LOGO
Lecture 3:
Chemical Reactions
Chemistry - SPRING 2015
Course lecturer :
Jasmin Šutković
18th March 2015
Contents
International University of Sarajevo
1. What is a Chemical Reaction ?
1.2 The mol and molar masses
2.
3.
4.
5.
6.
Determining Empirical and Molecular formulas
Chemical equations
Mass relationships in chemical equations
Types of Chemical equations
Earths atmosphere
1. Chemical reactions
A process where a substance is converted to one or
more other substances with different compositions and
properties.
The MOLE
 To analyze the transformations that occur
between individual atoms or molecules in a
chemical reaction, it is therefore absolutely
essential for chemists to know how many atoms
or molecules are contained in a measurable
quantity in the laboratory.
 The unit that provides this link is the mole
(mol)!!
The MOLE cont...
 Abbreviated - mol
The mole is a unit of measurement used in chemistry to
express amounts of a chemical substance, defined as
the amount of any substance that contains as many
elementary entities (e.g., atoms, molecules, ions,
electrons) as there are atoms in 12 grams of pure
carbon-12 (12C), the isotope of carbon with atomic
weight 12.
 In 12g of C there is 6.022×1023 atoms of carbon,


This corresponds to "Avogadro constant"
(6.022×1023 ) elementary entities of the
substance.
1.2 Atomic, molecular and
Molar masses
Atomic mass
The mass number is approximately or relativeley equal to
the numerical value of the atomic mass (Ar = relative
atomic mass )..check PSE !
 Exp of atomic mass (Ar) : for H it is 1 or for O it is 16
Molecular mass
Molecular mass or relative molecular mass represents the
sum of atomic masses in a compuond (the mass of the
molecule).
2.1 Molecular mass (Mr)
 The molecular mass is the mass of a molecule,
sometimes shown as Mr
 It is calculated by adding together the atomic
masses of the elements in the substance, each
multiplied by its subscript (written or implied) in
the molecular formula.
 Expample 1 : Mr of H2O= 2(1)+ (16) = 18
 Mr of a compound is the number that represents how
much the mass of this coumpound is higher than the
atomic mass unit (amu) =(1/12 mass of C12 isotope)
 SO the unit for Mr is amu or u .
EXAMPLE 2
The procedure for calculating molecular masses is
illustrated in Example 2
Find the molecular mass of table sugar (sucrose),
which has a molecular formula C12H22O11
molecular mass(Mr) of C12H22O11
= 12(12.01) + 22(1.008) + 11(16.00)
= 342.30
Empirical formula
 Unlike molecules, which have covalent bonds, ionic compounds do
not have a readily identifiable molecular unit.
 So for ionic compounds we use the formula masses (also called the
empirical formula masses) of the compound rather than the
molecular mass.
 The formula mass is the sum of the atomic masses of all the
elements in the empirical formula, each multiplied by its subscript
(written or implied).
 It is directly analogous to the molecular mass of a covalent
compound.
Molar Mass
 Molar mass is measured in grams/mole,
and is the mass of one mole of any
substance.
 In each case, the number of grams in 1 mol is the same as the
number of atomic mass units that describe the atomic mass, the
molecular mass, or the formula mass, respectively.
 Exp: Relative atomic mass if Natrium is 23 (Ar) and Molar mass
of Natrium is 23 g/mol
Example ...
The molar mass of ethanol is the mass of ethanol
(C2H5OH) that contains 6.022 x 1023 ethanol molecules.
Using the procesure in Example 1 and 2 , the molecular
mass (M) of ethanol would be 46.069 g/mol.
•How ..?
Because 1 mol of ethanol contains 2 mol of carbon
atoms (2 × 12.011 g), 6 mol of hydrogen atoms (6 ×
1.0079 g), and 1 mol of oxygen atoms (1 × 15.9994
g), its molar mass is 46.069 g/mol.
Flowchart for converting
between units.
2. Determining molecular and
empirical formulas
 When a new chemical compound, such as a
potential new pharmaceutical, is synthesized in
the laboratory or isolated from a natural source,
chemists determine its elemental composition, its
empirical formula, and its structure to understand
its properties.
 In this section, we focus on how to determine the
empirical formula of a compound .
Laws and order in Chemical
binding
 Dalton's Atomic Theory - Dalton
 Law of Conservation of Mass - Lavosier
 Law of Definite Proportion – J.Proust
 Law of Multiple Proportion - Dalton's
Law
 http://sciencepark.etacude.com/chemistry/l
aw.php
Calculating mass
percanteges
 The law of definite proportions states that a
chemical compound always contains the
same proportion of elements by mass; that
is, the percent composition — the
percentage of each element present in a
pure substance — is constant (although
we now know there are exceptions to this
law).
Calculating mass
percanteges cont...
For example, today we know that sucrose (cane sugar) is
42.11% carbon, 6.48%hydrogen, and 51.41% oxygen by
mass. This means that 100.00 g of sucrose always
contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41
g of oxygen.
 How to prove this ?
 First we will use the molecular formula of sucrose (C12H22O11) to
calculate the mass percentage of the component elements; then we
will show how mass percentages can be used to determine an
empirical formula.
 According to its molecular formula, each molecule of sucrose
contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen
atoms.
 We can then use these masses to calculate the percent composition of
sucrose. To three decimal places, the calculations are the following:
Sucrose example cont...
Sum = 342.297g = 342.297 g/mol = Molar Mass (M)
The mass percentage of each element in sucrose is the mass of the
element present in 1 mol of sucrose divided by the molar mass of
sucrose, multiplied by 100 to give a percentage. The result is shown to
two decimal places:
Sucrose example cont...
You can check your work by verifying that the sum of the percentages of all the
elements in the compound is 100%:
42.12% + 6.48% + 51.41% = 100.01%
If the sum is not 100%, you have made an error in your calculations. (Rounding to
the correct number of decimal places can, however, cause the total to be slightly
different from 100%.)
Thus 100.00g of sucrose contains 42.12 g of carbon, 6.48g of hydrogen, and
51.41 g of oxygen; to two decimal places, the percent composition of sucrose is
indeed 42.12% carbon, 6.48% hydrogen, and 51.41% oxygen.
Percentage = Compositions
Empirical formula of
Penicilin
 Antibiotics are chemical compounds that selectively kill
microorganisms, many of which cause diseases.
 Although we may take antibiotics for granted today,
penicillin was discovered only about 80 years ago. The
discovery of penicillin is a historical detective story in
which the use of mass percentages to determine
empirical formulas played a key role.
 In 1928, Alexander Fleming discovered penicilin…
Story of Penicilin ….
 It was Alexander Fleming at St Mary’s Hospital
Medical School who noticed a mould growing on
a culture plate of bacteria one day in September
1928. Around the Penicillium notatum mould was
a clear area where the colonies of bacteria
appeared to have been dissolved.
 Fleming set about establishing the identity of the
substance that was killing the bacteria, giving it
the name ‘penicillin’
Penicillum
Penicilin empirical formula
cont...
 Although Fleming was unable to isolate penicillin in pure
form, the medical importance of his discovery stimulated
researchers in other laboratories.
 Finally, in 1940, two chemists at Oxford University,
Howard Florey and Ernst Chain isolated penicilin and its
use greatly increased the survival rate of wounded
soldiers in World War II.
 As a result of their work, Fleming, Florey, and Chain
shared the Nobel Prize in Medicine in 1945
Penicilin empirical formula
cont...
They discovered that a typical sample of penicillin G contains
53.9% carbon, 4.8% hydrogen, 7.9% nitrogen, 9.0% sulfur,
and 6.5% sodium by mass. The sum of these numbers is
only 82.1%, rather than 100.0%, which implies that there
must be one or more additional elements.
 A reasonable candidate is oxygen, which is a common component of
compounds that contain carbon and hydrogen! Hovewre for techniqual
reasons it is difficult to analyze the oxygen directly. If we assume that all
the missing mass is due to oxygen, then penicillin G contains (100.0% 82.1%) = 17.9% oxygen. From these mass percentages, the empirical
formula and eventually the molecular formula of the compound can be
determined.
The Way...
 In order to determine the
empirical formula we need
to convert the mass
percentages to relative
number of atoms.
 We assume that we deal
with 100g of compounds ...
100-82.1 = 17,9
The way...cont ....
 But ..moles are not whole number ration that we
need for empirical formula ...
 The empirical formula expresses the relative
numbers of atoms in the smallest whole numbers
possible. To obtain whole numbers, we do it as
follows :
Penicilin Empirical formula
C16H17N2NaO4S
Combustion analysis
 Analytical procedure where we determine the
composition of an unknown hydrocarbon . This is
done in an OXYGEN atmosphere by one of
several possible methods.
Book page number 156
STEPS IN OBTAINING AND
EMPIRICAL FORMULA VIA
COMBUSTION ANALYSIS
Determine the empirical
formula of naphthalene ?
Complete combustion of a 20.10 mg sample of naphthalene
in oxygen yielded 69.00 mg of CO2 and 11.30 mg of HO.
Given: mass of sample and mass of combustion products
Asked for: empirical formula
Example shown in page 157 !!!
From Empirical to
Molecular Formula
 Empirical formula = smallest possible ratio –
shortes representation of a chemical formula
 For covalent compound we usually use
Molecular formula, giving the reall number of
atoms in a compound.
 Without additional info it is not easy to detemine
wether a compound is shown via empirical or
molecular formula .
 Exp :Pinicilin
 C16H17N2NaO4S or C32H34N4Na2O8S2
Example with Glucose ...
 Combustion analysis report that Glucose contains
39.68% carbon and 6.58% hydrogen ..where is O ?
Combustion analysis occurs in Oxygen atmosphere ..it
can not calculate O % !!
 If we assume that the remaining percentage is due to
oxygen, then glucose would contain 53.79% oxygen
 A 100.0 g sample of glucose would therefore contain
39.68 g of carbon, 6.58 g of hydrogen, and 53.79 g of
oxygen. To calculate the number of moles of each
element in the 100.0 g sample, we divide the mass of
each element by its molar mass (M)
Example with Glucose cont....
The oxygen : carbon ratio is 1.018, or approximately 1, and the hydrogen :
carbon ratio is approximately 2. The empirical formula of glucose is therefore
CHO2, but what is its molecular formula?
Example with Glucose cont....
Each glucose contains 6 CH2O formula units, which gives a molecular formula
for glucose of (CH2O) x 6, which is more commonly written as
C6H12O6
3.Chemical equations
 We can describe chemical reaction as chemical
equations, an expression that gives the identities and
quantities of the substances in a chemical reaction.
 Chemical formulas and other symbols are used to
indicate the starting material(s), or reactant(s), which by
convention are written on the left side of the equation,
and the final compound(s), or product(s), which are
written on the right. An arrow points from the reactant
to the products:
Example book page 165
How to interpret a chemical
equation ?
Ammonium Dichromate
Volcano
Balancing simple chemical
equations
 When a chemist faces a new chemical reaction
it is not balanced not he/she knows whether what
is a reactant or product!
 Exp: combustion of n-heptane (C7H16)
 Equation is not balanced!!!
Steps in Balancing a
C.Equation
The following method is used , called “inspection,” or better known as
trial and error.
n-heptane balancing sample
4.Mass relationships in
chemical equations
 Stoichiometry is a collective term for the quantitative relationships
between the masses, the numbers of moles, and the numbers of particles
(atoms, molecules, and ions) of the reactants and the products in a balanced
chemical equation.

A stoichiometric quantity is the amount of product or reactant specified by the
coefficients in a balanced chemical equation
 Why Stoichiometry? TO ANSWER problems such as:
 How much oxygen is needed to ensure complete combustion of a given amount
of isooctane?
 How many grams of pure gold can be obtained from a ton of lowgrade gold ore?
 If an industrial plant must produce a certain number of tons of sulfuric acid per
week, how much elemental sulfur must arrive by rail each week?
Example ...again Glucose ...
Limiting reactants problem
 The reactant that restricts the amount of product obtained is called
the limiting reactant. Example:
5. Clasifying Chemical reactions
Oxidation-Reduction
reactions
•
The term oxidation was first used to describe
reactions in which metals react with oxygen in air
to produce metal oxides.
– Metal acquires a positive charge by transferring electrons to
the neutral oxygen atoms of an oxygen molecule.
– Oxygen atoms acquire a negative charge and form oxide
ions (O2-).
– Metals lose electrons to oxygen and have been
oxidized—oxidation is the loss of electrons.
– Oxygen atoms have gained electrons and have been
reduced—reduction is the gain of electrons.
Oxidation-Reduction
reactions cont..

Oxidation and reduction reactions are now characterized by a
change in the oxidation states of one or more elements in the
reactants.

Oxidation states of each atom in a compound is the charge that
atom would have if all of its bonding electrons were transferred to
the atom with the greater attraction for electrons. Atoms in their
elemental form are assigned an oxidation state of zero.

Oxidation-reduction reactions are called redox reactions, in which
there is a net transfer of electrons from one reactant to another.
The total number of electrons lost must equal the total number of
electrons gained.
Oxidation-Reduction
reactions cont..
Rules for assigning oxidation states
1. The sum of the oxidation states of all the atoms in a
neutral molecule or ion must equal the charge on the
molecule or ion.
2. The oxidation state of an atom in any pure element,
whether monatomic, diatomic, or polyatomic, is zero.
3. The oxidation state of a monatomic ion is the same as its
charge.
Oxidation-Reduction
reactions cont..
Oxidants and Reductants
Oxidants
– Compounds that are capable of accepting electrons are called
oxidants, or oxidizing reagents, because they can oxidize
other compounds.
– An oxidant is reduced in the process of accepting electrons.
Reductants
– Compounds that are capable of donating electrons are called
reductants, or reducing agents, because they can cause the
reduction of another compound.
– A reductant is oxidized in the process of donating electrons.
Oxidation-Reduction
reactions cont..
Oxidants and Reductants
Relationships of oxidants and reductants are
summarized in the following equation:
oxidant + reductant  oxidation – reduction
gains eis reduced
loses eis oxidized
redox reaction
OXIDATION STATE
(number)
 In chapter 2 several examples are shown
how chemical bonds are formed (how
electrons are shared) like MgO,CaCl2
etc...
 There are some rule that we may
memorize to know the basic elements
oxidation states such as Oxigen, Hydrigen,
Chlor .
EXAMPLE (page 193)
 In the following example we can apply
those rules: reduction of copper I oxide
Condensation reactions
 Condensation reactions have the general form: A + B 
AB
 Some condensation reactions are redox reactions.
 Reaction of an amine (-NH2) with a carboxylic
acid(COOH) is a variant of a condensation reaction, in
which –OH from the carboxylic acid group and –H from
the amine group are eliminated as water. The reaction
forms an amide or peptide bond, the essential structural
unit of proteins and many polymers.
Example
 The reaction of bromine with ethylene to
give 1,2-dibromoethane, which is used in
agriculture to kill nematodes in soil, is as
follows:
This is a condensation reaction because it has the general
form A + B => AB
This reaction, however, can also be viewed as an
oxidation–reduction reaction, in which electrons are
transferred from carbon (-2 => -1) to bromine (0 => -1)
Is this a REDOX reaction ?
Catalysts
A substance that speeds up a chemical reaction,
but is not consumed by the reaction
Catalysts are classified as
1. homogeneous — uniformly dispersed throughout the
reaction mixture to form a solution(for exp : Sulfuric acid)
2. heterogeneous — in a different physical state from the
reactants (iron oxides)
6. The Earths atmosphere
 The atmosphere of Earth is a layer of gases
surrounding the planet Earth that is retained by
Earth's gravity.
 The atmosphere protects life on Earth by
absorbing ultraviolet solar radiation, warming the
surface through heat retention (greenhouse
effect), and reducing temperature extremes
between day and night.
Layers of the atmosphere
Troposphere
– Lowest layer of the atmosphere extending from Earth’s
surface to an altitude of 11–13 km (7–8 miles)
– Temperature of troposphere decreases steadily with
increasing altitude. Hot air rises, so this temperature
gradient leads to continuous mixing of the upper and
lower regions within the layer, which produces
fluctuations in temperature and precipitation, called
weather.
Layers of the atmosphere
cont...
Stratosphere
– Lies above the troposphere and extends from an altitude
of 13 km (8 miles) to 44 km (27 miles)
– In the stratosphere, ultraviolet light reacts with O2
molecules to form atomic oxygen, which then reacts
with an O2 molecule to produce ozone, O3
–Ozone molecules make up the ozone layer, which acts as a
protective screen that absorbs ultraviolet light that
would otherwise reach the surface of the earth
– Protects the earth’s surface from the sun’s harmful effects
Ozone hole
 Organic compounds that contain chlorine and fluorine
(chlorofluorocarbons, CFCs) are capable of reaching the
stratosphere, where they react with ultraviolet light to generate
chlorine atoms and other chlorine-containing species that catalyze
the conversion of ozone to O2, thereby decreasing the amount of
ozone in the stratosphere.
 Chlorofluorocarbon gases (CFCs) eat the ozone layer alive where we use it
everyday for industrial processes although the ozone layer can repair Itself.
 Replacing chlorofluoro-carbons with hydrochlorofluoro-carbons
(HCFCs) or with hydrofluorocarbons (HFCs) has been developed to
minimize further damage to the earth’s ozone layer.
The Ozone hole sample...
Readings....
 Book pages :135 -215 ...
 Good luck ...
 Dont forget tutorials at Friday