chapter 2
DESCRIPTION
Chapter 2. Atoms, Molecules & Ions. Quantum Corral. http://www.almaden.ibm.com/vis/stm/corral.html. Scanning Tunneling Microscope. Scanning Tunneling Microscope. Scanning Tunneling Microscope. http://www.cbu.edu/~mcondren/SeeAtoms.htm. http://mrsec.wisc.edu/. http://mrsec.wisc.edu/. - PowerPoint PPT PresentationTRANSCRIPT
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Dr. S. M. Condren
Atoms, Molecules& Ions
Chapter 2
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Dr. S. M. Condren
Quantum Corral
http://www.almaden.ibm.com/vis/stm/corral.html
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Dr. S. M. Condren
Scanning Tunneling Microscope
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Dr. S. M. Condren
Scanning Tunneling Microscope
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Dr. S. M. Condren
Scanning Tunneling Microscope
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Dr. S. M. Condren
http://www.cbu.edu/~mcondren/SeeAtoms.htm
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Dr. S. M. Condren
http
://m
rsec
.wis
c.ed
u/
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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Dr. S. M. Condren
Pull Probe StripProbe
Sample
Pull Probe Strip
http://ww
w.nsf.gov/m
ps/dmr/m
rsec.htm
http://www.nsf.gov/mps/dmr/mrsec.htm
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Dr. S. M. Condren
(a) (b)
North South
(c)
Which best represents the poles?
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Dr. S. M. Condren
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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Dr. S. M. Condren
Law of Constant Composition
A chemical compound always contains the same elements in the same proportions by mass.
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Dr. S. M. Condren
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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Dr. S. M. Condren
Dalton’s Atomic Theory
Postulates• proposed in 1803• know at least 2 for first exam
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Dr. S. M. Condren
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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Dr. S. M. Condren
Dalton’s Atomic Theory
Postulate 2• Atoms of different elements have different
properties.
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Dr. S. M. Condren
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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Dr. S. M. Condren
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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Dr. S. M. Condren
Radioactivity
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Dr. S. M. Condren
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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Dr. S. M. Condren
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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Dr. S. M. Condren
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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Dr. S. M. Condren
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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Dr. S. M. Condren
Cathode Ray Tube
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Dr. S. M. Condren
Thompson’s Charge/Mass Ratio
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Dr. S. M. Condren
Millikin’s Oil Drop
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Dr. S. M. Condren
Rutherford’s Gold Foil
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Dr. S. M. Condren
Rutherford’s Model of the Atom
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Dr. S. M. Condren
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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Dr. S. M. Condren
Structure of the Atom Composed of:
• protons• neutrons• electrons
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Dr. S. M. Condren
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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Dr. S. M. Condren
Structure of the Atom
Composed of:• protons• neutrons• electrons
• neutrons– found in nucleus– neutral charge– relative mass of 1.0087 amu
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Dr. S. M. Condren
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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Dr. S. M. Condren
Size of Nucleus
If the nucleus were1” in diameter,
the atom would be 1.5 miles in diameter.
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Dr. S. M. Condren
Ions
• charged single atom• charged cluster of atoms
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Dr. S. M. Condren
Ions
• cations– positive ions
• anions– negative ions
• ionic compounds– combination of cations and anions– zero net charge
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Dr. S. M. Condren
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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Dr. S. M. Condren
Isotopes
• atoms of the same element which differ in the number of neutrons in the nucleus
• designated by mass number
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Dr. S. M. Condren
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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Dr. S. M. Condren
Masses of Atoms
Carbon-12 Scale
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Dr. S. M. Condren
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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Dr. S. M. Condren
Isotopes of Hydrogen H-2 or D, 2H, deuterium
• 1 proton and 1 neutron in nucleus
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Dr. S. M. Condren
Isotopes of Hydrogen H-3 or T, 3H, tritium
• 1 proton and 2 neutrons in nucleus
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Dr. S. M. Condren
Isotopes of Oxygen
O-16• 8 protons, 8 neutrons, & 8 electronsO-17• 8 protons, 9 neutrons, & 8 electronsO-18• 8 protons, 10 neutrons, & 8 electrons
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Dr. S. M. Condren
The radioactive isotope 14C has how many neutrons?
6, 8, other
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Dr. S. M. Condren
The identity of an element is determined by the number of which particle?
protons, neutrons, electrons
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Dr. S. M. Condren
Mass Spectrometer
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Dr. S. M. Condren
Mass Spectra of Neon
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Dr. S. M. Condren
Measurement of Atomic Masses
Mass Spectrometer
a simulation is available athttp://www.colby.edu/chemistry/OChem/DEMOS/MassSpec.html
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Dr. S. M. Condren
Atomic Masses andIsotopic Abundances
natural atomic masses =sum[(atomic mass of isotope)
*(fractional isotopic abundance)]
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Dr. S. M. Condren
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) = 35.453Thus:34.96885*x + 36.96590*y = 35.453
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Dr. S. M. Condren
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
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Dr. S. M. Condren
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
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Dr. S. M. Condren
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 = 35.45334.96885*x + 36.96590*(1-x) = 35.453
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Dr. S. M. Condren
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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Dr. S. M. Condren
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 = 35.453
34.96885*x + 36.96590*(1-x) = 35.453
(34.96885 - 36.96590)x + 36.96590 = 35.453
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Dr. S. M. Condren
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(34.96885 - 36.96590)x = (35.453 - 36.96590)
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Dr. S. M. Condren
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 = 35.453
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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Dr. S. M. Condren
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 = 35.453
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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Dr. S. M. Condren
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
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Dr. S. M. Condren
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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Dr. S. M. Condren
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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Dr. S. M. Condren
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 TableNewlands - English
1864 - Law of Octaves - every 8th element has similar
properties
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Development of Periodic TableDmitri Mendeleev - Russian
1869 - Periodic Law - allowed him to predict properties of
unknown elements
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Mendeleev’s Periodic Tablethe 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 Jeffreys1887–1915, 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 Tablethe 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 20.18311 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 39.94819 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 * 145 150.36 151.96 157.25 158.93 162.51 164.93 167.26 168.93 173.04 174.97radioactive 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 theElementsPeriodic 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 NamesGroup IA alkali metalsGroup IIA alkaline earth metalsGroup VIIA halogensGroup VIIIAnoble gasestransition metalsinner transition metals• lanthanum series rare earths• actinium series trans-uranium series
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Types of Elementsmetals
nonmetalsmetalloids - semimetals
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Elements, Compounds, and Formulas
Elements• can exist as single atoms or moleculesCompounds• 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
waterNH3
ammoniaN2O
nitrous oxideCO
carbon monoxideCS2
carbon disulfide
SO3
sulfur trioxideCCl4
carbon tetrachloridePCl5
phosphorus pentachlorideSF6
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 BondingCharacteristics 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 IonsMonatomic IonsGroup IA => +1Group IIA => +2
Maximum positive valenceequals
Group A #
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Valence of Non-Metal IonsMonatomic IonsGroup VIA => -2Group 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
+
perchlorate ClO41-
cyanide CN1-
hydroxide OH1-
nitrate NO31-
sulfate SO42-
carbonate CO32-
phosphate PO43-
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Names of Ionic Compounds1. 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 chlorideFe2O3
iron(III) oxideN2O4
dinitrogen tetroxide
KIpotassium iodide
Mg3N2
magnesium nitrideSO3
sulfur trioxide
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NomenclatureNH4NO3
ammonium nitrateKClO4
potassium perchlorateCaCO3
calcium carbonateNaOH
sodium hydroxide
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Nomenclature DrillAvailable for PCs:
– on your disk to use at home or in the dorm
– in the Chemistry Resource Center– off the web under Chapter 2, Links
http://www.cbu.edu/~mcondren/c115lkbk.html
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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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ExampleHydrogen gas reacts with oxygen gas to
produce water.Step 1.hydrogen + oxygen -----> waterStep 2.H2 + O2 -----> H2O
Step 3.2 H2 + O2 -----> 2 H2O
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ExampleIron(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