INTRODUCTION

In the pervious chapters, we have discussed the physical as well as the chemical classification of matter.
According to the physical classification, matter has three states. These are solid, liquid and gaseous states.
The know that the compounds, also known as chemical compounds are formed as a result of the
combination of the elements. The combinations must be based on certain guidelines because we find that
in a compound like carbon dioxide (CO2 ) , the element carbon and oxygen are combined in certain fixed
ratio by mass. Carbon dioxide can have no other formula except CO2 .
In the present chapter, we shall discuses the basis of the combination between the atoms resulting in the
formation of compounds. We shall represent the elements and compound by chemical symbols and
formulae respectively. In addition to these, we shall discuss the carouse ways in which the masses of the
elements and compound have been expressed.

LAWS OF CHEMICAL COMBINATION

Since early seventeenth century, the scientists were trying to establish the basis of the chemical
combination between the atoms of the elements to form different compounds. As a result of the vast
researches carried by them, they came to the conclusion that these chemical combination are based upon
number of laws have been proposed after through experimental studies. Out of these two laws of chemical
combination are given These are :
 Law of Conservation of Mass
 Law of Constant Proportions.
Let us study the details of these laws of chemical combination.

LAW OF CONSERVATION OF MASS

We know that in a chemical reaction, the reactants combine to form the products also known as the
products of the reaction. The law deals with the mass of the reactants and of the products participating in
chemical reactions. 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 law may also be stated in another form.
The mass can neither be created nor 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 Indestructibility of Matter
Explanation :
Let us try to analyse as to what happens in a chemical reaction. To understand the same let us consider a
chemical reaction between barium chloride and
The reactants simply change partners in chemical sodium sulphate. When the solutions of these
reactions. No change in mass takes place. reactants prepared separately in water are mixed,
the following chemical reaction takes place:
Barium chloride + sodium sulphte  barium sulphte + sodium chloride
(White precipitate)

1
 If we look at this reaction, we find that the exchange of constituents has taken place between the reactions
i.e., chloride part of barium chloride has been exchanged by sulphate part of sodium sulphate. As such, no
loss or gain in mass is expected. In other words, the mass remains conserved.

ACTIVITY
In order to illustrate the law of conservation of mass, let
us carry the same chemical reaction in the laboratory.
The procedure is quite simple. In a graduated cylinder,
prepare about 10% solution of one of the reactants say
barium chloride in water. For that, weigh accurately 10
gram of the sample and dissolve in water with stirring in
the graduated cylinder. Add more of water as to touch
100 mL mark. This is 10 percent solution of the reactant.
Transfer a portion of this solution I a beaker as shown in
figure 3.1. n similar manner, prepare 10% solution of
sodium a small glass tube. Place the tube car fully in the beaker without disturbing. Even a thread can be
used for the purpose. Now weight the beaker on a balance and note the weight. Shake the beaker for
sometime by hand so that the two reactants mix together and react to form the products. Barium sulphate
is in the form of a white precipitate the other product sodium chloride is water soluble. Weigh the beaker
again and notice if there is any change in weight. No change will be noticed. This means that the mass of
reactants is the same as the mass of the products and this verifies the law of conservation of mass.

Q.1. In a reaction,4.0 g of sodium carbonate were reacted with 10 g of hydrochloric acid. The product was a
mixture of 2 5 g of carbon dioxide and 11.5 g of sodium chloride solution. Is this data in agreement
.
with the law of conservation of mass ?
Sol.: The chemical reaction leading to products is :
sodium carbonate + hydrochloric acid  sodium chloride solution + carbon dioxide
Mass of reactants  (4.0  10.0)  14.0 g
Mass of products  (11.5  2.5)  14.0 g
Since the reactant and products have the same mass, this means that there was no loss in mass as a result
of the reaction. The data is in agreement with the law of conservation of mass.

Q.2. If 6.3 g 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 ?
Sol.: The chemical reaction leading to products is :
Sodium bicarbonate + ethanoic acid  sodium ehtanoate solution +carbon dioxide

(residue left) (released)

Mass of reactants  (6 3  15 0)  12 3g
. . .
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
.

LAW OF CONSTANT PROPORTIONS

Law of constant proportions is also known as law of Definite Composition. This was given by a French
Chemist, Joseph Louis Proust I the year 1799. it deals with the composition of the various elements that
are present in a particular compound. According to the law :
Apure chemical compound always consist of the same elements that are combined together in a fixed (or
definite) proportion by mass.

2
 This means that a particular compound may be formed or obtained form a number of different sources. In
case it is pure then the ratio of the different elements in all the sample of that compound will remain the
same.

ACTIVITY
Water ( H 2O) contains the elements hydrogen and oxygen.
Please note that the law is applicable
These are present in the ratio of 2 :16 or 1 : 8 by mass. Now, only to the pure compounds. In case
water may be obtained form number of success such as rain some impurities are present the ratio
water, river water, wall water, tap water etc. it can also be may change. Under these conditions the
made chemically by the combination of hydrogen and law cannot be applied.
oxygen elements under certain specified conditions. The law
a state that the ratio by mass of hydrogen and oxygen in all these samples of water will be the same i.e., 1
: 8 by mass.

Q.3. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen will he
required to react completely with 4 g of hydrogen ?
Sol.: According to available data,
Mass of oxygen combining with 1 g of hydrogen = 8 g
(8 g )  (4 g )
 Mass of oxygen combining with 4 g of hydrogen   32 g
(1g )
Q.4. The percentage of the three elements calcium, carbon and oxygen in a given sample of calcium carbonate
is given as :
Calcium  40.0% ; Carbon  12.0% ; Oxygen  48.0%
If the law of constant proportions is true. what weights of these elements will be present in 1.5 g of
another sample of calcium carbonate ?
Sol.: We have learnt that according to the law of constant proportions the percentages of the elements present
in different samples of a pure substance always remains the same. This means that in the second sample of
calcium carbonate, the ratio of the different elements present will remain the same or will remain
unchanged. Thus,
40
Mass of calcium in 1.5 g of the sample   (1.5 g )  0.6 g
100
12
Mass of carbon in 1.5 g of the sample   (1.5 g )  0.18 g
100
48
Mass of oxygen in 1.5 g of the sample   (1.5 g )  0.72 g
100
Q.5. 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.
Sol.: 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
Mass of boron (0.096 g )
Mass of boron  100  100  40
Mass of compound (0.24 g )
Mass of oxygen (0.144 g )
Percentage of boron  100  100  60
Mass of compound (0.24 g )

DALTON'S ATOMIC THEORY

We have studied that anything which has mass and occupies some space is known as matter. People,
particularly philosophers all over the world even from ancient times were trying to find out the composition
or the constituents of matter. Indian philosopher, Maharishi Kanad, studied a number of religious books
particularly Vedas thoroughly. He stated that if we go on dividing a particular matter, a stage will be
ultimately reached when it may not be possible to break it further. He named the smallest indivisible

3
 portion of matter as Parmanu. His views were also shared by some Greek philosophers like Leuppus and
Democritus. However, they did not have any scientific support to their views. In the mean time, the laws if
chemical combination were given by scientists like Lavoisier and Proust. John Dalton, a British chemist
was also working in this field for a pretty long time. On the basis of the studies and investigations carried,
he came out with a statement that the smallest portion of matter which cannot be divided any further is an
atom. In the year 1808, he gave the famous theory known as Dalton's Atomic Theory of Matter.

FEATURES OF DALTON'S ATOMIC THEORY

The important features of the Dalton's Atomic Theory are listed :
1. Every matter is made up of very small particles known as atoms.
2. Atoms are the ultimate particles of matter and cannot be further sub-divided into smaller particles.
3. Atoms can neither be created nor destroyed during a physical change or a chemical reaction.
4. All atoms of a particular element are identical in all respects.

Dalton’s theory has given a satisfactory explanation for the laws of chemical combination and was very
popular at that time.
This means that they have same mass, size and also same chemical properties.
5. Atoms of different elements have different masses, sizes and also chemical properties.
6. Atoms are the smallest particles of matter which can take part in chemical combination.
7. Atoms of the same or different elements combine in small whole number ratios to form molecules of a
compound.

DRAWBACKS OR LIMITATIONS OF THE THEORY

The theory as pointed above, brought a revolution in the field of science, particularly chemistry. It gave for
the first time a definite or concrete information about the composition of matter. However, it could not
stand for a very long time. The later studies and researches carried by the scientists revealed that most of
the postulates of the Dalton's Atomic Theory are defective. Some important limitations of the theory are
listed
• Atoms have been found to be made of sub-atomic particles. The important out of these are electrons,
protons and neutrons. We shall study about the In spite of the limitations or drawbacks the
same in the next chapter. contribution by Dalton regarding the composition of
• Atoms of the same element may have different matter was marvelous and cannot be under
masses (Isotopes*) estimated.
• Atoms of different elements may have same
masses (Isobars*)
• Atoms of different elements may not always combine in simple whole number ratios to form molecules
of a compound.
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 H 22O11 which shows that the constituting atoms are not present
in simple whole number ratio. The actual ratio is 12 : 22 :11 .
• In a chemical reaction, the reactants simply change partners and form products.
• There is no change in mass during a chemical reaction which means that the mass of the products of
reaction is the same as that of the reactants.
• Law of constant composition is applicable only to the pure chemical compounds.
• In a chemical compound, the constituting atoms are always combined in a fixed ratio by mass.
• According to Dalton's Atomic Theory, atom is the smallest portion of the matter which cannot be further
sub-divided into the smaller particles.
• Though electrons, protons and neutrons have been found to be present in an atom. but it is still the
smallest particle of matter which can take part in chemical combination.

TEST YOUR ABILITY

1. Can matter be created ?
2. Name the scientist who stated the law of conservation of mass.
3. Suggest an alternate name to the law of conservation of mass.
4. To what types of substances is the law of constant proportions applicable ? State the law.
5. The name 'parmanu' was given by which Indian philosopher ?

4
6. Enlist the main features of the Dalton's atomic theory.
7. Point out three drawbacks of the Dalton's atomic theory.
8. Which postulate of Dalton's atomic theory is the basis of the law of conservation of mass ?
9. Do atoms always combine in small whole numbers to form molecules ?
10. Do you agree with the statement that atom is still the smallest particle of matter which can take part in
chemical combination ?
11. Give a suitable illustration to support the law of conservation of mass.
12. In one experiment, carbon dioxide was formed by burning coke in air. In an another experiment, lime
stone was burnt to give the same gas. Will the ratio by mass of carbon and oxygen in the two samples of
the gas be different ? What will be the actual ratio ?

ANSWERS TO SELECTIVE PROBLEMS

1. No 2. Lavoisier 3. Law of indestructibility of matter 4. Pure substances 5. Kanad 8. Atom can neither be
created nor destroyed 9. No, there are many exceptions 10. Yes, the statement holds good even to-day. 12. No, 3
: 8 by mass.

WHAT IS AN ATOM ? HOW BIG ARE ATOMS ?

According to Dalton's Atomic theory, an atom is the
smallest particle of an element. It maintains its identity in
all physical changes and chemical reactions. The size of
an atom is extremely small. In fact, it may not be
possible to think about anything smaller than an atom.
Just imagine that more than million atoms constitute the
page of the book that you are reading. For your
knowledge the radius of an atom of hydrogen is only
1010 m. If we try to compare it with the radius of a
4 2
grain of and ( 10 m) and an ant ( 10 m) which is a
very small insect, we can imagine about the size of an atom. Same is the case with the mass of the atoms
of different elements. For example, an atom of hydrogen has a mass nearly 1.6 10 kg. It was more or
27

less a dream to imagine how atoms look like. We must thank our scientists for their contributions in
promoting scientific knowledge. They have developed a sophisticated microscope known as scanning
tunneling microscope (STM) in 1981. With the help of this microscope, it has become possible to take the
photographs of some atoms. In the figure 3.2., atoms of silicon are shown (in magenta colour) on the
surface.

CHEMICAL SYMBOLS
Short names or abbreviations have become very common these
days. These are used for the sake of simplicity and also to have
less strain on our memory. For example, USA for united state of
America, UK for United Kingdom, PM for prime minister and so
on. People tried their hands to find suitable short names for the
elements also. In chemistry, the shortened names of the
elements are known as symbols.
Ancient alchemists were using difficult and even complicated
symbols for a few common elements and compounds. These are
given below :
The modem representation of the name of the element i.e. 'symbol' may be defined as :
short hand representation of the name of an element.
The symbols of the elements are based on certain guidelines given by Berzelius, another noted sweedish
chemist. These are briefly discussed.

5
 In some elements the first letter of their
English names represents their symbols.

There are number of elements whose names

begin with the same letter e.g. barium
bromine bismuth boron all begin with the
letter ‘b’. the symbols of such elements are
represented by the first letter followed by
some prominent letter from their English
names. The first letter is capital while the
second letter is small. Symbols of some of
these elements are given.
Name of elements Symbol Name of elements Symbol
Boron B Oxygen O
Carbon C Phosphorus P
Fluorine F Sulphur S
Hydrogen H Uranium U
Iodine I Vanadium V
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 Ci Palladium Pd
Cobalt Co Platinum Pt
Chromium Cr Zinc Zn

Name of element Latin Symbol

Silver Argentums Ag
Copper Cuprum Cu
Gold Aurum Au
Iron Ferrum Fe
Mercury Hydrargyrum Hg
Potassium Kalium K

6
Some elements have both English as well as
Latin/German names. For these elements symbol
include first letter from the Latin/German names
expressed as capital along with some other prominent
letter. Symbols of some of the elements belonging to
this category are given
Sodium Natrium Na
Lead Plumbum Pb
Antimony Stibium Sb
Tungsten Wolfram (Geramn W
name)
In writing the symbols of the elements potassium and tungsten, only the first letter of the Lain/German
name represents the symbol. these may be regarded as the excerptions. From the discussion, we conclude
that there are no hard and fast rules for framing the symbols of the elements. As pointed out ealier the
symbols of some of the elements are given just to honor certain famous scientists. For example. Es
(Einsteinium), No (Nobelium), Cm (Curium).
It is for the information of all of you, that the all 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 every where.
We have seen that no proper rules were followed for assigning symbols to different elements, the only
consideration was that not two elements may have the same symbols. Over the yeas are, the number of
the elements has vastly increased. At present about 118 elements are known. Their number may further
increase. The international Union of pure and Applied Chemistry (IUPAC) have now adopted a uniform
system for assigning names and symbols to the elements. I writing the IUPAC name of an element, the
Latin names of the digits (0 to 9) are put together in the order in which they appear in the atomic number
of a particular element. The suffix ‘ium’ is added at the end.
For example
Digit Latin Name Abbreviation
0 nil n
1 un u
2 bi b
3 tri t
4 quad q
5 pent p
6 hex h
7 sept s
8 oct o
9 enn e
To illustrate this system, let us assign name to an element with atomic number of 112. The IUPAC name
has been built from un (1) + un(1) + bi(2) + ium and is ununbium. The symbol of the element is 'uuh'. in
fact, the IUPAC name and symbol of any element can be written by following this procedure. The 1UPAC
has however approved the existing names and the symbols of the elements with atomic numbers upto
110. The systematic IUPAC names are used for the elements with atomic numbers more than 110.

ATOMIC MASS
We know that the smallest portion of matter is atom. It is so small in size that it may not be possible to
isolate a single atom and then weigh it. Moreover, the mass of the atom of an element is also extremely
24 27
small. For example, an atom of hydrogen has mass equal to 1.67 10 g (or 1.67 10 kg).
What does it show ? It indicates that a gram or kilogram is a very big unit to express the mass of an atom
of a particular element.

ATOMIC MASS AS RELATIVE MASS

To solve this problem, it was suggested that the mass of an atom should be expressed as the relative mass.
It could be done by fixing the mass of some atom of a Atomic mass of an element is not its
particular element as the standard mass. The masses of the actual mass but relative mass.
other atoms could be compared relative to it. In the
beginning, hydrogen was chosen to be standard element because it happens to be the lightest of all the
elements. The relative atomic mass of an element with respect to hydrogen may be expressed as :

7
 Mass of 1 atom of the element
Relative atomic mass of an element 
Mass of 1 atom of hydrogen
The relative atomic mass is expressed in units known as a.m.u. (atomic mass units). However, it can also
be simply represented as u (unified mass). An atom of oxygen is sixteen (16) times heavier than the atom
of hydrogen. This means that if atomic mass of hydrogen is 1u, then the atomic mass of oxygen is 16u.
There is no need to consider the actual masses of the elements for expressing their atomic masses.
The problem with hydrogen was that it exists in three forms with different masses*. Which form should be
considered for comparing the atomic masses was a big problem. Then hydrogen was replaced by oxygen
as the standard element. But the same problem persisted with oxygen also.
In 1961, ILIPAC body selected most abundant isotope of carbon with mass 12u and l/12th of this mass
i.e., 1u was considered as the standard.
In the light of this,
Mass of 1 atom ofthe element
Relative atomic mass of an element 
1
 Mass of 1 atom of C  12 1u 
2
An atomic mass unit may be defined as :
the moss of one twelfth (1/12) of the moss of one atom of carbon taken as 12 (or C-12 atom)
The different forms in which an element exists are known as isotopes. They have different masses also. For
the details of the isotopes, please consult chapter 4 on Structure of the Atom.
For example, atomic mass of oxygen is 16u. this means that an atom of oxygen is sixteen times heavier
than 1/12th of the mass of carbon atom taken as 1u. Thus,
1 1
Atomic mass of oxygen  16  th of atomic mass of carbon  16  12u  16u
12 12
AVERAGE ATOMIC MASS
Most of the elements exist in a number of forms having different atomic masses. As stated earlier, these are
known as isotopes. This means that the actual atomic masses of the elements are the average atomic
masses. The average ahs been found by taking into account the masses of the different isotopes and the
ratio or proportion in which they exist. The concept of average atomic mass will be discussed in detail in
Chapter 4. The atomic masses of most of the elements have been rounded off to the nearest whole
numbers. For example, atomic mass of hydrogen is taken as 1u where as actually it is 1.008 u . The
atomic masses of some common elements are given for reference. They can be helpful in the calculations
involved in creation numerical problems.
Element Atomic mass (u) Element Atomic mass (u)
Hydrogen 1.8 Sodium 23.0
Carbon .
12 0 Potassium 39.0
Oxygen .
16 0 Calcium 40.0
Nitrogen 14.0 Sulphur 32.0
Magnesium .
24 0 Chlorine 35.5
Phosphorus 31.0 Iron 56.0

GRAM ATOMIC MASS

The gram atomic mass of an element may be defined as :
Please note that the gram atomic mass of an The atomic mass of an element expressed in grams
elements is also known as its gram atom. Thus which is numerically equal to the mass in ‘u’.
1 gram atom of nitrogen (N) is the same as its For example,
gram atomic mass i.e. 14 g.

atomic mass of nitrogen (N) = 1 u and its gram

atomic mass = 14 g
Similarly,
atomic mass sulphur (S) = 32 u and its gram
atomic mass = 32 g

8
HOW DO ATOMS EXIST ?
We have studied that the smallest protein of matter is known as atom and its has an externally small size.
In general, the atoms of most of the elements do not exist independently of their own. The elements of
inert gases (also called noble gases are the exceptions. For example. The atoms helium (He0), neon (Ne)
and argon (Ar) etc. can exist independently. What about the atoms of other
elements ? The atoms An atom is the smallest portion of an elements while
of the same or molecule is the smallest portion of a compound.
different elements are Atoms in general do not exist independently but
bonded together molecules can exist independently.
tightly by strong forces
of attraction also called chemical bonds. The new species which are formed as a
result of this chemical combination are called molecules.
Molecule represents a group of two or more atoms (same or different)
chemically bonded to each other and held tightly by strong attractive forces.
Molecules are of two types.
These are :
Molecules of elements and Molecules of compounds.
Let us briefly study these two types of molecules.

MOLECULES OF ELEMENTS
Molecules of elements are formed by the combination of two or more atoms of the same element. The
number of the atoms present in the molecule represent its atomicity.
For example,
(i) A molecule of hydrogen is made from two atoms of hydrogen. Its atomicity is two and is represented as
H2 .
The atoms of noble gas elements are quite stable and
(ii) A molecule of oxygen is also made from two
can exist independently. These are there fore
atoms of oxygen.
monatomic in nature. For example helium (He) neon
Its atomicity is two and is represented as O2 . (Ne) argon (Ar) krypton (kr) and xenon (Xe).
(iii) A molecule of ozone is made from three
atoms of oxygen. Its
atomicity is three and is represented as O3 .
(iv) A molecule of phosphorus is made from four atoms of phosphorus. Its atomicity is four and is
represented as P4 .
Based upon the above discussion, the atomicity of a few common elements in their molecules is given
Name of the element Symbol of the Atomicity Representation of
element molecule
Argon Ar Monoatomic Ar
Helium He Monoatomic He
Hydrogen H Diatomic H 2
Oxygen O Diatomic O2
Chlorine CI Diatomic CI 2
Ozone O Triatomic O3
Phosphorus P Tetraatomic P4
Sulphur S Octaatomic S8
Sodium Na Monoatomic Na
Copper Cu Monoatomic Cu
Iron Fe Monoatomic Fe
Silver Ag Monoatomic Ag

Recently a form of the element carbon has been found in which sixty atoms of the element are combined
to form a very big molecule (C60 ) . It is known as Buckminister fullerene. It is also called fiillerene or

9
 bucky-balls. It has been found to be extremely useful in semi-conductors, super-conductors and has many
more applications.

MOLECULES OF COMPOUNDS
In the molecules of compounds, the atoms of different elements are combined or bonded together by
chemical bonds. These are present in definite proportion by mass according to law of constant
proportions. The molecules of compounds may be also diatomic, triatomic, tetra-atomic and polyatomic in
nature depending upon the number of the atoms linked or combined by chemical bonds. For example,

Compound Combining elements Nature Ratio by mass

Hydrogen chloride (HCI) Hydrogen, chlorine Diatomic 1: 35.5
Water ( H 2O) Hydrogen, oxygen Triatomic 1: 8
Ammonia ( NH 3 ) Hydrogen, nitrogen Tetra – 14 : 3
atomic
Carbon dioxide (CO2 ) Carbon, oxygen Triatomic 3: 8
Please note that the molecules of elements are homoatomic in nature which means that the atoms present
in them are the same. The molecules of the compounds are heteroatomic in nature in the sense that
different atoms are resent in them. For example, a molecule of nitrogen element is homoatomic in nature
( N 2 ) . A molecule of methane (CH 4 ) which Is a compound is heteroatomic in nature.
• Both the size and the mass of the atoms are extremely small.
• Atoms can be seen only by a special microscope called scanning tunneling microscope (STM).
• Elements are identified by symbols. Please note that the symbols of no. two elements can be the same.
• There are no hard and fast rules for framing the symbols of the elements. But IUPAC has adopted a
uniform system to assign names and symbols to the elements.
• Atomic mass of an element is not its actual mass. It is only the relative mass.
• According to IUPAC ; 1/12th of mass of carbon (12u) i.e. 1u(unified mass) is taken as standard (or
comparing the atomic masses of the elements.
• Atomic masses of most of the elements have been rounded off to the nearest whole numbers.
• Atoms in general do not exist independently but molecules have always an independent existence.
• Molecules of elements are homoatomic which means that the atoms present in them are the same (e.g.
H 2 , O2 , CI 2 ).
• Molecules of compounds are heteroatomic which means that the atoms present in them are different
(e.g. H 2O, NH 3 , CO2 ).

WHAT ARE IONS ?

So far we have discussed atoms and molecules. The atoms are represented by symbols. Molecule? are also
shown by certain symbols but these are
known as formulae. We shall discuss The ion which has one or more positive charges is called
CATION. At the sometime the ion carrying one or more
them a little later. H 2O, NH 3 and CO2
negative charges is known as ANION.
are the formulae of the molecules that we
have mentioned. These molecules are neutral in nature. This means that they do not have any charge. But
in some compounds certain charged species are present. These are known as ions.
An ion may be defined as :
an atom or group of atoms having positive or negative charge.

CHEMICAL FORMULAE OF COMPOUNDS

We know that atoms of elements combine to form a molecule. We represent the atoms with the help of
symbols. In the same way, the molecules can also be represented by the symbols of the constituent atoms.
This is known as the chemical formula of the molecule.
For example,
Formula of water : H 2O
Formula of ammonia : NH 3

10
Formula of sodium chloride : NaCI
Formula of sulphuric acid : H 2 SO4
We have seen that in the molecules, the atoms are combined with each other. Different elements have
different combining capacity of their atoms. This is known as valency. The exact picture of valency will be
clear only in the next chapter on Structure of Atom. However, it may be defined as : the combining
capacity of an element. It is expressed in terms of the number of hydrogen atoms, number of chlorine
atoms or double the number of oxygen atoms which combine with one atom of that element.
For example,
In sodium chloride (NaCl), the valency of Na is 1.
In ammonia ( ( NH 3 ), the valency of N is 3.
In methane (CH 4 ), the valency of C is 4.
In water ( H 2O), the valency of O is 2.
In addition to the atoms, the ions which are charged species have also some valencies. Positive ions or
cations have positive valencies. Negative ions or anions have negative valencies. The valencies of the
polyvalent ions are expressed by enclosing them in bracket and putting the positive or negative signs
outside it. Let us write the valenries of some commonly used positive and negative ions.
Valency of positive ions. Positive ions may be monovalent, bivalent, trivalent, tetravalent etc.
depending upon the charge present on them. These are listed.

List of some common positive ions (Cations)

Monovalent Bivalent Trivalent

Hydrogen H Barium Ba 2
Aluminium AI 3
Potassium K Calcium Ca 2 Chromium Cr
3

Sodium Na  Magnesium Mg 2 Iron Fe3


Silver Ag Zinc Zn2 Gold Au 3
Copper Cu  Cobalt Co2
Gold Au  Copper Cu 2
Ammonium ( NH 4 ) Iron Fe2
Some elements have shown more than one valency. In such cases, Roman Numerals are used to denote
the valencies. These are put in bracket. For example, copper (I) and copper (II) ; similarly, iron (II) and
iron (III).
Valency of negative ions. Like positive ions, negative ions may also be monovalent, bivalent, trivalent,
tetravalent etc. in nature. These are also listed ;

List of some common negative ions (Anions)

Monovalent Bivalent Trivalent
 2
Chloride CI Sulhide S Nitride N 3
Bromide Br  Oxide O 2 Phosphide P 3
Hydroxide I Carbonate (CO3 )2 Phosphate ( PO4 )3
Nitrate (OH ) 
sulphate ( SO4 ) 2
Borate ( BO3 )3
Nitrite ( NO3 ) Sulphite ( SO3 )2 Arsenate ( AsO3 )3
Bicarbonate ( HCO3 ) Manganate ( MnO4 )2
Cyanide (CN ) Oxalate (C2O4 )2
Permanganate ( MnO4 ) Chromate (CrO4 )2
Chlorate (CIO3 ) Dichromat (Cr2O7 )2
Now let us write the chemical formulae of some molecular compounds.

11
CHEMICAL FORMULAE OF SIMPLE MOLECULAR COMPOUNDS
In the formation of simple molecules, atoms of the elements participate. The symbols of these atoms are
written and their valencies are placed below them. These valencies are shifted cross-wise to the lower right
of these atoms as shown. This is known cross-over or criss-cross of valencies. Let us write the chemical
formulae of simple molecular compounds.
• In the formula of the compound the valency
‘1’ is normally omitted e.g. NH 3 and not as
N1 H 3 .
• In case there is some common factor. it has
to be taken out e.g. , CO2 and not C2O4 .

CHEMICAL FORMULAE OF SIMPLE IONIC COMPOUNDS

Simple ionic compounds are of binary nature. It means that both the positive and negative ions have one
atom only. The symbols of these ions are written side by side with their valencies at their bottom. A
common
factor if any, is removed to get a simple ratio of the valencies of the combining atoms. The criss-cross
method is then applied to arrive at the final chemical formula of the compound. Let us write the formulae
of a few simple ionic compounds.
For example,

In the final formulae of the compound the

positive and negative signs are omitted.

CHEMICAL FORMULAE OF IONIC COMPOUNDS CONTAINING POLYATOMIC IONS

If we look at the list of positive and negative ions, we find that mostly the negative ions (or anions) are
polyvalent while the positive ions (or cations) are not. However ( NH 4 ) ion is an exception. These are
enclosed in bracket as mentioned earlier. The rest of the procedure remains the same.
For example

12
Q.6. write down the formulae of (i) aluminium hydroxide, (ii) hydrogen, sulphide, (iii) ammonium sulphate, (iv)
calcium phosphate, (vi) potassium chromate.

Sol.
• Elements are represented by symbols and molecules by chemical formulae.
• Positive ions or cations are formed from the metals. Negative ions or anions are formed from the non-
metals.
• Cations have positive valencies while the valencies of anions are negative.
• In writing the formula of the compound, the valency 1 is omitted e.g. NH 3 (and not N1 H 3 ).
• In case, there is some common factor, it has to be taken out e.g. CO2 (and not C2O4 ).
• In the final formula of a compound the positive and negative signs are not shown e.g. NaCI (and not
Na CI  ).

TEST YOUR ABILITY

1. What is the name given to the short hand representation of an element ?
2. Give the symbols of the elements (i) Bismuth, iii) Sulphur, (iii) Magnesium, (iv) Manganese, (v) Iron.
3. Give the latin name of (i) Sodium (ii) Potassium. Also write the symbols of these elements.
4. Which microscope can help in seeing the atom of an element ?
5. Is the atomic mass of an element its actual mass ?
6. Which is the standard for comparing the atomic masses of different elements ?
7. What does IUPAC represent ?
8. The atomic mass of an element is in fraction. What does it mean ?
9. Define a molecule. How does it differ from an atom ?
10. Atoms are represented by symbols. What is the name given to the representation of a molecule ?
11. Give the names of two monoatomic and two diatomic molecules.
12. Give the formulae of (i) calcium oxide (ii) calcium carbonate (iii) ammonia (iv) copper nitrate.
13. Given the names of the compounds having the fonnulae (i) AI 2 ( SO4 )3 , (ii) MgCI 2 , (iii) K 2 SO4 , (iv)
KNO3 , (v) CaCO3 .
14. Give the examples of two polyatomic ions.
15. What is the chemical names of the following compounds ?
(i) quick lime, (ii) baking soda, (iii) table salt.
16. Name the elements present in the following compounds
(i) Potassium sulphate, (ii) ammonia, (iii) quick lime.

ANSWERS TO SELECTIVE PROBLEMS

1. Symbol 2. (i) Bi (ii) S (iii) Mg (iv) Mn (v) Fe. 3. (i) Natrium (Na) (ii) Kalium (K) 4. STM (Stem
Tunneling Microscope) 5. No, it is only the relative mass. 6. 1/12th of mass of C  12 and is 1u . 7.
Intemational union of pure and applied chemistry. 8. It means that the element exists as isotopes. 10.
Chemical formula 11. Monoatomic : Helium. Neon : Diatomic : Hydrogen. Nitrogen. 12. (i) CaO (ii)
CaCO3 (iii) NH 3 (iv) Cu ( NO3 )2 13. (i) Aluminium sulphate (ii) Magnesium chloride (iii) Potassium
sulphate (iv) Potassium nitrate (v) Calcium carbonate. 14. ( HCO3 ) and ( NH 4 ) 15. (i) Calcium oxide
(ii) Sodium bicarbonate (iii) Sodium chloride. 16. (i) K, S, 0 (ii) N, H (iii) Ca. O.

13
MOLECULAR MASS
We have studied that the mass of the atoms of an element is known as their atomic mass. It is in fact the
relative mass and is expressed in ‘u’ In the same way, the mass of the molecules of a chemical compound
is termed as the molecular mass. It is also the relative mass compared to 1/12 of the mass of carbon atom
taken as 1u. Thus, molecular mass of a compound may be defined as :
the average relative mass of its molecule as compared to the 12 of the mass of carbon taken as 1u.

CALCULATION OF MOLECULAR MASS

The molecular mass of a substance can be calculated as the
sum of the atomic masses of all the atoms which constitute a Molecular mass of a substance is also
molecule of that substance. known as its molar mass.
For example,
Molecular mass of NH 3  Atomic mass of N  3  Atomic mass of H
 14  3 1  17u
Q.7. Calculate the molar mass of the following substances :
(i) Water ( H 2O) (ii) Sulphur dioxide ( SO2 )
(iii) Oxygen molecule (O2 ) (iv) Carbon monoxide (CO)
Sol. : (i) Water ( H 2O)
Molar mass of H 2O  (2  Atomic mass of H) 1 Atomic mass of O)
 (2 1u)  (116u)  18 u
(ii) Sulphur dioxide ( SO2 )
Molar mass of SO2  (1 Atomic mass of S) (2  Atomic mass of O)
(iii) Oxygen molecule (O2 )
Molar mass of O2  2  Atomic mass O
 (2 16 u)  32 u
(iv) Carbon monoxide (CO)
Molar mass of CO  (1 Atomic mass of C) (1 Atomic mass of O)
 (112 u)  (116 u)  28 u ’

Gram Molecular Mass.

The gram molecular mass of a substance may be defined as :
The molecular mass of the substance expressed in grams which is numerically equal to the molecular mass
in u.
For example,
gram molecular mass of NH 3  17 g
Gram molecular mass of C2 H 6  30 g
Please note that gram molecular mass of a substance is also called its one gram molecule.

Q.8. Calculate the mass of 0.72 gram molecule of carbon dioxide (CO2 )
Sol. : Molecular mass of CO2  Atomic mass of C  2  Atomic mass of O
 (12u  2 16u)  44u
Gram molecular mass of CO2  44 g
1 gram molecule of CO2  44 g
(44 g )
0.72 gram molecule of CO2   (0.72 g )  31.68 g
(1g )

14
FORMULA UNIT MASS
The molecules of some substances are made up of ions. For example, sodium chloride is made up of
In the ionic compounds. we can use both molecular Na  and CI  ions. It may be represented as
mass and formula unit mass. (e.g. molecular mass of Na CI  . In these compounds, we can also
NaCI  58.5u ). However for the compounds in use the term formula unit mass in place of
which ions are not present the term formula unit mass molecular mass. In order to represent a
cannot be used. This means that we cannot say that molecule, we use the term formula unit. For
formula unit mass of NH 3 is 17 u , example,
A formula unit of sodium chloride ( NaCI )
has the formula unit mass  (23  35.5)  58.5u .
Similarly, a formula unit of potassium carbonate ( K 2CO3 ) has the formula unit mass
 (2  39  12  3 16)  1438u.

MOLE CONCEPT
The gram atomic mass of carbon is 12 g. The mass of one atom of carbon has been calculated as
1.9924 1023 g. The number of carbon atoms in 12g of carbon can be calculated as :
Gram atomic mass of carbon
No of carbon atoms 
Mass of one carbon atom
12( g )
Gram atomic mass of an element contains
 23
 6.022 1023
1 9924 10 ( g )
.
6.022 1023 atoms. Similarly the gram Similarly, the gram molecular mass of oxygen (O )
molecular mass of a compound also contains 2
23
the same number of molecules. is 32g. The mass of one molecule is 5.313 10 g.
The number of oxygen atoms in 32 g of oxygen can be calculated as :
Gram molecular mass of oxygen 32( g )
No. of oxygen molecules    6.022 1023
Mass of one oxygen molecule 5 313 1023 ( g )
.
This number is known as Avogadro's number or Advogadro’s constant and is denoted either as "No' or as
'NA'. It may be noted that the Avogadro’s number of particles of any substance are expressed in the form of
a term ‘mole’’. In other words.
A mole denotes Avogadro's number ( N 0 or N1 ) of particles.
Please note that these particles may be any thing i.e. atoms, molecules, ions etc. Thus,
One mole of hydrogen atoms  6.022 1023 hydrogen atoms
One mole of oxygen molecules  6.022 10 oxygen molecules.
23


One mole of sodium ions ( Na )  6.022 10 sodium ions.
23

NECESSITY OF MOLE CONCEPT

We know that the atoms and molecules of any substance (element and compound respectively) are very
small in size. This is quite evident from the
There is a very close analogy of similarity between the
fact that 12 g of carbon contain 6.022 10
23
terms mole and dozen. A dozen always represents twelve
atoms of carbon. This means that it may not
articles. They may be apples oranges pens pencils not
to be possible to count these atoms
books etc. In a similar way a mole represents
individually. However, they can collectively
be represented as one mole. This is a very 6.022 1023 (Avogadro’s number N 0 ) paricles. They
convenient method to represent different may be atoms molecules ions electrons protons etc. A
particles. For example, No. of oxygen atoms mole is quite often called Chemist’s dozen.
in 3 moles  (3  N0 )  3  6.022 1023
 1.81  1024 atoms

15
 Similarly, the number of carbon and oxygen atoms in 1 mole of carbon dioxide (CO2 ) may be calculated
as
( N0 )  2  Avogadro's number of oxygen atoms ( N 0 )
= Avogadro's no. of carbon atoms
 (6.022 10  2  6.022 10 ) atoms  1.811024 atoms
23 23

We can say that 1 mole of CO contains 1 mole atoms of carbon (6.022 10 ) and 2 mole atoms of
23
2

oxygen ( 1.20 10 atoms). This clearly shows that mole is a better option out of the two. It is therefore,
24

preferred.

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 :
Given no. of paricles (N)
Number of moles 
Avogardro's no. of particle (N0 )
Q.9. 1023 atoms of irons.
Calculate the number of moles of iron in a sample contain
No of aotms of iron (N) N
Sol. : Number of moles  
Avogardro's no. of atoms (N0 ) N 0
(1.0 10 atoms)
23
  0.0166 mol.
(6.022 1023 atoms)
• In using tem term mole, it is essential to specify the kind of particles involved. For example, one mole of
hydrogen atoms (H) contains 6.022 1023 atoms of hydrogen. But one mole of hydrogen molecules
( H 2 ) contains 6.022 1023 molecules of hydrogen.
• In the text, one use the word mole but when used as a unit, it is expressed as mol.

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 a have molecular mass or gram molecular mass. Both
can be expressed in terms of moles as follows :
given mass (m)
For elements : The no. of moles 
gram molar mass (M)
given mass (m)
For compounds : The no. of moles 
gram molar mass (M)
Q.10. Calculate the number of moles in the following :
(i) 28 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.
Sol. : (i) 28 g of He
Mass of Hein grams m (28 g )
The no. of moles     7 mol
Gram atomic mass M (4 g )
(ii) 46 g of Na
Mass of Nain grams m (46 g )
The no. of moles     2 mol
Gram atomic mass M (23 g )
(iii) 60 g of Ca
Mass of Cain grams m (60 g ) .
The no. of moles     1 5 mol
Gram atomic mass M (40 g )

Q.11. Calculate the mass of the following :

(i) 0.5 mole of O2 gas (ii) 0.5 mole of O atoms

16
 (iii) 3.0111023 atoms of (iv) 6.022 1023 molecules of O2 .
(Given : Gram atomic mass of oxygen =16 g,
Gram molecular mass of oxygen (O2 )  32 g )
Sol. : (i) 0.5 mole of O2 gas
Mass of O2 gram m
No. of moles  
Gram molecular mass M
 Mass of O2 in gram (m) =No. of mole  M  0.5  (32 g )  16 g
(ii) 0.5 mole of oxygen (O) atoms
Mass of oxygen (O) in grams m
No. of moles  
Gram atomic mass M
Mass of oxygen (O) in gram (m)= No. of moles M  0.5  (16 g )  8 g
(iii) 3.01110 atoms of oxygen (O)
23

No. of atoms of oxygen N

Step I : Calculate of no. atoms  
Avogadro ' s no. of atoms N0
3.01110 23
  0.5 gram atom
6.022 1023
Step II : Calculation of mass of oxygen (O) atoms
Mass of oxygen (0) atoms = Gram atomic mass of oxygen  No. of gram atoms of oxygen
 (16 g )  0.5  8 g
(iv) 6.022 10 molecules of oxygen (O )
23
2
Step I : Calculation of no. of gram moles of oxygen.
No. of molecules of oxygen N
No. of gram moles  
Avogadro 's no. of molecules N0
Step II : Calculation of mass of oxygen (O) 2 molecules.
Mass of oxygen (O) 2 molecules = Gram molecular mass of oxygen  No. of gram moles of oxygen.
 (32 g ) 1  32 g

Q.12. What is the mass in 'u' of :

(a) 1 mole of nitrogen atoms (b) 4 moles of aluminium atoms
(c) 10 moles of sodium sulphate (d) 5 moles of calcium carbonate.
(Atomic mass of N = 14u, Al = 27 u; Na = 23 u ; S = 32 u ; O = 16 u ; Ca = 40 u, C = 12 u)
Sol. : (a) 1 mole of nitrogen atoms
Mass of 1 mole of nitrogen (N) atoms = 14 u
(b) 4 moles of aluminium atoms
Mass of 1 mole of aluminium (Al) atoms = 27 u
 Mass of 4 moles of aluminium (Al) atoms = 4 x 27 = 108 u
(c) 10 moles of sodium sulphate ( Na2 SO4 )
Molar mass of Na2 SO4  2  Atomic mass of Na + Atomic mass of S  4  Atomic mass of O.
 2  23  32  4 16  46  32  64  142u
1 mole of sodium sulphate have mass = 142 u
10 moles of sodium sulphate have mass  10 142  1420u
(d) 5 moles of calcium carbonate (CaCO3 )
Molar mass of CaCO3 = Atomic mass of Ca + Atomic mass of C + 3 x Atomic mass of oxygen
 40  12  3 16  100u
1 mole of calcium carbonate have mass =  100u

17
 5 moles of calcium carbonate have mass  5 100  500u

Q.13. Calculate the weight of carbon monoxide having the same number of oxygen atoms as are present in 22 g
of carbon dioxide.
Sol. : Step I. No. of the oxygen atoms in 22 g of carbon dioxide (CO2 )
Molar mass of CO2  12  2 16  44 g
44 g of CO2 represen  1mol
(22 g )
22 g of CO2 represent  1(mol )   0.5 mol
(44 g )
Now, 1 mole of CO2 contain oxygen atoms  2  N0  2  6.022 10
23

0.5 mole of CO2 contain oxygen atoms  2  6.022 1023  0.5  6.022 1023 atoms
Step II . Weightofcarbonmorw3dde(CO)
6.022 1023 atoms of oxygen are present in 6.022 1023 molecules of carbon monoxide (CO).
By definition,
6.022 1023 molecules of CO have mass = Molar mass of CO= 12 + 16 = 28 g.
• Molecular mass of a substance is the sum of the atomic masses of the constituting atoms.
• Gram molecular mass of a substance is also called its one gram molecule.
• A mole represents Avogadro's number of particles (6.022 10 ) of a substance.
23

• A mole is quite often known as chemist's dozen.

• Gram molar mass (or molar mass) is the mass of Avogadro's number of molecules expressed in grams.
• No. of moles of a compound can be calculated by dividing its mass in grams by its gram molar mass.
• In using the term mole, it is essential to specify the kind of particles involved.
For example, one mole of oxygen atoms (0) contains 6.022 1023 atoms of oxygen. But one mole of
oxygen molecules (O2 ) contains 6.022 molecules of oxygen.

A SUMMARY OF DIFFERENT RELATIONSHIPS IN TERMS OF MOLE CONCEPT.

TEST YOUR ABILITY

1. When do we use the term formula unit ?
2. How do we represent formula units in terms of mass ?
3. Define the term mole.
4. Avogadro’s number of particle represent how many particles ?
5. A mole is often called chemist’s dozen. Discuss.
6. Define gram atomic mass and gram molecular mass in terms of mole concept.
7. How many atoms of oxygen are present in 8g of oxygen ?
8. 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 ?

18
 (iii) What is the name given to 6.022 10 ?
23

9. What is the mass of two moles of oxygen atoms ?

10. How many atoms of N and H are present in 0.5 mole of ammonia ( NH 3 ) ?
11. How many moles re present in :
(i) 10g of Ca (ii) 2.3g of Na (iii) 3.012 1023 atoms of oxygen (iv) 32 g of oxygen gas.
What is the weight of 1.20 10 atoms of Na (Atomic of Na = 23)
24
12.
13. Calculate the number of Mg atoms in 0.024 g of Mg.

ANSWER TO SELECTIVE PROBLEMS

1. In case of ionic compound 2. Formula unit mass 4. 6.022 1023 particles
7.3.012 1023 atoms 8. (i) 6.022 1023 atoms (ii) 6.022 10 molecules (iii) Avogadro’s
23

number ( N0 or N A )
2 16  32 g 10. 3.012 10 atoms of N and 9.035 10 atoms of H.
23 23
9.
11. (i) 0.25 mol (ii) 0.1 mol (iii) 0.5 mol (iv) 1 mol. 12. 45.82 g 13. 6.022 10 Mg atoms.
20

19
LET US RECAPITUALTE
 The combination between the elements is governed by certain laws known as laws of chemical
combination.
 The law of conservation of mass was given by A. Lavoisier in 1774.
 The law of constant proportions was stated by Louis Proust in 1799.
 The law of constant proportions is applicable only to pure chemical compounds or substances.
 Dalton was the first to state that the smallest indivisible portion of matter is an atom.
 Dalton’s Atomic theory has certain limitations or defects.
 Atom is the smallest portion of matter which can take part in chemical combination. It may or
may not
exist independently.
 Scanning tunneling microscope (STM) can be used to have a pictorial view of the atoms.
 Symbols give the shorthand representation of the name of elements.
 No two elements can have the same symbols.
 Atomic masses of the elements are their relative masses and not actual masses.
 1 u represents 1/12 of the mass of C-12 isotope of carbon. It is used as the standard for
representing
the relative atomic masses of different elements.
 All developments and researches in chemistry are governed by an organisation called
International
Union of Pure and Applied Chemistry (I.U.P.A.C.).
 The elements having fractional atomic masses exist as isotopes.
 The atomic masses of most of the elements are rounded off to nearest whole numbers.
 The gram atomic mass of an element is numerically equal to its atomic mass expressed as "u'.
 In molecule, the atoms of same or different elements are chemically bonded to each other.
 The number of atoms present in a molecule represents its atomicity.
 The molecules of noble gas elements are monatomic in nature.
 An ion is an atom or group of atoms having positive or negative charge.
 A positive ion is known as cation while the ion which has negative charge is called anion.
 Chemical formula is the representation of a molecule in terms of the symbols of the constituting
atoms.
 Valency of an element is the combining capacity of its atoms.
 The molecular mass of a substance (compound) can be calculated as the sum of the atomic
masses
of aII the atoms which constitute a molecule of that substance.
 The gram molecular mass of a substance is also called its one gram molecule.
 For molecules of ionic compounds, the term formula unit is used. Their molecular masses are also
known as formula unit masses.
 Avogadro's number represents 6 . 022x1023 particles of atoms, ions or molecules.
 Avogadro's number of particles (N0 or NA ) represent one mole of a substance.
 Gram atomic mass is the mass of Avogadro's no. of atoms in grams.
 Gram molecular mass is the mass of Avogadro's no. of molecules in grams.
 Atomic masses of a few common elements are :
hydrogen (1), carbon (12) nitrogen (14), oxygen (16), chlorine (35 . 5), sulphur (32), sodium (23)
Magnesium (24), potassium (39), calcium (40), phosphorus (31).

N.C.E.R.T. IN TEXT PROBLEMS

Q.1. In a reaction, 5 . 3 g of sodium carbonate reacted with 6 . 0 g of ethanoic acid. The products
were 2 . 2 g of carbon dioxide, 0 . 9 g of water and 8 . 2 g of sodium ethanoate. Show that these
observations are in agreement with the law of conservation of mass.
Ans. The chemical reaction leading to products is :

20
 sodium carbonate + ethanoic acid  sodium ethanoate + carbon dioxide + water.
Mass of reactants = (5 . 3 + 6 . 0) = 11 . 3 g
Mass of products = (2 . 2 + 0 . 9 + 8 . 2) = 11 . 3 g.
The reactants and products have same mass. This means that there was no loss of mass during the
reaction. Therefore, the data is in agreement with law of conservation of mass.

Q.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 with 3 g of hydrogen gas?
Ans. According to available data,
Mass of oxygen combining with 1g of hydrogen= 8 g.
(8 g )
Mass of oxygen combining with 3 g of hydrogen   (3 g )  24 g
(1 g )

Q. 3. Which postulate of Dalton's Atomic theory is the basis of law of conservation of mass?
Ans. The law of conservation mass is based on the following postulate of Dalton's atomic Theory.
"Atoms can neither be created nor destroyed during a physical change or a chemical reaction".

Q.4. Which postulate of Dalton's Atomic theory can explain the law of definite proportions?
Ans. The law of definite proportions is based on the following postulate of Dalton's Atomic theory.
"All atoms of a particular element are identical in every respect. This means that they have same mass,
same size and also same chemical properties."

Q.5. Define atomic mass unit

Ans. Atomic mass unit may be defined as :
The mass of one twelfth (1/12) of the mass of one atom of carbon taken as 12u. It is represented
as 1u.

Q.6. Why is not possible to see an atom with naked eye ?

Ans. It is not possible to see an atom with naked eye because of its extremely small size. For example
the radius of an atom of hydrogen is of the order of 1010 m. Actually an atom is regarded as a
microscopic particle. These microscopic particles cannot be seen with naked eye.

Q.7. Write down the formulae of : (i) Sodium oxide (ii) aluminium chloride (iii) sodium sulphide
(iv) magnesium hydroxide.
Ans.

Q.8. The names of the compounds represented by the following formulae :

(i) AI 2 (SO4 )3 (ii) Ca CI 2 (iii) K 2SO4 (iv) KNO3 (v) CaCO3 .
Ans. (i) Aluminiuin sulphate (ii) Calcium chloride (iii) Potassium sulphate (iv) Potassium nitrate
(v) Calcium carbonate.

Q.9. What is meant by the term chemical formula ?

Ans. For answer, consult text part.

Q.10. How many atoms are present in (i) H 2S molecule (ii) P PO43- ion ?
Ans. (i) Three (ii) Five.

21
Q.11. Calculate the molar mass of (a) water (H 2O) (b) nitric acid (HNO3 ).
Ans. (a) Water (H2 O)
Molar mass of H2 O = (2 x Atomic mass of H) + (1 x Atomic mass of oxygen)
=(2 1u) + (16 1u) = 18 u. (b) Nitric acid (HNO3 )
Molar mass of HNO3 = (1 x Atomic mass of H) + (1 x Atomic mass of N) + (3 x Atomic mass of O)
= (1 x 1 u) + (1 x 14 u) + (3 x 16 u) = 63 u.

Q.12. Calculate the formula unit mass of CaCI 2 .

Ans. Formula unit mass of CaCI2 (Calcium chloride)
= (1 x Atomic mass of Ca) + (2 x Atomic mass of CI) = (1 x 40 u) + (2 x 35 . 5u) = 111 u.

Q.13. Calculate the molar masses of :

(i) H 2 (ii) O 2 (iii) CI 2 (iv) CO2 (v) CH4 (vi) C2 H6 (vii) C2 H4 (viii) NH 3 (ix) CH3OH.
Ans. Hydrogen (H 2 )
Molar mass of H 2 = 2 x Atomic mass of H = (2 x 1u) = 2u
(ii) Oxygen (O2 )
Molar mass of O 2 = 2 x Atomic mass of 0 = ( 2 x 16 u ) = 32 u.
(iii) Chlorine (CI2 )
Molar mass of CI2 = 2 x Atomic mass of CI = (2 x 35 . 5u) = 71u.
(iv) Carbon dioxide (CO2 )
Molar mass of CO2 = (1 x Atomic mass of C) + (2 x Atomic mass of O)
= (1 x 12u) + (2 x 16u) = 12u + 32u = 44u
(iv) Methane (CH4 )
Molar mass of CH 4 = (1 x Atomic mass of C) + (4 x Atomic mass of H)
= ( 1 x 12u) + (4 x 1u) = 16u
(vi) Ethane ( (C2 H6 )
Molar mass of C2 H6 = (2 x Atomic mass of C) + (6 x Atomic mass of H)
= (2 x 12u) + (6 x 1u) = 30 u.
(vii) Ethylene (C2 H4 )
Molar mass of C2 H 4 = (2 x Atomic mass of C) + (4 x Atomic mass of H)
= (2 x 12u) + (4 x 1u) = 28u.
(viii) Ammonia (NH3 )
Molar mass of NH3 = (1 x Atomic mass of N) + (3 x Atomic mass of H)
= (1 x 14u) + ( 3 x 1u) = 17u.
(ix) Methyl alcohol ( (CH3OH)
Molar mass of CH3OH = (1 x Atomic mass of C) + (4 x Atomic mass of H) + (1 x Atomic mass
of 0)
= (1 x 12u) + (4 x 1u) + (1 x 16u) = 32u

Q.14. Calculate the formula unit mass of : (i) ZnO (ii) Na2O (iii) K 2CO3 .
Given : Atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u and 0 = 16u
Ans. (i) Formula unit mass of ZnO (Zinc oxide)
=(1 x Atomic mass of Zn) + (1 x Atomic mass of 0)
= (1 x 65u) + (1 x 16u) = 81u.
(ii) Formula unit mass of Na 2 O (Sodium oxide)
= (2 x Atomic mass of Na) + (1 x Atomic mass of O)
= (2 x 23u) + (1 x 16u) = 62u.
(iii) Formula unit mass of K 2 CO3 (Potassium carbonate).
= (2 x Atomic mass of K) + (1 x Atomic mass of C)

22
 + (3 x Atomic mass of O)
= (2 x 39u) + (1 x 12u) + (3 x 16u) = 138u

Q.15. Calculate the number of moles of the following :

(i) 52 g of He (ii) 12 . 044 x 1023 atoms of He.
Ans. (i) 52 g of He
Given mass (m) (52 g )
No. of moles    13 mol.
Gram atomic mass(M) (4 g )
(ii) 12 . 044 x 1023 atoms of He
No. of moles
No. of atoms of helium N 12.044 1023
    2 mol.
Avogadro's no. of atoms No 6.022 1023

Q.16. Calculate the number of particles in each of the following :

(i) 46 g of sodium atoms (ii) 8 got oxygen (O2 ) (iii) 0 . 1 mole of carbon atoms.
Ans. (i) 46 g of sodium

No. of particles (atoms)

Given mass (m)

=  Avogadro's no. (No)
Molar mass (M)
(46 g )
  (6.022 1023 )  1.205 1024 atoms
(23 g )
(ii) 8 g of oxygen (O2 )
No. of particles (molecules)
Given mass(m)
 x Avogadro s no. (No)
Molar mass(M)
(8 g )
  (6.022 1023 )  1.52 1023 molecules
(32 g )
(iii) 0 . 1 mole of carbon atoms
No. of particles (atoms) = No. of moles (n) x Avogadro's no. (No)
= (0 . 1) x (6 . 022 x 1023 ) = 6 . 022 x 1022 atoms.

Q.17. If one mole of carbon weighs 12 grams, what is the mass (in gram) of one atom of carbon ?
Ans. Molar mass of carbon = 12 g
Now, 6 . 022 x 1 1023 atoms of carbon have mass = 12 g
 one atom of carbon has mass
(1 atom)
= (12 g) x
(6.022×1023 atoms)
= 1 . 99 x 1023 g

Q.18. Which has more number of atoms ?

(a) I00 grams of sodium (b) 100 grains of iron
(Given : atomic mass of Na = 23u ; Fe = 56u)
Ans. (a) 100 grams of sodium
(Given mass)
No. of sodium atoms =  (Avogadro’s no.)
(Gram atomic mass)

23
 (100 g )
  (6.022 1023 )  2.618 1024 atoms
(23 g )

(b) 100 grams of iron

No. of iron atoms

(Given mass)
  (Avogadro’s no.)
(Gram atomic mass)

(100 g )
  (6.002 1023 )  1.075 1024 atoms
(56 g )

100 g of sodium have more number of atoms

N.C.E.R.T. EXERCISE
Q.19. 0.24 g of sample of a compound of oxygen and boron was found by analysis to contain 0.096
g boron and and 1.144 g of oxygen. Calculate the percentage composition of the compound
by weight.
Ans. Please consult solved example 3.5.

Q.20. When 3 . 0 g of carbon is burnt in 8 . 0 g of oxygen, 11 . 0 g of carbon dioxide is formed. What

mass of carbon dioxide will be formed when 3 . 0 g of carbon is burnt in 50 . 0 g of oxygen ?
Which law of chemical combination will govern your answer ?
Ans. Carbon and oxygen react to form carbon dioxide according to the equation
Carbon (C) + Oxygen (O2 )  Carbon dioxide (CO2 )
In CO2 , the element carbon and oxygen are combined as follows :
C + 2O  CO2
12g 2×16=32g 12+32=44g
(after simplifying) 3 g 8 g 11 g.
In the first case :
3 . Og of carbon are burnt in 8 . 0 g of oxygen to form 11 . 0 g of CO2
In the second case :
3 . Og of carbon must also combine with 8 . 0 g of oxygen only. This means that (50  8) 42 g of oxygen
will remain unreacted.

24
 The mass of CO2 in this case must be also 11 g.
The answer is based on Law of constant proportions. For the statement of the law, consult text-part.

Q.21. What are polyatomic ions ? Give examples.

Ans. Polyatomic ions are the group of atoms which carry either positive charge (cations) or negative charge
(anions). For example,
(i) Carbonate ion (CO3 )2 (ii) Nitrate ion (NO3 )
(iii) Ammonium ion (NH4 ) (iv) Phosphate ion (PO4 )3 .

Q.22. Write the chemical formulae of the following :

(a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium
chloride (e) Calcium carbonate (f) Zinc sulphate.
Ans.

Q.23. Give the names of the elements present in the following compounds :
(a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate.
Ans. The names of elements present can be given only if the chemical formula of the compound is known. For
example,
(a) Quick lime : It is the commercial name of the compound. Its chemical name is oxide and the
chemical formula is CaO.
Elements present: calcium (Ca) : oxygen (0)
(b) Hydrogen bromide : The chemical formula of the compound is HBr
Elements present : hydrogen (H) ; bromine (Br)
(c) Baking powder : It is the commercial name of the compound. Its chemical name is sodium
bicarbonate and the chemical formula is NaHCO3
Elements present : sodium (Na), hydrogen (H), carbon (C), oxygen (0).
(d) Potassium sulphate : The chemical formula of the compound is K 2SO4
Elements present : potassium (K), sulphur (S), oxygen (0).

Q.24. Calculate the molar mass of the following substances :

(a) Ethyne, C2 H 2 (b) Sulphur molecule, S 8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31) (d) Nitric acid, HNO3
(e) Hydrochloric acid, HCI.
Ans. (a) Ethyne, C2 H 2
Molar mass of C2 H 2 = (2 x Atomic mass of C) + (2 x Atomic mass of H) = (2 x 12 u) + (2 x 1u) = 26 u.
(b) Sulphur molecule, S 8
Molar mass of S8 = 8 x Atomic mass of S = (8 x 32 u) = 256 u
(c) Phosphorus molecule, P4
Molar mass of P4 =4 x Atomic mass of P = ( 4 x 31 u) = 124 u
(d) Nitric acid, HNO3
Molar mass of HNO3 = (1 x Atomic mass of H) + (1 x Atomic mass of N) + ( 3 x Atomic mass of 0)
= (1 x 1u) + (1 x 14 u) + (3 x 16 u) = 63 u.
(e) Hydrochloric acid, HCI.
Molar mass of HCI = (1 x Atomic mass of H) + (1 x Atomic mass of CI)

25
 = (1 x 1 u) + (1 x 35 . 5u) = 36 . 5u.

Q.25. What is the mass of :

(a) 1 moles of aluminium atoms
(b) 4 moles of aluminium atoms (Atomic mass of aluminum =27)
(c) 10 moles of sodium sulphite (Na2SO3 ) ?
Ans. (a) 1 mole of nitrogen atoms
Mass of 1 mole of nitrogen (N) atoms = 14 u
(b) 1.4 moles of aluminium atoms
Mass of 1 mole of aluminium (All atoms = 27u
Mass of 4 moles of aluminium (Al) atoms = 4 x 27 = 108 u
(c) 10 moles of sodium sulphite (Na2SO3 )
Molar mass of Na 2SO3 = 2 x Atomic mass of Na + Atomic mass of S + 3 x Atomic mass of O
= 2 x 23 + 32 + 3 x 16 = 126u
1 mole of sodium sulphite has mass = 126u
10 moles of sodium sulphite have mass = 10 x 126 = 1260u.

Q.26. Convert into moles

(a) 12g of oxygen gas (b) 20g of water (c) 22g of carbon dioxide.
Ans. (a) 12g of oxygen gas (O2 )
Molar mass of oxygen (O2 ) =(2 x 16u)=32u=32 g
32 g of oxygen (O2 ) = 1 mol
(12 g )
12 g of oxygen (O2 ) = (1 mol) x  0 . 375 mol
(32 g )
(b) 20g of water (H 2O)
Molar mass of water (H2 O) = (2 x 1u) + (1 x 16u) = 18u = 18g
18 g of water (H2 O) = 1 mol
(20 g)
20 g of water (H2 O) = (1 mol) x  1.11 mol
(18g)
(c) 22g of carbon dioxide (CO2 )
Molar mass of carbon dioxide (CO2 ) = (1 x 12 u) + (2 x 16u) = 44 u = 44g
44 g of carbon dioxide (CO2 ) = 1 mol
22 g of carbon dioxide (CO2 ) = (1 mol)
(22 g )
x  0.5 mol
(44 g )

Q.27. What is the mass of :

(a) 0 . 2 mole of oxygen atoms ?
(b) 0 . 5 mole of water molecules ?
Ans. (a) 0 . 2 mole of oxygen atoms
Mass of 1 mole of oxygen (0) atoms = 16 u
Mass of 0 . 2 mole of oxygen (0) atoms = 0 . 2 x 16 = 3 . 2 u
(b) 0 . 5 mole of water molecules
Mass of 1 mole of water (H2 O) molecules
=2 x 1 + 16 =18 u
Mass of 0 . 5 mole of water (H2 O) molecules
= 0 . 5 x 18 = 9 u.

Q.28. Calculate the number of molecules of sulphur (S8 ) present in 16 g of solid sulphur.
Ans. Gram molecular mass of sulphur (S8 ) = 8 x Gram atomic mass of sulphur
= 8 x 32 g = 256 g.
No. of Ss molecules in 256 g of sulphur = 6 . 022 x 1023

26
 (16 g)
No. of SB molecules in 16 g of sulphur  x 6 . 022 x 1023 = 3 . 76 x 1022 molecules.
(256 g)

Q.29. Calculate the number of aluminium ions in 0 . 051 g of aluminium oxide (AI 2O3 ).
Ans. Step I : Calculation of no. of moles in 0 . 051 g of AI O2 3
Gram molecular mass of AI2 O3 = 2x Gram atomic mass of Al + 3 x Gram atomic mass of O
= (2 x 27 g) + (3 x 16 g) = 102 g.
Now, 102 g of AI2 O3 = 1 mol
(0.051 g )
 0 . 051 g of AI2 O3 = x(1 mol)
(102 g )
= 0 . 0005 mol
Step II : Calculation of no. of Al ions in 0 . 001 mole of AI2 O3
1 mole of AI2 O3 contain Al atoms = 2 x No 0 . 0005 mole of AI2 O3 contain Al atoms = 2 x 0 . 005 x N 0
= 2 x 0 . 0005 x 6 . 022 x 1023 = 6 . 022 x 1020 atoms In AI O the valency of Al = 3 +
2 3
No. of aluminium ions (Al3+) present is the same as the no. of Al atoms
No. of Al3+ ions - 6 . 022 x 1020 ions.

CONCEPT BASED QUESTIONS

Q.1. Why are chemical reactions according to law of conservation of mass ?

Ans. In all chemical reactions, there is only exchange of reactants taking place when products are formed. Since
there is no loss or gain of mass, the chemical reactions are according to law of conservation of mass.

Q.2. Give one limitation of law of constant composition.

Ans. The law of constant composition does not hold when different isotopes of an element take part in the
formation of a particular compound. For
example,
In carbon dioxide (CO2 )
By using C-12 isotope, the ratio of C : O is 12 : 32.
By using C-14 isotope; the ratio of C : O is 14 : 32.
This means that the two ratios are not the same.

Q.3. In what respect does Dalton's Atomic theory hold good even today ?
Ans. Atom has been found to be made up of sub-atomic particles like electrons, protons and neutrons. But it is
still the smallest particle of matter which can take part in chemical combination.

Q.4. Atoms are not always combined in simple whole number ratios to form molecules. Given
one example.
Ans. The chemical formula of sucrose (cane sugar) is C12 H22O11. In this, the ratio of the elements present is 12 :
22 : 11. It is not a simple whole number ratio.

Q.5. What is basic difference between atoms and molecules ?

Ans. Atoms except those of noble or inert gas elements cannot exist of their own. However all molecules can
have independent existence.

Q.6. Why does not the atomic mass of an element represent the actual mass of its atom ?
Ans. Atoms of different elements are very small in size and their actual masses are extremely small. For
example, the mass of an atom of hydrogen is 1 . 67 x 10 - 27 kg. To solve this problem, we consider the
relative atomic masses of the elements. The relative atomic mass of hydrogen is 1 u and corresponding
gram atomic mass is 1 g.

Q.7. The atomic mass of an element is in fraction. What does it mean ?

27
Ans. If the atomic mass of an element is in fraction, this means that it exists in the form of isotopes. The atomic
mass is the average atomic mass and is generally fractional.

Q.8. Atoms of inert gas elements are monoatomic while of the other elements are not. Assign
reason.
Ans. The atoms of inert gas (also called noble gas) elements have stable electronic configuration (we shall
discuss in chapter 4). Therefore, they can exist independently and are monoatomic in nature. The atoms
of all other elements are yet to have stable configuration. To achieve it, they combine with the atoms of
the other elements.

Q.9. What is difference between the mass of a molecule and molecular mass ?
Ans. Mass of a molecule is that of a single molecule also known as its actual mass. But molecular mass is the
mass of Avogadro's number (6 . 022 x 1023) of molecules.

Q.10. Why is the value of Avogadro's number 6 . 022 x 1023 . and not any other value ?
Ans. It represents the number of atoms in one gram atom of an element or the number of molecules in one
gram mole of a compound. If we divide the atomic mass of an element by actual mass of its atom, the
value is 6 . 022 x 1023. Similarly, by dividing the molecular mass of a compound by the actual mass of its
molecule, the same result is obtained.

Q.11. Where do we use the wordy mole and mol ?

Ans. In the text part, we use the word mole while as a unit, we call it mol.

Q.12. Why is it necessary to balance a chemical equation ?

Ans. A chemical equation has to be balanced in order to satisfy the law of conservation of mass. According to
the law, there is no change in mass when the reactants change into the products. Therefore, the chemical
equation has to be balanced.

VERY SHORT ANSWER QUESTIONS

Q.1. Do atoms in some of the elements have actually fractional mass?

Ans. No, the atoms do not have fractional mass. Only their average comes out to be in fraction in elements
which exist as isotopes. For details, consult text part.

Q.2. Out of atoms and molecules which can exist independently ?

Ans. Molecules can exist independently. However, the atoms of noble gases (He, Ne, Ar, Kr, Xe) can also exist
independently.

Q.3. What does the symbol 'u' represent ?

Ans. The symbol 'u' represents unified mass.

Q. 4. What is the atomicity of ammonia ?

Ans. Atomicity of ammonia (NH3 ) is 4 because one molecule of NH3 has four atoms.

Q.5. Calculate the molar mass of ethyl alcohol (C2 H5OH).

Ans. Molar mass of C2 H5OH = (2 x Atomic mass of C) + (6 x Atomic mass of H)
+ (1 x Atomic mass of 0)
= (2 x 12u) + (6 x 1u) + (1 x 16u) = 46u.

Q.6. Does CO represent the symbol of element cobalt ?

Ans. No, the symbol of element cobalt is Co.

Q.7. Suggest an instrument with the help of which electrons can be seen ?
Ans. It is a microscope called scanning tunneling microscope (STM) which helps in noticing electrons.

28
Q.8. What is the valency of calcium is CaCO3 ?
Ans. The valency of Ca in CaCO3 is 2 + (i.e. Ca2+).

Q.9. What happens to an element 'A' if its atom gains two electrons ?
Ans. It changes to a divalent anion (A2-).

Q.10. Why is a cation so named ?

Ans. When electric current is passed through the solution of a salt like sodium chloride (NaCl), the positive ion
(Na+) migrates towards cathode (negative electrode). It is therefore, called cation. Please remember that
 Positive ion migrating towards cathode on passing electric current is known as cation.
 Negative ion migrating towards anode on passing electric current is known as anion.

Q.11. An element Z forms an oxide with formula Z2O 3 . What is its valency ?
Ans. The valency of the element in Z2 O 3 is 3+.

Q.12. The valency of an element A is 4. Write the formula of its oxide.

Ans. The formula of the oxide is A 2 O4 or AO2 .

Q.13. An element X has valency 3 while the element Y has valency 2. Write the formula of the
compound between X and Y.
Ans. The formula of the compound between X and Y is X2 Y3 .

Q.14. Formula of the carbonate of a metal M is M2CO3 . Write the formula of its chloride.
Ans. The valency of the metal (M) in M2 CO3 is (2+) i.e. metal exists as M2+. Therefore, the formula of metal
chloride is MCI2 .

SHORT ANSWER QUESTIONS

Q.1. How many moles are present in 11- . 5 g of sodium ?
Ans. Gram atomic mass of Na =23 g
23 g of Na represent = 1 mol
(1 mol)×(11.5g) = .
11 . 5 g of Na represent = 0 5 mol.
(23.0g)

Q.57. The mass of an atom of element (X) is 2 . 0 x 10-23 g. Calculate its atomic mass.
Ans. One atom of element (X) has mass = 2 . 0 x 10-23 g.
6 . 022 x 1023 atoms of element (X) will have mass = (2 . 0 x 10-23 g) x (6 . 022 x 1023)
= 12 . 044 g  12 . 0 g.
Q.58. How many particles are represented by 0 . 25 mole of an element ?
Ans. 1 . 0 mole of the element represents particles = 6 . 022 x 1023
6.022 x l023 x (0.25 mol)
0 . 25 mole of the element represents particles =
(1.0 mol)
= 1 505 x 10 particles
. 23

Q.59. Out of 4g of methane (CH4 ) and 11 g of CO2 which has more molecules ?
(1 mol)  (4 g)
Ans. 4g of methane CH 4 represent moles = = 0 . 25 mol.
(16 g )
(1 mol)  (11 g)
11 g pf CO2 represent moles =  0 . 25 mol
(44 g )
Since both CH 4 and CO2 have same number of moles (0 . 25 mol), they have also same number of
molecules.

29
Q.60. What is the number of molecules present 1 . 5 mole of ammonia (NH 3 ) ?
Ans. 1 . 0 mole of ammonia (NH ) contains molecules = 6 . 022 x 1023
3

(6.022  1023 )  (1.5 mol)

1 . 5 mole of ammona (NH3 ) contain molecules 
(1.0 mol)
= 9 . 033 x 1023 molecules.

Q.61. Four samples of carbon dioxide (CO2 ) were prepared by using different methods. Each
sample on analysis was found to contain 37 . 5%, carbon. Name the law which is in
agreement with this observation.
Ans. Carbon dioxide consists of elements carbon and oxygen. Since the percentage of carbon in each sample is
fixed, that of oxygen must be also fixed. This is according to law of constant proportions.

Q.62. Explain why the number of atoms in one mole of hydrogen gas is double the number of
atoms in one mole of helium gas.
Ans. Hydrogen gas is diatomic in nature (H 2 ) while helium gas is monoatomic (He). As a result, the number of
atoms in one mole of hydrogen. (2×NA ) are expected to be double as compared to number of atoms in
one mole of helium (N A )

Q.63. If the valency of carbon is 4 and that of sulphur is 2, write the formula of the compound
formed between carbon and sulphur atoms. Also name the compound.
Ans. The formula of compound can be written by exchanging the valencies (cross-over). Therefore, the
expected formula is C2S4 or CS2 . The compound is called carbon disulphide.

Q.64. What Is wrong the statement ‘1 mole of hydrogen' ?

Ans. The statement is not correct. We must always write whether hydrogen is in atomic form or molecular form.
The correct statement is : 1 mole of hydrogen atoms or one mole of hydrogen molecules.

LONG ANSWER QUESTIONS

Q.65. ‘The ratio by mass of the elements which constitute a particular compound always
remains fixed; whatever may be its source of formation. ‘’Do you agree with the
statement. Justify your answer.
Ans. For answer consult text part.

Q.66. 'Most of the postulates of the Dalton's atomic theory were found to be incorrect at a later
stage'
Support this statement.
Ans. For answer consult text part.

Q.67. A mole is quite of then known as chemist's. dozen. Why it is so named ?

Ans. A dozen represents a fixed number of articles i.e. 12. Similarly, a mole represents a fixed number of
particles i.e. Avogadro's number (N A ) or 6.022 x 1023. For details, consult text part.

Q.68. (a) The mass of one molecule of a substance is 4.65 x 10-23 g. What is its molecular mass ?
What could this substance be ?
(b) Which have more molecules ? 10 g of sulphur dioxide (SO2 ) or 1O g of oxygen (O2 ) ?
Ans. (a) Mass of one molecule of substance = 4.65 x 10-23
Mass of 6.022 x 1023 molecules of substance
= 6.022 x 1023 x (4. 66 x 10-23) =28 g
The substance can be carbon monoxide (CO) with molecular mass
= 12 + 16 = 28 u or 28 g
(b) No of molecules in 10 g of sulphur dioxide (SO2 )

30
 Molar mass of SO2 = 32 g + 2 x 16 g = 64 g
64 g of sulphur dioxide represent molecules = 6.022 x 1023
6.022 1023  (10 g)
1O g of sulphur dioxide represent molecules = = 9.40 x 1022
(64 g )
No. of molecules in 10 g of oxygen (O2 ) = 9.40 x 1022
Molar mass of oxygen (O2 ) = 32 g
32g of oxygen represent molecules = 6.022 x 1023
6.022  1023  (10 g )
10 g of oxygen represent molecules =
(32 g )
= 18.8 x 1022 '
Thus, 10 g of oxygen (CO2 ) has more molecules present.

Q.69. (a) An element shows variable valencies 4 and 6. Write the formulae of its two oxides.
(b) An element forms an oxide A2O5 .
(i) What is the valency of the element A ?
(ii) What will be the formula of the chloride of the element ?
Ans. (a) Let the element be represented by the symbol E.
Formula of oxide in which valency of E is 4 = E 2 O4 or EO2
Formula of oxide in which valency of E is 6 = E 2 O6 or EO3
(b) Formula of oxide of the element = A 2 O5
(i) The valency of the element A in the oxide = 5 +
(ii) The formula of the chloride of the element A = ACI5 .

HINGHER ORDER THINKING QUESTIONS

(based on new c.b.s.e. sample papers)

Q.70. 0 . 44 g of a hydrocarbon on complete. combustion with oxygen gave 0 . 88 g of carbon

dioxide and 1 . 8 g of water. Show that the results are in agreement with the law of
conservation of mass.
Ans. Hydrocarbon is a compound of carbon and hydrogen.
The amount of hydrocarbon reacted = 0 . 44 g.
Let us calculate the amount of carbon and hydrogen which are present in carbon dioxide and water
respectively. These are the products.
Calculation of mass of carbon (C)
Mass of carbon dioxide (CO2 ) formed = 0 . 88 g
CO2 C
(12  2 16)  44 g (112)  12 g
Now, 44 . 0 g of carbon dioxide (CO2 ) contain
C = 12 . 0 g
 0 . 88 g of carbon dioxide (CO2 ) contain
(0.88 g )
C = (12 . 0 g) x = 0 . 24 g
(44.0 g )
Calculation of mass of hydrogen (H)
Mass of water (H2 O) formed = 1 . 8 g
H2O  2H
(2 x 1 + 16) = 18 g (2 x 1) 2g
Now, 18 g of water ( (H2 O) contain H = 2 . 0 g
 1 . 8 g of water (H2 O) contain

31
 (1.8 g )
H = (2 . Og) x 
(18.0 g )
= 0 . 20 g
Total mass of C and H in the products = (0 . 24 + 0 . 20) = 0 . 44 g
This mass comes out to be the same as the mass of the hydrocarbon. The data is in agreement with the
law of conservation of mass.

Q.71. On analysing an impure sample of sodium chloride, the percentage of chlorine was found to
be 45 . 5. What is the percentage of pure sodium chloride in the sample ?
Ans. Molecular mass of pure NaCI = Atomic mass of Na + Atomic mass of Cl
= 23 + 35 . 5 = 58 . 5u
Percentage of chlorine in pure NaCI
35.5
= x 100 = 60 . 6
58.5
Now, if chlorine is 60 . 6 parts, NaCI = 100 parts
45.5
if chlorine is 45 . 5 parts, NaCI = x 100 = 75
60.6
Thus, percentage of pure NaCI = 75%.

Q.72. A flask P contains 0 . 5 mole of oxygen gas. Another flask Q contains O . 4 rnole of ozone
gas. Which of the two flasks contains greater number of oxygen atoms ?
Ans. 1 molecule of oxygen (O2 ) = 2 atoms of oxygen
1 molecule of ozone (O3 ) = 3 atoms of oxygen
In flask P : 1 mole of oxygen gas = 6 . 022 x 1023 molecules
0 . 5 mole of oxygen gas = 6 . 022 x 1023 x 0 . 5 molecules
= 6 . 022 x 1023 x 0 . 5 x 2 atoms
= 6 . 022 x 1023 atoms
In flask Q : 1 mole of ozone gas = 6 . 022 x 1023 molecules
0 . 4 mole of ozone gas = 6 . 022 x 1023 x 0 . 4 molecules
= 6 . 022 x 1023 x 0.4 x 3 atoms
= 7 . 23 x 1023 atoms
 Flask Q has a greater number of oxygen atoms as compared to flask P.

Q.73. What weight of calcium the same number of atoms as are present in 3 . 2 g of sulphur ?
Ans. Step I. No. of atoms in 3 . 2 g of sulphur
Gram atomic mass of S = 32 g
32 g of sulphur contain atoms = 6 . 022 x 1023
6.022 1023
3 . 2 g of sulphur contain atoms = x 3.2
32
= 6 . 022 1022
Step II. Weight of 6 . 022 x 1022 atoms of calcium.
Gram atomic mass of Ca = 40 g
6 . 022 x 1023 atoms of Ca weigh = 40 g
40
6 . 022 x 1022 atoms of Ca weigh = x 6 . 022 x 1022 . 4 g.
6.022 1023

MULTIPLE CHOICE QUESTIOS

Select the correct answer :

1. One 'u' stands for :
1
(a) An atom of carbon (C—12) (b) th of mass of carbon atom (C-12)
12
1
(c) th of hydrogen atom. (d) one atom of all the elements.
12

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2. How many times an atom of sulphur is heavier than an atom of carbon ?
(a) 32 times (b) 12 times (c) 8/3 times (d) 12/32 times.
3. 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.
4. .
A sample of CaCO3 contains 3 01 x 10 ions of Ca and Ca 32- . The mass of the sample is :
23 2+

(a) 100 g (b) 50 g (c) 200 g (d) 5 g.

5. Which has maximum number of molecules ?
(a) 1 g of CO2 (b) 1 g of N 2 (c) 1 g of H 2 (d) 1 g of CH 4 .
6. The law of constant proportion was given by
(a) Dalton (b) Berzelius (c) Proust (d) Lavoisier.
7. The law of constant proportion is applied to
(a) Any element (b) Any chemical compound
(c) Pure chemical compound (d) None of these.
8. In carbon disulphide (CS2 ), the mass of sulphur in combination with 3 . 0 g of carbon is :
(a) 4 . 0 g (b) 6 . 0 g (c) 64 . 0 g (d) 16 . 0 g.
9. 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.
10. The indivisibility of matter was proposed by :
(a) Rutherford (b) Dalton (c) Bohr (d) Einstein.
11. The value of Avogadro's constant is :
(a) 6 . 0 x 1024 (b) 6 . 01 x 1022 (c) 6 . 022 x 1023 (d) 6 . 022 x 10-23.
12. The atomic mass of calcium is 40u. The number of calcium atoms in 0 . 4u of calcium is :
(a) 6 . 022 x 1010 (b) 6 . 022 x 1022 (b) 6 . 022 x 1023 (d) 6 . 022 x 1021.
13. A sample of pure water irrespective of its source contains 88 89% oxygen and 11 . 11% hydrogen by
.
mass. The data supports :
(a) Law of conservation of mass (b) Law of constant composition
(c) Dalton's Atomic theory. (d) Avogadro's Law.
14. The atomic mass of calcium (Ca) is 40. The number of moles in 60 g of calcium are :
(a) 0 . 5 mol (b) 2 . 0 mol (c) 1 . 5 mol (d) 0 . 75 mol
15. Which is not correct according to Dalton's atomic theory?
(a) Atoms are indivisible. (b) Atoms combine in simple whole number ratios.
(c) All atoms of an element may not have same mass.
(d) Atoms of different elements have same masses.
16. Percentage of calcium in calcium carbonate is:
(a) 40 (b) 30 (c) 48 (d) 36
17. Correct formula of ferric sulphate is :
(a) FeSO4 (b) Fe(SO4 )2 (c) Fe3 (SO4 )2 (d) Fe2 (SO4 )3
18. A chemical equation is always balanced to fulfill the condition of:
(a) Dalton's atomic theory (b) Law of constant composition
(c) Law of multiple proportions. (d) Law of conservation of mass.
19. No. of moles of 0 . 6 g of SO2 is :
(a) 100 (b) 10 (c) 0 . 01 ( d) 0 . 1.
2O. The number of molecules in 4 . 25 g of NH3 approximately :
(a) 1 . 5 x 1023 (b) 2 x 1023 (c) 4 x 1023 (d) 6 x 1023

ANSWERS

1. (b) 2. (c) 3. (d) 4. (b) 5. (c) 6. (c) 7. (c) 8. (d) 9. (b) 10. (b) 11. (c) 12. (d) 13. (b) 14.
(c)
15. (c) 16. (a) 17. (d) 18. (d) 19. (c) 20. (a)

HINTS & EXPLANATIONS

1. (b) is the correct answer.
2. (c) 32/12 = 8/3 times.

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3. (d) 44 g of CO2 has oxygen atoms  2  6.022 1023
4.4 g of CO has oxygen atoms
2

 2  6.022 1023  4.4

 1.2 1023.
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4. (b) 6.022 1023 of Ca 2+ and Ca 32- ions are present in CaCO3  100 g
3.011023 Ca 2+ and Ca 2- ions are present in
3

(100 g )  3(3.011023 )
CaCO3   50 g
(6.022 1023 )
5. (c) 1g of Hg (0 . 5 mol) has max. no. of molecules i.e. 3 . 011 x 1023 molecules.
6. (d) is the correct answer
7. (c) is the correct answer
8. (d) In compound CS2
12 g of carbon are combined with sulphur = 64 g
(64 g )  (3 g )
3g of carbon are combined with sulphur   16 g
(12 g )
9. (b) 6 . 022 x 1023 molecules of CO correspond to mass = 44g
2
3 . 011 x 1023 molecules of CO2 correspond to mass = 22 g
10. (b)is the correct answer
11. (c) is the correct answer
12. (d) is the correct answer.
13. (b) The data supports Law of constant composition.
(1mol)×(60g) .
14. (c) No. of moles in 60 g of calcium = =1 5mol.
(40g)
15. (c) All atoms of an element have same mass according to Dalton's atomic theory.
(40g)
16. (a) Percentage of Ca = = 100 = 40.
(100g)
17. (d) is the correct answer.
18. (d) The equation must be balanced according to law of conservation of mass.
(0.64 g)
19. (c) No. of moles in 0 . 64 g of SO2 = = 0 . 01 mol.
(64g mol-1 )
20. (a) 4 . 25 g NH (0 . 25 mole) has appoximately 1 . 5 x 1023 molecules.
3

TEST YOUR KNOWLEADGE

VERY SHORT ANSWER QUESTIONS
1. What is the smallest particle of matter according to Dalton ?
2. A sample of water from a well was analysed. What will be the ratio of hydrogen and oxygen in it by mass?
3. Do all atoms of a particular element have always the same mass ?
4. Write the symbol of the following of elements :
(i) Boron (ii) Bismuth (iii) Barium (iv) Bromine.
5. The formula of a compound is Ca 3 (PO4 )2 . What is the valency of Ca in it ?
6. Which is used as the standard for comparing the atomic and molecular masses of substances ?
7. What is the value of Avogadro's number of particles ?
8. Give the formula of the compound formed by combining Fe2+ and CI- ions.
9. Give one example each of polyatomic cation and polyatomic anion.
10. Define the term mole.
11. What do the following denote (i) N (ii) N 2 ?
12. Write the chemical formulae of (i) sodium carbonate (ii) calcium sulphate ?
13. Explain gram atomic mass in terms of mole concept.
14. Why do mixtures not obey the law of constant composition ?
15. What do we call the particles which have less electrons than the normal atoms ?

34
16. What is the valency of PO4 in the compound Ca 3 (PO4 )2 ?
17. Calculate the molar mass of sulphuric acid (H2SO4 ).
18. What is atomicity ? Give one example each of diatomic and triatomic substances.
19. An element B shows the valency of 4+ and 6+. Write the formulae of its oxides.
20. How many atoms are present in 0 . 25 mole of hydrogen (H) ?

SHORT ANSWER QUESTIONS

21. State law of conservation of mass. Give an evidence in support of the law.
22. Point out the limitations of the law of constant composition if any.
23. Give three postulates of Dalton's atomic theory.
24. Most of the postulates of Dalton's Atomic theory have been found to be defective. Justify.
25. Give two points of distinction between atom and molecule.
26. Explain why some elements occur in atomic form while others in molecular form.
27. How many molecules and atoms of phosphorus are present in 0 . 1 mole of P4 molecule ?
28. How many atoms of carbon and oxygen are present in 1 . 5 mole of CO ? 2
29. A mole is quite often called chemist's dozen. Explain.
30. How many moles and atoms of gold are present in 49 . 25 g of gold ? (Atomic mass of gold = 97u)
31. Calculate the number of moles in 17 g of hydrogen peroxide (H2 O2 ).
32. Convert into moles (a) 12 g of oxygen gas (b) 20 g of water.
33. An element B forms an oxide B2 O3 . Find its valency and write the formula of its bromide.

LONG ANSWER QUESTIONS

34. Write the chemical formulae of :

(i) Barium chloride (ii) Magnesium sulphate (iii) Sodium sulphate (iv) Ammonium
nitrate (v) Calcium phosphate (vi) Potassium dichromate.
35. Name the elements which constitute ?
(i) Washing soda (ii) Baking soda (iii) Quick lime Also find their percentages in each compound.
36. Enlist the main postulates of Dalton's Atomic theory. Give two limitations of the theory.
37. Explain the following : (i) Law of conservation of mass (ii) Gram atomic mass
(iii) Gram molecular mass.
38. The mass of copper oxide obtained by reacting 2 . 16 g of metallic copper with nitric acid followed by
strong heating was found to be 2 . 7 g. In another experiment, 1 . 15 g of copper oxide on reduction
yielded 0 . 92 g of copper. Show that this data illustrates the law of constant composition.
39. What is the necessity of mole ? Describe gram atomic mass and gram molar mass in terms of mole
concept. Give one example in each case to support your answer.
40. What weight of calcium will contain the same number of atoms as are present in 3 . 2 g of sulphur.

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