of conservation of mass.

2. Designing Experiments: Design a procedure thatwould test the law of conservation of mass forthe burning log described in Extension item 1.

TEACHER‘S NOTES

Graduated pipetCut here Cut here

Thread loop through hole in pipet stem

Open end of plunger

Hot needle

Thread

FIGURE 2

A T O M S : T H E B U I L D I N G B L O C K S O F M A T T E R 95

(before reaction) to the nearest 0.01 g. Recordthe mass in your data table.

5. Add the vinegar to the baking soda a little at a time to prevent the reaction from getting outof control, as shown in Figure A. Allow the vinegar to slowly run down the inside of the cup. Observe and record your observationsabout the reaction.

6. When the reaction is complete, place both cupson the balance, and determine the total final massof the system to the nearest 0.01 g. Calculate anychange in mass. Record both the final mass andany change in mass in your data table.

7. Examine the plastic bottle and the hook-insertcap. Try to develop a modified procedure thatwill test the law of conservation of mass moreaccurately than the procedure in Part I.

8. In your notebook, write the answers to items 1through 3 in Analysis and Interpretation—Part I.

PROCEDURE—PART II9. Your teacher should approve the procedure you

designed in Procedure—Part I, step 7. Implementyour procedure with the same chemicals andquantities you used in Part I, but use the bottleand hook-insert cap in place of the two cups.Record your data in your data table.

10. If you were successful in step 9 and your resultsreflect the conservation of mass, proceed to complete the experiment. If not, find a lab groupthat was successful, and discuss with them whatthey did and why they did it. Your group shouldthen test the other group’s procedure to deter-mine whether their results are reproducible.

CLEANUP AND DISPOSAL11. Clean your lab station. Clean all equip-

ment, and return it to its proper place.Dispose of chemicals and solutions inthe containers designated by your teacher.Do not pour any chemicals down the drain or throw anything in the trash unless yourteacher directs you to do so. Wash your handsthoroughly after all work is finished and beforeyou leave the lab.

Continued from page 95

TECHNIQUES TO DEMONSTRATEShow students how to fill the microplunger with a plugof baking soda and how to manipulate the plunger todeliver the baking soda in the plastic cup. Demonstratehow to control the rapid fizzing of the baking soda bythe slow addition of vinegar down the inside of thecup. Students’ modified procedures to step 7 inProcedure—Part I will vary but should demonstratetheir understanding of the law of conservation of massas it relates to the experiment.

SAMPLE DATAStudent data will vary.

DISPOSALAll of the solutions and chemicals used in this lab maybe washed down the sink with water.

ANALYSIS AND INTERPRETATION—PART I—ANSWERS1. The reactants effervesce with the formation of a newgaseous substance. The baking soda seemed to disap-pear eventually.2. Students’ answers will vary, but typical results showa mass loss of 1.0–1.5 g for Part I.3. Students’ answers should indicate that although themass loss appears to violate the law of conservation ofmass, the law probably still holds. The reason is that notall of the products were present for the second measure-ment of mass. The bubbles that formed and poppedwere a gaseous product that escaped into the air.

ANALYSIS AND INTERPRETATION—PART II—ANSWERS1. Again, there was vigorous bubbling for the first fewseconds. However, the fizzing noise was not very audible.The sides of the bottle became tight and hard to push in,and when the cap was removed, the hissing noise of gasescaping under pressure could be heard.

CHAPTER LAB

95A

CONTINUATION OF ANSWERS AND TEACHER’S NOTES

CONTINUATION OF ANSWERS AND TEACHER’S NOTES

2. The reactants were a liquid (vinegar) and a solid(baking soda). The products of the reaction were a gas(the carbon dioxide seen forming the bubbles) and aliquid (a solution of excess vinegar and sodium acetate).

CONCLUSIONS—ANSWERS1. In Part I the system was left open, but it was closedin Part II so that not even gas could escape.2. Part II worked better because one of the productswas a gas and the bottle-and-cap system kept the gastrapped so that its mass could be measured with theother products. In Part I, the gas escaped.

EXTENSIONS—ANSWERS1. Just as the reaction studied in this experiment pro-duced a product that escaped into the environment,the burning of a log produces many products, not justashes. The smoke, CO2, and water vapor escape as thelog burns. Presumably, if one had a way to measure thetotal mass of the products, it would be the same as thetotal mass of the reactants.2. Students’ answers will vary, but they should describea way to burn a log in a closed, rigid container thatcould withstand heat and flames without reacting.Students may also recognize that the amount of oxygenneeded to burn the log would require a very large con-tainer unless the oxygen were purified from the air orkept under high pressure. You may want to point out to students that this experiment would be very difficultand potentially dangerous because of the pressuresthat would build up as a result of the carbon in the log being converted to carbon dioxide gas.

95B