Why do atoms bond

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Why do atoms bond?
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Introduction to Bonding

• Atoms are generally found in nature in
  combination held together by chemical bonds.
• A chemical bond is a mutual electrical attraction
  between the nuclei and outer electrons of
  different atoms that binds the atoms together.
• There are two types of
  chemical bonds ionic,
  and covalent.

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Introduction to Bonding

• What determines the type of bond that forms?
• The outer electrons of the two atoms involved are
  redistributed to the most stable arrangement.
• The interaction and rearrangement of the outer
  electrons determines which type of bond that
  forms.
• Before bonding the atoms are at their highest
  possible potential energy

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Introduction to Bonding

• There are 2 philosophies of atom to atom
  interaction
• One deals with balancing the opposing forces of
  repulsion and attraction
• As the atoms approach repulsion occurs
  between the negative
  e- clouds of
  each atom
• And attraction occurs
  between the positive
  nuclei and the
  negative electron
  clouds

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Introduction to Bonding

• As the optimum distance is achieved that balances
  these forces, there is a release of potential
  energy
• The atoms vibrate within the window of maximum
  attraction/minimum repulsion
• The more energy
  released the stronger
  the connecting bond
  between the atoms

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Introduction to Bonding

• Another chemical bond philosophy between two
  atoms centers on achieving the most stable
  arrangement of the atoms valence electrons
• By rearranging the electrons so that each atom
  achieves a noble gas-like arrangement of its
  electrons creates a pair of stable atoms (only
  occurs when bonded)

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Introduction to Bonding

• Sometimes to establish this arrange-ment one or
  more valence electrons are transferred between
  two atoms
• Basis for ionic
  bonding
• Sometimes valence
  electrons are shared between
  two atoms
• Basis for covalent
  bonding

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Introduction to Bonding

• A good predictor for which type of bonding will
  develop between a set of atoms is the difference
  in their electronegativities.
• The more extreme the difference between the two
  atoms, the less equal the exchange of electrons
• This leaves us with three different levels of
  interaction pure covalent, polar covalent, and
  ionic

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Introduction to Bonding

• Lets consider the compound Cesium Fluoride, CsF.
• The electronegativity value (EV)
  for Cs is .70 the EV for F is 4.00.
• The difference between the two is 3.30, which
  falls within the scale of ionic
  character.
• When the electronegativity difference between two
  atoms is greater than 1.7 the bond is mostly
  ionic.

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Ionic ?1.7
.3ltPolar Covalent lt1.7
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Introduction to Bonding

• The take home lesson on electro-negativity and
  bonding is this
• The closer together the atoms are on
  the P.T., the more evenly their e- interact, and
  so are more likely to form
  a pure covalent bond
• The farther apart they are on the P.T., the less
  evenly their e- interact, and
  are therefore more likely
  to form an ionic bond.
• In between exists
  the polar covalent interactions

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Rule of Thumb
metal w/nonmetal usually ionic
nonmetal w/nonmetal
usually covalent
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Introduction to Covalent Bonding

• In a co-valent bond
• The electronegativity difference between the
  atoms involved is not extreme
• So the interaction between the involved electrons
  is more like a sharing relationship
• It may not be an equal sharing relationship, but
  at least the electrons are being shared.

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Covalent Bonds
Lets look at the molecule Cl2

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Covalent Bonds
How about the molecule HCl?

(Polar Covalent) shared, but not evenly
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So whats the bottom line?
To be stable the two atoms involved in the
covalent bond share their electrons in order to
achieve the arrangement of a noble gas.
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Introduction to Ionic Bonding

• In an ion - ic bond
• The electronegativity difference is extreme,
• So the atom with the stronger pull doesnt really
  share the electron
• Instead the electron is essentially transferred
  from the atom with the least attraction to the
  atom with the most attraction

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An electron is transferred from the sodium atom
to the chlorine atom

Na
Cl
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Both atoms are happy, they both achieve the
electron arrangement of a noble gas.
Notice 8 e- in each valence shell!!!
-1
1
Cl
Na
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Very Strong Electrostatic attraction established
IONIC BONDS
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Hydrated vs. Anhydrous

• In the construction of a crystal lattice,
  depending on the ions involved there can be small
  pores develop between ions in the ionic
  crystal.
• Some ionic compnds have enough space between the
  ions that water molecules can get trapped in
  between the ions
• Ionic compounds that absorb water into their
  pores form a special type of ionic compound
  called a hydrate.

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Trapped Water Molecules
Hydrated Crystal
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Hydrate Formation

• Hydrates typically have somewhat different
  properties than their dry versions - A.K.A.
  anhydrate or anhydrous
• Anhydrous copper sulfate is nearly colorless
• The hydrated version is a bright blue
  color
• When Copper (II) Sulfate is fully hydrated there
  are 5 water molecules trapped for every Copper
  sulfate.

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Hydrate Formation

• These hydrate-able ionic compounds are sometimes
  used to indicate the presence of water.
• For example, Cobalt Chloride
  is a compound that is
  blue in its anhydrous
  
  version, and magenta
  when it is hydrated.

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So whats the bottom line?
To be stable the two atoms involved in the ionic
bond will either lose or gain their valence
electrons in order to achieve a stable
arrangement of electrons.
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Percent Composition

• An important quantitative measurement that can be
  made for any chemical substance is a
  Percent Composition.
• The percent composition of a compound is a
  relative measure of the mass of each
  different element present in the compound.
• It gives you a rough comparison of the
  masses of the each component in the total
  sample

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Percent Composition

• percent composition in a compnd can be
  determined in 2 ways
• The 1st is by calculating the percent
  composition by mass from a chemical formula.
• The 2nd is a lab scenario where an unknown
  compound is chemically broken up into its
  individual components and percent composition is
  determined by analyzing the results.

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What is the percent composition of Hydrogen
Oxygen in Water (H2O)?
1st Assume you have a mole of the compound in
question, and calculate its molar mass
(21.008) (115.99)
18 g H2O
2nd Use the MM of each component and the MM
of the compound to calculate the percent by
mass of each component
(21.008) 2 g/mol
H
x 100
11.1
O
100 11.1 88.9
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Calculating PC Using Analysis Data

• In this method, the mass of the sample is
  measured, then the sample is decomposed or
  separated into the component elements
• The masses of the component elements are then
  determined and the percent composition is
  calculated as before
• divide the mass of each element by the total
  mass of the sample and multiply by 100.

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Find the percent composition of a compound that
contains 1.94g of carbon, 0.48g of Hydrogen, and
2.58g of Sulfur in a 5.0g sample of the compound.

• Calculate the percents for each component by the
  equation (Component Mass/Total Sample Mass) x
  100

C 1.94g/5.0g x 100 38.8
H 0.48g/5.0g x 100 9.6
S 2.58g/5.0g x 100 51.6
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Classroom Practice 1
Calculate the percent composition of Mg(NO3)2.
Mg 16.2 N 18.9 O 64.0
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Empirical Formulas

• Scientists communicate the atoms involved in a
  compound through symbolic formulas.
• There are three types of formulas that chemistry
  use empirical, molecular, and structural
• The simplest formula is called an empirical
  formula
• simplest ratio of the atoms in a compnd
• Ionic compounds are always written as empirical
  formulas

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Empirical Formulas

• Procedure for calculating empirical formulas
• convert the percent compositions into moles
• compare the mols of each compo-nent to calculate
  the simplest whole number ratio
• divide each amount in moles by the smallest of
  the mole amounts
• This sets up a simple ratio

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Calculate the empirical formula of a
compound that is 80.0 Carbon and 20.0 Hydrogen
by mass
Since we dont know the original mass of the
sample, we can assume a 100 g sample

• We have 80 grams of Carbon and 20 grams of
  Hydrogen
• We need to calculate the number of moles of each
  element that we have.

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Calculating Empirical Formulas
1 mole C
80.0g C
12.01 g C
1
CH3
1 mole H
20.0g H
2.97
1.008 g H

• Now we need to calculate the smallest whole
  number ratio in order to find the empirical
  formula.
• Divide each component by the smallest
  number in moles

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Calculating Empirical Formulas
Determine the empirical formula of a compound
containing 2.644g of Au and 0.476g of Cl.
1 mol Au
2.664g Au
.01352mol Au
197 g Au
1
1 mol Cl
.476g Cl
.01345mol Cl
1
35.4 g Cl
AuCl
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Classroom Practice 2
Determine the empirical formula for a compound
which is 54.09 Ca, 43.18 O, and 2.73 H.
CaO2H2 or Ca(OH)2
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Molecular Formulas

• The empirical formula for a compound provides the
  simplest ratio of the atoms in the compound
• However, it does not tell you the actual numbers
  of atoms in each molecule of the compound
• For instance the empirical formula for glucose
  is CH2O (121)
• While the molecular formula for glucose is C6H12O6

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Molecular Formulas

• A molecular formula indicates the numbers of each
  atom involved in the the compound
• The molecular formula is always a multiple of the
  empirical formula
• To calculate the molecular formula you must have
  2 pieces of info.
• Empirical formula
• Molar mass of the unknown compound (always given)

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Calculating Molecular Formula
Find the molecular formula of a compound that
contains 56.36 g of O and 54.6 g of P. The molar
mass of the compound is 189.5 g/mol.
1st find the Emp. Formula
1 mol O
56.36 g O
3.525mol O
15.99 g O
1.99
PO2
1 mol P
54.6g P
1.763mol P
1
30.97 g P
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Calculating Molecular Formula
Now determine the mass of the empirical formula
PO2 (130.97g P)(215.99g O) 62.95g
MM Given in the problem 189.5 g/mol
189.5 g/mol
3.01
P3O6
Molecular Formula 3(PO2)
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One More A Good 1
Methyl acetate is a solvent commonly used in some
paints, inks, and adhesives. Determine the
molecular formula for methyl acetate, which has
the following chemical analysis 48.64 C, 8.16
H, and 43.20 O. The Molar Mass of the compound
in question is reported as 74g/mol.
1st determine the empirical formula 2nd determine
the molecular formula
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1 mole C
48.64g C
4.050mol C
1.50
12.01 g C
2.702 mol
1 mole H
8.16g H
8.095 mol H
2.99
1.008 g H
2.702 mol
1 mole O
43.20g O
2.702 mol O
15.99 g O
1.00
2.702 mol
74g
C1.5H3O1
2
C3H6O2
36632
C9H18O6
222/74 3
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Classroom Practice 3
A hydrocarbon is 84.25 carbon and 15.75
hydrogen and has a molecular weight of 114. What
is its molecular formula?
C8H18