prentice-hall © 2002 general chemistry: chapter 2 slide 1 of 25 chapter 2: atoms and the atomic...

Prentice-Hall © 2002General Chemistry: Chapter 2Slide 1 of 25

Chapter 2: Atoms and the Atomic Theory

Philip DuttonUniversity of Windsor, Canada

Prentice-Hall © 2002

General ChemistryPrinciples and Modern Applications

Petrucci • Harwood • Herring

8th Edition


Contents

• Early chemical discoveries • Electrons and the Nuclear Atom• Chemical Elements• Atomic Masses• The Mole


Early Discoveries

Lavoisier 1774Law of conservation of mass

Proust 1799 Law of constant composition

Dalton 1803-1888 Atomic Theory


Dalton’s Atomic Theory

Each element is composed of small particles called atoms.

Atoms are neither created nor destroyed in chemical reactions.

All atoms of a given element are identical

Compounds are formed when atoms of more than one element

combine

General Chemistry: Chapter 2Slide 5 of 25

Consequences of Dalton’s theory

In forming carbon monoxide, 1.33 g of oxygen combines with 1.0 g of carbon.

In the formation of carbon dioxide 2.66 g of oxygen combines with 1.0 g of carbon.

Law of Definite Proportions: combinations of elements are in ratios of small whole numbers.


Behavior of charges


Cathode ray tube


Properties of cathode rays

Electron m/e = -5.6857 x 10-9 g coulomb-1


Charge on the electron

From 1906-1914 Robert Millikan showed ionized oil drops can be balanced against the pull of gravity by an electric field.

The charge is an integral multiple of the electronic charge, e.


Radioactivity

Radioactivity is the spontaneous emission of radiation from a substance.

X-rays and g-rays are high-energy light.

a-particles are a stream of helium nuclei, He2+.

b-particles are a stream of high speed electrons

that originate in the nucleus.


The nuclear atom

Geiger and Rutherford1909


The a-particle experiment

Most of the mass and all of the positive charge is concentrated in a small region called the nucleus .

There are as many electrons outside

the nucleus as there are units of positive charge on the nucleus


The nuclear atom

Rutherfordprotons 1919

James Chadwickneutrons 1932


Scale of Atoms

Useful units:

1 amu (atomic mass unit) = 1.66054 x 10-27 kg 1 pm (picometer) = 1 x 10-12 m 1 Å (Angstrom) = 1 x 10-10 m = 100 pm = 1 x 10-8 cm

The heaviest atom has a mass of only 4.8 x 10-22 g

and a diameter of only 5 x 10-10 m.

Biggest atom is 240 amu and is 50 Å across.

Typical C-C bond length 154 pm (1.54 Å)

Molecular models are 1 Å /inch or about 0.4 Å /cm


Atomic Diameter 10-8 cm Nuclear diameter 10-13 cm

Nuclear Structure

Particle Mass Chargekg amu Coulombs (e)

Electron 9.109 x 10-31 0.000548 –1.602 x 10-19 –1Proton 1.673 x 10-27 1.0073 +1.602 x 10-19 +1Neutron 1.675 x 10-27 1.0087 0 0

1 Å

The nuclei of most atoms consist of protons and neutrons, which are called the nucleons.

Atomic number: the number of protons defines the type of the element.

The number of neutrons determines the isotope of an element.

The stability of the neutron

• In the stable nuclei the bound neutrons are stable.

• The free neutrons are unstable; they undergo beta decay with a half-life of just about 10.2 minutes (614 s). Quark structure: Feynmann diagram:

Prof G. I. Csonka © 2009General Chemistry: Chapter 2Slide 16 of 25

Half life of the decaying free neutron

Prof. G.I. Csonka © 2009General Chemistry: Chapter 2Slide 17 of 25

0 10 20 30 40 50 600

25

50

75

100

t [min]

N

Half-life: t ½ = 10.2 min

2/1

21

)0()(t

t

NtN

n0 → p+ + e− + ūe

Free neutrons decay by emission of an electron and an electron antineutrino to become a proton, a process known as beta decay:


Energy – mass conversion

Einstein (1905): E = mc2

• All energy changes are accompanied by mass changes (Δm)– In chemical reactions ΔE is too small to notice:

Δm < 0.0000001 u (negligible). – In nuclear reactions ΔE > 1.1 MeV/nucleon.

1 MeV = 1.6022x10-13 J

If ΔE =1.4924x10-10 J or 931.5 MeV then Δm = 1.0 u

Gabor I. Csonka © 2008 General Chemistry: Chapter 26Slide 19 of 47

Nuclear Binding Energy

Total mass defect

Positive binding energy:

0.0304 x 931.5 = 28.4 MeV

Per nucleon:

28.4 / 4 = 7.1 MeV/nucleon


Isotopes, atomic numbers and mass numbers

To represent a particular atom we use the symbolism:

A= mass number Z = atomic number

Isotopes of an element

• Isotopes are different types of atoms (nuclides) of the same chemical element, Z, each having a different atomic mass (A: mass number)

• Same number of protons, but different number of neutrons.

• About 339 nuclides occur naturally on Earth, of which 256 (~75%) are stable. 3100 are known currently


Az anyag22

The isotopes of the hydrogen2H=D1

3H=T1

1H=H1

(t1/2) : 12.33 years

Mass defect of the isotopes vs. energy

Gábor I. Csonka © 2008 General Chemistry: Chapter 26Slide 23 of 47

Z A Atomic mass Mass defect MeV/nucleon

H 1 1 1.00783 [amu]D 1 2 2.01410 0.00239 1.11T 1 3 3.01605 0.00911 2.83He 2 3 3.01603 0.00829 2.57

2 4 4.00260 0.03038 7.07Li 3 6 6.01512 0.03435 5.33

3 7 7.01600 0.04213 5.61Be 4 9 9.01218 0.06244 6.46B 5 10 10.01294 0.06951 6.47

5 11 11.00931 0.08181 6.93C 6 12 12.00000 0.09894 7.68

6 13 13.00335 0.10425 7.476 14 14.00324 0.11303 7.52


Average Binding Energy as a Function of Mass Number


Neutron-to-Proton Ratio (Segré diagram)


Measuring atomic masses

The relative atomic mass

• Unit: rest mass of carbon-12 is exactly12.• It is the mean of the atomic masses over all

the isotopes of the element weighted by isotopic abundance as found in a particular environment (standard: the earth crust).

• This number may be a fraction which is not close to a whole number, due to the averaging process.


Isotopic abundance in the earth crust

Az anyagKészítette: Dr. Csonka Gábor egyetemi tanár, [email protected]

28

Z A Atomic mass abundance% t1/2

H 1 1 1.00783 99.9885D 1 2 2.01410 0.0115T 1 3 3.01605 12.33 yHe 2 3 3.01603

2 4 4.00260 99.9999Li 3 6 6.01512 7.59

3 7 7.01600 92.41Be 4 9 9.01218 100.0000B 5 10 10.01294 19.9

5 11 11.00931 80.1

Isotopic abundance in the earth crust (2)

Az anyagKészítette: Dr. Csonka Gábor egyetemi tanár, [email protected]

29

Z A Atomic mass abundance% t1/2

C 6 12 12.00000 98.93006 13 13.00335 1.07006 14 14.00324 5730 y

N 7 14 14.00307 99.63207 15 15.00011 0.3680

O 8 16 15.99491 99.75708 17 16.99913 0.03808 18 17.99916 0.2050

Cl 17 35 34.96885 75.780017 37 36.96590 24.2200

30

Boron (Z=5)

Isotope composition

19.8 % 10B m = 10.013 u

80.2 % 11B m = 11.009 u

(0.198 10.013 u) + (0.802 11.009 u) = 10.811 uThe last digit is uncertain due to the variable isotope compositionof boron in the earth crust (±0.007 u)

Definition:An element consists of the mixture of its isotopes in the earth crust

The relative atomic mass (atomic weight) of the boron element:


The Periodic tableAlkali Metals

Alkaline Earths

Transition Metals

Halogens

Noble Gases

Lanthanides and Actinides

Main Group

Main Group


The Periodic Table

• Read relative atomic masses.• Read the ions formed by main group elements.• Read the electron configuration.• Learn trends in physical and chemical properties.

We will discuss these in detail in Chapter 10.


The Mole

• Physically counting atoms is impossible.• We must be able to relate measured mass to

numbers of atoms.– buying nails by the pound.– using atoms by the gram


Avogadro’s number

The mole is an amount of substance that contains the same number of elementary entities as there are carbon-12 atoms in exactly 12 g of carbon-12.

NA = 6.02214199 x 1023 mol-1


Molar Mass

• The molar mass, M, is the mass of one mole of a substance.

M (g/mol 12C) = A (g/atom 12C) x NA (atoms 12C /mol 12C)


Combining Several Factors in a Calculation—Molar Mass, the Avogadro Constant, Percent Abundance.

Potassium-40 is one of the few naturally occurring radioactive isotopes of elements of low atomic number. Its percent natural abundance among K isotopes is 0.012%. How many 40K atoms do you ingest by drinking one cup of whole milk containing 371 mg of K?

Want atoms of 40K, need atoms of K,

Want atoms of K, need moles of K,

Want moles of K, need mass and M(K).

Example 2-9


Convert strategy to plan

mK(mg) x (1g/1000mg) mK (g) x 1/MK (mol/g) nK(mol)

Convert mass of K(mg K) into moles of K (mol K)

Convert moles of K into atoms of 40K

nK(mol) x NA atoms K x 0.012% atoms 40K

nK = (371 mg K) x (10-3 g/mg) x (1 mol K) / (39.10 g K)

= 9.49 x 10-3 mol K

and plan into action

atoms 40K = (9.49 x 10-3 mol K) x (6.022 x 1023 atoms K/mol K)

x (1.2 x 10-4 40K/K)= 6.9 x 1017 40K atoms


Chapter 2 Questions

3, 4, 11, 22, 33, 51, 55, 63, 83.

prentice-hall © 2002 general chemistry: chapter 2 slide 1 of 25 chapter 2: atoms and the atomic...

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