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