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1styear CHEMISTRY Notes Chapter-3

Chapter-3
ATOMIC STRUCTURE.

Atomic Structure
Introduction
About the structure of atom a theory was put on by John Dalton in 1808. According to this theory matter was made from small indivisible particles called atoms.
But after several experiments many particles have been discovered with in the atom which are electrons, protons, neutrons, positrons etc. For the discovery of these fundamental particles the experiments are as follows.
1. Faraday's experiment indicates the existence of electron.
2. Crook's tube experiment explains the discovery of electron and proton.
3. Radioactivity also confirms the presence of electrons and protons.
4. Chadwick's experiment shows the presence of neutrons.
The details of these experiments are given below.

Faraday's Experiment


Passage of Electricity Through Solution
In this experiment Faraday passed the electricity through an electrolytic solution. He observed that when two metal plates called electrodes are placed in an electrolytic solution and electricity is passed through his solution the ions present in the solution are moves towards their respective electrodes. In other words these ions are moves towards the oppositely charge electrodes to give up their charge and liberated as a neutral particles.
Faraday also determined the charges of different ions and the amount of elements liberated from the electrolytic solution. Due to this experiment presence of charge particles in the structure of atoms is discovered. The basic unit of electric charge was later named as electron by Stoney in 1891.

Crook's Tube Or Discharge Tube Experiment
Passage of Electricity Through Gases Under Low Pressure
Introduction
The first of the subatomic particles to be discovered was electron. The knowledge about the electron was derived as a result of the study of the electric discharge in the discharge tube by J.J. Thomson in 1896. This work was later extended by W. Crooke

Working of Discharge Tube
When a very high voltage about 10,000 volts is applied between the two electrodes, no electric discharge occurs until the part of the air has been pumped out of the tube. When the pressure of the gas inside the tube is less than 1 mm, a dark space appears near the cathode and thread like lines are observed in the rest of 0.01 mm Hg it fills the whole tube. The electric discharge passes between the electrodes and the residual gas in the tube begins to glow. These rays which proceed from the cathode and move away from it at right angle in straight lines are called cathode rays.

Properties of Cathode Rays
1. They travel in straight lines away from the cathode and produce shadow of the object placed in their path.
2. The rays carry a negative charge.
3. These rays can also be easily deflected by an electrostatic field.
4. The rays can exert mechanical pressure showing that these consist of material particle which are moving with kinetic energy.
5. The produce fluorescence when they strike the glass wall of the discharge tube.
6. Cathode rays produce x-rays when they strike a metallic plate.
7. These rays consists of material particle whose e/m resembles with electron.
8. These rays emerge normally from the cathode and can be focused by using a concave cathode.

Positive Rays
In 1890 Goldstein used a discharge tube with a hole in the cathode. He observed that while cathode rays were emitting away from the cathode, there were coloured rays produced simultaneously which passed through the perforated cathode and caused a glow on the wall opposite to the anode. Thomson studied these rays and showed that they consisted of particles carrying a positive charge. He called them positive rays.

Properties of Positive Rays
1. These rays travel in a straight line in a direction opposite to the cathode.
2. These are deflected by electric as well as magnetic field in the way indicating that they are positively charged.
3. The charge to mass ratio (e/m) of positive particles varies with the nature of the gas placed in the discharge tube.
4. Positive rays are produced from the ionization of gas and not from anode electrode.
5. Positive rays are deflected in electric field. This deflection shows that these are positively charged so these are named as protons.

The Information Obtained From Discharge Tube Experiment
The negatively charge particles electrons and the positively charge particles protons are the fundamental particle of every atom.

Radioactivity
In 1895, Henry Becqueral observed that uranium and its compounds spontaneously emitted certain type of radiation which affected a photographic plate in the dark and were able to penetrate solid matter. He called these rays as radioactivity rays and a substance which possessed the property of emitting these radioactivity rays was said to be radioactivity element and the phenomenon was called radioactivity.
On further investigation by Maric Curic, it was found that the radiation emitted from the element uranium as well as its salts is independent of temperature and the source of the mineral but depend upon the mineral but depend upon the quantity of uranium present e.g. Pitchblende U3O8 was found to be about four times more radioactive than uranium.

Radioactive Rays
Soon after the discovery of radium it was suspected that the rays given out by radium and other radioactive substance were not of one kind. Rutherford in 1902 devised an ingenious method for separating these rays from each other by passing them between two oppositely charged plate. It was observed that the radioactive rays were of three kinds, the one bending towards the negative plate obviously carrying positive charge were called a-rays and those deflected to the positive plate and carrying -ve charge were named as ß-rays. The third type gamma rays, pass unaffected and carry no charge.

Properties of a - RAYS
1. These rays consists of positively charged particles.
2. These particles are fast moving helium nuclei.
3. The velocity of a-particles is approximately equal to 1/10th of the velocity of light.
4. Being relatively large in size, the penetrating power of a-rays is very low.
5. They ionize air and their ionization power is high.

Properties of ß - RAYS
1. These rays consists of negatively charged particles.
2. These particles are fast moving electron.
3. The velocity of ß-particles is approximately equal to the velocity of light.
4. The penetrating power of ß-rays is much greater than a-rays.
5. These rays ionizes gases to lesser extent.

Properties of Y - RAYS
1. Gamma rays do not consist of particles. These are electromagnetic radiations.
2. They carry no charge so they are not deflected by electric or magnetic field.
3. Their speed is equal to that of light.
4. These are weak ionizer of gases.
5. Due to high speed and non-material nature they have great power of penetration.

Chadwick Experiment (Discovery of Neutron)
When a light element is bombarded by a-particles, these a-particles leaves the nucleus in an unstable disturbed state which on settling down to stable condition sends out radioactivity rays. The phenomenon is known as "Artificial Radioactivity".
In 1933, Chadwick identified a new particle obtained from the bombardment of beryllium by a-particles. It had a unit mass and carried no charge. It was named "Neutron".

Spectroscopic Experiment
After the discovery of fundamental particles which are electrons, protons & neutron, the next question concerned with electronic structure of atom.
The electronic structure of the atom was explained by the spectroscopic studies. In this connection Plank's Quantum theory has great impact on the development of the theory of structure of atom.

Planck's Quantum Theory
In 1900, Max Planck studied the spectral lines obtained from hot body radiations at different temperatures. According to him,
When atoms or molecules absorb or emit radiant energy, they do so in separate units of waves called Quanta or Photons.
Thus light radiations obtained from excited atoms consists of a stream of photons and not continuous waves.
The energy E of a quantum or photon is given by the relation
E = h v
Where v is the frequency of the emitted radiation and h the Planck's constant. The value of h = 6.62 x 10(-27) erg. sec.
The main point of this theory is that the amount of energy gained or lost is quantized which means that energy change occurs in small packets or multiple of those packets, hv, 2 hv, 3 hv and so on.

Spectra
A spectrum is an energy of waves or particles spread out according to the increasing or decreasing of some property. E.g. when a beam of light is allowed to pass through a prism it splits into seven colours. This phenomenon is called dispersion and the band of colours is called spectrum. This spectrum is also known as emission spectrum. Emission spectra are of two types.
1. Continuous Spectrum
2. Line Spectrum

1. Continuous Spectrum
When a beam of white light is passed through a prism, different wave lengths are refracted through different angles. When received on a screen these form a continuous series of colours bands: violet, indigo, blue, green, yellow and red (VIBGYOR). The colours of this spectrum are so mixed up that there is no line of demarcation between different colours. This series of bands that form a continuous rainbow of colours is called continuous spectrum.

2. Line Spectrum
When light emitted from a gas source passes through a prism a different kind of spectrum may be obtained.
If the emitted from the discharge tube is allowed to pass through a prism some discrete sharp lines on a completely dark back ground are obtained. Such spectrum is known as line spectrum. In this spectrum each line corresponds to a definite wave length.

Identification of Element By Spectrum
Each element produces a characteristics set of lines, so line spectra came to serve as "finger prints" for the identification of element. It is possible because same element always emit the same wave length of radiation. Under normal condition only certain wave lengths are emitted by an element.

Rutherford's Atomic Model
Evidence for Nucleus and Arrangement of Particles
Having known that atom contain electrons and a positive ion, Rutherford and Marsden performed their historic "Alpha particle scattering experiment" in 1909 to know how and where these fundamental particles were located in the structure of atom.
Rutherford took a thin of gold with thickness 0.0004 cm and bombarded in with a-particles. He observed that most of the a-particles passed straight through the gold foil and thus produced a flash on the screen behind it. This indicated that old atoms had a structure with plenty of empty space but some flashes were also seen on portion of the screen. This showed that gold atoms deflected or scattered a-particles through large angles so much so that some of these bounced back to the source.
Based on these observations Rutherford proposed a model of the atom which is known as Rutherford's atomic model.

Assumption Drawn From the Model
1. Atom has a tiny dense central core or the nucleus which contains practically the entire mass of the atom leaving the rest of the atom almost empty.
2. The entire positive charge of the atom is located on the nucleus. While electrons were distributed in vacant space around it.
3. The electrons were moving in orbits or closed circular paths around the nucleus like planets around the sun.
4. The greater part of the atomic volume comprises of empty space in which electrons revolve and spin.

Weakness of Rutherford Atomic Model
According to the classical electromagnetic theory if a charged particle accelerate around an oppositely charge particle it will radiate energy. If an electron radiates energy, its speed will decrease and it will go into spiral motion finally falling into the nucleus. Similarly if an electron moving through orbitals of ever decreasing radii would give rise to radiations of all possible frequencies. In other words it would given rise to a continuous spectrum. In actual practise, atom gives discontinuous spectrum.

X-Rays and Atomic Number
In 1895, W.Roentgen discovered that when high energy electrons from cathode collide with the anode in the Crook's tube, very penetrating rays are produced. These rays were named as X-rays.

Explanation
When an electron coming from the cathode strike with the anode in the crook's tube, it can remove an electron from the inner shell of the atom. Due to removal of t his electron the electronic configuration of this ion is unstable and an electron from an orbital of higher energy drops into the inner orbital by emitting energy in form of a photon. This photon corresponds to electromagnetic radiations in the x-rays region.

Relationship Between Wave Length and Nuclear Charge
In 1911, Mosley stablished a relationship between the wave length and nuclear charge. He found that when cathode rays struck elements used as anode targets in the discharge tube, characteristic x-rays were emitted. The wave length of the x-rays emitted decreases regularly with the increase of atomic mass. On careful examination of his data Mosely found that the number of positive charges on the nucleus increases from atom to atom by single electronic unit. He called the number of positive charges as the atomic number.
Diagram Coming Soon

Bohr's Theory
Rutherford's model of atom fails to explain the stability of atom and appearance of the line spectra. Bohr in 1913 was the first to present a simple model of the atom which explained the appearance of line spectra.
Some of the postulates of Bohr's theory are given below.
1. An atom has a number of stable orbits or stationary states in which an electron can reside without emission or absorption of energy.
2. An electron may pass from one of these non-radiating states to another of lower energy with the emission of radiations whose energy equals the energy difference between the initial and final states.
3. In any of these states the electrons move in a circular path about the nucleus.
4. The motion of the electron in these states is governed by the ordinary laws of mechanics and electrostatic provided its angular momentum is an integral multiple of h/2p
It can be written as
mvr = nh / 2p
Here mvr becomes the angular momentum of the electron. Thus Bohr's first condition defining the stationary states could be stated as
"Only those orbits were possible in which the angular momentum of the electrons would be an integral multiple of h/2p". These stationary states correspond to energy levels in the atom.

Calculation of Radius of Orbits
Consider an electrons of charge e revolving.
Atomic number and e the charge on a proton.
Let m be the mass of the electro, r the radius of the orbit and v the tangential velocity of the revolving electron.
The electrostatic force of attraction between the nucleus and the electron according to Coulomb's law
= Z e x e / r2
Diagram Coming Soon
The centrifugal force acting on the electron.
= mv2 / r
Bohr assumed that these two opposing forces must be balanced each other exactly to keep the electron in an orbit.
Therefore
Ze2 / r2 = m v2 / r
Multiply both sides by r
r x Ze2 / r2 = r x m v2 / r
Ze2 / r = m v2
OR
r = Ze2 / m v2 .................. (1)
The Bohr's postulate states that only those orbits are possible in which
mvr = nh / 2p
Therefore,
V = nh / 2pmr
Substituting the value of V in eq (1)
r = Ze2 / m(nh/2pmr)2
or
r = Ze2 x 4p2 mr2/n2h2
or
1/r = 4p2mZe2/n2h2
cr
r = n2h2 / 4p2mZe2 ............... (2)
This equation gives the radii of all the possible stationary states. The values of constants present in this equation are as follows.
H = 6.625 x 10(-27) ergs sec OR 6.625 x 10(-37) J.s
Me = 9.11 x 10(-28) gm OR 9.11 x 10(-31) kg
E = 4.802 x 10(-10) e.s.u OR 1.601 x 10(-19) C
By substituting these values we get for first shell of H atom
r = 0.529 x 10(-8) m OR 0.529
The above equation may also be written as
r = n2 (h2 / 4p2mZe2) x n2 a0 .................... (3)
For the first orbit n = 1 and r = 0.529. This is the value of the terms in the brackets sometimes written as a0 called Bohr's Radius. For the second shell n = 2 and for 3rd orbit n = 3 and so on.

Hydrogen Atom Spectrum
Balmer Series
The simplest element is hydrogen which contain only one electron in its valence shell.
Balmer in 1885 studied the spectrum of hydrogen. For this purpose he used hydrogen gas in the discharge tube. Balmer observed that hydrogen atom spectrum consisted of a series of lines called Balmer Series. Balmer determined the wave number of each of the lines in the series and found that the series could be derived by a simple formula.

Lyman Series
Lyman series is obtained when the electron returns to the ground state i.e. n = 1 from higher energy level n(2) = 2, 3, 4, 5, etc. This series of lines belongs to the ultraviolet region of spectrum.

Paschen Series
Paschen series is obtained when the electron returns to the 3rd shell i.e. n = 3 from the higher energy levels n2 = 4, 5, 6 etc. This series belongs to infrared region.

Bracket Series
This series is obtained when an electron jumps from higher energy levels to 4th energy level.

Heisenberg Uncertainty Principle
According to Bohr's theory an electron was considered to be a particle but electron also behaves as a wave according to be Broglie.
Due to this dual nature of electron in 1925 Heisenberg gave a principle known as Heisenberg Uncertainty Principle which is stated as,
It is impossible to calculate the position and momentum of a moving electron simultaneously.
It means that if one was known exactly it would be impossible to known the other exactly. Therefore if the uncertainty in the determination of momentum is ?px and the uncertainty in position is ?x then according to this principle the product of these two uncertainties may written as
?px . ?x ˜ h
So if one of these uncertainties is known exactly then the uncertainty in its determination is zero and the other uncertainty will become infinite which is according to the principle.

Energy Levels and Sub-Levels
According to Bohr's atomic theory, electrons are revolving around the nucleus in circular orbits which are present at definite distance from the nucleus. These orbits are associated with definite energy of the electron increasing outwards from the nucleus, so these orbits are referred as Energy Levels or Shells.
These shells or energy levels are designated as 1, 2, 3, 4 etc K, L, M, N etc.
The spectral lines which correspond to the transition of an electron from one energy level to another consists of several separate close lying lines as doublets, triplets and so on. It indicates that some of the electrons of the given energy level have different energies or the electrons belonging to same energy level may differ in their energy. So the energy levels are accordingly divided into sub energy levels which are denoted by letters s, p, f (sharp, principle, diffuse & fundamental).
The number of sub levels in a given energy level or shell is equal to its value of n.
e.g. in third shell where n = 3 three sub levels s, p, d are possible.

Quantum Numbers
There are four quantum numbers which describe the electron in an atom.

1. Principle Quantum Number
It is represented by "n" which describe the size of orbital or energy level.
The energy level K, L, M, N, O etc correspond to n = 1, 2, 3, 4, 5 etc.
If
n = 1 the electron is in K shell
n = 2 the electron is in L shell
n = 3 the electron is in M shell

2. Azimuthal Quantum Number

This quantum number is represented by "l" which describes the shape of the orbit. The value of Azimuthal Quantum number may be calculated by a relation.
l = 0 ----> n - 1
So for different shell the value of l are as
n = 1 K Shell l = 0
n = 2 L Shell l = 0, 1
n = 3 M Shell l = 0, 1, 2
n = 4 N Shell l = 0, 1, 2, 3
when l = 0 the orbit is s
when l = 1 the orbit is p
when l = 2 the orbit is d
when l = 3 the orbit is f

3. Magnetic Quantum Number
It is represented by "m" and explains the magnetic properties of an electron. The value of m depends upon the value of l. It is given by
m = + l ----> 0 ----> l
when l = 1, m has three values (+1, 0, -1) which corresponds to p orbital. Similarly when l = 2, m has five values which corresponds to d orbital.

4. Spin Quantum Number
It is represented by "s" which represents spin of a moving electron. This spin may be either clockwise or anticlockwise so the values for s may be +1/2 or -1/2.

Pauli's Exclusion Principle
According to this principle
No two electrons in the same atom can have the same four quantum number.
Consider an electron is present in 1s orbital. For this electron n = 1, l = 0, m = 0. Suppose the spin of this electron is s = +1/2 which will be indicated by an upward arrow ?. Now if another electron is put in the same orbital (1s) for that electron n = 1, l = 0, m = 0. It can occupy this orbital only if the direction of its spin is opposite to that of the first electron so s = -1/2 which is symbolized by downward arrow ?. From this example, we can observe the application of Pauli's exclusion principle on the electronic structure of atom.

Electronic Configuration
The distribution of electrons in the available orbitals is proceeded according to these rules.
1. Pauli Exclusion Principle
2. Aufbau Principle
3. (n + l) Rule
4. Hund's Rule
The detail of these rules and principles is given below.

1. Aufbau Principle
It is states as
The orbitals are filled up with electrons in the increasing order of their energy.
It means that the orbitals are fulled with the electrons according to their energy level. The orbitals of minimum energy are filled up first and after it the orbitals of higher energy are filled.

2. Hund's Rule
If orbitals of equal energy are provided to electron then electron will go to different orbitals and having their parallel spin.
In other words we can say that electrons are distributed among the orbitals of a sub shell in such a way as to give the maximum number of unpaired electrons and have the same direction of spin.

3. (n + l) Rule
According to this rule
The orbital with the lowest value of (n + l) fills first but when the two orbitals have the same value of (n + l) the orbital with the lower value of n fills first.
For the electronic configuration the order of the orbital is as follows.
1s, 2s, 2p, 3s, 4s, 3d, 4p, 5s, 4d, 5p, 6s etc.

Atomic Radius
For homonuclear diatomic molecules the atomic radius may be defined as
The half of the distance between the two nuclei present in a homonuclear diatomic molecules is called atomic radius.
It may be shown as
In case of hetronuclear molecular like AB, the bond length is calculated which is (rA + rB) and if radii of any one is known the other can be calculated.
For the elements present in periodic table the atomic radius decreases from left to right due to the more attraction on the valence shell but it increases down the group with the increase of number of shells.

Ionic Radius
Ionic radius is defined as
The distance between nucleus of an ion and the point up to which nucleus has influence of its electron cloud.
When an electron is removed from a neutral atom the atom is left with an excess of positive charge called a cation e.g
Na ----> Na+ + c-
But when an electron is added in a neutral atom a negative ion or anion is formed.
Cl + e- ----> Cl-
As the atomic radius, the ionic radii are known from x-ray analysis. The value of ionic radius depends upon the ions that surround it.
Ionic radii of cations have smaller radii than the neutral atom because when an electron is removed. The effective charge on the nucleus increases and pulls the remaining electrons with a greater force.
Ionic radii of anions have a large radii than the neutral atom because an excess of negative charge results in greater electron repulsion.
Radius of Na atom = 1.57
Radius of Na+ atom = 0.95 (smaller than neutral atom)
Radius of Cl atom = 0.99
Radius of Cl- atom = 1.81 (larger than neutral atom)

Ionization Potential
Definition
The amount of energy required to remove most loosely bounded electron from the outermost shell of an atom in its gaseous state is called is called ionization potential energy.
It is represented as
M(gas) ----> M+(gas) + e- ................... ?E = I.P
The energy required to remove first electron is called first I.P. The energy required to remove 2nd or 3rd electron is called 2nd I.P or 3rd I.P
M(gas) ----> M+(gas) + e- ................... ?E = 1st I.P
M+(gas) ----> M++(gas) + e- ................?E = 2nd I.P
M++(gas) ----> M+++(gas) + e- ............ ?E = 3rd I.P
The units of I.P is kilo-Joule per mole.

Factors on which I.P Depends
1. Size of the Atom
If the size of an atom is bigger the I.P of the atom is low, but if the size of the atom is small then the I.P will be high, due to fact if we move down the group in the periodic table. The I.P value decreases down the group.

2. Magnitude of Nuclear Charge
If the nuclear charge of atom is greater than the force of attraction on the valence electron is also greater so the I.P value for the atom is high therefore as we move from left to right in the periodic table the I.P is increased.

3. Screening Effect

The shell present between the nucleus and valence electrons also decreases the force of attraction due to which I.P will be low for such elements.

Electron Affinity
Definition
The amount of energy liberated by an atom when an electron is added in it is called electron affinity.
It shows that this process is an exothermic change which is represented as
Cl + e- ----> Cl- ............ ?H = -348 kJ / mole
Factors on which Electron Affinity Depends

1. Size of the Atom
If the size of atom is small, the force of attraction from the nucleus on the valence electron will be high and hence the E.A for the element will also be high but if the size of the atoms is larger the E.A for these atoms will be low.

2. Magnitude of the Nuclear Charge
Due to greater nuclear charge the force of attraction on the added electron is greater so the E.A of the atom is also high.

3. Electronic Configuration
The atoms with the stable configuration has no tendency to gain an electron so the E.A of such elements is zero. The stable configuration may exist in the following cases.
1. Inert gas configuration
2. Fully filled orbital
3. Half filled orbital

Electronegativity
Definition
The force of attraction by which an atom attract a shared pair of electrons is called electronegativity.

Application of Electronegativity
1. Nature of Chemical Bond
If the difference of electronegativity between the two combining atoms is more than 1.7 eV, the nature of the bond between these atoms is ionic but if the difference of electronegativity is less than 1.7 eV then the bond will be covalent.

2. Metallic Character
If an element possesses high electronegativity value then this element is a non-metal but if an element exist with less electronegativity, it will be a metal.

Factors for Electronegativity
1. Size of the Atom
If the size of the atom is greater the electronegativity of the atom is low due to the large distance between the nucleus and valence electron.

2. Number of Valence Electrons
If the electrons present in the valence shell are greater in number, the electronegativity of the element is high.

1styear CHEMISTRY Notes Chapter-2

Chapter-2
THREE STATES OF MATTER


Three States Of Matter
Matter

It is defined as any thing which has mass and occupies space is called matter.
Matter is composed of small and tiny particles called Atoms or molecules. It exist in three different states which are gaseous, liquid & solid.

Properties of Gas

1. It has no definite shape.
2. It has no definite volume, so it can be compressed or expanded.
3. A gas may diffuse with the other gas.
4. The molecules of a gas are in continuous motion.

Properties of Liquids

1. A liquid has no definite shape.
2. It has a fixed volume.
3. The diffusion of a liquid into the other liquid is possible if both of the liquids are polar or non-polar.
4. It can be compressed to a negligible.

Properties of Solids

1. A solid has a definite shape.
2. It has a fixed volume.
3. The rate of diffusion of solid with each other is very slow.
4. It cannot be compressed easily.

1styear CHEMISTRY Notes Chapter-1

Chapter-1
INTRODUCTION TO FUNDAMENTAL CONCEPTS OF CHEMISTRY


Introduction to Fundamental Concepts of Chemistry 
Atom
It is the smallest particle of an element which can exist with all the properties of its own element but it cannot exist in atmosphere alone.


Molecule
When two or more than two atoms are combined with each other a molecule is formed. It can exist freely in nature.

Formula Weight
It is the sum of the weights of the atoms present in the formula of a substance.

Molecular Weight
It is the sum of the atomic masses of all the atoms present in a molecule.

Chemistry
It is a branch of science which deals with the properties, composition and the structure of matter.

Empirical Formula
Definition
It is the simplest formula of a chemical compound which represents the element present of the compound and also represent the simplest ratio between the elements of the compound.
Examples
The empirical formula of benzene is "CH". It indicates that the benzene molecule is composed of two elements carbon and hydrogen and the ratio between these two elements is 1:1.
The empirical formula of glucose is "CH2O". This formula represents that glucose molecule is composed of three elements carbon, hydrogen and oxygen. The ratio between carbon and oxygen is equal but hydrogen is double.

Determination of Empirical Formula
To determine the empirical formula of a compound following steps are required.
1. To detect the elements present in the compound.
2. To determine the masses of each element.
3. To calculate the percentage of each element.
4. Determination of mole composition of each element.
5. Determination of simplest ratio between the element of the compound.

Illustrated Example of Empirical Formula
Consider an unknown compound whose empirical formula is to be determined is given to us. Now we will use the above five steps in order to calculate the empirical formula.

Step I - Determination of the Elements
By performing test it is found that the compound contains magnesium and oxygen elements.

Step II - Determination of the Masses
Masses of the elements are experimentally determined which are given below.
Mass of Mg = 2.4 gm
Mass of Oxygen = 1.6 gm

Step III - Estimation of the Percentage
The percentage of an element may be determined by using the formula.
% of element = Mass of element / Mass of compound x 100
In the given compound two elements are present which are magnesium and oxygen, therefore mass of compound is equal to the sum of the mass of magnesium and mass of oxygen.
Mass of compound = 2.4 + 1.6 = 4.0 gm
% Mg = Mass of Mg / Mass of Compound x 100
= 2.4 / 4.0 x 100
= 60%
% O = Mass of Oxygen / Mass of Compound x 100
= 1.6 / 4.0 x 100
= 40%

Step IV - Determination of Mole Composition
Mole composition of the elements is obtained by dividing percentage of each element with its atomic mass.
Mole ratio of Mg = Percentage of Mg / Atomic Mass of Mg
= 60 / 24
= 2.5
Mole ratio of Mg = Percentage of Oxygen / Atomic Mass of Oxygen
= 40 / 16
= 2.5

Step V - Determination of Simplest Ratio
To obtain the simplest ratio of the atoms the quotients obtained in the step IV are divided by the smallest quotients.
Mg = 2.5 / 2.5 = 1
O = 2.5 / 2.5 = 1
Thus the empirical formula of the compound is MgO

Note
If the number obtained in the simplest ratio is not a whole number then multiply this number with a smallest number such that it becomes a whole number maintain their proportion.

Molecular Formula
Definition
The formula which shows the actual number of atoms of each element present in a molecule is called molecular formula.
OR
It is a formula which represents the element ratio between the elements and actual number of atoms of each type of elements present per molecule of the compound.
Examples
The molecular formula of benzene is "C6H6". It indicates that
1. Benzene molecule is composed of two elements carbon and hydrogen.
2. The ratio between carbon and hydrogen is 1:1.
3. The number of atoms present per molecule of benzene are 6 carbon and 6 hydrogen atoms.

The molecular formula of glucose is "C6H12O6". The formula represents that


1. Glucose molecule is composed of three elements carbon, hydrogen and oxygen.
2. The ratio between the atoms of carbon, hydrogen and oxygen is 1:2:1.
3. The number of atoms present per molecule of glucose are 6 carbon atoms. 12 hydrogen atoms and 6 oxygen atoms.

Determination of Molecular Formula
The molecular formula of a compound is an integral multiple of its empirical formula.
Molecular formula = (Empirical formula)n
Where n is a digit = 1, 2, 3 etc.
Hence the first step in the determination of molecular formula is to calculate its empirical formula by using the procedure as explained in empirical formula. After that the next step is to calculate the value of n
n = Molecular Mass / Empirical Formula Mass

Example
The empirical formula of a compound is CH2O and its molecular mass is 180.
To calculate the molecular formula of the compound first of all we will calculate its empirical formula mass
Empirical formula mass of CH2O = 12 + 1 x 2 + 16
= 30
n = Molecular Mass / Empirical Formula Mass
= 180 / 30
= 6
Molecular formula = (Empirical formula)n
= (CH2O)6
= C6H12O6

Molecular Mass
Definition
The sum of masses of the atoms present in a molecule is called as molecular mass.
OR
It is the comparison that how mach a molecule of a substance is heavier than 1/12th weight or mass of carbon atom.
Example
The molecular mass of CO2 may be calculated as
Molecular mass of CO2 = Mass of Carbon + 2 (Mass of Oxygen)
= 12 + 2 x 16
= 44 a.m.u
Molecular mass of H2O = (Mass of Hydrogen) x 2 + Mass of Oxygen
= 1 x 2 + 16
= 18 a.m.u
Molecular mass of HCl = Mass of Hydrogen + Mass of Chlorine
= 1 + 35.5
= 36.5 a.m.u

Gram Molecular Mass
Definition
The molecular mass of a compound expressed in gram is called gram molecular mass or mole.
Examples
1. The molecular mass of H2O is 18. If we take 18 gm H2O then it is called 1 gm molecular mass of H2O or 1 mole of water.
2. The molecular mass of HCl is 36.5. If we take 36.5 gm of HCl then it is called as 1 gm molecular mass of HCl or 1 mole of HCl.

Mole
Definition
It is defined as atomic mass of an element, molecular mass of a compound or formula mass of a substance expressed in grams is called as mole.
OR
The amount of a substance that contains as many number of particles (atoms, molecules or ions) as there are atoms contained in 12 gm of pure carbon.
Examples
1. The atomic mass of hydrogen is one. If we take 1 gm of hydrogen, it is equal to one mole of hydrogen.
2. The atomic mass of Na is 23 if we take 23 gm of Na then it is equal to one mole of Na.
3. The atomic mass of sulphur is 32. When we take 32 gm of sulphur then it is called one mole of sulphur.

From these examples we can say that atomic mass of an element expressed in grams is called mole.
Similarly molecular masses expressed in grams is also known as mole e.g.
The molecular mass of CO2 is 44. If we take 44 gm of CO2 it is called one mole of CO2 or the molecular mass of H2O is 18. If we take 18 gm of H2O it is called one mole of H2O.
When atomic mass of an element expressed in grams it is called gram atom
While
The molecular mass of a compound expressed in grams is called gram molecule.
According to the definition of mole.
One gram atom contain 6.02 x 10(23) atoms
While
One gram molecule contain 6.02 x 10(23) molecules.

Avagadro's Number
An Italian scientist, Avagadro's calculated that the number of particles (atoms, molecules) in one mole of a substance are always equal to 6.02 x 10(23). This number is known as Avogadro's number and represented as N(A).

Example
1 gm mole of Na contain 6.02 x 10(23) atoms of Na.
1 gm mole of Sulphur = 6.02 x 10(23) atoms of Sulphur.
1 gm mole of H2SO4 = 6.02 x 10(23) molecules H2SO4
1 gm mole of H2O = 6.02 x 10(23) molecules of H2O
On the basis of Avogadro's Number "mole" is also defined as
Mass of 6.02 x 10(23) molecules, atoms or ions in gram is called mole.

Determination Of The Number Of Atoms Or Molecules In The Given Mass Of A Substance
Example 1
Calculate the number of atoms in 9.2 gm of Na.
Solution
Atomic mass of Na = 23 a.m.u
If we take 23 gm of Na, it is equal to 1 mole.
23 gm of Na contain 6.02 x 10(23) atoms
1 gm of Na contain 6.02 x 10(23) / 23 atoms
9.2 gm of Na contain 9.2 x 6.02 x 10(23) /23
= 2.408 x 10(23) atoms of Na

Determination Of The Mass Of Given Number Of Atoms Or Molecules Of A Substance
Example 2
Calculate the mass in grams of 3.01 x 10(23) molecules of glucose.
Solution
Molecular mass of glucose = 180 a.m.u
So when we take 180 gm of glucose it is equal to one mole So,
6.02 x 10(23) molecules of glucose = 180 gm
1 molecule of glucose = 180 / 6.02 x 10(23) gm
3.01 x 10(23) molecules of glucose = 3.01 x 10(23) x 180 / 6.02 x 10(23)
= 90 gm

Stoichiometry
(Calculation Based On Chemical Equations)
Definition
The study of relationship between the amount of reactant and the products in chemical reactions as given by chemical equations is called stoichiometry.
In this study we always use a balanced chemical equation because a balanced chemical equation tells us the exact mass ratio of the reactants and products in the chemical reaction.
There are three relationships involved for the stoichiometric calculations from the balanced chemical equations which are
1. Mass - Mass Relationship
2. Mass - Volume Relationship
3. Volume - Volume Relationship

Mass - Mass Relationship
In this relationship we can determine the unknown mass of a reactant or product from a given mass of teh substance involved in the chemical reaction by using a balanced chemical equation.
Example
Calculate the mass of CO2 that can be obtained by heating 50 gm of limestone.
Solution
Step I - Write a Balanced Equation
CaCO3 ----> CaO + CO2
Step II - Write Down The Molecular Masses And Moles Of Reactant & Product
CaCO3 ----> CaO + CO2

Method I - MOLE METHOD
Number of moles of 50 gm of CaCO3 = 50 / 100 = 0.5 mole
According to equation
1 mole of CaCO3 gives 1 mole of CO2
0.5 mole of CaCO3 will give 0.5 mole of CO2
Mass of CO2 = Moles x Molecular Mass
= 0.5 x 44
= 22 gm

Method II - FACTOR METHOD
From equation we may write as
100 gm of CaCO3 gives 44 gm of CO2
1 gm of CaCO3 will give 44/100 gm of CO2
50 gm of CaCO3 will give 50 x 44 / 100 gm of CO2
= 22 gm of CO2

Mass - Volume Relationship
The major quantities of gases can be expressed in terms of volume as well as masses. According to Avogardro One gm mole of any gas always occupies 22.4 dm3 volume at S.T.P. So this law is applied in mass-volume relationship.
This relationship is useful in determining the unknown mass or volume of reactant or product by using a given mass or volume of some substance in a chemical reaction.
Example
Calculate the volume of CO2 gas produced at S.T.P by combustion of 20 gm of CH4.
Solution
Step I - Write a Balanced Equation
CH4 + 2 O2 ----> CO2 + 2 H2O
Step II - Write Down The Molecular Masses And Moles Of Reactant & Product
CH4 + 2 O2 ----> CO2 + 2 H2O

Method I - MOLE METHOD
Convert the given mass of CH4 in moles
Number of moles of CH4 = Given Mass of CH4 / Molar Mass of CH4
From Equation
1 mole of CH4 gives 1 moles of CO2
1.25 mole of CH4 will give 1.25 mole of CO2
No. of moles of CO2 obtained = 1.25
But 1 mole of CO2 at S.T.P occupies 22.4 dm3
1.25 mole of CO2 at S.T.P occupies 22.4 x 1.25
= 28 dm3

Method II - FACTOR METHOD
Molecular mass of CH4 = 16
Molecular mass of CO2 = 44
According to the equation
16 gm of CH4 gives 44 gm of CO2
1 gm of CH4 will give 44/16 gm of CO2
20 gm of CH4 will give 20 x 44/16 gm of CO2
= 55 gm of CO2
44 gm of CO2 at S.T.P occupy a volume 22.4 dm3
1 gm of CO2 at S.T.P occupy a volume 22.4/44 dm3
55 gm of CO2 at S.T.P occupy a volume 55 x 22.4/44
= 28 dm3

Volume - Volume Relationship
This relationship determine the unknown volumes of reactants or products from a known volume of other gas.
This relationship is based on Gay-Lussac's law of combining volume which states that gases react in the ratio of small whole number by volume under similar conditions of temperature & pressure.
Consider this equation
CH4 + 2 O2 ----> CO2 + 2 H2O
In this reaction one volume of CH4 gas reacts with two volumes of oxygen gas to give one volume of CO2 and two volumes of H2O
Examples
What volume of O2 at S.T.P is required to burn 500 litres (dm3) of C2H4 (ethylene)?
Solution
Step I - Write a Balanced Equation
C2H4 + 3 O2 ----> 2 CO2 + 2 H2O

Step II - Write Down The Moles And Volume Of Reactant & Product
C2H4 + 3 O2 ----> 2 CO2 + 2 H2O

According to Equation
1 dm3 of C2H4 requires 3 dm3 of O2
500 dm3 of C2H4 requires 3 x 500 dm3 of O2
= 1500 dm3 of O2

Limiting Reactant
In stoichiometry when more than one reactant is involved in a chemical reaction, it is not so simple to get actual result of the stoichiometric problem by making relationship between any one of the reactant and product, which are involved in the chemical reaction. As we know that when any one of the reactant is completely used or consumed the reaction is stopped no matter the other reactants are present in very large quantity. This reactant which is totally consumed during the chemical reaction due to which the reaction is stopped is called limiting reactant.
Limiting reactant help us in calculating the actual amount of product formed during the chemical reaction. To understand the concept the limiting reactant consider the following calculation.

Problem
We are provided 50 gm of H2 and 50 gm of N2. Calculate how many gm of NH3 will be formed when the reaction is irreversible.
The equation for the reaction is as follows.
N2 + 3 H2 ----> 2 NH3
Solution
In this problem moles of N2 and H2 are as follows
Moles of N2 = Mass of N2 / Mol. Mass of N2
= 50 / 28
= 1.79
Moles of H2 = Mass of H2 / Mol. Mass of H2
= 50 / 2
= 25
So, the provided moles for the reaction are
nitrogen = 1.79 moles and hydrogen = 25 moles
But in the equation of the process 1 mole of nitrogen require 3 mole of hydrogen. Therefore the provided moles of nitrogen i.e. 1.79 require 1.79 x 3 moles of hydrogen i.e. 5.37 moles although 25 moles of H2 are provided but when nitrogen is consumed the reaction will be stopped and the remaining hydrogen is useless for the reaction so in this problem N2 is a limiting reactant by which we can calculate the actual amount of product formed during the reaction.
N2 + 3 H2 ----> 2 NH3
1 mole of N2 gives 2 moles of NH3
1.79 mole of N2 gives 2 x 1.79 moles of NH3
= 3.58 moles of NH3
Mass of NH3 = Moles of NH3 x Mol. Mass
= 3.58 x 17
= 60.86 gm of NH3