 Protons
and electrons are attracted to
each other because of opposite charges
 Electrically charged particles moving in
a curved path give off energy
 Despite these facts, atoms don’t collapse
 Electrons and energy move in a wave
motion which contributes to the chemical
properties of an element
 Electromagnetic
radiation propagates
through space as a wave moving at the
speed of light.
c = 
C = speed of light, a constant (3.00 x 108 m/s)
 = frequency, in units of hertz (hz, sec-1)
 = wavelength, in meters
 The
wavelength of the green light from a
traffic signal is centered at 522nm. What
is the frequency of this radiation?
 3.0
x 108 m/s = v x (522nm x 1 m )
1
109 nm
 5.47 x 1014 1/s or 5.47 x 1014 hz
 The
energy (E ) of electromagnetic
radiation is directly proportional to the
frequency () of the radiation.
E = h
 E = Energy, in units of Joules
(kg·m2/s2)
 h = Planck’s constant (6.626 x 10-34 J·s)
  =frequency, in units of hertz (hz, sec1)
 Long Wavelength
= Low Frequency =
Low ENERGY
 Short Wavelength = High Frequency =
High ENERGY
C = 
E = hv
Common rearrangements:
E = hc

 = hc
E
Example

•
Calculate the energy (in joules) of
a) A photon with a wavelength of 5.00 x 10-4nm
(infrared region)
b) A photon with a wavelength of 5.0 x 10-2nm (x-ray
region)
 Physical
and chemical properties of
elements and compounds are dependent
upon their valence electron
configurations
 Regular variation of properties is
observed – called periodic law or
periodicity
 Elements in columns have similar outer
electron configurations and behaviors
 An
orbital is a region within an atom
where there is a probability of finding an
electron. This is a probability diagram for
the s orbital in the first energy level…
 Orbital
shapes are defined as the surface
that contains 90% of the total electron
probability.
“One cannot
simultaneously determine
both the position and
momentum of an electron.”



You can find out where the
electron is, but not where it
is going.
OR…
You can find out where the
electron is going, but not
where it is!
 Orbitals
of the same shape (s, for instance)
grow larger as n increases…
 Nodes
are regions of low probability within
an orbital.
 The
s orbital has a spherical shape
centered around the origin of the three
axes in space.
 There
are three dumbbell-shaped p
orbitals in each energy level above n =
1, each assigned to its own axis (x, y
and z) in space.
Things get a bit more
complicated with
the five d orbitals
that are found in the
d sublevels
beginning with n =
3. To remember the
shapes, think of:
“double dumbells”
…and a “dumbell
with a donut”!
 1s
 2s
2p
 3s 3p 3d
 4s 4p 4d 4f
 5s 5p 5d 5f
 6s 6p 6d 6f
 7s 7p 7d
 8s 8p 8d
Start at 1s, fill
diagonally from
lower left to
upper right
 Box
diagrams
for several
elements.
 What would the
configuration
notations and
the noble gas
notations be for
each element?
• Ex Write an electron configuration for a barium
atom (Z=56)
 Long, shorthand
 Write
an electron configuration for
molybdenum (Z=42)
• Long, shorthand
 Write
an electron configuration for a BROMIDE
ION (atomic number 35)
 Write possible configurations for ferrous and
ferric ions
 Chromium
steals a 4s electron to half
fill its 3d sublevel
 Copper steals a 4s electron to FILL its
3d sublevel
 Each
electron in an atom has a unique
set of 4 quantum numbers which
describe it.
• Principal quantum number
• Angular momentum quantum number
• Magnetic quantum number
• Spin quantum number
No two electrons in an atom can
have the same four quantum
numbers.
Wolfang Pauli
 Generally
symbolized by n, it denotes the
shell (energy level) in which the electron is
located.
 Number
2n2
of electrons that can fit in a shell:
 The
angular momentum quantum
number, generally symbolized by l,
denotes the orbital (subshell) in which
the electron is located.
l=0 l=1 l=2 l=3
s p
d
f
 The
magnetic quantum number,
generally symbolized by m, denotes
the orientation of the electron’s orbital
with respect to the three axes in space.
 The
three quantum numbers (n, l, and m)
are integers.
 The principal quantum number (n) cannot
be zero.
• n must be 1, 2, 3, etc.
 The
angular momentum quantum number (l
) can be any integer between 0 and n - 1.
• For n = 3, l can be either 0, 1, or 2.
 The
magnetic quantum number (ml) can be
any integer between -l and +l.
• For l = 2, m can be either -2, -1, 0, +1, +2.
 Spin
quantum number denotes the
behavior (direction of spin) of an
electron within a magnetic field.
 Possibilities for electron spin:
+1
-1
2
2
 Ex.
Complete the table for all electrons
in Neon, (Z=10)
• n
• l
• ml
• ms
 The
energy required to remove one mole
electrons from a mole of gaseous atoms to
produce one mole of gaseous ions
 M(g)  M+(g) + e Second ionization energy – energy change
accompanying
M+(g)  M2+(g) + e Measured in kJ/mol
 Positive energy values – requires energy to
remove electrons (endothermic)
 Magnitude
is determined by the
attraction of the positive nucleus for the
negative electrons that are being
removed
 Attraction is dependant on
• Nuclear energy
• Shielding effect of inner electrons
Mg + 738 kJ  Mg+ + eMg+ + 1451 kJ  Mg2+ + eMg2+ + 7733 kJ  Mg3+ + e-
 Tends
to increase across a period
• Electrons in the same quantum level do
not shield as effectively as electrons in
inner levels
• Irregularities at half filled and filled
sublevels due to extra repulsion of
electrons paired in orbitals, making them
easier to remove
 Tends
to decrease down a group
• Outer electrons are farther from the
nucleus
 The
energy change when one mole of
gaseous atoms gains one mole of
electrons to form one mole of gaseous
ions
X(g) + e-  X Measured in kJ/mol
 Positive
values indicates that energy
may have to be put in to cause this to
happen (endothermic)
 Negative values indicates that energy is
released in this process (exothermic)
 Affinity
 Affinity
tends to increase across a period
tends to decrease as you go down
in a period
• Electrons farther from the nucleus
experience less nuclear attraction
• Some irregularities due to repulsive
forces in the relatively small p orbitals
 Periodic Trends
in Atomic Radius
• Radius decreases from left to right
across a period
 Increased effective nuclear charge
due to decreased shielding
• Radius increases down a group
 Addition of principal quantum
levels
 Cations
• Positively charged ions
• Lose electrons
• Smaller than the corresponding atom
 Anions
• Negatively charged ions
• Gains electrons
• Larger than the corresponding atom
A
measure of the ability of an atom in
a chemical compound to attract
electrons
 Electronegativities tend to increase
across a period
 Electronegativities tend to decrease
down a group or remain the same
 Have
complete outer shells
 Stable, difficult to alter chemically
 Relatively inert
 Don’t form ions
 Group
1
• ns1 electron structure
• Lose s1 electron when they form ions (+1 charge)
 Group
2
• ns2 electron structure
• Lose 2 s electrons when they form ions (+2
charge)
 Group
16
• ns2 np4 electron structure
• Gains 2 electrons when they form ions (-2
charge)
 Group
17
• ns2 np5 electron structure
• Gains 1 electron when they form ions (-1 charge)
 Amu – used to describe the mass of atoms in whole numbers
 1 amu = 1.66 x 10-24g
 Atoms are very small and have extremely small masses
 Example – 1 atom of Cl35 contains 17 protons and 18
neutrons. This is a total of 35 amu and has a mass of 5.81 x
10-23g. This is a very small number so we use the concept of
the mole to overcome the problem of handling such small
quantities. As you will see 1 mole contains 6.022 x 1023
particles. So if we take 1 mole of Cl35 atoms they will have a
mass of (5.81 x 10-23)(6.022 x 1023) = 35.0g
 Chemical formulas are also molecular formulas
 Chemical formulas show exact ratios of the elements present
in a molecule
 Ratios are used as conversion factors
 Relative atomic mass (RAM) – weighted average of the
masses of all the atoms in a normal isotopic sample based
upon the scale on the last slide
 Average RAM is on the periodic table
 Relative atomic mass (molar mass)
• Found by adding all the individual RAMs together in one molecule of a
compound
 Relative formula mass (molar mass) – found by adding all the
individual RAMs together in one formula unit of an ionic
compound
Calculate the formula mass of magnesium carbonate, MgCO3.
1 dozen = 12
1 gross = 144
1 mole = 6.022 x 1023
There are exactly 12 grams of carbon-12 in one mole of carbon12.
6.022 x 1023 is called “Avogadro’s Number” in honor of the
Italian chemist Amadeo Avogadro (1776-1855).
I didn’t discover it. Its just named after
me!
number of moles = mass of sample
molar mass
Amadeo Avogadro
How many grams of lithium are in 3.50 moles of lithium?
Hint  Use conversion factors!!
How many moles of lithium are in 18.2 grams of lithium?
Hint use conversion factors!!
How many atoms of lithium are in 3.50 moles of lithium?
Hint  Use conversion factors!!
How many atoms of lithium are in 18.2 g of lithium?
Hint Use three conversion factors!!
Chemical change involves a reorganization of the atoms in one
or more substances.
C2H5OH + 3O2  2CO2 + 3H2O
Reactants
Products
1 mole of ethanol reacts with 3 moles of oxygen to produce
2 moles of carbon dioxide and 3 moles of water
 Shorthand way to describe a chemical reaction
 Steps to write them
• Write down the equation in words
• Fill in correct molecular formulas for each substance & make sure they
are balanced
• Balance the equation using coefficients
• Add state symbols
Make a list of each element present in the equation.
2.
Change the coefficients to make the equation balanced.
Examples: Balance the following equations
C2H5OH(l) + O2(g)  CO2(g) + H2O
(NH4)2Cr2O7(s)  Cr2O3(s) + N2(g) + H2O(g)
1.
 Write equations for the following reactions
•
•
•
•
Hydrogen + copper(II) oxide  copper + water
H2(g) + CuO(s)  Cu(s) + H2O(l)
Carbon + oxygen  carbon monoxide
C(s) + O2(g)  CO(g)
 Balance the following
•
•
•
•
Na + O2  Na2O
C4H8 + O2  CO2 + H2O
Pb(NO3)2  PbO + NO2 + O2
Fe + Br2  FeBr3
 Synthesis
A+B→C
 H2 + O2 → 2H2O
 Decomposition
A→B+C
 2H2O → H2 + O2
 Double
Replacement
 AB + CD → AD + CB
 MgO + 2NaCl → MgCl2 + Na2O
 Some
ionic compounds are more soluble in water
than others
 Precipitation occurs when cations and anions
combine to form insoluble compounds (don’t
dissolve in water)
 Driving force – forms a solid (precipitate)
 These are also called double displacement
reactions
 AB + CD  AD + CB
 You must memorize the solubility rules for these
reactions
 Examples
• AgNO3(aq) + KCl(aq) 
• Na2SO4(aq) + Pb(NO3)2(aq) 
• KOH(aq) + MgBr2(aq) 
• (NH4)2SO4(aq) + ZnCl2(aq) 
 White
 barium sulfate, silver chloride,
aluminum hydroxide, zinc hydroxide
 Cream  silver bromide
 Pale yellow  silver iodide, lead(II) iodide
 Brown gas  nitrogen dioxide
 Green  nickel(II) hydroxide, iron(II)
hydroxide
 Blue  copper(II) hydroxide
 Red/brown  iron(III) hydroxide
 Acid
 donates H+ in aqueous solutions
 Base  accepts H+ in aqueous solutions
• HNO3(aq) + NaOH(aq)  H2O(l) + NaNO3(aq)
 Driving
force – formation of water
 Produces heat, a salt (ionic compound)
and water
 Also
called redox reactions
 Driving force – transfer of electrons;
formation of a gas
 Occurs in metal-nonmetal reactions – ionic
compounds are formed
 Includes single-replacement reactions,
combustion reactions, decomposition
reactions, and synthesis reactions
 Single
replacement (displacement)
• A + BC  B + AC
 Combustion
• Always includes O2
• Carbon compound + O2  CO2 + H2O
• Always makes CO2 and H2O
 Driving
force  formation of a gas
 General reactions
• Acid + metal  salt + hydrogen
 Zn + H2SO4  ZnSO4 + H2
• Acid + carbonate  salt + water + CO2
 H2SO4 + CaCO3  CaSO4 + H2O + CO2
• Production of oxygen by decomposition of
hydrogen peroxide with MnO2 catalyst
 2H2O2  2H2O + O2
Express the mass of each elements as a percent of the mass of a
compound
Calculate the percentage composition of magnesium carbonate,
MgCO3.
From previous slide:
 Percent
Composition can help obtain the
empirical and molecular formula for an
unknown molecule.
Empirical formula: the lowest whole number ratio of atoms
in a compound.
Molecular formula: the true number of atoms of each
element in the formula of a compound.
molecular formula = (empirical formula)n [n =
integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH
Formulas for ionic compounds are ALWAYS empirical
(lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas for molecular compounds MIGHT be empirical
(lowest whole number ratio).
Molecular:
Empirical:
H2O
C6H12O6 C12H22O11
1. Base calculation on 100 grams of compound (change
% to g).
2. Determine moles of each element in 100 grams of
compound.
3. Divide each value of moles by the smallest of the
values.
4. Multiply each number by an integer to obtain all
whole numbers.
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What
is the empirical formula of adipic acid?
Elements
present
% by mass
RAM
% by
mass/RAM
Divide by
smallest
Empirical
formula
C
H
O
 Empirical
Formula of Adipic Acid
 C1.5 H2.5 O1
 There
cannot be 1.5 atoms of carbon in a
molecule, so we must multiply ALL
subscripts by 2
 Final (True) Empirical Formula C3 H5 O2
Calculate the empirical formula of a compound
that is 77.78% Fe and 22.22% O.
 A hydrocarbon has the composition by mass, hydrogen 7.69%
and carbon 92.31%. Calculate its empirical formula.
The empirical formula for adipic acid is C3H5O2. The molecular mass
of adipic acid is 146 g/mol. What is the molecular formula of
adipic acid?
1.
Find the formula mass of C3H5O2
2.
Divide the molecular mass by the mass given by the empirical
formula.
3.
Multiply the empirical formula by this number to get the
molecular formula.
1.
2.
3.
4.
Balance the equation.
Convert masses to moles.
Use the stoichiometric coefficients in the equation to
find the reacting ratio of the moles. Use this
relationship to find the number of moles of the
unknown substance.
Convert from moles to grams.
6.50 grams of aluminum reacts with an excess of oxygen. How many
grams of aluminum oxide are formed?
1. Identify reactants and products and write the balanced equation.
Al + O2 Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
 3.5g
of Hydrogen reacts with nitrogen to
form Ammonia. How much ammonia is
made?
The limiting reactant is the reactant that is consumed first,
limiting the amounts of products formed.
C + O2  CO2
So if there are 6 Carbon atoms and 1 oxygen molecule, the
oxygen will be consumed first and is the limiting reactant.
It is important to remember this is an atomic ratio, not a ratio
based upon grams!
 The two nonmetals, sulfur and chlorine, react according to the
following equation: S(s) + 3Cl2(g)  SCl6(l)
If 202g of sulfur are allowed to react with 303g of Cl2 in the
reaction above, which is the limiting reactant, how much
product will be produced and what mass of excess reactant
will be left over?
Stoichiometric calculations give theoretical yeilds
The amount you get in an experiment is the actual yeild
To determine percent yield
actual yield/theoretical yield x 100
To determine percent error
|actual – theoretical|/theoretical x 100
 Aluminum will react with oxygen gas according to the
following equation
4Al + 3O2  2Al2O3
In one such reaction, 23.4g of Al are allowed to burn in excess
oxygen. 39.3g of aluminum oxide are formed. What is the
percent yield and the percent error?