Dalton’s Atomic Theory
• Elements - made up of atoms
• Same elements, same atoms.
• Different elements, different
atoms.
• Chemical reactions involve
bonding of atoms
The Atom
• Made up of:
–Protons – (+) charged
–Electrons – (-) charged
–neutrons
Periodic Table
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Alkaline Metals – Grps. I & II
Transition Metals
Non-metals
Halogens – Group VII
Noble Gases –Group VIII - little
chemical activity
Periodic Table
• Atomic Mass - # at bottom
• how much element weighs
• Atomic Number - # on top
• gives # protons = # electrons
Periodic Table
• Atomic Mass
–number below the element
–not whole numbers because
the masses are averages of the
masses of the different
isotopes of the elements
Ions
• Are charged species
• Result when elements gain
electrons or lose electrons
2 Types of Ions
• Anions – (-) charged
• Example:
F-
• Cations – (+) charged
• Example:
Na+
Highly Important!
• Gain of electrons makes
element (-) = anion
• Loss of electrons makes
element (+) = cation
Isotopes
• Are atoms of a given element
that differ in the number of
neutrons and consequently in
atomic mass.
Example
Isotopes
12C
13C
14C
11C
% Abundance
98.89 %
1.11 %
–For example, the mass of C =
12.01 a.m.u is the average of
12
13
14
the masses of C, C and C.
Determination of Aver. Mass
• Ave. Mass =
[(% Abund./100) (atomic
mass)] + [(% Abund./100)
(atomic mass)]
Take Note:
• If there are more than 2
isotopes, then formula has to
be re-adjusted
Sample Problem 1
• Assume that element Uus is synthesized
and that it has the following stable
isotopes:
– 284Uus (283.4 a.m.u.)
34.6 %
– 285Uus (284.7 a.m.u.)
21.2 %
– 288Uus (287.8 a.m.u.)
44.20 %
Solution
• Ave. Mass of Uus =
• [284Uus]
(283.4 a.m.u.)(0.346)
• [285Uus] +(284.7 a.m.u.)(0.212)
• [288Uus] +(287.8 a.m.u.)(0.4420)
• = 97.92 + 60.36 + 127.21
• = 285.49 a.m.u (FINAL ANS.)
Oxidation Numbers
• Is the charge of the ions (elements in their
ion form)
• Is a form of electron accounting
• Compounds have total charge of zero
(positive charge equals negative charge)
Oxidation States
• Are the partial charges of the
ions. Some ions have more than
one oxidation states.
Oxidation States
• - generally depend upon the how
the element follows the octet
rule
• Octet Rule – rule allowing
elements to follow the noble gas
configuration
Nomenclature
• - naming of compounds
Periodic Table
• Rows (Left to Right) - periods
• Columns (top to bottom) groups
Determination of Aver. Mass
• Ave. Mass =
[(% Abund./100) (atomic
mass)] + [(% Abund./100)
(atomic mass)]
Sample Problem 1
• Assume that element Uus is
synthesized and that it has the
following stable isotopes:
–284Uus (283.4 a.m.u.)
34.6 %
–285Uus (284.7 a.m.u.)
21.2 %
288
– Uus (287.8 a.m.u.)
44.20
%
Solution
• Ave. Mass of Uus =
• [284Uus]
(283.4 a.m.u.)(0.346)
• [285Uus] +(284.7 a.m.u.)(0.212)
• [288Uus] +(287.8 a.m.u.)(0.4420)
• = 97.92 + 60.36 + 127.21
• = 285.49 a.m.u (FINAL ANS.)
STOICHIOMETRY
–For example, the mass of C =
12.01 a.m.u is the average of
12
13
14
the masses of C, C and C.
Formula Weight & Molecular Weight
• The FORMULA WEIGHT of a compound
equals the SUM of the atomic masses of the
atoms in a formula.
• If the formula is a molecular formula, the
formula weight is also called the MOLECULAR
WEIGHT.
The MOLE
• Amount of substance that
contains an Avogadro’s
number (6.02 x 10 23)of
formula units.
The MOLE
• The mass of 1 mole of atoms,
molecules or ions = the formula
weight of that element or
compound in grams. Ex. Mass of 1
mole of water is 18 grams so molar
mass of water is 18 grams/mole.
Formula for Mole
Mole = mass of element
formula weight of
element
Sample Mole Calculations
1 mole of C = 12.011 grams
»
12.011 gm/mol
• 0.5 mole of C = 6.055 grams
»
12.011 gm/mol
Avogadro’s Number
• Way of counting atoms
• Avogadro’s number
23
= 6.02 x 10
Point to Remember
One mole of anything is 6.02
23
x 10 units of that
substance.
And……..
• 1 mole of C has the same
number of atoms as one mole
of any element
Formula Weight & Molecular Weight
• The FORMULA WEIGHT of a compound
equals the SUM of the atomic masses of the
atoms in a formula.
• If the formula is a molecular formula, the
formula weight is also called the MOLECULAR
WEIGHT.
Summary
• Avogadro’s Number gives the
number of particles or atoms in a
given number of moles
• 1 mole of anything = 6.02 x 10
atoms or particles
23
Sample Problem 2
• Compute the number of
atoms and moles of atoms in
a 10.0 gram sample of
aluminum.
Solution
• PART I:
• Formula for Mole:
–Mole = mass of element
atomic mass of element
Solution (cont.)
• Part II:
atoms
To determine # of
• # atoms = moles x Avogadro’s
number
Problem # 2
• A diamond contains 5.0 x 1021
atoms of carbon. How many
moles of carbon and how
many grams of carbon are in
this diamond?
Molar Mass
• Often referred to as molecular mass
–Unit = gm/mole
• Definition:
–mass in grams of 1 mole of the
compound
Example Problem
• Determine the Molar Mass of
C6H12O6
Solution
• Mass of 6 mole C = 6 x 12.01 = 72.06 g
• Mass of 12 mole H = 12 x 1.008 = 12.096 g
• Mass of 6 mole O = 6 x 16
= 96 g
• Mass of 1 mole C6H12O6
180.156 g
=
Problem #3
• What is the molar mass of
(NH4)3(PO4)?
Molar Mass
• Often referred to as molecular mass
–Unit = gm/mole
• Definition:
–mass in grams of 1 mole of the
compound
Sample Problem
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•
•
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•
•
•
•
•
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Given 75.99 grams of (NH4)3(PO4), determine the ff:
1. Molar mass of the compound
2. # of moles of the compound
3. # of molecules of the compound
4. # of moles of N
5. # of moles of H
6. # of moles of O
7. # of atoms of N
8. # of atoms of H
9. # of atoms of O