1

Atoms and the Atomic Theory 

All the matters can be broken down into elements. Is

matter continuously divisible into ever smaller and smaller

pieces, or is there an ultimate limit? What is an element

made of?

Greeks

•Aristotle- Continuous Theory of Matter

•Democritus- Discontinous Theory of Matter

•Atomos- “indivisible”

Early Chemical Discoveries and Atomic Theory

Three important fundamental laws in chemistry

1) Law of conservation of mass

2) Law of constant composition

3) Law of multiple proportions

2

Law of Conservation of Mass

1774 Antoine Lavoisier –showed heating the red power HgO causes it to decompose into the silvery liquid mercury and the colorless gas oxygen. 2HgO 2Hg +O2 then show that oxygen

is the key substance involved in combustion.

Furthermore, results of combustion reactions•Total mass of products = total mass of reactants(tin + air+ sealed glassed vessel) (tin oxide + remaining air + glass

vessel)

Law of Conservation of Mass~ The total mass of substances present after a chemical reaction is the same as the total mass of substances before the reaction. Matter is neither created nor destroyed in a chemical reaction.

3

E.g. A 0.382g sample of magnesium reacts with 2.652g of nitrogen gas. The sole product is magnesium nitride. After reaction, the mass of unreacted nitrogen is 2.505g. What mass of magnesium nitride is produced?

Mass before reaction =0.382g Mg + 2.652g N2 gas =

3.034g

Mass after reaction = ?g Mg3N2 gas + 2.505 N2 gas

3.034g – 2.505g = 0.529g

4

Law of Constant Composition

1799 Joseph Proust – Law of Constant Composition (Definite Proportion)~ All samples of a compound have the same composition- the same proportion by mass of the constituent elements. This means that the relative amount of each element in a particular compound is always the same, regardless of the source of the compound or how the compound is prepared.

E. g. Water is made up of two elements H and O. The two sample of water below have the same proportions of the two elements, expressed as percentages by mass. Every sample of water contains 1 part hydrogen and 8 parts oxygen by mass._________________________________________

Sample A Composition Sample B10.000g 27.000g1.119g H %H = 11.19 3.031g H8.881g O %O = 88.81 23.979g O

5

Dalton’s Atomic Theory

How can the Law of conservation of mass and Law of constant composition be explain? Why do element behave as they do?

1803-1808 John Dalton : proposed a new theory of matter. 1. Each chemical element is composed of minute, indestructible particles called atoms. 2. All atoms of a given element are identical to one another in mass and other properties, but the atoms of one element are different from the atoms of other elements.

6

Dalton’s Atomic Theory

3. Atoms of an element are not changed into atoms of a different element by chemical reactions; atoms are neither created nor destroyed in chemical reactions. Chemical compounds are formed when atoms combine with each other. – If atoms of an element are indestructible, then

the same remains unchanged. This explains the law of conservation of mass.

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Dalton’s Atomic Theory

4. In each of their compounds, different elements combine in a simple numerical ratio: e.g. one atom of A to one of B (AB) or one atom of A to two of B (AB2). – If all atoms of an element are alike in mass

(assumption 2) and if atoms unite in fixed numerical ratio (assumption 3), the percent composition of a compound must have a unique value, regardless of the origin of the sample analyzed. This explains the law of constant composition.

8

Law of Multiple Proportions

Dalton’s theory leads to a prediction- the law of multiple proportions.~ If two elements form more than a single compound, the masses of one element combined with a fixed mass of the second are in the ratio of small whole numbers. Same elements to combine in different ratios to give different substances.

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E.g. Oxygen and carbon can combine either in a 1: 1.333 mass ratio to make a substance or in a 1: 2.667 mass ratio to make a substance.

first 1 g carbon per 1.333 g oxygen C:O mass ratio = 1: 1.333second 1 g carbon per 2.667 g oxygen C:O mass ratio = 1: 2.667

comparison C:O mass ratio in second sample = (1 g C)/(2.667g O) = 2 of C:O ratios C:O mass ratio in first sample (1 g C)/(1.333g O)

• Compare two substances clearly the second substance contains exactly twice as much oxygen as the first for a given number of carbon. If the first oxide has the molecular formula CO then the

second oxide will be CO2.

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E.g.There are two compounds, both contain nitrogen and hydrogen. Compound A contains 1.50g of N and 0.216g H. Compound B contains 2.00g of N and 0.144g H. If the formula of compound B is N2H2, what is the formula of

compound A?

N:H ratio in A = 1.50: 0.216g = 1.00: 0.144N:H ratio in B = 2:00: 0.144 = 1.00: 0.0720H in A is (0.144/0.0720 = 2 ) twice as much in

B

IF B is N2H2 then A is N2H4

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Atomic Mass RatioDalton’s theory enables us to set up a scale of relative atomic masses. He cannot measure the exact mass of atoms but relative mass.E.g. Consider calcium sulfide, which consists of 55.6% calcium by mass and 44.4% sulfur by mass. Suppose there is one calcium atom for each sulfur atom in calcium sulfide. Because we know that the mass of a calcium atom relative to that of a sulfur atom must be the same as the mass % in calcium, we know that the ratio of the mass of a calcium atom to that of a sulfur atom is

mass of Ca atom = 55.6 = 1.25mass of a S atom 44.4 or mass of a Ca atom = 1.25 x mass of a sulfur atom

By continuing in this manner with other compounds, it is possible to build up a table of relative atomic masses. We define a quantity called atomic mass ratio, which is the ratio of the mass of a given atom to the mass of some particular reference atom.

12

The Structure of AtomsWhat is an atom made of ?

Discovery of subatomic particleThe Discovery of Electrons*1897 J.J. Thomson- cathode ray experiment Thomson’s experiment involved the use of cathode-ray tube. When a sufficiently high voltage is applied across the electrode, an electric current flows through the tube from negatively charged electrode ( the cathode) to the positively charged electrode (the anode).

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Thomson’s Experiment

Voltage source

+-

Vacuum tube

Metal Disks

14

Voltage source

Thomson’s ExperimentThomson’s Experiment

By adding an electric field By adding an electric field

+

-

15

Voltage source

Thomson’s ExperimentThomson’s Experiment

By adding an electric field he found By adding an electric field he found that the moving pieces were that the moving pieces were negative negative

+

-

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Thomson’s Model• Found the electron• Couldn’t find positive

(for a while) • Said the atom was like

plum pudding• A bunch of positive

stuff, with the electrons able to be removed

• established the ratio of mass to electric charge for cathode ray m/e = -5.6857x10-9 g/coulomb.

Spherical cloud of positive charge

Electrons

17

Millikan’s Oil-Drop Experiment: Mass of Electron

1909 Robert Millikan determined the electronic charge through a series of oil-drop experiments. The currently accepted value of the charge of the e is –1.6022x10-19C. Substituting into Thomson’s mass to charge ratio then gives the mass of electron as 1/1836(= 9.1094x10-28g).

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X-Ray and RadioactivityErnest Rutherford identified two type of radiation from radioactive materials, alpha () and beta (). -particles (He2+)carry two fundamental units of positive charge and have essentially the same mass as He atoms. -particles are negatively charged particles produced by changes occurring within the nuclei of radioactive atoms and have the same properties as electrons.

A third form of radiation, that is not affected by an electric field was discovered in 1900 by Paul Villard. This radiation, called -ray, is not made up of particles; it is electromagnetic radiation of extremely high penetrating power. Properties of the three radioactive emissions discovered Original name Modern name Mass (amu)

Charge-ray -particle 4.00 +2-ray -particle (electron) 5.49x10-4 -1-ray -ray 0 0_______

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Lead block

Uranium

Gold Foil

Florescent Screen

1909 Ernest Rutherford Scattering Experiment ~ used particle to study the inner structure of atoms. ~directed a beam of -particles at a thin gold foil

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Rutherford Expected

• The alpha particles to pass through without changing direction very much

• WHY?• The positive charges were spread

out evenly. Alone they were not enough to stop the alpha particles

21

What he expected

22

Because

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Rutherford thought the mass was evenly distributed in the atom

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Rutherford thought the mass was evenly distributed in the atomparticles should pass through the low + density model.

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What he got

The majority of -particles penetrated the foil undeflected. Some particles experienced slightly deflections. A few (about one in every 20,000) suffered rather

serious deflections as they penetrated the foil.

A similar number did not pass through the foil at all, but bounced back in the direction from which they had come.

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How he explained it

+

• Atom is mostly empty• Small dense,

positive piece at center

• Alpha particles are deflected by it if they get close enough

27

+

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The Nuclear Atom: Protons and Neutrons

1911 Rutherford explained his results by proposing a model of the atom known as the nuclear atom and having these features.

1. Most of the mass and all of the positive charge of an atom are centered in a very small region called the nucleus. The atom is mostly empty space.

2. The magnitude of the positive charge is different for different atoms and is approximately one-half the atomic weight of the element.

3. There are as many electrons outside the nucleus as there are units of positive charge on the nucleus. The atom as a whole is electrically neutral.

 Rutherford’s nuclear atom suggested the existence of positively

charged fundamental particles of matter in the nuclei of atoms- called protons. He predicted the existence in the nucleus of electrically neutral particles.

* 1932 James Chadwick~ verified that there is another type of particles in atom called

neutron.

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The Structure of AtomsTherefore•Modern picture of an atom, then, consist of three types of particles-electrons, protons and neutron.

Electric Charge Mass Particle SI (C ) Atomic SI (g) amu LocatedElectron -1.602x10-19 -1 9.109x10-28

5.49x10-4 outside nucleusProton +1.602x10-19 +1 1.673x10-24

1.0073 in nucleusNeutron 0 0 1.675x10-24

1.0087 in nucleus

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Size of an atom• Atoms are small ~10-10 meters• Hydrogen atom, 32 pm radius• Nucleus tiny compared to atom• IF the atom was the size of a stadium, the

nucleus would be the size of a marble.

• Radius of the nucleus near 10-15m.

• Density near 1014 g/cm

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Conclusion:

• Matter is composed, on a tiny scale, of particles called atoms. Atoms are in turn made up of minuscule nuclei surrounded by a cloud of particles called electrons. Nuclei are composed of particles called protons and neutrons, which are themselves made up of even smaller particles called quarks. Quarks are believed to be fundamental, meaning that they cannot be broken up into smaller particles.

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Chemical Elements

Atomic numberWhat is that makes one atom different from

another?Elements differ from one another according to the

number of protons in their nucleus• atomic number (Z) = Number of proton in

atom’s nucleus• mass number (A) = # of protons (Z) + # of

neutrons (N)

C12

6

Mass #Symbol of element

Atomic #

E.g.

33

Isotopes• Contrary to what Dalton thought, we know that atoms

of an element do not necessarily all have the same mass.

• Isotope- atoms of the same element containing different numbers of neutrons and therefore having different masses.

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Isotopes of Hydrogen

35

Mass Spectrometer- The most accurate means of determining atomic and

molecular weights.

•

36

Mass Spectrum of Elemental Carbon

This small peak represents the relative abundance of C13 in nature.

When 12C and 13C are analyzed in a mass spectrometer, the ratio of their masses is found to be : Mass13C = 1.0836129

Mass12C

Since the atomic mass unit is defined such that the mass of 12C is exactly 12 amu, then on this same scale,

Mass13C = (1.0836129)(12amu) = 13.003355 amu

37

Average Atomic MassWhen considering atomic masses from the P-Table,

recall that reported values are actually weighted averages of all the naturally occurring isotopes.

Average atomic mass = (% of each isotope)(atomic mass of each isotope)

100

Boron has two isotopes 10B and 11B. They have the abundance 18.7% and 81.3% respectively. Determine the average atomic mass for Boron.

38

When natural copper is vaporized and injected

into a mass spectrometer, the results

shown below are obtained. Use these data to compute the average

mass of copper. The mass values for 63Cu and 65Cu are 62.93 amu and 64.93 amu respectively.

Computing Average Mass from Mass Spectrometer Data = Atomic Weight

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Isotopes and Average Atomic Mass Questions

1. Do either of the following pairs represent isotopes of one another?

a. 40K19 and 40Ar18 b. 90Sr38 and 94Sr38

2. The nobel gas Neon, has three isotopes of masses, 22, 21and 20. If the isotopes have the abundance of 8.01%, 1.99% and 90.00% respectively, what is the average atomic mass of neon atoms?

3. A naturally occurring sample of an element consists of two isotopes, one of mass 85 and one of mass 87. The abundance of these isotopes is 71% and 29%. Calculate the average mass of an atom of this element.

4. If 69Ga and 71Ga occur in the %’s 62.1 and 37.9, calculate the average atomic mass of gallium atoms.

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Mass Spectrum of Chlorine Molecule

(35Cl-35Cl)+, (35Cl-37Cl)+, or (37Cl-37Cl)+

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Ions

Ion= an electrically charged particle obtained from an atom or chemically bonded group of atoms lose or gain electrons. The charge on an ion is equal to the # of protons minus the # of electrons. An atom that gains extra electrons becomes a negatively charged ion, called an anion. An atom that loses electrons becomes positively charged ion, called a cation.

EAZ

number p + number n

number p

+ ? number p - number e

E.g. Determine numbers of electrons in Mg2+ cation and the S2- anion?

Mg2+ number e =?

S2- number e =?

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Introduction to Periodic Table

With discovery of many elements1869 Mendeleev and Meyer ~ independently proposed periodic table organized the elements In modern periodic table, The periodic table of the elements is organized into 18 groups and 7 periods. Elements are represented by one or two-letter symbols and are arranged according to atomic number.

* a horizontal row of elements- a period* a vertical row of elements- a group or family

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Periodic Table of ElementsGroup 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Period 1A 8A

1 1 H

1.008

2A 3A 4A 5A 6A 7A

2 He

4.003

2 3 Li

6.941

4 Be

9.012

5 B

10.81

6 C

12.01

7 N

14.01

8 O

16.00

9 F

19.00

10 Ne

20.18

3 11 Na

22.99

12 Mg

24.30

3B 4B 5B 6B 7B --- --8B -- --- 1B 2B

13 Al

26.98

14 Si

28.09

15 P

30.97

16 S

32.07

17 Cl

35.45

18 Ar

39.95

4 19 K

39.10

20 Ca

40.08

21 Sc

44.96

22 Ti

47.88

23 V

50.94

24 Cr

52.00

25 Mn

54.94

26 Fe

55.85

27 Co

58.93

28 Ni

58.69

29 Cu

63.55

30 Zn

65.39

31 Ga

69.72

32 Ge

72.61

33 As

74.92

34 Se

78.96

35 Br

79.90

36 Kr

83.80

5 37 Rb

85.47

38 Sr

87.62

39 Y

88.91

40 Zr

91.22

41 Nb

92.91

42 Mo

95.94

43 Tc (98)

44 Ru

101.1

45 Rh

102.9

46 Pd

106.4

47 Ag

107.9

48 Cd

112.4

49 I n

114.8

50 Sn

118.7

51 Sb

121.8

52 Te

127.6

53 I

126.9

54 Xe

131.3

6 55 Cs

132.9

56 Ba

137.3 *

71 Lu

138.91

72 Hf

178.5

73 Ta

180.9

74 W

183.8

75 Re

186.2

76 Os

190.2

77 I r

192.2

78 Pt

195.1

79 Au

197.0

80 Hg

200.6

81 Tl

204.4

82 Pb

207.2

83 Bi

209.0

84 Po

(209)

85 At

(210)

86 Rn

(222)

7 87 Fr

(223)

88 Ra

226.0 **

103 Lr

227.0

104 Rf

(216)

105 Db

(262)

106 Sg

(266)

107 Bh

(264)

108 Hs

(269)

109 Mt

(268)

110 Ds

111 Rg

112 Uub

113 Uut

114 Uuq

115 Uup

116 Uuh

117 Uus

118 Uuo

* Lanthanoids * 57 La

58 Ce

59 Pr

60 Nd

61 Pm

62 Sm

63 Eu

64 Gd

65 Tb

66 Dy

67 Ho

68 Er

69 Tm

70 Yb

* *Actinoids ** 89 Ac

90 Th

91 Pa

92 U

93 Np

94 Pu

95 Am

96 Cm

97 Bk

98 Cf

99 Es

100 Fm

101 Md

102 No

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It is customary also to divide the elements into broad categories known as Metals: Except mercury (liquid), metals are solid s at room temperature. They are generally malleable, ductile, good conductors of heat and electricity, and have a lustrous or shiny appearance. Nonmetals: generally have opposite properties of metals; e.g. poor conductors of heat and electricity.Metalloid (semimetal): is an element having both metallic and nonmetallic properties.Or into three groupsMain group elements are those in groups 1, 2 and 13-18. when form ions, group 1, 2 lose the same # e as their group #; group 13 lose group #-10; group 14-18 gain 18-group #.Transition elements: from group 3 to 12, and because all of them are metals, they are also called the transition metals. The # of electrons lost in TM is not related to their group #.Inner transition metals which include Lanthanides and Actinindes.

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Nuclear Chemistry

• Nuclear reactions involve changes that originate in the nucleus of the atom.

• Chemical changes involve changes in the electron cloud.

• Uses: – 60Co- gamma ray emitter- ionizing

radiation for treatment of cancerous tumors.

– 201Thallium stress test of heart muscle

– Radiocarbon dating 14C ½ life 5730 years

– Nuclear power ~ 20% of US electricity production

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Radioactivity

• Recall that all atoms of the same element have the same number of protons. The number of neutrons in the atoms nucleus, however, may be different from one atom to the next= Isotopes.

Uranium- 234 Uranium-235 Uranium-23892 protons 92 protons 92 protons142 neutrons 143 neutrons 146 neutronsTrace 0.7% 99.3%• Different isotopes have different abundances• Different isotopes have different stabilities

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Patterns of Nuclear Stability

As the atomic number increases, the neutron to proton ratio of the stable nuclei increases. The stable nuclei are located in the shaded area of the graph known as the belt of stability. The majority of radioactive nuclei occur outside this belt.

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Nuclear Equations• Radionuclides are unstable nuclei

that emit particles and electromagnetic radiation to transform into a stable nucleus.

U23892 Th

234

90 + He4

2

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Nuclear Equations

• Mass numbers and atomic numbers must be balanced in all nuclear equations.

50

What product is formed when thorium-232

undergoes alpha decay?

51

Types of Radioactive Decay

Alpha decay- nucleus emits 2 protons and 2 neutrons (He nucleus)

Beta decay- a neutron in the nucleus decays into a proton and an electron, the electron is emitted

Gamma- high energy, short wavelength electromagnetic radiation- accompanies other radioactive emissions.

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Types of Radioactive DecayElectron Capture- capture

by the nucleus of an electron from the electron cloud surrounding the nucleus.

Positron- particle with the same mass as an electron, but an opposite charge collides with an electron and produces gamma radiation.

53

Penetrating Power of Radioactive Decay

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Radioactive Decay ParticlesParticle Nuclear

EquationExample

Alpha= 2 protons and 2 neutrons

Nucleus 4He 226Ra 222Rn +4He

Beta =neutron converts to proton and a high energy electron

1n 1p + 0e 131I131Xe + 0e

Electron Capture= electron captured by nucleus

1p + 0e 1n 81Rb + 0e 81Kr

88 86 22

0 1 -1 53 54 -1

1 -1 037 -1 36

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Radioactive Decay Particles

Particle Nuclear Equation

Example

Positron=proton converted to a neutron and an electron

1p 1n + 0e 11C 11B +0e

Gamma= electromagnetic radiation

Not shown in equations, but almost always accompanies other decay.

1 0 1

Remember a positron has the same mass as an electron, but the opposite charge

6 5 1

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Radioactive Decay Particles

Particle Effect

Alpha Decrease atomic mass by ___ and atomic number by _____.

Beta Atomic number _______________.

Electron Capture

Atomic number _______________.

Positron Atomic number _______________.

Gamma

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Half Life- the time required for half of any given quantity of a substance to react /

decay. (independent of initial quantity of atoms)

Number of Th-232 atoms in a sample initially containing 1 million atoms as a function of time. Th-232 has a half-life of 14 billion years.

Half Life Simulation

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Half Life Problems

 Example:An isotope of cesium (cesium-137) has

a half-life of 30 years. If 1.0 mg of cesium-137 disintegrates over a period of 90 years, how many mg of cesium-137 would remain?

 

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1. A 2.5 gram sample of an isotope of strontium-90 was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific Test Site. The half-life of strontium-90 is 28 years. In what year will only 0.625 grams of this strontium-90 remain?

2. Actinium-226 has a half-life of 29 hours. If 100 mg of actinium-226 disintegrates over a period of 58 hours, how many mg of actinium-226 will remain?

3. Thallium-201 has a half-life of 73 hours. If 4.0 mg of thallium-201 disintegrates over a period of 6.0 days and 2 hours, how many mg of thallium-201 will remain?

4. Sodium-25 was to be used in an experiment, but it took 3.0 minutes to get the sodium from the reactor to the laboratory. If 5.0 mg of sodium-25 was removed from the reactor, how many mg of sodium-25 were placed in the reaction vessel 3.0 minutes later if the half-life of sodium-25 is 60 seconds?

5. Selenium-83 has a half-life of 25.0 minutes. How many minutes would it take for a 10.0 mg sample to decay and have only 1.25 mg of it remain?

Half Life Problems

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Uranium-238: an example of an unstable nucleus decaying to form

other unstable nucleiUranium-238 is radioactive, undergoing alpha decay. But, the daughter nuclide is also radioactive, undergoing beta decay, to produce yet another radioactive nuclide, which decays. The atom goes through a rather involved sequence of radioactive decays (both alpha and beta), until a stable isotope (lead-206) is reached.

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Fission Reaction

Collision of a neutron with a U-235 nucleus can cause the nucleus to split, creating two smaller nuclides and three free neutrons. The three neutrons may travel outward from the fission, colliding with nearby U-235 nuclei, causing them to split as well. Each split (fission) is accompanied by a large quantity of energy.

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Fission Chain ReactionCollision of a neutron with a U-235 nucleus can cause the nucleus to split, creating two smaller nuclides and three free neutrons. The three neutrons may travel outward from the fission, colliding with nearby U-235 nuclei, causing them to split as well. Each split (fission) is accompanied by a large quantity of energy. If sufficient neutrons are present, we may achieve a chain reaction. If only one neutron were produced with each fission, no chain reaction would occur, because some neutrons would be lost through the surface of the uranium sample.

Mousetrap Chain Reaction

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Fission Reaction

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Fusion ReactionTremendous energy needed to overcome the repulsion between nuclei. Heat required for this reaction is on the order of 40,000,000 K. The energy from an atomic bomb could generate this heat (hydrogen or thermonuclear weapon).