Mole Concept

All of the objects around you - this book, your pen or pencil, and the things of nature such as rocks, water and plant
and animal substances - constitute the matter of the universe. Each of the particular kinds of matter, such as a certain
kind of paper or plastic or metal, is referred to as a material. We can define chemistry as the science of the
composition and structure of materials and of the changes that material undergoes.
Because chemistry deals with all materials, it is a subject of enormous breadth. It would be difficult to exaggerate
the influence of the chemistry on modern science and technology. In the section that follows, we will take a brief
glimpse at basic concepts of chemistry.

Classification of matter

Matter: Matter is anything that has mass and occupies space.
Two ways of classifying matter.
I. Physical classification
II. Chemical classification
I. Physical classification:

MATTER

SOLID LIQUID GAS

Particles are very closely Particles are less closely Particles are farthest apart
(i) packed in ordered manner. packed.
(ii) No freedom of movement Particles can move around to Movement of particles is
of particles some extent very easy and fast
(iii) Definite shape and volume Definite volume, indefinite indefinite shape and
shape volume
(iv) Exists at low T and high P Exists at intermediate P & T Exists at high T and low P

Note: For same substance:

 Solid and Liquid co-exist at MELTING POINT.
 Liquid and gas co-exist at BOILING POINT.
 Solid and gas co-exist at SUBLIMATION POINT.
 Solid, liquid and gas co-exist at TRIPLE POINT.

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Mole Concept

II. Chemical classification:

MATTER

Pure substances Mixtures

 Have fixed composition  Variable composition

 Cannot be separated by  Can be separated by physical
physical methods methods to pure substances

Homogeneous Heterogeneous
Elements Compounds (solutions)
 Made of two or more  Are not uniform in
 Made of only one  Have uniform
types of atoms/ions composition and/or
type of atoms composition and
density
Ex. : S8,P4,O3 density through
Ex. : Graphite + Diamond
C(diamond) out
or Fe + FeO
C(graphite) Ex. : Mixture of gases
or Sand + H2O
Hg(), I2(s) or NaCl + H2O
Ionic Covalent or Glucose + H2O
Br2()
or Alloys
Ex. : NaCl(s) Ex. : CH4(g)
Note : PHASE : It is the state of matter uniform in density and composition.
 Homogeneous mixtures have single phase while heterogeneous mixtures are multi-phase.
Ex : NaCl + H2O mixture has one phase
Ex : Graphite + Diamond mixture has 2 phases.

Illustration 1.
Match the column :
Column-I Column-II
Can't be decomposed into simpler
(A) Element (P)
substances by physical decomposition
Can be decomposed into simpler
(B) Compound (Q)
substances by chemical methods.
Can't be decomposed into simpler
(C) Mixture (R)
substances by chemical methods
(D) H2O (S) Uniform composition
Fe(s) + S(s) in powdered
(E) (T) Consists of one type of atoms
form
(F) Sea water (U) Consists of 2 or more type of atoms.

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Mole Concept

Solution:
(A) (R) & (T), (P), (S)
(B) (Q) & (U), (S), (P)
(C) (U), (Q)
(D) (Q) & (U), (S), (P)
(E) (U), (Q)
(F) (U), (Q)

Note :
(i) Mixtures can be separated by both physical and chemical methods.
(ii) Compounds can only be broken by chemical methods.

Illustration 2.

2HgO   2Hg + O2 
How many phases do you observe in the above given reaction at room temperature ?
Solution:
3
HgO solid, Hg liquid, O2 gas

Some specific properties of substances :

Deliquescence :
The property of certain compounds of taking up the moisture present in atmosphere and becoming wet when
exposed, is known as deliquescence. These compounds are known as deliquescent. Sodium hydroxide, potassium
hydroxide, anhydrous calcium chloride, anhydrous magnesium chloride, anhydrous ferrice chloride, etc., are the
examples of deliquescent compounds.

Hygroscopicity :
Certain compounds combine with the moisture of atmosphere and are converted into hydroxides or hydrates. Such
substances are called hygroscopic. Anhydrous copper sulphate, quick lime (CaO), anhydrous sodium carbonate, etc.,
are of hygroscopic nature.
Efflorescence :
The property of some crystalline substances of losing their water of crystallisation on exposure and becoming
powdery on the surface is called efflorescence and such salts are known as efflorescent. The examples are : Ferrous
sulphate (FeSO4.7H2O), sodium carbonate (Na2CO3.10H2O), sodium sulphate (Na2SO4.10H2O), potash alum
[K2SO4.Al2(SO4)3.24H2O], etc.
Malleability :
This property is shown by metals. When metallic solid is being beaten, it does not break but is converted into thin
sheet. It is said to possess the property of malleability. Copper, gold, silver, aluminium, lead, etc., can be easily
hammered into sheets. Gold is the most malleable metal.
Ductility :
The property of metal to be drawn into wires is termed ductility. Copper, silver, gold, aluminium, iron, etc., are ductile
in nature. Platinum is the most ductile metal.

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Mole Concept

Brittleness :
The solid materials which break into small pieces on hammering are called brittle. The solids of non-metals are
generally brittle in nature.
Ex: Ice, Diamond etc.

Dalton's Atomic Theory

Ancient Indian and Greek philosophers have always wondered about the unknown and unseen form of matter. The
idea of divisibility of matter was considered long back in India, around 500 BC. An Indian philosopher Maharishi
Kanad, postulated that if we go on dividing matter (padarth), we shall get smaller and smaller particles. Ultimately,
a time will come when we shall come across the smallest particle beyond which further division will not be possible.
He named these particles Parmanu. Another Indian philosopher, Pakudha Katyayama, elaborated this doctrine and
said that these particles normally exist in a combined form which gives us various forms of matter. Around the same
era, the Greek philosopher Democritus expressed the belief that all matter consists of very small, indivisible particles,
which he named atomos (meaning uncuttable or indivisible).
An Englishman, began teaching at a Quaker
school when he was 12. His fascination with
science included an intense interest in
meteorology (he kept careful daily weather
records for 46 years), which led to an
interest in the gases of the air and their
ultimate components, atom. Dalton is best
known for his atomic theory, in which he
postulated that the fundamental
differences among atoms are their masses.
He was the first to prepare a table of relative
John Dalton (1766 - 1844) atomic weight.

Although Democritus' ideal was not accepted by many of his contemporaries (notably Plato and Aristotle), somehow
it endured. Experimental evidence from early scientific investigations provided support for the notion of "atomism"
and gradually gave rise to the modern definitions of elements and compounds. It was in 1808, John Dalton,
formulated a precise definition of the indivisible building blocks of matter that we call atoms. Dalton's work marked
the beginning of the modern era of chemistry. The hypotheses about the nature of matter on which Dalton's atomic
theory is based can be summarized as follows :
(i) Elements are composed of extremely small particles called atoms.
(ii) All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of
one element are different from the atoms of all other elements.
(iii) Compounds are composed of atoms of more than one element. In any compound, the ratio of the numbers of
atoms of any two of the elements present is either an integer or a simple fraction.
(iv) A chemical reaction involves only the separation, combination or rearrangement of atoms ; it does not result in
their creation or destruction.

Limitations of Dalton’s atomic theory
According to Dalton’s atomic theory, an atom is the ultimate, discrete and indivisible particle of matter. Later
researches proved that Dalton’s atomic theory was not wholly correct.

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Mole Concept

Dalton’s atomic theory suffered from the following drawbacks :

(i) Atoms of the same or different types have a strong tendency to combine together to form a new ‘group of
atoms’. For example, hydrogen, nitrogen, oxygen gases exist in nature as ‘group of two atoms’. This indicates
that the smallest unit capable of independent existence is not an atom, but a ‘group of atoms’.
(ii) With the discovery of sub-atomic particles, e.g., electrons, neutrons and protons, the atom can no longer be
considered indivisible.
(iii) Discovery of isotopes indicated that all atoms of the same element are not perfectly identical. At least, they
differ in their masses. Atoms of the same element having different masses are called isotopes.
Dalton’s atomic theory could not explain why certain substances, all containing atoms of the same element,
should differ in their properties. For example, charcoal, graphite and diamond all are made up of only Carbon-
atoms, but still their properties are quite different.

Some basic concepts

At the center of each atom is a "nucleus" that consists of two types of particles called "protons" and "neutrons"
(and known collectively as "nucleons"). The number of protons is called the "atomic number". The total number of
protons and neutrons in the nucleus of the atom is called the "mass number" and is determined by the type of
atom (i.e. which element it is) and also by the number of neutrons present in that particular atom. In many cases,
different atoms of the same element can have different numbers of neutrons in their nuclei ("nuclei" is the plural
word for "nucleus"; one nucleus - many nuclei). The number of neutrons in an atom of a specific element therefore
determines its "mass number".
For example, most atoms of the element phosphorus have a mass number of 31 and so may be referred to as
phosphorus-31, however, some atoms of phosphorus have an extra neutron and therefore a mass number of 32.
They are therefore atoms of the isotope phosphorus-32.
Although there are three types of particles that form atoms (protons, neutrons and electrons), electrons have very
low mass compared with protons and neutrons so the mass of atoms is mainly due to the numbers of protons and
neutrons that make-up the atom, i.e. the "mass number" of the atom.
The atomic number of an atom is the same as the atomic number of the element of which that atom is a particle
because the atomic number of an atom determines which element it is an atom of.
However, the mass number of an atom is not the same as the mass number of an element because atoms of the
same element can have different masses due to having different numbers of neutrons (i.e. exist in the form of
different isotopes, which therefore have slightly different physical properties e.g. boiling points).
That is why Atomic Number of an element is meaningful, but the Mass Number "of an element" is not so meaningful.
When dealing with elements (macroscopic scale) rather than atoms (individual particles), chemists use Relative
Atomic Mass because "mass number" only applies to specific atoms of one particular isotope of the element.

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Mole Concept

Atom is actually made of 3 fundamental particles :

1. Electron
2. Proton
3. Neutron

Fundamental particle Discovered By Charge Mass  charge 

 
 mass 
(Specific Charge)
Electron J. J. Thomson –1.6 × 10 –19
coloumb 9.1 × 10 kg
–31
1.76 × 108 C/g
(e– or ) –4.8 × 10–10 esu 9.1 × 10–28g
–1 Unit 0.000544 a.m.u.
Proton (P) Rutherford +1.6 × 10–19coloum 1.672 × 10–27kg 9.58 × 104C/g
+4.8 × 10 –10
esu 1.672 × 10 –24
g

(Ionized H atom, 9.58 × 104 C/g 1.00727 a.m.u.

H+) +1 Unit
Neutron James Chadwick 0 1.675 × 10–27kg 0
(0n )
1
1.675 × 10 –24
g
1.00867 a.m.u.

Representation of atom : ZXA

A  Mass number : (total number of protons + total number of neutrons present in an atom.)
Z  Atomic number : (total number of protons present in an atom.)
 Isotope : Atoms of given element which have same atomic number but different mass number are called isotope :
e.g. 1H1 , 1H2 , 1H3 etc.
 Isobar : Atoms of different elements with the same mass number but different atomic number .
e.g. 18Ar40, 19K40 and 20Ca40
 Iso-electronic species : Species (atom, molecules or ions) having same number of electrons are called iso-electronic
e.g H¯, He, Li+ and Be2+ have 2 valence electrons each.
Note : Now a days this concept is extended to consider the same valence shell electron also.
 Iso-sters : Species having same number of electrons & same number of atoms. eg. N2O, CO2
 Iso-diaphers : Species having same difference in number of neutrons and protons or same number of excess of
neutron. eg. 199F, 23 11Na

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Mole Concept

The law of chemical combination

Atoine Lavoisier, John Dalton and other scientists formulated certain laws concerning the composition of matter
and chemical reactions. These laws are known as the law of chemical combination.

I. Law of indestructibility of matter or conservation of Mass :

 This law was proposed by Lavosier in 1774.
 The experimental certification was given by Landolt.
 According to this law in all chemical changes the total mass of the
system remains constant or in a chemical change mass is neither
created nor destroyed. Thus, in chemical change-
Total mass of reactant = Total mass of products
Ex. H2O(s)  H2O ()
Above reaction shows the physical change and the wt. of H2O (s) =
wt. of (H2O) ()
Antoine Lavosier
(1743-1794)
In case the reacting materials are not completely consumed, the Antoine-Laurent de Lavosier, the
relationship will be Total masses of reactants = Total masses of "Father of modern chemistry", was a
product + masses of unreacted reactants French nobleman prominent in the
histories of chemistry and biology.
 In nuclear reactions (Mass + energy) is conserved not the mass
He named both oxygen and
separately. hydrogen and predicted silicon.

Illustration 3.
When 4.2 g NaHCO3 is added to a solution of CH3COOH weighing 10.0 g, it is observed that 2.2 g CO2 is released into
atmosphere. The residue is found to weigh 12.0 g. Show that these observations are in agreement with the law of
conservation of mass.
Solution:
NaHCO3 + CH3COOH  CH3COONa + H2O + CO2
Initial mass = 4.2 + 10 = 14.2
Final mass = 12 + 2.2 = 14.2
Thus, during the course of reaction law of conservation of mass is obeyed.

II. Law of constant or definite proportion :

 This law was given by Joseph Louis proust. in 1799.

Proust was born the son of an

apothecary at Angers in north-
west France. He studied in Paris.
He lived in poverty for some years
before being awarded a pension
by Louis XVIII.
Joseph Proust (1754 - 1826)
 Chemical composition of a compound remains constant whether it is obtained by any method or any source.



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Mole Concept

  Example :
In water (H2O), Hydrogen and Oxygen combine in 2 : 1 molar ratio, the ratio remains constant whether it is tap
water, river water or sea water or produced by any chemical reaction.

Illustration 4.
1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated in steam gave 2.50 g
of its oxide. Show that these results illustrate the law of constant proportion.
Solution:
In the first sample of the oxide,
wt. of metal = 1.80 g, wt. of oxygen = (3.0 – 1.80) g = 1.2 g

 wt.of metal = 1.80g = 1.5

wt.of oxygen 1.2g
In the second sample of the oxide,
wt. of metal = 1.50 g, wt. of oxygen = (2.50 – 1.50) g = 1 g


wt.of metal = 1.50g = 1.5
wt.of oxygen 1g
Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen are fixed. Hence, the results
follow the law of constant proportion.

Note :
This law is not applicable in case of isotopes.

III. The law of multiple proportion :

 This law was given by Dalton in 1804.
 If two elements combine to form more than one compound, then the different masses of one element which
combine with a fixed mass of the other element, bear a simple ratio to one another.
Ex. Nitrogen and oxygen combine to form five stable oxides –
N2O Nitrogen 28 parts Oxygen 16 parts
N2O2 Nitrogen 28 parts Oxygen 32 parts
N2O3 Nitrogen 28 parts Oxygen 48 parts
N2O4 Nitrogen 28 parts Oxygen 64 parts
N2O5 Nitrogen 28 parts Oxygen 80 parts
The masses of oxygen which combine with same mass of nitrogen in the five compounds bear a ratio 16 : 32 :
48 : 64 : 80 or 1 : 2 : 3 : 4 : 5.

Note :
This law is not applicable in case of isotopes.

IV. Law of reciprocal proportion (or law of equivalent wt.) :

This law was put forward by Richter in 1792. It states as follows :
The ratio of the weights of two elements A and B which combine separately with a fixed weight of the third
element C is either the same or some simple multiple of the ratio of the weights in which A and B combine
directly with each other. This law may be illustrated with the help of the following example.

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Mole Concept

The elements C and O combine separately with the third element H to form CH4 and H2O
4 2
and they combine directly with each other to form CO 2, as shown in fig. H
In CH4, 12 parts by weight of carbon combine with 4 parts by weight of hydrogen. In H2O,
CH 4 H2O
2 parts by weight of hydrogen combine with 16 parts by weights of oxygen. Thus the 12 16
C O
weight of C and O which combine with fixed weight of hydrogen (say 4 parts by weight) 12 CO2 32
are 12 and 32 i.e. they are in the ratio 12 : 32 or 3 : 8.
Now in CO2, 12 parts by weight of carbon combine directly with 32 parts by weight of oxygen i.e. they combine
directly in the ratio 12 : 32 or 3 : 8 which is the same as the first ratio.

Special Note :
This law is also called as law of equivalent wt. due to each element combined in their equivalent wt. ratio.

 Example based on Law of Reciprocal proportion :
Illustration 5.
Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains 88.90% of oxygen and 11.10% of
hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of nitrogen. Show that these data illustrate the law
of reciprocal proportions.
Solution:
In NH3, 17.65g of H combine with N = 82.35g
82.35
 1 g of H combine with N = g = 4.66 g
17.65
In H2O, 11.10 g of H combine with O = 88.90 g
88.90
 1 g of H combine with O = g = 8.00 g
11.10
 Ratio of the weights of N and O which combine with fixed weight (=1g) of H = 4.66 : 8.00 = 1 : 1.7
In N2O3, ratio of weights of N and O which combine with each other = 36.85 : 63.15 = 1 : 1.7
Thus, the two ratios are the same. Hence it illustrates the law of reciprocal proportions.

V. Law of Gaseous volumes :

 This law was given by Gay-Lussac. in 1808.

Joseph Louis Gay-Lussac also; 6 December

1778 – 9 May1850) was a French chemist
and physicist. He is known mostly for two
laws related to gases, and for his work on
alcohol-water mixtures, which led to the
degrees Gay- Lussac used to measure
alcoholic beverages in many countries.

Joseph Louis Gay Lussac

(1778 - 1850)
 According to this law, gases react with each other in the simple ratio of their volumes. If products are also gases
then they are also in simple ratio of volume provided that all volumes are measure at same temp. & pressure.
eg. N2(g) + 3H2(g) 2NH3(g)
1 vol. 3vol. 2vol.

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Mole Concept

Illustration 6.
For the gaseous reaction, H2 + Cl2  2HCl
If 40 ml of hydrogen completely reacts with chlorine then find out the required volume of chlorine and volume of
produced HCl ?
Solution:
According to Gay Lussac's Law :
H2 + Cl2  2HCl
 1 ml of H2 will react will 1 ml of Cl2 and 2 ml of HCl will produce.
 40 ml of H2 will react with 40 ml of Cl2 and 80 ml of HCl will produce.
required vol. of Cl2 = 40 ml, produced vol. of HCl = 80 ml

VI. Berzelious Hypothesis and Avo-Gadro's Hypothesis :

(A) Berzelious Hypothesis : Equal volumes of all gases under similar conditions of temperature and pressure
contain equal number of atoms.

Berzelius
(1779 - 1848)
Jöns Jacob Berzelius was a Swedish
chemist. He worked out the modern
technique of chemical formula
notation, and is together with John
Dalton, Antoine Lavoisier, and
Robert Boyle considered a father of
modern chemistry.
The above statements was incorrect and later it was modified by Avo-Gadro.
(B) Avo-Gadro's Hypothesis : Equal volumes of all gases under similar conditions of temperature and pressure
contain equal number of molecules.

Lorenzo Romano Amedeo Carlo

Avogadro Di Quareqa edi Carreto
(1776-1856)
Italian mathematical physicist. He practiced law for
many years before he became interested in science.
His most famous work, now known as Avogadro's
law, was largely ignored during his lifetime,
although it became the basis for determining
atomic masses in the late nineteenth century.

Application of Avogadro's hypothesis :
(a) In finding the atomicity
(b) Relation between molecular weight and vapour density

• Vapour density 
density of gas (constant P, T)
density of H 2
• Molecular weight = 2 × vapour density
• Density of H2 = 0.000089 gm/cm3 at 0°C & 1 atm
 0.089 gm/lit.
(c) Relation between molecular weight and volume.

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Mole Concept

• 1 molecular weight = 22.4 lit. volume of gas at 0°C & 1 atm

• Weight of 1 gm mole gas = weight of 22.4 lit gas at 0°C & 1 atm
• Gram molecular volume = 22.4 lit. at 0°C & 1 atm
(d) Finding the molecular formula of gas.

Introduction to Mole

Atoms and molecules are extremely small in size and their numbers in even a small amount of any substance is really
very large. To handle such large numbers, a unit of similar magnitude is required. The 14th Geneva conference on
weight and measures adopted mole as a seventh basic SI unit of the amount of a substance. Mole concept is
essential tool for the fundamental study of chemical calculations. This concept is simple but its application requires
a thorough practice. There are many ways of measuring the amount of substance, weight and volume being the most
common, but basic unit of chemistry is the atom or a molecule and measuring the number of molecule is more important.

Definition of mole and molar mass

 A mole is the amount of a substance that contains as many entities (Atoms, Molecules, Ions or any other
particles) as there are atoms in exactly 12 g of C-12 isotope.
 A mole of a substance contains Avogadro's number (6.022 × 1023) of molecules (or formula units). The term
mole, like a dozen or a gross, thus refers to a particular number of things. A dozen eggs equals 12 eggs, a gross
of pencils equals 144 pencils, and a mole of ethanol equal 6.022 × 1023 ethanol molecules.
 The molar mass of a substance is the mass of one mole of the substance. Carbon-12 has a molar mass of exactly
12 g/mol, by definition.

Methods to calculate moles :

(i) If number of particles (molecules or atoms) is given then number of mole

mole =
Given no.of molecule / atom
NA

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Mole Concept

Given mass of substance (in gm)

(ii) If mass is given then, number of mole =
GAM / GMM
(iii) If volume of gas is given then, mole =
Volume of gas at STP (in lit.) 
22.7 L
(Standard molar volume is the volume occupies by 1 mole of any gas at NTP or STP, which is equal to 22.7 L)
 1 gram-atom = 1 mole atoms = NA atoms
 1 gram-molecule = 1 mole molecules = NA molecules
 1 gram-ion = 1 mole ions = NA ions

Illustration 7.
Calculate the atomic mass (average) of chlorine using the following data :
% Natural Abundance Molar Mass
Cl
35
75.77 34.9689
Cl
37
24.23 36.9659
Solution:
M 1x 1  M 2 x 2
(At. wt.) avg =
x1  x 2
34.9689  75.77  36.9659  24.23
 = 35.45
100

Illustration 8.
Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes :
Isotope Isotopic molar mass Abundance
36
Ar 35.96755 g mol -1
0.337%
38
Ar 37.96272 g mol-1 0.063%
Ar40
39.9624 g mol -1
99.600%
Solution:
M 1x1  M 2 x 2  M 3 x 3
(At wt.) avg =
x1  x 2  x 3
35.96755  0.337  37.96272  0.063  39.9624  99.6

100
= 39.94

Atomic and molecular masses

 Different types of Atomic Masses

The mass of an atom depends on the number of electrons, protons, and neutrons it contains. Knowledge of an
atom's mass is important in laboratory work. But atoms are extremely small particles - even the smallest speck of
dust that our unaided eyes can detect contains as many as 1 × 1016 atoms, Clearly, we cannot weigh a single atom,
but it is possible to determine the mass of one atom relative to another experimentally. The first step is to assign a
value to the mass of one atom of a given element so that it can be used as a standard.

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Mole Concept

Relative atomic mass :

Hydrogen, being lightest atom was arbitrarily assigned a mass of 1 (without any units) and other elements were
assigned masses relative to it. However, the present system of atomic masses is based on carbon - 12 as the standard
and has been agreed upon in 1961. Here, Carbon - 12 is one of the isotopes of carbon and can be represented as
C. In this system, 12C is assigned a mass of exactly 12 atomic mass unit (a.m.u.) and masses of all other atoms are
12

given relative to this standard.

It is defined as the number which indicates how many times the mass of one atom of an element is heavier in
comparison to 1/12th part of the mass of one atom of C-12.

mass of one atom of an element Mass of one atom of an element

Relative atomic mass (of elements) = =
1 1 amu
[mass of C - 12 atom]
12

Atomic mass unit (a.m.u. or u) :

The quantity 1/12th mass of an atom of C12 is known as atomic mass unit.
Since mass of 1 atom of C - 12 = 1.9924 × 10–23 g
1.9924  10 23 g
  1/12th part of the mass of 1 atom = = 1.67 × 10–24 g
12
1
 1 a.m.u. = g
6.022  10 23
It may be noted that the atomic masses as obtained above are the relative atomic masses and not the actual masses
of the atoms. These masses on the atomic mass scale are expressed in terms of atomic mass units (abbreviated as
a.m.u.). Today, 'a.m.u.' has been replaced by 'u' which is known as unified mass.
One atomic mass unit (a.m.u.) is equal to 1/12th of the mass of an atom of carbon - 12 isotope.
Thus the atomic mass of hydrogen is 1.008 a.m.u. while that of oxygen is 15.9994 a.m.u. (or taken as 16 a.m.u.).

 Gram atomic mass or mass of 1 gram atom :
When numerical value of atomic mass of an element is expressed in grams then the value becomes gram atomic
mass or GAM.
gram atomic mass (GAM) = mass of 1 gram atom = mass of 1 mole atoms
= mass of NA atoms = mass of 6.022 × 1023 atoms.
Ex. GAM of oxygen = mass of 1 g atom of oxygen = mass of 1 mol atoms of oxygen.
= mass of NA atoms of oxygen.
 16 
=  g   NA = 16 g
 NA 
Ex. Mass of one atom of Oxygen = 16 a.m.u. or 16 × 1.67 × 10–24 g
Mass of NA atoms of Oxygen = 16 × 1.67 × 10–24 × 6.022 × 10–23 g = 16 g
Now see the table given below and understand the definition given before.
Element R.A.M. Atomic mass Gram Atomic
(Relative Atomic Mass) (mass of one atom) mass/ weight
N 14 14 a.m.u. 14 gm
He 4 4 a.m.u. 4 gm
C 12 12 a.m.u. 12 gm

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Mole Concept

Average atomic mass :

If an element exists in different isotopic forms (or allotropic forms) having relative abundance X1%, X2% ..... Xn%, with
relative atomic masses M1, M2 ..... Mn respectively then,
X1 X X
Avg. Atomic mass of element = (M1 )  2 (M 2 )  .....  n (M n ) =  X i (M i )
100 100 100 i  1 to n 100


 Relative Molecular mass
The number which indicates how many times the mass of one molecule of a substance is heavier in comparison to
1/12th part of the mass of an atom of C–12.
OR
The molecular mass of a substance is the sum of atomic masses of the elements present in a molecule. It is obtained
by multiplying the atomic mass of each element by the number of its atoms and adding them together.
Ex. molecular mass of oxygen (O2) = 32 a.m.u.
molecular mass of (O3) = 48 a.m.u.
molecular mass of HCl = 1 + 35.5 = 36.5 a.m.u.
molecular mass of H2SO4 = 2 + 32 + 64 = 98 a.m.u.

Gram molecular mass (mass of 1 gram molecule) :

When numerical value of molecular mass of the substance is expressed in grams then the value becomes gram
molecular mass or GMM.
gram molecular mass (GMM) = mass of 1 gram molecule = mass of 1 mole molecules
= mass of NA molecules = mass of 6.022 × 1023 molecules
Ex. GMM of H2SO4 = mass of 1 gram molecule of H2SO4
= mass of 1 mole molecules of H2SO4
= mass of NA molecules of H2SO4
 98 
=  g   NA = 98 g
 NA 
Ex. Molecular Mass of N2 = 28 a.m.u. = 28 × 1.67 × 10–28 g
Mass of NA molecules of N2 = 28 × 1.67 × 10–24 × 6.022 × 1023 g = 28 g
Thus molecular mass can be defined as the absolute mass in grams of 6.022 × 1023 molecules of a substance.

 Average molecular mass of gas mixture :

=
Total mass of mixture of gases
Total mole
M A n A  M Bn B  M Cn C
Ex. Mavg =
na  nB  nC
Where nA, nB, nC  moles of gases A, B, C
MA, MB, MC  molecular mass of gases A, B, C
   Mavg = MAXA + MBXB + MCXC
   Mavg = M X i i

where Xi = mole fraction of ith gas, Mi = molecular mass of ith gas.

Where XA, XB & XC are the mole fraction of gases A, B & C.

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Mole Concept

Illustration 9.
Find :
(i) No. of moles in 1020 atoms of Cu.
16
(ii) Mass of 200 8 O atoms in a.m.u.
14
(iii) Mass of 100 atoms of 7 N in gm.
(iv) No. of molecules & atoms in 54 gm H2O.
(v) No. of atoms in 88 gm CO2.
Solution:
1020
(i) moles
NA
16
(ii) Mass of 1 8 O atom = 16 a.m.u.
16
Mass of 200 8 O atom = 16 × 200 = 3200 a.m.u.
14
(iii) Mass of 1 7 N atom = 14 a.m.u. = 14 × 1.66 × 10–24 g
14
Mass of 100 7 N atom = 1400 × 1.66 × 10–24 g
(iv) Molar mass of H2O = 18 g mol–1
54
Moles of H2O molecule = 3
18
no. of molecules in H2O = 3NA
no. of atoms in H2O = 3 × 3NA = 9NA
88
(v) Moles of CO2 =  2 (Molar mass of CO2 = 44)
44
No. of atoms in CO2 = 6NA

Illustration 10.
In three moles of ethane (C2H6). Calculate the following :
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Solution:
(i) Gram atoms of C = 3 × 2 = 6
(ii) Gram atoms of Hydrogen = 3 × 6 = 18
(iii) Number of molecules of C2H6 = 3 NA.

Illustration 11.
Calculate the number of atoms in each of the following :
(i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.
Solution:
(i) Number of atoms of Ar = 52NA
(ii) Wt. of He = 52 × 1.66 × 10–24 gm.
52
Moles of He =  1.66  1024
4

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Mole Concept

52  1.66  10 24
Number of atoms of He =  N A  13 atoms
4
52
(iii) Moles of He =  14 mol
4
Number of atoms of He = 13 NA.

Illustration 12.
What will be the mass of one 12C atom in g ?
Solution:
Wt. of 1 atom of C = 1.66 × 10–24 gm. × (At wt.)
= 1.66 × 10–24 gm × 12 = 19.92 × 10–24 gm

Illustration 13.
Calculate mass of O atoms in 6 gm CH3COOH ?
Solution:
Molar mass of CH3COOH = 60
6
Moles of CH3COOH =  0.1
60
1 mole CH3COOH has 2 mol O atoms
0.1 mole CH3COOH has 0.2 mol O atoms
Mass of oxygen = moles × Atomic mass = 0.2 × 16 = 3.2 gm.

Illustration 14.
How many sucrose molecules (C12H22O11) are present in 3.42 g sucrose ?
Solution:
Molar mass of sucrose = 342
3.42
Moles of sucrose = = 0.01
342
Molecules of sucrose = 0.01 NA = 6.022 × 1021

Illustration 15.
Calculate mass of water present in 499 gm CuSO4.5H2O ?
(Atomic mass – Cu = 63.5, S = 32, O = 16, H = 1)

Solution:
Molar mass of CuSO4.5H2O = 249.5
499
moles of CuSO4.5H2O = 2
249.5
Each mole CuSO4 .5H2O has 5 mol H2O
2 mole CuSO4 .5H2O has 10 mol H2O
Mass of 10 mol H2O = 10 × 18 = 180 gm.

Illustration 16.
Find mass of 12.046 × 1023 atoms 12C6 sample ?

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Mole Concept

Solution:
Mass of 12C6 atom = 12 × 1.66 × 10–24 g
Mass of 12.046 × 1023 atoms = 12 × 1.66 × 10–24 × 12.046 × 1023 = 24 gm.

Illustration 17.
What is no. of O2 molecules in 3.2 × 10–15 g sample of oxygen ?
Solution:
3.2  10 15
Moles of O2 = = 10–16; Molecules of O2 = 10–16 NA = 6.022 × 107
32

Illustration 18.
Find no. of protons in 180 ml H2O. Density of water = 1 gm/ml.
Solution:
Mass of water = density × volume = 180 g
180
Moles of water =  10
18
1 mol water has 10 mol protons
10 mol water has 100 mol protons
10 mol water has 100 NA protons
10 mol water has 6.022 × 1025 protons

Illustration 19.
What mass of Na2SO4.7H2O contains exactly 6.022 × 1022 atoms of oxygen ?
Solution:
Molar mass of Na2SO4.7H2O = 275 gm.
1 mole Na2SO4.7H2O has 11 mol O-atoms.
 11 NA O – atoms are in 275 g Na2SO4.7H2O
275
 6.022 × 1022 O – atoms are in =  6.022  10 22 g = 2.5 g
11  6.022  1023

Illustration 20.
What is number of atoms and molecules in 112 L of O3(g) at STP ?
Solution:
112
Moles of molecules = 5
22.4
Moles of atoms = 5 × 3 = 15
No. of molecules = 5 NA
No. of atoms = 15NA.

Illustration 21.
16
Oxygen exists as three isotopes 8 O, 178 O , 188 O with relative abundance 90%, 7% and 3% respectively. What is average
atomic mass of oxygen ?
Solution:
Mavg. = x M = 0.9 × 16 + 0.07 × 17 + 0.03 × 18 = 16.13 a.m.u. or 16.13 g/mol
i i

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Mole Concept

Illustration 22.
If an elements M has Mavg.= 51.7 find relative abundances of 50M and 52M isotopes in nature ? [Assume M exists in only
two allotropic forms in nature]
Solution:
Let 50M isotope be x % and 52M isotope be (100 – x) % in nature
x 100  x
 51.7 = Mavg. = (50) + (52)  x = 15%  15% 50M and 85% 52M in nature.
100 100

(iv) Under any condition of temperature and pressure, moles of gases may be calculated using IDEAL GAS
EQUATION : PV = nRT,
where, R = Universal Gas Constant
= 0.082 L-atm/K-mol
= 8.314 J/K-mol
2 cal/K-mol
Units of pressure and their relation:
1 atm = 76 cm Hg
= 760 mm Hg
= 760 torr (1 torr = 1 mm Hg)
= 1.01325 106 dyne/cm2
= 1.01325 105 N/m2 or Pa
= 1.01325 bar (1 bar = 105 Pa)
1 bar = 75 cm Hg
Units of Volume and their relation:
1 ml = 1 cm3 = 1 c.c.
1 Litre = 1000 ml = 1 dm3
1 m3 = 1000 L
Units of Temperature and their relation:
T = 273.15 + t
where, T = Absolute temperature (in Kelvin) and t = temperature in oC
(Normally, we take 273K in calculation)
(v) Sometimes gas is collected over water. In this case, the measured pressure is sum of pressure of gas and the
vapour pressure of water (also called Aqueous Tension). In order to calculate moles of gas, the vapour pressure
of water should be deducted from the measured pressure.

If volume of gas is given then, mole

Volumeof gasat 0 C and1atm
= Volume of gas at STP = 
22.7 L 22.4L
(Standard molar volume is the volume occupied by 1 mole of an ideal gas at STP (Standard temperature and pressure
which is 273.15 K & 1 bar respectively), which is equal to 22.7 L).
1 mole of an ideal gas occupy 22.4 L at 0ºC and 1 atm.

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Mole Concept

Illustration 23.
Calculate the number of g-molecules (mole of molecules) in the following :
(i) 3.2 gm CH4
(ii) 70 gm nitrogen
(iii) 4.5 × 1024 molecules of ozone
(iv) 2.4 × 1021 atoms of hydrogen
(v) 11.2 L ideal gas at 0ºC and 1 atm
(vi) 4.54 ml SO3 gas at STP
(vii) 8.21 L C2H6 gas at 400K and 2atm
(viii) 164.2 ml He gas at 27ºC and 570 torr [NA = 6 × 1023]
Solution:
(i) 3.2 gram CH4
w 3.2
number of moles (CH4) =  = 0.2 moles
M 16
(ii) 70 gram N2
w 70
Number of moles = =  2.5
M 28
(iii) 4.5 × 1024 molecules of O3
4.5  10 24
Number of moles = no. of m olecules =  7.5
NA 6  10 23
(iv) 2.4 × 1021 atoms of hydrogen
21
2.4  10
Number of gram molecules of H2 = no. of m olecules = = 0.002
NA 2  6  10 23
(v) 11.2 litre ideal gas at 0ºC and 1 atm
11.2
Number of moles = Volume at 0º C &1 atm = = 0.5
22.4 litre 22.4
(vi) 4.54 ml SO3 gas at STP
VSTP (ml) 4.54
Number of moles = = = 2 × 10–4
22700ml 22700
(vii) 8.21 litre C2H6 at 400 K and 2 litre
PV 2  8.21
n= = = 0.5
R.T 0.0821  400
(viii) 164.2 ml He gas at 27ºC and 570 torr
3
PV  570  164.2 10 litre
n= =  atm   = 0.005
RT  760  0.0821  300

Illustration 24.
Find no. of protons in 180 ml H2O. Density of water = 1 g/ml.
Solution:
Mass of water = density × volume = 180 g
180
Moles of water =  10
18
1 mol water has 10 mol protons

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Mole Concept

 10 mol water has 100 mol protons = 100 × 6.022 × 1023 protons
= 6.022 × 1025 protons.

Illustration 25.
What mass of Na2SO4.7H2O contains exactly 6.022 × 1022 atoms of oxygen ?
Solution:
Molar mass of Na2SO4.7H2O = 275 g.
1 mole Na2SO4.7H2O has 11 mol O-atoms.
 11 NA O – atoms are in 275 g Na2SO4.7H2O
275
 6.022 × 1022 O – atoms are in =  6.022  10 22 g = 2.5 g
11  6.022  10 23

Illustration 26.
What is number of atoms and molecules in 112 L of O3(g) at 0ºC and 1atm ?
Solution:
112
Moles of molecules = 5
22.4
Moles of atoms = 5 × 3 = 15
No. of molecules = 5 NA
No. of atoms = 15NA.

Density :

It is of two types.
I. Absolute density
II. Relative density

For liquids and solids :

mass
Absolute density =
volume
density of the subs tance
Relative density or specific gravity =
density of water at 4C (1 gm ml –1 )
For gases :
mass PM
Absolute density = =
volume RT
where P is pressure of gas, M = molar mass of gas, R is the gas constant, T is absolute temperature.
Vapour Density :
Vapour density is defined as the density of the gas with respect to hydrogen gas at the same temperature and
pressure.

d gas PM gas
Vapour density =  RT

dH2 PM H2
RT

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Mole Concept

M M gas
V.D. = gas
=  M gas = 2× V.D.
M H 2
2

Illustration 27.
A gaseous mixture of H2 and NH3 gas contains 68 mass % of NH3. The vapour density of the mixture is –
Solution:
68
No. of moles of NH3 in 100g mixture = 4
17
32
No. of moles of H2 in 100g mixture =  16
2
Maverage =
Total m ass 100
 5
Total m oles 4  16
5
VD= = 2.5
2

Stoichiometry
Stoichiometry is the calculation of amounts of reactants and products involved in a reaction. Stoichiometric
calculations require a balanced chemical equation of the reaction.
• Remember a balanced chemical equation is one which contains an equal number of atoms of each element on
both sides of equation.

Significance of stoichiometric coefficients

Stoichiometric coefficients of chemical equation tells us about the ratio in which moles of reactants react and moles
of products form.

Illustration
2H2(g) + O2(g)  2H2O(g)
1 interpretation
st
2 moles 1 mole 2 moles
2nd interpretation 2 NA molecules NA molecules 2 NA molecules
3rd interpretation 2 molecules 1 molecules 2 molecules
4 interpretation
rd
2 volume 1 volume 2 volume
(at constant P & T)
5th interpretation 2 atm 1 atm 2 atm
(at constant V & T)
6th interpretation 4g 32 g 36 g

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Mole Concept

Divide by molar mass multiply by molar

of reactant (MR) mass of product (MP)

multiply by P/RT multiply by RT/P

Here : S.C.R. - Stoichiometric coefficient of reactant,

S.C.P.- Stoichiometric coefficient of product
• Consider a balanced Chemical reaction :
aA + bB  c C + dD
 mole of A reacting mole of B reacting mole of C produced mole of D produced 
 = = = 
 a b c d 
 If mole of A is nA, then -
b
• mole of B consumed = nA ×  
a
c
• mole of C produced = nA ×  
a
d 
• mole of D produced = nA ×  
a
Mass - mass analysis :
This relates the mass of a species (reactant or product) with the mass of another species (reactant or product)
involved in a chemical reaction.
In general, the following steps are adopted in making necessary calculations :
(A) Write down balanced molecular equation for the chemical change.
(B) Write down the relative masses of the reactants and the products with the help of formula below the respective
formula. These shall be the theoretical amounts of reactants and product.
(C) Calculate the number of moles below the formula of each of the reactant and product.
(D) By the applications of mole concept, mass co-relation through unitary method or proportionality method the
unknown quantity can be determined.
Let us consider a balanced chemical reaction,
aA + bB c C
WA gm ?
If MA, MC Molar mass of A and C respectively.
WA = Mass of A reacted

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Mole Concept

WC = Mass of C formed (to be found).

Method aA + bB  cC
cW
WA gm WC =   A M C gm
 a  MA
 
 WA   c  WA
   moles    moles
 MA  aMA

Illustration 28.
What mass of CaO is formed by heating 50 g CaCO3 in air ?
[molar mass : CaCO3 = 100, CaO = 56].
Solution:
CaCO3 (s)  CaO (s) + CO2 (g)
50 gm
50
= mol
100
1 1
= mol mol
2 2
1
=  56 = 28 gm
2

Mass-volume Analysis :
This establishes the relationship between the mass of a species (reactant or product) and the volume of a gaseous
species (reactant or product) involved in a chemical reaction.
For example the volume of CO2 at STP formed by heating 200 g CaCO3 :-
CaCO3 (s) CaO(s) + CO2 (g)
200 gm
200
= mol = 2 mol 2 mol
100
Volume of gas at STP = No. of moles × 22.7 L = 2 × 22.7 = 45.4 L.

Volume-volume Analysis :
This relationship deals with the volume of a gaseous species (reactant or product) with the volume of another
gaseous species (reactant or product) involved in a chemical reaction.
Gay Lussac's law of combining volumes :
aA(g) + b(B)(g)  cC(g)
VA mL VB mL VC mL
reacted reacted formed
VA : VB : VC = a : b : c
At constant P, T :
Volume of gaseous species reacting or formed is in ratio of their stoichiometric coefficients.

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Mole Concept

Illustration 29.
What volume of H2 (g) is produced by decomposition of 2.4 L NH3 (g) ?
Solution:
1 3
NH3 (g)  N2(g) + H2 (g)
2 2
t=0 2.4 L 0 0
t= 0 v1 v2
1 3 2.4 2
Gay-Lussac's law, 2.4 : v1 : v2 = 1 : :    v2 = 3.6 L
2 2 v2 3

Limiting Reagent (L.R.)

(i) The reactant which is completely consumed when a reaction goes to completion is called Limiting Reactant or
Limiting reagent.
(ii) The reactant whose mole ratio is least is limiting reactant.
Given moles of reactant
Where; Mole ratio =
Stoichiometric coefficient of reactant
(iii) For calculation of moles of product LR should be used. When amounts of more than two reactants are given :
aA + bB cC + dD
Initial reacting nA mol nB mol
mixture
nA nB
mole ratio
a b
nA nB
If <   A is limiting reagent.
a b
nA a
 then reaction occurs to completion & no reactant is left at the end.
nB b
nA a
If  then one of reactants is in excess & left at end. The other one which is completely consumed is called
nB b
Limiting Reagent (LR).
(iv) To find out LR :
aA + bB + cC  dD + eE
t=0 nA mol nB mol nC mol
Divide nA by a, nB by b & nC by c, which ever gives smallest no. is LR & all calculations are done using its moles.

Illustration
2A + B + 3C  D + 2E
t=0 5 mol 6 mol 6 mol 0 0
5 6 6
 2.5 6  2 (smallest)
2 1 3
 C is LR
12 6
At t =  5 = 1mol 6 = 4 mol 0 2mol 4mol
3 3

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Mole Concept

Illustration
2A(g) + B(g)  3C(g)
At t = 0 2L 3L 0 [ At constant P and T : V  n]
2 3
=1 =3
2 1
 A is LR
2
At t =  0 3   2L 3L
2

Percentage Composition

Here we are going to find out the percentage of each element in the compound by knowing the molecular formula
of compound.
We know that according to law of definite proportions any sample of a pure compound always possesses constant
ratio of their combining elements.
Example :
Every molecule of ammonia always has formula NH3 irrespective of method of preparation or sources. i.e. 1 mole of
ammonia always contains 1 mol of N and 3 moles of H. In other words, 17 g of NH3 always contains 14 g of N and
3 g of H. Now find out % of each element in the compound.
Mass of N in 1 mol NH 3 14g
Mass % of N in NH3 = × 100 = × 100 = 82.35 %
Mass of 1 mol of NH 3 17
Mass of H is 1 mol NH 3 3
Mass % of H in NH3 = × 100 = × 100 = 17.65%
Mass of 1 mol of NH 3 17
Percentage composition of a compound refers to the amount of various constituent elements present per
100 parts by mass of the compound. It can be calculated from the formula of the compound by using the following
relation :
( No. of atoms of element )  ( At . mass of element )  100
Percentage mass of an element 
Molecular mass of subs tan ce
In case of minerals or ores the mass percentage of a particular constituent species in a formula can also be calculated
using the similar relation as follows.
[Total mass of constituent species in one formula unit ]  100

Formula mass
For example, the percentage of iron Fe and FeO can be calculated in magnetite (Fe3O4) as follows :
Formula mass of Fe3O4 = 3 × 56 + 4 × 16 = 232
Atomic mass of Iron Fe = 56
Formula mass of FeO = 56 + 16 = 72
3  56  100 72 100
Therefore percentage of Fe  = 72.4 % and percentage of FeO  = 31 %
232 232
Note :
• Ratio of mass of two elements in a compound is a constant (independent of amount of compound). 
• Mass % of an element in a compound is a constant.

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Mole Concept

Illustration
In KMnO4
MK mK
with respect to K   M  Molar mass
M KMnO4 mKMnO4
4M O mO
with respect to O   m  Mass or mass % of element present in compound
M KMnO4 mKMnO4

Illustration 30.
How many grams of KClO4 contain 40 gm 'O'.
Solution:
40 4
  x = 86.5 g
x 35.5  39  64

Illustration 31.
Calculate mass of Cu in 3.67 × 103 g CuFeS2 ? (Atomic mass Cu = 63.5, Fe = 56, S = 32)
Solution:
Molar mass of CuFeS2 = 183.5
wCu
 = constant
wCuFeS2

63.5 wCu
    wCu = 1270 g.
183.5 3.67  103

Illustration 32.
What mass of H2SO4 contains 32 g oxygen ?
Solution:
wO 64 32
= constant    wH 2 SO 4
= 49 g.
wH 2 SO4 98 wH 2 SO4

Illustration 33.
A sample of MgSO4 has 6.022 × 1020 O atoms. What is mass of Mg in sample ?
Solution:
mass of O 16  4 8
In MgSO4;   = constant
mass of Mg 24 3
6.022 1020
Mass of 6.022 × 1020 O-atoms = 16 = 16 × 10–3 g
6.022 1023
8 16  10 3
   Mass of Mg = 6 × 10–3 g
3 mass of Mg

Illustration 34.
If mass % of oxygen in monovalent metal 'M' carbonate (M2CO3) is 48%, find atomic mass of metal ?
Solution:
Let A be atomic mass of M
3  16
Then × 100 = 48 A = 20
2 A  12  3(16)

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Mole Concept

Illustration 35.
Find value of X if 36.6 g BaCl2.XH2O on strong heating loses 5.4 g moisture ?
[At. mass Ba = 137, Cl = 35.5]
Solution:
5.4 g H2O is in 36.6 g BaCl2.XH2O
WH 2O 5.4 18 X
= const. = = X = 2
W BaCl2  H 2 O 36.6 137  71  18 X

Illustration 36.
0.492 g sample of haemoglobin has 0.34% by mass of Fe. If each molecule of haemoglobin has 4 Fe atoms find its
molecular mass ?
Solution:
0.34
Let M = Molecular mass of haemoglobin  × M = 4 × 56
100
M = 65882.3 a.m.u..

Percentage Yield
In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of the
product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual yield and
theoretical yield the percentage yield can be calculate as :
Actual yield
%yield = ×100
Theoretical yield

Illustration 37.
What is mass of O2 required to produce 960 gm of O3 if % yield of reaction 3O2  2O3 is 50% ?
Solution:
3O2 
 2O3
960
= 20mol = Actual yield
48
Let X mol O2 be required
2   20 
Theoretical yield of O3 =  X   50 =  (100)
 3  2 
 X
3 
X = 60 mol = 60 × 32 = 1920 g O2 required

Illustration 38.
What is mass of C obtained on reacting 20 moles of A with excess B by reaction :
2A + B 
 3C
If % yield of reaction is 80% ?
Solution:
2A + B 
 3C
20mol 30 mol = Theoretical yield
Actual yield = X mol = ?

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Mole Concept

X
 × 100 = 80 X = 24 moles
30

Percentage purity
The percentage of a specified compound or element in an impure sample.

Actual massof compound

%purity = 100
Total massof sample

Illustration 39.
A 120g CaCO3 sample having inert impurities on heating produced 56g of residue. Find % purity of sample.
Solution:

Balanced Reaction CaCO3(s)   CaO(s) + CO2(g)
1mol 56g
= 100g =1mol
100
% purity = × 100 = 83.3%
120

Illustration 40.
Which of following reactions occur to completion ?
2A + 3B 
 C + 2D
(a) Initial moles 20 30 0 0
(b) Initial moles 40 60 10 0
(c) Initial moles 40 90 0 0
(d) Initial moles 50 75 0 10
Solution:
(a, b, d)

Illustration 41.
For the reaction, A(s) + 2B(g) 
 3C(g) + D(g)
Molar mass [g mol–1] A = 100; B = 50; C = 60; D =20
(a) How many grams of C are produced by reaction of 250g A ? [Hint : Mass-mass relation]
(b) What mass of B reacts to give 500g D ?
(c) How many grams of A will produce 11.2L of C at STP ?
Solution:
(a) A(s) + 2B(g) 
 3C(g) + D(g)
250
=2.5mol excess 7.5mol
100
= 7.5 × 60g
= 450g
(b) A(s) + 2B(g) 
 3C(g) + D(g)
500
50 mol = 25 mol
20
= 50 × 50

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Mole Concept

= 2500 g
(c) [Hint : mass-volume relation]
A(s) + 2B(g) 
 3C(g) + D(g)
0.5 11.2
mol = 0.5 mol
3 22.4
0.5
= × 100
3
= 16.67g

Illustration 42.
18.625g KCl is formed due to decomposition of KClO4 in reaction
KClO4(s) 
 KCl(s) + 2O2(g)
Find volume of O2 obtained at STP. [Atomic mass K = 39, Cl = 35.5]
Solution:
KClO4(s) 
 KCl(s) + 2O2(g)
18.625 2 18.626
mol
74.5 74.5
= 0.25 = 0.5 mol
Vol. of O2 at STP = 0.5 × 22.4 = 11.2 L

Illustration 43.
What volume of CO2 at STP is obtained by thermal decomposition of 20g KHCO3 to CO2 & H2O. [Atomic mass K = 39]
Solution:

Write balanced reaction : 2KHCO3(s)   K2O(s) + 2CO2(g) + H2O(g)
20g=0.2mol 0.2mol
Vol. of CO2 = 0.2 × 22.4 = 4.48 L

Illustration 44.
What mass of CH3OH (methanol) will be produced by reacting 1120 L of H2(g) at STP with excess CO(g) ?
Solution:
Write balanced reaction : CO(g) + 2H2 (g) 
 CH3OH ()

1120
mol 25 mol
22.4
= 50 mol
Mass of CH3OH = 25 × 32 = 800 g

Illustration 45.
2A (s) + 3B (g) 
 4 C (g)
t=0 10 mol 10 mol
Find final moles of A, B and C.
Solution:
LR is B
2A + 3B 
 4C

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Mole Concept

2 4
t= 10 – 10   0 10  
3 3
10  40 
t= mol 0   mol
3  3 

Illustration 46.
If 2.4 gm Mg is treated with 0.64 gm O2. Find composition of final product mixture ?
[Atomic Mass Mg = 24]
Solution:
1
Mg(s) + O2(g)  MgO (s)
2
t=0 0.1 mol 0.02 mol 0
[O2 is LR]
t= 0.1 –0.02 × 2 0 0.02 × 2
= 0.06 mol = 0.04 mol

Illustration 47.
20g of impure NaCl sample is added aqueous solution having excess AgNO3. AgCl precipitate is filtered washed, dried
& its mass is 28.7g. Find % purity of NaCl.
[Atomic mass Ag = 108, Cl = 35.5]
Solution:
AgNO3(aq) + NaCl(aq) 
 AgCl(s) + NaNO3(s) [Precipitation reaction]
28.7
0.2moles moles
143.5
= 0.2 moles
Mass of NaCl = 0.2 × 58.5 = 11.7 gm
11.7
% purity = × 100
20
= 58.8 %

Principle of Atom Conservation (POAC)

POAC is nothing but the conservation of atoms of reactants and products involved in a chemical reaction, as
expressed before in the concepts of atomic theory. And if atoms are conserved, moles of atoms shall also be
conserved. The principle is fruitful for the students when they don't get the idea of balanced chemical equation in
the problem using POAC we do not need to balance a reaction and we can even add two or more reactions. This
principle can be under stood by the following example.
Consider the decomposition of KClO3 (s)  KCl (s) + O2 (g) (unbalanced chemical reaction)
Apply the principle of atom conservation (POAC) for K atoms.
or moles of K atoms in KClO3 = moles of K atoms in KCl
Now, since 1 molecule of KClO3 contains 1 atom of K
Thus, moles of K atoms in KClO3 = 1 × moles of KClO3
and moles of K atoms in KCl = 1 × moles of KCl

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Mole Concept

wt.of KClO 3 in g wt.of KCl in g

 moles of KClO3 = moles of KCl or =
mol.wt.of KClO 3 mol. wt.of KCl

• The above equation gives the mass-mass relationship between KClO3 and KCl which is important in
stoichiometric calculations. Again, applying the principle of atom conservation for O atoms,
moles of O in KClO3 = 3 × moles of KClO3
moles of O in O2 = 2 × moles of O2
 3 × moles of KClO3 = 2 × moles of O2
wt . of KClO3 vol. of O2 at NTP
or 3 × =2×
mol . wt . of KClO3 standard molar vol .(22.4 lt )
• The above equations thus give the mass-volume relationship of reactants and products.

Illustration 48.
27.6 g K2CO3 was treated by a series of reagents so as to convert all of its carbon to K2Zn3 [Fe(CN)6]2. Calculate the weight
of the product. [mol. wt. of K2CO3 = 138 and mol. wt. of K2Zn3 [Fe(CN)6]2 = 698]
Solution:
Here we have no knowledge about series of chemical reactions but we know about initial reactant and final product,
accordingly.
K2CO3  Several
  K2Zn3 [Fe(CN)6]2
Steps

Since C atoms are conserved, applying POAC for C atoms,

moles of C in K2CO3 = moles of C in K2Zn3 [Fe(CN)6]2
1 × moles of K2CO3 = 12 × moles of K2Zn3 [Fe(CN)6]2
( 1 mole of K2CO3 contains 1 moles of C)
wt . of K 2 CO3 wt . of the product
= 12 ×
mol. wt. of K 2 CO3 mol. wt. of product

27.6 698
wt. of K2Zn3 [Fe(CN)6]2 = × = 11.6 g
138 12

Degree of Dissociation, () :

It represents the mole of substance dissociated per mole of the substance taken.
Mo  M
A n particles;  
(n  1).M
where, n = number of product particles per particle of reactant
Mo = Molar mass of 'A'
M = Molar mass of final mixture
Dissociation decreases the average molar mass of system while association increases it.
Same formula is applicable for association, taking the correct value of 'n'.

Illustration 49.
For the reaction 2NH3(g)  N2(g) + 3H2(g)
Calculate degree of dissociation () if observed molar mass of mixture is 13.6

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Mole Concept

Solution:

M0 = MNH3 = 17, M = 13.6 (given), n = 4 =2

2
Mo  M 17  13.6
 =  0.25
(n  1).M (2  1)  13.6

Illustration 50.
Find value of  if % dissociation is 25%
Solution:
0.25

Illustration 51.
'A' dissociate into 'B' and 'C' according reaction.
A  2B + C
If 5 moles of 'A' is 40% dissociated, then find moles of 'A' left.
Solution:
3 mol

Problems based on mixture :

The composition of any mixture may be determined by reacting the mixture with some substance, by which either
one or more component of mixture may react.

Illustration 52.
1.5 gm mixture of SiO2 and Fe2O3 on very strong heating leave a residue weighing 1.46 gm. The reaction responsible for
loss of weight is
Fe2O3 (s) Fe3O4 (s) + O2(g)
What is the percentage by mass of Fe2O3 in original sample.
Solution:
1
3Fe2O3 (s) 2Fe3O4 + O2
2
1
3 × 160  32
2
= 480 gm  =16 gm
loss of 16 gm 480 gm Fe2O3
480
loss of 0.04 gm 0.04 × = 1.2 gm Fe2O3
16
1.2
% by mass =  100 = 80%
1.5

Illustration 53.
19 gm mixture of Na2CO3(s) and NaHCO3(s) on heating gives 2.2 gm CO2 gas.
Find % NaHCO3 (by mass) in mixture.
Solution:
44.21 %

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Mole Concept

Illustration 54.
Write decomposition reaction of CaCO3(s) and Na2CO3 (s) on heating.
Solution:

CaCO3(s)  CaO(s) + CO2(g)

NaCO3(s)   (No decomposition)

Problem Related with Parallel Reaction :

When same two reactants form two or more products by independent reactions.

Illustration 55.
Carbon reacts with oxygen forming carbon monoxide and/or carbon dioxide depending an availability of oxygen. Find
moles of each product obtained when 160 gm oxygen reacts with (a) 12 g carbon (b) 120 g carbon (c) 72 g carbon.
Solution:
1
(a) C + O2 
 CO [initially use a reaction using lesser amount of oxygen]
2
t=0 1mol 5mol
t = 0 5–0.5 = 1mol
(LR) 4.5mol
Since CO & O2 are left, CO2 will also formed.
1
CO + O2 
 CO2
2
t=0 1mol 4.5mol 0
t= 0 4 mol 1 mol
At end, 1 mole CO2 & no CO present
1
(b) C + O2 
 CO
2
t=0 10mol 5mol 0
t = 0 0 10mol
At end only 10 mol CO present.
1
(c) C + O2 
 CO
2
t=0 6mol 5mol 0
t = 0 2mol 6mol
[LR]
1
CO + O2 
 CO2
2
t=0 6mol 2mol 0
t = 2mol 0 [LR] 4mol
At end [2mol CO + 4mol CO2] left.

Illustration 56.
25.4 gm of iodine and 14.2 gm of chlorine are made to react completely to yield mixture of ICl and ICl3. Ratio of moles
of ICl & ICl3 formed is (Atomic mass : I = 127, Cl = 35.5)

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Mole Concept

Solution:
I2 + Cl2 ICl + ICl3
0.1mol 0.2 mol x mol y mol
Applying conservations of I and Cl -atom
0.2 = x+y  x = 0.1
0.4 = x+3y  y = 0.1
 nICl : nICl3 = x : y = 1 : 1

Illustration 57.
2 moles carbon and 1.5 moles of oxygen gas are reacted in a container to produce CO or CO2 or both. Find moles of
CO, CO2 produced.
Solution:
CO = 1 mole, CO2 = 1 mole

Problems Related with Sequential Reaction :

When one of products formed in previous reaction is consumed in the next one.

Illustration 58.
How many grams H2SO4 can be obtained from 1320 gm PbS as per reaction sequence ?
2PbS + 3O2  
 2PbO + 2SO2
3SO2 + 2HNO3 + 2H2O  
 3H2SO4 + 2NO
[At. mass : Pb = 208, S = 32]
Solution:
1320
Moles of PbS = = 5.5 mol
240
Moles of SO2 = 5.5 mol = moles of H2SO4
Mass of H2SO4 = 5.5 × 98 = 539 gm
[When amount of only one reactant is given generally other is assumed in excess.]

Illustration 59.
Calcium phosphide Ca3P2 formed by reacting magnesium with excess calcium orthophosphate Ca3(PO4)2, was hydrolysed
by excess water. The evolved phosphine PH3 was burnt in air to yield phosphorus pentoxide (P2O5). How many grams of
magnesium metaphosphate would be obtain if 192 gram Mg were used (Atomic weight of Mg = 24, P = 31)

Ca3(PO4)2 + Mg 
Ca3P2 + MgO
Ca3P2 + H2O 
Ca(OH)2 + PH3
PH3 + O2 
P2O5 + H2O
MgO + P2O5 
Mg(PO3)2
magnesium metaphosphate.
Solution:
Balanced chemical reaction :

Ca3(PO4)2 + 8Mg 
 Ca3P2 + 8MgO

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Mole Concept

192 1
excess  8mole mole 8mole
24 8
Ca3P2 + 6H2O 
3Ca(OH)2 + 2PH3
1 1
mole mole
8 4
2PH3 + 4O2 
 P2O5 + 3H2O
1 1
mole mole
4 8
MgO + P2O5 
 Mg(PO3)2
8mole 1/8mole (LR) 1/8 mole obtained
WMg PO3  = 1/8 × 182 = 22.75 Mg
2

Illustration 60.

A 
 2B + C
3B 
 2D
Find moles of D produced if initially 3 moles of A are taken -
Solution:
(4)

Illustration 61.

2A + 3B 
 4C + D
(excess)

3C 
 2E
If in the above reaction 6 moles of E are produced, find moles of 'A' initially taken.
Solution:
(4.5 mole)

Empirical and Molecular Formula

We have just seen that knowing the molecular formula of the compound we can calculate percentage composition
of the elements. Conversely if we know the percentage composition of the elements initially, we can calculate the
relative number of atoms of each element in the molecules of the compound. This gives us the empirical formula of
the compound. Further if the molecular mass is known then the molecular formula can be easily determined.
Thus, the empirical formula of a compound is a chemical formula showing the relative number of atoms in the
simplest ratio, the molecular formula gives the actual number of atoms of each element in a molecule.
i.e. Empirical formula : Formula depicting constituent atoms in their simplest ratio.
Molecular formula : Formula depicting actual number of atoms in one molecule of the compound.
The molecular formula is generally an integral multiple of the empirical formula.
i.e. molecular formula = empirical formula × n

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Mole Concept

molecular formula mass

where n =
empirical formula mass
Example :
Molecular Formula H2O2 C6H6 C2H6 C2H4O2
2:2 6:6 2:6 2:4:2
Simplest ratio 1:1 1:1 1:3 1:2:1
Empirical Formula HO CH CH3 CH2O

Determination of Empirical formula :

Following steps are involved in determining the empirical formula of the compounds –
(i) First of all, find the % by wt. of each element present in the compound.
(ii) The % by wt of each element is divided by its atomic weight. It gives atomic ratio of elements present in the
compounds.
(iii) Atomic ratio of each element is divided by the minimum value of atomic ratio so as to get simplest ratio of
atoms.
(iv) If the value of simplest atomic ratio is fractional then raise the value to the nearest whole number or multiply
with suitable coefficient to convert it into nearest whole number
(v) Write the Empirical formula as we get the simplest ratio of atoms.

Determination of molecular formula :

(i) Find out the empirical formula mass by adding the atomic masses of all the atoms present in the empirical
formula of compound.
(ii) Divide the molecular mass (determined experimentally by some suitable method) by the empirical formula mass
and find out the value of n.
(iii) Multiply the empirical formula of the compound with 'n' so as to find out the molecular formula of the
compound.

Illustration 66.
An organic compound contains 49.3% carbon, 6.84% hydrogen and its vapour density is 73. Molecular formula of
compound is :-
Solution:
V.D. = 73  M = 2 × 73 = 146
49.3 71.978
C = 146 × = 71.978 g = ~ 6 mole
100 12
6.84 71.978
H = 146 × = 9.9864 g = ~ 10 mole
100 12
100  (49.3  6.84) 64.86
O = 146 × = = 64.86 g  4 mol
100 16
M.F. = C6H10O4

Illustration 67.
The empirical formula of an organic compound containing carbon & hydrogen is CH2. The mass of 1 litre of organic gas
is exactly equal to mass of 1 litre N2 therefore molecular formula of organic gas is.

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Mole Concept

Solution:
Empirical Mass of CH2 = 12 + 2 = 14
 Mass of 1 litre of organic gas = Mass of 1 litre of N2

Since V, P, T, n is same.
m
Therefore, from PV = RT implies that molar mass should also be same.
M
 Molecular mass of organic compound will be 28 g
Molecular mass 28
n= = =2
Empirical mass 14
So molecular formula = 2 × CH2 = C2H4

Illustration 68.
In which of following has same empirical formula -
C2H2, C2H4, C2H6, C6H6
Solution:
C2H2, C6H6,

Illustration 69.
Determine the empirical formula of an oxide of iron, which has 70% Fe and 30% 'O' by mass. (Fe = 56)
Solution:
Fe2O3

Eudiometry :

Eudiometry or gas analysis involves the calculations based on gaseous reactions or the reactions in which at least
two components are gaseous, in which the amounts of gases are represented by their volumes, measured at the
same pressure and temperature. Some basic assumptions related with calculations are:
(i) Gay-Lussac's law of volume combination holds good. According to this law, the volumes of gaseous reactants
reacted and the volumes of gaseous products formed, all measured at the same temperature and pressure, bear
a simple ratio.
N2(g) + 3H2(g)  2NH3 (g)
1 vol. 3 vol. 2 vol.
Problem may be solved directly is terms of volume, in place of mole. The stoichiometric coefficients of a balanced
chemical reactions gives the ratio of volumes in which gaseous substances are reacting and products are formed,
at same temperature and pressure.
(ii) The volumes of solids or liquids is considered to be negligible in comparison to the volume of gas. It is due to
the fact that the volume occupied by any substance in gaseous state is even more than thousand times the
volume occupied by the same substance in solid or liquid states.
2H2(g) + O2 (g) 2H2O (l)
2 mole 1 mole 2 mole
2 vol. 1 vol. 0 vol.
(iii) Air is considered as a mixture of oxygen and nitrogen gases only. It is due to the fact that about 99% volume of
air is composed of oxygen and nitrogen gases only.

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Mole Concept

(iv) Nitrogen gas is considered as a non-reactive gas. It is due to the fact that nitrogen gas reacts only at very high
temperature due to its very high thermal stability. Eudiometry is performed in a eudiometer tube and the tube
cannot withstand very high temperature. This is why, nitrogen gas cannot participate in the reactions occurring
in the eudiometer tube.
(v) The total volume of non-reacting gaseous mixture is equal to sum of partial volumes of the component gases
(Amagat's law).
V = V1 + V2 +..............
Partial volume of gas in a non-reacting gaseous mixture is its volume when the entire pressure of the mixture is
supposed to be exerted only by that gas.
(vi) The volume of gases produced is often given by certain solvent which absorb contain gases.
Solvent Gases absorb
KOH CO2, SO2, Cl2
Ammonical Cu2Cl2 CO
Turpentine oil O3
Alkaline pyrogallol O2
water NH3, HCl
CuSO4/CaCl2 H2O
(vii) EUDIOMETER
A eudiometer is a laboratory device that measures the change in volume of a gas mixture following a physical
or chemical change.

Illustration 70.
10 ml of CO is mixed with 25 ml air (20% O2 by volume). Find final volume (in ml) after complete combustion.
Solution:
5 ml
10 ml
1
CO  O 2 
 CO 2
2 10 ml

Vf = VCO2 + Volume of remaining air = 10 + 20 = 30 ml

Illustration 71.
A 3 L gas mixture of propane (C3H8) and butane (C4H10) on complete combustion at 25°C produced 10 L CO2. Assuming
constant P and T conditions what was volume of butane present in initial mixture ?
Solution:

C3H8(g) + 5O2 
 3CO2(g) + 4H2O(l)
xL 3x L
13
C4H10(g) + O2 (g) 
 4CO2(g) + 5H2O(l)
2
(3-x) L 4(3 – x) L
from question 3x + 4 (3 –x) = 10  x = 2
 Volume of butane, C4H10 = (3 – x) = 1 L

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Mole Concept

Illustration 72.
CH3

100 ml gaseous meta Xylene undergoes combustion with excess of oxygen at room temperature and

CH3
pressure. Volume contraction / expansion (in ml) during reaction is
Solution:
21
C4H10(g) + O2 (g) 
 8CO2(g) + 5H2O(l)
2
21
 100
100 ml 2 800 ml 0
 1050ml
 Contraction in volume = (100 + 1050) – 800 = 350 ml

Illustration 73.
An alkene upon combustion produces CO2(g) and H2O(g). In this combustion process if there is no volume change occurs
then the no. of C atoms per molecule of alkene will be :
Solution:
(2)
3n
CnH2n(g) + O2(g)  nCO2(g) + nH2O(g)
2
if there no volume changes i.e. ng = 0
 3n 
(n + n) – 1    0 n=2
 2 

Illustration 74.
A gaseous hydrocarbon (CxHy) requires 6 times of its own volume of O2 for complete oxidation and produces 4 times of
its volume of CO2. Find out the volume of x + y.
Solution:
(012)
y 
 y
CXHy + (x + ) O2 xCO2 + H 2 O (l )
4 2
 y
a ax  ax
 4
Given that : a(x + y/4) = 6a
and ax = 4 (a)
x = 4 and y = 8 ...(2)
 x + y = 4 + 8 = 12

Illustration 75.
On heating 60 ml mixture containing equal volume of chlorine gas and it's gaseous oxide, volume becomes 75 ml due
complete decomposition of oxide. On treatment with KOH volume becomes 15 ml. What is the formula of oxide of
chlorine ?

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Mole Concept

Solution:
Cl2O
Let oxide of Cl is ClxOy
So, in 60 mL  30 mL ClxOy and 30 mL Cl2.
Now,
x y
ClxOy  Cl 2  O2
2 2
30mL
30.x 30.y
mL mL
2 2
Given :
30x 30y
75 = 30 +  x+y=3 ..........(i)
2 2
KOH absorbs Cl2 and volume becomes 15 mL so,
30x
(75 – 15) = VCl  30  x = 2 and y = 1
2
2
So the oxide : Cl2O

Illustration 76.
5 L of A (g) & 3 L of B(g) measured at same T & P are mixed together which react as follows
2A(g) + B(g)  C (g)
What will be the total volume (in litre) after the completion of the reaction at same T & P.
Solution:
(3)

2A(g) + B(g) 
 C(g)
5L 3L
L.R. is A
1
So, volume of C produced =  5 = 2.5 L
2
1
and, volume of B reacted =  5 = 2.5 L
2
So, volume of B remained = 3 – 2.5 = 0.5 L
Hence, Vtotal = VC + VB = 2.5 + 0.5 = 3 L

Solutions

A solution is a homogenous mixture of two or more pure substances whose composition may be altered within
certain limits. Though the solution is homogenous in nature, yet it retains the properties of its constituents.
Generally, solution is composed of two components, solute and solvent. Such type of solution is known as binary
solution. Solvent is that component in solution whose physical state is the same as that of the resulting solution
while another component is called as solute. If the physical state of both components is same, then the component
in excess is known as solvent and other one is called as solute. Each component in a binary solution can be in any
physical state such as liquid, solid and gaseous state.

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Mole Concept

Types of Solutions
Type of Solutions Solute Solvent Common Example
Gas Gas Mixture of oxygen and nitrogen gases
Gaseous Solutions Liquid Gas Chloroform mixed with nitrogen gas
Solid Gas Camphor in nitrogen gas

Gas Liquid Oxygen dissolved in water

Liquid Solutions Liquid Liquid Ethanol dissolved in water
Solid Liquid Glucose dissolved in water

Gas Solid Solution of hydrogen in palladium

Solid Solutions
Liquid Solid Amalgam of mercury with sodium
Solid Solid Copper dissolved in gold

Concentration or strength of solution :

The concentration of a solution is the amount of solute dissolved in a known amount of the solvent or solution.
Solution can be described as dilute or concentrated solution as per their concentration. A dilute solution has a very
small quantity of solute while concentrated solution has a large quantity of solute in solution. Various concentration
terms are as follows.
Mass percentage (W/W %):
It may be defined as the number of parts of mass of solute per hundred parts by mass of solution.
 w wt . of solute
% by mass   : = × 100
W  wt . of solution
[X % by mass means 100 g solution contains X g solute and hence = and hence (100 – X) g solvent]

Mass-volume percentage (W/V %) :

It may be defined as the mass of solute (in gm) present in 100 cm3 of solution. For example, If 100 cm3 of solution
contains 5 g of sodium hydroxide, then the mass-volume percentage will be 5% solution.
 w wt . of solute
%   =  100
V
  volume of solution
 w
[X %   means 100 ml solution contains X gm solute]
V 

Volume Percent (V/V %):

It can be represented as % v/v or % volume and normally used for the solutions in which both components are in
liquids state. It is the number of parts of by volume solute per hundred parts by volume of solution. Therefore,
v  volume of solute
%  =  100
V  volume of solution

Parts per million (ppm) :

The very low concentration of solute in solution can be expressed in ppm. It is the number of parts by mass of solute
per million parts by mass of the solution.
Mass of solute Mass of solute
Parts per million (ppm) = 106  × 106
Mass of solvent Mass of solution
Get yourselves very much comfortable in their inter conversion. It is very handy.

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Mole Concept

Molarity (M) :
Molarity is most common unit for concentration of solution. It is defined as the number of moles of solute present
in one litre or one dm3 of the solution or millimole of solute present in one mL of solution.
Mole of solute
Molarity (M) =
volume of solution in litre

Molality (m) :
The number of mole of the solute present in 1000 g of the solvent is known as molality of solution. It represented
by letter ‘m’.
Moles of solute
Molality (m) =
Mass of solvent (in kg )
The unit of molality is mol/kg and it does not affect by temperature.

Mole Fraction (X) :

Mole fraction may be defined as the ratio of number of moles of one component to the total number of moles of
all the components (solute and solvent) present in solution. It is denoted by letter X and the sum of all mole fractions
in a solution is always equals to one.
Moles of solute
Mole fraction (X) =
Total moles
Mole fraction does not depend upon temperature and can be extended to solutions having more than two
components.

Concentration Type Mathematical Formula Concept

Percentage by mass  w  Mass of solute × 100 Mass (in g) of solute
% = present in 100 gram of
w Mass of solution solution.
Volume percentage  v  Volume of solute × 100 Volume (cm3) of solute
% = present in 100 cm3 of
 v Volume of solution solution.
Mass-volume  w  Mass of solute × 100 Mass (in g) of solute
percentage % = present in 100 cm3 of
 v  Volume of solution solution.
Parts per million Mass of solute × 106 Parts by mass of solute
ppm = per million parts by
Mass of solution
mass of the solution
Mole fraction Mole of A Ratio of number of
XA = moles of one
Mole of A  Mole of B  Mole of C  ....
component to the total
Mole of B number of moles.
XB=
Mole of A  Mole of B  Mole of C  ....
Molarity Mole of solute Moles of solute in one
M=
Volume of solution (in L) litre of solution.

Molality Mass of solute × 1000 Moles of solute in one

m= kg of solvent.
Molar mass of solute Mass of solvent(g)

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Mole Concept

Illustration 77.
Calculate the mole fractions of the components of the solution composed by 92 g glycerol and 90 g water ? (M (water)
= 18; M (glycerol) = 92)
Solution:
Moles of water = 90 g/18 g = 5 mol water
Moles of glycerol = 92 g/92 g = 1 mol glycerol
Total moles in solution = 5 + 1 = 6 mol
Mole fraction of water = 5 mol/6 mol = 0.833
Mole fraction of glycerol = 1 mol/6 mol = 0.167

Illustration 78.
What will be the molarity of solution when water is added to 16.4g Ca(NO3)2 to make 100 mL of solution?
Solution:
Mol of Ca(NO3)2 = 16.4/164 = 0.1
Molarity = Mole of solute/Volume of solution (L) = 0.10 mol/0.10 L
Therefore, Molarity of given solution = 1.0 M

Illustration 79.
Calculate the molality of a solution containing 20 g of sodium hydroxide (NaOH) in 250 g of water?
Solution:
Moles of sodium hydroxide = 20/40 = 0.5 mol NaOH
250 g = 0.25 kg of water
Hence molality of solution = Mole of solute/Mass of solvent (kg)= 0.5 mol/0.25 kg
or Molality(m) = 2.0 m

Illustration 80.
Calculate the gram of copper sulphate (CuSO4) needed to prepare 250.0 mL of 1.00 M CuSO4?
Solution:
250
Moles of CuSO4 = M × V = 1 ×
1000
Molar mass of copper sulphate = 159.6 g/mol
Hence Mass of copper sulphate (g) = Moles of CuSO4 × Molar mass of copper sulphate.
250
=1× × 159.6 g/mol = 39.9 g of Copper sulphate
1000

Illustration 81.
How many gram of H2SO4 are present in 500 ml of 0.2M H2SO4 solution ?
Solution:
moles 500
M =  moles of H2SO4 = M × V = 0.2 × L = 0.1
vol. 1000
Mass of H2SO4 = 0.1 × 98 = 9.8 g

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Mole Concept

Illustration 82.
Calculate the ppm of mercury in water in a sample containing 30 mg of Hg in 500 ml of solution.
Solution:
Mass of solute × 106
Parts per million =
Mass of solution
Mass of Hg = 30 mg
Mass of water = 500/1 = 500g = 50 × 104 mg
v
(density = mass/volume; density of water 1 g/ml) w 
d
30 × 106
Therefore, ppm of mercury = = 60 ppm of mercury
50 104

Illustration 83.
A 100g NaOH solution has 20g NaOH. Find molality.
Solution:
20 / 40 500
m= × 1000 = = 6.25 mol / kg
100  20 80

Illustration 84.
Find molality of aqueous solution of CH3COOH whose molarity is 2M and density d = 1.2 g/mL.
1000×M
Hint :
1000×d - MMs
where d = density in gL–1, M = Molarity, m = molality, MS = molar mass of solute.
Solution:
2
m= × 1000 = 1.85 m
1200  2  60

Illustration 85.
w
A solution has 80% NaOH with density 2gL–1. Find (a) Molarity, (b) Molality of solution.
w
Solution:
Let V be vol. of solution, in L
 w
%  80
 w
Mass of solute = (d × V) × =2×V× = 1.6V
100 100
1.6V / 40
(a) M = = 0.04 m
V
1.6V / 40
(b) m = × 1000 = 100 mol kg 1
2V  1.6V

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Mole Concept

Illustration 86.
4.450 g sulphuric acid was added to 82.20 g water and the density of the solution was found to be 1.029 g/cc at 25°C
and 1 atm pressure. Calculate
(a) the weight percent,
(b) the mole fraction,
(c) the mole percent,
(d) the molality,
(e) the molarity of sulphuric acid in the solution under these conditions.
Solution:
Sulphuric acid = 4.450 g, Water = 82.20 g  Wt. of solution = 86.65 g
 Density of solution = 1.029 g/cc.

(a) Weight percent = wt . of solute 4.450

 100   100  5.14
wt . of solution 86.65
(b) Mole fraction :

Mole of solute =
wt. of solute 4.45
  0.0454
mol wt . of solute 98
82.20
Mole of solvent =  4.566
18
Total moles in solution = 0.0454 + 4.566 = 4.6114
0.0454
Mole fraction of solute =  0.0098
4.6114
(c) Mole percent = mole fraction of solute × 100 = 0.0098 × 100 = 0.98

(d) Molality =
moles of solute
 1000
mass of solvent (in gm )
0.0454  1000
=  0.552
82.2
(e) Molarity = m oles of solute
litre of solution
Mass 86.65 86.65
Volume of solution =  ml = litre
Density 1.029 1.029  1000
0.0454 0.0454  1000  1.029
Molarity = =  0.539
86.54 86.65
1.029  1000

Illustration 87.
Find number of Na+ & PO4–3 ions in 250 ml of 0.2M Na3PO4 solution.
Solution:
Na3PO4 + aq. 
 3Na+(aq) + PO4–3 (aq) [Ionic compound when added to water ionize completely].
50 millimoles (m.m.) 150 mm 50 mm
No. of Na ions = 150 × 10 × NA; No. of PO4–3 ions = 50 × 10–3 × NA
+ –3

Illustration 88.
80g NaOH was added to 2L water. Find molality of solution if density of water = 1g/mL

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Mole Concept

Solution:

m=
moles of NaOH × 1000 = 80 / 40 × 1000 = 1 molal
mass of H 2O 2  1000

Illustration 89.
The average concentration of Na+ ion in human body is 3.0 to 3.9 gm per litre. The molarity of Na+ ion is about.
Solution:
0.15 M
3  3.9
6.9
= 2 
n solute
M Na    0.15M
volume of solution in Lt 23 46

Mixing of and Dilution Solutions :

It is based on law of conservation of moles of solute

(i) Two solutions having same solute

Total moles = M1 V1  M 2 V2
Final molarity =
Total volume V1  V2
(ii) Dilution Effect : When a solution is diluted, the moles of solute do not change but molarity changes while on
taking out a small volume of solution from a larger volume, the molarity of solution do not change but moles
change proportionately.
M 1 V1
Final molarity =
V1  V2
n-fold or n-times dilution
   Final volume = V1 + V2 = n(V1)

Illustration 90.
50 ml 0.2 M H2SO4 is mixed with 50 ml 0.3M H2SO4. Find molarity of final solution.
Solution:

Total m oles of H 2SO 4 500.2103  50103 0.3

Mf = = = 0.25 M.
Total volum e (50  50)103

Illustration 91.
Find final molarity in each case :
(i) 500 ml 0.1 M, HCl + 500 ml 0.2M HCl
(ii) 50 ml, 0.1M HCl + 150 ml, 0.3MHCl + 300 ml H2O
(iii) 4.9g H2SO4 + 250 ml, H2O + 250 ml 0.1 M H2SO4

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Mole Concept

Solution:
500  0.1  500  0.2
(i) Mf = = 0.15 M.
500  500
50  0.1  150  0.3 50
(ii) Mf = = = 0.1 M
50  150  300 500
4.9 250
  0.1
50  25
(iii) Mf = 98 1000 = = 0.15 M
 250  250  500
 
 1000 

Illustration 92.
How much water should be added to 2M HCl solution to form 1 litre of 0.5 M HCl ?
Solution:
Let V be initial volume
Then mol of HCl = constant
2 × V = 1 × 0.5  V = 0.25 L
Volume of water added = 1 – 0.25 = 0.75 L

Illustration 93.
1.11g CaCl2 is added to water forming 500 ml of solution. 20 ml of this solution is taken and diluted 10 folds. Find moles
of Cl– ions in 2 ml of diluted solution.
Solution:

1.11
= 0.01 mol CaCl2
111
0.01 0.01
Moles of CaCl2 in 20ml solution = × 20 =
500 25
0.01
In 200 ml solution, moles of CaCl2 = [Note : Dilution does not change moles of solute]
25
0.01/25 0.01
In 2 ml of dilute solution moles of CaCl2 = ×2= = 4 × 10–6
200 2500
 moles of Cl– = 2 × 4 × 10–6 = 8 × 10–6

Illustration 94.
What volumes of 1M & 2M H2SO4 solution are required to produce 2L of 1.75M H2SO4 solution?
Solution:
Let XL be vol. of 1M solution.
 (2 – X)L is vol. of 2M solution.
Moles of H2SO4 : 2 × 1.75 = 1(X) + (2 – X)2  X = 0.5 L
i.e. 0.5L of 1M & 1.5 L of 2M solution are required.

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Mole Concept

Illustration 95.
A solution is made by mixing 300 ml 1.5M Al2(SO4)3 + 300 ml 2M CaSO4 + 400 ml 3.5M CaCl2.
Find final molarity of (1) SO4–2, (2) Ca2+, (3) Cl–. [Assume complete dissociation of these compounds].
Solution:
3001.5103 3 3002103
(1) [SO4 ]f = Total moles =
–2
= 1.95M
Total volume (300  300  400)103
300  2  400  3.5
(2) [Ca+2]f = = 2M
1000
400  3.5  2
(3) [Cl–]f = = 2.8M
1000

Illustration 96.
A solution of KCl has a density of 1.69 g mL–1 and is 67% by weight. Find the density of the solution if it is diluted so that
the percentage by weight of KCl in the diluted solution is 30%.
Solution:
Let the volume of the KCl solution be 100 mL,
Weight of KCl solution = 100 × 1.69 = 169 g
100 g of solution contains = 67 g of KCl
67
169 g of solution =  169  113.23g
100
Lex x mL of H2O be added.
New volume of solution = (100 + x) mL
New weight of solution = (169 + x) g
(Since x mL of H2O = x g of H2O, dH2O = 1)
New percentage of the solution = 30%

% by weight =
weight of solute  100
weight of solution
113.23
30 =  100
(169  x)
x = 208.43 mL = 208.43 g
(169  x)
New density = New weight of solution =
New volume of solution (100  x)
 d = 1.224 g/ml

Illustration 97.
Calculate the amount of the water "in m" which must be added to a given solution of concentration of 40 mg silver

nitrate per ml, to yield a solution of concentration of 16 mg silver nitrate per ml ?

Solution:
1.5 ml
Before dilution After dilution
(nsolute) = (nsolute)
Mi Vi = Mf Vf

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Mole Concept

40 16
170 1 = 170 (1  V)
1 1
 V = 1.5 mL

Illustration 98.
100 ml, 3%(w/v) NaOH solution is mixed with 100 ml, 9%(w/v) NaOH solution. The molarity of final solution is-
Solution:
(1.5)
Total NaOH in 100 ml (1st solution) = 3 g
Total NaOH in 100 ml (2nd solution) = 9 g
 12/ 40 
 Molarity =    1.5M
 200/1000 

Illustration 99.
1120 gm of 2 'm' urea solution is mixed with 2480 gm of 4 'm' urea solution. Calculate the molality of the resulting
solution?
Solution:
3.33 m
Let 2 m, 1120 g solution have mass of solute = w g
 solvent = (1120 – w) g
& Let 4 m, 2480 g solution have mass of solute = w1 g
 solvent = (2480 – w) g
n solute
molality =
wt. of solvent in kg
w / 60
2=  1000
1120  w
w = 120 g
w'/60
& 4=  1000
2480  w'
w' = 480 g
120  480
resulting molality m = 60  1000 = 3.33 m
1120  120  2480  480

Volume Strength of H2O2 Solution :

Labelled as 'volume H2O2, it means volume of O2 (in litre) at STP that can be obtained from 1 litre of H2O2 solution,
when H2O2 when it decomposes according to
1
H2O2  H2O + O2
2
Volume Strength of H2O2 Solution = 11.35 × molarity

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Mole Concept

Illustration 100.
Find the % w/v of "10 V" H2O2 solution-
Solution:
volume strength 10
Molarity (M) of solution = 
11.35 11.35
 w  M  mol. wt. of solute 10 34
%   =  = 3%
 v  10 11.35 10

Illustration 101.
2H2O2(aq) 
 2H2O(l) + O2(g)
1
Under conditions where 1 mole of gas occupies 24 dm3, X L of M solution of H2O2 produces 3 dm3 of O2. Thus X is :-
24
Solution:
(6)
1
moles of H2O2 = ×X
24
3 1
moles of O2 = = .
24 8
1 X
moles of H2O2 = =
4 24
X=6

Percentage Labelling of Oleum :

Labelled as '% oleum', it means maximum amount of H2SO4 that can be obtained from 100 gm of such oleum
(mixture of H2SO4 and SO3) by adding sufficient water. For ex. 109 % oleum sample means, with the addition of
sufficient water to 100 gm oleum sample 109 gm H2SO4 is obtained.
% labelling of oleum sample = (100 + x)%
x = mass of H2O required for the complete conversion of SO3 in H2SO4.

Illustration 102.
Find the mass of free SO3 present in 100 g, 109 % oleum sample.
Solution:
109 % means, 9 gm of H2O is required for 100 gm oleum
SO3 + H2O  H2SO4
9g
1/2mole 1/2mole
40g
 Mass of free SO3 = 40 g, Mass of H2SO4 = 60 g

Illustration 103.
Find the % labelling of 100 g oleum sample if it contains 20 g SO3.
Solution:
% labelling of oleum sample = (100 + x)%

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Mole Concept

SO3 + H2O  H2SO4

20g
1/4mole 1/4mole
 4.5g
 % labelling of oleum sample = (100 + 4.5) % = 104.5%

Illustration 104.
An oleum sample is labelled as 118 %, Calculate composition of mixture (mass of components) if 40 g water is added to
30 g given oleum sample.
H2SO4= 35.4 g, H2O = 34.6g
Solution:
In 100 g sample requires water = 18 g
18
30 g sample will require water =  30 = 5.4 g
100
Mass of H2O = 40 – 5.4 = 34.6 g and mass of H2SO4 = 70 – 34.6 = 35.4 g.

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