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1
Review
Chemistry 11 Notes
Bonding Stoechiometry
Organic
Review Unit
1. Matter
a) Elements and Compounds
 All matter is made up of about 100 elements. Elements are pure substances that cannot
be broken down into simpler parts by ordinary chemical means. Each element is
composed of a fundamental particle called an atom. Each element has a unique atom and
is represented by a symbol (memorize the sheet “Some Common Symbols”).
 The elements with their symbols are organized into the Periodic Table.
 Elements combine to form compounds (atoms combine to form molecules).
Element 1
+
Element 2

Compound
(atom)
(atom)
(molecule)
The physical and chemical properties of a compound are completely different from the
elements it is made from.
+

two hydrogen atoms
+
one oxygen atom  one water molecule
 Notice that the water molecule can only be made by joining together two hydrogen
atoms (symbol = H) with one oxygen atom (symbol = O). The formula for water will be
H2O.
Note: If there is no number after a symbol in a formula, assume it is a one.
Example: CaO means Ca1O1
 In summary, pure substances are organized this way:
Matter
Elements Compounds
Particles
Atoms  Molecules
Naming
Symbols  Formulas
 Impure substances are called mixtures. They are made by mixing different types of
compounds and/or elements together. Only mixtures have a variable composition.
1b) Measurement
 Measurements are fundamental to the study of matter. There are two types of
measurements:
(i)
quantitative – describes matter using numbers.
(ii)
qualitative – describes matter in a descriptive, non-numeric way.
 Measurements are reported in significant figures.
© Mr. E. Therrien Chemistry 11 Notes ©
2
Significant Figures = all measured digits + one estimated digit
(exact)
(approximate)
Example:
43.52 km
known
(measured)
uncertain
(estimated)
 All of the digits in a measurement are significant unless it is a zero that is used as a
place marker.
for whole numbers
for decimal numbers
6700
0.003
place markers (not significant)
place markers (not significant)
 To identify significant digits, try this:
(i) for whole numbers:
67 490 =
count
(ii) for decimal numbers:
(iii) for mixed numbers:
4 sig. fig.
0.004 80 =
count
3 sig. fig.
680.420 = 6 sig. fig.
count everything
If the zeros in the number were actually measured, they must be shown to be significant.
This is done by marking them with a bar or a decimal point.
6700 = 2 sig. fig. (6.7 x 103)
6700 = 3 sig. fig. (6.70 x 103)
6700. = 4 sig. fig. (6.700 x 103)
 Remember that significant figures are used only with measurements. They are not
applied to defined quantities (100 cm = 1 m) or to pure numbers.
 The use of significant figures allows us to change the units of a measurement without
changing the accuracy of the measurement.
1.24 m
=
1240 mm
=
0.00124 km
3 sig. fig.
 All measurements use significant figures. When using an instrument:
- make sure you understand the scale first.
- Remember that the last digit of your measurement must be an estimate
(if the spacing is very small, the only practical estimate is either exactly on the line
[estimate a 0] or halfway between the lines [estimate a 5].)
- be sure the scale is read at eye level.
- the level of a liquid is read at the bottom of its meniscus.
- Vernier scales can be used to increase the accuracy of the estimated digit.
© Mr. E. Therrien Chemistry 11 Notes ©
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 When significant figures are used in calculations, the answer cannot show more
accuracy than the least accurate measurement. It must be rounded off to the proper digit.
(1) Addition and Subtraction – round off to the highest place value (the least accurate
measurement).
6.4 cm + 8.425 cm + 2.81 cm = 17.635 cm = 17.6 cm
the least accurate measurement (tenths)
(2) Multiplication and Division – round off to the least number of
significant digits used (the least accurate measurement).
6.4 cm x 8.425 cm  2.81 cm = 19.188 612 1 cm = 19 cm
the least accurate measurement (2 sig. fig.)
2. Parts of an Atom
 The atom has two main parts:
(i)
the nucleus (or center) of the atom contains positively charged particles called
protons (p+) and a neutral particle with no charge called a neutron (n0). It
occupies very little space but contains the mass of the atom.
(ii)
The space in an atom is occupied by negative particles called electrons (e-)
traveling around the nucleus at regular intervals. Their have almost no mass.
 Atoms are different because their nuclei contain different numbers of protons.
Carbon  6 p+
nitrogen  7 p+
The number of protons in the nucleus of an atom is called its atomic number.
Oxygen  atomic number = 8  has 8 p+
 Almost all of the mass of an atom is in its nucleus. The mass number of an atom, then,
is the mass of the particles in the nucleus (protons and neutrons).
Oxygen  mass number = 16  has 16 p+ + n0
A Simplified Lithium Atom
= protons
= neutrons
= electrons
atomic number = 3
mass number = 7
atomic symbol = 73Li
Nucleus
Energy levels
 The first 20 electrons are placed in four energy levels (2e-, 8e-, 8e-, 2e-) . Their
arrangements can be shown using energy level diagrams.
Example: show the energy level diagram for potassium (19K)
K 19p+ 2e- 8e- 8e- 1e Note: (i) Since atoms have no overall charge, then
number of p+ = number of e© Mr. E. Therrien Chemistry 11 Notes ©
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0
(ii) number of n = mass number – atomic number
(iii) Electrons travel around the nucleus in increasing intervals called energy levels.
3. Periodic Table
◘ The modern periodic table arranges the elements in order of increasing atomic number.
◘ Metals are separated from nonmetals by the “staircase line”.
metals - shiny, malleable, ductile, conductors of heat and electricity.
◘ The columns are families (groups) of elements having similar chemical properties.
Some of these families have names to memorize (see sheet).
◘ The rows are called periods. The elements gradually change from metallic to nonmetallic from left to right.
The number of the row tells us the number of energy levels for electrons in that atom.
◘ The Group A elements are called representative elements.
The Group B elements are called transition elements.
◘ Note that the number given under the symbol is not the mass number of that particular
atom. It is an averaged mass number called the atomic mass and will be used later.
4. Ions
a) Creating Ions
 We know that atoms are electrically neutral because they have equal numbers of
protons (p+) and electrons (e-).
 When atoms join together, though, they can lose or gain electrons. This causes the
charges in the atom to become unbalanced. Ions are atoms that have a charge.
atom
electron
ion
metal (M)
loses one
M+
non-metal (N)
gains one
N-
 Some atoms will gain or lose more than one electron:
- Atoms of metals (M) tend to lose one, two or three electrons and form positive ions
(cations).
M - 1e- = M+
M - 2e- = M2+
M - 3e- = M3+
Example Mg - 2- (lost) = Mg2+ (called a magnesium ion).
- Atoms of non-metals (N) tend to gain one, two or three electrons and form negative
ions (anions).
N + 1e- = NN + 2e- = N2N + 3e- = N3Example
O + 2e- (gain) = O2- (called an oxide ion).
Note that anion names end in “-ide”.
© Mr. E. Therrien Chemistry 11 Notes ©
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4b) Predicting Ionic Charges
▪ It is possible to predict the ion that an atom will form:
(i) Representative Elements - these are the Group A elements in the periodic table. The
ionic charge for these elements can be determined by their position in the periodic table.
1A
M+
2A
M2+
Let M = metal and N = non-metal
3A
4A
3+
transition
M
metals
5A
N3-
6A
N2-
7A
N-
8A
(ii) Transition Elements - most transition (Group B) elements are metals that can form
more than one positive ion.
iron = Fe2+ and Fe3+
lead = Pb2+ and Pb4+
Since it is difficult to predict these ions, you will be given a list of them to work with (see
“Common Ions” sheet).
Two methods of naming these ions are used:
a) Stock System - a Roman numeral is used in brackets after the name
to indicate the charge.
Pb2+ = lead(II)
b) Classical System - the classical name of the element is used with the
endings “-ous” (for the lower charge) and “-ic” (for the higher
charge).
If
the
symbol
is
Fe2+
Fe3+
Pb2+
Pb4+
then
its
name
is
iron(II)
iron(III)
lead(II)
lead(IV)
and its
classical
name
was
ferrous
ferric
plumbous
plumbic
Try This Predict the ion that will be formed from these atoms:
1) phosphorus
6) lead(IV)
11) chromium(III)
2) aluminum
7) oxygen
12) calcium
3) cobalt(III)
8) magnesium
13) sulfur
4) nitrogen
9) lithium
14) potassium
5) mercury(II)
10) chlorine
15) tin(IV)
4c) Polyatomic Ions
 Poly = many
 Polyatomic ions are tightly bound groups of atoms that behave as a unit and carry a
charge.
Sulfate Ion (SO42-)
One sulfur atom and four
oxygen atoms forming one
strongly bonded unit with
two extra electrons.
© Mr. E. Therrien Chemistry 11 Notes ©
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 There are many of these ions. Since we cannot predict their charges, we will be given a
list of them to work with (see “Common Ions” sheet).
 Most polyatomic groups contain oxygen and their names end in “-ate”.
An “-ite” ending means one less oxygen than the “-ate” ending.
sulfate = SO42sulfite = SO32-
chlorate = ClO3 chlorite = ClO2 -
 A few poyatomic groups do not contain oxygen. Their names end
in “-ide”.
cyanide = CN5. Formulas
a) Creating Formulas
 Ionic compounds are created when one kind of positive cation joins with one kind of
negative anion. The positive ion is always written first.
 The sum of the ionic charges in the formula must be zero. This is done by adjusting the
numbers of each ion.
Li+ + O2-  Li2O
(+1) (-2)
(sum=0)
Note that the formula for the compound (Li2O) shows no charge. It is neutral overall.
How to Balance Ionic Charges
- When the charges are opposite numbers, their sum will automatically equal zero.
Ca2+ + O2-  CaO
(+2)
(-2)
(sum=0)
- When the charges are different, the number of the ions must be adjusted until the
sum of the charges becomes zero.
Al3+ + Cl-  AlCl3
(+3) (-1)
(sum=0)
One convenient way of balancing charges is to cross-exchange the digits.
Al3+ + O2-  Al3+ 2O2- 3
Note that when you want to show more than one polyatomic group, you must put
brackets around the group.
Ca2+ + NO3-  Ca(NO3)2
(+2)
(-1)
(sum=0)
5b) Nomenclature
 Let M = Metal N = Non-metal P = Polyatomic Group H = Hydrogen
(1) Ionic Compounds - always begin with a metal
- use the “Common Ions” sheet.
© Mr. E. Therrien Chemistry 11 Notes ©
7
M
+
N
=
M
+
P
=
M (stock number)
N -ide
if necessary
M ( ) P -ate (or -ite)
NaCl = sodium chloride
FeClO3 = iron(III) chlorate
(2) Molecular Compounds - no metals present
- use prefixes (not the “Common Ions” sheet).
N1 +
N2
=
prefix N1
prefix
except mono
N2 -ide
1 = mono
2 = di
3 = tri
4 = tetra
5 = penta
6 = hexa
7 = hepta
8 = octa
9 = nona
10 = deca
CO = carbon monoxide
P2O5 = diphosphorus pentoxide
(3) Acids - always begin with hydrogen
- there are two types.
a) Binary Acids contain only two elements
H
+
N
=
hydro
N -ic
acid.
HCl = hydrochloric acid
b) Oxo-acids contain oxygen in a polyatomic group
ite
ous
H + P
ate = P
ic
acid .
H2SO4 = sulfuric acid
H2SO3 = sulfurous acid
Nomenclature Summary
Formula begins with
Metal
Ionic substance
Use the
“Common Ions”
Sheet
Non-metal
Hydrogen
Molecular Substance
Use Prefixes
© Mr. E. Therrien Chemistry 11 Notes ©
Acid
binary
oxo-
name begins
with “hydro-“
“ate-ic
ite-ous”
acids
8
6. Reactions
a) Balancing Equations
 A chemical equation is an expression for a chemical reaction. It is a quantitative
statement indicating the number of moles of each reactant and of each product.
Reactants  Products
 In chemical equations, matter must be conserved. The number of atoms of each kind
on the reactant side must equal those on the product side.
Consider this reaction: carbon + oxygen gas  carbon monoxide
C + O2  CO

+
(Matter is not conserved)
However,
+
2C + O2  2CO

(Matter is conserved)
This equation is balanced.
 All equations must be balanced.
Balanced equations have 1. the chemical facts
2. correct formulas
3. atoms conserved
 Hints: 1) Try balancing O and H last
2) Never change a formula to make the equation balance. Change coefficients, not
subscripts!
2CO2
coefficient
subscript
3) Equations that will not balance probably contain an incorrect formula
All chemical reaction involves changes to substances. The starting chemicals (reactants)
are chemically changed into new chemicals (products)
Reactants  Products
Chemical equations must obey the Law of Conservation of Matter. Each side of the
chemical equation must contain equal amounts of each element.
Consider this equation:
Carbon + oxygen gas  carbon monoxide gas
C
+ O2

CO

Since the amounts of each element are not balanced, we use coefficients in front to the
formulas to make it balanced.
© Mr. E. Therrien Chemistry 11 Notes ©
9
2C
+ O2


2CO
Note: i. the coefficients tell us the amount of each chemical: 2 H2O
means 2 mol or 2 molecules of water
ii. Equations are balanced only wit coefficients.
DO NOT CHANGE THE FORMULAS!
Try This Balance these equations
a) H2 + O2
 H2O
b) Ca + O2
 CaO
c) Mg(ClO3)2
 MgCl2 + O2
d) Ca + H2O
 Ca(OH)2 + H2
e) HNO3 + Ba(OH)2  Ba(NO3)2 + H2O
Symbol
+

=

(s)

(l)
(aq)
(g)



Pt

heat
Symbols used in Equations
Explanation
Used to separate two reactants or two products
“Yields”, separates reactants from products
An alternative to 
Used in place of a  for reversible reactions
Designates a reactant or product in the solid state; placed after the
formula
Alternative to (s); used only for a solid product (precipitate)
Designates a reactant or product in the liquid state; placed after the
formula
Designates an aqueous solution; the substance is dissolved in water
Designates a reactant or product in the gaseous state; placed after the
formula
Alternative to (g); used only for a gaseous product
Indicates that heat is supplied to the reaction
A formula written above or below the yield sign indicates its use as a
catalyst (in this example, platinum)
6 b) Types of Reactions
a) Combination Reactions
- Two or more reactants combine to form a single product. R
Note:
i.
Two elements must form a binary compound.
S + O2  SO2
H2 + Cl2  2HCl
ii.
Non-metal oxides in water form an acid.
SO3 + H2O
 H2SO4
CO2 + H2O
 H2CO3
© Mr. E. Therrien Chemistry 11 Notes ©
+
S

RS
10
Try This
a) C + O2

b) aluminum + oxygen gas 
c) copper (I) + sulfur

d) N2O5 + H2O 
b) Decomposition Reactions (Opposite of Combination)
- A single reactant is broken down into two or more products:
RS

R
+ S
decomposes

Note: i. Binary compound
element
+
element
into
2H2O
2HgO


2H2 + O2
2Hg + O2
ii. for more complicated compounds, try this:
If the compound contains
Example
C + O
NiCO3
H + O
NH4NO3
O
KClO3
NiCO3
heat


heat
NH4NO3 

heat
2KClO3 

One product might be
CO2
H2O
O2
NiO + CO2
N2O + 2H2O
2KCl + 3O2
Try This
a) H2O 
b) mercury (II) oxide heat


c) magnesium carbonate heat


c) Single Replacement Reactions
- The atoms of one element can replace the atoms of a different element that is part of an
ionic compound.
For Metals:
M1 + M2X  M1X + M2
Note: that the first metal (M1) can only replace the second metal (M2) if it is more
reactive, as listed in the activity series of metals.
Examples:
i. Ca + 2NaOH v Ca(OH)2 + 2Na
ii. Mg + NaOH  no reaction (Mg is not more reactive)
For Non-Metals: (mostly the halogen gases) N1 + XN2  XN1 + N2
Note: that the first non-metal (N1) can only replace the second non-metal (N2) if it is
more reactive.
© Mr. E. Therrien Chemistry 11 Notes ©
11
Examples:
i. Cl2 + BaBr2
ii. I2 + BaBr2


BaCl2 + Br2
no reaction (I is not more reactive)
Try This
a) Fe + CuSO4 
b) Mg + HCl 
c) Cl2 + KI

d) Double Replacement Reactions
- This involves the reaction of two compounds to form two new
compounds.
RS + TU  RU + TS
2KI
+ Pb(NO3)2  PbI2 + 2KNO3
Note:
i.
These reactions involve ionic compounds or acids in aqueous
solution.
ii.
The two reactants just exchange their positive ions.
iii.
Two compounds will not always react. There are three “driving
forces” which cause these reactions to occur:
a) a molecular compound can be formed
b) a gas can be formed
c) a precipitate can be formed
- One special type of a double replacement reaction is a neutralization
reaction:
Acid + Hydroxide
 salt +
water
HCl
+
NaOH
 NaCl +
H2 O
Try This
a) Barium chloride + sodium sulfate 
BaCl2
+
Na2SO4 
b) Silver nitrate + sodium chromate 
c) Hydrochloric acid + calcium hydroxide 
HCl
+
Ca(OH)2 
Reactants
E1 + E2
C
E1 + C1
C1 + C2
SUMMARY
Let E= element and C = Compound
Reaction Type
Products
Combination
C
Decomposition
E1 + E2
Single replacement
E2 + C2
Double replacement C3 + C4
© Mr. E. Therrien Chemistry 11 Notes ©
Activity Series of
the Elements
Metals Non-metals
Li
F
Rb
Cl
K
Br
Na
I
Sr
S
Ba
Ca
Mg
Al
MN
Zn
Cr
Fe
Cd
Co
Ni
Sn
Pb
H
Sb
Bi
As
Cu
Hg
Ag
Pt
Au
General Equation
R + S  RS
RS  R + S
T + RS  TS + R
RS + TURU + TS
12
Other Equation Terms
a) Heat
- Exothermic reactions are accompanied by the release of heat into their surroundings.
CH4 + 2O2  CO2 + 2H2O + energy
The release of energy is shown as a product in the equation.
We can show this in an energy level diagram:
- Endothermic reactions take place only when heat is continuously to the reactants.
CaCO3 + energy  CaO + CO2
The heat is much like a reactant – without it, the reaction cannot take place.
Remember that the unit for energy is the Joule (J).
b) Physical State
- Chemical reactions often depend on the physical state of the chemicals involved. This
information can be included in an equation by using these symbols:
(s) = solid (l) = liquid (g) = gas (a) = aqueous (dissolved in water)
For example, water may be H2O(s), H2O(l) or H2O(g).
- Double replacement reactions generally take place between ionic compounds in
aqueous solution: PbCl2(aq) + Na2CrO4(aq)
These two chemicals will now react if one of the products will form a precipitate that is
insoluble in water. PbCl2(aq) + Na2CrO4(aq)  PbCrO4(s) + 2NaCl(aq)
The precipitate PbCrO4(s) may now be removed from the solution by filtration.
- To write a balanced equation that includes physical states, then, we must be able to
predict if a chemical is soluble in water (aq) or insoluble in water (s). To do this, we will
use a solubility chart.
© Mr. E. Therrien Chemistry 11 Notes ©
13
Note: “s” on a solubility chart = soluble and “(s)” in a equation = solid (insoluble)
Try This Write a balanced equation for each reaction in solution. Include the physical
state symbols.
1. Ba(OH)2 + Na3PO4 
5. (NH4)S + CuCl2 
2. AgNO3 + K2S 
6. NaOH + CuSO4 
3. BaCl2 + Na3PO4 
7. Na2S + Pb(NO3)2 
4. K2CrO4 + Pb(NO3)2 
8. PbBr2 + K2CrO4 
Ionic Reactions
a) Aqueous Solutions
- The particles of a substance, when dissolving, separate from each other and disperse
throughout the solution.
+
water molecule (
)
- When NaCl is dissolved in water, the resulting solution contains ions:
NaCl(s)  Na+(aq) + Cl-(aq). The water simply causes the Na+ and Cl- ions to become
separated. The process of separating the ions in a solid is called dissociation.
- When a crystal contains more than one ion at a particular type, all of the ions will
dissociate.
CaCl2(s)  Ca2+(aq) + 2Cl-(aq)
Al2(SO4)3(s)  2Al3+(aq) + 3SO42-(aq)
- The presence of ions in a solution permits the solution to conduct electricity; the more
ions present, the better the ability to conduct.
- Many soluble substances, such as HCl(g), are made up of molecules, not ions. These
substances, though, create ions when dissolved in water:
HCl(g)  H+(aq) + Cl-(aq)
This process of splitting a molecule into ions is called ionization.
- Precipitation is the reverse of the dissociation process.
Ag+(aq) + Cl-(aq)  AgCl(s)
b) Ionic Equations
A reaction between two chemicals dissolved in water is shown three ways:
i. Molecular Equation - the reaction is written as if all of the chemicals consist of
molecules:
BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2 NaCl(aq)
ii. Total Ionic Equation - the equation is written to show that the dissolved
chemicals are actually in the forms of ions.
Ba2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + SO42-(aq)  BaSO4(s) + 2 Na+(aq) + 2 Cl-(aq)
iii. Net Ionic Equation - the equation is written to show only the ions or molecules
that are actively involved in the reaction.
© Mr. E. Therrien Chemistry 11 Notes ©
14
Ba2+(aq)
SO42-(aq)
+
 BaSO4(s)
The Na+ and Cl- ions were eliminated since they did not take part in the reaction. Ions
like this that occur in the same form on both sides of the equation are called spectator
ions.
Note: i. Ionic equation should be written from molecular equations.
ii. All solution equation must contain the physical state symbols.
iii. Ionic Equations are balanced in terms of mass and charge.
Try This Write the total and net ionic equations for the following:
a) AgNO3 + K2CO3 
b) Na3PO4 + MgSO4 
c) SnSO4 + Na2S

d) FeBr2 + KOH

e) SnCl2 + K3PO4

© Mr. E. Therrien Chemistry 11 Notes ©
15
Bonding Unit
1. Valence Electrons
a) Electron Theory
 Families of elements have similar chemical properties because they have the same
number of valence electrons. These are the electrons found in the outer energy level of an
atom.
Energy Level Diagrams (Family 6)
O 8p+ 2e- 6evalence electrons
+
S 16p 2e 8e 6e
 Note that electrons are located in a series of energy levels around the nucleus of the
atom. An orbital is a region within the energy level in which electrons are likely to be
found.
- the outer energy level always contains four valence orbitals.
- each valence orbital can hold two electrons.
- electrons occupy an empty valence orbital before forming electron pairs.
- there is a maximum of eight valence electrons in an atom. This is called the octet rule.
 Valence electrons are classified in terms of how they occupy an orbital:
bonding electron - single electron in an orbital
lone pair - pair of electrons in a filled orbital
1b) Electron Dot Symbols
 The number of valence electrons in a representative element (Family A element) is
given by the family number.
calcium = Family 2A = 2 valence electrons
oxygen = Family 6A = 6 valence electrons
 Electron dot symbols (Lewis Models) will be used to display valence electrons.
- write the symbol
- use a dot to represent each valence electron
- place on dot, in order, on each of the four sides (representing the
four valence orbitals)
- place a second electron on each side (fill the orbitals)
Examples:
..
Oxygen Atom 2 lone pairs
:O.
2 bonding electrons
.
Magnesium Atom
Mg :
© Mr. E. Therrien Chemistry 11 Notes ©
16
Note that there can be a maximum of eight valence electrons (octet rule) placed in four
valence orbitals.
 Try This Give the energy level diagram and the electron dot diagram for all of the
elements in a) Group 2 b) Period 2
2. Ionic Bonding
 Ions are formed when neutral atoms either gain or lose valence electrons.
Positive
Cation
electron loss
Neutral
Atom
electron gain
Negative
Anion
 The driving force behind the formation of ions is the tendency of atoms to reach a
more stable electron configuration – to have a stable octet of electrons in its valence
energy level. This means that the ions will have the same stable electron configuration as
a noble gas.
..
..
Example (16S): : S . + 2e-  [ : S : ]2- (same as the noble gas 18Ar)
.
..
 The loss of electrons from metal atoms (Families I, II, III) produces a positive ion.
.
Ca . - 2e-  [ Ca ]2+
The gain of electrons by nonmetal atoms (Families V, VI, VII) produces a negative ion.
..
..
: Cl . + 1e  [ : Cl : ]..
..
 Ionic compounds are formed when one type of positive ion and one type of negative
ion join together in regular patterns to form a crystal array.
The crystal array 
 has the formula unit  NaCl  which means  “one
+
Na ion for every one Cl ion in the crystal.”
We can use electron dot symbols to show this bonding process.
..
+
NaCl:
[Na ] [ : Cl : ]..
..
..
CaCl2:
[ Ca ]2+ [ : Cl : ]- [ : Cl : ]..
..
 Try This Use electron dot diagrams to show the bonding process between these
elements:
a) potassium + oxygen b) calcium + sulfur c) magnesium + nitrogen
d) sodium + phosphorus e) aluminum + bromine f) potassium + iodine
© Mr. E. Therrien Chemistry 11 Notes ©
17
3) Covalent Bonding
a) Types of covalent Bonds
- Some atoms share electrons in order to attain stable octets. This creates a unit molecule
held together by covalent bonds. Substances formed this way are called molecular
compounds.
The molecule 
 has the molecular formula  CH4  which
means  “five atoms bonded together into a single unit molecule”.
Molecular compounds contain only nonmetals. There are only a few of these:
Solids = C, P, S
Gases = H2, O2, F2, Br2, I2, N2, Cl2
..
Why are the gases diatomic?
Let’s consider the fluorine atom - . F :
..
By sharing one electron each, two fluorine atoms can complete their octets and become
more stable. This process can be shown with electron dot diagrams (Lewis Models).
..
..
.. ..
:F. + .F:  :F:F:
..
..
.. ..
a single covalent bond
 A structural diagram may also be used (F-F). Note that the shared pair is represented
by a line and the lone pairs are not shown but are assumed.
 Atoms can share more than one pair of electrons. Double covalent bonds involve two
shared pairs.
..
..
.. ..
: O . + . O :  : O :: O :
or
O=O
.
.
 Nitrogen atoms combine with a triple bond.
.
.
: N . + . N :  : N ::: N :
or
NN
.
.
 Molecular compounds (made from different elements) can also be shown:
..
..
H
H
water : O : H
ammonia H : N : H
ethene
C :: C
..
..
H
H
H
H
3b) Electron Dot Diagrams
Example: Show the covalent bonds present in HCN
Step 1 Begin with the central atom. It will be the one(s) that need the most electrons.
C
© Mr. E. Therrien Chemistry 11 Notes ©
18
Step 2 Add the remaining atoms in order.
. .
.C:N:
. .
. .
H:C:N:
. .
Step 3 Create multiple bonds until all atom (except H) have completed octets.
H : C ::: N:
Step 4 Rearrange the lone pairs, if necessary, so that they are at a maximum distance
from each other.
H : C ::: N :
or
H-CN
Try This Draw the electron dot diagrams for these compounds.
1. H2S
2. PH3
3. H2O2
4. OF2
5. HOCl
6. CO2
4. VSEPR Theory
7. H2CO
8. N2F4
9. C2N2
10. C2F2
11. CH4O
12. SiO2
(Valence Shell Electron Pair Repulsion Theory)
States that because electron pairs repel, molecules adjust their shapes (stereochemistry)
so that the valence-electron pairs are as far apart as possible
Proposes that the arrangement of atoms around a central atom in a molecule depends
upon the repulsions between all of the electron pairs in the valence shell of the central
atom
Repulsions between electron pairs in a valence shell decrease in this order:
Lone pair – lone pair
Lone pair – shared pair
Shared pair – shared pair
5. Polar Bonds
a) Polar Bonding
 When identical atoms are covalently bonded, the electrons are equally shared between
the atoms.
.. ..
: Cl : Cl :
.. ..
© Mr. E. Therrien Chemistry 11 Notes ©
19
 When different atoms are covalently bonded, though, the electrons are not shared
equally. One atom will have a stronger attraction for the shared pair of electrons.
In the HCl molecule, chlorine attracts the electron pair more than the hydrogen. Thus, the
shared electrons tend to spend more time near the chlorine atom and this causes it to
become slightly negative. The hydrogen atom, in turn, becomes slightly positive.
..
δ+
H : Cl : δ..
This is called a polar molecule (or dipole). It is created when a covalent bond has an
unequal sharing of electrons.
Note that the polar molecule as a whole remains electrically neutral.
5b) Measuring Polarity
 The amount of attraction that an atom has for a shared pair of electrons is called its
electronegativity (En). It is the electron gathering ability of the atoms in a molecule. Each
element has been assigned a number to match this ability
(see handout “Electronegativities of the Elements”).
 Note the periodic pattern of the En values:
Increasing Electronegativity
 The greater the difference in the electronegativities (ΔEn) between two atoms, the more
polar a bond becomes.
ΔEn
0
pure
covalent
(X-Y)
0.4
1.7
polar
covalent
(δ+ X-Y δ-)
3.4
ionic
( [X]+ [Y]-)
 Note: (i) This scale is a continuum. There are no real definite boundaries between the
types of bonds.
(ii) Even ionic bonds will share electrons to some extent.
(iii) En values are constants. They have no units.
Try This Classify the bond type expected to form between these atoms:
a) N and H
e) H and F
i) Cl and Ca
b) F and F
f) K and I
j) H and C
c) Ca and O
g) Li and O
k) N and F
d) Al and Cl
h) N and N
l) S and Na
© Mr. E. Therrien Chemistry 11 Notes ©
20
5c) Polar Molecules
The presence of any polar bond within a molecule can make the entire molecule polar.
If the molecule is symmetrical, however, the bond dipole forces will cancel each other
and the molecule will actually be non-polar.
have at least one polar bond (ΔEn > 4)
Polar Molecules must
be asymmetrical (not symmetrical)
Examples
H2 O
H2S
O-H
H
S-H
H
Polar
non-polar
(ΔEn = 0.4)
CO2
O=C=O
non-polar
(symmetrical)
6. Intermolecular Forces
¤ The Weak Forces (van der Waals forces) that exist between separate covalent
molecules are also known as intermolecular forces.
¤ We cannot measure the strengths of these forces directly but boiling points (b.p.) can
provide us with an indirect measure of their strength.
higher b.p. = greater forces
¤ Molecular substances are held together by two weak forces that are created in the
electronic fields of atoms:
(1) Dispersion Forces – are very weak Weak Forces that exist between all covalent
molecules. They result from the presence of small temporary molecular dipoles that are
caused by the momentary shifting of electrons away from parts of the molecule. This
causes two molecules to have a very brief attraction between the parts with opposite
charges.
Substances with larger molecules tend to have higher boiling points.
Larger molecules  more electrons  more dipoles  greater force  higher b.p.
(2) Dipole-dipole Forces – are stronger Weak Forces that occur when molecules with
permanent dipoles are attracted to one another.
δ+ δδ+ δlarger ΔEn  stronger dipoles  greater force  higher b.p.
© Mr. E. Therrien Chemistry 11 Notes ©
21
Hydrogen Bonding is a powerful type of dipole-dipole force. It occurs between a
hydrogen atom in one dipole and a very electronegative atom (usually O, N, halogen) in a
second polar molecule.
Hydrogen atoms are special because they exist mainly as a proton (p+) when bonded into
a dipole. Thus it will have a strong attraction to negative parts of other dipoles.
Bonded hydrogen  naked proton  unshielded + charge  greater force  higher b.p.
Try This 1) Explain the forces that exist between molecules of
a) O2 b) CO2 c) NH3
2) Consider three similar small molecules – CH4(g), NH3(g), H2O(l). Why is water the only
liquid?
3) Explain Cl2(g), Br2(l), I2(s) at SATP.
4) Match the formula with the name
C3H8 propane gas
C8H18 liquid gasoline
C14H30 solid paraffin wax
© Mr. E. Therrien Chemistry 11 Notes ©
22
Mole & Stoichiometry
1. Defining the Mole
◙ Particles (atoms, molecules, ions) can be counted three ways by using a unit called the
mole (mol).
The term  mole  can tell us  number of particles
 mass of the particles
 volume of gas particles
Thus, there are three relationships that can define the mole.
a) Avogadro’s Number
◙ One mole of any substance (element or compound) always contains the same number
of particles.
1 mole = 6.02 x 1023 particles = Avogadro’s Number
memorize me !
Note: (i) 1 mole of an element = 6.02 x 1023 atoms
(i) 1 mole of a compound = 6.02 x 1023 molecules
(ii) The unit is particles/mol
Try This Use factor label to solve the following problems:
a) How many atoms are present in 1.75 mole of calcium?
b) How many moles are in 2.42 x 1021 molecules of water?
c) How many molecules of carbon dioxide (CO2) are needed to make
0.15 mol of the gas?
1b) Molar Mass
 Remember that the atomic mass of an atom is given in the periodic table and has the
unit a.m.u. Similarly, the molar mass of an element is the number of grams of the element
that is equivalent to its atomic mass.
Carbon  12.0 a.m.u. (atomic mass)
 12.0 g (molar mass)
 In a compound, the molar mass is the sum of the masses of all the elements:
H2SO4
Molecular Mass
H x 2 = 2.0 a.m.u.
S x 1 = 32.1 a.m.u.
O x 4 = 64.0 a.m.u.
98.1 a.m.u.
one molecule of H2SO4 has a
mass of 98.1 a.m.u.

1 mole = molar mass in grams
Its unit is g/mol
© Mr. E. Therrien Chemistry 11 Notes ©
Molar Mass
H x 2 = 2.0 g
S x 1 = 32.1 g
O x 4 = 64.0 g
98.1 g
one mole of H2SO4 has a
mass of 98.1 g
23
Try This 1. How many grams of each element should you measure out (to the nearest
tenth) in order to get one mole of:
a) oxygen b) calcium c) nitrogen d) carbon e) phosphorus f) chlorine
2. How many grams of each compound should you measure out (to the nearest tenth) in
order to get one mole of: a) CaCl2 b) CaCO3 c) Ca(NO3)2 d) Ca3(PO4)2
3. Solve the following problems using factor-label:
a) How many grams are in 6.95 mol of magnesium oxide (MgO)?
b) How many moles are in 375 g of nitric acid (HNO3)?
c) What is the mass of 4.2 mol of silver nitrate?
d) How many moles are in 2.15 g of lithium oxalate?
1c) Molar Volume
● Volumes of gases can be measured at standard ambient temperature and
pressure SATP.
25oC
SATP
100 kPa
● At SATP, one mole of any gas always occupies a volume of 24.8 L. This is called its
molar volume.
1 mole (gas) = 24.8 L at SATP
memorize me too!
The unit for molar volume is L/mol.
.
Try This Solve the following using factor-label:
a) Find the volume of 0.505 mole NH3 gas at SATP.
b) How many moles are in 567 mL of any gas at SATP?
c) What is the density (g/L) of NO2(g) at SATP?
(Hint: use molar mass and molar volume).
2. Mole Conversions
The mole can now be defined in terms of amount (in particles), mass (in grams) or
volume of a gas (in liters). To change from one of these units into another, the mole unit
must be used as an intermediate step.
particles
(atoms or
molecules)
6.02 x 1023 particles/mole
mass
(of a formula)
MOLE
g/mol
volume
gaz at SATP
24.8 L/mol
© Mr. E. Therrien Chemistry 11 Notes ©
24
22
Example What volume of CO2 gas at SATP will contain 8.14 x 10 molecules?
x mole = 8.14 x 1022 molecules x 1 mole/ 6.02 x 1023 molecules = 0.1352 mole
x L = 0.1352 mole x 24.8 L/mole = 3.35 L
Try This 1. Use a two-step factor-label to solve the following:
a) What is the volume of 2.8 g of CO2(g) at SATP?
b) What is the mass of 5.2 x 1025 molecules of solid NaC2H3O2?
c) How many molecules are in 2.61 g of CaCl2 ?
d) What is the mass of 1.52 L of CH4 gas at SATP?
e) How many molecules of NO2 gas at SATP will make a volume of 6.45 L?
3. Concentration
● Concentration is a measure of the amount of solute that is dissolved in a given amount
of solvent. The most important unit of concentration is Molarity (M).
Molarity is the number of moles of a solute dissolved in one liter of a solution.
concentration (c) = Molarity (M) = n/V = moles of solute / liter of solution
S.I. term
non-S.I. term
● Molar solutions are prepared in volumetric flasks.
To Make a 1M Solution:
Add some solvent
add 1 mole of solute
Fill to the 1 liter mark
Try This Use factor-label or the molarity formula to solve these problems:
solute
g
mL
150
Molarity
(M)
(a)
no. of moles
(n)
(b)
liters
(L)
(c)
NaCl
6.0
CaCl2
1.52
(d)
(e)
(f)
1.21
Mg(OH)2
(g)
75
0.15
(h)
(i)
4. Ideal Gas Law
● We know that all gases exhibit the same behavior:
1 mole = 24.8 L at SATP (25oC and 100 kPa)
● Actual gases, though, are seldom at SATP. The volume of a gas at other temperatures
and pressures will be different.
© Mr. E. Therrien Chemistry 11 Notes ©
25
action
Tx2
effect
Vx2
Tx½
Vx½
Px2
Vx½
Px½
Vx2
statement
volume is directly related to
Kelvin temperature.
volume is inversely related to
pressure.
variation
VαT
V α 1/P
● Note: (i) All temperatures (T) are in Kelvin degrees (K = oC + 273).
(ii) The main unit for measuring pressure (P) is the kiloPascal (kPa).
● These ideas are combined into the Ideal Gas Law:
PV = nRT
pressure volume
(kPa)
(L)
number constant temperature
of moles (8.31)
(K)
Try This 1) Find the number of moles of helium gas that will be needed in a 3.54 L
balloon in order to create a pressure of 152 kPa at 22.3oC.
2) What pressure would result from adding 15 g of O2 gas into a 12 L container at 60oC?
3) What temperature would be needed in order for 6.2 mole of a gas to create a pressure
of 305 kPa in a 5.6 L container?
5. Using The Mole
a) Percent Composition
 This is the percentage, by mass, of each element that makes up a compound, for a total
of 100%.
Let m = mass (g) or molar mass (g/mol)
% composition = m(element) / m(compound) x 100
Example 1 What is the percent composition of the compound H2CO3?
m(compound)
H x 2 = 2.0 g/mol % H = 2.0/78.0 x 100 = 2.6 %
C x 1 = 12.0 g/mol % C = 12.0/78.0 x 100 = 15.4 %
O x 4 = 64.0 g/mol % O = 64.0/78.0 x 100 = 82.1 %
78.0 g/mol
Example 2 What is the percent composition of a compound that contains 9.03 g of Mg
that has combined with 3.48 g of N?
Mg = 9.03 g
% Mg = 9.03/ 12.51 x 100 = 72.2 %
N = 3.48 g
% N = 3.48/ 12.51 x 100 = 27.8 %
12.51 g
© Mr. E. Therrien Chemistry 11 Notes ©
26
Example 3 What is the mass of the carbon found in 6.56 g of CO2?
m(compound) = 6.56 g
% C = 12.0/ 44.0 x 100 = 27.3 %
% comp = m(element) / m(compound) x 100
27.3 = m(element) / 6.56 g x 100
1.79 g = m(element)
Try This 1) Find the percent composition of a) CaCO3 b) HNO3
2) What is the percent composition of a molecule that contains
a) 3.70 g Al and 3.30 g O
b) 4.35 g Ag and 0.65 g S
3) What is the mass of the hydrogen found in a) 64 g C4H8
b) 12.45 g H2SO4
5b) Empirical Formulas
► An empirical formula gives the lowest whole-number ratio of the elements in a
compound.
Molecular (actual) Formula
Empirical (reduced) Formula
H2O
H2O
H2O2
HO
C2H6
CH3
Note that the molecular formula may or may not be the same as the empirical formula.
► A math review of lowest whole-number ratio:
ratio
4 boys : 8 girls
C4H8
►
simplify with G.C.F.
4 boys : 8 girls
C4H8
molecular
formula
empirical
formula
lowest whole-number ratio
1 boy : 2 girls
CH2
%
composition
Our strategy for this step will be
% Comp.  grams  moles  l w-n ratio of moles  emp. Formula
Try This 1. What is the empirical formula for a compound that is:
a) 75% carbon and 25% oxygen
b) 67.6% Hg, 10.8% S and 21.6% O
2. For the empirical formula for a substance that was found to contain:
a) 8.20 g of magnesium and 5.40 g of oxygen.
b) 29.0 g Ag and 4.30 g S
© Mr. E. Therrien Chemistry 11 Notes ©
27
6. Interpreting Chemical Equations
Calculations using balanced equations are called stoichiometric calculations. The starting
point for any problem involving quantities of chemicals in a reaction is the balanced
equation.
Consider the reaction in which the reactants are nitrogen gas and hydrogen gas. They
produce the product ammonia gas.
N2 + H2 
NH3
The balanced equation shows us that 1 mole of nitrogen gas joins with three moles of
hydrogen gas to produce two moles of ammonia gas.
N2 + 3H2  2NH3
Atoms
Molecules
Moles
Volume (L@SATP)
Mass (g)
Reactants
2+6
1+3
1+3
24.8 + 74.4
28.0 + 6.0
Product
8
2
2
49.6
34.0
Conserved?
Yes
No
No
No
Yes
Thus, we have the Law of Conservation of Matter (Mass) - Matter (mass) is conserved
during any chemical change.
Try This Prove that atoms and their mass are conserved during the following reaction:
2HCl + Ca(OH)2

CaCl2 + 2H2O
7. Mole Ratios
Balanced equations tell us the number of moles of each chemical taking part in the
reaction. The ratio that compares the number of moles of one chemical to another
chemical is called a mole ratio. There are six mole ratios that can be made from this
reaction: N2 + 3H2 
2NH3
(1:3:2 mole ratio)
1) 1 mole N2 2) 3 mole H2 3) 1 mole N2 4) 2 mole NH3 5) 3 mole H2 6) 2 mole NH3
3 mole H2
1 mole N2
2 mole NH3 1 mole N2
2 mole NH3 3 mole H2
These mole ratios are used to help solve mole-mole problems. It will be a one-step
Solution:
Moles of Chemical A

Moles of Chemical B
Mole ratio
Try This
Methanol fuel burns in air: 2CH3OH + 3O2 
2CO2 + 4H2O
If 3.50 mole of methanol (CH3OH) are burned in lots of oxygen, then
a) How many moles of oxygen are used?
b) How many moles of water are produced?
c) How many moles of carbon dioxide gas are produced?
8. Gravimetric Stoichiometry
The procedure for calculating the masses, in grams, of reactants or products from a
balanced chemical equation is called gravimetric stoichiometry. One example of this is
© Mr. E. Therrien Chemistry 11 Notes ©
28
the mass-mass problem. The measured mass of one chemical (A) is used to find the
unknown mass of a second chemical (B).
1. Mass of A

Moles of A
formula mass (A)
2. Moles of A 
Moles of B
mole ratio
3. Moles of B

Grams of B
formula mass (B)
Note that our mass-mass problems have a three-step solution.
Example:
How many grams of hydrogen gas will be required to react with 50.0 grams of nitrogen
gas in the following reaction
N2(g) + H2(g) 
NH3(g)
Step 1: g

mole (refer to given info)
X mole N2 = 50.0 g N2 x 1 mole N2 = 1.773 mole N2
28.2 g N2
Step 2: mole N2 
mole H2
X mole H2 = 1.773 mole N2 x 3 mole H2 = 5.352 mole H2
1 mole N2
Step 3: mole H2 
g H2
X g H2
= 5.352 mole H2 x 2.02 g H2 = 10.7 g H2
1 mole H2
Try This
1. Suppose 2.85g of hydrogen gas are reacted with oxygen gas in the reaction
2H2(g) + O2(g) 
2H2O(l)
a) How many grams of oxygen gas will be needed?
b) How many grams of water will be made?
2. The equation for the roasting of iron(II) sulfide is as follows:
4FeS + 7O2 
2Fe2O3 + 4SO2
If we started with 95g of iron(II) sulfide, then
a) What mass of oxygen gas will be needed?
b) What mass of sulfur dioxide will be produced?
3. One type of antacid is magnesium hydroxide:
Mg(OH)2 + 2HCl 
MgCl2 + 2H2O
If an antacid tablet has a mass of 4.56g, then
a) What mass of hydrochloric acid will it react with?
b) What mass of water will it make?
9. Gas Stoichiometry
9a) Standard Conditions
(At SATP)
The three steps of the stoichiometry calculation are the same for solids and gases.
When gases at SATP are involved in the calculation, we can use the molar volume as a
conversion factor
1 mole of any gas = 24.8 L at SATP = molar volume
© Mr. E. Therrien Chemistry 11 Notes ©
29
Our three-step solution will be:
1. L (at SATP) of A
2. mol of A
3. mol of B

mole of A
molar volume

mole of B
mole ratio

L (at SATP) of B
molar volume
Note: (i) The second step will always be a “mole flip”.
(ii) A problem may involve both volume and mass, but the solution will still be in threesteps.
Try This
1. Tin (II) fluoride, formerly found in many toothpastes, is formed in the reaction Sn(s) +
2HF(g) 
SnF2(s) + H2(g)
a) How many liters of H2 will be formed from 6.5 L HF at SATP?
b) What mass of tin will react with 3.45 L HF at SATP?
2. Match heads contain potassium chlorate to help burning:
2KClO3(s) 
2KCl(s) + 3O2(g)
a) What mass of KCl is formed when 45 mL O2 is also formed at SATP?
b) What volume of oxygen at SATP can be made from 3.12 g KClO3 ?
9b) Non-Standard Conditions (Not at SATP)
If the conditions in an experiment are not at SATP, then the Ideal Gas Law rather than
molar volume is used in calculations involving a gas.
pv = nRT
This equation can be rearranged into n = pv/RT or v = nRT/p
p = pressure (kPa/atm)
v = volume (L)
n = moles
R = gas constant * 8.31
T = temp (kelvin)
Our solutions for these problems (involving gases at non-SATP) will follow these three
steps:
1. volume (L) of A

mole of A
n = pv/RT
2. mole of A 
mole of B
mole ratio
3. mole of B

L of B
v = nRT/p
Try This
1. What volume of ammonia gas (NH3) at 450 kPa and 80oC can be made from the
complete reaction under the same conditions of 64 L H2?
N2(g) + 3H2(g)  2NH3(g)
2. Photosynthesis is 6CO2(g) + 6H2O(l) 
C6H12O6(s) + 6O2(g)
Suppose a green plant produces 312 g of glucose (C6H12O6).
a) What volume of CO2 at 23 oC and 102 kPa is used?
b) What volume of 6O2 is made at 10 oC and 101 kPa?
© Mr. E. Therrien Chemistry 11 Notes ©
30
10. Solutions
Most reactions take place in solutions. The concentration (Molarity) of a solution is
found with this formula:
M=n/V
Molarity
(mol/L)
moles
of solute
(mole)
volume
of solution
(L)
This formula will be used in the three-step mole method whenever solutions (aq) are
involved in the problem.
(aq = dissolved in water)
The formula can be rearranged into n = MV and V = n/M
Titration is a common method used to determine an unknown concentration from the
known concentration of another chemical that it reacts with.
Chemical 1
Known V1
Unknown M1
Chemical 2
add  Known V2
Known M2
Endpoint
stop at  (sudden
change)

Calculate
M1
Try This
1. Suppose 12.0 g of zinc metal is to be reacted with 0.800 M HCl:
Zn(s) + 2HCl(aq)

ZnCl2(aq) + H2(g)
What volume, in L, of HCl will be needed?
2. A can of pop contains 280 mL of 0.150 M solution of carbonic acid (H2CO3), which
decomposes when pop goes “flat”:
H2CO3(aq) 
CO2(g) + H2O(l)
How much CO2 gas at 37 oC and 101 kPa will you burp after drinking the can?
3. Calculate the molarity (mol/L) of the sulfuric acid that is made by reacting 10.0 L of
SO2(g) at SATP with water to make 0.845 L of the acid:
SO3(g) + H2O(l) 
H2SO4(aq)
11. Applications
a) Limiting Reactant
During chemical reactions, one reactant is used up first. This is the limiting reactant
and it actually determines the amount of product formed.
Let’s assume A + B

C
If A = 1 mole but B = ½ mole, then B is the limiting reactant because it is used up first.
(Note that only ½ mole of C can be made during this reaction.)
© Mr. E. Therrien Chemistry 11 Notes ©
31
Any problem that gives a mass for all of the reactants will require us to identify the
limiting reactant. Our calculations will then involve only the limiting reactant. We will
ignore the reactant(s) in excess.
Example:
1. Solid ammonium chloride can be made from two gases:
NH3(g) + HCl(g) 
NH4Cl(s)
If 1.00 g NH3 is mixed with 1.00 g HCl, calculate
the actual mass of the ammonium chloride that will be formed.
a. find moles of reactants:
x mole NH3 = 1.00 g NH3 x 1 mole/17.0 g NH3
= 0.0588 mole NH3
x mole HCl = 1.00 g HCl x 1 mole/ 36.5 g HCl
= 0.0274 mole HCl
b. identify limiting reactant:
(actual) mole NH3 : mole HCl = 1 : 1
(measured) mole NH3 : mole HCl = 0.0588 : 0.0274 = 1 : 0.47
c. solve:
x mole HCl = 1.00 g HCl x 1 mole / 36.5 g HCl
= 0.0274 mole HCl
x mole NH4Cl = 0.0274 mole HCl x 1 mole NH4Cl / 1 mole HCl
= 0.0274 mole NH4Cl
x g NH4Cl = 0.0274 mole NH4Cl x 53.5 g NH4Cl / 1 mole NH4Cl
= 1.47 g NH4Cl
Try This
1. Calcium metal burns to form CaO(g) : 2Ca + O2  2CaO
Calculate the mass of CaO that can be formed when 0.48 g Ca is burned in 0.32 g O2.
2. What mass of HCl(g) is produced when 4.50 g H2(g) and 140.0 g Cl2(g) are reacted
according to the equation: H2(g) + Cl2(g) 
2HCl(g)
11b. Percentage yield
Most chemical reactions do not produce the amount of product that is predicted by the
balanced equation. The actual yield is usually less than the theoretical yield.
% yield = actual yield (unit) / theoretical yield (unit) x 100
Note that yield is usually measured as mass, in g.
Example: Ethanol can be made from sugar: C6H12O6(s) 2C2H5OH + 2CO2
In one experiment 10.0 g of sugar produced 0.664 g of ethanol (C2H5OH). What was the
percentage yield?
1. actual yield (given): 0.664 g C2H5OH
2. theoretical yield: x mole sugar = 10.0 g sugar x 1 mole sugar / 180.0 g sugar
= 0.05556 mole sugar
x mole C2H5OH = 0.05556 mole sugar x 2 mole C2H5OH / 1 mole sugar
© Mr. E. Therrien Chemistry 11 Notes ©
32
= 0.1111 mole C2H5OH
x g C2H5OH = 0.1111 mole C2H5OH x 46.0 g C2H5OH / 1 mole C2H5OH
= 5.11 g C2H5OH
3. solve: % yield = actual yield / theoretical yield x 100
= (0.664 g) / (5.11 g) x 100
= 13.0 %
Try This
1. One step in the formation of acid rain is the oxidation of sulfur dioxide (SO2):
2SO2 + O2 
2SO3. If 3.55 kg of SO2 is burned to make 4.20 kg of SO3, calculate
the percentage yield of the process.
2. Bromine gas can be made from acids: HBrO3 + 5HBr  3H2O + 3Br2. If 10.0 g of
HBrO3 was reacted and 26.3 g of Br2 was produced, what was the percentage yield of the
reaction?
© Mr. E. Therrien Chemistry 11 Notes ©
33
Organic Chemistry
1. Organic Chemistry
- Organic chemistry is the study of carbon compounds.
- The properties of organic compounds are determined by two factors:
i.
A molecular skeleton called a hydrocarbon. This is a series of carbon atoms
linked together to form a stable framework that is almost completely unreactive.
The unbonded electron positions are then filled (saturated) with hydrogen atoms.
ii.
A series of different reactive groups, called functional groups, can then be added.
They determined the basic “chemistry” of the molecule.
- Carbon atoms have unique properties that allow them to make many compounds:
i.
Carbon atoms can form four bonds ( C )
ii.
They link together with covalent bonds to form chains and rings of different sizes,
C C
C
C C C
Electron dot formula Structural formula
iii.
They form single, double and triple covalent bonds.
C C C
C C C
C C C
- Hydrocarbons tend to have some unique properties that are determined by their covalent
bonds:
i.
Hydrocarbons are easily decomposed by heat.
ii.
Hydrocarbons reactions are generally slow.
iii.
Most hydrocarbons are insoluble in water. Their molecules are nonpolar and are
not attracted to the polar water molecules.
2. Alkanes
a) Continuous –Chain Alkanes
- The simplest hydrocarbons are the alkanes that contain only hydrogen and carbon atoms
with atoms with single covalent bonds.
H H
H C C H
H H
- The formula for and alkane can be given several ways:
C5H12 (molecular formula)
H H H H H
1) H-C – C – C – C – C – H
(complete structural formula)
2 C
+ 6 H

H H H H H
2) CH3-CH2-CH2-CH2-CH3
(condensed structural formula or semi-structural)
© Mr. E. Therrien Chemistry 11 Notes ©
34
3) C-C-C-C-C
(skeletal structural formula)
4)
(line structural formula)
- The simplest alkanes form a chain. The first ten compounds in this homologous series
are to be memorized:
1
CH4 Methane
2 C2H6
Ethane
3 C3H8 Propane
4 C4H10 Butane
5 C5H12 Pentane
6 C6H14 Hexane
7 C7H16 Heptane
8 C8H18 Octane
9 C9H20 Nonane
10 C10H22 Decane
Note:
i.
The names for the alkanes end in “-ane”.
ii.
The general formula for alkanes is CnH2n+2
b) Branched – Chain Alkanes
- Sometimes and atom or group of atoms can replace a hydrogen atom in an alkane. The
added atom is called a substituent.
CH3-CH2-CH3  CH3-CH(Br)-CH3
If the substituent is another hydrocarbon, it is called and alkyl group.
CH3-CH2-CH3  CH3-CH(CH3)-CH3
An alkyl group (methyl) joined to the parent chain (propane)
-When an alkyl group is attached to a parent chain, it creates a branched-chain alkane.
The names of the alkyl groups are based on its alkane name but ending in “yl”.
Examples:
Methyl (-CH3)
Ethyl (-CH2-CH3) or (-C2H5)
Propyl (-CH2-CH2-CH3) or (-C3H7)
Note:
i.
An alkyl group cannot exist by itself; it must be attached to some molecule.
ii.
The general formula for alkyl groups is CnH2n+1
Example: CH3-CH2-CH-CH2-CH3
CH2
© Mr. E. Therrien Chemistry 11 Notes ©
35
CH2-CH3
This shows an ethyl group attached to a heptane molecule.
The formula can also be given as CH3-CH2-CH(C2H5)-CH2-CH2-CH3
Nomenclature for Branched-Chain Alkanes
1. For One Branch
1. Find the longest chain and name it.
2. Number the carbons in order to locate the alkyl group. (Begin from the end that
gives the branch the lower number).
3. The alkyl group name is placed in front of the parent chain name.
4. The number must be separated from the name with a hyphen.
Example: CH3-CH2-CH2
- longest chain = hexane
CH-CH2-CH3
- branch name = methyl
CH3
- branch location = 3rd atom
NAME: = 3 - methylhexane
2. For More Than One Branch
- Each branch is again located on the main chain.
- The branch names are listed alphabetically.
- Prefixes are used if an alkyl group appears more than once.
Example: CH3-CH2-CH2
CH-CH3
- longest chain = octane
- branches
= methyl at 4 and 5
= ethyl at 4
CH3-C-CH2-CH3
CH2-CH2-CH3
NAME: = 4,5-dimethyl, 4-ethyl- octane
Note:
i.
Branches can be identified in condensed formulas by placing them in brackets.
From the previous example (4-ethyl-4,5-dimethyloctane)
CH3-CH2-CH2-CH(CH3)-C(CH3)(C2H5)-CH2-CH2-CH3
ii.
Don’t forget: Count from the end that gives the lowest numbers
2c) Structural Isomers
- It is possible to draw the structures of two or more hydrocarbons that have the same
molecular formula but different structural formulas. These compounds are called
structural isomers.
C4H10
CH3-CH2-CH2-CH3
© Mr. E. Therrien Chemistry 11 Notes ©
CH3-CH(CH3)-CH3
36
butane
2-methylpropane
- Isomers will only be a different chemical only if they can be given a different name.
- Isomers have different chemical and physical properties.
Try This
1. How many structural isomers are possible for each of the first five straight-chain
alkanes.
2. Give the condensed structural formula and name for
a) all of the isomers for C5H12
b) the five isomers of C6H14
3. Optional Challenge – Name all 18 isomers of C8H18.
3. Alkenes and Alkynes
- Remember that carbon can also form multiple bonds. Hydrocarbons containing a double
bond are called alkenes
( C=C )
- Alkenes are unsaturated hydrocarbons because they contain fewer hydrogen atoms.
These atoms were removed in order to create the double bond.
- The other unsaturated hydrocarbon is the alkyne which contains a triple bond ( -CC- ).
Nomenclature:
1. Find the longest chain containing the multiple bonds (Its name will now end in “ene” or “-yne”).
2. The main chain must now be numbered to give the multiple bonds the lowest
number (instead of the branched chains). This number will be used in front of the
name of the main chain.
CH3-CH=C(CH3)-CH(CH3)-CH3
(3,4-dimethyl-2-pentene)
CH3-CH2-CH(C2H5)-CCH
(3-ethyl-1-pentyne)
Note:
i.
The general formula for an alkene is CnH2n and for an alkyne is CnH2n-2.
ii.
The physical properties of the unsaturated substances are similar to those of the
alkanes.
iii.
The most important difference in chemical properties is that multiple bonds are
quite reactive.
Try This Name or provide the structural formula for these compounds:
i.
CH3-C(CH3)=CH-CH(CH3)-CH3
ii.
CH3-CH2-CH(C2H5 )-CCH
iii.
5,6-dimethyl-2-octyne
iv.
2-methyl-1-propene
© Mr. E. Therrien Chemistry 11 Notes ©
37
Summary
Number
of C
1
2
3
4
5
6
7
8
9
10
Name of
the
prefix
for
Meth
Eth
Prop
But
Pent
Hex
Hept
Oct
Non
Dec
Hydrocarbons
molecule alkane
CnH2n + 2
Hydrocarbons
molecule alkene
CnH2n
Hydrocarbons
molecule alkyne
CnH2n -2
CH4
C2H6
C3H8
C4H10
C5H12
C6H14
C7H16
C8H18
C9H20
C10H22
nil
C2H4
C3H6
C4H8
C5H10
C6H12
C7H14
C8H16
C9H18
C10H20
nil
C2H2
C3H4
C4H6
C5H8
C6H10
C7H12
C8H14
C9H16
C10H18
4. Cycloalkanes (Geometric Shapes)
- In some alkanes, the two ends of the chain have attached to form a ring. These are
called cycloalkanes.
Examples:
1) cyclobutane
2) cyclohexane
H2C – CH2
H2
C
H2C – CH2
H2C
CH2
or
or
H2C
CH2
C
H2
Note:
i.
Cyclohexanes have only single bonds and that there general formula is C nH2n.
Cycloalkenes and cycloalcynes also exist.
ii.
Cycloalkanes have strained bonds that lead to a less stable molecule than its
equivalent alkane.
Nomenclature: The ring carbons are numbered so as to give the substituents the lowest
possible number (One substituent will always be 1).
CH2CH3
CH3
H3C
CH3
1,3-dimethylcyclooctane
© Mr. E. Therrien Chemistry 11 Notes ©
1-ethyl-2-methylcyclopropane
38
Try This Give the structural formula for these compounds.
a) 1-ethylcyclobutane
b) 1,3-dimethylcyclopentane
c) 1,1-dimethylcyclopropane
d) cyclopropylcyclohexane (!)
5) Benzene
- The most important group of cyclic hydrocarbons is the aromatic compounds, the
simplest of which is benzene (C6H6). After benzene forms a ring, each carbon atom has
one extra electron that is free to make half of a double bond.
HC
CH
HC
CH
C
H
HC
HC

CH
C
H
C
H
The benzene ring can be shown as
CH
C
H
or
Resonance occurs when two valid structures can be drawn for the same molecule.
Resonance is the result of the combination of blending of the two possible structures. The
actual substance is an average of all the possible arrangements.
Molecules that exhibit resonance are very stable.
Nomenclature:
1) One substituent Added:
(branch)
CH3 or C6H5(CH3)
Methylbenzene (or toluene)
2) When Used as a Substituent – named phenyl (-C6 H5) *(CnHn-1)
CH3-CH-CH2-CH3 or CH3-CH(C6H5)-CH2-CH3
2-phenylbutane
3) Two Substituent Added (note the positions)
H3C
CH3
H3C
H3C
CH3
1,2-dimethylbenzene
or
o-dimethylbenzene
(“ortho” position)
1,3-dimethylbenzene
or
m-dimethylbenzene
(“meta” position)
© Mr. E. Therrien Chemistry 11 Notes ©
CH3
1,4-dimethylbenzene
or
p-dimethylbenzene
(“para” position)
39
Note: that one Substituent must be in the “l” position.
6 a) Fractional Distillation
Directions: Use the key words and phrases with its matching step in the diagram to give a
step-by-step explanation of the fractional distillation of oil.
Tower
a) Crude oil, fraction, alkanes
b) Heater, 350oC, boiling point, vaporizing
c) Bubble cap, rising gases, condensation, trays
Fractions
1.
2.
3.
4.
5.
6.
Residue, (tars, asphalt), highest boiling point (400oC), C30
Lubricants and waxes, 3500C, approximately C20 – C30
Heating/diesel oil, 250oC, C16 – C20
Kerosene, 180oC, approximately C12 – C16
Gasoline, 110oC, approximately C5 – C12
Natural gases, lowest boiling points, C1 – C4, LPG
6 b) Hydrocarbon Reaction
1. Cracking
Large fractions form the fractionation process is chemically broken with heat and a
catalyst to produce more valuable fuel hydrocarbons with 5-12 carbon atoms per
molecule.
Large molecule  small molecules
Example:
C17H36  C8H18 + C9H18
alkane alkene
2. Reforming
This is the opposite of cracking. It can convert low-grade gasolines into higher grades
with the use of heat and a catalyst. Small molecules  large molecule
Example:
2 C5H12(l)  C10H22(l) + H2(g)
© Mr. E. Therrien Chemistry 11 Notes ©
40
3. Combustion
Most hydrocarbons are used as fuels (95%).
Hydrocarbon + O2(g)  CO2(g) +18H2O(g) + heat
Example:
2C8H18 + 25O2  16CO2 + 18H2O + heat
4. Hydrogenation
This process converts unsaturated hydrocarbons to saturated ones.
Alkene + H2(g)  alkane
Alkyne + H2(g)  alkene
Example:
CH2=CH-CH2-CH3(g) + H2(g)  CH3-CH2-CH2-CH3(g)
1-butene
butane
Try This
1. Identify and complete each reaction. You must not break the Law (of Conservation of
Matter)
a) C14H30 into octane b) ethane into octane c) 1-octene into octane d) 1-octyne into octane
2. Show the complete combustion of:
a) pentane b) hexane c) hexene
7. Functional Groups
- The hydrocarbon skeleton of an organic molecule is chemically inert. Most organic
chemistry, then, involves the atoms and molecules that are attached to this main chain.
a) Structures
- Functional groups are the atoms in an organic compound that have been added to a
hydrocarbon chain. They are the only part of the molecule that is capable of reacting
chemically.
- Compounds are classified according to their functional groups. Since these groups can
be attached to any chain, we will represent the inert hydrocarbon chain with “R”.
Family Name
Condensed Formula
Example (3C’s)
General Formula
1. Halocarbons
R(X)
CH3-CH2-CH2(Cl)
CnH2n+1X
2. Alcohols
R(OH)
CH3-CH2-CH2(OH)
CnH2n+2O
3. Ethers
R1 – O – R2
CH3-O-CH2-CH3
CnH2n+2O
4. Aldehydes
R – C(=O)H
CH3-CH2-C(=O)H
CnH2nO
R1 – C(=O) – R2
CH3-C(=O)-CH3
CnH2nO
5. Carboxylic acids
R – C(=O)(OH)
CH3-CH2-C(=O)(OH)
CnH2nO2
6. Esters
R1 – C(=O) – O – R2
CH3-C(=O)-O-CH3
CnH2nO2
7. Polymers
( - R1 – R2 - )n
(-CH2-CH2-CH2-)n
Ketones
© Mr. E. Therrien Chemistry 11 Notes ©
41
Try This
1. Why does oxygen have to bond twice in a molecule?
2. How many times would these atoms bond? a) S b) Cl c) N d) F e) P
3. Which families cannot be made from methane?
4. Do any groups have the same general formula?
b) Two Physical Properties
- The presence of a functional group has a major effect on the physical properties of a
compound. Intermolecular forces determine many physical properties, such as solubility
and boiling point:
- hydrocarbons – have only weak dispersion forces.
- functional groups – may have strong hydrogen bonds.
- Remember that hydrogen bonds are created between two molecules:
a) One molecule contains a hydrogen atom bonded to a highly
electronegative atom (especially oxygen). These hydrogen atoms now can be
considered to be a naked proton (+).
b) The other molecule will be polar with a - section (especially on oxygen).
The hydrogen bonding, then, occurs between two highly polar molecules.
- Boiling Points: Compounds with hydrogen bonding have higher than expected boiling
points: at SATP, C2H6 = gas but CH3OH = liquid.
- Solubilities: Compounds with hydrogen bonds are usually soluble in water (a polar
molecule): C6H14 = insoluble in water but C5H11OH = soluble in water.
Solubility decreases, as the molecule gets longer. (C12H25OH =slightly soluble in water).
This is because the hydrogen bonding –OH group has less effect on the larger molecule.
8. 1. Halocarbons (R – X)
- Halocarbons are produced by the substitution of a halogen (Family VIIA) for hydrogen
in the hydrocarbon chain;
Cl = chloro Br = bromo I = Iodo F = Fluoro
as well as
NO2 = nitro NH2 = amino
Nomenclature: The halocarbon is treated as a substituent. Don’t forget that substituents
are named alphabetically.
CH3-C(Br)(CH3)-CH3
2-bromo-2-methypropane
CH3-CH(Cl)-C(NO2)(C2H5)-CH2-CH3
2-chloro-3-ethyl-3-nitropentane
Try This Name or provide the formula for the following:
a) CH3-CH2(I)
f) trichloromethane
b) CH2(Cl)-CH2(Cl)
g) 2-chloro-3-iodo-1-butene
c) CH(F)(F)-CH(Cl)(Cl)
h) 1,5-dinitro-2-pentyne
© Mr. E. Therrien Chemistry 11 Notes ©
42
d) CH3-CH2-CC-CH2(NH2)
i) 1,3-difluorocyclopentane
e) CH3-CH(NO2)-CH2(Br)
j) 1,3-dibromo-3-phenylhexane
Reactions:
a) Addition Reactions – A halogen (X) is added to an unsaturated hydrocarbon at the
double/triple bond. These reactions are quite rapid because there are no strong
covalent bonds to be broken.
R1 = R2 + X2  RX – RX
Ethyne + chlorine gas  1,2-dichlorethane
CHCH + Cl2  CH(Cl)=CH(Cl) which can be further reacted
CH(Cl)=CH(Cl) + Cl2  CH(Cl)(Cl)-CH(Cl)(Cl)
b) Substitution Reactions – A halogen replaces hydrogen in a saturated hydrocarbon.
These reactions occur quite slowly at room temperature because C – H bonds are
quite stable.
R – H + X2  R – X + HX
Ethane + chlorine  chloroethane + hydrogen chloride
C2H6 + Cl2  C2H5Cl + HCl
More complicated molecules will form isomers.
CH2(Cl)-CH2-CH2-CH3 + HCl
CH3-CH2-CH2-CH3 + Cl2
CH3-CH(Cl)-CH2-CH3 + HCl
c) Elimination Reactions – This is the most common method for preparing alkenes.
A halocarbon reacts with a hydroxide to produce an alkene.
R – R – X + OH-  R = R + H2O + XCH3-CH(Br)-CH3 + NaOH  CH2 = CH-CH3 + H2O + NaBr
9. 2. Alcohols
- Alcohols contain the hydroxyl (-OH) group: R – O – H
Nomenclatrue: Add “-ol” to the stem of the name of the parent chain. Number the
position of the hydroxyl group.
If more than one –OH group is present, use these endings:
a)“-diol” (two hydroxyls)
b) “-triol” (three hydroxyls)
Examples:
CH3-CH(OH)-CH3
2-propanol
CH3-CH(CH3)-CH2-CH2(OH)
3-methyl-1-butanol
CH2(OH)-CH2-CH(OH)-CH3
1,3-butandiol
Phenols are alcohols in which the –OH group is attached to benzene.
OH
[C6H5(OH)]
Try This Name or give the formulas for the following:
a) CH3-CH(OH)-CH3
e) 2-pentanol
b) CH2(Cl)-CH(OH)-CH3
f) 2,2,4-pentatriol
© Mr. E. Therrien Chemistry 11 Notes ©
43
c) CH2(OH)-CH2(OH)-CH2-CH3
g) 3-ethyl-2-pentanol
d) C6H4(OH)(CH3)
h) 2-ethylphenol
Reactions:
1. MAKING ALCOHOLS
i. Fermentation – producing ethanol from sugars by the action of yeast.
C6H12O6(aq)  2CH3CH2OH(aq) + 2CO2(g)
ii. Substitution – the halogen in a halocarbon is replaced by a hydroxide.
R – X + M – OH
 R – OH + MX
CH3-CH2(Cl) + NaOH
 CH3-CH2(OH) + NaCl
iii. Addition – a water molecule (H-OH) is combined with an unsaturated hydrocarbon.
R1 = R2 + H – OH
 R(OH)
CH2=CH2 + H-OH
 CH3-CH2(OH)
i.
2. REACTING ALCOHOLS
Elimination – Alcohols can be used to make alkenes when catalyzed by
concentrated sulfuric acid
acid
R – OH  R1 = R2 + H – OH
CH3-CH2(OH)  CH2=CH2 + H2O
Properties: Alcohols tend to have high boiling points due to the strong hydrogen
bonding that occurs between these molecules.
- Smaller molecules are also soluble in water as they can hydrogen bond with the polar
water molecules.
- Longer chain molecules tend to be insoluble as the hydroxyl group has less effect on the
molecule. They can be good solvents for non-polar molecular compounds (“like dissolves
like”).
Try This
1. Write a balanced equation for the following situations:
a) reacting 2-iodopropane with Ca(OH)2
b) making 1-butene from 1-butanol
c) making 1-propanol from propene
2. Why does wine bubble as it is fermenting?
10. 3. Ethers
- These are compound in which oxygen is bonded between two carbon groups.
O
R1
Properties:
Halocarbons
R2 Notice that this is actually a bent molecule.
Summary of Hydrogen Bonding
Hydrogen
donor
X
Hydrogen
acceptor
X
© Mr. E. Therrien Chemistry 11 Notes ©
Boiling point
increases
Solubility
increases
44
Ethers
X

Alcohols


Nomenclature: The two-alkyl groups are named in alphabetical order and followed by
the word “ether”.
CH3-CH2-CH2-O-CH3
CH3-CH2-O-CH2-CH3
methylpropyl ether
diethyl ether
Try This
1. The molecular formula C3H8O can represent either two alcohols or an ether. Give
their formulas and names.
2. Would an ether be soluble in a) a halocarbon? b) an alcohol?
11. 4. Aldehydes and Ketones
- These two families share certain structural features and chemical properties.
- They both contain the carbonyl group
C = O, which we will show as (-C(=O)-) in
O
our condensed formulas.
- Aldehydes = chain + carbonyl group + hydrogen = R – C – H
O
- Ketones = chain 1 + carbonyl group + chain 2 = R1 – C – R2
Nomenclature:
Aldhydes: change the “_e” ending to “_al”
O
H–C–H
CH3-CH2-CH(=O)
methanal
propanal
Ketones: count the total number of carbons present. Change the “_e” ending to “_one”.
Number the carbonyl location.
O
CH3-C-CH3
propanone
CH3-CH2-C(=O)-CH2-CH2-CH3
3- hexanone
Properties: - They have low boiling points because they have no O – H bonds for
hydrogen bonding.
- When added to water, though, they can hydrogen bond to the water molecules so they
are quite soluble in water.
Try This
C4H8O.
Draw and name the aldehyde and the ketone with the molecular formula of
12. 5. Carboxylic Acids
- These organic acids contain the carboxyl functional group, -C(=O)-OH which include
both the carbonyl group and the hydroxyl group. In molecular formulas, the carboxyl
group is often presented as –COOH.
© Mr. E. Therrien Chemistry 11 Notes ©
45
Properties: Carboxylic acids create the sour taste in foods and have distinctive odors.
- They are polar molecules and are both hydrogen acceptors and hydrogen donors. Thus
they will readily form hydrogen bonds, and smaller molecules are easily dissolved in
water.
- They have all of the properties of acids – react with metals, make indicators change
color, etc.
Nomenclature: Replace the “_e” ending with “_oic acid”.
structural formula
O
CH3-CH2-C-OH
propanoic acid
molecular formula
condensed formula
C2H5COOH
CH3-CH2-C(=O)-OH
Reactions: 1. Fermentation: glucose  ethanol  ethanoic acid
(grape juice) (wine)
(vinegar)
2. They undergo typical acid reactions (neutralization, for example) but at a slower rate.
Try This
1. Name a) CH3-CH2-CH2-C(=O)OH b) CH3-C(CH3)(CH3)-C(=O)OH c) COOH
2. Give the formula for a) hexanoic acid b) 3-methylpentanoic acid
3. Use the example to write a balanced equation for the organic reaction:
a) KOH + HCl  KCl + HOH
KOH + CH3-COOH  ?
b) 2Na + 2HCl  2NaCl + H2
Na + CH3COOH  ?
13. 6. Esters (Acid + Alcohol  Ester)
- Esters are derivatives of carboxylic acids in which the – OH of the carboxyl group has
been replaced by an – OR from an alcohol.
O
R1-C
or
R1COOR2 or
R1-C(=O)-O-R2
O-R2
Reactions: Esters are prepared by combining a carboxylic acid with an alcohol in a
process called esterification:
O
O
R1-C
+
HO-R2  R1-C
+ HOH
OH
O-R2
carboxylic acid
alcohol
ester
water
Nomenclature: The name of an ester has two parts:
i.
Locate the alcohol branch and name it as an alkyl group.
ii.
Locate the acid branch. The ending of the acid name is changed from “_oic acid”
to “_oate”
O
alcohol
acid
© Mr. E. Therrien Chemistry 11 Notes ©
46
CH3-C
CH3-CH2-C(=O)-O-CH2-CH3
O-CH3
methyl ethanoate
ethyl propanoate
Properties: -Esters are “odor” chemical (fruits and flowers).
- are added to foods to enhance taste + odor.
Try This Give the esterification reaction (with formulas and names) for
i.
Propanoic acid and butanol
ii.
Butanoic acid + propanol
iii.
Ethanol and methanoic acid
14. 7. Polymerization
- A polymer is a giant molecule formed by the covalent bonding together of repeating
smaller molecules called monomers.
…….. + monomer + monomer + ……..  polymer
- One method of creating these polymers by addition polymerization – the joining
together of unsaturated (multiple bond) monomers.
If A = the monomer, the polymer will appear as -A-A-A-Acatalyst
n CH2=CH2  (-CH2-CH2-)n
ethane (ethylene)
polyethylene
catalyst
n CH3CH=CH2  (-CH2(CH3)-CH2-)n
propene (propylene)
polypropylene
- A condensation polymer is formed by reacting two different compounds together. In
order to do this, a small molecule (usually water) must be split off from the functional
groups.
If A = first monomer and B = second monomer, then the polymer will appear as
-A-B-A-BHere is a general example:
HO-X-OH + HO-Y-OH  (-O-X-O-Y) + 2H2O
“di” acid or “di” alcohol
Note: polyethylene = ice cream plastic containers,
polypropylene = nylon rope
Try This Write the formula for the following polymers, which were formed by addition
polymerization.
1. Polyvinyl chloride (PVC) which is made form chloroethane.
2. Teflon which is made form tetrafluoroethene.
© Mr. E. Therrien Chemistry 11 Notes ©