3 Atoms and Molecules

◼ Introduction
The smallest unit of any element is called an atom. The atoms of different elements combine
with one another to form new substances called compounds. The compounds so formed are
neutral in character, i.e., they have no electric charges on them. All matter is made by the
combination of atoms of different elements combined together in some fixed ratio. In a way,
atoms are the basic building blocks of the matter.
There is a beautiful description of atomic theory in the treaties, “Vaisheshik Darshan”
written by Kanad, in which he has proposed that the whole universe is made up of atoms.
By scientific study substance is classified into pure substance and mixture. Then substance
is viewed as element and compound. Why one element differ from another? Similarly, why
one compound is different from another? How new compounds are formed by the mutual
action of elements or compounds?
By the late 1700s, scientists began to realize that the concept of atoms provided an
explanation for many experimental observations.
◼ Laws of chemical combination
One of the most important aspects of the subject of chemistry is the study of chemical
reactions. These chemical reactions take place according to the certain laws, called the 'laws
of chemical combination'.
(A) Law of conservation of mass (B) Law of constant proportion
The laws of chemical combination are the experimental laws which led to the idea of atoms
being the smallest unit of matter. The laws of chemical combination played a significant role
in the development of Dalton's atomic theory of matter.
⚫ Law of conservation of mass
This law is given by Antoine Lavoisier.
According to this law, matter is neither created nor destroyed in the course of chemical
reaction although it may change from one form to other.
The total mass of materials present after a chemical reaction is the same as the total
mass before reaction. The law was tested by Heydweiller and Landolt (1901-1906) who
performed fifteen different reactions in a closed H-tube so that nothing was allowed to
escape. It was noticed that there was hardly any change in the weights of the tube.
Consider any reaction, say combination of AgNO3 and KI in H-tube (see figure). Let
AgNO3 and KI are filled in limb A and B, respectively and then tube is tilted to allow these
reactants to react as AgNO3 + KI → AgI + KNO3. The mass of tube before and after the
reaction comes out to be same.
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Olympiads : Class 9

A B
H-tube

6.3 g of sodium bicarbonate was added to 15.0 g of acetic acid solution. Carbon dioxide
gas was produced which was allowed to escape while the residue (of sodium acetate
and water) was found to weigh 18.0 g. What is the mass of carbon dioxide which
escaped out into the atmosphere?
Solution:
NaHCO3 + CH3COOH ⎯→ CH3COONa + H2O + CO2
According to law of conservation of mass,
Total mass before reaction = Total mass after reaction
Mass of NaHCO3 + Mass of CH3COOH = Mass of CH3COONa + Mass of H2O + Mass of CO2

6.3gm + 15.0gm 18.0gm

Mass of CO2 = 21.3 – 18.0 = 3.3 gm

How much mass of silver nitrate will react with 5.85 g of sodium chloride to produce
14.35 g of silver chloride and 8.5 g of sodium nitrate if law of conservation of mass is
followed?
Answer:
AgNO3 + NaCl → NaNO3 + AgCl
x 5.85 g 8.5 g 14.35g
x + 5.85 = 8.5 + 14.35  x = 17 g
⚫ Law of constant (or definite) proportion: [Proust, 1799]
According to this law, a chemical compound always contains the same elements
combined in the same proportion by mass.
This law was proved within the limits of experimental error by the work of Stas, who
prepared the compounds in several different ways and showed that their composition
was the same.

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For example, CO2 can be prepared by following ways:

(i) by heating CaCO3 : CaCO3 ⎯⎯→ CaO + CO2

(ii) by heating NaHCO3 : 2NaHCO3 ⎯⎯→ Na2CO3 + H2O + CO2

(iii) by burning carbon in O2 : C + O2 ⎯⎯→ CO2

(iv) by CaCO3 and HCl : CaCO3 + 2HCl ⎯⎯→ CaCl2 + H2O + CO2
CO2 is collected separately as a product of each reaction and the analysis of CO2 of each
collection /reveals that it has the combination ratio of carbon and oxygen as 12: 32 or 3: 8.
Important aspects of law of constant proportion
(i) The discovery of isotopes, however, led to slight modification in this law. Since the
isotopes of an element have different atomic weights, it is possible to have the same
chemical compound with different composition according to the isotope used in its
formation.
For example, CO2 using C12 isotope has C : O :: 12 : 32
CO2 using C14 isotope has C : O :: 14 : 32
(ii) Also, the elements combining in the same ratio of their masses may give different
compounds under different experimental conditions. For example, combination of
carbon, hydrogen, and oxygen in the ratio 12 : 3 : 8 may give C2H5OH or CH3OCH3 under
different experimental conditions.

In an experiment, 2.4g of iron oxide on reduction with hydrogen gave 1.68 g of iron.
In another experiment, 2.69 g of iron oxide gave 1.88 g of iron on reduction. Which
law is illustrated from the above data?
Solution:
In first experiment,
2.4 g of iron oxide gave 1.68 g of iron
Mass of iron oxide = 2.4 g
Mass of iron = 1.68 g
 Mass of oxygen = 2.4 – 1.68 = 0.72 g
Ratio of masses of iron and oxygen = 1.68/0.72 = 7 : 3
In second experiment,
Mass of iron oxide = 2.69 g
Mass of iron = 1.88 g
Mass of oxygen = 2.69 – 1.88 = 0.81g
Ratio of masses of iron and oxygen = 1.88/0.81 = 7 : 3
The same ratio confirms that these experiments clarify law of constant proportions.
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Olympiads : Class 9

What mass of hydrochloric acid is needed to decompose 50 g of limestone?

Solution:
CaCO3 + 2HCl ⎯⎯ → CaCl2 + H2O + CO2
100gm 2×36.5
= 73gm
100 g of CaCO3 requires 73 of HCl
73
 50 g of CaCO3 requires × 50 = 36.5 of HCl
100

Copper oxide was prepared by the following methods:

(a) In the first case, 1.75 g of the metal was dissolved in nitric acid and igniting the
residual copper nitrate yielded 2.19 g of copper oxide.
(b) In the second case, 1.14 g of metal dissolved in nitric acid were precipitated as
copper hydroxide by adding caustic alkali solution. The precipitated copper
hydroxide after washing, drying and heating yielded 1.43 g of copper oxide.
(c) In the third case, 1.46 g of copper when strongly heated in a current of air yielded
1.83 g of copper oxide. Show that the given data illustrate the law of constant
composition.
Answer:
In the first experiment.
2.19 g of copper oxide contain 1.75 g of Cu.
 100 g of copper oxide contains
1.75
= × 100 = 79.91 %
2.19
In the second experiment, 1.43 g of copper oxide contain 1.14 g of copper
 100 g of copper oxide contains
1.14
= × 100 = 79.72 %
1.43
In the third experiment, 1.83 g of copper oxide contain 1.46 g of copper
 100 g of copper oxide contains
1.46
= × 100 = 79.78 %
1.83
Thus, the percentage of copper in copper oxide derived from all the three experiments is
nearly the same.
Hence, the above data illustrate the law of constant composition.

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 Chemistry
◼ Dalton's atomic theory
Keeping in view the laws of chemical combinations and the work of Greek philosophers, a
meaningful atomic theory was finally proposed by an English school teacher John Dalton in
1803. The basic postulates of Dalton’s theory are as follows:
(i) Each element is composed of extremely small particles called atoms.
(ii) All atoms of a given element are identical i.e., atoms of a particular element are all alike
but differ from atoms of other elements.
(iii) Atoms of different elements possess different properties (including different masses).
(iv) Atoms are indestructible i.e., atoms are neither created nor destroyed in chemical
reactions.
(v) Atoms of elements take part to form molecules i.e. compounds are formed when atoms
of more than one element combine.
(vi) In a given compound, the relative number and kind of atoms are constant.
Advantages of Dalton’s atomic theory
(i) Dalton’s theory provides us a conceptual picture of matter. One can visualize an element
as being comprised of tiny particles called atoms. Atoms are the basic building blocks of
matter. They are the smallest units of an element that can combine with other elements
in a chemical reaction. In compounds, the atoms of two or more elements combine in
definite arrangements. Mixtures do not involve the intimate interaction between atoms
that are found in compounds.
(ii) It is thus evident that Dalton’s theory embodies several simple laws of chemical
combinations that were known at that time. Postulate (iv) indicates for the law of
conservation of mass. Postulate (vi) indicates the law of definite proportion.
(iii) Dalton’s theory also explains the law of multiple proportion. It also explains that what
makes an element to differ from each other.
Limitations of Dalton’s atomic theory
(i) Distinction between atoms and molecules: According to Dalton, the smallest particle
of an element as well as of compound was atom. However, he called ‘atoms’ as the
smallest particle of compound. Later on Avogadro used the term molecule for the
‘compound atom’ i.e. molecule is the smallest particle of compound.
(ii) It could not justify Berzelius hypothesis: Berzelius, a Swedish chemist, proposed that
under similar conditions of temperature and pressure, equal volume of gases contain
equal number of atoms. But Dalton’s atomic theory could not explain this.
(iii) It could not explain why atoms combine to form a molecule.

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Olympiads : Class 9
(iv) It could not explain the nature of forces which hold the atoms and molecules in solids,
liquid and gaseous state.
(v) It could not explain that why atoms of an element differ in their masses.

◼ 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
STM picture of
all physical changes and chemical reactions. For example, He, Ne, Ar, etc. silicon atoms
Atoms cannot be divided using chemicals. They do consist of parts, which
include protons, neutrons, and electrons but an atom is a basic chemical building block of
matter.
Atomic size
The size of an atom is extremely small. 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.

◼ Symbols of Atoms of Different Elements

Before 1600 AD, alchemists tried to represent the substances that they used to their
experiments by various kinds of pictorial symbols such as a triangle representing the earth
for silver and so on.
Dalton used some other types of symbols to represent the elements. For instance, he drew a
circle for oxygen or a circle with a dot for hydrogen.
Later, Johann Berzelius, a Swiss chemist suggested that the initial capital letter should
represent a particular element such as O for oxygen, H for hydrogen and C for carbon and so
on. But in some cases, the symbols or the alphabet he had suggested did not agree with the
English names of elements. This was because some of his symbols were based on the Latin
names of the elements as well as certain other elements on their Greek names. But the
method suggested by him forms the basis on which the modern system of chemical symbols
and formulae are taken into account.

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Symbols obtained from other languages:
Common name Latin (L) or
Symbol
(of the element) Greek (G) name
Gold Au Aurum (L)
Silver Ag Argentum (L)
Copper Cu Cuprum (L)
Lead Pb Plumbum (L)
Mercury Hg Hydrargyrum (L)
Iron Fe Ferrum (L)
Potassium K Kalium (L)
Tin Sn Stannum (L)
Nitrogen N Nitrogenium (L)
Carbon C Carbonium (L)
Oxygen O Oxygenium (L)
Silicon Si Silex (L)
Calcium Ca Calx (L)
Hydrogen H Hydrogenium (L)
Antimony Sb Stibium (L)
Sodium Na Natrium (L)
Chromium Cr Chromos (G)
Krypton Kr Kryptos (G)

A symbol is a short form that stands for the atom of an element. Each element is denoted by
a symbol. This is usually the first letter of its name in English or Latin or Greek. For example,
the symbol O represents the element oxygen, Latin name for copper is Cuprum, so the letters
Cu represents copper. The symbol of an element also represents the mass of the element
which contains one Avogadro’s number of atoms of that element.
Thus,
H2 represents one molecule of hydrogen.
H2O represents one molecule of water.
Na2CO3 denotes one formula unit of sodium carbonate.
CH3COOH denotes one molecule of acetic acid.
CuSO4.5H2O denotes one formula unit of hydrated copper sulphate.
◼ Atomic Mass
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 the 12C atom 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.
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Olympiads : Class 9
Atomic masses of some common elements (in amu or u)
Element Symbol Atomic mass
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
◼ Molecules
The atoms of the same or different elements are bonded together tightly by strong forces of
attraction also called chemical bonds to form molecules.
Molecules may be classified as molecules of elements and molecules of compounds
Molecules of elements
These 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,
Diatomic → H2, O2
Triatomic → O3
Molecules of compounds
In these 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
proportion. For example,
Diatomic → Hydrogen chloride (HCl)
Triatomic → Water (H2O)
◼ Ions
An ion can be a molecule or atom with a charge (+ and –) due to loss or gain of electron.
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.

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2. On the basis of nature of charge
The ions carrying positive charge are called cations while ions that carry 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.
Cations
The cation is named first and the charge of the cation is expressed as a Roman numeral.
Occasionally suffixes, -‘ic’ and –‘ous’ are used; however, we will not be using them. -ic is for
the higher charged cation, -ous is for the lower charged cation.
For example, ferric is (Fe3+) and ferrous is (Fe2+), Sulphurous acid (H2SO3), Sulphuric acid
(H2SO4).
Metal Ion Name Old Name
Chromium Cr2+ Chromium (II) Chromous
Cr3+ Chromium (III) Chromic
Iron Fe 2+ Iron (II) Ferrous
Fe 3+ Iron (III) Ferric
Cobalt Co2+ Cobalt (II) Cobaltous
Co 3+ Cobalt (III) Cobaltic
Copper Cu+ Coper (I) Cuprous
Cu2+ Coper (II) Cupric
Tin Sn 2+ Tin (II) Stannous
Sn4+ Tin (IV) Stannic
Mercury Hg22+ Mercury (I) Mercurous
Mercury (II) Mercuric
Hg2+
Lead Pb2+ Lead (II) Plumbous
Pb 4+ Lead (IV) Plumbic
Positive ions (Cations)
+1 Charge
Name Formula
Ammonium NH 4+

Copper (I) Cu+

Lithium Li+
Potassium K+
Silver Ag+
Sodium Na+

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+2 Charge
Name Formula
Barium Ba2+
Calcium Ca2+
Copper (II) Cu2+
Iron (II) Fe2+
Lead (II) Pb2+
Magnesium Mg2+
Nickel (II) Ni2+
Strontium Sr2+
Tin (II) Sn2+
Zinc Zn2+
+3 Charge
Name Formula
Aluminium Al3+
Iron (III) Fe3+
+4 Charge
Name Formula
Lead (IV) Pb4+
Tin (IV) Sn4+
◆ Oxo cations
Oxo cations are the polyatomic cations containing one or more oxygen atoms or they are
polyatomic ions with a positive charge that contains oxygen. Some examples are NO+, O2+,
VO2+, VO+ etc.
Anions
(1) Monoatomic anions are named by adding ‘ide’ to the stem name of the element.
Examples are H– hydride; F– fluoride; Se2– selenide; N3– nitride; P3– phosphide.
(2) Certain polyatomic anions have their names ending in ‘ide’:
– –
OH Hydroxide CN Cyanide N 3− Azide
O22− Peroxide S 22− Disulphide −
I3 Triiodide
2–
O2– Superoxide C22− Acetylide NH Imide
O3– Ozonide NH2− Amide
Ions such as HS– and HF2– are named hydrogensulphide and hydrogendifluoride.

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 Chemistry
(3) Oxyacid anions: Oxyanions are named using the name characteristic of the central atom
ending in 'ate', with the oxidation number of the central atom being shown in roman
numerals put in parentheses after the name of the anion. Examples,
SO24− Tetraoxosulphate (VI) SO32− Trioxosulphate(IV)

PO33− Trioxophosphate (III) S2O32− Trioxothiosulphate (VI)

(4) Certain anions have well established trivial names in which prefixed such as per, hypo,
etc. are used. This is in conformity with the name of the corresponding acids: the 'ous'
acids give 'ite' anions and the 'ic' acids give the 'ate' anions. Examples are:
NO3− Nitrate N2O22− Hyponitrite

SO32− Sulphite ClO2− Chlorite

SeO32− Selenite S2O52− Bisulphite

PO33− Phosphite AsO33− Arsenite

NOO2− Peroxonitrite ClO4− Perchlorate

S2O22− Thiosulphite
(5) Polyatomic cations derived from the addition of protons to mono-atomic anions are
named by adding-onium to the root name of the anion, e.g. phosphonium, arsonium,
iodonium ions. Exceptions are ammonium, hydroxyl, hydrazinium, anilinium,
pyridinium, etc.
The following trivial names are acceptable:
H2O Water NH3 Ammonia BH3 Borane
AsH3 Arsine SiH4 Silane SbH3 Stibine
PH3 Phosphine Si2H6 Disilane N2H4 Hydrazine
Radicals : The ions formed after removal of hydrogen ion (H+ ion) from an acid is called acid
radical. The ion formed after removal of hydroxide ions (OH– ions) from a base is called basic
radical.
The positive radical present in the salt comes from the corresponding base and the negative
radical comes from the corresponding acid.
The name of the salts starts with the name of the metal present as positive radical which is
followed by the name of the negative radical. The name of the negative radical is determined
by the name of the acid from which the salt is produced.
◼ Writing the Chemical Formulae
The molecular formula of a compound is electrically neutral, that is, the positive and the
negative valencies of the ions or the radicals present in the compounds add up to zero.
The following steps should be taken while attempting to write a formula:

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Olympiads : Class 9
(a) Write the symbols side by side, usually the one with the positive valency first.
The ion containing more than one atom should be written within brackets.
(b) Write the valency of each ion on the top of its symbol.
(c) Divide the valency numbers by their Highest Common Factor to get the simple ratio.
Ignore the (+) or (–) signs of the radicals. If the valency number is one, it need not be
written.
(d) Write the interchanged valency numbers to the lower right of the radicals, if the radical
is a group of atoms and receives a valency number of more than one then encloses it
within brackets.
Take the case of aluminium sulphate and barium carbonate.
Step 1: Al3+ (SO4)2– Ba2+ (CO3)2–
3 2 2 2
Step 2: Al (SO4) Ba (CO3)
2 3 2 2
Step 3: Dividing the valencies by their HCF.
Al2(SO4)3, Ba1(CO3)1
Step 4: Valency number 1 is not written.
Al2(SO4)3 BaCO3
Hence, we can see that in Al2(SO4)3 molecule, 3 × 2 = +6 charges on two Al3+ ions are
balanced by –2 × 3 = –6 charges on three SO42– ions. Similarly, in BaCO3, molecule, 2 × 1
= +2 charges on one Ba+2 ion balanced by –2 charges on one CO32– ion.
Name of Symbols with Exchanges of Formula
compound valencies valency
Magnesium Mg Cl
2+ 1– Mg Cl MgCl2
Chloride 2 1

Mg1 Cl2
Calcium Oxide Ca2+ O2– Ca O CaO
2 2 [cancelling
common
Ca2 O2
factor]
Aluminium Al3+ (OH)1– Al (OH) Al (OH)3
Hydroxide 3 1

Al1 (OH)3
Phosphorus P3+ O2– P O P2O3
trioxide 3 2

P2 O3

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 Chemistry
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. For example, Al2(SO4)3 is called
aluminium sulphate and not dialuminium trisulphate.
Naming of molecular compounds: They 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'.
For example, H2S is named as hydrogen sulphide.
When there is more than one atom of an element present in the formula of the compound,
then the number of atoms is indicated using appropriate prefixes (mono for 1, di for 2, tri
for 3, tetra for 4 atoms etc.) in the name of the compound.
For example, CO2 is named as carbon dioxide, CCl4 is named as carbon tetra chloride.
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).
For example, two non-metals, nitrogen and oxygen, combine to form different compound
like nitrogen monoxide (NO), nitrogen di-oxide (NO2), di-nitrogen trioxide (N2O3) etc.
But, if two non-metals form only one compound, then prefixes are not used in naming such
compounds.
For example, Hydrogen and sulphur combine to form only one compound H 2S, So, H2S is
named as hydrogen sulphide and not hydrogen mono sulphide.
Importance of molecular formula
The molecular formula of a compound has quantitative significance. It represents:
(1) The representative numbers of different atoms in one molecule of the compound.
(2) The ratios of the respective masses of the elements present in the compound.
(3) The molecular formula represents one molecule of the substance, for example, CO2
represents one molecule of carbon dioxide.
(4) Molecular formula gives the name and the actual number of atoms of all kinds present is
one molecule of the substance. Thus, the formula CO2, indicates that one molecule of
carbon dioxide contains one atom of carbon and two atoms of oxygen.
Let us illustrate. The formulae CO2
(a) This means that the molecular formula of carbon dioxide is CO2.
(b) Each molecule contains one carbon atom joined by chemical bonds with two oxygen
atoms.
(c) The molecular mass of carbon dioxide is 44 given that the atomic mass of carbon is 12
and that of oxygen is 16. (12+ 16+16 = 44)

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Olympiads : Class 9
Similarly,
1. H2 is the molecular formula of hydrogen. This formula shows that one molecule of
hydrogen contains two atoms of hydrogen.
2. H2O is the molecular formula of water. This formula shows that one molecule of water
contains two atoms of hydrogen and one atom of oxygen.
Naming of salts
Acids from which the salt is
Suffix and name of the salt
produced
1. 'ous' acid ‘ite’
For Example: For Example:
Sulphurous acid (H2SO3) CaSO3 Calcium sulphite
Nitrous acid (HNO2) Zn(NO2) 2 Zinc nitrite
Phosphorous acid (H3PO3) Mg3(PO3)2 Magnesium
Phosphite
2. 'ic' acid ‘ate’
For Example: For Example:
Sulphuric acid (H2SO4) ZnSO4 Zinc sulphate
Nitric acid (HNO3) NaNO3 Sodium nitrate
Phosphoric acid (H3PO4) AlPO4 Aluminium phosphate

◼ Molecular mass
We have studied that the mass of the atom of an element is known as its atomic mass. In the
same way, the mass of a molecule of a chemical compound is known as the molecular mass.
Thus, molecular mass of a compound may be defined as
The average relative mass of its molecule as compared to the 1/12 of the mass of
carbon –12 taken as 1 u.
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 molecule of that substance. For example,
Molecular mass of NH3
= atomic mass of N + 3 × atomic mass of H
= 14 + 3 × 1 = 17 u
Formula unit mass
Formula unit: Ionic compounds consist of a very large but equal number of cations and
anions arranged in a definite order in the crystal lattice. The formula of ionic compound
represents only the simplest formula and not the actual formula, which is known as one
formula unit. eg. Na+Cl– is 1 formula unit.

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 Chemistry
In these compounds, we can also use the term formula unit mass in place of molecular mass.
For example, the formula unit mass of NaCl = (23 + 35.5) = 58.5 u.
Formula unit mass of K2CO3 = (2 × 39 + 12 + 3 × 16) = 138 u.
For example, Calculate the mass percentage of different elements present in ethyl alcohol
(C2H5OH)
Molar mass of ethanol = 2 × 12 + 6 × 1 + 16 = 46 g
24
% of C = × 100 = 52.17%
46
6
% of H = × 100 = 13.04%
46
16
% of O = × 100 = 34.7%
46
◼ Mole concept and Avogadro's hypothesis
There are many ways of measuring the amount of a substance, weight and volume being the
most common. But the basic unit of chemistry is the atom or a molecule and to measure the
number of atoms or molecules is, therefore, of foremost importance.
Mole in Latin means heap or mass or pile. A mole of atoms is a collection of atoms whose
total weight is the number of grams equal to the atomic weight. As equal number of moles
of different elements contain equal number of atoms, it is convenient to express amounts of
the elements in terms of moles. Just as a dozen means twelve objects, a score means twenty
objects, chemists have defined a mole as a ‘definite number’ of particles, viz., atoms,
molecules, ions or electrons, etc. This ‘definite number’ is called the Avogadro constant,
equal to 6.023 × 1023, in honour of Amedeo Avogadro.
The value of the Avogadro constant depends on the atomic-mass scale. At present the mole
is defined as the amount of a substance containing as many atoms, molecules, ions, electrons
or other elementary entities as there are carbon atoms in exactly 12 g of 12C.
The following are the definitions of ‘mole’ represented in the form of equations:
Weight in g
(i) Number of moles of molecules =
Molecular mass
Weight in g
(ii) Number of moles of atoms =
Atomic mass
Volume at NTP
(iii) Number of moles of gases =
Standard molar volume
(Standard molar volume is the volume occupied by 1 mole of any gas at NTP, which is
equal to 22.4 litres.)
NTP [ Normal temperature (T = 298 K) and pressure (P = 1 atm)]
(iv) Number of moles of atoms / molecules / ions / electrons =
Number of atoms / molecules / ions / electrons
Avogadro constant
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Olympiads : Class 9

The simplest formula of a compound containing 50% of element X (gram atomic mass
= 10g/mol) and 50% of element Y (gram atomic mass 20g/mol) is:
Solution:
50
Moles of X = =5
10
50
Moles of Y = = 2.5
20
X: Y = 2 : 1
Hence, X2Y.

Calculate the weight of 6.023 × 1023 molecules of CaCO3.

Solution:
Number of molecules
No. of moles of CaCO3 =
Avogadro constant
6.023×  1023
= =1
6.023  1023
Weight of CaCO3 = no. of moles X molecular wt. = 1 × 100 = 100 g.

What will be the number of oxygen atoms in 1 mole of O2?

Solution:
1 mole of O2 contains 6.023 × 1023 molecules of oxygen, and since the oxygen molecule is
diatomic, i.e., 1 molecule contains 2 atoms, the no. of oxygen atoms in 1 mole of O 2 is equal
to 2 × 6.023 × 1023 i.e., 12.046 × 1023 Oxygen atoms.

A piece of Cu weighs 0.635 g. How many atoms of Cu does it contain?

Solution:
No. of moles of Cu = wt. of Cu/ atomic mass of Cu
0.635
atomic wt. of Cu = = 0.01
63.5
No. of atoms of Cu= no. of moles of atoms × Avogadro Constant
0.01 × 6.023 × 1023 = 6.023 × 1021
[16]
 Chemistry

Calculate the number of molecules in 11.2 litres of SO2 gas at NTP.

Solution:
Volume at NTP (Litres) 11.2
No. of moles of SO2 = = = 0.5
Standard molar volume (litres) 22.4
No. of molecules of SO2 = No. of moles × Av. Const.
= 0.5 × 6.023 × 1023 = 3.011 × 1023

From 200 mg of CO2, 1021 molecules are removed. How many moles of CO2 are left?
Solution:
wt. in g 0.2
Total no. of moles of CO2 = = = 0.004545
Mol.wt. 44
1021
No. of moles removed = = 0.00166
6.022  1023
No. of moles of CO2 left = 0.004545–0.00166 = 0.00288

10

Calculate the volume of 20 g of hydrogen gas at NTP.

Solution:
20
Moles of hydrogen gas = = 10
2
Volume of the gas at NTP = No. of moles × 22.4
= 10 × 22.4 = 224 litres

11

Calculate the number of moles and the number of atoms of H, S and O in 5 moles of
H2SO4.
Solution:
1 molecule of H2SO4 contains 2 atoms of H or 1 mole of H2SO4 contains 2 moles of H or 5
moles of H2SO4 contains 10 moles of H.
1 mole of H2SO4 contains 1 mole of S or 5 moles of H2SO4 contains 5 moles of S and again,

[17]
Olympiads : Class 9
1 mole of H2SO4 contains 4 moles of O or 5 moles of H2SO4 contains 20 moles of O.
No. of atoms of H = 10 × 6.023 × 1023
No. of atoms of S = 5 × 6.023 × 1023
No. of atoms of O = 20 × 6.023 × 1023

12

Calculate the volume occupied by 1 mole of He, H and O atoms at NTP.

Solution:
As He is monoatomic, 1 mole of it will occupy 22.4 litres at NTP.
But hydrogen and oxygen being diatomic gases, 1 mole of their respective atoms will occupy
a volume of 11.2 litres at NTP.

Memory Map
1 mole of
carbon atoms

6.022 × 10 23

atoms of c 12 g of carbon
1 mole of
hydrogen atoms

6.022 ×10 23

1 g of hydrogen
atoms of h

1 mole of any particle

(atoms, molecule, ions)

6.022× 10 number
23
relative mass of
those particle in grams
of that particle

1 mole of
molecules

6.022 × 10 23 molecular mass

number of molecules in grams

[18]
 Chemistry
Laws of Chemical Combination

Law of Law of
Conservation Constant
of mass Proportion

Dalton’s Atomic Theory

(atom is indivisible)

Advantage Limitation
1. Atom can combine 1. It could not explain why
2. Laws of chemical atom combines.
combination 2. Why atoms of an element
differ in their masses.

Atom
(Smallest particle of an element)

Symbol Atomic Mass

Shorthand representation Mass of 1/12 of C12 = 1a.m.u.
of an element

+ atom

Molecules

Types of Molecular Mass Formula Atomicity Molecular

Molecules Unit Mass Number of atoms formula
in a molecule

Molecule of element Molecule of compounds

(N2, H2) (CO2, H2O)
Ions
Charged particle

On the basis of charge On the basis of number of atoms

Cation Anion Monoatomic Diatomic Triatomic Polyatomic

Mole concept
Definite number of particles,
chemist dozen

Avogadro number Relative of those

(6.023 × 1023) particles in grams

[19]