Atoms and Elements - Higher Educationesminfo.prenhall.com/tro1e_chps/tro_ch02.pdf · 46 Chapter 2...

Atoms and Elements2.1 Imaging and Moving Individual Atoms

2.2 Early Ideas about the Building Blocks of Matter

2.3 Modern Atomic Theory and the Laws That Led to It

2.4 The Discovery of the Electron

2.5 The Structure of the Atom

2.6 Subatomic Particles: Protons, Neutrons, and Electrons in Atoms

2.7 Finding Patterns: The Periodic Law and the Periodic Table

2.8 Atomic Mass: The Average Mass of an Element’s Atoms

2.9 Molar Mass: Counting Atoms by Weighing Them

2

44

If you cut a piece of graphite from the tip

of a pencil into smaller and smaller pieces, how far could you go?

Could you divide it forever? Would you eventually run into

some basic particles that were no longer divisible, not because of

their sheer smallness, but because of the nature of matter? This

fundamental question about matter has been asked by thinkers for

over two millennia. The answer given, however, has varied over time. On the scale of everyday

objects, matter appears continuous, or infinitely divisible. Until about 200 years ago, many

scientists thought that matter was indeed continuous—but they were wrong. Eventually, as you

divide the graphite from your pencil tip into smaller and smaller pieces, you end up with carbon

atoms. The word atom comes from the Greek atomos, meaning “indivisible.” You cannot divide a

carbon atom into smaller pieces and still have carbon. Atoms compose all ordinary matter—if you

want to understand matter, you must begin by understanding atoms.

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2.1 Imaging and Moving Individual AtomsOn March 16, 1981, Gerd Binnig and Heinrich Rohrer worked late into the night in theirlaboratory at IBM in Zurich, Switzerland. They were measuring how an electricalcurrent—flowing between a sharp metal tip and a flat metal surface—varied with the sep-aration between the tip and the surface. The results of that night’s experiment and subse-quent results over the next several months won Binnig and Rohrer a share of the 1986Nobel Prize in Physics. They had discovered scanning tunneling microscopy (STM), a tech-nique that can image, and even move, individual atoms and molecules.

A scanning tunneling microscope works by moving an extremely sharp electrode(often tipped with a single atom) over a surface and measuring the resulting tunnelingcurrent, the electrical current that flows between the tip of the electrode and the surfaceeven though the two are not in physical contact (Figure 2.1, page 46).

These observations have tacitly led to the conclusion which seems universallyadopted, that all bodies of sensible magnitude . . . are constituted of a vast number of extremely small particles, or atoms of matter . . . .—JOHN DALTON (1766–1844)

The tip of a scanning tunneling microscope (STM) moves across an atomic surface.

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46 Chapter 2 Atoms and Elements

Movement of tip is used to createan image with atomic resolution.

Tunneling current is extremelysensitive to distance.

Tip is scanned across surface andmoved up and down to maintain

constant tunneling current.

� FIGURE 2.1 Scanning TunnelingMicroscopy In this technique, anatomically sharp tip is scanned across asurface. The tip is kept at a fixed distancefrom the surface by moving it up anddown so as to maintain a constant tunnel-ing current. The motion of the tip isrecorded to create an image of the surfacewith atomic resolution.

The tunneling current, as Binnig and Rohrer found that night in their laboratory atIBM, is extremely sensitive to distance, making it possible to maintain a precise separationof approximately two atomic diameters between the tip and the surface simply by movingthe tip so as to keep the current constant. By measuring the up-and-down movement ofthe tip as it is scanned horizontally across a surface, an image of the surface—with atomicresolution—can be created, like the one shown in Figure 2.2(a) �.

In other words, Binnig and Rohrer had discovered a type of microscope that could“see” atoms. Later work by other scientists showed that the STM could also be used topick up and move individual atoms or molecules, allowing structures and patterns to bemade one atom at a time. Figure 2.2(b), for example, shows the Kanji characters for theword “atom” written with individual iron atoms on top of a copper surface. If all of thewords in the books in the Library of Congress—19 million books on 530 miles ofshelves—were written in letters the size of this Kanji character, they would fit in an areaof about 5 square millimeters.

(a) (b)

� FIGURE 2.2 Imaging Atoms(a) A scanning tunneling microscopeimage of iodine atoms (green) on a plat-inum surface. (b) The Japanese charactersfor “atom” written with atoms.

As we saw in Chapter 1, it was only 200 years ago that John Dalton proposedhis atomic theory. Today we can not only image and move individual atoms, we are evenbeginning to build tiny machines out of just a few dozen atoms (an area of research callednanotechnology).

These atomic machines, and the atoms that compose them, are almost unimaginablysmall. To get an idea of the size of an atom, imagine picking up a grain of sand at your fa-vorite beach. That grain contains more atoms than you could count in a lifetime. In fact,the number of atoms in one sand grain far exceeds the number of grains on the entirebeach. If every atom within the sand grain were the size of the grain itself, the sand grainwould be the size of a large mountain range.

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2.3 Modern Atomic Theory and the Laws That Led to It 47

Despite their size, atoms are the key to connecting the macroscopic and microscopicworlds. An atom is the smallest identifiable unit of an element. There are about 91 differ-ent naturally occurring elements. In addition, scientists have succeeded in making over20 synthetic elements (not found in nature). In this chapter, we learn about atoms: whatthey are made of, how they differ from one another, and how they are structured. We alsolearn about the elements made up of these different kinds of atoms, and about some ofthe characteristics of those elements. We will find that the elements can be organized in away that reveals patterns in their properties and helps us to understand what underliesthose properties.

2.2 Early Ideas about the Building Blocks of Matter

The first people to propose that matter was composed of small, indestructible particleswere Leucippus (fifth century B.C., exact dates unknown) and his student Democritus(460–370 B.C.). These Greek philosophers theorized that matter was ultimately composedof small, indivisible particles called atomos. Democritus wrote, “Nothing exists exceptatoms and empty space; everything else is opinion.” Leucippus and Democritus proposedthat many different kinds of atoms existed, each different in shape and size, and that theymoved randomly through empty space. However, other influential Greek thinkers of thetime, such as Plato and Aristotle, did not embrace the work of Leucippus and Democri-tus. Instead, they held that matter had no smallest parts and that different substances werecomposed of various proportions of fire, air, earth, and water. Since there was no experi-mental way to test the relative merits of the two ideas, Aristotle’s view prevailed, partly be-cause he was so influential. The idea that matter was composed of atoms took a back seatin intellectual thought for nearly 2000 years.

In the sixteenth century modern science began to emerge. A greater emphasis on ob-servation led Nicolaus Copernicus (1473–1543) to publish On the Revolution of the Heav-enly Orbs in 1543. That book—which proposed that the sun, not Earth, was at the centerof the universe—marks the beginning of what we now call the scientific revolution. Thenext 200 years—with the works of scientists such as Francis Bacon (1561–1626), JohannesKepler (1571–1630), Galileo Galilei (1564–1642), Robert Boyle (1627–1691), and IsaacNewton (1642–1727)—brought rapid advancement as the scientific method became theestablished way to learn about the physical world. By the early 1800s certain observationsled the English chemist John Dalton (1766–1844) to offer convincing evidence that sup-ported the early atomic ideas of Leucippus and Democritus.

2.3 Modern Atomic Theory and the Laws That Led to It

Recall the discussion of the scientific method from Chapter 1. The theory that all matter iscomposed of atoms grew out of observations and laws. The three most important laws thatled to the development and acceptance of the atomic theory were the law of conservationof mass, the law of definite proportions, and the law of multiple proportions.

The Law of Conservation of MassIn 1789, as we saw in Chapter 1, Antoine Lavoisier formulated the law of conservation ofmass, which states the following:

In a chemical reaction, matter is neither created nor destroyed.

In other words, when you carry out any chemical reaction, the total mass of the sub-stances involved in the reaction does not change. For example, consider the reaction be-tween sodium and chlorine to form sodium chloride.

The exact number of naturally occurringelements is controversial, because someelements that were first discovered whenthey were synthesized are believed to bepresent in trace amounts in nature.

We will see in Chapter 19 that this law is aslight oversimplification. However, thechanges in mass in ordinary chemicalprocesses are so minute that they can beignored for all practical purposes.

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The combined mass of the sodium and chlorine that react exactly equals the mass of thesodium chloride that results from the reaction. This law is consistent with the idea thatmatter is composed of small, indestructible particles. The particles rearrange during achemical reaction, but the amount of matter is conserved because the particles are inde-structible (at least by chemical means).

7.7 g Na 11.9 g Cl2

Total mass � 19.6 g

Na(s) Cl2(g)

Mass of reactants � Mass of product

19.6 g NaCl

NaCl(s)

Conceptual Connection 2.1 The Law of Conservation of Mass

When a small log completely burns in a campfire, the mass of the ash is much less thanthe mass of the log. What happened to the matter that composed the log?

Answer: Most of the matter that composed the log underwent a chemical change by reacting withoxygen molecules in the air and was released as gases into the air, mostly carbon dioxide and water.

The law of definite proportions is sometimescalled the law of constant composition.

The Law of Definite ProportionsIn 1797, a French chemist named Joseph Proust (1754–1826) made observations on thecomposition of compounds. He found that the elements composing a given compoundalways occurred in fixed (or definite) proportions in all samples of the compound.In contrast, the components of a mixture could be present in any proportions whatsoever.He summarized his observations in the law of definite proportions, which states thefollowing:

All samples of a given compound, regardless of their source or how they wereprepared, have the same proportions of their constituent elements.

For example, the decomposition of 18.0 g of water results in 16.0 g of oxygen and 2.0 g ofhydrogen, or an oxygen-to-hydrogen mass ratio of:

or 8:1

This ratio holds for any sample of pure water, regardless of its origin. The law ofdefinite proportions applies not only to water, but also to every compound. Consider

Mass ratio =

16.0 g O

2.0 g H= 8.0

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2.3 Modern Atomic Theory and the Laws That Led to It 49

ammonia, a compound composed of nitrogen and hydrogen. Ammonia contains 14.0 gof nitrogen for every 3.0 g of hydrogen, resulting in a nitrogen-to-hydrogen mass ratio of:

or 4.7:1

Again, this ratio is the same for every sample of ammonia. The law of definite propor-tions also hints at the idea that matter might be composed of atoms. Compounds havedefinite proportions of their constituent elements because they are composed of adefinite ratio of atoms of each element, each with its own specific mass. Since the ratioof atoms is the same for all samples of a particular compound, the ratio of masses isalso the same.

EXAMPLE 2.1 Law of Definite ProportionsTwo samples of carbon dioxide are decomposed into their constituent elements. Onesample produces 25.6 g of oxygen and 9.60 g of carbon, and the other produces 21.6 g ofoxygen and 8.10 g of carbon. Show that these results are consistent with the law of defi-nite proportions.

Solution

Mass ratio =

14.0 g N

3.0 g H= 4.7

To show this, compute the mass ratio of one element to theother for both samples by dividing the mass of one element bythe mass of the other. It is usually more convenient to dividethe larger mass by the smaller one.

For the first sample:

or 2.67:1

For the second sample:

or 2.67:1Mass oxygen

Mass carbon=

21.6

8.10= 2.67

Mass oxygen

Mass carbon=

25.6

9.60= 2.67

The ratios are the same for the two samples, so these results are consistent with the lawof definite proportions.

For Practice 2.1Two samples of carbon monoxide were decomposed into their constituent elements.One sample produced 17.2 g of oxygen and 12.9 g of carbon, and the other sample pro-duced 10.5 g of oxygen and 7.88 g of carbon. Show that these results are consistent withthe law of definite proportion.

The Law of Multiple ProportionsIn 1804, John Dalton published his law of multiple proportions, which asserts the fol-lowing principle:

When two elements (call them A and B) form two different compounds, themasses of element B that combine with 1 g of element A can be expressed as aratio of small whole numbers.

Dalton already suspected that matter was composed of atoms, so that when two elementsA and B combined to form more than one compound, an atom of A combined with one,two, three, or more atoms of B (AB1, AB2, AB3, etc.). Therefore the masses of B that react-ed with a fixed mass of A would always be related to one another as small whole-numberratios. For example, consider the compounds carbon monoxide and carbon dioxide,which we discussed in the opening section of Chapter 1 as well as in Example 2.1 and itsFor Practice problem. Carbon monoxide and carbon dioxide are two compounds com-posed of the same two elements: carbon and oxygen. We saw in Example 2.1 that the mass

Answers to For Practice and For More Practice Problems can be foundin Appendix IV.

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The ratio of these two masses is itself a small whole number.

With the help of the molecular models, we can see why the ratio is 2:1—carbon dioxidecontains two oxygen atoms to every carbon atom while carbon monoxide containsonly one.

Mass oxygen to 1 g carbon in carbon dioxide

Mass oxygen to 1 g carbon in carbon monoxide=

2.67

1.33= 2

Conceptual Connection 2.2 The Laws of Definite and Multiple Proportions

Explain the difference between the law of definite proportions and the law of multipleproportions.

Answer: The law of definite proportions applies to two or more samples of the same compound andstates that the ratio of one element to the other in the samples will always be the same. The law ofmultiple proportions applies to two different compounds containing the same two elements(A and B) and states that the masses of B that combine with 1 g of A are related as a small whole-number ratio.

Mass oxygen that combineswith 1 g carbon � 2.67 g

Mass oxygen that combineswith 1 g carbon � 1.33 g

Carbon dioxide

Carbon monoxide

EXAMPLE 2.2 Law of Multiple ProportionsNitrogen forms several compounds with oxygen, including nitrogen dioxide and dini-trogen monoxide. Nitrogen dioxide contains 2.28 g oxygen to every 1.00 g nitrogenwhile dinitrogen monoxide contains 0.570 g oxygen to every 1.00 g nitrogen. Show thatthese results are consistent with the law of multiple proportions.

Solution

To show this, simply compute the ratio of the mass of oxygenfrom one compound to the mass of oxygen in the other.Always divide the larger of the two masses by the smaller one.

Mass oxygen to 1 g

nitrogen in nitrogen dioxide

Mass oxygen to 1 gnitrogen in dinitrogen monoxide

=

2.28

0.570= 4.00

The ratio is a small whole number (4); therefore, these results are consistent with the lawof multiple proportions.

For Practice 2.2Hydrogen and oxygen form both water and hydrogen peroxide. A sample of water is de-composed and forms 0.125 g hydrogen to every 1.00 g oxygen. A sample of hydrogenperoxide is decomposed and forms 0.250 g hydrogen to every 1.00 g oxygen. Show thatthese results are consistent with the law of multiple proportions.

ratio of oxygen to carbon in carbon dioxide is 2.67:1; therefore, 2.67 g of oxygen wouldreact with 1 g of carbon. In carbon monoxide, however, the mass ratio of oxygen to car-bon is 1.33:1, or 1.33 g of oxygen to every 1 g of carbon.

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2.4 The Discovery of the Electron 51

John Dalton and the Atomic TheoryIn 1808, John Dalton explained the laws just discussed with his atomic theory, which in-cluded the following concepts:

1. Each element is composed of tiny, indestructible particles called atoms.

2. All atoms of a given element have the same mass and other properties that distin-guish them from the atoms of other elements.

3. Atoms combine in simple, whole-number ratios to form compounds.

4. Atoms of one element cannot change into atoms of another element. In a chemicalreaction, atoms change the way that they are bound together with other atoms toform a new substance.

Today, as we have seen in the opening section of this chapter, the evidence for theatomic theory is overwhelming. Matter is indeed composed of atoms.

Chemistry in Your DayAtoms and Humans

We are all composed of atoms. We get those atoms from thefood we eat over the years. Yesterday’s cheeseburger contributesto today’s skin, muscle, and hair. The carbon atoms in our ownbodies have been used by other living organisms before we gotthem and will be used by still others when we are done withthem. In fact, it is likely that at this moment, your body containssome carbon atoms (over one trillion*) that were at one timepart of your chemistry professor.

The idea that humans are composed of atoms acting in ac-cord with the laws of chemistry and physics has significant im-plications and raises important questions. If atoms composeour brains, for example, do they determine our thoughts andemotions? Are our feelings caused by atoms acting according tothe laws of chemistry and physics?

Richard Feynman (1918–1988), a Nobel Prize–winningphysicist, asserted “The most important hypothesis in all of

biology is that everything that animals do, atoms do. In otherwords, there is nothing that living things do that cannot be un-derstood from the point of view that they are made of atomsacting according to the laws of physics.” Indeed, biology has un-dergone a revolution throughout the last 50 years, mostlythrough the investigation of the atomic and molecular basis forlife. Some have seen the atomic view of life as a devaluation ofhuman life. We have always wanted to distinguish ourselvesfrom everything else, and the idea that we are made of the samebasic particles as all other matter takes something away fromthat distinction . . . . Or does it?

* This calculation assumes that all of the carbon atoms metabolized by yourprofessor over the last 40 years have been uniformly distributed into atmos-pheric carbon dioxide, and then incorporated into plants that you eat.

QuestionsDo you find the idea that you are made of recycled atoms dis-turbing? Why or why not? Reductionism refers to the explanationof complex systems by reducing them to their parts. Is reduction-ism a good way to understand humans? Is it the only way?

2.4 The Discovery of the ElectronBy the end of the nineteenth century, scientists were convinced that matter was composed ofatoms, the permanent, supposedly indestructible building blocks that composed everything.However, further experiments revealed that the atom itself was composed of even smaller,more fundamental particles.

Cathode RaysIn the late 1800s an English physicist named J. J. Thomson (1856–1940), working at Cam-bridge University, performed experiments to probe the properties of cathode rays. Cath-ode rays are produced when a high electrical voltage is applied between two electrodeswithin a partially evacuated tube called a cathode ray tube shown in Figure 2.3 (page 52).

Cathode rays travel from the negatively charged electrode, called the cathode, to thepositively charged electrode, called the anode. Thomson found that these rays were actu-ally streams of particles with the following properties: they traveled in straight lines; theywere independent of the composition of the material from which they originated; and

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Positive (red) and negative (yellow)electrical charges attract one another.

�1 (�1) 0

�

� �

� �

�

�

� ��

� �

Properties of Electrical Charge

�

Positive and negative chargesof exactly the same magnitudesum to zero when combined.

Positive charges repel one another.Negative charges repel one another.


�

�

Charge-to-Mass Ratio of the Electron

Magnet

Electricallycharged plates

Evacuated tube

Electric and magnetic fieldsdeflect electron beam.

N

S��

Undeflectedelectron beam

Anode

Cathode

they carried a negative electrical charge. Electrical charge is a fundamental property ofsome of the particles that compose the atom and results in attractive and repulsiveforces—called electrostatic forces—between them. The characteristics of electrical chargeare summarized in the figure in the margin.

You have probably experienced excess electrical charge when brushing your hair on adry day. The brushing action causes the accumulation of charged particles in your hair,which repel each other, causing your hair to stand on end.

J. J. Thomson measured the charge-to-mass ratio of the particles within cathode raysby deflecting them using electric and magnetic fields, as shown in Figure 2.4 �.

The value he measured, coulombs (C) per gram, implied that the cath-ode ray particle was about 2000 times lighter than hydrogen, the lightest known atom.These results were incredible—the indestructible atom could apparently be chipped! J. J.Thomson had discovered the electron, a negatively charged, low mass particle presentwithin all atoms. Thomson wrote, “We have in the cathode rays matter in a new state, astate in which the subdivision of matter is carried very much further . . . a state in whichall matter . . . is of one and the same kind; this matter being the substance from which allthe chemical elements are built up.”

-1.76 * 108

� FIGURE 2.4 Thomson’s Measurement of the Charge-to-Mass Ratio of the ElectronJ. J. Thomson used electric and magnetic fields to deflect the electron beam in a cathode ray tube. Bymeasuring the field strengths at which the two effects canceled exactly, leaving the beam undeflected,he was able to calculate the charge-to-mass ratio of the electron.

The coulomb (C) is the SI unit for charge.

High voltage

�

�

Cathode rays

Partially evacuatedglass tube

� FIGURE 2.3 Cathode Ray Tube

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2.4 The Discovery of the Electron 53

Millikan’s Oil Drop Experiment: The Charge of the ElectronIn 1909, American physicist Robert Millikan (1868–1953), working at the University ofChicago, performed his now famous oil drop experiment in which he deduced the chargeof a single electron. The apparatus for the oil drop experiment is shown in Figure 2.5 �.

Oil was sprayed into fine droplets using an atomizer. Thedroplets were allowed to fall under the influence of gravitythrough a small hole into the lower portion of the apparatuswhere they could be viewed with the help of a light sourceand a viewing microscope. During their fall, the dropswould acquire electrons that had been producedby the interaction of high-energy radiationwith air. These charged drops now interact-ed with two electrically charged plateswithin the apparatus. (Remember that likecharges repel each other.) The negativelycharged plate at the bottom of the appara-tus repelled the negatively charged drops.By varying the voltage on the plates, the fallof the charged drops could be slowed,stopped, or even reversed.

From the voltage required to halt thefree fall of the drops, and from the massesof the drops themselves (determined fromtheir radii and density), Millikan calculated thecharge of each drop. He then reasoned that, sinceeach drop must contain an integral (or whole) numberof electrons, the charge of each drop must be a whole-number multiple of the electron’s charge. Indeed, Mil-likan was correct; the measured charge on any drop wasalways a whole-number multiple of C, thefundamental charge of a single electron. With this number in hand, and knowingThomson’s mass-to-charge ratio for electrons, we can deduce the mass of an electronas follows:

As Thomson had correctly deduced, this mass is about 2000 times lighter than hydrogen,the lightest atom.

Why did scientists work so hard to measure the charge of the electron? Since the elec-tron is a fundamental building block of matter, scientists want to know its properties, in-cluding its charge. The magnitude of the charge of the electron is of tremendousimportance because it determines how strongly an atom holds its electrons. On the onehand, imagine how matter would be different if electrons had a much smaller charge, sothat atoms held them more loosely. Many atoms might not even be stable. On the otherhand, imagine how matter would be different if electrons had a much greater charge, sothat atoms held them more tightly. Since atoms form compounds by exchanging andsharing electrons (more on this in Chapter 3), the result could be fewer compounds ormaybe even none. Without the abundant diversity of compounds, life would not be pos-sible. So, the magnitude of the charge of the electron—even though it may seem like aninsignificantly small number—has great importance.

-1.60 * 10-19 C *

g

-1.76 * 108 C

= 9.10 * 10-28 g

Charge *

mass

charge

= mass

-1.60 * 10-19

Viewingmicroscope

Ionizingradiation

Lightsource

Atomizer

Positivelycharged plate

Negativelycharged plate

Charged oil droplets aresuspended in electric field.

� FIGURE 2.5 Millikan’s Measurement of the Electron’sCharge Millikan calculated the charge on oil droplets falling in anelectric field. He found that it was always a whole-number multiple of

C, the charge of a single electron.-1.60 * 10-19

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Electron

Sphere of positivecharge


2.5 The Structure of the AtomThe discovery of negatively charged particles within atoms raised a new question. Sinceatoms are charge-neutral, they must contain positive charge that neutralized the negativecharge of the electrons—but how did the positive and negative charges within the atomfit together? Were atoms just a jumble of even more fundamental particles? Were theysolid spheres? Did they have some internal structure? J. J. Thomson proposed that thenegatively charged electrons were small particles held within a positively charged sphere,as shown here.

This model, the most popular of the time, became known as the plum-puddingmodel (plum pudding is an English dessert). The picture suggested by Thomson, to thoseof us not familiar with plum pudding, was more like a blueberry muffin, where the blue-berries are the electrons and the muffin is the positively charged sphere.

The discovery of radioactivity—the emission of small energetic particles fromthe core of certain unstable atoms—by scientists Henri Becquerel (1852–1908) and MarieCurie (1867–1934) at the end of the nineteenth century allowed the structure of the atomto be experimentally probed. At the time, three different types of radioactivity had beenidentified: alpha ( ) particles, beta ( ) particles, and gamma ( ) rays. We will discuss theseand other types of radioactivity in more detail in Chapter 19. For now, just know that �particles are positively charged and that they are by far the most massive of the three.

In 1909, Ernest Rutherford (1871–1937), who had worked under Thomson and sub-scribed to his plum-pudding model, performed an experiment in an attempt to confirmit. His experiment, which employed � particles, proved it wrong instead. In the experi-ment, Rutherford directed the positively charged � particles at an ultrathin sheet of goldfoil, as shown in Figure 2.6 �.

gbaAlpha particles are about 7000 times moremassive than electrons.

A few a particles are deflectedthrough large angles.

Most a particles pass through with little or no deflection.

DetectorLead

Source

Gold foil

Alpha particles

Rutherford’s Gold Foil Experiment

� FIGURE 2.6 Rutherford’s GoldFoil Experiment Alpha particles weredirected at a thin sheet of gold foil. Mostof the particles passed through the foil,but a small fraction were deflected, and afew even bounced backward.

These particles were to act as probes of the gold atoms’ structure. If the gold atoms wereindeed like blueberry muffins or plum pudding—with their mass and charge spreadthroughout the entire volume of the atom—these speeding probes should pass rightthrough the gold foil with minimum deflection.

Rutherford performed the experiment, but the results were not as he expected.A majority of the particles did pass directly through the foil, but some particles weredeflected, and some (1 in 20,000) even bounced back. The results puzzled Rutherford,who wrote that they were “about as credible as if you had fired a 15-inch shell at a pieceof tissue paper and it came back and hit you.” What must the structure of the atom be inorder to explain this odd behavior?

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2.5 The Structure of the Atom 55

Plum-pudding model Nuclear model

NucleusProtonNeutronAlpha-particles

� FIGURE 2.7 The Nuclear AtomRutherford’s results could not be ex-plained by the plum-pudding model.Instead, they suggested that the atommust have a small, dense nucleus.

Rutherford created a new model—a modern version of which is shown in Figure 2.7 �alongside the plum-pudding model—to explain his results.

He realized that to account for the deflections he observed, the mass and positivecharge of an atom must all be concentrated in a space much smaller than the size of theatom itself. He concluded that, in contrast to the plum-pudding model, matter must notbe as uniform as it appears. It must contain large regions of empty space dotted withsmall regions of very dense matter. Using this idea, he proposed the nuclear theory of theatom, with three basic parts:

1. Most of the atom’s mass and all of its positive charge are contained in a small corecalled the nucleus.

2. Most of the volume of the atom is empty space, throughout which tiny, negativelycharged electrons are dispersed.

3. There are as many negatively charged electrons outside the nucleus as there are posi-tively charged particles (named protons) within the nucleus, so that the atom is elec-trically neutral.

Although Rutherford’s model was highly successful, scientists realized that it was in-complete. For example, hydrogen atoms contain one proton and helium atoms containtwo, yet the helium-to-hydrogen mass ratio is 4:1. The helium atom must contain someadditional mass. Later work by Rutherford and one of his students, British scientist JamesChadwick (1891–1974), demonstrated that the missing mass was composed of neutrons,neutral particles within the nucleus. The dense nucleus contains over 99.9% of the massof the atom, but it occupies very little of its volume. For now, we can think of the electronsthat surround the nucleus in analogy to the water droplets that make up a cloud—although their mass is almost negligibly small, they are dispersed over a very large volume.Consequently, the atom, like the cloud, is mostly empty space.

Rutherford’s nuclear theory was a success and is still valid today. The revolutionarypart of this theory is the idea that matter—at its core—is much less uniform than it ap-pears. If the nucleus of the atom were the size of the period at the end of this sentence, theaverage electron would be about 10 meters away. Yet the period would contain almost theentire mass of the atom. Imagine what matter would be like if atomic structure brokedown. What if matter were composed of atomic nuclei piled on top of each other likemarbles? Such matter would be incredibly dense; a single grain of sand composed of solidatomic nuclei would have a mass of 5 million kilograms (or a weight of about 11 millionpounds). Astronomers believe there are some objects in the universe composed of suchmatter—they are called neutron stars.

If matter really is mostly empty space, as Rutherford suggested, then why does it appearso solid? Why can we tap our knuckles on a table and feel a solid thump? Matter appearssolid because the variation in its density is on such a small scale that our eyes cannot see it.

Imagine a scaffolding 100 stories high and the size of a football field as shown in themargin. It is mostly empty space. Yet if you viewed it from an airplane, it would appear asa solid mass. Matter is similar. When you tap your knuckle on the table, it is much like onegiant scaffolding (your finger) crashing into another (the table). Even though they areboth primarily empty space, one does not fall into the other.

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2.6 Subatomic Particles: Protons, Neutrons,and Electrons in Atoms

We have just seen that all atoms are composed of the same subatomic particles:protons, neutrons, and electrons. Protons and neutrons have nearly identical masses.In SI units, the mass of the proton is , and the mass of the neutronis . A more common unit to express these masses, however, is theatomic mass unit (amu), defined as 1/12 the mass of a carbon atom containing sixprotons and six neutrons. Expressed in this unit, the mass of a proton or neutron isapproximately 1 amu. Electrons, in contrast, have an almost negligible mass of

or 0.00055 amu.The proton and the electron both have electrical charge. We know from Millikan’s

oil drop experiment that the electron has a charge of . In atomic orrelative units, the electron is assigned a charge of and the proton is assigned a chargeof +1. The charge of the proton and the electron are equal in magnitude but oppositein sign, so that when the two particles are paired, the charges sum to zero. The neutronhas no charge.

Notice that matter is usually charge-neutral (it has no overall charge) because pro-tons and electrons are normally present in equal numbers. When matter does acquirecharge imbalances, these imbalances usually equalize quickly, often in dramatic ways.For example, the shock you receive when touching a doorknob during dry weather isthe equalization of a charge imbalance that developed as you walked across the carpet.Lightning, as shown below, is simply an equalization of charge imbalances that developduring electrical storms.

-1-1.60 * 10-19 C

0.00091 * 10-27 kg

1.67493 * 10-27 kg1.67262 * 10-27 kg


Negative charge builds up on clouds.

Positive charge builds up on ground.

Electrical discharge equalizescharge imbalance.

� When the normal charge balance ofmatter is disturbed, as happens during anelectrical storm, it quickly equalizes, oftenin dramatic ways.

TABLE 2.1 Subatomic Particles

Mass (kg) Mass (amu) Charge (relative) Charge (C)

Proton 1.00727 1

Neutron 1.00866 0 0

Electron 0.00055 -1 .60218 * 10-19-10.00091 * 10-27

1.67493 * 10-27

+1 .60218 * 10-19+1.67262 * 10-27

If a proton had the mass of abaseball, an electron would have

the mass of a rice grain.

If you had a sample of matter—even a tiny sample, such as a sand grain—that wascomposed only of protons or only of electrons, the repulsive forces inherent in thatmatter would be extraordinary, and the matter would be unstable. Luckily, that is nothow matter is. The properties of protons, neutrons, and electrons are summarized inTable 2.1.

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2.6 Subatomic Particles: Protons, Neutrons, and Electrons in Atoms 57

Elements: Defined by TheirNumbers of ProtonsIf all atoms are composed of the same subatomicparticles, then what makes the atoms of one ele-ment different from those of another? The answeris the number of these particles. The number thatis most important for the identity of an atom isthe number of protons in its nucleus. In fact, thenumber of protons defines the element. For ex-ample, an atom with 2 protons in its nucleus is ahelium atom; an atom with 6 protons in its nucle-us is a carbon atom; and an atom with 92 protonsin its nucleus is a uranium atom (Figure 2.8 �).The number of protons in an atom’s nucleus iscalled the atomic number and is given the symbolZ. The atomic numbers of known elements rangefrom 1 to 116 (although additional elements maystill be discovered), as shown in the periodic tableof the elements (Figure 2.9 �). In the periodictable, described in more detail in Section 2.7, theelements are arranged so that those with similar properties occur in the same column. Anelement, identified by a unique atomic number, is represented with a unique chemicalsymbol, a one- or two-letter abbreviation or the element that is listed directly below itsatomic number on the periodic table. The chemical symbol for helium is He; for carbon,

The Number of Protons Defines the Element

Carbonnucleus:6 protons

Heliumnucleus:2 protons

� FIGURE 2.8 How Elements Differ Each element is defined by a unique atom-ic number (Z), the number of protons in the nucleus of every atom of that element.

lithium

3Li

sodium

11Na

potassium

19K

rubidium

37Rb

cesium

55Cs

francium

87Fr

beryllium

magnesium

12Mg

calcium

20Ca

strontium

38Sr

barium

56Ba

radium

88Ra Rf Db Sg

Be4

scandium

21Sc

titanium

22Ti

yttrium

39Y

rutherfordium

104

vanadium

23V

chromium

24Cr

Atomic number (Z)

Chemical symbol

Name

zirconium

40Zr

niobium

41Nb

72Hf

hafnium tantalum

73Ta

tungsten

74W

dubnium

105

seaborgium

106Bh

bohrium

107Hs

hassium

108Mt

meitnerium

Ds Rg ** ** **

109

darmstadtium

110

roentgenium

111 112 114 116

manganese

25Mn

iron

26Fe

molybdenum

42Mo

technetium

43Tc

ruthenium

44Ru

rhenium

75Re

76Os

cobalt

27Co

rhodium

45Rh

iridiumosmium

77Ir

argon

18Ar

neon

10Ne

helium

2He

36Kr

xenon

54Xe

radon

86Rn

chlorine

17Cl

fluorine

9F

35Br

iodine

53I

85At

sulfur

16S

oxygen

8O

34Se

tellurium

52Te

84Po

phosphorus

15P

nitrogen

7N

33As

antimony

51Sb

83Bi

silicon

14Si

carbon

6C

32Ge

tin

50Sn

82Pb

aluminum

13Al

boron

5B

31Ga

indium

49In

thallium lead bismuth polonium astatine

81Tl

30Zn

cadmium

48Cd

mercury

80Hg

copper zinc gallium germanium arsenic selenium bromine krypton

29Cu

silver

47Ag

gold

79Au

nickel

28Ni

palladium

46Pd

platinum

78Pt

lutetium

71Lu

lathanum

57La

actinium

89Ac

cerium

58Ce

praseodymium

59Pr

neodymium

60Nd

91

Paprotactinium uranium

92

Uneptunium

93

Np

promethium

61Pm

samarium

62Sm

europium

63Eu

plutonium

94

Pu95

Am

gadolinium

64Gd

curiumamericium

96

Cm

ytterbium

70Yb

102

No

thulium

69Tm

101

Md

erbium

68Er

fermium mendelevium nobelium

100

Fm

holmium

67Ho

einsteinium

99

Es

dysprosium

66Dy

californium

98

Cf

terbium

65Tb

berkelium

97

Bkthorium

90

Thlawrencium

103

Lr

hydrogen

1H

beryllium

Be4

The Periodic Table

� FIGURE 2.9 The Periodic Table Each element is represented by its symbol and atomicnumber. Elements in the same column have similar properties.

TROMC02_044-081-hr 12/1/06 9:43 AM Page 57

96

Cm

Curium

� Element 96 is named curium, afterMarie Curie, co-discoverer of radioactivity.


it is C; and for uranium, it is U. The chemical symbol and the atomic number always gotogether. If the atomic number is 2, the chemical symbol must be He. If the atomic num-ber is 6, the chemical symbol must be C. This is just another way of saying that thenumber of protons defines the element.

Most chemical symbols are based on the English name of the element. For example,the symbol for sulfur is S; for oxygen, O; and for chlorine, Cl. Several of the oldest knownelements, however, have symbols based on their original Latin names. Thus, the symbol forsodium is Na from the Latin natrium, and the symbol for tin is Sn from the Latin stannum.The names of elements were often given to describe their properties. For example,argon originates from the Greek word argos meaning inactive, referring to argon’s chemi-cal inertness (it does not react with other elements). Chlorine originates from the Greekword chloros meaning pale green, referring to chlorine’s pale green color (see p. 48).Other elements, including helium, selenium, and mercury, were named after figures fromGreek or Roman mythology or astronomical bodies. Still others (such as europium,polonium, and berkelium) were named for the places where they were discovered or wheretheir discoverer was born. More recently, elements have been named after scientists; forexample, curium for Marie Curie, einsteinium for Albert Einstein, and rutherfordiumfor Ernest Rutherford.

Isotopes: When the Number of Neutrons VariesAll atoms of a given element have the same number of protons; however, they do not nec-essarily have the same number of neutrons. Since neutrons have nearly the same mass asprotons (1 amu), this means that—contrary to what John Dalton originally proposed inhis atomic theory—all atoms of a given element do not have the same mass. For example,all neon atoms contain 10 protons, but they may have 10, 11, or 12 neutrons. All threetypes of neon atoms exist, and each has a slightly different mass. Atoms with the samenumber of protons but different numbers of neutrons are called isotopes. Some ele-ments, such as beryllium (Be) and aluminum (Al), have only one naturally occurring iso-tope, while other elements, such as neon (Ne) and chlorine (Cl), have two or more.

Fortunately, the relative amount of each different isotope in a naturally occurringsample of a given element is usually the same. For example, in any natural sample of neonatoms, 90.48% of them are the isotope with 10 neutrons, 0.27% are the isotope with11 neutrons, and 9.25% are the isotope with 12 neutrons. These percentages are called thenatural abundance of the isotopes. Each element has its own characteristic natural abun-dance of isotopes.

The sum of the number of neutrons and protons in an atom is called the mass numberand is given the symbol A:

For neon, with 10 protons, the mass numbers of the three different naturally occurringisotopes are 20, 21, and 22, corresponding to 10, 11, and 12 neutrons, respectively.

Isotopes are often symbolized in the following way:

A = number of protons (p) + number of neutrons (n)

XAZ

Mass number

Atomic numberChemical symbol

where X is the chemical symbol, A is the mass number, and Z is the atomic number.Therefore, the symbols for the neon isotopes are

Notice that the chemical symbol, Ne, and the atomic number, 10, are redundant: if theatomic number is 10, the symbol must be Ne. The mass numbers, however, are differentfor different isotopes, reflecting the different number of neutrons in each one.

2010Ne 21

10Ne 2210Ne

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2.6 Subatomic Particles: Protons, Neutrons, and Electrons in Atoms 59

X-AMass numberChemical symbol

or name

A second common notation for isotopes is the chemical symbol (or chemical name)followed by a dash and the mass number of the isotope.

(a) From the periodic table, we find that the atomic number (Z) of chlorine is17, so chlorine atoms have 17 protons.

Z = 17

The mass number (A) for the isotope with 18 neutrons is the sum of thenumber of protons (17) and the number of neutrons (18).

A = 17 + 18 = 35

The symbol for the chlorine isotope is its two-letter abbreviation with theatomic number (Z) in the lower left corner and the mass number (A) inthe upper left corner.

3517Cl

(b) For , the number of protons is the lower left number. Since this is aneutral atom, there are an equal number of electrons.

5224Cr

Number of electrons = 24 (neutral atom)

Number of protons = Z = 24

The number of neutrons is equal to the upper left number minus thelower left number.

Number of neutrons = 52 - 24 = 28

For Practice 2.3(a) What are the atomic number, mass number, and symbol for the carbon isotope

with 7 neutrons?

(b) How many protons and neutrons are present in an atom of ?3919K

In this notation, the neon isotopes are

We can summarize what we have learned about the neon isotopes in the following table:

Symbol Number of Protons Number of Neutrons A (Mass Number) Natural Abundance(%)

Ne-20 or 10 10 20 90.48

Ne-21 or 10 11 21 0.27

Ne-22 or 10 12 22 9.25

Notice that all isotopes of a given element have the same number of protons (otherwisethey would be different elements). Notice also that the mass number is the sum of thenumber of protons and the number of neutrons. The number of neutrons in an isotope isthe difference between the mass number and the atomic number ( ). The differentisotopes of an element generally exhibit the same chemical behavior—the three isotopesof neon, for example, all exhibit the same chemical inertness.

EXAMPLE 2.3 Atomic Numbers, Mass Numbers, and Isotope Symbols(a) What are the atomic number (Z), mass number (A), and symbol of the chlorine

isotope with 18 neutrons?

(b) How many protons, electrons, and neutrons are present in an atom of ?

Solution

5224Cr

A - Z

2210Ne

2110Ne

2010Ne

neon-22neon-21neon-20Ne-22Ne-21Ne-20

TROMC02_044-081-hr 12/4/06 12:32 PM Page 59


Ions: Losing and Gaining ElectronsThe number of electrons in a neutral atom is equal to the number of protons in its nucle-us (given by the atomic number Z). During chemical changes, however, atoms often loseor gain electrons to form charged particles called ions. For example, neutral lithium (Li)atoms contain 3 protons and 3 electrons; however, in many chemical reactions lithiumatoms lose one electron ( ) to form Li+ ions.

The charge of an ion is indicated in the upper right corner of the symbol. The Li+ ion con-tains 3 protons but only 2 electrons, resulting in a charge of 1+. The charge of an ion de-pends on how many electrons were gained or lost in forming the ion. Neutral fluorine (F)atoms contain 9 protons and 9 electrons; however, in many chemical reactions fluorineatoms gain one electron to form ions.

The ion contains 9 protons and 10 electrons, resulting in a charge of . For manyelements, such as lithium and fluorine, the ion is much more common than the neu-tral atom. In fact, virtually all of the lithium and fluorine in nature are in the form oftheir ions.

Positively charged ions, such as Li+, are called cations and negatively charged ions,such as , are called anions. Ions act very differently than the atoms from which they areformed. Neutral sodium atoms, for example, are extremely unstable, reacting violentlywith most things they contact. Sodium cations (Na+), in contrast, are relatively inert—weeat them all the time in sodium chloride (table salt). In ordinary matter, cations andanions always occur together so that matter is charge-neutral overall.

F-

1-F-

F + 1 e- : F-

F-

Li : Li++ 1 e-

e-

2.7 Finding Patterns: The Periodic Law and the Periodic Table

The modern periodic table grew out of the work of Dmitri Mendeleev (1834–1907),a nineteenth-century Russian chemistry professor. In his time, about 65 different ele-ments had been discovered. Through the work of a number of chemists, many of theproperties of these elements—such as their relative masses, their chemical activity, andsome of their physical properties—were known. However, there was no systematic wayof organizing them.

� Dmitri Mendeleev, a Russian chem-istry professor who proposed the periodiclaw and arranged early versions of the pe-riodic table, was honored on a Russianpostage stamp.

Conceptual Connection 2.3 The Nuclear Atom, Isotopes, and Ions

In light of the nuclear model for the atom, which of the following statements is most like-ly to be true?

(a) The isotope of an atom with a greater number of neutrons is larger than one with asmaller number of neutrons.

(b) The size of an anion is greater than the size of the corresponding neutral atom.

(c) The size of a cation is greater than the size of the corresponding neutral atom.

Answer: (b) The number of neutrons in the nucleus of an atom does not affect the atom’s size be-cause the nucleus is miniscule compared to the atom itself. The number of electrons, however, doesaffect the size of the atom because most of the volume of the atom is occupied by electrons, andelectrons repel one another. Therefore, an anion, with a greater number of electrons, is larger thanthe corresponding neutral atom.

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2.7 Finding Patterns: The Periodic Law and the Periodic Table 61

Before 1800

Time of Discovery

1800–1849 1850–1899

1900–1949 1950–1999lithium

Li

sodium

Na

potassiumK

rubidium

Rb

cesium

Cs

franciumFr

beryllium

magnesium

Mg

calciumCa

strontium

Sr

barium

Ba

radiumRa Rf Db Sg

Be

scandiumSc

titaniumTi

yttrium

Y

rutherfordium

vanadiumV

chromiumCr

zirconium

Zrniobium

Nb

Hfhafnium tantalum

Tatungsten

W

dubnium seaborgiumBh

bohriumHs

hassiumMt

meitneriumDs Rg

darmstadtium roentgenium

manganeseMn

ironFe

molybdenum

Motechnetium

Tcruthenium

Ru

rhenium

Re Os

cobaltCo

rhodium

Rh

iridiumosmium

Ir

argon

Ar

neon

Ne

helium

He

Kr

xenon

Xe

radon

Rn

chlorine

Cl

fluorine

F

Br

iodine

I

At

sulfur

S

oxygen

O

Se

tellurium

Te

Po

phosphorus

P

nitrogen

N

As

antimony

Sb

Bi

silicon

Si

carbon

C

Ge

tin

Sn

Pb

aluminum

Al

boron

B

Ga

indium

In

thallium lead bismuth polonium astatine

Tl

Zn

cadmium

Cd

mercury

Hg

copper zinc gallium germanium arsenic selenium bromine kryptonCu

silver

Ag

gold

Au

nickelNi

palladium

Pd

platinum

Pt

lutetium

Lu

lathanum

La

actiniumAc

cerium

Cepraseodymium

Prneodymium

Nd

Paprotactinium uranium

Uneptunium

Np

promethium

Pmsamarium

Smeuropium

Eu

plutoniumPu Am

gadolinium

Gd

curiumamericiumCm

ytterbium

Yb

No

thulium

Tm

Md

erbium

Er

fermium mendelevium nobeliumFm

holmium

Ho

einsteiniumEs

dysprosium

Dy

californiumCf

terbium

Tb

berkeliumBk

thoriumTh

lawrenciumLr

hydrogenH

H1

He2

Li3

Be4

B5

C6

N7

O8

F9

Ne10

Na11

Mg12

Al13

Si14

P15

S16

Cl17

Ar18

K19

Ca20

Elements with similar properties recur in a regular pattern.

The Periodic Law

� Many of the elements that we know today were discovered during Mendeleev’s lifetime.

In 1869, Mendeleev noticed that certain groups of elements had similar properties.Mendeleev found that when he listed elements in order of increasing mass, their proper-ties recurred in a periodic pattern (Figure 2.10 �).

Mendeleev summarized these observations in the periodic law, which states thefollowing:

When the elements are arranged in order of increasing mass, certain sets ofproperties recur periodically.

Mendeleev then organized all the known elements in a table consisting of a series ofrows in which mass increased from left to right. The rows were arranged so that elementswith similar properties aligned in the same vertical columns (Figure 2.11 �).

H1

He2

Li3

Be4

B5

C6

N7

O8

F9

Ne10

Na11

Mg12

Al13

Si14

P15

S16

Cl17

Ar18

K19

Ca20

A Simple Periodic Table

Elements with similar propertiesfall into columns.

Periodic means exhibiting a repeatingpattern.

� FIGURE 2.11 Making a PeriodicTable The elements in Figure 2.10 canbe arranged in a table in which atomicnumber increases from left to right andelements with similar properties (as repre-sented by the different colors) are alignedin columns.

� FIGURE 2.10 Recurring Properties These elements are listed in order of increasing atom-ic number. Elements with similar properties are shown in the same color. Notice that the colors forma repeating pattern, much like musical notes, which form a repeating pattern on a piano keyboard.

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Since many elements had not yet been discovered, Mendeleev’s table contained somegaps, which allowed him to predict the existence of yet undiscovered elements and some oftheir properties. For example, Mendeleev predicted the existence of an element he calledeka-silicon, which fell below silicon on the table and between gallium and arsenic. In 1886,eka-silicon was discovered by German chemist Clemens Winkler (1838–1904), who namedit germanium, after his home country.

Mendeleev’s original listing has evolved into the modern periodic table shown inFigure 2.12 �. In the modern table, elements are listed in order of increasing atomicnumber rather than increasing relative mass. The modern periodic table also containsmore elements than Mendeleev’s original table because more have been discovered sincehis time. Mendeleev’s periodic law was based on observation. Like all scientific laws, theperiodic law summarized many observations but did not give the underlying reason forthe observations—only theories do that. For now, we accept the periodic law as it is, butin Chapters 7 and 8 we examine a powerful theory—called quantum mechanics—thatexplains the law and gives the underlying reasons for it.

As shown in Figure 2.12, the elements in the periodic table can be broadly classifiedas metals, nonmetals, or metalloids. Metals, found on the lower left side and middle ofthe periodic table, have the following properties: they are good conductors of heat andelectricity; they can be pounded into flat sheets (malleability); they can be drawn intowires (ductility); they are often shiny; and they tend to lose electrons when they under-go chemical changes. Good examples of metals include chromium, copper, strontium,and lead.

Nonmetals are found on the upper right side of the periodic table. The dividing linebetween metals and nonmetals is the zigzag diagonal line running from boron to asta-tine. Nonmetals have more varied properties—some are solids at room temperature,

Eka means the one beyond or the next onein a family of elements. So, eka-siliconmeans the element beyond silicon in thesame family as silicon.

1A1

1

2

3

4

5

6

7

2A2

3B3

4B4

5B5

6B6

7B7 8 9 10

3A13

4A14

5A15

6A16

7A17

8A18

1B11

2B12

8B

Metals Metalloids Nonmetals1H

4Be

3Li

11Na

19K

21Sc

37Rb

55Cs

87Fr

12Mg

20Ca

38Sr

56Ba

88Ra

23V

41Nb

73Ta

105Db

39Y

22Ti

40Zr

72Hf

104Rf

24Cr

42Mo

74W

106Sg

25Mn

43Tc

75Re

5B

13Al

31Ga

49In

81Tl

6C

14Si

32Ge

50Sn

82Pb

7N

15P

33As

51Sb

83Bi

8O

16S

34Se

52Te

84Po

9F

17Cl

35Br

53I

85At

2He

10Ne

18Ar

36Kr

54Xe

86Rn

57La

89Ac

26Fe

44Ru

76Os

28Ni

46Pd

78Pt

29Cu

47Ag

79Au

30Zn

48Cd

80Hg

27Co

45Rh

77Ir

107Bh

108Hs

109Mt

110Ds

111Rg

112 114 116

Lanthanides

Actinides

70Yb

102No

71Lu

103Lr

69Tm

101Md

68Er

100Fm

67Ho

99Es

66Dy

98Cf

65Tb

97Bk

64Gd

96Cm

63Eu

95Am

62Sm

94Pu

61Pm

93Np

60Nd

92U

59Pr

91Pa

58Ce

90Th

Major Divisions of the Periodic Table

� FIGURE 2.12 Metals, Nonmetals, and Metalloids The elements in the periodic table fall into these three broad classes.

TROMC02_044-081-hr 12/1/06 9:43 AM Page 62


others are liquids or gases—but as a whole they tend to be poor conductors of heat andelectricity and they all tend to gain electrons when they undergo chemical changes.Good examples of nonmetals include oxygen, carbon, sulfur, bromine, and iodine.

Many of the elements that lie along the zigzag diagonal line that divides metals andnonmetals are called metalloids and show mixed properties. Several metalloids are alsoclassified as semiconductors because of their intermediate (and highly temperature-dependent) electrical conductivity. The ability to change and control the conductivity ofsemiconductors makes them useful in the manufacture of the electronic chips and cir-cuits central to computers, cellular telephones, and many other modern devices. Goodexamples of metalloids include silicon, arsenic, and antimony.

The periodic table, as shown in Figure 2.13 on page 64, can also be broadly dividedinto main-group elements, whose properties tend to be largely predictable based on theirposition in the periodic table, and transition elements or transition metals, whose prop-erties tend to be less predictable based simply on their position in the periodic table.Main-group elements are in columns labeled with a number and the letter A. Transitionelements are in columns labeled with a number and the letter B. An alternative number-ing system does not use letters, but only the numbers 1–18. Both numbering systems areshown in most of the periodic tables in this book. Each column within the main-groupregions of the periodic table is called a family or group of elements.

The elements within a group usually have similar properties. For example, the group8A elements, called the noble gases, are mostly unreactive. The most familiar noble gas isprobably helium, used to fill buoyant balloons. Helium is chemically stable—it does notcombine with other elements to form compounds—and is therefore safe to put into bal-loons. Other noble gases include neon (often used in electronic signs), argon (a smallcomponent of our atmosphere), krypton, and xenon.

Metalloids are sometimes calledsemimetals.

Arsenic

Carbon

GoldChromium

StrontiumCopper

Silicon

Sulfur

Iodine

Lead

Metals Metalloids Nonmetals

Bromine

Sr

Cr

C

Si

Pb

As

S

Br

I

Cu

Au

� Some representative metals, metalloids, and nonmetals.

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The group 1A elements, called the alkali metals, are all reactive metals. A marble-sized piece of sodium explodes violently when dropped into water. Other alkali metals in-clude lithium, potassium, and rubidium.

The group 2A elements, called the alkaline earth metals, are also fairly reactive,although not quite as reactive as the alkali metals. Calcium for example, reacts fairlyvigorously when dropped into water but will not explode as dramatically as sodium.Other alkaline earth metals include magnesium (a common low-density structuralmetal), strontium, and barium.

1A

Main-groupelements

Transitionelements

Main-groupelements

1

2

3

4

Peri

ods

5

6

7

2A

3B 4B 5B 6B 7B

3A 4A 5A 6A 7A

8A

1B 2B8B

1H

4Be

3Li

11Na

19K

21Sc

37Rb

55Cs

87Fr

12Mg

20Ca

38Sr

56Ba

88Ra

23V

41Nb

73Ta

105Db

39Y

22Ti

40Zr

72Hf

104Rf

24Cr

42Mo

74W

106Sg

25Mn

43Tc

75Re

5B

13Al

31Ga

49In

81Tl

6C

14Si

32Ge

50Sn

82Pb

7N

15P

33As

51Sb

83Bi

8O

16S

34Se

52Te

84Po

9F

17Cl

35Br

53I

85At

2He

10Ne

18Ar

36Kr

54Xe

86Rn

57La

89Ac

26Fe

44Ru

76Os

28Ni

46Pd

78Pt

29Cu

47Ag

79Au

30Zn

48Cd

80Hg

27Co

45Rh

77Ir

107Bh

108Hs

109Mt

110Ds

111Rg

112 114 116

Groupnumber

He

Ne

Ar

Kr

Xe

Noble gases Alkali metals

Li

Na

K

Rb

Cs

� FIGURE 2.13 The Periodic Table: Main-Group and Transition Elements The elements in the periodic table fall into columns.The two columns at the left and the six columns at the right comprise the main-group elements. Each of these eight columns is a group orfamily. The properties of main-group elements can generally be predicted from their position in the periodic table. The properties of theelements in the middle of the table, known as transition elements, are less predictable.

TROMC02_044-081-hr 12/1/06 9:43 AM Page 64


Be

Mg

Ca

Sr

Ba

Alkaline earth metals

F

Cl

Br

I

At

Halogens

The group 7A elements, called the halogens, are very reactive nonmetals. The mostfamiliar halogen is probably chlorine, a greenish-yellow gas with a pungent odor. Becauseof its reactivity, chlorine is often used as a sterilizing and disinfecting agent. Other halo-gens include bromine, a red-brown liquid that easily evaporates into a gas; iodine, a pur-ple solid; and fluorine, a pale-yellow gas.

Ions and the Periodic TableWe have learned that, in chemical reactions, metals tend to lose electrons (thus formingcations) and nonmetals tend to gain them (thus forming anions). The number of elec-trons lost or gained, and therefore the charge of the resulting ion, is often predictable fora given element, especially main-group elements. Main-group elements tend to form ionsthat have the same number of electrons as the nearest noble gas (the noble gas that has thenumber of electrons closest to that of the element).

• A main-group metal tends to lose electrons, forming a cation with the samenumber of electrons as the nearest noble gas.

• A main-group nonmetal tends to gain electrons, forming an anion with thesame number of electrons as the nearest noble gas.

For example, lithium, a metal with three electrons, tends to lose one electron to form a 1+cation having two electrons, the same number of electrons as helium. Chlorine, a non-metal with 17 electrons, tends to gain one electron to form a anion having 18 elec-trons, the same number of electrons as argon.

1-

TROMC02_044-081-hr 12/7/06 5:17 PM Page 65


In general, the alkali metals (group 1A) tend to lose one electron and therefore form1+ ions. The alkaline earth metals (group 2A) tend to lose two electrons and thereforeform 2+ ions. The halogens (group 7A) tend to gain one electron and therefore form 1–ions. The oxygen family nonmetals (group 6A) tend to gain two electrons and thereforeform 2– ions. More generally, for main-group elements that form predictable cations,the charge of the cation is equal to the group number. For main-group elements that formpredictable anions, the charge of the anion is equal to the group number minus eight.Transition elements may form different ions with different charges. The most commonions formed by main-group elements are shown in Figure 2.14 �. In Chapters 7 and 8,when we learn about quantum-mechanical theory, you will understand why these groupsform ions as they do.

EXAMPLE 2.4 Predicting the Charge of IonsPredict the charges of the monoatomic ions formed by the following main-group elements.

(a) Al (b) S

Solution(a) Aluminum is a main-group metal and will therefore tend to lose electrons to form a

cation with the same number of electrons as the nearest noble gas. Aluminum atomshave 13 electrons and the nearest noble gas is neon, which has 10 electrons. Thereforealuminum will tend to lose 3 electrons to form a cation with a +3 charge ( ).

(b) Sulfur is a nonmetal and will therefore tend to gain electrons to form an anionwith the same number of electrons as the nearest noble gas. Sulfur atoms have 16electrons and the nearest noble gas is argon, which has 18 electrons. Thereforesulfur will tend to gain 2 electrons to form an anion with a charge .

For Practice 2.4Predict the charges of the monoatomic ions formed by the following main-groupelements.

(a) N (b) Rb

(S2-)-2

Al3+

1A

2A 3A 4A 5A 6A

7A 8A

Noble

Gases

H�

Li�

Na�

K�

Rb�

Cs�

Mg2�

Ca2�

Sr2�

Ba2�

Al3�

N3� O2�

S2�

Se2�

Te2�

F�

H�

Cl�

Br�

I�

Elements That Form Ions with Predictable Charges

Transition metals

� FIGURE 2.14 Elements That Form Ions with Predictable Charges

TROMC02_044-081-hr 12/1/06 9:43 AM Page 66

2.8 Atomic Mass: The Average Mass of an Element’s Atoms

An important part of Dalton’s atomic theory was that all atoms of a given element havethe same mass. However, in Section 2.6, we learned that because of isotopes, the atoms ofa given element often have different masses, so Dalton was not completely correct.We can, however, calculate an average mass—called the atomic mass—for each element.

The atomic mass of each element is listed directly beneath theelement’s symbol in the periodic table and represents the averagemass of the isotopes that compose that element, weighted according tothe natural abundance of each isotope. For example, the periodic tablelists the atomic mass of chlorine as 35.45 amu. Naturally occurringchlorine consists of 75.77% chlorine-35 atoms (mass 34.97 amu) and24.23% chlorine-37 atoms (mass 36.97 amu). Its atomic mass iscomputed as follows:

Notice that the atomic mass of chlorine is closer to 35 than 37. Naturally occurring chlo-rine contains more chlorine-35 atoms than chlorine-37 atoms, so the weighted averagemass of chlorine is closer to 35 amu than to 37 amu.

= 35.45 amuAtomic mass = 0.7577(34.97 amu) + 0.2423(36.97 amu)

2.8 Atomic Mass: The Average Mass of an Element’s Atoms 67

Oxygen: 65%

Carbon: 18%

Hydrogen: 10%

Nitrogen: 3%

Calcium: 1.5%

Phosphorus: 1%

Other: 1.5%

TABLE 2.2 Approximate Percent Elemental Composition of Humans

Element % by Mass % by Number of Atoms

Oxygen 65 26.4

Carbon 18 9.2

Hydrogen 10 62.4

Nitrogen 3 1.4

Calcium 1.5 0.2

Phosphorus 1 0.3

Other 1.5 0.2� FIGURE 2.15 Elemental Compositionof Humans (by Mass)

Chemistry and MedicineThe Elements of Life

What are the atoms that compose living things? That composehumans? The chemistry of life revolves around the chemistry ofcarbon, an element that forms a disproportionately large num-ber of compounds with a few other elements such as hydrogen,oxygen, and nitrogen. We will first encounter the chemistry ofcarbon—called organic chemistry—in Chapter 3 and then ex-

17

Cl35.45

chlorine

Atomic mass is sometimes called atomicweight, average atomic mass, or averageatomic weight.

amine it in more detail in Chapter 20. By mass, our bodies aremostly oxygen atoms because of the large amount of water inour bodies. Carbon comes in second and hydrogen third, asshown in Figure 2.15 � and Table 2.2.

Because the atoms of different elements have differentmasses (more on this in Sections 2.8 and 2.9), the elementalcomposition of humans looks different when expressed bynumber of atoms. When expressed this way, hydrogen comes infirst (because hydrogen atoms are so light), with oxygen secondand carbon third.

TROMC02_044-081-hr 12/4/06 12:36 PM Page 67


In general, the atomic mass is calculated according to the following equation:

where the fractions of each isotope are the percent natural abundances converted to theirdecimal values. Atomic mass is useful because it allows us to assign a characteristic massto each element, and as we will see shortly, it allows us to quantify the number of atoms ina sample of that element.

� (fraction of isotope 3 : mass of isotope 3) � Á

� (fraction of isotope 2 : mass of isotope 2)

� (fraction of isotope 1 : mass of isotope 1)

Atomic mass � an

(fraction of isotope n) : (mass of isotope n)

When percentages are used in calculations,they are converted to their decimal valueby dividing by 100.

EXAMPLE 2.5 Atomic MassCopper has two naturally occurring isotopes: Cu-63 with mass 62.9396 amu and a nat-ural abundance of 69.17%, and Cu-65 with mass 64.9278 amu and a natural abundanceof 30.83%. Calculate the atomic mass of copper.

Solution

Convert the percent natural abundances intodecimal form by dividing by 100.

Fraction Cu-65 =

30.83

100= 0.3083

Fraction Cu-63 =

69.17

100= 0.6917

Compute the atomic mass using the equationgiven in the text.

For Practice 2.5Magnesium has three naturally occurring isotopes with masses of 23.99 amu, 24.99 amu,and 25.98 amu and natural abundances of 78.99%, 10.00%, and 11.01%, respectively.Calculate the atomic mass of magnesium.

For More Practice 2.5Gallium has two naturally occurring isotopes: Ga-69 with a mass of 68.9256 amu and anatural abundance of 60.11%, and Ga-71. Use the atomic mass of gallium listed in theperiodic table to find the mass of Ga-71.

= 43.5353 amu + 20.0172 amu = 63.5525 = 63.55 amu

Atomic mass = 0.6917(62.9396 amu) + 0.3083(64.9278 amu)

Mass Spectrometry:Measuring the Mass of Atoms and MoleculesThe masses of atoms and the percent abundances of isotopes of elements are measuredusing mass spectrometry. In a mass spectrometer, such as the one diagrammed in Figure 2.16 �, the sample (containing the atoms whose mass is to be measured) is inject-ed into the instrument and vaporized. The vaporized atoms are then ionized by an elec-tron beam. The electrons in the beam collide with the vaporized atoms, removingelectrons from the atoms and creating positively charged ions. Charged plates with slits inthem accelerate the positively charged ions into a magnetic field, which deflects them. Theamount of deflection depends on the mass of the ions—lighter ions are deflected morethan heavier ones.

In the diagram shown here, you can see three different paths, each corresponding toatoms of different mass. Finally, the ions strike a detector and produce an electrical signal

TROMC02_044-081-hr 12/1/06 9:43 AM Page 68

2.9 Molar Mass: Counting Atoms by Weighing Them 69

Magnetic field separates particlesbased on ratio of mass to charge.

Heatervaporizes

sample

Sample

Electric fieldaccelerates particles

Electricallychargedplates

Magnets

Beam ofcharged particles

Heaviestparticles

Lightestparticles

Detector

Electron beamionizes sample

Mass Spectrometer

� FIGURE 2.16 The Mass Spectrometer Atoms are converted to positively charged ions,accelerated, and passed through a magnetic field that deflects their path. The heaviest ions undergothe least deflection.

that is recorded. The result is the separation of the atoms in the sample according to theirmass, producing a mass spectrum such as the one shown in Figure 2.17 �. The position ofeach peak on the x-axis gives the mass of the isotope that was ionized, and the intensity(indicated by the height of the peak) gives the relative abundance of that isotope.

Mass spectrometry can also be used on molecules. Because molecules often fragment(break apart) during ionization, the mass spectrum of a molecule usually contains manypeaks representing the masses of different parts of the molecule, as well as a peak repre-senting the mass of the molecule as a whole. The fragments that form upon ionization, andtherefore the corresponding peaks that appear in the mass spectrum, are specific to themolecule, so that a mass spectrum is like a molecular fingerprint. In other words, massspectroscopy can be used to identify an unknown molecule and to determine how much ofit is present in a particular sample. For example, mass spectrometry has been used to detectorganic compounds present in meteorites, a puzzling observation which some scientistsspeculate may be evidence of life outside of our planet. Most scientists, however, think thatthe compounds probably formed in the same way as the first organic molecules on Earth,indicating that the formation of organic molecules may be common in the universe.

Since the early 1990s, mass spectrometry has also been successfully applied to biolog-ical molecules, including proteins (the workhorse molecules in cells) and nucleic acids(the molecules that carry genetic information). For a long time, these molecules could notbe analyzed by mass spectrometry because they were difficult to vaporize and ionize with-out destroying them, but modern techniques have overcome this problem. A tumor, forexample, can now be instantly analyzed to determine whether it contains specific proteinsassociated with cancer.

2.9 Molar Mass: Counting Atoms by Weighing Them

Have you ever bought shrimp and noticed a quantity called count? Shrimp is normallysold by count, which tells you the number of shrimp per pound. For example, 41–50count shrimp means that there are between 41 and 50 shrimp per pound. The smaller thecount, the larger the shrimp. The big tiger prawns have counts as low as 10–15, whichmeans that each shrimp can weigh up to 1/10 of a pound. The nice thing about categoriz-ing shrimp in this way is that you can count the shrimp by weighing them. For example,two pounds of 41–50 count shrimp contains between 82 and 100 shrimp.

100%

75%

50%

25%

Cl-35

Rel

ativ

e ab

un

dan

ce

Cl-37

35

Relative mass

37

� FIGURE 2.17 The Mass Spec-trum of Chlorine The position of eachpeak on the x-axis indicates the mass ofthe isotope. The intensity (or height) ofthe peak indicates the relative abundanceof the isotope.

TROMC02_044-081-hr 12/1/06 9:43 AM Page 69


A similar (but more precise) concept exists for atoms. Counting atoms is much moredifficult than counting shrimp, yet we often need to know the number of atoms in a givenmass of atoms. For example, intravenous fluids—fluids that are delivered to patients bydirectly dripping them into their veins—are saline solutions that must have a specificnumber of sodium and chloride ions per liter of fluid. The result of using an intravenousfluid with the wrong number of sodium and chloride ions could be fatal.

Atoms are far too small to count by any ordinary means. As we saw earlier, even if youcould somehow count atoms, and counted them 24 hours a day as long as you lived, youwould barely begin to count the number of atoms in something as small as a sand grain.Therefore, if we want to know the number of atoms in anything of ordinary size, we mustcount them by weighing.

The Mole: A Chemist’s “Dozen”When we count large numbers of objects, we often use units such as a dozen (12 objects)or a gross (144 objects) to organize our counting and to keep our numbers smaller. Withatoms, quadrillions of which may be in a speck of dust, we need a much larger number forthis purpose. The chemist’s “dozen” is called the mole (abbreviated mol) and is defined asthe amount of material containing particles.

This number is also called Avogadro’s number, named after Italian physicist Amedeo Avo-gadro (1776–1856), and is a convenient number to use when working with atoms, mole-cules, and ions. In this book, we will usually round Avogadro’s number to four significantfigures or . Notice that the definition of the mole is an amount of a substance.We will often refer to the number of moles of substance as the amount of the substance.

The first thing to understand about the mole is that it can specify Avogadro’s num-ber of anything. For example, 1 mol of marbles corresponds to marbles,and 1 mol of sand grains corresponds to sand grains. One mole of anythingis units of that thing. One mole of atoms, ions, or molecules, however,makes up objects of everyday sizes. For example, 22 copper pennies contain approxi-mately 1 mol of copper atoms and a tablespoon of water contains approximately 1 molof water molecules.

The second, and more fundamental, thing to understand about the mole is how itgets its specific value.

The numerical value of the mole is defined as being equal to the number ofatoms in exactly 12 grams of pure carbon-12 (12 g

).

The definition of the mole gives us a relationship between mass (grams of carbon) andnumber of atoms (Avogadro’s number). This relationship, as we will see shortly, allows usto count atoms by weighing them.

Converting between Number of Moles and Number of AtomsConverting between number of moles and number of atoms is similar to converting be-tween dozens of shrimp and number of shrimp. To convert between moles of atoms andnumber of atoms we simply use the conversion factors:

or

The following example shows how to use these conversion factors.

6 .022 * 1023 atoms

1 mol atoms

1 mol atoms

6.022 * 1023 atoms

6 .022 * 1023 C atoms=C = 1 mol C atoms

6 .022 * 10236.022 * 1023

6.022 * 1023

6.022 * 1023

1 mol = 6.0221421 * 1023 particles

6 .0221421 * 1023

One tablespoon of watercontains approximately one

mole of water molecules.

Twenty-two copper penniescontain approximately 1 mol

of copper atoms.

One tablespoon is approximately 15 mL;one mole of water occupies 18 mL.

Beginning in 1982, pennies became almostall zinc, with only a copper coating. Before1982, however, pennies were mostly copper.

TROMC02_044-081-hr 12/1/06 9:43 AM Page 70


EXAMPLE 2.6 Converting between Number of Moles and Number of AtomsCalculate the number of copper atoms in 2.45 mol of copper.

Check Since atoms are small, it makes sense that the answer is large. The number ofmoles of copper is almost 2.5, so the number of atoms is almost 2.5 times Avogadro’snumber.

For Practice 2.6A pure silver ring contains silver atoms. How many moles of silver atomsdoes it contain?

2.80 * 1022

Sort You are given the amount of copper in molesand asked to find the number of copper atoms.

Strategize Convert between number of moles andnumber of atoms by using Avogadro’s number as aconversion factor.

Given 2.45 mol Cu

Find Cu atoms

Conceptual Plan

Solve Follow the conceptual plan to solve the prob-lem. Begin with 2.45 mol Cu and multiply by Avo-gadro’s number to get to Cu atoms.

Solution

2.45 mol Cu *

6.022 * 1023 Cu atoms

1 mol Cu= 1.48 * 1024 Cu atoms

6.022 � 1023 Cu atoms

1 mol Cu

Cu atomsmol Cu

Converting between Mass and Amount (Number of Moles)To count atoms by weighing them, we need one other conversion factor—the mass of1 mol of atoms. For the isotope carbon-12, we know that this mass is exactly 12 grams,which is numerically equivalent to carbon-12’s atomic mass in atomic mass units. Sincethe masses of all other elements are defined relative to carbon-12, the same relationshipholds for all elements.

The mass of 1 mol of atoms of an element is called the molar mass.

The value of an element’s molar mass in grams per mole is numerically equal tothe element’s atomic mass in atomic mass units.

For example, copper has an atomic mass of 63.55 amu and a molar mass of 63.55 g/mol.One mole of copper atoms therefore has a mass of 63.55 g. Just as the count for shrimpdepends on the size of the shrimp, so the mass of 1 mol of atoms depends on the element:1 mol of aluminum atoms (which are lighter than copper atoms) has a mass of 26.98 g;1 mol of carbon atoms (which are even lighter than aluminum atoms) has a mass of12.01 g; and 1 mol of helium atoms (lighter yet) has a mass of 4.003 g.

26.98 g aluminum � 1 mol aluminum � 6.022 � 1023 Al atoms Al

C

He

12.01 g carbon � 1 mol carbon � 6.022 � 1023 C atoms

4.003 g helium � 1 mol helium � 6.022 � 1023 He atoms

Relationships Used6.022 * 1023

= 1 mol (Avogadro’s number)

The lighter the atom, the less mass it takes to make 1 mol.

TROMC02_044-081-hr 12/1/06 9:43 AM Page 71


1 dozen marbles

1 dozen peas

� The two pans contain the same num-ber of objects (12), but the masses aredifferent because peas are less massivethan marbles. Similarly, a mole of lightatoms will have less mass than a mole ofheavier atoms.

EXAMPLE 2.7 Converting between Mass and Amount (Number of Moles)Calculate the amount of carbon (in moles) contained in a 0.0265-g pencil “lead.”(Assume that the pencil lead is made of pure graphite, a form of carbon.)

Sort You are given the mass of carbon and asked to findthe amount of carbon in moles.

Given 0.0265 g C

Find mol C

1 mol

12.01 g

mol Cg C

Strategize Convert between mass and amount (in moles)of an element by using the molar mass of the element.

Conceptual Plan

Relationships Used12.01 g C = 1 mol C (carbon molar mass)

Solve Follow the conceptual plan to solve the problem. Solution

0.0265 g C *

1 mol C

12.01 g C= 2.21 * 10-3 mol C

Check The given mass of carbon is much less than the molar mass of carbon. Thereforethe answer (the amount in moles) is much less than 1 mol of carbon.

For Practice 2.7Calculate the amount of copper (in moles) in a 35.8-g pure copper sheet.

For More Practice 2.7Calculate the mass (in grams) of 0.473 mol of titanium.

Therefore, the molar mass of any element becomes a conversion factor between themass (in grams) of that element and the amount (in moles) of that element. For carbon:

The following example shows how to use these conversion factors.

12.01 g C = 1 mol C or 12.01 g C

mol C or

1 mol C

12.01 g C

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molar massof element

mol element

Avogadro’snumber

number of atomsg element

mol Cu number of Cu atomsg Cu

1 mol Cu

63.55 g Cu

6.022 � 1023 Cu atoms

1 mol Cu

We now have all the tools to count the number of atoms in a sample of an element byweighing it. First, obtain the mass of the sample. Then convert it to amount in molesusing the element’s molar mass. Finally, convert to number of atoms using Avogadro’snumber. The conceptual plan for these kinds of calculations takes the following form:

The examples that follow demonstrate these conversions.

EXAMPLE 2.8 The Mole Concept—Converting between Mass and Number of Atoms

How many copper atoms are in a copper penny with a mass of 3.10 g? (Assume that thepenny is composed of pure copper.)

Sort You are given the mass of copper and asked to findthe number of copper atoms.

Given 3.10 g Cu

Find Cu atoms

Strategize Convert between the mass of an element ingrams and the number of atoms of the element by firstconverting to moles (using the molar mass of the ele-ment) and then to number of atoms (using Avogadro’snumber).

Conceptual Plan

Relationships Used

6.022 * 1023= 1 mol (Avogadro’s number)

63.55 g Cu = 1 mol Cu (molar mass of copper)

Solution

= 2.94 * 1022 Cu atoms

3.10 g Cu *

1 mol Cu

63.55 g Cu*

6.022 * 1023 Cu atoms

1 mol Cu

Check The answer (the number of copper atoms) is less than (one mole).This is consistent with the given mass of copper atoms, which is less than the molar massof copper.

For Practice 2.8How many carbon atoms are there in a 1.3-carat diamond? Diamonds are a form ofpure carbon.

For More Practice 2.8Calculate the mass of tungsten atoms.2 .25 * 1022

(1 carat = 0.20 grams)

6.022 * 1023

Solve Finally, follow the conceptual plan to solve theproblem. Begin with 3.10 g Cu and multiply by the ap-propriate conversion factors to arrive at the number ofCu atoms.

Notice that numbers with large exponents, such as , are deceptivelylarge. Twenty-two copper pennies contain or 1 mol of copper atoms, but

pennies would cover Earth’s entire surface to a depth of 300 m. Even objectssmall by everyday standards occupy a huge space when we have a mole of them. For ex-ample, a grain of sand has a mass of less than 1 mg and a diameter of less than 0.1 mm,yet 1 mol of sand grains would cover the state of Texas to a depth of several feet. For everyincrease of 1 in the exponent of a number, the number increases by a factor of 10, so 1023

is incredibly large. One mole has to be a large number, however, if it is to have practicalvalue, because atoms are so small.

6 .022 * 10236.022 * 1023

6.022 * 1023

TROMC02_044-081-hr 12/1/06 9:43 AM Page 73

Given Al atoms

Find radius (r) of sphere

d = 2.70 g>cm3

8.55 * 1022

p r3V �4

3

rV (in cm3)


6.022

1 cm3

2.70 g Al

1 mol Al

1023 Al atoms

26.98 g Al

1 mol Al

mol Al g Al V (in cm3)Number of Al atoms

�

Sort You are given the number of aluminum atoms in asphere and the density of aluminum. You are asked tofind the radius of the sphere.

Conceptual PlanStrategize The heart of this problem is density, whichrelates mass to volume, and though you aren’t given themass directly, you are given the number of atoms, whichyou can use to find mass.

(1) Convert from number of atoms to number of molesusing Avogadro’s number as a conversion factor.

(2) Convert from number of moles to mass using molarmass as a conversion factor.

(3) Convert from mass to volume (in cm3) using densi-ty as a conversion factor.

(4) Once you compute the volume, find the radius fromthe volume using the formula for the volume of asphere.

Solve Finally, follow the conceptual plan to solve theproblem. Begin with Al atoms and multiplyby the appropriate conversion factors to arrive at volumein cm3.

Then solve the equation for the volume of a sphere for rand substitute the volume to compute r.

8 .55 * 1022

Relationships and Equations Used

V =

4

3 pr3 (volume of a sphere)

2.70 g>cm3 (density of aluminum)

26.98 g Al = 1 mol Al (molar mass of aluminum)

6.022 * 1023= 1 mol (Avogadro’s number)

Solution

r = 3C3V

4p= 3C3(1.4187 cm3)

4p= 0.697 cm

V =

4

3 pr3

*

26.98 g Al

1 mol Al*

1 cm3

2.70 g Al= 1.4187 cm3

8 .55 * 1022 Al atoms *

1 mol Al

6 .022 * 1023 Al atoms

EXAMPLE 2.9 The Mole ConceptAn aluminum sphere contains aluminum atoms. What is the radius of thesphere in centimeters? The density of aluminum is .2.70 g>cm3

8.55 * 1022

Check The units of the answer (cm) are correct. The magnitude cannot be estimatedaccurately, but a radius of about one-half of a centimeter is reasonable for just over one-tenth of a mole of aluminum atoms.

For Practice 2.9A titanium cube contains atoms. What is the edge length of the cube? Thedensity of titanium is .

For More Practice 2.9Find the number of atoms in a copper rod with a length of 9.85 cm and a radius of1.05 cm. The density of copper is .8.96 g>cm3

4.50 g>cm32.86 * 1023

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Chapter in Review 75

Conceptual Connection 2.4 Avogadro’s Number

Why is Avogadro’s number defined as and not a simpler round numbersuch as ?

Answer: Remember that Avogadro’s number is defined with respect to carbon-12—it is the numberequal to the number of atoms in exactly 12 g of carbon-12. If Avogadro’s number were defined as

(a nice round number), it would correspond to 1.99 g of carbon-12 atoms (an inconve-nient number). Avogadro’s number is defined with respect to carbon-12 because, as you recall fromSection 2.6, the amu (the basic mass unit used for all atoms) is defined relative to carbon-12. Therefore,the mass in grams of 1 mol of any element is equal to its atomic mass. As we have seen, these two defi-nitions together make it possible to determine the number of atoms in a known mass of any element.

1 .00 * 1023

1 .00 * 10236.022 * 1023

Conceptual Connection 2.5 The Mole

Without doing any calculations, determine which of the following contains the most atoms.

(a) a 1-g sample of copper(b) a 1-g sample of carbon(c) a 10-g sample of uranium

Answer: (b) The carbon sample contains more atoms than the copper sample because carbon has alower molar mass than copper. Carbon atoms are lighter than copper atoms, so a 1-g sample of carboncontains more atoms than a 1-g sample of copper. The carbon sample also contains more atoms thanthe uranium sample because, even though the uranium sample has 10 times the mass of the carbonsample, a uranium atom is more than 10 times as massive (238 g/mol for U versus 12 g/mol for carbon).

Chapter in Review

Key TermsSection 2.3law of conservation

of mass (47)law of definite

proportions (48)law of multiple

proportions (49)atomic theory (51)

Section 2.4cathode rays (51)cathode ray tube (51)electrical charge (52)electron (52)

Section 2.5radioactivity (54)nuclear theory (55)nucleus (55)proton (55)neutron (55)

Section 2.6atomic mass unit (amu) (56)atomic number (Z) (57)chemical symbol (57)isotope (58)natural abundance (58)mass number (A) (58)

ion (60)cation (60)anion (60)

Section 2.7periodic law (61)metal (62)nonmetal (62)metalloid (63)semiconductor (63)main-group elements (63)transition elements (transition

metals) (63)family (group) (63)

noble gases (63)alkali metals (64)alkaline earth metals (64)halogens (65)

Section 2.8atomic mass (67)mass spectrometry (68)

Section 2.9mole (mol) (70)Avogadro’s number (70)molar mass (71)

Key ConceptsImaging and Moving Individual Atoms (2.1)Although it was only 200 years ago that John Dalton proposed hisatomic theory, technology has progressed to the level where individualatoms can be imaged and moved by techniques such as scanningtunneling microscopy (STM).

The Atomic Theory (2.2, 2.3)The idea that all matter is composed of small, indestructible particlescalled atoms dates back to the fifth century B.C.; however, at the time the

atomic idea was rejected by most Greek thinkers. At about A.D. 1800certain observations and laws including the law of conservation ofmass, the law of constant composition, and the law of multiple propor-tions led John Dalton to reformulate the atomic theory with the follow-ing postulates: (1) each element is composed of indestructible particlescalled atoms; (2) all atoms of a given element have the same mass andother properties; (3) atoms combine in simple, whole-number ratios toform compounds; and (4) atoms of one element cannot change intoatoms of another element. In a chemical reaction, atoms change the waythat they are bound together with other atoms to form a new substance.

TROMC02_044-081-hr 12/1/06 9:43 AM Page 75


Key Skills

Using the Law of Definite Proportions (2.3)

• Example 2.1 • For Practice 2.1 • Exercises 33, 34

Using the Law of Multiple Proportions (2.3)

• Example 2.2 • For Practice 2.2 • Exercises 37–40

Working with Atomic Numbers, Mass Numbers, and Isotope Symbols (2.6)


Predicting the Charge of Ions (2.7)


Calculating Atomic Mass (2.8)

• Example 2.5 • For Practice 2.5 • For More Practice 2.5 • Exercises 73, 74, 76–78

Converting between Moles and Number of Atoms (2.9)

• Example 2.6 • For Practice 2.6 • Exercises 79, 80

Converting between Mass and Amount (in Moles) (2.9)

• Example 2.7 • For Practice 2.7 • For More Practice 2.7 • Exercises 81, 82

Using the Mole Concept (2.9)

• Examples 2.8, 2.9 • For Practice 2.8, 2.9 • For More Practice 2.8, 2.9 • Exercises 83–90, 106, 107

The Electron (2.4)J. J. Thomson discovered the electron in the late 1800s through experi-ments examining the properties of cathode rays. He deduced that electrons were negatively charged, and then measured their charge-to-mass ratio. Later, Robert Millikan measured the charge of the electron,which—in conjunction with Thomson’s results—led to the calculationof the mass of an electron.

The Nuclear Atom (2.5)In 1909, Ernest Rutherford probed the inner structure of the atom byworking with a form of radioactivity called alpha radiation and there-by developed the nuclear theory of the atom. This theory states that theatom is mainly empty space, with most of its mass concentrated in atiny region called the nucleus and most of its volume occupied by therelatively light electrons.

Subatomic Particles (2.6)Atoms are composed of three fundamental particles: the proton(1 amu, +1 charge), the neutron (1 amu, 0 charge), and the electron(~0 amu, charge). The number of protons in the nucleus of theatom is called the atomic number (Z) and defines the element. The sumof the number of protons and neutrons is called the mass number (A).

-1

Key Equations and Relationships

Relationship between Mass Number (A), Number of Protons (p),and Number of Neutrons (n) (2.6)

A = number of protons (p) + number of neutrons (n)

Avogadro’s Number (2.9)

1 mol = 6.0221421 * 1023 particles

Atoms of an element that have different numbers of neutrons (andtherefore different mass numbers) are called isotopes. Atoms that havelost or gained electrons become charged and are called ions. Cations arepositively charged and anions are negatively charged.

The Periodic Table (2.7)The periodic table tabulates all known elements in order of increasingatomic number. The periodic table is arranged so that similar elements aregrouped together in columns. Elements on the left side and in the center ofthe periodic table are metals and tend to lose electrons in their chemicalchanges. Elements on the upper right side of the periodic table are non-metals and tend to gain electrons in their chemical changes. Elements lo-cated on the boundary between these two classes are called metalloids.

Atomic Mass and the Mole (2.8, 2.9)The atomic mass of an element, listed directly below its symbol in theperiodic table, is a weighted average of the masses of the naturally oc-curring isotopes of the element.

One mole of an element is the amount of that element that containsAvogadro’s number ( ) of atoms. Any sample of an elementwith a mass (in grams) that equals its atomic mass contains one mole ofthe element. For example, the atomic mass of carbon is 12.011 amu,therefore 12.011 grams of carbon contains 1 mol of carbon atoms.

6 .022 * 1023

Atomic Mass (2.8)

Atomic mass = an

(fraction of isotope n) * (mass of isotope n)

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Exercises 77

ExercisesReview Questions

1. What is scanning tunneling microscopy? How does a scanningtunneling microscope work?

2. Why is it important to understand the nature of atoms?

3. Summarize the history of the atomic idea. How was Dalton ableto convince others to accept an idea that had been controversialfor 2000 years?

4. State the law of conservation of mass and explain what it means.

5. State the law of definite proportions and explain what it means.

6. State the law of multiple proportions and explain what it means.

7. Explain the difference between the law of definite proportionsand the law of multiple proportions.

8. What are the main ideas in Dalton’s atomic theory? How did theyhelp explain the laws of conservation of mass, of constant compo-sition, and of definite proportions?

9. How and by whom was the electron discovered? What basic prop-erties of the electron were reported with its discovery?

10. Explain Millikan’s oil drop experiment and how it led to the mea-surement of the electron’s charge. Why is the magnitude of thecharge of the electron so important?

11. Explain the plum-pudding model of the atom.

12. Describe Rutherford’s gold foil experiment. How did the experi-ment show that the plum-pudding model of the atom was wrong?

13. Describe Rutherford’s nuclear model of the atom. What was revo-lutionary about his model?

14. If matter is mostly empty space, as suggested by Rutherford, thenwhy does it appear so solid?

15. List the three subatomic particles that compose atoms and givethe basic properties (mass and charge) of each.

16. What defines an element?

17. Explain the difference between Z (the atomic number) and A (themass number.

18. Where do elements get their names?

19. What are isotopes? What is percent natural abundance?

20. Describe the two different notations used to specify isotopes andgive an example of each.

21. What is an ion? A cation? An anion?

22. What is the periodic law? How did it lead to the periodic table?

23. What are the characteristic properties of metals, nonmetals, andmetalloids?

24. What are the characteristic properties of each of the followinggroups?

a. noble gases b. alkali metalsc. alkaline earth metals d. halogens

25. How do you predict the charges of ions formed by main-groupelements?

26. What is atomic mass? How is it computed?

27. Explain how a mass spectrometer works.

28. What kind of information is contained in a mass spectrum?

29. What is a mole? How is it useful in chemical calculations?

30. Why is the mass corresponding to a mole of one element differentfrom the mass corresponding to a mole of another element?

Problems by Topic

Note: Answers to all odd-numbered Problems, numbered in blue, can be found in Appendix III. Exercises in the Problems by Topic section are paired,with each odd-numbered problem followed by a similar even-numbered problem. Exercises in the Cumulative Problems section are also paired, butsomewhat more loosely. (Challenge Problems and Conceptual Problems, because of their nature, are unpaired.)

The Laws of Conservation of Mass, Definite Proportions,and Multiple Proportions

31. A hydrogen-filled balloon was ignited and 1.50 g of hydrogen re-acted with 12.0 g of oxygen. How many grams of water vaporwere formed? (Assume that water vapor is the only product.)

32. An automobile gasoline tank holds 21 kg of gasoline. Whenthe gasoline burns, 84 kg of oxygen is consumed and carbon diox-ide and water are produced. What is the total combined mass ofcarbon dioxide and water that is produced?

33. Two samples of carbon tetrachloride were decomposed into theirconstituent elements. One sample produced 38.9 g of carbon and448 g of chlorine, and the other sample produced 14.8 g of carbonand 134 g of chlorine. Are these results consistent with the law ofdefinite proportions? Show why or why not.

34. Two samples of sodium chloride were decomposed into theirconstituent elements. One sample produced 6.98 g of sodiumand 10.7 g of chlorine, and the other sample produced 11.2 g ofsodium and 17.3 g of chlorine. Are these results consistent withthe law of definite proportions?

35. The mass ratio of sodium to fluorine in sodium fluoride is 1.21:1.A sample of sodium fluoride produced 28.8 g of sodium upon de-composition. How much fluorine (in grams) was formed?

36. Upon decomposition, one sample of magnesium fluoride pro-duced 1.65 kg of magnesium and 2.57 kg of fluorine. A secondsample produced 1.32 kg of magnesium. How much fluorine (ingrams) did the second sample produce?

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37. Two different compounds containing osmium and oxygen havethe following masses of oxygen per gram of osmium: 0.168 and0.3369 g. Show that these amounts are consistent with the law ofmultiple proportions.

38. Palladium forms three different compounds with sulfur. The massof sulfur per gram of palladium in each compound is listed below:

Compound Grams S per Gram Pd

A 0.603

B 0.301

C 0.151

Show that these masses are consistent with the law of multipleproportions.

39. Sulfur and oxygen form both sulfur dioxide and sulfur trioxide.When samples of these were decomposed the sulfur dioxide pro-duced 3.49 g oxygen and 3.50 g sulfur, while the sulfur trioxideproduced 6.75 g oxygen and 4.50 g sulfur. Calculate the mass ofoxygen per gram of sulfur for each sample and show that these re-sults are consistent with the law of multiple proportions.

40. Sulfur and fluorine form several different compounds includingsulfur hexafluoride and sulfur tetrafluoride. Decomposition of asample of sulfur hexafluoride produced 4.45 g of fluorine and1.25 g of sulfur, while decomposition of a sample of sulfur tetra-fluoride produced 4.43 g of fluorine and 1.87 g of sulfur. Calcu-late the mass of fluorine per gram of sulfur for each sample andshow that these results are consistent with the law of multipleproportions.

Atomic Theory, Nuclear Theory, and Subatomic Particles

41. Which of the following statements are consistent with Dalton’satomic theory as it was originally stated? Why?

a. Sulfur and oxygen atoms have the same mass.b. All cobalt atoms are identical.c. Potassium and chlorine atoms combine in a 1:1 ratio to form

potassium chloride.d. Lead atoms can be converted into gold.

42. Which of the following statements are inconsistent with Dalton’satomic theory as it was originally stated? Why?

a. All carbon atoms are identical.b. An oxygen atom combines with 1.5 hydrogen atoms to form a

water molecule.c. Two oxygen atoms combine with a carbon atom to form a

carbon dioxide molecule.d. The formation of a compound often involves the destruction

of one or more atoms.

43. Which of the following statements are consistent with Ruther-ford’s nuclear theory as it was originally stated? Why?

a. The volume of an atom is mostly empty space.b. The nucleus of an atom is small compared to the size of the

atom.c. Neutral lithium atoms contain more neutrons than protons.d. Neutral lithium atoms contain more protons than electrons.

44. Which of the following statements are inconsistent with Ruther-ford’s nuclear theory as it was originally stated? Why?

a. Since electrons are smaller than protons, and since a hydrogenatom contains only one proton and one electron, it must followthat the volume of a hydrogen atom is mostly due to the proton.

b. A nitrogen atom has seven protons in its nucleus and sevenelectrons outside of its nucleus.

c. A phosphorus atom has 15 protons in its nucleus and 150 elec-trons outside of its nucleus.

d. The majority of the mass of a fluorine atom is due to its nineelectrons.

45. A chemist in an imaginary universe, where electrons have a differ-ent charge than they do in our universe, performs the Millikan oildrop experiment to measure the electron’s charge. The charges ofseveral drops are recorded below. What is the charge of the elec-tron in this imaginary universe?

Drop # Charge

A

B

C

D

46. Imagine a unit of charge called the zorg. A chemist performs theoil drop experiment and measures the charge of each drop inzorgs. Based on the results below, what is the charge of the elec-tron in zorgs (z)? How many electrons are in each drop?

Drop # Charge

A

B

C

D

47. On a dry day, your body can accumulate static charge from walk-ing across a carpet or from brushing your hair. If your body de-velops a charge of (microcoulombs), how many excesselectrons has it acquired? What is their collective mass?

48. How many electrons are necessary to produce a charge of ?What is the mass of this many electrons?

49. Which of the following statements about subatomic particlesare true?

a. If an atom has an equal number of protons and electrons, itwill be charge-neutral.

b. Electrons are attracted to protons.c. Electrons are much lighter than neutrons.d. Protons have twice the mass of neutrons.

50. Which of the following statements about subatomic particlesare false?

a. Protons and electrons have charges of the same magnitude butopposite sign.

b. Protons have about the same mass as neutrons.c. Some atoms don’t have any protons.d. Protons and neutrons have charges of the same magnitude but

opposite sign.

51. How many electrons would it take to the equal the mass of aproton?

52. A helium nucleus has two protons and two neutrons. Howmany electrons would it take to equal the mass of a heliumnucleus?

-1.0 C

-15 mC

-12 .8 * 10-9 z

-6 .4 * 10-9 z

-9 .6 * 10-9 z

-4 .8 * 10-9 z

-4 .6 * 10-19 C

-11 .5 * 10-19 C

-9 .2 * 10-19 C

-6 .9 * 10-19 C

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Exercises 79

Isotopes and Ions53. Write isotopic symbols of the form for each of the following

isotopes.

a. the sodium isotope with 12 neutronsb. the oxygen isotope with 8 neutronsc. the aluminum isotope with 14 neutronsd. the iodine isotope with 74 neutrons

54. Write isotopic symbols of the form X-A (e.g., C-13) for each ofthe following isotopes.

a. the argon isotope with 22 neutronsb. the plutonium isotope with 145 neutronsc. the phosphorus isotope with 16 neutronsd. the fluorine isotope with 10 neutrons

55. Determine the number of protons and neutrons in each of thefollowing isotopes.

a. b. c. d.

56. Determine the number of protons and neutrons in each of thefollowing isotopes.

a. b. c. d.

57. The amount of carbon-14 in artifacts and fossils is often used toestablish their age. Determine the number of protons and neu-trons in a carbon-14 isotope and write its symbol in the form .

58. Uranium-235 is used in nuclear fission. Determine the number ofprotons and neutrons in uranium-235 and write its symbol in theform .

59. Determine the number of protons and electrons in each of thefollowing ions.

a. b. c. d.

60. Determine the number of protons and electrons in each of thefollowing.

a. b. c. d.

61. Predict the charge of the ion formed by each of the followingelements.

a. O b. K c. Al d. Rb

62. Predict the charge of the ion formed by each of the followingelements.

a. Mg b. N c. F d. Na

63. Fill in the blanks to complete the following table.

Number of Number ofSymbol Ion Formed Electrons in Ion Protons in Ion

Ca _____ _____

_____ 2 _____

Se _____ _____ 34

In _____ _____ 49


Number of Number of Symbol Ion Formed Electrons in Ion Protons in Ion

Cl _____ _____ 17

Te _____ 54 _____

Br _____ _____

_____ _____ 38Sr2 +

Br-

Be2 +

Ca2 +

Sr2+Ga3+Se2-Al3+

Cr3+Br-S2-Ni2+

AZX

AZX

3315P99

43Tc22688Ra40

19K

20882Pb222

86Rn2311Na14

7N

AZX

The Periodic Table and Atomic Mass65. Write the name of each of the following elements and classify it as

a metal, nonmetal, or metalloid.

a. Na b. Mg c. Br d. N e. As

66. Write the symbol for each of the following elements and classify itas a metal, nonmetal, or metalloid.

a. lead b. iodine c. potassiumd. silver e. xenon

67. Which of the following elements are main-group elements?

a. tellurium b. potassium c. vanadium d. manganese

68. Which of the following elements are transition elements?

a. Cr b. Br c. Mo d. Cs

69. Classify each of the following elements as an alkali metal, alkalineearth metal, halogen, or noble gas.

a. sodium b. iodine c. calciumd. barium e. krypton

70. Classify each of the following elements as an alkali metal, alkalineearth metal, halogen, or noble gas.

a. F b. Sr c. K d. Ne e. At

71. Which of the following pairs of elements do you expect to bemost similar? Why?

a. N and Ni b. Mo and Sn c. Na and Mgd. Cl and F e. Si and P

72. Which of the following pairs of elements do you expect to bemost similar? Why?

a. nitrogen and oxygen b. titanium and galliumc. lithium and sodium d. germanium and arsenice. argon and bromine

73. Rubidium has two naturally occurring isotopes with the follow-ing masses and natural abundances:

Isotope Mass (amu) Abundance (%)

Rb-85 84.9118 72.15

Rb-87 86.9092 27.85

Calculate the atomic mass of rubidium and sketch its mass spectrum.

74. Silicon has three naturally occurring isotopes with the followingmasses and natural abundances:

Isotope Mass (amu) Abundance (%)

Si-28 27.9769 92.2

Si-29 28.9765 4.67

Si-30 29.9737 3.10

Calculate the atomic mass of silicon and sketch its mass spectrum.

75. The atomic mass of fluorine is 18.998 amu and its mass spectrumshows a large peak at this mass. The atomic mass of chlorine is35.45 amu, yet the mass spectrum of chlorine does not show apeak at this mass. Explain the difference.

76. The atomic mass of copper is 63.546 amu. Do any copper isotopeshave a mass of 63.546 amu? Explain.

77. An element has two naturally occurring isotopes. Isotope 1 has amass of 120.9038 amu and a relative abundance of 57.4%, andisotope 2 has a mass of 122.9042 amu. Find the atomic mass ofthis element and, by comparison to the periodic table, identify it.

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Cumulative Problems

91. A 7.83-g sample of HCN is found to contain 0.290 g of H and4.06 g of N. Find the mass of carbon in a sample of HCN with amass of 3.37 g.

92. The ratio of sulfur to oxygen by mass in SO2 is 1.0:1.0.

a. Find the ratio of sulfur to oxygen by mass in SO3.b. Find the ratio of sulfur to oxygen by mass in S2O.

93. The ratio of oxygen to carbon by mass in carbon monoxide is1.33:1.00. Find the formula of an oxide of carbon in which theratio by mass of oxygen to carbon is 2.00:1.00.

94. The ratio of the mass of a nitrogen atom to the mass of an atom of12C is 7:6 and the ratio of the mass of nitrogen to oxygen in N2Ois 7:4. Find the mass of 1 mol of oxygen atoms.

95. An � particle, , has a mass of 4.00151 amu. Find the valueof its charge-to-mass ratio in .

96. Naturally occurring iodine has an atomic mass of 126.9045. A12.3849-g sample of iodine is accidentally contaminated with1.00070 g of 129I, a synthetic radioisotope of iodine used in thetreatment of certain diseases of the thyroid gland. The mass of129I is 128.9050 amu. Find the apparent “atomic mass” of the con-taminated iodine.

97. Nuclei with the same number of neutrons but different massnumbers are called isotones. Write the symbols of four isotones of236Th.


Symbol Z A # p # # n Charge

Si 14 _____ _____ 14 14 _____

_____ 32 _____ _____ _____

_____ _____ _____ _____ 24 2+

_____ 15 _____ _____ 15 16 _____

Cu2 +

2-S2-

e-

C>kg.

4He2+


Symbol Z A # p # # n Charge

__ 8 _____ _____ _____ 8

20 _____ _____ _____ 20 _____

_____ 25 _____ _____ 13 2+

_____ 14 _____ 10 _____ _____

100. Neutron stars are believed to be composed of solid nuclear matter,primarily neutrons. Assume the radius of a neutron to be approxi-mately cm, and calculate the density of a neutron.[Hint: For a sphere .] Assuming that a neutron starhas the same density as a neutron, calculate the mass (in kg) of asmall piece of a neutron star the size of a spherical pebble with a ra-dius of 0.10 mm.

101. Carbon-12 contains 6 protons and 6 neutrons. The radius of thenucleus is approximately 2.7 fm (femtometers) and the radius ofthe atom is approximately 70 pm (picometers). Calculate the vol-ume of the nucleus and the volume of the atom. What percentageof the carbon atom’s volume is occupied by the nucleus? (Assumetwo significant figures.)

102. A penny has a thickness of approximately 1.0 mm. If you stackedAvogadro’s number of pennies one on top of the other on Earth’ssurface, how far would the stack extend (in km)? [For compari-son, the sun is about 150 million km from Earth and the neareststar (Proxima Centauri) is about 40 trillion km from Earth.]

103. Consider the stack of pennies in the previous problem. Howmuch money (in dollars) would this represent? If this moneywere equally distributed among the world’s population of 6.5 bil-lion people, how much would each person receive? Would eachperson be a millionaire? Billionaire? Trillionaire?

104. The mass of an average blueberry is 0.75 g and the mass of an au-tomobile is . Find the number of automobiles whosetotal mass is the same as 1.0 mol blueberries.

2 .0 * 103 kg

V = (4/3)pr31 .0 * 10-13

N3-

Mg2 +

Ca2 +

2-

e-

78. Bromine has two naturally occurring isotopes (Br-79 and Br-81)and has an atomic mass of 79.904 amu. The mass of Br-81 is80.9163 amu, and its natural abundance is 49.31%. Calculate themass and natural abundance of Br-79.

The Mole Concept

79. How many sulfur atoms are there in 3.8 mol of sulfur?

80. How many moles of aluminum do aluminum atomsrepresent?

81. What is the amount, in moles, of each of the following?

a. 11.8 g Ar b. 3.55 g Zn c. 26.1 g Ta d. 0.211 g Li

82. What is the mass, in grams, of each of the following?

a. b. 0.0355 mol Bac. 43.9 mol Xe d. 1.3 mol W

83. How many silver atoms are there in 3.78 g of silver?

84. What is the mass of platinum atoms?

85. How many atoms are there in each of the following?

a. 5.18 g P b. 2.26 g Hg c. 1.87 g Bi d. 0.082 g Sr

4.91 * 1021

2.3 * 10-3 mol Sb

5.8 * 1024

86. Calculate the mass, in grams, of each of the following.a. b.c. d.

87. How many carbon atoms are there in a diamond (pure carbon)with a mass of 52 mg?

88. How many helium atoms are there in a helium blimp containing536 kg of helium?

89. Calculate the mass, in grams, of a single platinum atom.

90. Using scanning tunneling microscopy, scientists at IBM wrotethe initials of their company with 35 individual xenon atoms (asshown below). Calculate the total mass of these letters in grams.

7 .9 * 1021 uranium atoms1.8 * 1023 lead atoms2.82 * 1022 helium atoms1.1 * 1023 gold atoms

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Exercises 81

105. Suppose that atomic masses were based on the assignment of amass of 12.000 g to 1 mol of carbon, rather than 1 mol of 12C.Find the atomic mass of oxygen.

106. A pure titanium cube has an edge length of 2.78 in. How many ti-tanium atoms does it contain? Titanium has a density of

.

107. A pure copper sphere has a radius 0.935 in. How many copperatoms does it contain? [The volume of a sphere is andthe density of copper is .]8.96 g>cm3

(4/3)pr3

4.50 g>cm3

108. Boron has only two naturally occurring isotopes. The massof boron-10 is 10.01294 amu and the mass of boron-11 is11.00931 amu. Use the atomic mass of boron to calculate therelative abundances of the two isotopes.

109. Lithium has only two naturally occurring isotopes. The massof lithium-6 is 6.01512 amu and the mass of lithium-7 is7.01601 amu. Use the atomic mass of lithium to calculate therelative abundances of the two isotopes.

Challenge Problems

110. In Section 2.9, it was stated that 1 mol of sand grains would coverthe state of Texas to several feet. Estimate how many feet by as-suming that the sand grains are roughly cube-shaped, each onewith an edge length of 0.10 mm. Texas has a land area of 268,601square miles.

111. Use the concepts in this chapter to obtain an estimate for the num-ber of atoms in the universe. Make the following assumptions:(a) Assume that all of the atoms in the universe are hydrogen atomsin stars. (This is not a ridiculous assumption because over three-fourths of the atoms in the universe are in fact hydrogen. Gas anddust between the stars represent only about 15% of the visiblematter of our galaxy, and planets compose a far tinier fraction.)(b) Assume that the sun is a typical star composed of pure hydro-gen with a density of and a radius of . (c) As-sume that each of the roughly 100 billion stars in the Milky Waygalaxy contains the same number of atoms as our sun. (d) Assumethat each of the 10 billion galaxies in the visible universe containsthe same number of atoms as our Milky Way galaxy.

7 * 108 m1.4 g>cm3

Conceptual Problems

113. Which of the following is an example of the law of multiple pro-portions? Explain.

a. Two different samples of water are found to have the sameratio of hydrogen to oxygen.

b. When hydrogen and oxygen react to form water, the mass ofwater formed is exactly equal to the mass of hydrogen andoxygen that reacted.

c. The mass ratio of oxygen to hydrogen in water is 8:1. The massratio of oxygen to hydrogen in hydrogen peroxide (a com-pound that only contains hydrogen and oxygen) is 16:1.

114. The mole is defined as the amount of a substance containing thesame number of particles as exactly 12 grams of C-12. The amu is

defined as 1/12 of the mass of an atom of C-12. Why is it impor-tant that both of these definitions reference the same isotope?What would be the result, for example, of defining the mole withrespect to C-12, but the amu with respect to Ne-20?

115. Without doing any calculations, determine which of the follow-ing samples contains the greatest amount of the element in moles.Which contains the greatest mass of the element?

a. 55.0 g Cr b. 45.0 g Ti c. 60.0 g Zn

116. The atomic radii of the isotopes of an element are identical to oneanother. However, the atomic radii of the ions of an element aresignificantly different from the atomic radii of the neutral atom ofthe element. Explain this behavior.

a. Assuming that the sample is statistically representative of anaturally occurring sample, calculate the percent naturalabundance of each Wt isotope.

b. Draw the mass spectrum for a naturally occurring sampleof Wt.

c. The mass of each Wt isotope is measured relative to C-12 andtabulated below. Use the mass of C-12 to convert each of themasses to amu and calculate the atomic mass of Wt.

Isotope Mass

Wt-296

Wt-297

Wt-298 24 .8312 * Mass(12C)

24 .7490 * Mass(12C)

24 .6630 * Mass(12C)

112. Below is a representation of 50 atoms of a fictitious element calledwestmontium (Wt). The red spheres represent Wt-296, the bluespheres Wt-297, and the green spheres Wt-298.

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