Atoms and Molecules

ATOMS AND MOLECULES (AM-1)

(AM-1)
1.0 INTRODUCTION
In the previous chapters, we have discussed the physical as well as the chemical classification of matter. We
know that the chemical compounds are formed as a result of the combination of elements.
The combination must be based on certain guidelines. As we see in a compound like carbon dioxide (CO 2),
the element carbon & oxygen are combined in certain fixed ratio by mass. Carbon dioxide can have no
other formula except CO2.
Here we shall discuss the basis of the combination. We shall represent the elements & compounds by chemical
symbols & formula respectively. In addition to these, we shall discuss to various ways in which the masses
of the elements & compounds have been expressed.
The structure of matter has been a subject of speculation from very early times. The idea of divisibility of
matter was considered long back in India, around 500 BC. An Indian philosopher Maharishi Kanad, postulated
that if we go on dividing matter (padarth), a time will come when we shall come across the smallest particles
beyond which further divisions will not be possible . He named these particles Parmanu.
Around the same era, ancient Greek philosophers : Democritus and Leucippus suggested that if we go
on dividing matter, a stage will come when particles obtained cannot be divided further. Democritus called
these indivisible particles, atoms (meaning indivisible).
1.1 LAWS OF CHEMICAL COMBINATION
By studying the results of quantitative measurements of many reactions it was observed that whenever substances
react, they do so according to certain laws. These laws are called the Laws of chemical combination.
There are two important laws of chemical combination. These are :
1. Law of conservation of mass. 2. Law of constant proportions.
1.1.1 Law of Conservation of Mass
This law was stated by the French chemist Antoine Laurent Lavoisier (1774). This law states that :
During any physical or chemical change, the total mass of the products remains equal to the
total mass of the reactants.
ACTIVE CHEMISTRY 1
Aim : To demonstrate the law of conservation of mass.
Method
(i) Take one of the following sets, X and Y of chemicals,
X Y
(i) Copper sulphate Sodium carbonate
(ii) Barium chloride Sodium sulphate
(iii) Lead nitrate Sodium chloride
(ii) Prepare separately a 5% solution of any one pair of substance listed under X and Y in water.
(iii) Take a little amount of solution of Y in a conical flask and some solution of X in an ignition tube.
(iv) Hang the ignition tube in the flask carefully, see that the solutions do not get mixed. Put a cork on the flask
(see fig.)
Cork
Thread
Conical flask

Solution of X

Solution of Y

Precipitation reaction.

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Class IX : Chemistr y
(v) Weigh the flask with its contents carefully.
(vi) Now tilt and swirl the flask, so that the solution X and Y get mixed.
(vii) Weigh again.

Now answer
(i) What happens in the reaction flask?
(ii) Do you think that a chemical reaction has taken place?
(iii) Why should we put a cork on the mouth of the flask?
(iv) Does the mass of the flask and its contents change?

Discussion and conclusion

It is observed that on mixing two solution chemical reaction takes place which is indicated by the formation
of a white precipitate in this case.
Barium chloride + Sodium sulphate Barium sulphate (white ppt) + Sodium chloride.
The mass of the flask and its contents remains constant. Thus, during a chemical reaction mass is neither
created nor destroyed.
From the above activity we conclude that according to law of conservation of mass,
The mass can neither be created nor destroyed in a chemical reaction.
The total mass of the products of a chemical reaction is equal to the total mass of the reactants that have
combined. This law is also known as the law of indestructibility of matter.

1.1.2 Law of Constant Composition or Definite Proportion

This law deals with the composition of chemical compounds. It was discovered by the French chemist, Joseph
Proust (1799). This law states that :
A chemical compound always contains same elements combined together in same proportion
by mass.
It implies from this law that in a chemical compound the elements are present in fixed and not arbitrary ratio
by mass. For example, pure water obtained from different sources such as river, well, spring, sea, etc., always
contains hydrogen and oxygen combined together in the ratio 1:8 by mass. Similarly, carbon dioxide can be
obtained by different methods such as :
(a) burning of carbon,
(b) heating limestone, or
(c) the action of dilute hydrochloric acid on marble pieces.
It can be shown experimentally that different samples of carbon dioxide contain carbon and oxygen in the
ratio of 3 : 8 by mass.

Illustration 1. What mass of silver nitrate will react with 5.85 g of sodium chloride to produce 14.35 g of silver
chloride & 8.5 g of sodium nitrate if the law of conservation of mass is true ?
Solution The reaction is : silver nitrate + sodium chloride silver chloride + sodium nitrate
According to law of conservation of mass.
Total mass of reactants = Total mass of products
Mass of AgNO3 + 5.85 g = 14.35 g + 8.5 g
Mass of AgNO3 = 22.85 – 5.85 = 17.0 g

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 Atoms and Molecules
Illustration 2. Copper oxide was prepared by 2 different methods. In one case 1.74 g of the metal gave 2.19 g
of oxide. In the 2nd case, 2.9 g of the metal gave 3.65 g of the oxide. Show that the given data
illustrate the law of constant proportion.
Mass of copper
Solution Case-I : % of copper in the oxide = × 100
Mass of copper oxide
1.74
= × 100 = 79.4%
2.19
% of oxygen = 100 – 79.4 = 20.6 %

2.9
Case-II : % of copper in the oxide = × 100 = 79.4%
3.64
% of oxygen = 100 – 79.4 = 20.6%

1.2 DALTON'S ATOMIC THEORY

The next problem faced by scientists was to give appropriate explanations of these laws. British chemist John
Dalton provided a theory about the nature of matter in 1808.
His theory was based on the laws of chemical combination. Dalton's atomic theory provided an explanation
for the law of conservation of mass and the law of definite proportions.
On the basis of the studies & investigations carried, he came out with a statement that the smallest portion
of matter which cannot be divided any further is an atom.
1.2.1 Postulates of Dalton's atomic theory
The important features of the Dalton's Atomic theory are listed below:
1. Every matter is made up of very small particles known as atoms.
2. Atoms are the ultimate particles of matter which cannot be created or destroyed in a chemical reaction
and cannot be further subdivided into smaller particles.i.e.atoms are “indivisible”.
3. All atoms of a particular element are identical in all respects. This means that they have same mass,
size and chemical properties
4. Atoms of different elements have different masses, sizes and chemical properties.
5. Atoms are the smallest particles of matter which can take part in chemical combination.
6. Atoms of the same or different elements combine in small whole number ratios to form molecules of
a compound.
7. The relative number and kinds of atoms are constant in a given compound.
8. Atoms of two different elements may combine in different ratios to form more than one compound.
For example, carbon and oxygen may combine to form carbon monoxide (CO) and carbon dioxide (CO2)
in which the ratios of the combining atoms (C and O) are 1 : 1 and 1 : 2 respectively.
1.2.2 Drawbacks of Dalton's atomic theory
Some of the drawbacks of the Dalton's atomic theory of matter are given below :
1. According to Dalton's atomic theory, atoms were thought to be indivisible. But is now known that atoms
can be further divided into still smaller particles called electrons, protons & neutrons.
2. Dalton's atomic theory said that all the atoms of an element have exactly the same mass. But it is now
known that atoms of the same element can have slightly different masses, as in case of isotopes.
3. Dalton's atomic theory said that atoms of different elements have different masses. But it is now known
that even atoms of different elements can have the same mass as in case of isobars.
4. Substances made up of the same kind of atoms may have different properties. For example, charcoal,
graphite & diamond are all made up of carbon atoms but have different physical properties.
5. The ratio in which the different atoms combine to form compound may be fixed and integral but not
be simple. For example, sugar (sucrose) molecule (C12H22O11) contains C, H & O in the ratio 12 : 22
: 11 which is integral and fixed but not simple.
The modifications, as given above, are called the postulates of "modern atomic theory".

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Illustration 3. Dalton’s atomic theory could not explain :

(A) Why atoms of same element have different mass numbers
(B) Why atoms of different elements have same mass number
(C) Both A and B
(D) Neither A nor B
Solution (C) Dalton’s atomic theory could not explain that atom of same element may have different mass
numbers (concept of isotopes) and atoms of different element have same mass number (concept of
isobar)

QUICK CHECK - 1
Objective Questions
1. Which of the following data illustrates the law of conservation of mass :
(A) 56g CO react with 32g of oxygen to produce 44g CO 2
(B) 1.70 g of AgNO3 react with 0.365g HCl to produce 1.435 g AgCl and 0.63g of HNO 3
(C) 12g C is heated in vacuum and on cooling there is no change in mass
(D) None of the above

2. If law of conservation of mass was to hold true, then 20.8 g of BaCl2 on reacting with 9.8 g of H2SO4 will
produce 7.3 g of HCl and BaSO4 equal to :
(A) 11.65 g (B) 23.3 g (C) 25.5 g (D) 30.6 g

3. Hydrogen and oxygen combine in the ratio 1 : 8 by mass to form water. What mass of oxygen will be required
to react completely with 4g of hydrogen : [NCERT]
(A) 10 gm (B) 48 gm (C) 32 gm (D) 64 gm

4. 0.24g of 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 of boron : [NCERT]
(A) 40% (B) 60% (C) 70% (D) 80%

5. Choose the correct option :

Statement-I-During any chemical change, the total mass of products is equal to the total mass of the reactants
Statement-II-According to Dalton, atoms can neither be created nor destroyed in a chemical reaction :
(A) Statement I is true but II is false
(B) Statement I is false but statement II is true
(C) Statement I and II are true but statement II is not the correct explanation of statement I
(D) Statement I and II are true and statement II is the correct explanation of statement I

Subjective Questions
6. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon
dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law
of conservation of mass. [NCERT]

7. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of hydrogen gas would be
required to react completely with 32 g of oxygen gas? [NCERT]

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 Atoms and Molecules

8. 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? [NCERT]

9. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass? [NCERT]

10. Which postulate of Dalton’s atomic theory can explain the law of definite proportions? [NCERT]

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ATOMS AND MOLECULES

(AM-2)

2. 0 WHAT IS AN ATOM ?
An atom is defined as the smallest particle of an element which may or may not be capable of free existence.
However, it is the smallest particle that takes part in a chemical reaction. An atom maintains its identity
in all physical changes and chemical reactions. For example, He, Ne, H, O, Fe, Cu, Zn etc.
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.
(i) Atoms of an element are different from those of any other element.
(ii) Free atoms, except those of noble gases, do not exist under normal conditions.

2.1 How big are atoms : atomic size

The size of an atom is extremely small. For your knowledge, the radius of an atom of hydrogen is only
10–10m. If we try to compare it with the radius of a grain of sand (10 –4 m) we can imagine about the size
of an atom. Same is the case with mass of the atoms of different elements. For example, an atom of hydrogen
has a mass nearly 1.6 × 10–27 kg.

Atomic radius is measured in nanometres.

1/109 m = 1 nm
1m = 109 nm

Relative R adii (in m) Example

–10
10 Atom of hydrogen
–9
10 Molecule of wa ter
–8
10 Molecule of haemoglobin
–4
10 Gra in of sand
–2
10 Ant
–1
10 Watermelon

2.2 ATOMIC SYMBOL

Symbol means a short hand method of representing the full name of an element.

2.2.1 Modern symbols

Modern symbols for the elements were introduced by J.J. Berzelius. These are also known as chemical symbols.
The symbol of an element are generally either the first letter or the first two letter or the first and the third
letters of the name of the element. For example the symbol of the following elements are the first letter of
the name of that element.

S.No. Element Symbol

1 Hydrogen H
2 Carbon C
3 Nitrogen N
4 Oxygen O
5 Fluorine F

Some symbols derived from the first two letters of the names of the element

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 Atoms and Molecules

S.No. Element Symbol

1 Aluminium Al
2 Barium Ba
3 Lithium Li
4 Neon Ne
5 Calcium Ca

Some symbols are derived from the first and the third letter of the name of the elements.
S.No. Element Symbol
1 Arsenic As
2 Magnesium Mg
3 Chlorine Cl
4 Zinc Zn
5 Chromium Cr
Though the names of most of the elements have been taken from English, there are some elements which
have been named from Latin and Greek.

Name of element La tin name Symbol

Silver Argentum Ag
Copper Cup rum Cu
Gold Aurum Au
Iron Ferrum Fe
Mercury Hydrargyrum Hg
Potassium Kalium K
Sodium Natrium Na
Lead Plumbum Pb
Antimony Stibium Sb
Tungsten Wolfram W
(German name)

2.2.2 Significance of Atomic Symbols

The symbol of an element has both qualitative as well as quantitative significance. These are ,
(i) the symbol stands for the name of the element.
(ii) the symbol stands for one atom of the element.
(iii) the symbol represents quantity of the element equal in mass to its atomic mass, or gram-atomic mass.
(iv) the symbol also represents mass of the element which contains one Avogadro’s number of atoms of that
element.
For example, the symbol O stands for
(i) the element oxygen.
(ii) one atom of oxygen.
(iii) the mass of oxygen equal to its gram-atomic mass, i.e., O represents 16 g of oxygen.
(iv) the mass of oxygen which contains one Avogadro’s number (= 6.023 × 10 23) of oxygen atoms.

2.3 HOW DO ATOMS EXIST

The atoms of only a few elements called noble gases (such as helium, neon, argon, krypton, etc.) are chemically
unreactive and exist in the free state (as single atoms). Atoms of most of the elements are
chemically very reactive and do not exist in the free state (as single atoms).
Atoms usually exist in two ways :
1. in the form of molecules, and
2. in the form of ions.

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Class IX : Chemistr y
2.3.1 Molecule
Atoms are usually not capable of independent or free existence but group of atoms of same or different
elements exist as one species e.g. H2, O2, P4, S8, H2O, NH3, NaCl etc.
Definition of a molecule
A group of atoms held by some force (known as bond) existing together as one species and having characteristic
properties is called a molecule.
A molecule is the smallest particle of an element or of a compound which can exist alone or
freely under ordinary conditions and shows all the properties of that substance (element or
compound).

Atoms of the same or different elements can join together to form molecules.
Molecules are of two types :(i) Molecules of an element and (ii) Molecules of compound.

(A) Molecules of Elements

Many elements such as Argon (Ar), Helium (He), sodium (Na), Iron (Fe). etc., contain single atoms and are
represented by their symbols. This is not true with non-metals. Though molecules of all the non-metal elements
are made up of identical atoms.

For example, a molecule of oxygen is made up of two atoms of oxygen. This is denoted by O 2. Some of the
molecules of other elements that exist in-group of atoms with their atomicity.

A molecule may be made up of two or more atoms. Accordingly it may be diatomic, triatomic, tetratomic,
and so on.
Generally any molecule containing more than four atoms is called polyatomic such as S 8.

Symbols and atomicity of some elements

Type of Element Name Chemical symbol Molecular Formula Atomicity

Non-Metal Argon Ar Ar Monoatomic
Helium He He Monoatomic
Oxygen O O2 Diatomic
Hydrogen H H2 Diatomic
Nitrogen N N2 Diatomic
Chlorine Cl Cl2 Diatomic
Phosphorus P P4 Tetra-atomic
Sulphur S S8 Poly-atomic

Metals
Metals exist as atomic crystals and do not form molecules. They are therefore, monoatomic.

Symbols and atomicity of metals

Name of metal Chemical symbol Atomicity
Sodium Na Monoatomic
Iron Fe Monoatomic
Aluminium Al Monoatomic
Copper Cu Monoatomic

(B)Molecules of The Compounds

A molecule of a compound contains two or more atoms of different elements. For example, a molecule of
water (H2O) consists of two atoms of hydrogen and one atom of oxygen. Similarly a molecule of glucose
(C6H12O6) contains six atoms of carbon, twelve atoms of hydrogen and six atoms of oxygen.

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 Atoms and Molecules
Symbols and atomicity of some compounds

Compound Combining Elements Molecular Formula Type of molecule

Water H,O H2O Triatomic
Hydrogen peroxide H,O H2O2 Tetra-atomic
Ammonia N,H NH3 Tetra-atomic
Carbondioxide C,O CO2 Triatomic

A molecular formula of a substance gives us the number of atoms of each kind present in its one molecule.
The number of atoms present in one molecule of an element is called its atomicity.

2.3.2 Ion
An ion is a species carrying either positive or negative charge.
Classification of ion
1. On the basis of number of atoms.
The ion consisting of only single atom are called monoatomic ions, whereas an ion consisting of a group
of atoms having some definite charge on them are called polyatomic ion. The compounds consisting
of cations and anions are called ionic compounds.
2. On the basis of nature of charge
The ions carrying positive charge are called cations while ions that carrying negative charge are called anions.
3. On the basis of number (amount) of charges.
If an ion contains +1 or –1 charge then it is monovalent, if it contains +2 or –2 it is divalent similarly
for +3 or –3 ion is called trivalent ion. The ions which carry 3 or more charge can also be called polyvalent
ions.
Naming of ionic compounds : Cation is always named 1st followed by the anion. The number of cations
and anions are not written in the name. Eg. Al2 (SO4)3 is called aluminium sulphate and not dialuminium
trisulphate.
Some Ionic Compounds :

S.No. Ionic compound Cations Anions

1 Sodium chloride Sodium ion (Na+) Chloride ion (Cl–)
+ 2–
2 Potassium sulphide Potassium ion (K ) Sulphide ion (S )
3 Calcium sulphate Calcium ion (Ca2+) Sulphate ion (SO 42–)

List of Common Ions /Radicals with Positive Valency

Positive Valency 1 Symbol
1. Ammonium NH4
2. Hydrogen H+
3. Lithium Li+
+
4. Sodium Na .
+
5. Potassium K
6. Cuprous [Copper (I)] Cu+
7. Argentous [Silver (I)] Ag+
8. Mercurous [Mercury (I)] Hg+
+
9. Aurous [Gold (I)] Au

Positive Valency 2 Symbol

1. Magnesium Mg2+
2. Calcium Ca2+
2+
3. Zinc Zn
4. Barium Ba2+
5. Nickel Ni2+
2+
6. Uranium U

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Class IX : Chemistr y
7. Cupric [Copper (II)] Cu2+
8. Argentic [Silver (II)] Ag2+
9. Mercuric [Mercury (II)] Hg2+
10. Ferrous [Iron (II)] Fe2+
11. Plumbous [Lead (II)] Pb2+
12. Stannous [Tin (II)] Sn2+
13. Platinous (Platinum (II)] Pt 2+

Positive Valency 3 Symbol

1. Aluminium Al3+
2. Chromium Cr3+
3. Bismuth Bi3+
4. Arsenic As3+
5. Ferric [Iron (III)] Fe3+
6. Auric [Gold (III)] Au3+

Positive Valency 4 Symbol

1. Stannic [Tin (IV)] Sn4+
2. Plumbic (Lead (IV)] Pb4+
3. Platinic [Platinum (IV)] Pt 4+

Radicals with Negative Valency

Negative Valency 1 Symbol
1. Fluoride F–
2. Chloride Cl –
3. Bromide Br –
4. Iodide I–
5. Hypochlorite ClO –
6. Chlorate ClO3

7. Bicarbonate or hydrogen carbonate HCO3

8. Bisulphite or hydrogen sulphite HSO3

9. Bisulphide or hydrogen sulphide HS –
10. Bisulphate or hydrogen sulphate HSO4
11. Hydride H–
12. Hydroxide OH–
13. Aluminate AlO2

14. Permanganate MnO 4

15. Cyanide CN –
16. Nitrite NO2

17. Nitrate NO3

18. Acetate CH3COO–

Negative Valency 2 Symbol

1. Sulphate SO24

2. Sulphite SO23
2–
3. Sulphide S
4. Thiosulphate S2 O23

5. Zincate ZnO22

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 Atoms and Molecules

6. Plumbate PbO22
7. Oxide O2–
8. Peroxide O22

9. Manganate MnO24

10. Dichromate Cr2 O72

11. Carbonate CO23

12. Silicate SiO23

13. Stannate SnO23

14. Oxalate (COO)22

Negative Valency 3 Symbol

1. Nitride N3–
2. Phosphide P3–
3. Phosphite PO33

4. Phosphate PO34

Negative Valency 4 Symbol

1. Carbide C4–

Illustration 4. Some element exist as single atom. Where as others can’t why ?
Solution. The atoms of only a few elements called noble gases (such as helium, neon, argon, krypton, etc.) are
chemically unreactive and exist in the free state (as single atoms). Atoms of most of the elements are
chemically very reactive and do not exist in the free state (as single atoms).

Illustration 5. Match the following elements & compounds given in column-A with column-B
Column-A Column-B
Elements/Compound Atomicity
(1) Argon (a) 8
(2) Sulphur (b) 4
(3) Oxygen (c) 2
(4) Phosphorous (P4) (d) 1
(5) Ozone (O3) (e) 3
(6) Bromine (Br2) (f) 5
(7) Carbon monoxide (CO) (g) 6
(8) Hydrogen peroxide (H2O2) (h) 7
(9) Lime water Ca(OH)2
(10) Ammonia (NH3)
(11) Quick Lime (CaO)
(12) Baking Soda (NaHCO3)
(13) Lime Stone (CaCO3)
(14) Common salt (NaCl)
(15) Sodium Sulphate (Na2SO4)
Solution. (1) d ;(2) a ;(3) c ; (4) b ; (5) e ; (6) c ; (7) c ; (8) b ; (9) f ; (10) b ; (11) c;
(12) g ; (13) f ; (14) c ; (15) h

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Class IX : Chemistr y

QUICK CHECK - 2
Objective Questions

1. A particle which maintains its chemical identity even after physical and chemical changes is :
(A) Atom (B) Molecule (C) Compound (D) None of these

2. The atomic symbols for mercury and potassium are respectively :

(A) Mr & P (B) K & Hg (C) Hg & P (D) Hg & K

3. Which of the following is not divalent ?

(A) Sulphate (B) Sulphite (C) Phosphate (D) Peroxide

4. Which of the following statement is correct :

(A) Atoms of same element only combine to form compounds
(B) Atoms of different element only combine to form compounds.
(C) Atoms of same or different element may combine to form compounds.
(D) All are wrong

5. Molecules of phosphorus and ammonia are respectively :

(A) Monoatomic & Triatomic (B) Tetra-atomic & Triatomic
(C) Tetra-atomic & Tetra-atomic (D) Monoatomic

Subjective Questions

6. What is the atomicity of Argon ?

7. Give the names of the elements present in the following compounds: [NCERT]
(a) Quick lime (b) Hydrogen bromide
(c) Baking soda (d) Potassium sulphate

8. What are poly atomic ions? Give example. [NCERT]

9. Why is it not possible to see an atom with naked eyes? [NCERT]

3
10. How many atoms are present in a (i) H2S molecule and (ii) PO 4 ion? [NCERT]

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 Atoms and Molecules

ATOMS AND MOLECULES

(AM-3)
3.1 WRITING CHEMICAL FORMULA
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, or in other
words we can say, chemical formula of a molecular compound represents the actual number & kind of atoms
of different elements present in one molecule of the compound, e.g. H2O.

Some ionic compounds

Ionic Compound Constituting Elements Ratio by Mass
Calcium oxide Calcium and oxygen 5:2
Magnesium sulphide Magnesium and sulphur 3:4
Sodium chloride Sodium and chlorine 23 : 35.5

Chemical formula of an ionic compound simply represents the ratio of the cations & anions present in the
structure of the compound, eg. : NaCl. However, in both cases, the writing of chemical formula is based on
the concept of "Valency".
Valency of an element is defined as the combining capacity of the element.
It is also equal to the number of hydrogen atoms or number of chlorine atoms or double the number of oxygen
atoms with which one atom of the element combines.
Important points
1. While writing the formula of an ionic compound the metal is written on the left hand side while the non-
metal is written on the right hand side. The name of the metal remains as such but that of the non-metal
is changed to have the ending 'ide'.
Example : MgO is named as magnesium oxide, KCl is named potassium chloride etc.
2. Molecular compounds, formed by the combination between two different non-metals, are written in such
a way that the less electronegative element is written on the left hand side while the more electronegative
element is written on the right hand side. In naming molecular compounds, the name of the less electronegative
non-metal is written as such but the name of the more electronegative element is changed to have the
ending 'ide'.
Example : H2S is named as hydrogen sulphide.
3. When there is more than one atom of an element present in the formula of the compound, then the number
of atoms are indicated by the use of appropriate prefixes (mono for 1, di for 2, tri for 3. tetra for 4 atoms
etc.) in the name of the compound.
Example : CO2 is named as carbon dioxide, CCl4 is named as carbon tetra chloride.
4. The prefixes are also needed in naming those binary compounds in which the two non-metals form more
than one compound (by having different number of atoms).
Example : Two non-metal, nitrogen and oxygen, combine to form different compound like nitrogen
monoxide (NO), nitrogen di-oxide (NO2), Nitrogen tri oxide (N2O3) etc.
5. But, if two non-metals form only one compound, then prefixes are not used in naming such compounds.
Example : Hydrogen and sulphur combine to form only one compound H2S, So, H2S is named as hydrogen
sulphide and not hydrogen monosulphide.

3.2 Writing of formula of molecular compound

Steps
The steps to be followed for writing the formula of molecular compound are,
1. First, write the symbols of the elements contributing in the compound.
2. Then, below each symbol, write its corresponding valency
3. Finally, we exchange the valencies of the combining atoms that is, with first atom, we write the valency
of the second atom and with second atom, we write the valency of the first atom, the valencies are tobe
written as subscripts to the symbols.
4. If the valencies have any common factor, then the formula is divided by the common factor. This gives
the required formula of the compound.

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Class IX : Chemistr y
Example : To work out for the formula of hydrogen sulphide
(i) Hydrogen sulphide compound is made up of hydrogen and sulphur elements. So first we write down
the symbol of hydrogen and sulphur.
(ii) The valency of hydrogen is 1 and the valency of sulphur is 2. So below the symbol H we write 1
and below the symbol S we write 2.
H S
Symbol H S

1 2
Valencies 1 2 Cross-over valencies
5. We now cross-over the valencies of H and S atoms. With H atom we write the valency of S (which is
2) so that it becomes H2 with S atom we write the valency of H (which is 1) so that it becomes S 1. Now,
joining together H2 and S1 the formula of hydrogen sulphide becomes H2S1 or H2S (This is because we
don't write the subscript 1 with an atom in a formula).

3.3 Writing the formula of Ionic compound

Steps

1. First, write the symbols of the ions from which the ionic compound is made. As a convention, the cation
is written on the left side while the anion is written on the right side.

2. Then, the valencies of the respective cation and anion are written below their symbols.
3. The valencies of the cation and anion are exchanged. The number of cation and anions in the formula
of the compound are adjusted in such a way that total positive charge of cation become equal to the
total negative charge of the anion making the ionic compound electrically neutral.

4. The final formula of the ionic compound is then written but the charges present on the cation and the
anion are not shown.
Examples
(a) Molecular compounds
Ammonia Methane Carbon dioxide
N H C H C O

3 1 4 1 4 2
NH3 CH 4 C 2O4 or CO2

(b) Ionic compounds

Sodium chloride Sodium phosphate
Na Cl Na (PO)4

+1 –1 +1 –3
NaCl Na 3 PO4

3.4 Significance of molecular formula of a substance

(i) It tells the name of the substance.
(ii) It tells about the names of the different elements present in the substance.
(iii) It represents one molecule of the substance.
(iv) It tells about the number of atoms of each element present in 1 molecule of substance.
(v) It tells about atomicity of the substance.

14
 Atoms and Molecules
Chemical formula of some compounds
Positive ion (cation) Negative ion (anion)
Name of the Chemical
Valency Valency
compound Name Formula Name Formula Formula
number number
Hydrogen
Hydrogen H 1 Chloride Cl 1 HCl
chloride
Hydrogen
Hydrogen H 1 Sulphide S 2 H2S
sulphide
Sulphuric acid
H2(SO4)1,
(Hydrogen Hydrogen H 1 Sulphate SO4 2
H2(SO4)
sulphate)
Sodium Na 1(NO3)1,
Sodium Na 1 Nitrate NO3 1
nitrate NaNO3
Aluminium Al3(PO4)3,
Aluminium Al 3 Phosphate PO4 3
Phosphate AlPO4
Aluminium
Aluminium Al 3 Sulphate SO4 2 Al2(SO4)3
sulphate
Ferrous Fe2(SO4)2,
Ferrous Fe 2 Sulphate SO4 2
sulphate FeSO4
Ferric
Ferric Fe 3 Sulphate SO4 2 Fe2(SO4)3
sulphate
Potassium Dichromat K2(Cr2O7)1,
Potassium K 1 Cr2O7 2
dichromate e K2Cr2O7
Magnesium
Magnesium Mg 2 Nitrate NO3 1 Mg(NO3)2
nitrate
Barium Ba 2(CO3)2,
Barium Ba 2 Carbonate CO3 2
carbonate BaCO3
Potassium Permang
Potassium K 1 MnO4 1 KMnO4
permanganate -anate
Calcium
Calcium Ca 2 Hydroxide OH 1 Ca(OH)2
hydroxide
Aluminium oxide Aluminium Al 3 Oxide O 2 Al2O3
Magnesium
Magnesium Mg 2 Phosphate PO4 3 Mg3(PO4)2
phosphate
Ammonium
Ammonium NH4 1 Sulphite SO3 2 (NH4)2SO3
sulphate
Zinc
Zinc Zn 2 Phosphate PO4 3 Zn3(PO4)2
phosphate

Illustration 6. Write down the names of compounds represented by the following formulae: [NCERT]
(i) Al2(SO4)3 (ii) CaCl2 (iii) K2SO4 (iv) KNO3 (v) CaCO3
Solution. (i) Al2(SO4)3 : Aluminium sulphate
(ii) CaCl2 : Calcium chloride
(iii) K2SO4 : Potassium sulphate
(iv) KNO3 : Potassium nitrate
(v) CaCO3 Calcium carbonate

Illustration 7. What is meant by the term chemical formula? [NCERT]

Solution. The chemical formula of a compound means the symbolic representation of the composition of a
compound. For example, from the chemical formula CO 2 of carbon dioxide, we come to know that
one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of
the compound, carbon dioxide.

15
Class IX : Chemistr y

QUICK CHECK - 3
Objective Questions

1. Formula of Calcium phosphate is :

(A) Ca2PO4 (B) Ca2(PO4)3 (C) Ca3(PO4)2 (D) CaPO4

2. Negative radicals are also called :

(A) Acid radicals (B) Basic radicals (C) Neutral radicals (D) Amphoteric radicals

3. A radical which consist only one type of atom :

(A) Simple radical (B) Compound radical (C) Both A and B (D) None of these
4. Formula for Aluminium oxide is :
(A) AlO (B) AlO2 (C) Al3O2 (D) Al2O3

5. The Molecular formula of potassium nitrate is :

(A) KNO2 (B) KNO3 (C) KNO4 (D) KON

Subjective Questions

6. What is meant by the term chemical formula ? Give examples.

7. What are ionic and molecular compound ? Give examples.

8. Write the formula for the following.

(a) Caustic potash (b) Baking soda
(c) Lime stone (d) Caustic soda
(e) Common salt

9. Write down the formulae of [NCERT]

(i) sodium oxide
(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide

10. Write the chemical formulae of the following. [NCERT]

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

16
 Atoms and Molecules

ATOMS AND MOLECULES

(AM-4)

4. 1 ATOMIC MASS
Atom is so small in size that it may not be possible to isolate a single atom and then weigh it. For example,
an atom of hydrogen has mass equal to 1.67 × 10–24 g.
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 particular element as the standard mass. The masses
of the 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. Later, it was found that hydrogen gas in
its natural state has three isotopes. Thus the average mass of naturally occurring hydrogen works out as 1.008
amu rather than 1 amu.
However, using hydrogen as the reference, the masses of atoms of other elements came out to be fractional.
Hence, the reference was changed to oxygen taken as 16. In other words , 1/16 th of the mass of an atom
of naturally occurring oxygen was taken as one unit. This was selected because of the following two reasons.
(i) Oxygen combined with most of the elements.
(ii) By comparing with oxygen taken as 16, the relative atomic masses of most of the elements were found
to be whole numbers.
However, a difficulty arouse when it was found that naturally occurring oxygen is a mixture of atoms of
slightly different masses (called "isotopes").

Carbon-12 as standard reference

It was found that the atomic mass of the most common isotope of carbon, 12C is a whole number 12. Thus,
the mass of 1/12 of 12C is equivalent to 1 atomic mass unit (a.m.u.) or unified atomic mass.
Atomic mass of a substance when expressed in terms of grams is called gram atomic mass.
Atomic masses of some common elements (in amu or u)
Element Symbol Atomic mass (amu/u)
Hydrogen H 1
Carbon C 12
Lithium Li 7
Nitrogen N 14
Oxygen O 16
Fluorine F 19
Neon Ne 20
Sodium Na 23
Magnesium Mg 24
Phosphorous P 31
Sulphur S 32
Chlorine Cl 35.5
Calcium Ca 40
4.1.1 Average Atomic Mass
There are many cases where different atoms of the same element possess different relative masses. Such
atoms of the same element which have different relative masses are called isotopes. In such cases atomic
mass of the element is average of relative masses of different isotopes of the element. For example, Chlorine
contains two types of atoms having relative masses 35 u and 37 u. The relative abundance of these isotopes
in nature is in the ratio 3 : 1. Thus, atomic mass of Chlorine is the average of these different relative masses as
described below :
35 3 37 1
Atomic mass of Chlorine = = 35.5 u.
4

17
Class IX : Chemistr y
Thus, the atomic mass of an element may be defined as the average relative mass of an atom of the
element as compared with the mass of an atom of Carbon (C-12) taken as 12 u (or 12 amu).
The use of relative masses of atoms is preferred over absolute masses because gram is too big unit for
expressing their masses. If some how we express the mass of an atom in gram the numerical value which we
get is extremely small and hence, inconvenient to use. For example, mass of an atom of Carbon (C-12) is
1.99×10–23 gram.

4.1.2 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 u, therefore,
1 gram-atom of Mg = Gram atomic mass of Mg = 24 g
2 gram-atom of Mg = 2 × Gram atomic mass of Mg = (2 × 24) g = 48 g.
From the above discussion, we can get the relationship between mass of the element and its gram-atoms as :
Mass in gram Wg
Number of g-atom = Gram atomic mass =
GAM

4.2 MOLECULAR MASS

Like atoms, the molecules are also very small in size. Hence, their actual masses cannot be determined by
direct weighing. Therefore, masses of molecules are also determined relative to the mass of Carbon atom
(C– 12) taken as 12 u. These relative masses of molecules are called molecular masses. Molecular mass of
a substance (element or compound) may be defined as the average relative mass of a molecule of the
substance as compared with mass of an atom of Carbon (C-12) taken as 12 u.
Molecular mass of a substance tells us the number of times a molecule of the substance is heavier than 1/12th
of mass of a Carbon (C-12) atom. For example, molecular mass of water (H2O) is 18 u. It means that a
molecule of water is 18 times heavier than 1/12th of mass of a Carbon atom (C-12).
Molecular mass of a substance can be obtained by adding the atomic masses of all the atoms present in a
molecule of the substance. For example,
Molecular mass of NH3 = atomic mass of N + 3 (atomic mass of H)
= 14 + 3× 1 = 17 u.

4.2.1 Gram Molecular Mass (GMM)

It may be defined as that much quantity of the substance (element or compound) whose mass in
gram 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 u. Therefore,
1 gram-molecule of NH3 = gram molecular mass of NH 3 = 17 g
2 gram-molecule of NH3 = 2 × gram molecular mass of NH3 = (2× 17) g = 34 g.
From the above discussion it follows that
Mass in gram W g
Number of g-molecule = Gram molecular mass =
GMM

4.3 Formula Mass

Since ionic compounds do not consist of molecules, the use of term ‘molecular mass’ for them is not very
correct. So, we use the term ‘formula mass’ for ionic compounds in which individual molecules do not exist.
The formula mass of an ionic compound is the relative mass of its ‘formula unit’ as compared
with the mass of a Carbon-12 atom taken as 12 units. In other words, formula mass is the sum of the
atomic masses of the atoms (or ions) represented by its formula. In order to calculate the formula mass of an
ionic compound, we should know the formula of the ionic compound as well as the atomic masses of all the
atoms (or ions) present in the formula.

18
 Atoms and Molecules
The formula mass is then calculated in a way similar to the calculation of molecular mass. Please note that
the atomic mass of an atom and its ion is just the same (because the electrons which convert an
atom into an ion have negligible mass). For example, the formula mass of Sodium chloride (NaCl) will be
the sum of atomic masses of Sodium (Na) and Chlorine (Cl). Now, the atomic mass of Na is 23 u and the
atomic mass of Cl is 35.5 u, so the formula mass of Sodium chloride (NaCl) will be 23 + 35.5 = 58.5 u.

Illustration 8. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4,NH3,CH3OH. [NCERT]
Solution. Molecular mass of H2 = 2 × Atomic mass ( H) = 2 × 1 = 2 u
Molecular mass of O2 = 2 × Atomic mass (O) = 2 × 16 = 32 u
Molecular mass of Cl2 = 2 × Atomic mass (Cl) = 2 × 35.5 = 71 u
Molecular mass of CO2 = Atomic mass ( C) + 2 × Atomic mass (O) = 12 + 2 × 16 = 44 u
Molecular mass of CH4 = Atomic mass (C ) + 4 × Atomic mass (H) = 12 + 4 × 1 = 16 u
Molecular mass of C2H6 = 2 × Atomic mass of C + 6 × Atomic mass of H = 2 × 12+6×1= 30 u
Molecular mass of C2H4 = 2 × Atomic mass (C) + 4 × Atomic mass (H) = 2×12+4×1=28 u
Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H = 14 + 3 × 1 = 17 u
Molecular mass of CH3OH = Atomic mass (C) + 4 × Atomic mass (H) + Atomic mass (O)
= 12 + 4 × 1 + 16 = 32 u

QUICK CHECK - 4
Objective Questions
1. Molecular mass of H2SO4 is :
(A) 49 u (B) 98 u (C) 72 u (D) 96 u

2. The average atomic mass of a sample of an element X is 16.2u. What are the percentage of isotope 16
8 X and

18
8 X in the sample ?

(A) 16
8 X = 10% and 18
8 X = 90% (B) 16
8 X = 90% and 18
8 X = 10%

(C) 16
8 X = 52% and 18
8 X = 48% (D) None of these

3. Which of the following has highest molecular mass.

(A) C8H16O8 (B) C7H14O7 (C) C6H12O6 (D) C5H10O5
4. The molecular mass of hydrated copper sulphate is :
(A) 159.5 (B) 248.5 (C) 249.5 (D) 158.5

Subjective Questions
5. Define the atomic mass unit. [NCERT]
6. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65 u, Na = 23 u, K =
39 u, C = 12 u, and O = 16 u. [NCERT]
7. Calculate the molar mass of the following substances. [NCERT]
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3

8. Calculate the molecular masses of H2O2, HCl, CaCO3, C6H6, CH3COOH

19
Class IX : Chemistr y

ATOMS AND MOLECULES

(AM-5)

5.1 MOLE CONCEPT

In everyday life, we buy things in terms of number or in terms of mass (or weight). Further, generally a bigger
unit for counting numbers such as dozen (for 12), score (for 20) or gross (for 144) is used & similarly, a bigger
unit for mass like kilogram or quintal etc. is used. For example, we buy eggs, bananas, oranges etc. in dozens
whereas we buy wheat, rice, sugar etc. in kilograms.
There is a very close analogy or similarity between the terms mole & dozen. A dozen always represents 12
articles. They may be apples, oranges, pens etc. In a similar way, a mole represents 6.022 × 10 23 particles.
They may be atoms, molecules, ions, electrons, protons etc.
Gram atomic mass of an element contains 6.022 × 1023 atoms. Similarly the gram molecular mass of a
compound also contains the same number of molecules.
For example : The gram atomic mass of carbon = 12 g. The mass of one atom of carbon = 1.9924 ×
10–23 g. The number of C-atoms in 12 g of carbon will be,

Gram atomic mass of carbon

No. of C-atoms=
Mass of one carbon atom

12(g)
= 23
6.022 1023
1.9924 10 (g)

Similarly, gram molecular mass of oxygen (O2) = 32 g. The mass of one molecule is 5.313 × 10–23 g. The
number of oxygen molecules in 32 g of oxygen will be,
Gram molecular mass
No. of O2 molecules =
Mass of one oxygen molecule

32(g)
= 23
6.022 1023 .
5.313 10 (g)
This 6.022 × 1023 is known as Avogadro's number or Avogadro's constant & is denoted by either 'N 0' or 'NA'.
A mole denotes Avogadro's number of particles.
One mole of H-atoms = 6.022 × 1023 H-atoms
One mole of O2-molecules = 6.022 × 1023 O2-molecules
One mole of sodium ions (Na+) = 6.022 × 1023 sodium ions.
Necessity of mole concept
We know that the atoms & molecules of any substance (element or compound respectively) are very small
in size. This is quite evident from the fact that 12 g of carbon contains 6.022 × 10 23 atoms of carbon. This
means that it may not be possible to count these atoms individually. However, they can collectively be represented
as one mole. This is a very convenient method to represent different particles.
1 mole of particles is defined as that amount of the substance which contains Avogadro's number
of particles. Thus,
1 mole of C - atoms = 6.022 × 1023 atoms = 12 gm
1 mole of O - atoms = 6.022 × 1023 atoms = 16 gm
1 mole of O2 molecules = 6.022 × 1023 molecules of O2 = 32 gm
1 mole of H2O molecules = 6.022 × 1023 molecules of H2O = 18 gm
Thus , 1 mole of atoms = 6.022 × 1023 atoms.
1 mole of molecules = 6.022 × 1023 molecules.
For example : oxygen atom in elementary oxygen (O) and oxygen molecule is O2.
1 mole of oxygen atoms (O) = 6.022 × 1023 oxygen atoms
1 mole of oxygen molecules = 6.022 × 10 23 oxygen molecules.

20
 Atoms and Molecules

Memory chart
1 mole of
carbon atoms

6.022 × 10
23
12 gm of
atoms of C carbon atoms
1 mole of
hydrogen atoms

6.022 × 10
23
1 gm of
atoms of H H atoms
1 mole of any particle
(atoms, molecule, ions)

23
6.022 × 10 number Relative mass of those
of that particle particles in grams
1 mole of
molecules

23
6.022 × 10 Molecular mass
number of particle in grams

Fig. Relationship between mole, Avogadro number and mass.

23 22.4 L of a
6.022 × 10
Particles In of gas at N.T.P.
te
par rms o ms
ticl f ter e
es In olum
v
1 mole of oxygen atoms
23 1 MOLE
= 6.022 × 10 atoms
1 mole of oxygen molecules
23
= 6.022 × 10 molecules In terms of
mass

1 gram atom 1 gram molecule 1 gram formula

of element of substance weight of substance

12 g 23 g 18 g 44 g 58.5 g 100 g
C Na H2O CO2 NaCl CaCO3

Illustration 9. Covert 22 g of carbon dioxide (CO2) into moles. (Atomic masses : C = 12 u ; O = 16 u)

Solution. 1 mole of CO2 = Molecular mass of CO2 in grams
= Mass of C + Mass of O × 2
= 12 + 32
= 44 grams
So, the mass of 1 mole of carbon dioxide is 44 grams.
Now, 44 g of carbon dioxide = 1 mole
1 1
So, 22 g of carbon dioxide = 22 mole = = 0.5 mole
44 2
Thus, 22 grams of carbon dioxide are equal to 0.5 mole.
Note : The above problem can also be solved directly by using the formula :

21
Class IX : Chemistr y
Mass of subs tan ce in grams
Formula 1. Number of moles of molecules =
Gram molecular mass of subs tan ce
Mass of subs tan ce
Formula 2. Number of moles of molecules = Molar mass of subs tan ce

Illustration 10. What is the mass of 0.5 mole of water (H2O). (Atomic masses : H = 1 u ; O = 16 u)
Solution. In order to solve this problem, we should know the mass of 1 mole of water. This can be obtained by
using the given values of the atomic masses of hydrogen and oxygen as follows :
1 mole of water (H2O) = Molecular mass of H2O in grams
= Mass of 2H atoms + Mass of O atom
= 2 × 1 + 16
= 2 + 16 = 18 grams
Thus, the mass of 1 mole of water is 18 grams.
Now, Mass of 1 mole of water = 18 g
So, Mass of 0.5 mole of water = 18 × 0.5 g = 9 g
Thus, the mass of 0.5 mole of water (H2O) is 9 grams.
Note : The above problem can also be solved directly by using formula 1 and 2.

Illustration 11. What is the number of molecules in 0.25 moles of oxygen?

Solution. We know that :
1 mole of oxygen contains = 6.022 × 1023 molecules
So, 0.25 mole of oxygen contains = 6.022 × 1023 × 0.25
= 1.505 × 1023 molecules
Thus, 0.25 mole of oxygen contains 1.505 × 1023 molecules.

Illustration 12. A piece of copper weighs 0.635 g. How many atoms of copper does it contain?
Solution. Gram atomic mass of copper = 63.5 g

0.635
Number of moles in 0.635 g of copper = = 0.01
63.5
Number of copper atoms in one mole = 6.02 × 1023
Number of copper atoms in 0.01 moles = 0.01 × 6.02 × 1023 = 6.02 × 1021

Illustration 13. How many molecules of water and oxygen atoms are present in 0.9 g of water?
Solution. Gram molecular mass of water = 18 g
0.9
Number of moles in 0.9 g of water = 0.05
18
Number of water molecules in one mole of water = 6.02 × 10 23
Number of molecules of water in 0.05 moles = 0.05 × 6.02 × 10 23 = 3.010 × 1022
As one molecule of water contains one oxygen atom.
So, number of oxygen atoms in 3.010 × 10 22 molecule of water = 3.010 × 1022

Illustration 14. What is the mass of 3.01 × 1022 molecules of ammonia ?

Solution. Gram-molecular mass of ammonia = 17 g
Number of molecules in 17 g (one mole) of NH3 = 6.02 ×1023
Let the mass of 3.01 × 1022 molecules of NH3 be = x g

3.01 1022 x
So,
6.02 1023 17

17 3.01 1022
or x = 0.85g
6.02 1023

22
 Atoms and Molecules

Illustration 15. How many electrons are present in 1.6 g of methane?

Solution. Gram-molecular mass of methane
(CH4) = 12 + 4 = 16 g
1.6
Number of moles in 1.6 g of methane = 0.1
16
Number of molecules of methane in 0.1 mole
= 0.1 × 6.02 × 1023
= 6.02 × 1022
One molecule of methane has = 6 + 4 = 10 electrons
So, 6.02 × 1022 molecules of methane have
= 10 × 6.02 × 1022 electrons
= 6.02 × 1023 electrons

QUICK CHECK - 5
Atomic masses of some common elements
Element Atomic mass Element Atomic mass
H 1 P 31
C 12 S 32
Li 7 Cl 35.5
N 14 K 39
O 16 Ca 40
F 19 Fe 56
Ne 20 Cu 63
Na 23 Zn 65
Mg 24 Ag 108
AI 27 I 127
Objective Questions
1. Number of gram molecules in 63 gm HNO3 :
(A) 1 (B) 2 (C) 3 (D) 4
2. Number of gram atoms in 72g of Mg :
(A) 1 (B) 2 (C) 3 (D) 4
3. Weight in gms of 2 gram molecules of H2O :
(A) 18 (B) 36 (C) 9 (D) 72
4. How many moles of NaOH are present in 16.0 g of it ?
(A) 3 (B) 4 (C) 0.4 (D) 5
5. Which of the following contains the largest number of molecules ?
(A) 0.2 mol H2 (B) 8.0 g H2 (C) 18 g H2O (D) 6.0 g CO2

Subjective Questions
6. What is the mass of : [NCERT]
(a) 1 mole of nitrogen atoms? (b) 4 moles of aluminium atoms
(c) 10 moles of sodium sulphite (Na 2SO3)?
7. Convert into mole. [NCERT]
(a) 12 g of oxygen gas (b) 20 g of water
(c) 22 g of carbon dioxide.
8. If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?[NCERT]
9. 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)? [NCERT]
10. What is the mass of: [NCERT]
(a) 0.2 mole of oxygen atoms? (b) 0.5 mole of water molecules?

23
Class IX : Chemistr y

ATOMS AND MOLECULES

(AM-6)

6.1 EMPIRICAL FORMULA AND MASS

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

Illustration 16. An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition.
C = 41% ; H = 5% and O = 54% The molecular weight of the compound is 118.
Calculate the molecular formula of the compound.
Solution Step-1
To calculate the empirical formula of the compound.
Percentage
Element Symbol % At. mass Relative no. Simplest whole of
At.mass
atoms
41 3.4
Carbon C 41 12 =3.4 1 2
12 3.4
5 5
Hydrogen H 5 1 =5 1.5 3
1 3.4
54 3.4
Oxygen O 54 16 =3.4 1 2
16 3.4
Empirical formula is C2H3O2
Step - 2
To calculate the empirical formula mass.
The empirical formula of the compound is C2H3O2.
Empirical formula mass
= (2 × 12) + (3 × 1) + (2 × 16) = 59.
Step - 3
To calculate the value of 'n'
Molecular mass 118
n = Empirical formula mass = =2
59
Step - 4
To calculate the molecular formula of the salt
Molecular formula = n × (Empirical formula)
= 2 × C2H3O2 = C4H6O4
Thus the molecular formula is C4H6O4.

24
 Atoms and Molecules

QUICK CHECK - 6
Objective Questions

1. The correct molecular formula of a compound in which Carbon and Hydrogen are present in the atomic ratio
1 : 1.5 is :

(A) C1H1.5 (B) C4H6

(C) C2H3 (D) C1.5 H1

2. A hydrocarbon contains 80% carbon. The empirical formula of the compound is :

(A) CH2 (B) CH3

(C) C5H2 (D) None of these

3. An organic compound has the empirical formula CH. Its molecular mass is 78. Molecular formula of the
compound will be

(A) CH4 (B) C6H6

(C) C4H10 (D) C6H8

4. The empirical formula of a compound of molecular mass 240 is CH2O. The molecular formula of the
compound is:

(A) C8H16O8 (B) C7H14O7

(C) C6H12O6 (D) C5H10O5

Subjective Questions

5. Given the following empirical formula and molecular weight, compute the true molecular formula.

Empirical formula Molecular weight

(a) CH2 84

(b) CH2O 150

(c) HO 34

(d) HgCl 472 [Hg = 200.5, Cl = 35.5]

(e) HF 80

6. An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition.

C = 52.17 % ; H = 13.04 % and O = 34.78 % The molecular weight of the compound is 46.
Calculate the molecular formula of the compound.

7. A compound has the following percentage composition by mass : Carbon 14.4%, hydrogen 1.2% and chlorine
84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. The relative molecular
mass of this compound is 168, so what is its molecular formula.

8. Determine the empirical formula of a compound containing 47.9% potassium, 5.5% beryllium and 46.6% fluorine
by mass. Work to one decimal place.

25
Class IX : Chemistr y

Indian philosopher Maharishi Kanad named the smallest particle of matter “Parmanu”

Another Indian philosopher, Pakudha Kayayana elaborated that these particles normally exist combined
form which is now called “molecule”

John Dalton was the first scientist who called these particles “atom” which means “indivisible”

Law of Conservation of Mass : According to “Lavoisier” during any physical or chemical change, the total
mass of the products remains equal to the total mass of the reactants.

Law of constant composition or definite proportion : According to “proust” a chemical compound always
contain same elements combined together in same proportion by mass.

All the matter is made up of extremely small indivisible and indestructible ultimate particles called atoms.

Atom: Smallest particle of an element that takes part in chemical reaction.

Atomic Mass : One twelfth of the actual mass of an atom of carbon – 12 isotopes. Unit amu / u

Gram Atomic mass : Mass in grams which is numerically equal to its atomic mass. Unit gram

Molecular mass : The average relative mass of a molecule of the substance as compared with mass of an
atom of carbon (C – 12) taken as 12 u Unit amu / u

Gram Molecular Mass: Mass in grams which is numerically equal to its molecular mass. Unit gm

Ions : Electrically changed atom (or group of atoms).

A positively charged ion is known as cation eg. Na +, Mg2+

A negatively charged ion is known as anion eg. Cl–, O2–.

Molecules : smallest particle of an element or of a compound which can exist alone or freely under ordinary
conditions and shows all the properties of that substance, eg H 2, O2 etc.

Valency : Combining capacity of an clement

Radicals : Charge particles (atom or group of atoms)

Positive radicals = basic radicals

Negative radicals = acid radicals
Mole Concept :

26
 Extended or Long Form of the Periodic Table
p-Block Elements
IA 0
Metals
(1) (18)
Period 1 2
Non metals
H IIA IIIA IVA VA VIA VIIA He
1 1.0079 4.0026
Hydrogen (2) (13) (14) (15) (16) (17) Helium
Metalloids
3 4 5 6 7 8 9 10
2 Li Be d-Block Elements B C N O F Ne
6.940 9.0122 10.811 12.011 14.007 15.999 18.998 20.180
Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluonne Neon
11 12 13 14 15 16 17 18
3 Na Mg IIIB VIB VB VIB VIIB VIII IB IIB Al Si P S Cl Ar
22.990 24.305 26.982 28.086 30.974 32.066 35.453 39.948
Sodium Magnesium (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) Aluminium Silicon Phosporus Sulphur Chlorine Argon

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
39.098 40.078 44.956 47.867 50.941 51.996 54.938 55.847 58.933 58.693 63.546 65.39 62.723 72.61 74.922 78.96 79.904 83.80
Potassium Calcium Scandium Titanium Vanadium Chrominum Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
85.468 87.62 88.906 91.224 92.906 95.94 98 101.07 102.91 106.42 107.87 112.41 114.82 118.71 121.76 127.60 126.90 131.29
Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon

55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
6 Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
132.91 137.33 138.91 178.49 180.95 183.84 186.21 190.23 192.22 195.08 196.97 200.59 204.38 207.2 208.98 210 210 222
Cesium Barium Lanthanum Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth Polonium Astatine Radon

87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
7 Fr Ra Ac** Unq Unp Unh Uns Uno Une Uun Uuu Uub Uut Uuq Uup Uuh Uus Uuo
223 226 227 261 262 266 264 269 268 269 272 277
Francium Radium Actinium Unnilquadium Unnilpentium Unnilhexium Unnilseptium Unniloctium Unnilennium Ununnilium Unununium Ununbium Ununtrium Ununquadium Ununpentium Ununhexium Ununseptium Ununoctium

58 59 60 61 62 63 64 65 66 67 68 69 70 71
6 Lanthanide Series Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
140.12 140.91 144.24 145 150.36 151.96 157.25 158.93 162.50 164.93 167.26 168.93 173.07 174.97
Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium

90 91 92 93 94 95 96 97 98 99 100 101 102 103

7 **Actinide Series Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
232.04 231.04 238.03 237 244 243 247 247 251 252 257 258 259 262

27
Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium Califonium Einsteinium Fermium Mendelevium Nobelium Lawrencium
Atoms and Molecules

IUPAC designations of groups of elements are given in brackets

Class IX : Chemistr y

EXERCISE-1 OBJECTIVE CHECK

1. Correct formula of ferric sulphate is :

(A) FeSO4 (B) Fe(SO4)2 (C) Fe3(SO4)2 (D) Fe2(SO4)3

2. 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) Law of multiple proportion

3. A chemical equation is always balanced to fulfil the condition of :

(A) Dalton’s atomic theory (B) Law of constant composition
(C) Law of conservation of mass (D) None of these

4. Study the following table :

Compound Mass of the compound
(in gram) taken
I. CO2 4.4
II. NO2 2.3
III. H2O2 6.8
IV. SO2 1.6
Which two compound have least mass of oxygen:
(A) II and IV (B) I and III (C) I and II (D) III and IV

5. Number of electrons and neutrons in CH4 respectively are -

(1) 16 and 12 (2) 10 and 10 (3) 10 and 16 (4) 10 and 6

6. SiO2 is a .............. :
(A) Monoatomic element (B) Diatomic element
(C) Triatomic compound (D) Tetratomic compound

7. If law of conservation of mass was to hold true, then 1.70 g of AgNO3 on reacting with 0.365 g of
HCl will produce 1.435 g of AgCl and HNO3 equal to :
(A) 6.33 g (B) 0.63 g (C) 63.3 g (D) 60.3 g

8. 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

9. The Gram atomic mass of calcium (Ca) is 40 g. 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

10. No. of moles of 0.64 g of SO2 is :

(A) 100 (B) 10 (C) 0.01 (D) 0.1

11. The number of molecules in 4.25 g of NH3 is approximately :

(A) 1.5 × 1023 (B) 2 × 1023 (C) 4 × 1023 (D) 6 × 1023

28
 Atoms and Molecules
12. The atomic mass of oxygen is 16 and the molecular mass of ozone is 48. What is the atomicity of
ozones if it is an allotrope of oxygen ?
(A) 1 (B) 2 (C) 3 (D) 4

13. The no. of oxygen atoms is 4.4 g of CO2 is approx.

22 23 23 23
(A) 6 × 10 (B) 6 × 10 (C) 12 × 10 (D) 1.2 × 10

14. How many grams of sodium will have the same number of atoms as 6 g of magnesium ? (Na = 23,
Mg = 24.)
(A) 10.75 g (B) 5.75 g (C) 8 g (D) 20.75 g

15. The total number of protons, electrons and neutrons in 12 g of 12

6 C is:
(A) 1.084 × 1025
(B) 6.022 × 10 23
(C) 6.022 × 10 22
(D) 18

16. The molecular mass of an organic compound is 78 and its percentage composition is 92.4% C and
7.6% H. Determine the molecular formula of the compound.
(A) C2H6 (B) C3H6 (C) C6H6 (D) C5H6

17. Which of the following elements shows variable valency ?

(A) Na (B) Mg (C) Fe (D) Zn

18. The formula of Ammonium Sulphate is :

(A) NH4SO4 (B) (NH4)2 SO4 (C) NH4 (SO4)2 (D) NH4 (SO4)3

19. Number of molecules in 14 g of carbon monoxide is

23 23 23 23
(A) 12.044 × 10 (B) 6.022 × 10 (C) 3.011 × 10 (D) 1.5050 × 10

20. Number of molecules present in 14 gm of N2 molecule is

23 23 23 22
(A) 6.022 × 10 (B) 3.011 × 10 (C) 1.51 × 10 (D) 6.022 × 10

29
Class IX : Chemistr y

EXERCISE-2

Very Short Answer type questions

1. Define atomic mass unit.
2. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur. [NCERT]

3. Give symbols for the following elements : Aluminium, Tin, Bromine, Neon.
4. What is meant by formula unit mass?
5. What do you understand by the 'atomicity' of the substance?
Short Answer type questions
6. 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. [NCERT]
7. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.(Hint: The mass of an ion is the
same as that of an atom of the same element. Atomic mass of Al = 27 u) [NCERT]

8. How many g of element are present in 35.125 g atom of Si. (Given at. wt. of Si = 28.)

9. Arrange the following in order of increasing masses:

(i) 0.1 gram–atom of silver (ii) 0.1 mole of H2SO4

(iii) 1023 molecule of CO2 gas (iv) 1 gram of carbon

(v) 1023 atoms of calcium.

(Atomic masses : Ag = 108 u, S = 32 u, N = 14 u, Ca = 40 u)

10. Calculate the no. of molecules in a drop of water weighing 0.07 g.

Long Answer type questions

11. If 1 g of SO 2 contains x molecules, what will be the number of molecule in 1 g of methane?

(S = 32 u, O = 16 u, C = 12 u, H = 1 u)

12. Find the number of atoms of each type present in 3.42 grams of canesugar (C 12H22O11).
13. What are the postulates and limitations of Dalton's atomic theory?

14. The density of liquid mercury is 13.6 / cm3. How many moles of mercury are there in 1 litre of the metal ?
(Atomic mass of Hg = 200)

15. The action of bacteria on meat and fish produces a poisonous compound called cadaverine. As its name and
origin imply, it stinks ! It is 58.77% C, 13.81 % H, and 27.42 % N. Its molar mass is 102 g/mol. Determine
the molecular formula of cadaverine.

30
 Atoms and Molecules
CHECK LIST
Suggested Time Total Doubts Chapter
Total Ques. Time Taken by completely
Exercise Name to Solve All taken in this
Given Student Solved ON :
Questions chapter
Date
Quick Check 1 10 15 minutes
Quick Check 2 10 15 minutes
CLASS WORK

Quick Check 3 10 20 minutes

Quick Check 4 8 15 minutes

Quick Check 5 10 25 minutes

Quick Check 6 10 25 minutes
HOME WORK

Objective Check 20 20 minutes

Subjective Check 15 30 minutes

ANSWERS
QUICK CHECK-1

1. (B) 2. (B) 3. (C) 4. (A) 5. (D)

QUICK CHECK-2

1. (A) 2. (D) 3. (C) 4. (B) 5. (C)

QUICK CHECK-3

1. (C) 2. (A) 3. (A) 4. (D) 5. (B)

QUICK CHECK-4

1. (B) 2. (B) 3. (A) 4. (C)

QUICK CHECK-5

1. (A) 2. (C) 3. (B) 4. (C) 5. (B)

QUICK CHECK-6

1. (C) 2. (B) 3. (B) 4. (A)

EXERCISE - 1 (OBJECTIVE CHECK)

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. D B C A D C B C C C A C D B A
Que. 16 17 18 19 20
Ans. C C B C B

31
Class IX : Chemistr y

IMPORTANT NOTES

32