class 9 science chapter 3 atoms and molecules pdf

ATOMS AND MOLECULES
3.1 INTRODUCTION
Maharishi kanad postulated thatifwe go on dividing matter (Padarth)
a limit will come when we will come across smallest particles
beyond which no further division is possible. He named these
particles as parmanu.
Question based on basic knowledge required to understand
this chapter
1. Kanad gave which theory
(A) Atomic theory (B) Electronic theory
(C) Parmanu theory (D) Oxide theory
2. An atom has
(A) Proton inside the nucleus whereas electrons and neutrons
outside the nucleus
(B) Proton and neutron inside the nucleus and electrons outside
the neucleus
(C) electrons inside the neucleus
(D) none of these
3. Valency electrons refers to the
(A) electrons of the outermost shell
(B) total electrons in an atom
(C) electrons of the inner shells
(D) none of these
4. Which of the following is a monoatomic molecule?
(A) HCl (B) CO2 (C) NH3 (D)He
5. When an electron from an outermost shell of an atom is removed
than what happens?
(A) a cation is formed (B) an anion is formed
(C) a neutral species is formed (D) no change is noticed
6. Na+
, Mg2+
are actually
(A) cations (B) anions (C) neutral species
(D) Na+
is a cation
7. “During a chemical reaction matter is neither created nor
destroyed”. The above statement is according to
(A) Law of conservation of mass (B) Law of constant
3.1 Introduction
3.2 Laws of chemical
combination
3.3 Postulates of Dalton’s
Atomic Theory
3.4 Atoms
3.5 Molecule
3.6 Ion
3.7 Writing Formulae of
Molecular Compounds
3.8 Atomic, Molecular &
Equivalent Masses
3.9 Formula Mass
3.10 Composition of
Compound
3.11 Mole Concept
3.12 Avogadro’s Law
3.13 Law of Multiple
Proportions
3.14 Expression
Concentration of
Solutions
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proportions
(C) Avogadro law (D) None of these
8. A molecule is defined as a
(A) the smallest particle of an element or of a compound which can exist freely under ordinary conditions
and shows all the properties of that substance
(B) smallest particle of an element
(C) the charged species obtained by the loss or gain of electron from an atom
(D) any uncharged species
9. If the molecular mass of H2O is 18 grams and that of hydrogen is 1 g the molecular mass of oxygen is
(A) 16 g (B) 17 g (C) 16–5 g (D) 18 g
10. Which of the following is a pair of atoms?
(A) H2 and O2 (B) NH3 and C (C) N2 and Xe (D) He and C
3.2 LAW OF CHEMICAL COMBINATION
A number of laws have been proposed after through experimental studies.Out of these, two laws of
chemical combinations are given. These are:
Law of Conservation of Mass.
Law of Constant Proportions
3.2.1 Law of Conservation of Mass
It was given by a French chemist, A. Lavoisier in 1774. He is known as the father of chemistry. The law
may be stated as:
The total mass of the products of a chemical reaction is equal to the total mass of the reactants that have
combined.
The mass can neither be created nor be destroyed in a chemical reaction.
In other words, the mass remains unchanged or conserved in a chemical reaction.The law is also known
as the Law of Industructibility of Mass.
Illustration 1
If 6.3g of sodium bicarbonate are added to 15.0 g of ethanoic acid (or acetic acid) solution, the
residue left is found to weigh 18.0 g. What mass of CO2
is released in the reaction?
Solution
The chemical reaction leading to products is:
Sodium bicarbonate + Ethanoic acid 
left)
(Residue
solution
ethanoate
Sodium +
(released)
dioxide
Carbon
Mass of reactants = (6.3 + 15.0) = 21.3 g
Mass of products = Mass of residue + Mass of carbon dioxide released.
= 18.0 g + x g
According to law of conservation of mass.
Mass of reactants = Mass of products
21.3 g = (18.0 + x) g or x = 21.3 – 18.0 = 3.3 g

 Mass of carbon dioxide released = 3.3 g
3.2.2 Law of constant proportions
The law of constant proportions is also known as the law of definite proportions. This law was postulated
by Proust in 1797.
“A chemical compound always consists of the same elements combined together in the same ratio,
independent of the method by which it is prepared, or the source from where it is obtained.”
For example, water (H2
O) is always found to contain only hydrogen and oxygen, in the ratio of 1 : 8 by
mass (or 2 : 1 by volume) independent of the sources of water.
Maharashi Kanad was a great Indian sage in the Vedic period. According to him, if a pure substance is
broken down into smaller pieces, a stage is reached when no further subdivision is possible. He called the
ultimate smallest particle of any pure substance anu. Each anu of a substance has the same properties.
He further belived that each anu may be made of two or more still smaller particles. He called these
particles as parmanu. According to him, a parmanu is the ultimate smallest unit of matter.
Illustration 2
0.24 g sample of a compound of oxygen and boron was found on analysis to contain 0.096 g of
boron and 0.144 g of oxygen. Calculate the percentage composition of the compound.
Solution
Mass of the compound = 0.24 g
Mass of boron in the compound = 0.096 g
Mass of oxygen in the compound = 0.144 g
Precentage of boron = 100
compound
of
Mass
boron
of
Mass

= 100
(0.24g)
(0.096g)
 = 40%
Percentage of oxygen = 100
compound
of
Mass
oxygen
of
Mass

= 100
(0.24g)
(0.144g)
 = 60%
Illustration 3
On analysis it was found that the black oxide of copper and the red oxide of copper contain 79.9%
and 88.8% metal, respectively. Establish the law of multiple proportions with the help of this data.
Solution
In the black oxide,
79.9 g copper combines with (100 – 79.9) i.e. 20.1 g oxygen
In the red oxide,
88.8 g copper combines with (100 – 88.8) i.e. 11.2 g oxygen
 79.9 g copper will combine with
11.2 79.9
88.8

= 10.08 g oxygen

Thus, the weights of oxygen that combine with the same, viz 79.9 g copper at 20.1 gand 10.08 g respectively.
These are in the ratio 20.1 : 10.08 = 2 : 1
It is a simple whole number ratio. Hence, the law of multiple proportional is established.
Try yourself
1. Explain the law of conservation of mass by giving a suitable example.
2. What do you mean by law of constant composition? Explain
3. Define law of multiple proportion.
4. When 4.2 g of NaHCO3 is added to a solution of CH3COOH (acetic acid) weighing 10 g, it is observed
that 2.2 g of CO2 is released to the atmosphere. The residue left is found to weigh 12 g. Show that these
observations are in agreement with the law of conservation of mass.
5. What weight of NaCl would react with 6.8 g of AgNO3, if 5.74 g of AgCl and 3.4 g of NaNO3 are
produced. Is the law of conservation of mass true?
3.3 DALTON’S ATOMIC THEORY
3.3.1 The major postulates of Dalton’s atomic theory are
(i) All forms of matter are made up of very small particles called atoms.
(ii) Atoms cannot be created, divided or destroyed as a result of a chemical change.
(iii) All atoms of an element are identical, and different from those of the other elements.
(iv) Atoms of elements combine in the ratio of whole numbers to produce a large number of compound
(v) Atoms of a particular substance are identical in all properties, and differ from those of other substances.
According to Dalton’s atomic theory, an atom is the smallest, discrete and indivisible particle of matter.
3.3.2 Drawbacks or Limitations of the Theory
1. An atom consists of sub-atomic particles. These are electrons, protons and neutrons.
2. Atoms of the same element may have different masses (isotopes) and similarly atoms of different
elements can have same mass (isobars).
3. Moreover, atoms may not always combine in simple whole number ratios to form molecules of
compounds. One such example is of sucrose or cane-sugar which all of us daily use in one form or the
other. The molecule of sucrose is represented as C12
H22
O11
which shows that the constituting atoms are
not present in simple whole number ratio. The actual ratio is 12 : 22 : 11.
3.4 ATOMS
An atom is the smallest particle of an element that takes part in chemical reactions and maintains its
chemical identity throughout all chemical and physical changes.
Atoms of an element are different from those of any other element.
Free atoms except those of noble gases do not exist under normal conditions
Atomic size: Atoms are very small. Hydrogen atom is the smallest The radius of a hydrogen atom is
nearly 1  10–10
m or 0.1 nm.
1 nanometre = 10–9
metre
1 nm = 10–9
m
Atomic symbols: Short hand representation of the name of an element.
In some elements, the first letter of their English names represents their symbols.

Name of element Symbol Name of element Symbol
Boron B Oxygen O
Carbon C Phosphorus P
Fluorine F Sulphur S
Hydrogen H Uranium U
Iodine I Vanadium V
There are number of elements whose names begin with the same letter. Frist letter followed by some
prominent letter from their English names.
Name of element Symbol Name of
element
Symbol
Aluminium Al Lithium Li
Argon Ar Magnesium Mg
Arsenic As Manganese Mn
Barium Ba Molybdenum Mo
Beryllium Be Nickel Ni
Bismuth Bi Neon Ne
Bromine Br Strontium Sr
Calcium Ca Silicon Si
Chlorine Cl Palladium Pd
Cobalt Co Platinum Pt
Chromium Cr Zinc Zn
Some elements have both English as well as Latin/German names.For these elements, symbols include
first letter from the Latin/German names expressed as capital along with some other prominent letter.
Name of element Latin Symbol
Silver Argentum Ag
Copper Cuprum Cu
Gold Aurum Au
Iron Ferrum Fe
Mercury Hydragyrum Hg
Potassium Kalium K
Sodium Natrium Na
Lead Plumbum Pb
Antimony Stibium Sb
Tungsten (German
name)
W
The developments and researches in the field of chemistry are guided by a body known as International
Union of Pure and Applied Chemistry (IUPAC). They have approved the symbols of the elements
and these are accepted everywhere.
Atomic mass: The average mass of an atom of an element in atomic mass unit is called its atomic mass.
Atomic mass unit: The mass equal to
12
1
th of the mass of a C
12
6 atom is called one atomic mass unit.

1 atomic mass unit =
12
6
Massof a C atom
12
Absolute mass of a C
12
6 atom is 1.9924  10–23
g. Therefore,
1 atomic mass unit =
12
g
10
1.9924 –23

1 atomic mass unit = 1.66  10–24
g = 1.66  10–27
kg
Relative atomic mass (Ar
) of an element is defined as the average relative mass of an atom of the
element compared with an atom of C
12
6 taken as 12 u. Thus Relative atomic mass of an element
(Ar
) =
atom/12)
C
one
of
(Mass
element
the
of
atom
1
of
mass
Average
12
6
(1)
The relative atomic mass is denoted byAr
. The relative atomic mass is a pure number, and hence it has no
unit.
The relative atomic mass of an element indicates the number of times one atom of that element is heavier
than
12
1
th of a C
12
6 atom.
Relative atomic masses of some common elements
Name of element Symbol Exact Common
value
Hydrogen H 1.008 1.0
Carbon C 12.010 12.0
Oxygen O 15.999 16.0
Nitrogen N 14.007 14.0
Chlorine Cl 35.45 35.5
Sulphur S 32.06 32.0
Sodium Na 22.99 23.0
Silver Ag 107.87 108.0
Copper Cu 63.54 63.5
Gram Atomic Mass: The gram atomic mass of an element may be defined as:
The atomic mass of an element expressed in grams which is numerically equal to the mass in ‘u’.
For example,
Atomic mass of nitrogen (N) = 14 u
Gram atomic mass of nitrogen (N) = 14 g
Similarly,
Atomic mass of sulphur (S) = 32 u
Gram atomic mass of sulphur (S) = 32 g

3.5 MOLECULE
A molecule is the smallest particle of an element or of a compound which can exist freely under ordinary
conditions, and shows all the properties of that substance (element or compound).
3.5.1 Types of Molecules
Molecules are of two types. These are:
(i) Molecules of elements (ii) Molecules of compounds
Let us briefly study these two types of molecules.
3.5.2 Molecules of Elements
Molecules of element are formed by the combination of two or more atoms of the same element. The
number of the atoms present in the molecules represent its atomicity. For example:
(i) Amolecule of hydrogen is made from two atoms of hydrogen. Its atomicity is two and is represented
as H2
.
(ii)Amolecule of oxygen is also made from two atoms of oxygen. Its atomicity is two and is represented
as O2
.
(iii)Amolecule of ozone is made from three atoms of oxygen. Its atomicity is three and is represented as
O3
.
(iv) A molecule of phosphours is made from four atoms of phosphorus. Its atomicity is four and is
represented as P4
.
3.5.3 Difference between Atoms and Molecules
The main points of distinction between the atoms and molecules are given:
(i) The atom is the smallest portion of an element while molecule is the smallest portion of a compound.
(ii)Atoms in general do not exist independently but molecules can exist independently.
Empirical formula of a substance is the simplest formula which gives the lowest whole -number ratio
between the number of atoms of different elements present in that substance.
For example,
(i) Crystalline (or solid) sodium chloride is a three -dimensional structure containing sodium (Na+
) and
chloride (Cl–
) ions. So, sodium chloride may be represented by a formula (Na+
Cl–
)w
where n is a large
number. The actual number of Na+
and Cl–
ions in a sample of sodium chloride depends upon the size of
the sample. But, in all samples, small or big, the ratio between the number of Na+
ions and Cl–
ions is
always 1 : 1. So, the simplest formula for sodiumchloride is Na+
Cl–
, or only NaCl. Threrfore, the empirical
formula (or stoichiometric formula) of sodium chloride is NaCl.
(ii) A covalent compound has a molecular formula C6
H6
, i.e., there are six carbon atoms and six hydrogen
atoms in its molecule. The lowest whole number ratio between the number of carbon and hydrogen atoms
is 1 : 1 (6 : 6 can be simplified to 1 : 1). Therefore, the empirical formula of a compound having molecular
formula of C6
H6
is CH.
Gases like hydrogen, nitrogen, oxygen chlorine form diatomic molecules. Thus, the molecules of hydrogen
(H2
), nitrogen (N2
), oxygen (O2
), chlorine (Cl2
) are diatomic molecule. That is, the atomicity of hydrogen,
nitrogen, oxygen, chlorine etc. is two.

3.5.4 Metallic elements (or metal)
In metals, atoms of the element are packed together. Thus, atoms of metallic elements form large
aggregates. There is no discrete molecule formed in such cases also. So, the metallic elements are also
described by their atomic symbol (or empirical formula).
3.6 ION
The charged species obatined when an atom loses or gains electrons is called an ion.
3.6. 1How is an ion formed
When an atom gains or loses one or more electrons, an ion is formed. Positively charged ion is called
cation, and the negatively charged ion is called anion. Thus,
 A cation (positively charged ion) is formed when an atom loses one or more electrons.
 An anion (negatively charged ion) is formed when an atom gains one or more electrons.
For examples,
(i) When sodium atom(Na) loses an electron, sodium ion (Na+
) is formed. Sodiumion (Na+
) is positively
charged. The Na+
ion is a cation.
atom
Sodium
Na(g) – electron
e–
 ion
Sodium
(g)
Na
* When chlorine atom(Cl) gains one electron, a chloride ion (Cl–
) is formed. Chloride ion (Cl–
) is negatively
charged. The Cl–
is an anion.
atom
Chlorine
Cl(g) + electron
e–
 on
i
Chlorine
–
(g)
Cl
3.6.2 Formation of sodium chloride from sodium and chlorine
When sodium metal reacts with chlorine gas (a nonmetal), an ionic compound sodium chloride is formed.
During this reaction, a sodium atom loses one electron from its outermost shell, and a chlorine atom gains
one electron. Actually, in this reaction, one electron is transferred from sodium atom to chlorine atom.
Polyatomic Ions : A group of atoms carrying net electrical charge on it is called polyatomic ions.
Examples : Sulphate ion –
2
4
SO
Nitrate ion –
3
NO

Carbonate ion –
2
3
CO
Chromate ion –
2
4
CrO
Dichromate ion –
2
7
2O
Cr
3.6.3 Valency of positive ions
Positive ions may be monovalent, divalent, trivalent, tetravalent etc. depending upon the charge present on
them. These are listed in the following table.
Monovalent Divalent Trivalent
Hydrogen H+
Barium Ba2+
Aluminium Al3+
Potassium K+
Calcium Ca2+
Chromium Cr3+
Sodium Na+
Magnesium Mg2+
Iron Fe3+
3.6.4 Valency of negative ions
Like positive ions, negative ions may also be monovalent, divalent, trivalent, tetravalent etc. in nature.
These are also listed:
Monovalent Divalent Trivalent
Chloride Cl–
Sulphide S2–
Nitride N3–
Bromide Br–
Oxide O2–
Phosphide P3–
Iodide I–
Carbonate (CO3
)2– Phosphate (PO4
)3–
Hydroxide (OH)–
Sulphate (SO4
)2– Borate (BO3
)3–
Illustration 5
Bring out clearly the difference between 4Cl, Cl2 and 2Cl2.
Solution
4 Cl means 4 atoms of chlorine
Cl2 means 1 molecule of chlorine
2Cl2 means 2 molecules of chlorine
Illustration 6
Classify the following into diatomic, triatomic, tetraatomic, polyatomic molecules
(i) caustic soda (ii) marble (iii) lime water (iv) NH3
(v) PCl3 (vi) H2O2 (vii) CH4 (viii) Cl2
(ix) NH4Cl (x) hydrochloric acid
Solution
(i) Casutic soda  NaOH  Triatomic
(ii) marble  CaCO3  Polyatomic
(iii) Lime water  Ca(OH)2  Polyatomic
(iv) NH3  Tetraatomic
(v) PCl3  Tetraatomic
(vi) H2O2  Tetraatomic

(vii) CH4  Polyatomic
(viii) Cl2  Diatomic
(ix) NH4Cl  Polyatomic
(x) Hydrochloric acid  HCl  Diatomic acid
Illustration 7
Define the term atomicity by giving proper example.
Solution
Atomicity is defined as total number of atoms present in 1 molecule of a substance.
For example, in H2SO4 total number of atom is 2 + 1 + 4 = 7. Hence the atomicity of H2SO4 is 7.
Try yourself
6. An element is made up of ________ atoms
7. ________ and ________ are called noble elements.
8. Write the symbols of the following elements
(i) Calcium (ii) Potassium (iii) Magnesium (iv) Mercury
(v) Lead (vi) Iron
9. Write names of the symbols given below
(i) Zn (ii) Na (iii) Al (iv) Pb
(v) Cu (vi) Ag
10. Give one example of each of the following elements
(i) Divalent (ii) Tetravalent (iii) Trivalent (iv) Monovalent
3.7 WRITING FORMULAE OF MOLECULAR COMPOUNDS
3.7.1 Chemical Formula
Each chemical compound is known by a specific name. Writing the full name of a compound repeatedly
is time-consuming and inconvenient. Therefore, in chemistry each substance is denoted by its chemical
formula.
There are two types of chemical formulae. These are:
(i) Molecular formula :Ashorthand notation of a molecule in terms of symbols of the various elements
present in it is called molecular formula.
(ii) Empirical formula (or stoichiometric formula) : Empirical formula of a substance is the simplest
formula which gives the lowest whole - number ratio between the number of atoms of different elements
present in that substance.
Ex.:- molecular formula of glucose is C6
H12
O6
but emperical formula is CH2
O.
3.7.2 How is the chemical formula of a molecular compound written
The chemical formula of a molecular compound is written as follows:
Step1:Write the symbols of the constituent elements side by side, in such a way that the less electronegative
element is on the left, and the more electronegative element is on the right.

Step 2: Write their valency numbers over the symbols and criss-cross the valency numbers to write as
subscripts to the symbols.
Step 3: Divide these numbers (written as subscripts) by a common factor, if needed.
This method of writing the chemical formula of a molecular compound is illustrated through the following
examples:
3.7.3 Water
Water contains hydrogen and oxygen. Oxygen is more electronegative than hydrogen. So,
Step 1: Write the symbols of hydrogen and oxygen side-by-side with symbol of oxygen on the right.
H O
Step 2: Write the valencies of hydrogen and oxygen in water.
H1+
O2–
Step 3: Criss-cross the valency numbers.
H
+1
O
2–
H O
2
So the molecular formula of water is H2
O.
3.7.4 Hydrogen chloride
Hydrogen chloride contains hydrogen and chlorine. Chlorine is more electronegative than hydrogen. Then
following the steps, one can write
HCl  H1+
Cl1-
 H1+
Cl1-
 H1
Cl1
 HCl
So, the molecular formula of hydrogen chloride is HCl.
3.7.5 Hydrogen Sulphide
Hydrogen sulphide contains hydrogen and sulphur. Sulphur is more electronegative than hydrogen. Then
by following the rules given above,
Step 1: H S
Step 2:
H+1
S2–
H S
2 1
Step 3: H2
S1
or H2
S
3.7.6 Carbon chloride
Carbon tetrachloride contains carbon and chlorine. Chlorine is more electronegative
than carbon. So,
Step 1: C Cl
Step 2:
C4+
C1–
C Cl
1 4
Step 3: C1
Cl4
or CCl4

It is not correct assign a molecular formula to an ionic compund. Instead, an ionic compound is described
by a formula which describes a simple ratio of the elements present in it.
The formula which describes the simplest atomic ratio of the elements present in a compound is called its
empirical formula, or stoichiometric formula.
For example, Sodium is an ionic compound in which sodium and chlorine are present in 1:1 ratio. So, the
stoichiometric formula of sodium chloride is Na+
Cl-
(or NaCl).
To write the stoichiometric formula of an ionic compound, proceed as follows:
(i) Write the symbol of the cation showing the charge number on it.
(ii) Write the symbol of the anion showing the charge number on it, on the right hand side of the cation.
(iii) Now, write the charge number (valency) of the cation at the bottom right of the anion, and the charge
number of the anion at the bottom of the cation.
Thus, The symbol of cation is subscripted with the charge number of the anion, and the anion is subscripted
with the charge number of the cation. This is called the criss-crossing of valencies.
(iv) If a compound contains polyatomic ions, then the formula of the ion is enclosed within brackets before
criss-crossing the valencies.
(v) If these subscripts are 1, these are not written in the final stoichiometric formulae. Otherwise these
subscripts are reduced to the lowest possible integers by dividing each by the highest common factor.
This method of writing the chemical formulae for ionic compounds is illustrated below.
(i) Aluminium sulphate.Aluminium sulphate containsAl3+
and S 
2
4
O ions. The formula for aluminium
sulphate can be obtained as follows:
Step 1: The symbols of the two ions are written side by side.
Al3+
S 
2
4
O
Because the negative ion (i.e., S 
2
4
O ) contains more than one atom, it is enclosed in brackets as shown
below.
Al3+
(SO4
)2-
Step 2: Criss-crossing of the valency numbers gives,
Al3+
(SO4
)2–
 Al2
(SO4
)3
Step 3: There is no common factor between 2 and 3. Therfore, the formula of aluminium
sulphate is Al2
(SO4
)3
.
(ii) Sodium bicarbonates. Sodium bicarbonate (or sodium hydrogencarbonate) contains Na+
and HC 
3
O
ions. The formula of sodium bicarbonate is obtained as follows:
Step 1: Writing the two ions side by side, viz.,
Na1+
HC 
1
3
O or Na1+
(HCO3
)1-
Step 2: Criss-crossing of the valency numbers gives
Na1+
(HCO3
)1-
 Na1
(HCO3
)1
 NaHCO3
Step 3: The formula of sodium bicarbonate is NaHCO3
.

The chemical formulae of some typical compounds
Positive ion (cation) Positive ion (anion) Chemical formula
Name of the
compound Name Formula Valency
No.
Name Formula Valency
No.
Hydrogen
chloride
Hydrogen H 1 Chloride Cl 1 H1

  H1Cl1 HCl
Hydrogen
sulphide
Hydrogen H 1 Sulphide S 2 H1

 S2
H2S1 H2S
Sulphuric acid
(hydrogen
sulphate)
Hydrogen H 1 Sulphate SO4 2 H1

 (SO4)2
H2(SO4)1
H2SO4
Sodium nitrate Sodium Na 1 Nitrate NO3 1 Na1

 NO3 Na1(NO3)1
NaNO3
Aluminium
phosphate
Aluminium Al 3 Phosphate PO4 3 Al3

 (PO4)3
Al3(PO4)3
AlPO4
Aluminium
sulphate
Aluminium Al 3 Sulphate SO4 2 Al3

 (SO4)2
Al2(SO4)3
Ferrous
sulphate
Ferrous Fe 2 Sulphate SO4 2 Fe2

 (SO4)2
Fe2(SO4)2
FeSO4
Ferric sulphate Ferric Fe 3 Sulphate SO4 2 Fe1

 (SO4)2
Fe2(SO4)3
Potassium
dichromate
Potassium K 1 Dichromate Cr2O7 2 K1

 (Cr2O7)2
K2(Cr2O7)1 K2Cr2O7
Sodiumchloride Potassium bromide Zinc oxide
1+ 1–
Cl
Na
NaCl
Hydrogen chloride Barium oxide Aluminiumoxide
Illustration 8
Write down the formulae of (i) aluminium hydroxide, (ii) hydrogen suphide, (iii) ammonium sulphate,
(iv) sodium carbonate, (v) calcium phosphate, (vi) potassium chromate.
Solution
(i)Almuniumhydroxide (ii) Hydrogen suphide (iii)Ammonium sulphate
1+ –2
(SO )
4
(NH )
4
(NH ) SO
4 2 4

(iv) Sodium carbonate (v) Calcium phosphate (vi) Poatssium chromate
Illustration 9
Calculate the empirical formula of a compound contains 27.3% C and 72.7% O by mass.
Solution
To calculate the empirical formula prepare table as shown below and then calculate according to the
above steps.
S. No. Element At. Mass % age % At. Weight Simple ratio
1 C 12 27.3 27.3
2.27
12

2.27
1
2.27

2 O 16 72.7 72.7
4.54
16

4.54
2
2.27

So the empirical formula of the compound is CO2
Illustration 10
Find the empirical formula of a compound having the following percentage composition.
(K = 31.84%, Cl = 28.98%, O = 39.18)
Solution
The calculations are as given below
1 K 31.84 39 31.84
0.81
39

0.81
1
0.81

2 Cl 29.98 35.5 28.98
0.81
35.5

0.81
1
0.81

3 O 39.18 16 39.18
2.44
16

2.44
3
0.81

So the empirical formula of the compound is KClO3
Try yourself
11 The empirical formula of a compound is CH2O, its molecular mass is 60. What will be its molecular
formula?
12. The percentage composition of an organic compound is as follows
C = 10.08%, H = 0.84%, Cl = 89.10%
If its vapour density is 60. Calculate its molecular formula.

3.7.7 Homo and Heteroatomic molecules
(a) Homoatomic molecules: These are molecules of the element. They are made of atoms of same
element. They may be further classified as monoatomic, diatomic and polyatomic molecules depending
upon whether they contain one, two or more than two atoms respectively. He, O2
, O3
, S8
, P4
are some
common examples.
(b) Heteroatomic molecules: These are the molecules of compound. They are made of atoms of different
elements. They may be further classified into diatomic and polyatomic molecules. Some examples are:
CO, BeF2
, NH3
, H2
SO4
, C12
H22
O11
etc.
3.8 ATOMIC, MOLECULAR AND EQUIVALENT MASSES
3.8.1 Atomic masses
An atom of an element is so small that its mass cannot be determined even with the help of most sensitive
balance. At the same time gram is too big a unit for expression the mass of the atom. This difficulty was
overcome by introducing the concept of relative masses.
Initially H atom, being the lightest, was chosen as standard. The number of times the atom of the element
is heavier than H atom was taken its atomic mass. Accordingly, the atomic mass could be expressed as
Mass of atom of theelement
Atomic mass=
Mass of H atom
In 1961 the most common, isotope of carbon (C-12 isotope) was chosen as standard and its mass was
fixed as 12 units. This scale in which relative atomic masses of different atoms are expressed is called
atomic mass unit scale or amu scale. The atomic mass unit is defined as one-twelfth of the actual mass of
an atom of carbon (carbon-12 isotope)
Now, the number of times the atom of an element is heavier than 12th part of C-12 isotope is called atomic
mass of the element. This can be put as
Mass of atom of theelement
Atomic mass=
1
mass of (C-12) atom
12

Atomic mass of a few common elements refered to 12
C = 12
Element Symbol Atomic mass Element Symbol Atomic mass
Aluminium Al 27 Lead Pb 207.2
Argon Ar 18 Lithium Li 6.94
Beryllium Be 9.01 Magnesium Mg 24.3
Boron B 10.8 Manganese Mn 54.93
Barium Ba 137.33 Neon Ne 20.2
Calcium Ca 40.1 Nitrogen N 14
Carbon C 12 Oxygen O 16
Chlorine Cl 35.5 Phosphorus P 31
Copper Cu 63.5 Potassium K 39.1
Gram-Atomic Mass (GAM)
It may be defined as that much quantity of the element whose mass in gram is numerically equal to its
atomic mass. Gram atomic mass is also called one gram-atom of the element. For example, atomic mass
of magnesium (Mg) is 24 amu, therefore,
1 gram atoms of Mg = gram atomic mass of Mg = 24 g
2 gram atoms of Mg = 2 × gram atomic mass of Mg = (2 × 24) g

From the above discussion, we can get the relationship between masses of the element and its grams-
atoms is
Mass of element ingram
Number of Gm-atoms of Element=
GAM of element
3.8.2 Molecular Mass
Molecular mass of a substance (element or compound) is defined as the average relative mass of the
molecule of the substance as compared with mass of an atom of carbon (C12
) taken as 12 amu.
It can be put as
Average relative mass of one molecule
Molecular mass =
1
×Mass of C-12atom
12
Gram Molecular Mass (GMM)
It is defined as that much quantity of the substance (element or compound) whose mass in grams is
numerically equal to its molecular mass. Gram molecular mass is also called one gram molecule of the
substance. For example, molecular mass of ammonia is 17 amu, therefore,
1 gram atoms of NH3
= gram molecular mass of NH3
= 17 g
2 gram atoms of NH3
= 2 × gram molecular mass of NH3
= (2 × 17) g
From the above discussion it follows that
Mass of substance ingrams
Number of Gm-atoms of Element=
GAM of the substance
3.8.3 Equivalent Mass
The number of parts by mass of the substance which combines or displaces directly or indirectly
1.008 parts by mass of hydrogen, or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine. For
example,
In the reaction, Mg + 2HCl  MgCl2
+ H2
2 parts by mass of hydrogen is displaced from Mg = 24 parts
1 part by mass of hydrogen is displaced from Mg = 12 parts
Thus, equivalent mass of Mg is 12.
3.8.4 Gram Equivalent Mass (GEM)
The quantity of a substance whose mass in grams is numerically equal to its equivalent mass is called its
one-gram equivalent or gram equivalent mass.
For example, eq. mass of oxygen = 8
 1 g equivalent of oxygen = GEM of oxygen = 8 g
4 g equivalent of oxygen = 4 × GEM of oxygen = 4 × 8 = 32 g
Mass in grams
Number of gm-equivalents =
GEM of the substance
3.9 FORMULA MASS
The sum of the atomic masses of all atoms in a formula unit of a substance is called the formula mass of
the substance.
The concept of formula mass is applicable only to ionic substances.

3.9.1 How is the formula mass of an ionic substance calculated
The formula mass of any ionic compound is expressed as the sum of the atomic masses of all the atoms
present in its formula. for example, NaCl,
Formula mass of NaCl = (1 x Atomic mass of Na) + (1 x Atomic mass of Cl)
We know that
Atomic mass of Na = 23 u
and Atomic mass of Cl = 35.5 u
So, Formula mass of NaCl = (1 x 23 u) + (1 x 35.5 u)
= (23 + 35.5) u = 58.5 u
Thus, the molecular mass (or empirical formula mass) of a substance is equal to the sum of the atomic
masses of all the elements present in its molecular (or empirical) formula, taking into account the number
of atoms of each element present in its molecular (or empirical) formula.
For example,
Formula mass of CO2
Formula mass of Al2
O3
1 C = 1 × 12 u = 12 u 2 Al = 2 × 27 u = 54 u
2 O = 2 × 16 u = 32 u 3 O = 3 × 16 u = 48 u
Molecular mass of CO 2
= 44 u Molecular mass of Al2
O3
= 102 u
Illustration 11
Calculate the molar mass of the following substances:
(i) Ethane (C2
H6
) (ii) Ethyne (C2
H2
)
(iii) Sulphur molecule (S8
) (iv) Phosphorus molecule (P4
)
Solution
(i) Ethane (C2
H6
) :
Molar mass of C2
H6
= 2  Atomic mass of C + 6  Atomic mass of H
= (2  12u) + (6  1u) = 30u.
(ii) Ethyne (C2
H2
) :
Molar mass of C2
H2
= 2  Atomic mass of C + 2  Atomic mass of H
= (2  12u) + (2  1u) = 26u.
(iii) Sulphur molecules (S8
) :
Molar mass of S8
= 8  Atomic mass S
= (8  32u) = 256u.
(iv) Phosphorus molecules (P4
) :
Molar mass of P4
= 4  Atomic mass of P
= (4  31u) = 124 u.
Illustration 12
Calculate the mass of 0.72 gram molecule of carbon dioxide (CO2
).
Solution
Molecular mass of CO2
= Atomic mass of C + 2  Atomic mass of O
= (12u + 2  16u) = 44u
Gram molecular mass of CO2
= 44g

1 gram molecule of CO2
= 44 g
0.72 gram molecule of CO2
=
 
 

1g
44g
(0.72g) = 31.68 g
3.9.2 Mass of one mole of any material is called its molar mass
Molar mass of hydrogen (H2
) = Mass of 6.023 x 1023
molecules of hydrogen
One mole of any gaseous substance at 273 K and 1 atmosphere pressure occupies a volume equal to
22.4 L or 22400 mL. The volume occupied by one mole of any gaseous substance is called molar volume
(Vm
). So,
Molar volume of any gaseous substance at 273 K and 1 atm pressure = 22.4 L mol–1
= 22400 mL mol–1
3.10 COMPOSITION OF COMPOUND
The composition of a compound can be described by the mass percentage of each element present in it.
To determined the mass percentage of an element in a compound
Let us suppose, W g of a compound contain w g of an element A.
Mass percentage of the element,
A =
compound
the
of
mass
Total
compound
the
of
mass
given
in the
A
element
of
Mass
× 100 =
g
W
w
× 100 =
W
w
× 100.
3.11 MOLE CONCEPT
3.11.1 Important
 1 mole = 6.023  1023
particles
 1 mole atoms = 6.023  1023
atoms
 One mole molecules = 6.023  1023
molecules
 Mass of one mole of atoms = Gram atomic mass
 Mass of one mole of molecules = Gram molecular mass
 Moles of a compound = Mass of compound
 Volume occupied by 1 mole of a gas at N.T.P. = 22.4 litres.
Thus, we conclude that
Gram atomic mass of an element contains 6.023  1023
atoms. Similarly the gram molecular mass of a
compound also contains the same number of molecules.

It may not to be possible to count the atoms individually. However, they can collectively be represented
as one mole. This is very convenient method to represent different particles. For example,
No. of oxygen atoms in 3 moles = (3  NO
) = 3 6.023  1023
= 1.81  1024
atoms.
Similarly, the number of carbon and oxygen atoms in 1 mole of carbon dioxide (CO2
) may be calculated
as
= Avogadron’s no. of carbon atoms (NO
) + 2  Avogadro’s number of oxygen atoms (NO
)
= (6.023  1023
+ 2  6.023  1023
) atoms
= 1.81  1024
atoms.
3.11.2 Expressing number of particles in terms of Moles
Number of particles of any substance (element or compound) are related to the number of moles by
the relation.
Number of moles =
)
(N
particles
of
no.
s
Avogadro'
(N)
particles
of
no.
Given
O
3.11.3 Expressing mass of a substance in terms of Moles
We have studied that the elements have atomic mass or gram
atomic mass. Similarly, the compounds have molecular mass or
gram molecular mass. Both can be expressed in terms of moles of as follows.
For elements : The no. of moles =
(M)
mass
atomic
gram
)
(
mass
given m
For compounds : The no. of moles =
(M)
mass
molar
gram
)
(
mass
given m
Illustration 13
Calculate the number of moles in the following (i) 52 g of He (ii) 46 g of Na. (iii) 60 g of Ca.
Given gram atomic mass of (i) He = 4 g (ii) Na = 23 g (iii) Ca = 40 g.
Solution
(i) 52 g of He
The no. of moles =
M
mass
atomic
Gram
grams
in
He
of
Mass m
 =
(4g)
(52g)
= 13 mol
(ii) 46 g of Na
The no. of moles =
M
mass
atomic
Gram
grams
in
Na
of
Mass m
 =
(23g)
(46g)
= 2 mol

(iii) 60 g of Ca
The no. of moles =
M
mass
atomic
Gram
grams
in
Ca
of
Mass m
 =
(40g)
(60g)
= 1.5 mol
Illustration 14
Calculate the mass of the following :
(i) 0.5 mole of O2
gas (ii) 0.5 mole of O atoms
(iii) 3.011  1023
atoms of O (iv) 6.023  1023
molecules of O2
(Given : Gram atomic mass of oxygen = 16g, Gram molecular mass of oxygen (O2
) = 32 g)
Solution
(i) 0.5 mole of O2
gas
No. of moles =
M
mass
molecular
Gram
grams
in
O
of
Mass 2 m

 Mass of O2
in grams (m) = No. of moles  M
= 0.5  (32 g) = 16 g
(ii) 0.5 mole of oxygen (O) atoms
No. of moles =
M
mass
atomic
Gram
grams
in
(O)
oxygen
of
Mass m

Mass of oxygen (O) in grams (m) = No. of moles  M
= 0.5  (16g) = 8 g.
(iii) 3.011  1023
atoms of oxygen (O)
Step I: Calculate of no. of gram atoms of oxygen
No. of gram atoms =
O
N
N
atoms
of
no.
s
Avogadro'
oxygen
of
atoms
of
No.

= 23
23
10
023
.
6
10
011
.
3


= 0.5 gram atom
Step II: Calculate of mas of oxygen (O) atoms
Mass of oxygen (O) atoms
= Gram atomic mass of oxygen  No. of gram atoms of oxygen
= 16 × 0.5 = 8 g.
(iv) 6.023  1023 molecules of oxygen (O2
)
Step I : Calculate of mass of oxygen.
No. of gram moles =
atoms
of
no.
s
Avogadro'
oxygen
of
molecules
of
No.
=
O
N
N
= 23
23
10
023
.
6
10
023
.
6


= 1 gram mol.

Step II : Calculate of mass of oxygen (O2
) molecules.
Mass of oxygen (O2
) molecules
= Gram molecular mass of oxygen  No.of gram moles of oxygen.
= 32 × 1 = 32 g
Try yourself
13. How many moles of calcium carbonate are present in 10 g of the substance?
(Ca = 40, C = 12, O = 16)
14. Calculate the mass of 0.2 mole of nitrogen dioxide. (NO2) (N = 14, O = 16)
15. How many molecules are there in 2 g MgO? (Mg = 24, O = 16)
16. Which contains more molecules, 4 g of methane (CH4) or 4 g of oxygen (O2)?
(C = 12, O = 16, H = 1)
17. Calculate the ratio of molecules present in 4.4 g of CO2 and 1.6 g of SO2.
(C = 12, O = 16, S = 32)
18. Calculate the mass of 1 molecule of sulphur dioxide (SO2) (S = 32, O = 16)
19. Calculate the number of atoms in
(i) 18 g of Glucose (C6H12O6) (ii) 0.20 mole molecule of oxygen
3.12 AVOGADRO’S LAW
(Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of
molecules)
On applying Avogadro’s law to the reaction between hydrogen and chlorine.
Hydrogen (1–Vol) + Chlorine (1 Vol)  Hydrochloric gas (2 Vol)
ApplyingAvogadro’s law
n molecules of hydrogen + n molecules of chlorine  2n molecules of hydrochloric gas
1 molecule + 1 molecule  2 molecules
½ molecule + ½ molecule  1 molecule
It means that one molecule of hydrochloric gas contains ½ molecule of hydrogen and ½ molecule of
chlorine. Since a molecule may contain more than one atoms. It is possible to divide it into atoms.
Application of Avogadro law
Avogadro’s law is also helpful in developing the relationship between
(a) Molecular mass and vapour density: By definition
2
Density of gas Mass of some volumeof gasat STP
Vapour density (V.D.)= =
Density of hydrogen Mass of same volume of H STP
2 2
Mass of N molecules of gas Mass of1 gasmolecule
= =
Mass of N molecules of H Mass of 1 H molecule
Mass of 1 molecule of gas 1
= = Molecular mass of gas
2×Mass of Hatom 2

Thus, Molecular mass = 2 × V. D.

(b) Mass and volume of gas:
2
2×Mass of 1 L of gas at STP
Molecular mass=2×V.D.
Mass of 1 L of H at STP
2×Mass of 1 L of gas at STP
=
0.089g = 22.4 × Mass of 1 L of gas at STP
= Mass of 22.4 L of gas at STP
Thus, Mass of 22.4 L of gas at NTP = Molecular mass (in g mol–1
)
3.13 LAW OF MULTIPLE PROPORTIONS
This law was discovered by John Dalton (1803). This law states that
When two elements combine with each other to form two or more than two compounds, the masses of
one of the elements which combine with fixed mass of the other, bear a simple whole number ratio to one
another.
For example, carbon and oxygen combine with each other to form carbon monoxide (CO) and carbon
dioxide (CO2
)
In carbon monoxide (CO): 12 parts by mass of carbon combine with 16 parts by mass of oxygen
In carbon dioxide (CO2
): 12 parts by mass of carbon combine with 32 parts by mass of oxygen.
Ratio of the masses of the oxygen which combine with fixed mass of carbon (12 parts) in these compounds
is 16 : 32 or 1 : 2, which is a simple whole number ratio.
Similarly, copper and oxygen combine to form two oxides, the red cuprous oxide (Cu2
O) and the black
cupric oxide (CuO).
In red oxide (Cu2
O): 16 parts by mass of oxygen combine with 63.5 × 2 parts by mass of copper
In black oxide (CuO): 16 parts by mass of oxygen combine with 63.5 parts by mass of copper
Illustration 15
Hydrogen and oxygen are known to form two compounds. The hydrogen content in one of these
5.93% while in the other it is 11.2%. Show that this data illustrates the law of multiple proportions
Solution
In the first compound,
Hydrogen = 5.93%
Oxygen = (100 – 5.93) = 94.07%
In the second compound,
Hydrogen = 11.2%
Oxygen = (100 – 11.2) = 88.8%
In the first compound the number of parts by mass of oxygen that combine with one part by mass of
hydrogen =
94.07
5.93
= 15.86 parts
In the second compound the number of parts by mass of oxygen that combine with one part by mass of
hydrogen =
88.8
11.2
= 7.9 parts
The ratio of masses of oxygen that combine with fixed mass (1 part) by mass of hydrogen is 15.86 : 7.9
or 2 : 1

Since the ratio is a simple whole number ratio, hence the given data illustrates the law of multiple propor-
tions.
3.14 EXPRESSION CONCENTRATION OF SOLUTIONS
3.14.1 Normality
Normality of a solution is defined as the number of gram equivalents of the solute dissolved per litre of the
solution. It is represented by N.
Mathematically,
Gram-equivalentsof solute
Normality(N)=
Volume of solution in litres
Mass of solute in grams
=
(Gm eq.mass of solute)×(volume of solution(L))
or
B B
B (L) B ml
W (g) W (g)×1000
N= or
GEM ×V GEM ×V
A solution having normality equal to one is called a normal solution. Such a solution contains one gram
equivalent of solute per litre of solution.Adeci normal solution contains 1/10 g equivalents of solute per
litre of solution. Similarly, a semi-normal solution contains 1/2 g equivalents and a centinormal solution
contains 1/100 g equivalents per litre of solution.
3.14.2 Molarity
Molarity of a solution is defined as the number of gram moles of the solute dissolved per litre of the
solution. It is represented by M.
Gram-moles of solute Mass of solute ingrams
Molarity(M)= =
Volume of solution in litres (GMM of solute) ×Volume of soln(L)
or
B B
B (L) B ml
W (g) W (g)×1000
M= or
GMM ×V GMM ×V
A solution havingmolarity one is called molar solution. Such a solution contains one mole of solute per litre
of solution. Molarity is expressed in units of moles per litre or moles per dm3
. It may be noted that both
normality as well as molarity of a solution change with change in temperature.
3.14.4 Molality
Molality of a solution may be defined as the number of gram moles of the solute dissolved in 1000 g (1 kg)
of the solvent. It is represented by m.
Grammolesof solute
Normality(M)=
Mass of solvent(kg)
Mass of solute (g)
=
(GMM of solute)×(Mass of solvent(kg))
or
B B
B A B
W (g) W (g)×1000
M= or
GMM ×W (kg) GMM ×W ( )
A g

A solution containing one mole of solute per 1000 g of solvent has molality equal to one and is called a
molal solution. Molality is expressed in units of moles per kilogram (mol kg–1
). The molality of a solution
does not change with temperature.
Illustration 16
Calculate the molarity and normality of a solution containing 40 g of NaOH dissolved in two litres
of the solution.
Solution
Step I: To calculate the mass of NaOH present in 1 litre of the solution.
Amount of NaOH present in 2 litres of the solution = 40 g
 Amount of NaOH present in 1 litre of the solution = 40/2 = 20 g
Step II: To calculate the normality and molarity of the solution
Molecular mass of NaOH = 40
Equivalent mass of NaOH = 40/1 = 40
Mass of solute in grams per litre 20
Normality= = =0.5N
Equivalent massof the solute 40
Mass of solute in grams per litre 20
Molarity= = =0.5M
Molecular massof the solute 40
Solved Examples
Example 1
Three oxides of lead on analysis were found to contain lead as under
(i) 3.45 g of yellow oxide contains 3.21 g of lead
(ii) 1.195 g of brown oxide contains 1.035 g of lead.
(iii) 1.77 g of red oxide contains 1.61 g lead.
Solution
The amounts of lead and oxygen in three oxides are
(i) Yellowoxide: Mass of lead = 3.21 g
Mass of oxygen = 3.45 – 3.21 = 0.24 g
(ii) Brown oxide: Mass of lead = 1.035 g
Mass of oxygen = 1.195 – 1.035 = 0.160 g
(iii) Red oxide: Mass of lead = 1.61 g
Mass of oxygen = 1.77 – 1.61 = 0.16 g
Let us fix the mass of lead as 1 g and calculate the different weights of oxygen which combine with 1 g
of lead in these oxides.
(i) Yellowoxide:
Mass of oxygen which combines with 3.21 g of lead = 0.24 g
Mass of oxygen which combines with 1 g of lead =
0.24
3.21
= 0.075 g
(ii) Brown oxide:
0.160
1.035
= 1.15 g

(iii) Red oxide:
0.16
1.61
= 0.10 g
The ratio of different masses of oxygen which combine with same mass of lead (1 g) in these oxides is
0.075 : 0.15 : 0.10
3 : 6 : 4
This is simple ratio
Hence, the data illustrate the law of multiple proportions
Example 2
Calculate the mass of 1 atom of C14
.
Solution
The atomic mass of carbon is 14 g
The mass of 6.023 × 1023
atoms of C14
= 14 g
The mass of 1 atom of C14
=
23
23
14
2.32 10
6.023 10
g

 

Example 3
Calculate the mass of an atom of silver (atomic mass = 108)
Solution
Mass of 6.023 × 1023
atoms of silver = 108 g
Mass of 1 atom of silver = 23
108
6.023 10

= 1.793 × 10–22
g
Example 4
Calculate the number of moles in the following
(i) 7.85 g of iron (ii) 4.68 mg of silicon (iii) 65.6 µg of carbon
Solution
(i) 7.85 g of iron
Atomic mass of Fe = 55.8 g
Mole of iron =
Mass of iron
Atomic mass =
7.85
0.141
55.8
 mol
(ii) 4.68 mg of silicon = 4.68 × 10–3
g of silicon
Atomic mass of silicon = 28.1
Mole of silicon =
3
Mass of silicon 4.68 10
Atomic mass 28.1


 = 1.66 × 10–4
mol
(iii) 65.6 µ g of carbon = 65.6 × 10–6
g of carbon
Atomic mass of carbon = 12
Moles of carbon =
-6
Massof carbon 65.6×10
=
Atomic mass 12
= 5.47 × 10–6
mol
Example 5
Calculate the ratio of molecules present in 16 g of methane (CH4) and 16 g of oxygen (O2)
(C = 12, H = 1, O = 16)
Solution
To calculate the ratio number of molecules, let us calculate the number of moles in both methane and
oxygen.
Molecular weight of CH4 = 12 + 4 × 1 = 12 + 4 = 16 g

 16 g of CH4 = 1 mole
Molecular weight of O2 = 2 × 16 = 32 g
 32 g of O2 = 1 mole
 1 g of O2 = 1/32 mole
 16 g of O2 = 16/32 = 0.5 mole
So, the ratio of molecules = ratio of moles = 1 : 0.5 = 10 : 5 = 2 : 1
Example 6
The mass of one molecule of an element is 4.65 × 10–23
g. What is its molecular mass? What could
be this substance?
Solution
Any substance will have 6.023 × 1023
molecules per mole.
 Mass of one molecule = 4.65 × 10–23
g
 Mass of 6.023 × 1023
molecules = 6.023 × 1023
× 4.65 × 10–23
= 28 g
Element wil be nitrogen.
Example 7
From the chemical analysis of a solid, following results were obtained
C = 40%, H = 6.67%, O = 53.33%
If the molecular weight of the solid compound is 180. Find its molecular formula (C = 12, H = 1,
O = 16)
Solution
First let us calculate the empirical formula
1 C 12 40 40
3.33
12

3.33
1
3.33

2 H 16 6.57 6.67
6.67
1

6.67
2
3.33

3 O 16 53.33 53.33
3.33
16

3.33
1
3.33

Empirical formula = CH2O
Empirical formula mass = CH2O = 12 + 2 × 1 + 16 = 12 + 2 + 16 = 30
Now,
Molecular mass 180
n= = =6
Empirical formula mass 30
So, Molecular mass = (Empirical formula)n
= (CH2O)6 = C6H12O6
Example 8
The empirical formula of a compound is CH and its molecular weight is 78. What will be its
molecular formula?
Solution
Empirical formula = CH
Empirical formula mass = 12 + 1 = 13
Now,
Molecular mass 78
n= = =6
Empirical formula mass 13
So, Molecular mass = (Empirical formula)n
= (CH)6 = C6H6
*****

EXERCISE-I
1. Can matter be created?
2. Name the scientist who stated the law of conservation of mass?
3. To what types of substances is the law of constant composition applicable? State the law.
4. The name ‘parmanu’ was given by which Indian philosopher?
5. Enlist the main features of the Dalton’s atomic theory?
6. Point out three drawbacks of the Dalton’s atomic theory?
7. Do atoms always combine in small whole numbers to form molecules?
8. Give a suitable illustration to support the law of conservation of mass.
9. What is the name given to the short hand representation of an element?
10. Give the symbols of the elements (i) Bismuth, (ii) Sulphur, (iii) Magnesium, (iv) Manganese, (v) Iron
11. Which is the standard for comparing the atomic masses of different elements?
12. What does IUPAC represent?
13. The atomic mass of an element is in fraction. What does it mean?
14. Define a molecule. How does it differ from an atom?
15. Atoms are represented by symbols. What is the name given to the representation of a molecule?
16. Give the name of two monoatomic and two diatomic molecules.
17. Give the formulae of (i) calcium oxide (ii) calcium carbonate (iii) ammonia (iv) copper nitrate.
18. Give the examples of two polyatomic ions.
19. What is the chemical names of the following compounds?
(i) quick lime, (ii) baking soda, (iii) table salt
20. Name the elements present in the following compounds.
(i) Potassium sulphate, (ii) ammonia, (iii) quick lime
21. When do we use the term formula unit?
22. How do we represent formula units in terms of mass?
23. Define the term mole.
24. Avogadro’s number of particles represent how many particles.
25. Answer the following:
(i) How many atoms are present in one gram atomic mass of a substance?
(ii) How many molecules are present in one gram molecular mass of a substance?
(iii) what is the name given to 6.023  1023
.
26. What is the mass of two moles of oxygen atoms?
27. How many moles are present in:
(i) 10g of Ca (ii) 2.3g of Na
(iii) 3.012  1023
atoms of oxygen (iv) 32 g of oxygen gas
28. Calculate the number of Mg atoms in 0.024 g of Mg.
29. Give one limitation of law of constant proportion.
30. What is basic difference between atoms and molecules?
31. Why does not the atomic mass of an element represent the actual mass of its atom?

32. What is difference between the mass of a molecule and molecular mass?
33. Why is it necessary to balance a chemical equation?
34. Define atom and molecule.
35. Differentiate between an atom and a molecule.
36. Classify the following into diatomic, triatomic, tetraatomic and pentatomic molecules. N2
, CH4
, NH3
,
H2
O, PCl3
, H2
O2
, Cl2
37. Define the term “atomicity”.
38. Define atomic mass and gram atomic mass.
39. Define molecular mass and gram molecular mass.
40. What is atomic mass unit? Name the isotope used as a standard for defining atomic mass unit.
41. Explain mole and gram mole.
42. Explain molar mass and gram molecular mass.
43. Calculate the mass of one mole of benzene.
44. Calculate the mass of 0.2 mole of water.
45. Name the laws of chemical combination.
46. Define and explain law of conservation of mass.
47. Define law of definite proportion. Give two examples.
48. Give one experiment to prove law of conservation of mass.
49. What is the numerical value of Avagadro’s number?
50. Define empirical and molecular formula of a compound.
51. How are the empirical and molecular formula related? Can the empirical and molecular formula of a
compound be the same?
52. A compound contains C, H and N in the atomic ratio of 3 : 4 : 1. The molecular mass of the compound
is 108. Calculate the molecular formula of the compound.
53. Give the definition of atomic mass unit.
54. Give two examples of tetraatomic.
55. Define law of conservation of mass.
56. Define atomic mass of an element.
57. Give one example to illustrate law of definite proportions.
58. How many atoms are present in one gram atomic mass of a substance?
59. How many molecules are present in one gram molecular mass of a substance?
60. Define empirical and molecular formula of a compound.
61. Differentiate between atoms and molecules.
EXERCISE-II
1. In a reaction 5.3 g of sodium carbonate reacted with 6 gof ethanoic acid. The products were 2.2 g of
carbon dioxide, 0.9 g water and 8.2 g of sodiumethanoate. Show that these observations are in agreement
with the law of conservation of mass.
Sodiam carbonate + ethanoic acid ® Sodium ethanoate + carbon dioxide + water.
2. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas
would be required to react completely with 3 g of hydrogen gas?

3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
5. Define the atomic mass unit.
6. Why is it not possible to see an atom with naked eyes?
7. Write down the formulae of
(i) sodium oxide (ii)aluniniumchloride (iii)sodiumsulphide (iv) magnesium hydroxide
8. Write down the names of compounds represented by the folloiwng formulae:
(i) Al2
(SO4
)3
(ii) CaCl2
(iii) K2
SO4
(iv) KNO3
(v) CaCO3
9. What is meant by the term chemical formula?
10. How many atoms are present in a :
(i) H2
S molecule and (ii) –
3
4
PO ion?
11. (a) Calculate the relative molecular mass of water (H2
O).
(b) Calculate the molecular mass of HNO3
.
12. Calculate the formula unit mass of CaCl2
.
13. Calculate the molecular masses of H2
, O2
, Cl2
, CO2
, C2
H6
, C2
H4
, NH3
, CH3
OH.
14. Calculate the formula unit masses of ZnO, Na2
O, K2
CO3
, given atomic masses of Zn = 65 u.
Na = 23 u. K = 39 u. C = 12 u. and O = 16 u.
15. Calculate the number of moles for the following:
(i) 52 g of He (finding mole from mass)
(ii) 12.044  1023
number of He atoms (fiding mole from number of particles)
16. Calculate the mass of the following :
(i) 0.5 mole of N2
gas (mass from mole of molecule)
(ii) 0.5 mole of N atoms (mass from mole of atom)
(iii) 3.011  1023
number of N atoms (mass from number)
(iv) 6.022  1023
number of N2
molecules (mass from number)
17. Calculate the number of particles in each of the following :
(i) 46 g of Na atoms (number from mass)
(ii) 8 g P2
molecules (number of molecules from mass)
(iii) 0.1 mole of carbon atoms (number from given moles).
18. If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?
19. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na
= 23 u, Fe = 56 u)?
20. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron
and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
21. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of
carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of
chemical combination will govern your answer?
22. What are polyatomic ions? Give examples.
23. Write the chemical formulae of the following:
(i) Magnesium chloride (ii) Calcium oxide (iii) Copper nitrate

(iv)Aluminiumchloride (v) Calcium carbonate
24. Give the names of the elements present in the following compounds.
(i) Quick lime (ii) Hydrogen bromide (iii) Baking powder (iv) Potassium sulphate
25. Calculate the molar mass of the following substances.
(i) Ethyne, C2
H2
(ii) Sulphur molecule, S8
(iii) Phosphorus acid, P4
(Atomic mass of phosphorus = 31)
(iv) Hydrochloric acid, HCl (v) Nitric acid, HNO3
26. What is the mass of –
(i) 1 mole of nitrogen atoms?
(ii) 4 moles of aluminium atoms (Atomic mass of aluminium= 27)?
(iii) 10 moles of sodium sulphate (Na2
SO3
)?
27. Convert into mole.
(i) 12 g of oxygen gas (ii) 20 g of water (iii) 22 g of carbon dioxide
28. What is the mass of:
(i) 0.2 mole of oxygen atoms? (ii) 0.5 mole of water molecules?
29. Calculate the number of molecules of sulphur (S) present 16 g of solid sulphur?
30. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
EXERCISE-III
SECTION-A
 Fill in the blanks
1. Law of conservation of mass was enunciated by __________
2. The number of oxygen atoms in 88 g of CO2
is __________
3. 1 mole of solute present in __________ g of solvent is called a molal solution.
4. The smallest particle of matter which is capable of independent existence is called __________
5. The chemical symbol of Tungsten is __________.
6. In water, the proportion of hydrogen and oxygen is __________ by mass.
7. The symbolic expression for a molecule is called a __________.
8. The elements present in compound ammonia is __________ and __________.
9. One atomic mass unit is a mass unit equal to exact one twelfth the mass of one atom of __________
10. The number of atoms present in one molecule of an element is called its __________.
11. 0.2 mol of ozone (O3
) at NTP will occupy volume _________
12. The moles of x atoms of a triatomic gas =
o
x
N × _________.

SECTION-B
 Multiple choice question with one correct answers
1. One ‘u’ stands for:
(A) An stom of carbon (C-12) (B)
12
1
th of carbon atom (C-12)
(C)
12
1
th of hydrogen atoms (D) one atom of all the elements
2. The no. of oxygen atoms in 4.4 g of CO2
is approx.
(A) 6 x 1022
(B) 6 x 1023
(C) 12 x 1023
(D) 1.2 x 1023
3. Which has maximum number of molecules?
(A) 1 g of CO2
(B) 1 g of N2
(C) 1g of H2
(D) 1 g of CH4
4. The law of constant composition is applied to
(A) Any element (B) Any chemical compound
(C) Pure chemical compound (D) None of these.
5. What mass of carbon dioxide (CO2
) will contain 3.011 x 1023
molecules?
(A) 11.0 g (B) 22.0 g (C) 4.4 g (D) 44.0 g
6. The value of Avogadro’s constant is:
(A) 6.0 x 1024
(B) 6.01 x 1022
(C) 6.023 x 1023
(D) 6.023 x 10-23
7. The law of constant proportion was proposed by
(A) Dalton (B) Avogadro (C) Joseph Proust (D) A. Lavoisier
8. The best standard for atomic mass is
(A) carbon - 12 (B) oxygen - 16 (C) hydrogen - 1 (D) chlorine - 35
9. One amu is
(A) 1.00758 g (B) 0.000549 g (C) 1.66 x 10-24
g (D) 6.023 x 10-23
g
10. The percentage of hydrogen in water is
(A) 1.11% (B) 11.11% (C) 8.89% (D)88.9%
11. The atom is indivisible was proposed by
(A) Einstein (B) Lavoisier (C) Dalton (D) Proust
12. The percentage of calcium in CaCO3
is
(A) 40% (B) 12% (C) 48% (D) none of these
SECTION-C
 Assertion & Reason
Instructions: In the following questions as Assertion (A) is given followed by a Reason (R). Mark your
responses from the following options.
(A) Both Assertion and Reason are true and Reason is the correct explanation of ‘Assertion’
(B) Both Assertion and Reason are true and Reason is not the correct explanation of ‘Assertion’
(C) Assertion is true but Reason is false
(D) Assertion is false but Reason is true

1. Assertion: Gram atomic mass of an element contains Avogadro’s number of atoms.
Reason: amu is called atomic mass unit
2. Assertion: It is not possible to seen an atom with naked eyes.
Reason: It is very small in size.
3. Assertion: Atomic radius is expressed in ‘nanometres’
Reason: 1 nm = 10–9
m
4. Assertion: The ratio by volume of H2
: Cl2
: HCl in a reaction, H2
+ Cl2
 2HCl is 1 : 1 : 2
Reason: Substances always react in such a way that their volume ratio is simple whole number
5. Assertion: 1 g atom of iron represent the number of iron atoms present in 1 g of it
Reason: 1 g atom of element weighs equal to GAM of element.
SECTION-D
 Match the following (one to one)
Column-I and column-II contains four entries each. Entries of column-I are to be matched with some
entries of column-II. Only One entries of column-I may have the matching with the same entries of column-
II and one entry of column-II Only one matching with entries of column-I
1. Column I Column II
(A) 1 amu (P) atom can not be divided
(B) Dalton’s law (Q) weight-weight unit
(C) Avogadro’s number (R) 6.02 × 1023
(D) Molality (S) (1/12)th mass one C12
(A) Molecular formula (P) 2 × vapour density
(B) Molecular weight (Q) sodium atom
(C) 1/2 mole of Cl2
(R) (Empirical formula)n
(D) Na (S) 35.5 g
(A) Quicklime (P) HBr
(B) Baking powder (Q) K2
SO4
(C) Potassium sulphate (R) NaHCO2
(D) Hydrogen bromide (S) CaO
(A) Molar volume (P) Mercury
(B) Moles of oxygen atom/mole (Q) 14
H2
SO4
(C) Hydragyrum (R) 22.4 litre
(D) Mass of 1 atom of Nitrogen in amu (S) 4

EXERCISE-IV
SECTION-A
 Multiple choice question with one correct answers
1. A seminormal solution is
(A) 0.5 N (B) 0.1 N (C) 0.01 N (D) 0.001 N
2. One and a half mole of oxygen combines with aluminiumto formAl2
O3
, then the weight of aluminiummetal
used in this reaction is (atomic weight of Al = 27)
(A) 27 g (B) 81 g (C) 108 g (D) 54 g
3. Weight of 6.02 × 1023
atoms of oxygen would be
(A) 8 amu (B) 16 g (C) 1 g (D) 1amu
4. Number of atoms in 40 g of 40
20 Ca is
(A) NA
(B) 0.1 NA
(C) 12 NA
(D) 32 NA
(where NA
: Avogadro’s number)
5. Gram atoms and number of atoms respectively in 60 gram of carbon will be
(A) 5, 30.1 × 1023
(B) 5, 12.05 × 1023
(C) 5, 6.02 × 1023
(D) 60, 12.04 × 1023
6. Mass of one atom of an element is 2.6578 × 10–23
g. Its mass in amu would be
(A) 14 amu (B) 16 amu (C) 13 amu (D) 12 amu
7. Molar mass of sulphuric acid is
(A) 98 g (B) 89 g (C) 86 g (D) 100 g
8. Which of the following is diatomic molecule?
(A) HCl (B) H2
O (C) NH3
(D) Xe
9. The symbol for iron is
(A) I (B) Ir (C) Fe (D) L
10. The chloride of a metal has the formula ACl3
. The formula of its phosphate will be
(A) APO4
(B) A(PO4
)2
(C) A2
PO4
(D) A3
PO4
11. Argon is
(A) Monoatomic (B) Diatomic (C) Tetraatomic (D) Polyatomic
12. Radii of molecule of water is
(A) 10–10
m (B) 10–9
m (C) 10–8
m (D) 10–11
m
13. Formula of chromium sulphate is
(A) Cr2
(SO4
)3
(B) CrSO4
(C) Cr2
SO4
(D) Cr2
SO6
14. How many electrons weighs 1 kg?
(A) 6.023 × 1023
(B)
31
1
10
9.108

(C)
54
6.023
10
9.108
 

 
 
(D)
8
1
10
9.108 6.023
 

 

 

15. Which has maximum number of oxygen atoms?
(A) 10 ml of H2
O (B) 0.1 mole of V2
O5
(C) 12 g of O3
gas (D) 12.044 × 1022
molecules of CO2
16. A molal solution is one that contains one mole of solute in
(A) 1000 g of the solvent (B) one litre of the solvent
(C) one litre of the solution (D) 22.4 litre of the solution
17. The number of gram moles of solute dissolved per litre of solution is known as
(A) molarity (B) molality (C) normality (D) mole fraction
18. The balancing of chemical equation is based upon
(A) Law of combining volumes (B) Law of multiple proportions
(C) Law of conservation of mass (D) Law of definite proportion
19. If NA
is Avogadro number, then the number of valence electrons in 4.2 g of N3–
ions
(A) 2.4 NA
(B) 4.2 NA
(C) 1.6 NA
(D) 3.2 NA
20. At certain temperature two volumes of A combines with fine volumes of B to produce two volumes of C.
If atomicity of A and B is 2, the formula of compound C is
(A) AB3
(B) A2
B3
(C) AB5
(D) A2
B5
SECTION-B
 Multiple choice question with one or more than one correct answers
1. In which mode of expression, the concentration of a solution remains dependent on temperature
(A) Molarity (B) Normality (C) Formality (D) Molality
2. Which of the following pairs of compound does not illustrate the law of multiple proportion
(A) NaOH, KOH (B) SO2
, SO3
(C) H2
O, D2
O (D) KCl, KI
3. Which of the following occupies a volume of 4.48 L at NTP?
(A) 0.2 mol of H2
(B) 3.2 g of oxygen (C) 12.8 g of SO2
(D) 800 mg of He
4. Which of the following represents 1 g molecule of the substance
(A) 6.02 × 1024
molecule of NH3
(B) 4 g of helium
(C) 40 g of calcium (D) 127 g of iodine
5. Purest form of carbon is obtained by dehydration of sucrose C12
H22
O11
with conc. H2
SO4
as
2 4
conc. H SO
12 22 11 2
C H O 12C+11H O


By starting with 34.2 g of sucrose the amount of carbon formed is
(A) 14.4 g (B) 12 g atom of C (C) 3.2 g atom of C (D) 1.2 g atom of C
6. Acompound contains 20% of X (atomic mass 10) and 50% Y (atomic mass 20), which formulae pertain to
about data?
(A) XY (B) X2
Y (C) X4
Y3
(D) (X2
)3
Y3

SECTION-C
 Comprehension
Passage-1
1. How many moles are 5 grams of calcium?
(A) 0.125 mole (B) 0.25 mole (C) 0.375 mole (D) 0.5 mole
2. What is the weight in gms of 6 moles of SO3
?
(A) 480 (B) 4.80 (C) 0.480 (D) 480
3. Mass of 1 atom of C14
in grams is
(A) 6.023 × 1023
(B) 2.32 × 10–32
(C) 1.66 × 10–4
(D) 1.793 × 10–22
Passage-2
Valency is defined as combining capacity of an element. The elements having valency 1,2,3,4 are monova-
lent, divalent, trivalent and tetravalent respectively.
For writing the chemical formula of a molecular compound write the valency numbers over the symbols of
elements and criss-cross the valency numbers to write as subscripts to the symbols.
1. An element (X) is tetravalent and another element is divalent. The formula of a compound formed of these
elements will be
(A) X2
Y2
(B) X2
Y (C) X2
Y3
(D) XY2
2. Select the divalent element from the following
(A) Ca (B) Na (C) Al (D) Cr
3. The formula of oxide of a divalent element is
(A) XO (B) XO2
(C) X2
O3
(D) X2
O
SECTION-D
 Match the following (one to many)
Column-I and column-II contains four entries each. Entries of column-I are to be matched with some
entries of column-II. One or more than one entries of column-I may have the matching with the some entries
of column-II and one entry of column-II may have one or more than one matching with entries of column-I
(A) Molecule (P) CO
(B) Homoatomic (Q) S8
(C) Heteroatomic (R) C12
H22
O11
(D) Compound (S) P4

(A) 1 mole (P)
weight of substance
Atomic weight
(B) No. of atoms in 1 g of O3
(Q) 22.4 litres at NTP
(C) No
(R) one gram mole of a compound
(D) 6.023 × 1023
(S) no. of atoms in 1 g of O2
(A) Normality (P) Molarity × n–factor
(B) Molarity (Q) g equivalent in V litre
(C) Molality (R) g moles in V litre
(D) Mole fraction (S) Temperature independent
(A) 0.5 mole of N2
gas (P) 7 g
(B) 0.5 mole of N atoms (Q) 15 g
(C) 6.023 × 1023
molecules of N2
(R) 28 g
(D) 1 mole of N2
gas (S) 22400 ml at NTP
*******

Answers
KNOWLEDGE BASE QUESTIONS
1. (B) 2. (B) 3. (A) 4. (D) 5. (A)
6. (A) 7. (A) 8. (A) 9. (A) 10. (D)
TRY YOURSELF
6. same kind of
7. He, Ne
8. (i) Ca (ii) K (iii) Mg (iv) Hg
(v) Pb (vi) Fe
9. (i) Zinc (ii) Sodium (iii)Aluminium (iv) Lead
(v) Copper (vi) Silver
10. (i) Mg2+
(ii) Si (iii)Al3+
(iv) Na+
11. C2H4O2 12. CHCl3
13. 0.1 mole 14. 9.2 g 15. 3.011 × 1022
16. 4 g of CH4 17. 4 : 1 18. 1.06 × 10–22
g
19. (i) 4.45 × 1024
(ii) 2.409 × 1023
g
EXERCISE-III
SECTION-A
1. Lavoisier 2. 2. 4092 × 1024
atoms.
3. 1000 4. molecule
5. W 6. 1 : 8
7. Formula 8. Nitrogen and Hydrogen
9. Carbon-12 10. Atomicity
11. 9.48 L 12. atomicity
SECTION-B
1. (B) 2. (D) 3. (C) 4. (C) 5. (B)
6. (C) 7. (C) 8. (A) 9. (C) 10. (B)
11. (C) 12. (A)
SECTION-C
1. (B) 2. (A) 3. (B) 4. (C) 5. (D)
SECTION-D
1. (A)-(S), (B)-(P), (C)-(R), (D)-(Q) 2. (A)-(R), (B)-(P), (C)-(S), (D)-(Q)
3. (A)-(S), (B)-(R), (C)-(S), (D)-(P) 4. (A)-(R), (B)-(S), (C)-(P), (D)-(Q)

EXERCISE-IV
SECTION-A
1. (A) 2. (D) 3. (B) 4. (A) 5. (A)
6. (B) 7. (A) 8. (A) 9. (C) 10. (A)
11. (A) 12. (B) 13. (A) 14. (D) 15. (C)
16. (A) 17. (A) 18. (C) 19. (A) 20. (D)
SECTION-B
1. (A,B,C) 2. (A,C,D) 3. (A,B,C) 4. (B,C) 5. (A,D)
6. (A,B,D)
SECTION-C
Passage-1
1. (A) 2. (A) 3. (C)
Passage-2
1. (D) 2. (A) 3. (A)
SECTION-D
1. (A)-(P,Q,R,S), (B)-(Q,S), (C)-(P,R), (D)-(P,R)
2. (A)-(P,Q,R), (B)-(S), (C)-(P,R,Q), (D)-(P,R,Q)
3. (A)-(P,Q), (B)-(R), (C)-(S), (D)-(S)
4. (A)-(Q), (B)-(P), (C)-(R,S), (D)-(R,S)
*****

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