HSTMr.WatsonDr. S. M. Condren

Atoms, Molecules

& Ions

Chapter 2

HST

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Quantum Corral

http://www.almaden.ibm.com/vis/stm/corral.html

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Scanning Tunneling Microscope

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Scanning Tunneling Microscope

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Scanning Tunneling Microscope

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HSTMr.Watsonht

tp:/

/mrs

ec.w

isc.

edu/

http://mrsec.wisc.edu/

Developed in collaboration with theInstitute for Chemical Education and the

Magnetic Microscopy CenterUniversity of Minnesota

http://www.physics.umn.edu/groups/mmc/

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Pull Probe StripProbe

Sample

Pull Probe Strip

http://w

ww

.nsf.g

ov/m

ps/dm

r/mrse

c.htm

http://www.nsf.gov/mps/dmr/mrsec.htm

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(a) (b)

North South

(c)

Which best represents the poles?

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Atoms & MoleculesAtoms

can exist alone or enter into chemical combination

the smallest indivisible particle of an element

Molecules

a combination of atoms that has its own characteristic set of properties

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Law of Constant Composition

A chemical compound always contains the same elements in the same proportions by mass.

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Law of Multiple Proportions

the same elements can be combined to form different compounds by combining the elements in different proportions

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Dalton’s Atomic Theory

Postulates

proposed in 1803

know at least 2 for first exam

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Dalton’s Atomic Theory

Postulate 1

An element is composed of tiny particles called atoms.

All atoms of a given element show the same chemical properties.

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Dalton’s Atomic Theory

Postulate 2Atoms of different elements have different properties.

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Dalton’s Atomic Theory

Postulate 3

Compounds are formed when atoms of two or more elements combine.

In a given compound, the relative number of atoms of each kind are definite and constant.

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Dalton’s Atomic Theory

Postulate 4

In an ordinary chemical reaction, no atom of any element disappears or is changed into an atom of another element.

Chemical reactions involve changing the way in which the atoms are joined together.

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Radioactivity

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Radioactivity

Alpha – helium-4 nucleus

Beta – high energy electron

Gamma – energy resulting from transitions from one nuclear energy level to another

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Alpha Radiation

composed of 2 protons and 2 neutrons

thus, helium-4 nucleus

+2 charge

mass of 4 amu

creates element with atomic number 2 lower

Ra226 Rn222 + He4()

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Beta Radiation

composed of a high energy electron which was ejected from the nucleus

“neutron” converted to “proton”

very little mass

-1 charge

creates element with atomic number 1 higher

U239 Np239 + -1

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Gamma Radiation

nucleus has energy levels

energy released from nucleus as the nucleus changes from higher to lower energy levels

no mass

no charge

Ni60* Ni60 +

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Cathode Ray Tube

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Thompson’s Charge/Mass Ratio

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Millikin’s Oil Drop

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Rutherford’s Gold Foil

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Rutherford’s Model of the Atom

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Rutherford’s Model of the Atom

atom is composed mainly of vacant space

all the positive charge and most of the mass is in a small area called the nucleus

electrons are in the electron cloud surrounding the nucleus

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Structure of the Atom Composed of:

protons

neutrons

electrons

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Structure of the Atom

Composed of:

protons

neutrons

electrons

protons– found in nucleus– relative charge of +1– relative mass of 1.0073 amu

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Structure of the Atom

Composed of:

protons

neutrons

electronsneutrons

– found in nucleus– neutral charge– relative mass of 1.0087 amu

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Structure of the Atom

Composed of:

protons

neutrons

electrons

electrons– found in electron cloud– relative charge of -1– relative mass of 0.00055 amu

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Size of Nucleus

If the nucleus were1” in diameter,

the atom would be 1.5 miles in diameter.

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Ions

charged single atom

charged cluster of atoms

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Ions

cations– positive ions

anions– negative ions

ionic compounds– combination of cations and anions– zero net charge

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Atomic number, Z

the number of protons in the nucleus

the number of electrons in a neutral atom

the integer on the periodic table for each element

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Isotopes

atoms of the same element which differ in the number of neutrons in the nucleus

designated by mass number

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Mass Number, A

integer representing the approximate mass of an atom

equal to the sum of the number of protons and neutrons in the nucleus

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Masses of Atoms

Carbon-12 Scale

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Isotopes of Hydrogen H-1, 1H, protium

1 proton and no neutrons in nucleus

only isotope of any element containing no neutrons in the nucleus

most common isotope of hydrogen

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Isotopes of Hydrogen H-2 or D, 2H, deuterium

1 proton and 1 neutron in nucleus

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Isotopes of Hydrogen H-3 or T, 3H, tritium

1 proton and 2 neutrons in nucleus

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Isotopes of Oxygen

O-16

8 protons, 8 neutrons, & 8 electrons

O-17

8 protons, 9 neutrons, & 8 electrons

O-18

8 protons, 10 neutrons, & 8 electrons

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The radioactive isotope 14C has how many neutrons?

6, 8, other

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The identity of an element is determined by the number of which particle?

protons, neutrons, electrons

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Mass Spectrometer

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Mass Spectra of Neon

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Measurement of Atomic Masses

Mass Spectrometer

a simulation is available at

http://www.colby.edu/chemistry/

OChem/DEMOS/MassSpec.html

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Atomic Masses andIsotopic Abundances

natural atomic masses =

sum[(atomic mass of isotope)

*(fractional isotopic abundance)]

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?

let x = fraction Cl-35 y = fraction Cl-37

x + y = 1 y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

Thus:34.96885*x + 36.96590*y = 35.453

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?

let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?

let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?

let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35 y = fraction Cl-37x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.45334.96885*x + 36.96590*(1-x) = 35.453(34.96885 - 36.96590)x + 36.96590 =

35.453

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?

let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

- 1.99705x = - 1.5129

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

- 1.99705x = - 1.5129

1.99705x = 1.5129

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35

y = fraction Cl-37

x + y = 1 <=> y = 1 - x

(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.453

34.96885*x + 36.96590*y = 35.453

34.96885*x + 36.96590*(1-x) = 35.453

(34.96885 - 36.96590)x + 36.96590 = 35.453

(34.96885 - 36.96590)x = (35.453 - 36.96590)

- 1.99705x = - 1.5129

1.99705x = 1.5129

x = 0.7553 <=> 75.53% Cl-35

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35 y = fraction Cl-37x + y = 1 <=> y = 1 - x(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.45334.96885*x + 36.96590*y = 35.45334.96885*x + 36.96590*(1-x) = 35.453(34.96885 - 36.96590)x + 36.96590 = 35.453(34.96885 - 36.96590)x = (35.453 - 36.96590)- 1.99705x = - 1.5129

1.99705x = 1.5129 x = 0.7553 <=> 75.53% Cl-35

y = 1 - x

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35 y = fraction Cl-37x + y = 1 <=> y = 1 - x(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.45334.96885*x + 36.96590*y = 35.45334.96885*x + 36.96590*(1-x) = 35.453(34.96885 - 36.96590)x + 36.96590 = 35.453(34.96885 - 36.96590)x = (35.453 - 36.96590)- 1.99705x = - 1.5129

1.99705x = 1.5129 x = 0.7553 <=> 75.53% Cl-35

y = 1 - x = 1.0000 - 0.7553

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?let x = fraction Cl-35 y = fraction Cl-37x + y = 1 <=> y = 1 - x(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.45334.96885*x + 36.96590*y = 35.45334.96885*x + 36.96590*(1-x) = 35.453(34.96885 - 36.96590)x + 36.96590 = 35.453(34.96885 - 36.96590)x = (35.453 - 36.96590)- 1.99705x = - 1.5129

1.99705x = 1.5129 x = 0.7553 <=> 75.53% Cl-35

y = 1 - x = 1.0000 - 0.7553 = 0.2447

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Example: Chlorine has two isotopes, Cl-35 and Cl-37, which have masses of 34.96885 and 36.96590 amu, respectively. The natural atomic mass of chlorine is 35.453 amu. What are the percent abundances of the two isotopes?

let x = fraction Cl-35 y = fraction Cl-37x + y = 1 <=> y = 1 - x(AW Cl-35)(fraction Cl-35) + (AW Cl-37)(fraction Cl-37) = 35.45334.96885*x + 36.96590*y = 35.45334.96885*x + 36.96590*(1-x) = 35.453(34.96885 - 36.96590)x + 36.96590 = 35.453(34.96885 - 36.96590)x = (35.453 - 36.96590)- 1.99705x = - 1.5129

1.99705x = 1.5129 x = 0.7553 <=> 75.53% Cl-35

y = 1 - x = 1.0000 - 0.7553 = 0.2447 24.47% Cl-37

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Development of Periodic Table

Newlands - English

1864 - Law of Octaves - every 8th element has similar

properties

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Development of Periodic Table

Dmitri Mendeleev - Russian

1869 - Periodic Law - allowed him to predict properties of

unknown elements

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Mendeleev’s Periodic Table

the elements are arranged according to increasing atomic weights

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Missing elements: 44, 68, 72, & 100 amu

Mendeleev’s Periodic Table

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Properties of Ekasilicon

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Modern Periodic TableMoseley, Henry Gwyn Jeffreys

1887–1915, English physicist.

Studied the relations among bright-line spectra of different elements.

Derived the ATOMIC NUMBERS from the frequencies of vibration of X-rays emitted by each element.

Moseley concluded that the atomic number is equal to the charge on the nucleus.

This work explained discrepancies in Mendeleev’s Periodic Law.

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Modern Periodic Table

the elements are arranged according to increasing atomic numbers

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I A II A III B IV B V B VI B VII B VIII B I B II B III A IV A V A VI A VII A VIII A1 1 2

1 H H He1.008 1.008 4.0026

3 4 5 6 7 8 9 10

2 Li Be B C N O F Ne6.939 9.0122 10.811 12.011 14.007 15.999 18.998 20.183

11 12 13 14 15 16 17 18

3 Na Mg Al Si P S Cl Ar22.99 24.312 26.982 28.086 30.974 32.064 35.453 39.948

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr39.102 40.08 44.956 47.89 50.942 51.996 54.938 55.847 58.932 58.71 63.54 65.37 69.72 72.59 74.922 78.96 79.909 83.8

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe85.468 87.62 88.906 91.224 92.906 95.94 * 98 101.07 102.91 106.42 107.9 112.41 114.82 118.71 121.75 127.61 126.9 131.29

55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

6 Cs Ba **La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn132.91 137.33 138.91 178.49 180.95 183.85 186.21 190.2 192.22 195.08 196.97 200.29 204.38 207.2 208.98 * 209 * 210 * 222

87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 118

7 Fr Ra ***Ac Rf Ha Sg Ns Hs Mt Uun Uuu Uub Uut Uuq Uup Uuh Uuo* 223 226.03 227.03 * 261 * 262 * 263 * 262 * 265 * 268 * 269 * 272 * 277 *284 *285 *288 *292 *294

Based on symbols used by ACS S.M.Condren 2007

58 59 60 61 62 63 64 65 66 67 68 69 70 71

* Designates that **Lanthanum Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Luall isotopes are Series 140.12 140.91 144.24 * 145 150.36 151.96 157.25 158.93 162.51 164.93 167.26 168.93 173.04 174.97

radioactive 90 91 92 93 94 95 96 97 98 99 100 101 102 103

*** Actinium Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Series 232.04 231.04 238.03 237.05 * 244 * 243 * 247 * 247 * 251 * 252 * 257 * 258 * 259 * 260

Periodic Table of theElements

Periodic Table of the ElementsPeriodic Table of the Elements

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Organization of Periodic Table

period - horizontal row

group - vertical column

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Family Names

Group IA alkali metalsGroup IIA alkaline earth metalsGroup VIIA halogensGroup VIIIAnoble gasestransition metalsinner transition metalslanthanum series rare earthsactinium series trans-uranium series

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Types of Elements

metals

nonmetals

metalloids - semimetals

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Elements, Compounds, and Formulas

Elements

can exist as single atoms or molecules

Compounds

combination of two or more elements

molecular formulas for molecular compounds

empirical formulas for ionic compounds

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Organic CompoundsOrganic Chemistry

branch of chemistry in which carbon compounds and their reactions are studied.

the chemistry of carbon-hydrogen compounds

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Inorganic Compounds Inorganic Chemistry

field of chemistry in which are studied the chemical reactions and properties of all the chemical elements and their compounds, with the exception of the hydrocarbons (compounds composed of carbon and hydrogen) and their derivatives.

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Molecular and Structural Formulas

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Bulk Substances

mainly ionic compounds– empirical formulas– structural formulas

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Models of Sodium Chloride

NaCl “table salt”

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How many atoms are in the formula Al2(SO4)3?

3, 5, 17

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Naming Binary Molecular Compounds

For compounds composed of two non-metallic elements, the more metallic element is listed first.

To designate the multiplicity of an element, Greek prefixes are used:

mono => 1; di => 2; tri => 3; tetra => 4; penta => 5; hexa => 6; hepta => 7; octa => 8

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Common CompoundsH2O

water

NH3

ammonia

N2O

nitrous oxide

CO

carbon monoxide

CS2

carbon disulfide

SO3

sulfur trioxide

CCl4carbon tetrachloride

PCl5phosphorus

pentachloride

SF6

sulfur hexafluoride

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Alkanes - CnH2n+2

methane - CH4

ethane - C2H6

propane - C3H8

butanes - C4H10

pentanes - C5H12

hexanes - C6H14

heptanes - C7H16

octanes - C8H18

nonanes - C9H20

decanes - C10H22

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Burning of Propane Gas

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Butanes

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Ionic Bonding

Characteristics of compounds with ionic bonding:

non-volatile, thus high melting points

solids do not conduct electricity, but melts (liquid state) do

many, but not all, are water soluble

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Ion Formation

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ValanceCharge on Ions

compounds have electrical neutrality

metals form positive monatomic ions

non-metals form negative monatomic ions

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Valence of Metal Ions

Monatomic Ions

Group IA => +1

Group IIA => +2

Maximum positive valence

equals

Group A #

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Valence of Non-Metal Ions

Monatomic Ions

Group VIA => -2

Group VIIA => -1

Maximum negative valence

equals

(8 - Group A #)

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Charges of Some Important Ions

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Polyatomic Ions

more than one atom joined together

have negative charge except for NH4+ and

its relatives

negative charges range from -1 to -4

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Polyatomic Ionsammonium NH4

+

perchlorateClO41-

cyanide CN1-

hydroxide OH1-

nitrate NO31-

sulfate SO42-

carbonate CO32-

phosphate PO43-

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Names of Ionic Compounds

1. Name the metal first.

If the metal has more than one oxidation state, the oxidation state is specified by Roman numerals in parentheses.

2. Then name the non-metal,

changing the ending of the non-metal to

-ide.

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NomenclatureNaCl

sodium chloride

Fe2O3

iron(III) oxide

N2O4

dinitrogen tetroxide

KI

potassium iodide

Mg3N2

magnesium nitride

SO3

sulfur trioxide

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NomenclatureNH4NO3

ammonium nitrate

KClO4

potassium perchlorate

CaCO3

calcium carbonate

NaOH

sodium hydroxide

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Nomenclature Drill

Available for PCs:–

http://science.widener.edu/svb/pset/nomen_b.html

– in the Chemistry Resource Center– , Links

http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry

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How many moles of ions are there per mole of Al2(SO4)3?

2, 3, 5

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Chemical Equation

reactants

products

coefficients

reactants -----> products

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Writing and BalancingChemical Equations

Write a word equation.

Convert word equation into formula equation.

Balance the formula equation by the use of prefixes (coefficients) to balance the number of each type of atom on the reactant and product sides of the equation.

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Example

Hydrogen gas reacts with oxygen gas to produce water.

Step 1.

hydrogen + oxygen -----> water

Step 2.

H2 + O2 -----> H2O

Step 3.

2 H2 + O2 -----> 2 H2O

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Example

Iron(III) oxide reacts with carbon monoxide to produce the iron oxide (Fe3O4) and carbon dioxide.

iron(III) oxide + carbon monoxide -----> Fe3O4 + carbon dioxide

Fe2O3 + CO -----> Fe3O4 + CO2

3 Fe2O3 + CO -----> 2 Fe3O4 + CO2