Dalton’s Atomic Theory
• Elements - made up of atoms
• Same elements, same atoms.
• Different elements, different
atoms.
• Chemical reactions involve
bonding of atoms
Law of Definite Composition
• A compound always
contains the same
proportion of elements by
mass
Law of Multiple Proportions
• Compounds form from
specific combinations of
atoms
• H2O
vs
H2O2
Chemical Bonds
• Holds compounds together
• Need to be broken for
chemical and physical
changes to occur
The Atom
• Made up of:
–Protons – (+) charged
–Electrons – (-) charged
–neutrons
Periodic Table
•
•
•
•
•
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
Charges
• When elements combine, they have to be in
the form of IONS.
• Cations and anions combine to form
compounds.
• For a neutral compound, the sum of the
charges must be ZERO.
• For a polyatomic ion, the sum of the
charges must equal the charge of the ION.
Examples
• In CO2, the charge of C is + 4
• In CO, the charge of C is +2.
• In KMnO4, since the charge of K is +1, O is
-2 so -2 x 4 = -8, Mn must be +7.
• In (PO4)3-, the charge of O is -2, so -2 x 4=
-8, then P must have a charge of +5, so the
sum when the charges are added will be -3.
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 the masses of 12C, 13C and
14C.
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.)
Periodic Table
• Mendeleev – arranged
elements in the (.) table
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
–For example, the mass of C
= 12.01 a.m.u is the average
of the masses of 12C, 13C and
14C.
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
Rule 1 – IONIC COMPOUNDS
• Metals w/ Fixed Oxidation States
–Name metal or first element as
is
- Anion always ends in “–ide”
Terminal element or anion
•
•
•
•
•
ON SF Br -
oxide
nitride
sulfide
fluoride
bromide
P - phosphide
Se - selenide
Cl - chloride
I - iodide
C - carbide
Note
• Only elements that come directly
from the periodic table WILL
end in –IDE.
• POLYATOMIC IONS will be
named AS IS.
Name the following:
• CaO • NaCl • MgO • CaS • Na3N -
Answers:
• CaO • NaCl • MgO • CaS • Na3N -
calcium oxide
sodium chloride
magnesium oxide
calcium sulfide
sodium nitride
Where do the subscripts come
from?
• Answer:
From the oxidation states
of the ions.
• Remember: Ions are the species that
combine.
• Target:
Compounds! (No charges!)
Second Rule
• II. Ionic Compounds - Metals with no
fixed oxidation states (Transition
Metals) except for Ag, Zn and Al
• Metal(Roman #) + 1st syllable + ide
– Use Roman numerals after the metal
to indicate oxidation state
Name the following:
•
•
•
•
Copper (I) sulfide
Iron (II) oxide
Tin (II) iodide
Iron (III) nitride
Answers:
•
•
•
•
Copper (I) sulfide
Iron (II) oxide
Tin (II) iodide
Iron (III) nitride
Cu2S
FeO
SnI2
FeN
What about…….?
•
•
•
•
•
•
Cesium hydroxide
Iron (III) acetate
Lithium phosphate
Aluminum Sulfite
Lead (II) sulfate
Silver nitrate
POLYATOMIC IONS
• Consist of more than 1 element.
• Have charges.
• Ex. SO4 2-, SO3 2-, PO4 3-,PO3 3-
Rule 3 – Covalent Compounds
• III. For Non-metals (grps IV, V, VI VII),
use prefixes.
Mono – 1
Hepta - 7
Di - 2
Octa - 8
Tri – 3
Nona - 9
Tetra – 4
Deca - 10
Penta – 5
Hexa - 6
Rule 3 – Covalent Compounds
(only have Non- Metals)
• Name 1st element as is. Use
prefix, if necessary.
• Prefix + 1st element + prefix + 1st
syllable of anion + ide
Name the following compounds
•
•
•
•
•
•
CO2 - carbon dioxide
N2O – dinitrogen oxide
SO3 – sulfur trioxide
N2O5 – dinitrogen pentoxide
P2S5 – diphosphorus pentasulfide
CO – carbon monoxide
Naming Acids
• I.
Acids without Oxygen
–Use hydro + 1st syllable + “- ic acid”
• Example: HCl = hydrochloric acid
HCN = hydrocyanic acid
HBr = hydrobromic acid
II. Acids with oxygen
• Polyatomic “ate” converts to “ic” + acid
• Polyatomic “ite” converts to “ous” + acid
- H2SO3
– H2SO4
– HNO3
– HNO2
– H3PO4
sulfurous acid
sulfuric acid
nitric acid
nitrous acid
phosphoric acid
Trick!
• If anion ends in “ – ate”, acid ends in “ –
ic”
• Example:
• HClO4 perchlorate
perchloric acid
• HClO3 chlorate
chloric acid
Trick!
• If anion ends in “ – ite”, acid ends in “ –
ous”
• Example:
• HClO2 chlorite
• HClO
hypochlorite
chlorous acid
hypochlorous
acid
Name the following:
• HBrO4
• HBrO3
• HBrO2
• HBrO
(perbromate)
(bromate)
(bromite)
(hypobromite)
Fundamental laws
• Law of Conservation of
Mass
• Mass is neither created or
destroyed
• Conversion from one form to
another
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
%
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.)
Chemical Formula
• Gives the combining whole
number ratios of the
elements in a compound
• C6H12O6
Structural Formula
• Gives the spatial
arrangement of atoms in
the compound
• Structural formula for H2O
is H – O – H
Empirical Formula
• Only gives the types of
elements in the compound
and the ratio of the
elements in the formula
Empirical Formula
• Does not tell exactly how
many of the elements are in
the compound
Molecular Formula
• Gives you the exact elemental
composition of the compound
• Formula of the compound as
it would actually exist.
EF vs. MF
Sucrose or table sugar:
Molecular Formula = C6H12O6
Empirical Formula = CH2O
Sample Problem
• The compound adrenaline
contains
% C = 56.79
% H = 6.56
% O = 28.37
% N = 8.28
by mass. Find the empirical
formula.
Empirical Formula
• EF Determination when %
Masses are given
Steps to Solve for EF
• Step 1: Sum up all given percentages. If
total equals 100%, go to step 2. If total
does not equal 100, the missing % is due to
one of the component elements.
• Step 2: Convert Mass % to grams.
• Step 3: Calculate moles using mole =
gram/molar mass
•
Empirical Formula
• Step 4. To get simplest ratios, divide
the moles calculated by the smallest
calculated mole. You must have a
ration of 1 for at least one of the
element. (Follow rule for rounding).
• Step 5. You now have the ratios or
subscripts for the EF.
Molecular Formula Detn.
Step 1. Obtain empirical formula
mass by adding atomic masses of
all elements in empirical formula
Molecular Formula Detn.
Step 2. Get ratio by applying
the formula below:
Molecular Formula =
given molar mass
Empirical formula mass
Molecular Formula Detn.
Step 3.
Multiply empirical formula
subscripts by obtained
ratio
Sample Problem
• Caffeine, a stimulant found in coffee,
contains 49.5 % C, 5.15% H, 28.9 %
N, and 16.5 % O by mass. The
molar mass of the compound is 195
g/mol. Determine the empirical and
molecular formula of caffeine.
Sample Problem
• Ibuprofen, a headache remedy,
contains 75.69 % C, 8.80% H, and
15.51 % O by mass. The molar mass
of the compound is 206 g/mol.
Determine the empirical and
molecular formula of ibuprofen.