UNIT 1

SOME BASIC CONCEPTS OF CHEMISTRY

Chemistry is the science of molecules and their

transformations. It is the science not so much of the one

hundred elements but of the infinite variety of molecules that

may be built from them.

Roald Hoffmann

Science can be viewed as a continuing human effort to

systematise knowledge for describing and understanding

nature. You have learnt in your previous classes that we come

across diverse substances present in nature and changes in

them in daily life. Curd formation from milk, formation of

vinegar from sugarcane juice on keeping for prolonged time

and rusting of iron are some of the examples of changes which

we come across many times. For the sake of convenience,

science is sub-divided into various disciplines: chemistry,

physics, biology, geology, etc. The branch of science that

studies the preparation, properties, structure and reactions

of material substances is called chemistry.

DEVELOPMENT OF CHEMISTRY

Chemistry, as we understand it today, is not a very old

discipline. Chemistry was not studied for its own sake, rather

it came up as a result of search for two interesting things:

i. Philosopher’s stone (Paras) which would convert

all baser metals e.g., iron and copper into gold.

ii.‘Elexir of life’ which would grant immortality.

People in ancient India, already had the knowledge of many

scientific phenomenon much before the advent of modern

science. They applied that knowledge in various walks of

life. Chemistry developed mainly in the form of Alchemy

and Iatrochemistry during 1300-1600 CE. Modern

chemistry took shape in the 18

century Europe, after a

few centuries of alchemical traditions which were

introduced in Europe by the Arabs.

After studying this unit, you will be

able to

•

••

• appreciate the contribution of

India in the development of

chemistry understand the role of

chemistry in different spheres of

life;

••

• explain the characteristics of

three states of matter;

•

••

• classify different substances into

elements, compounds and

mixtures;

••

• use scientific notations and

determine significant figures;

••

• differentiate between precision and

accuracy;

••

• define SI base units and convert

physical quantities from one

system of units to another;

••

• explain various laws of chemical

combination;

••

• appreciate significance of atomic

mass, average atomic mass,

molecular mass and formula

mass;

••

• describe the terms – mole and

molar mass;

••

• calculate the mass per cent of

component elements constituting

a compound;

••

• determine empirical formula and

molecular formula for a compound

from the given experimental data;

and

••

• perform the stoichiometric

calculations.

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2 CHEMISTRY

Other cultures – especially the Chinese and

the Indian – had their own alchemical traditions.

These included much knowledge of chemical

processes and techniques.

In ancient India, chemistry was called

Rasayan Shastra, Rastantra, Ras Kriya or

Rasvidya. It included metallurgy, medicine,

manufacture of cosmetics, glass, dyes, etc.

Systematic excavations at Mohenjodaro in

Sindh and Harappa in Punjab prove that the

story of development of chemistry in India is

very old. Archaeological findings show that

baked bricks were used in construction work.

It shows the mass production of pottery, which

can be regarded as the earliest chemical process,

in which materials were mixed, moulded and

subjected to heat by using fire to achieve

desirable qualities. Remains of glazed pottery

have been found in Mohenjodaro. Gypsum

cement has been used in the construction work.

It contains lime, sand and traces of CaCO

Harappans made faience, a sort of glass which

was used in ornaments. They melted and forged

a variety of objects from metals, such as lead,

silver, gold and copper. They improved the

hardness of copper for making artefacts by

using tin and arsenic. A number of glass objects

were found in Maski in South India (1000–900

BCE), and Hastinapur and Taxila in North

India (1000–200 BCE). Glass and glazes were

coloured by addition of colouring agents like

metal oxides.

Copper metallurgy in India dates back to

the beginning of chalcolithic cultures in the

subcontinent. There are much archeological

evidences to support the view that technologies

for extraction of copper and iron were developed

indigenously.

According to Rigveda, tanning of leather

and dying of cotton were practised during

1000–400 BCE. The golden gloss of the black

polished ware of northen India could not be

replicated and is still a chemical mystery. These

wares indicate the mastery with which kiln

temperatures could be controlled. Kautilya’s

Arthashastra describes the production of salt

from sea.

A vast number of statements and material

described in the ancient Vedic literature can

be shown to agree with modern scientific

findings. Copper utensils, iron, gold, silver

ornaments and terracotta discs and painted

grey pottery have been found in many

archaeological sites in north India. Sushruta

Samhita explains the importance of Alkalies.

The Charaka Samhita mentions ancient

indians who knew how to prepare sulphuric

acid, nitric acid and oxides of copper, tin and

zinc; the sulphates of copper, zinc and iron and

the carbonates of lead and iron.

Rasopanishada describes the preparation

of gunpowder mixture. Tamil texts also

describe the preparation of fireworks using

sulphur, charcoal, saltpetre (i.e., potassium

nitrate), mercury, camphor, etc.

Nagarjuna was a great Indian scientist. He

was a reputed chemist, an alchemist and a

metallurgist. His work Rasratnakar deals with

the formulation of mercury compounds. He has

also discussed methods for the extraction of

metals, like gold, silver, tin and copper. A book,

Rsarnavam, appeared around 800 CE. It

discusses the uses of various furnaces, ovens

and crucibles for different purposes. It

describes methods by which metals could be

identified by flame colour.

Chakrapani discovered mercury sulphide.

The credit for inventing soap also goes to him.

He used mustard oil and some alkalies as

ingredients for making soap. Indians began

making soaps in the 18

century CE. Oil of

Eranda and seeds of Mahua plant and calcium

carbonate were used for making soap.

The paintings found on the walls of Ajanta

and Ellora, which look fresh even after ages,

testify to a high level of science achieved in

ancient India. Varähmihir’s Brihat Samhita is

a sort of encyclopaedia, which was composed

in the sixth century CE. It informs about the

preparation of glutinous material to be applied

on walls and roofs of houses and temples. It

was prepared entirely from extracts of various

plants, fruits, seeds and barks, which were

concentrated by boiling, and then, treated with

various resins. It will be interesting to test such

materials scientifically and assess them for use.

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3SOME BASIC CONCEPTS OF CHEMISTRY

A number of classical texts, like

Atharvaveda (1000 BCE) mention some dye

stuff, the material used were turmeric, madder,

sunflower, orpiment, cochineal and lac. Some

other substances having tinting property were

kamplcica, pattanga and jatuka.

Varähmihir’s Brihat Samhita gives

references to perfumes and cosmetics. Recipes

for hair dying were made from plants, like

indigo and minerals like iron power, black iron

or steel and acidic extracts of sour rice gruel.

Gandhayukli describes recipes for making

scents, mouth perfumes, bath powders,

incense and talcum power.

Paper was known to India in the

century as account of Chinese traveller

I-tsing describes. Excavations at Taxila indicate

that ink was used in India from the fourth

century. Colours of ink were made from chalk,

red lead and minimum.

It seems that the process of fermentation

was well-known to Indians. Vedas and

Kautilya’s Arthashastra mention about many

types of liquors. Charaka Samhita also

mentions ingredients, such as barks of plants,

stem, flowers, leaves, woods, cereals, fruits and

sugarcane for making Asavas.

The concept that matter is ultimately made

of indivisible building blocks, appeared in

India a few centuries BCE as a part of

philosophical speculations. Acharya Kanda,

born in 600 BCE, originally known by the

name Kashyap, was the first proponent of the

‘atomic theory’. He formulated the theory of

very small indivisible particles, which he

named ‘Paramãnu’ (comparable to atoms). He

authored the text Vaiseshika Sutras.

According to him, all substances are

aggregated form of smaller units called atoms

(Paramãnu), which are eternal, indestructible,

spherical, suprasensible and in motion in the

original state. He explained that this individual

entity cannot be sensed through any human

organ. Kanda added that there are varieties of

atoms that are as different as the different

classes of substances. He said these

(Paramãnu) could form pairs or triplets, among

other combinations and unseen forces cause

interaction between them. He conceptualised

this theory around 2500 years before John

Dalton (1766-1844).

Charaka Samhita is the oldest Ayurvedic

epic of India. It describes the treatment of

diseases. The concept of reduction of particle

size of metals is clearly discussed in Charaka

Samhita. Extreme reduction of particle size is

termed as nanotechnology.

Charaka Samhita

describes the use of bhasma of metals in the

treatment of ailments. Now-a-days, it has been

proved that bhasmas have nanoparticles of

metals.

After the decline of alchemy, Iatrochemistry

reached a steady state, but it too declined due

to the introduction and practise of western

medicinal system in the 20

century. During

this period of stagnation, pharmaceutical

industry based on Ayurveda continued to

exist, but it too declined gradually. It took

about 100-150 years for Indians to learn and

adopt new techniques. During this time, foreign

products poured in. As a result, indigenous

traditional techniques gradually declined.

Modern science appeared in Indian scene in

the later part of the nineteenth century. By the

mid-nineteenth century, European scientists

started coming to India and modern chemistry

started growing.

From the above discussion, you have learnt

that chemistry deals with the composition,

structure, properties and interection of matter

and is of much use to human beings in daily

life. These aspects can be best described and

understood in terms of basic constituents of

matter that are atoms and molecules. That

is why, chemistry is also called the science of

atoms and molecules. Can we see, weigh and

perceive these entities (atoms and molecules)?

Is it possible to count the number of atoms

and molecules in a given mass of matter and

have a quantitative relationship between the

mass and the number of these particles? We

will get the answer of some of these questions

in this Unit. We will further describe how

physical properties of matter can be

quantitatively described using numerical

values with suitable units.

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4 CHEMISTRY

1.1 IMPORTANCE OF CHEMISTRY

Chemistry plays a central role in science and

is often intertwined with other branches of

science.

Principles of chemistry are applicable in

diverse areas, such as weather patterns,

functioning of brain and operation of a

computer, production in chemical industries,

manufacturing fertilisers, alkalis, acids, salts,

dyes, polymers, drugs, soaps, detergents,

metals, alloys, etc., including new material.

Chemistry contributes in a big way to the

national economy. It also plays an important

role in meeting human needs for food,

healthcare products and other material

aimed at improving the quality of life. This

is exemplified by the large-scale production

of a variety of fertilisers, improved variety of

pesticides and insecticides. Chemistry

provides methods for the isolation of life-

saving drugs from natural sources and

makes possible synthesis of such drugs.

Some of these drugs are cisplatin and taxol,

which are effective in cancer therapy. The

drug AZT (Azidothymidine) is used for

helping AIDS patients.

Chemistry contributes to a large extent

in the development and growth of a nation.

With a better understanding of chemical

principles it has now become possible to

design and synthesise new material having

specific magnetic, electric and optical

properties. This has lead to the production

of superconducting ceramics, conducting

polymers, optical fibres, etc. Chemistry has

helped in establishing industries which

manufacture utility goods, like acids,

alkalies, dyes, polymesr metals, etc. These

industries contribute in a big way to the

economy of a nation and generate

employment.

In recent years, chemistry has helped

in dealing with some of the pressing aspects

of environmental degradation with a fair

degree of success. Safer alternatives to

environmentally hazardous refrigerants, like

CFCs (chlorofluorocarbons), responsible for

ozone depletion in the stratosphere, have

been successfully synthesised. However,

many big environmental problems continue

to be matters of grave concern to the

chemists. One such problem is the

management of the Green House gases, like

methane, carbon dioxide, etc. Understanding

of biochemical processes, use of enzymes for

large-scale production of chemicals and

synthesis of new exotic material are some of

the intellectual challenges for the future

generation of chemists. A developing country,

like India, needs talented and creative

chemists for accepting such challenges. To

be a good chemist and to accept such

challanges, one needs to understand the

basic concepts of chemistry, which begin with

the concept of matter. Let us start with the

nature of matter.

1.2 NATURE OF MATTER

You are already familiar with the term matter

from your earlier classes. Anything which has

mass and occupies space is called matter

Everything around us, for example, book, pen,

pencil, water, air, all living beings, etc., are

composed of matter. You know that they have

mass and they occupy space. Let us recall the

characteristics of the states of matter, which

you learnt in your previous classes.

1.2.1 States of Matter

You are aware that matter can exist in three

physical states viz. solid, liquid and gas. The

constituent particles of matter in these three

states can be represented as shown in Fig. 1.1.

Particles are held very close to each other

in solids in an orderly fashion and there is not

much freedom of movement. In liquids, the

particles are close to each other but they can

move around. However, in gases, the particles

are far apart as compared to those present in

solid or liquid states and their movement is

easy and fast. Because of such arrangement

of particles, different states of matter exhibit

the following characteristics:

(i) Solids have definite volume and definite

shape.

(ii) Liquids have definite volume but do not

have definite shape. They take the shape

of the container in which they are placed.

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5SOME BASIC CONCEPTS OF CHEMISTRY

(iii) Gases have neither definite volume nor

definite shape. They completely occupy the

space in the container in which they are placed.

These three states of matter are

interconvertible by changing the conditions of

temperature and pressure.

Solid liquid Gas

On heating, a solid usually changes to a

liquid, and the liquid on further heating changes

to gas (or vapour). In the reverse process, a gas

on cooling liquifies to the liquid and the liquid

on further cooling freezes to the solid.

1.2.2. Classification of Matter

In class IX (Chapter 2), you have learnt that

at the macroscopic or bulk level, matter can be

classified as mixture or pure substance. These

can be further sub-divided as shown in Fig. 1.2.

When all constituent particles of a

substance are same in chemical nature, it is

said to be a pure substance. A mixture

contains many types of particles.

A mixture contains particles of two or more

pure substances which may be present in it in

any ratio. Hence, their composition is variable.

Pure sustances forming mixture are called its

components. Many of the substances present

around you are mixtures. For example, sugar

solution in water, air, tea, etc., are all mixtures.

A mixture may be homogeneous or

heterogeneous. In a homogeneous mixture,

the components completely mix with each other.

This means particles of components of the

mixture are uniformly distributed throughout

Fig. 1.2 Classification of matter

Fig. 1.1 Arrangement of particles in solid, liquid

and gaseous state

the bulk of the mixture and its composition is

uniform throughout. Sugar solution and air

are the examples of homogeneous mixtures.

In contrast to this, in a heterogeneous

mixture, the composition is not uniform

throughout and sometimes different

components are visible. For example, mixtures

of salt and sugar, grains and pulses along with

some dirt (often stone pieces), are

heterogeneous mixtures. You can think of

many more examples of mixtures which you

come across in the daily life. It is worthwhile to

mention here that the components of a

mixture can be separated by using physical

methods, such as simple

hand-picking, filtration, crystallisation,

distillation, etc.

Pure substances have characteristics

different from mixtures. Constituent particles

of pure substances have fixed composition.

Copper, silver, gold, water and glucose are

some examples of pure substances. Glucose

contains carbon, hydrogen and oxygen in a

fixed ratio and its particles are of same

composition. Hence, like all other pure

substances, glucose has a fixed composition.

Also, its constituents—carbon, hydrogen and

oxygen—cannot be separated by simple

physical methods.

Pure substances can further be

classified into elements and compounds.

Particles of an element consist of only one

type of atoms. These particles may exist as

atoms or molecules. You may be familiar

with atoms and molecules from the

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6 CHEMISTRY

previous classes; however, you will be

studying about them in detail in Unit 2.

Sodium, copper, silver, hydrogen, oxygen,

etc., are some examples of elements. Their

all atoms are of one type. However, the

atoms of different elements are different in

nature. Some elements, such as sodium or

copper, contain atoms as their constituent

particles, whereas, in some others, the

constituent particles are molecules which

are formed by two or more atoms. For

example, hydrogen, nitrogen and oxygen

gases consist of molecules, in which two

atoms combine to give their respective

molecules. This is illustrated in Fig. 1.3.

carbon atom. Thus, the atoms of different

elements are present in a compound in a fixed

and definite ratio and this ratio is characteristic

of a particular compound. Also, the properties

of a compound are different from those of its

constituent elements. For example, hydrogen

and oxygen are gases, whereas, the compound

formed by their combination i.e., water is a

liquid. It is interesting to note that hydrogen

burns with a pop sound and oxygen is a

supporter of combustion, but water is used

as a fire extinguisher.

1.3 PROPERTIES OF MATTER AND

THEIR MEASUREMENT

1.3.1 Physical and chemical properties

Every substance has unique or characteristic

properties. These properties can be classified

into two categories —

physical properties,

such as colour, odour, melting point, boiling

point, density, etc., and chemical properties,

like composition, combustibility, ractivity with

acids and bases, etc.

Physical properties can be measured or

observed without changing the identity or the

composition of the substance. The measurement

or observation of chemical properties requires

a chemical change to occur. Measurement of

physical properties does not require occurance

of a chemical change. The examples of chemical

properties are characteristic reactions of different

substances; these include acidity or basicity,

combustibility, etc. Chemists describe, interpret

and predict the behaviour of substances on the

basis of knowledge of their physical and chemical

properties, which are determined by careful

measurement and experimentation. In the

When two or more atoms of different

elements combine together in a definite ratio,

the molecule of a compound is obtained.

Moreover, the constituents of a compound

cannot be separated into simpler

substances by physical methods. They can

be separated by chemical methods.

Examples of some compounds are water,

ammonia, carbon dioxide, sugar, etc. The

molecules of water and carbon dioxide are

represented in Fig. 1.4.

Note that a water molecule comprises two

hydrogen atoms and one oxygen atom.

Similarly, a molecule of carbon dioxide

contains two oxygen atoms combined with one

Fig. 1.4 A depiction of molecules of water and

carbon dioxide

Water molecule

Carbon dioxide

molecule (CO

)

Fig. 1.3 A representation of atoms and molecules

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7SOME BASIC CONCEPTS OF CHEMISTRY

following section, we will learn about the

measurement of physical properties.

1.3.2 Measurement of physical properties

Quantitative measurement of properties is

reaquired for scientific investigation. Many

properties of matter, such as length, area,

volume, etc., are quantitative in nature. Any

quantitative observation or measurement is

represented by a number followed by units

in which it is measured. For example, length

of a room can be represented as 6 m; here, 6

is the number and m denotes metre, the unit

in which the length is measured.

Earlier, two different systems of

measurement, i.e., the English System and

the Metric System were being used in

different parts of the world. The metric system,

which originated in France in late eighteenth

century, was more convenient as it was based

on the decimal system. Late, need of a common

standard system was felt by the scientific

community. Such a system was established

in 1960 and is discussed in detail below.

1.3.3 The International System of Units (SI)

The International System of Units (in French

Le Systeme International d’Unités —

abbreviated as SI) was established by the

General Conference on Weights and

Measures (CGPM from Conference

Generale des Poids et Measures). The CGPM

is an inter-governmental treaty organisation

created by a diplomatic treaty known as

Metre Convention, which was signed in Paris

in 1875.

The SI system has seven

base units and

they are listed in Table 1.1. These units pertain

to the seven fundamental scientific quantities.

The other physical quantities, such as speed,

volume, density, etc., can be derived from these

quantities.

Table 1.1 Base Physical Quantities and their Units

Base Physical Symbol Name of Symbol

Quantity for SI Unit for SI

Quantity Unit

Length l metre m

Mass m kilogram kg

Time t second s

Electric current I ampere A

Thermodynamic T kelvin K

temperature

Amount of substance n mole mol

Luminous intensity I

candela cd

Maintaining the National

Standards of Measurement

The system of units, including unit

definitions, keeps on changing with time.

Whenever the accuracy of measurement

of a particular unit was enhanced

substantially by adopting new principles,

member nations of metre treaty (signed in

1875), agreed to change the formal

definition of that unit. Each modern

industrialised country, including India, has

a National Metrology Institute (NMI), which

maintains standards of measurements.

This responsibility has been given to the

National Physical Laboratory (NPL),

New Delhi. This laboratory establishes

experiments to realise the base units and

derived units of measurement and

maintains National Standards of

Measurement. These standards are

periodically inter-compared with

standards maintained at other National

Metrology Institutes in the world, as well

as those, established at the International

Bureau of Standards in Paris.

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8 CHEMISTRY

These prefixes are listed in Table 1.3.

Let us now quickly go through some of the

quantities which you will be often using in this

book.

The definitions of the SI base units are given

in Table 1.2.

The SI system allows the use of prefixes to

indicate the multiples or submultiples of a unit.

Table 1.2 Definitions of SI Base Units

Unit of length

Unit of mass

Unit of time

Unit of electric

current

Unit of

thermodynamic

temperature

Unit of amount

of substance

Unit of luminous

Intensity

The metre, symbol m is the SI unit of length. It is defined by

taking the fixed numerical value of the speed of light in

vacuum c to be 299792458 when expressed in the unit ms

–1

where the second is defined in terms of the caesium

frequency

∆

The kilogram, symbol kg. is the SI unit of mass. It is defined

by taking the fixed numerical value of the Planck constant

to be 6.62607015

–34

when expressed in the unit Js, which

is equal to kgm

–1

, where the metre and the second are defined

in terms of c and

∆

The second symbol s, is the SI unit of time. It is defined by

taking the fixed numerical value of the caesium frequency

∆

, the unperturbed ground-state hyperfine transition

frequency of the caesium-133 atom, to be 9192631770 when

expressed in the unit Hz, which is equal to s

–1

The ampere, symbol A, is the SI unit of electric current. It is

defined by taking the fixed numerical value of the elementary

charge e to be 1.602176634

–19

when expressed in the unit

C, which is equal to As, where the second is defined in terms

∆

The Kelvin, symbol k, is the SI unit of thermodynamic

temperature. It is defined by taking the fixed numerical value

of the Boltzmann constant k to be 1.380649

–23

when

expressed in the unit JK

–1

, which is equal to kgm

–2

–1

where

the kilogram, metre and second are defined in terms of h, c

and

∆

The mole, symbol mol, is the SI unit of amount of substance.

One mole contains exactly 6.02214076

elementary

entities. This number is the fixed numerical value of the

Avogadro constant, N

when expressed in the unit mol

–1

and

is called the Avogadro number. The amount of substance,

symbol n, of a system is a measure of the number of specified

elementary entities. An elementary entity may be an atom,

a molecule, an ion, an electron, any other particle or specified

group of particles.

The candela, symbol cd is the SI unit of luminous intensity

in a given direction. It is defined by taking the fixed numerical

value of the luminous efficacy of monochromatic radiation

of frequency 540×10

Hz, K

, to be 683 when expressed in the

unit lm·W

–1

, which is equal to cd·sr·W

–1

, or cd sr kg

–1

-2

, where the kilogram, metre and second are defined in

terms of h, c and

∆

metre

kilogram

second

ampere

kelvin

mole

Candela

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9SOME BASIC CONCEPTS OF CHEMISTRY

Fig. 1.5 Analytical balance

1.3.4 Mass and Weight

Mass of a substance is the amount of matter

present in it, while weight is the force exerted

by gravity on an object. The mass of a

substance is constant, whereas, its weight

may vary from one place to another due to

change in gravity. You should be careful in

using these terms.

The mass of a substance can be determined

accurately in the laboratory by using an

analytical balance (Fig. 1.5).

The SI unit of mass as given in Table 1.1 is

kilogram. However, its fraction named as gram

(1 kg = 1000 g), is used in laboratories due to

the smaller amounts of chemicals used in

chemical reactions.

1.3.5 Volume

Volume is the amont of space occupied by a

substance. It has the units of (length)

. So in

SI system, volume has units of m

. But again,

in chemistry laboratories, smaller volumes are

used. Hence, volume is often denoted in cm

or dm

units.

A common unit, litre (L) which is not an SI

unit, is used for measurement of volume of

liquids.

1 L = 1000 mL , 1000 cm

= 1 dm

Fig. 1.6 helps to visualise these relations.

Table 1.3 Prefixes used in the SI System

Multiple Prefix Symbol

–24

yocto y

–21

zepto z

–18

atto a

–15

femto f

–12

pico p

–9

nano n

–6

micro µ

–3

milli m

–2

centi c

–1

deci d

10 deca da

hecto h

kilo k

mega M

giga G

tera T

peta P

exa E

zeta Z

yotta Y

Fig. 1.6 Different units used to express

volume

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10 CHEMISTRY

In the laboratory, the volume of liquids or

solutions can be measured by graduated

cylinder, burette, pipette, etc. A volumetric

flask is used to prepare a known volume of a

solution. These measuring devices are shown

in Fig. 1.7.

Fig. 1.7 Some volume measuring devices

1.4 UNCERTAINTY IN MEASUREMENT

Many a time in the study of chemistry, one

has to deal with experimental data as well as

theoretical calculations. There are meaningful

ways to handle the numbers conveniently and

fahrenheit) and K (kelvin). Here, K is the SI

unit. The thermometers based on these scales

are shown in Fig. 1.8. Generally, the

thermometer with celsius scale are calibrated

from 0° to 100°, where these two

temperatures are the freezing point and the

boiling point of water, respectively. The

fahrenheit scale is represented between 32°

to 212°.

The temperatures on two scales are related

to each other by the following relationship:

° = ° +

( )

F C

The kelvin scale is related to celsius scale

as follows:

K = °C + 273.15

It is interesting to note that temperature

below 0 °C (i.e., negative values) are possible

in Celsius scale but in Kelvin scale, negative

temperature is not possible.

Fig. 1.8 Thermometers using different

temperature scales

1.3.6 Density

The two properties — mass and volume

discussed above are related as follows:

Mass

Density

Volume

Density of a substance is its amount of mass

per unit volume. So, SI units of density can be

obtained as follows:

SI unit of density =

SI unit of mass

SI unit of volume

or kg m

–3

This unit is quite large and a chemist often

expresses density in g cm

–3

, where mass is

expressed in gram and volume is expressed in

. Density of a substance tells us about how

closely its particles are packed. If density is

more, it means particles are more closely

packed.

1.3.7 Temperature

There are three common scales to measure

temperature — °C (degree celsius), °F (degree

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11SOME BASIC CONCEPTS OF CHEMISTRY

present the data realistically with certainty to

the extent possible. These ideas are discussed

below in detail.

1.4.1 Scientific Notation

As chemistry is the study of atoms and

molecules, which have extremely low masses

and are present in extremely large numbers,

a chemist has to deal with numbers as large

as 602, 200,000,000,000,000,000,000 for

the molecules of 2 g of hydrogen gas or as

small as 0.00000000000000000000000166

g mass of a H atom. Similarly, other constants

such as Planck’s constant, speed of light,

charges on particles, etc., involve numbers of

the above magnitude.

It may look funny for a moment to write

or count numbers involving so many zeros

but it offers a real challenge to do simple

mathematical operations of addition,

subtraction, multiplication or division with

such numbers. You can write any two

numbers of the above type and try any one of

the operations you like to accept as a

challenge, and then, you will really appreciate

the difficulty in handling such numbers.

This problem is solved by using scientific

notation for such numbers, i.e., exponential

notation in which any number can be

represented in the form N × 10

, where n is an

exponent having positive or negative values

and N is a number (called digit term) which

varies between 1.000... and 9.999....

Thus, we can write 232.508 as

2.32508 ×10

in scientific notation. Note that

while writing it, the decimal had to be moved

to the left by two places and same is the

exponent (2) of 10 in the scientific notation.

Similarly, 0.00016 can be written as

1.6 × 10

–4

. Here, the decimal has to be moved

four places to the right and (–4) is the exponent

in the scientific notation.

While performing mathematical operations

on numbers expressed in scientific notations,

the following points are to be kept in mind.

Reference Standard

After defining a unit of measurement such

as the kilogram or the metre, scientists

agreed on reference standards that make

it possible to calibrate all measuring

devices. For getting reliable measurements,

all devices such as metre sticks and

analytical balances have been calibrated by

their manufacturers to give correct

readings. However, each of these devices

is standardised or calibrated against some

reference. The mass standard is the

kilogram since 1889. It has been defined

as the mass of platinum-iridium (Pt-Ir)

cylinder that is stored in an airtight jar at

International Bureau of Weights and

Measures in Sevres, France. Pt-Ir was

chosen for this standard because it is

highly resistant to chemical attack and its

mass will not change for an extremely long

time.

Scientists are in search of a new

standard for mass. This is being attempted

through accurate determination of

Avogadro constant. Work on this new

standard focuses on ways to measure

accurately the number of atoms in a well-

defined mass of sample. One such method,

which uses X-rays to determine the atomic

density of a crystal of ultrapure silicon, has

an accuracy of about 1 part in 10

but has

not yet been adopted to serve as a

standard. There are other methods but

none of them are presently adequate to

replace the Pt-Ir cylinder. No doubt,

changes are expected within this decade.

The metre was originally defined as the

length between two marks on a Pt-Ir bar

kept at a temperature of 0°C (273.15 K). In

1960 the length of the metre was defined

as 1.65076373 × 10

times the

wavelength

of light emitted by a krypton laser.

Although this was a cumbersome number,

it preserved the length of the metre at its

agreed value. The metre was redefined in

1983 by CGPM as the length of path

travelled by light in vacuum during a time

interval of 1/299 792 458 of a second.

Similar to the length and the mass, there

are reference standards for other physical

quantities.

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12 CHEMISTRY

Multiplication and Division

These two operations follow the same rules

which are there for exponential numbers, i.e.

by an analytical balance is slightly higher

than the mass obtained by using a platform

balance. Therefore, digit 4 placed after

decimal in the measurement by platform

balance is uncertain.

The uncertainty in the experimental or the

calculated values is indicated by mentioning

the number of significant figures. Significant

figures are meaningful digits which are

known with certainty plus one which is

estimated or uncertain. The uncertainty is

indicated by writing the certain digits and the

last uncertain digit. Thus, if we write a result

as 11.2 mL, we say the 11 is certain and 2 is

uncertain and the uncertainty would be

in the last digit. Unless otherwise stated, an

uncertainty of +1 in the last digit is always

understood.

There are certain rules for determining the

number of significant figures. These are

stated below:

(1) All non-zero digits are significant. For

example in 285 cm, there are three

significant figures and in 0.25 mL, there

are two significant figures.

(2) Zeros preceding to first non-zero digit are

not significant. Such zero indicates the

position of decimal point. Thus, 0.03 has

one significant figure and 0.0052 has two

significant figures.

(3) Zeros between two non-zero digits are

significant. Thus, 2.005 has four

significant figures.

(4) Zeros at the end or right of a number are

significant, provided they are on the right

side of the decimal point. For example,

0.200 g has three significant figures. But,

if otherwise, the terminal zeros are not

significant if there is no decimal point. For

example, 100 has only one significant

figure, but 100. has three significant

figures and 100.0 has four significant

figures. Such numbers are better

represented in scientific notation. We can

express the number 100 as 1×10

for one

significant figure, 1.0×10

for two

significant figures and 1.00×10

for three

significant figures.

Addition and Subtraction

For these two operations, first the numbers are

written in such a way that they have the same

exponent. After that, the coefficients (digit

terms) are added or subtracted as the case

may be.

Thus, for adding 6.65 × 10

and 8.95 × 10

exponent is made same for both the numbers.

Thus, we get (6.65 × 10

) + (0.895 × 10

)

Then, these numbers can be added as follows

(6.65 + 0.895) × 10

= 7.545 × 10

Similarly, the subtraction of two numbers can

be done as shown below:

(2.5 × 10

–2

) – (4.8 × 10

–3

)

= (2.5 × 10

–2

) – (0.48 × 10

–2

)

= (2.5 – 0.48) × 10

–2

= 2.02 × 10

–2

1.4.2 Significant Figures

Every experimental measurement has some

amount of uncertainty associated with it

because of limitation of measuring instrument

and the skill of the person making the

measurement. For example, mass of an object

is obtained using a platform balance and it

comes out to be 9.4g. On measuring the mass

of this object on an analytical balance, the

mass obtained is 9.4213g. The mass obtained

5 6 10 6 9 10 5 6 6 9 10

5 6 6 9 10

5 8 5 8

. . . .

. .

( )

× ×

( )

(

)

= 38..64 10

= 3.864 10

( )

− −

− + −

9 8 10 2 5 10 9 8 2 5 10

2 6

. . . .×

2 6

9 8 2 5 10

2 7 10

( )

− −

−

( )

= 24.50 10

= 2.450 10

. .

.55 10

2 7 5 5 10 1

3 4

( )

− − −

−

= =0.4909 0

= 4.909 10

. .

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13SOME BASIC CONCEPTS OF CHEMISTRY

(5) Counting the numbers of object,

for example, 2 balls or 20 eggs, have

infinite significant figures as these

are exact numbers and can be

represented by writing infinite number of

zeros after placing a decimal i.e.,

2 = 2.000000 or 20 = 20.000000.

In numbers written in scientific notation,

all digits are significant e.g., 4.01×10

has three

significant figures, and 8.256 × 10

–3

has four

significant figures.

However, one would always like the results

to be precise and accurate. Precision and

accuracy are often referred to while we talk

about the measurement.

Precision refers to the closeness of various

measurements for the same quantity. However,

accuracy is the agreement of a particular value

to the true value of the result. For example, if

the true value for a result is 2.00 g and student

‘A’ takes two measurements and reports the

results as 1.95 g and 1.93 g. These values are

precise as they are close to each other but are

not accurate. Another student ‘B’ repeats the

experiment and obtains 1.94 g and 2.05 g as

the results for two measurements. These

observations are neither precise nor accurate.

When the third student ‘C’ repeats these

measurements and reports 2.01 g and 1.99 g

as the result, these values are both precise and

accurate. This can be more clearly understood

from the data given in Table 1.4.

Here, 18.0 has only one digit after the decimal

point and the result should be reported only

up to one digit after the decimal point, which

is 31.1.

Multiplication and Division of

Significant Figures

In these operations, the result must be

reported with no more significant figures as

in the measurement with the few significant

figures.

2.5×1.25 = 3.125

Since 2.5 has two significant figures, the

result should not have more than two

significant figures, thus, it is 3.1.

While limiting the result to the required

number of significant figures as done in the

above mathematical operation, one has to

keep in mind the following points for

rounding off the numbers

1. If the rightmost digit to be removed is more

than 5, the preceding number is increased

by one. For example, 1.386. If we have to

remove 6, we have to round it to 1.39.

2. If the rightmost digit to be removed is less

than 5, the preceding number is not

changed. For example, 4.334 if 4 is to be

removed, then the result is rounded upto

4.33.

3. If the rightmost digit to be removed is 5,

then the preceding number is not changed

if it is an even number but it is increased

by one if it is an odd number. For example,

if 6.35 is to be rounded by removing 5,

we have to increase 3 to 4 giving 6.4 as

the result. However, if 6.25 is to be

rounded off it is rounded off to 6.2.

1.4.3 Dimensional Analysis

Often while calculating, there is a need to

convert units from one system to the other. The

method used to accomplish this is called

factor

label method or unit factor method or

dimensional analysis. This is illustrated

below.

Example

A piece of metal is 3 inch (represented by in)

long. What is its length in cm?

Addition and Subtraction of

Significant Figures

The result cannot have more digits to the right

of the decimal point than either of the original

numbers. 12.11

18.0

1.012

31.122

1 2 Average (g)

Student A 1.95 1.93 1.940

Student B 1.94 2.05 1.995

Student C 2.01

1.99 2.000

Measurements/g

Table 1.4 Data to Illustrate Precision

and Accuracy

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14 CHEMISTRY

Solution

We know that 1 in = 2.54 cm

From this equivalence, we can write

2 54

in.

= =

Thus,

2 54

cm.

equals 1 and

2 54

. cm

also equals 1. Both of these are called unit

factors. If some number is multiplied by these

unit factors (i.e., 1), it will not be affected

otherwise.

Say, the 3 in given above is multiplied by

the unit factor. So,

3 in = 3 in ×

2 54

. cm

= 3 × 2.54 cm = 7.62 cm

Now, the unit factor by which multiplication

is to be done is that unit factor (

2 54

. cm

the above case) which gives the desired units

i.e., the numerator should have that part which

is required in the desired result.

It should also be noted in the above

example that units can be handled just like

other numerical part. It can be cancelled,

divided, multiplied, squared, etc. Let us study

one more example.

Example

A jug contains 2L of milk. Calculate the volume

of the milk in m

Solution

Since 1 L = 1000 cm

and 1m = 100 cm, which gives

100

= =

To get m

from the above unit factors, the

first unit factor is taken and it is cubed.

100

1 1

6 3













⇒ =

( )

Now 2 L = 2×1000 cm

The above is multiplied by the unit factor

2 1000

2 10

6 3

3 3

× × = = ×

−

Example

How many seconds are there in 2 days?

Solution

Here, we know 1 day = 24 hours (h)

day

= =

then, 1h = 60 min

hmin

min

= =

so, for converting 2 days to seconds,

i.e., 2 days – – – – – – = – – – seconds

The unit factors can be multiplied in

series in one step only as follows:

day

day h

× × ×

min

= 2 × 24 × 60 × 60 s

= 172800 s

1.5 LAWS OF CHEMICAL

COMBINATIONS

The combination of elements

to form compounds is

governed by the following five

basic laws.

1.5.1 Law of Conservation of Mass

This law was put forth by Antoine Lavoisier

in 1789. He performed careful experimental

studies for combustion reactions and reached

to the conclusion that in all physical and

chemical changes, there is no net change in

mass duting the process. Hence, he reached

to the conclusion that matter can neither be

created nor destroyed. This is called ‘Law of

Conservation of Mass’. This law formed the

basis for several later developments in

chemistry. Infact, this was the result of exact

measurement of masses of reactants and

products, and carefully planned experiments

performed by Lavoisier.

Antoine Lavoisier

(1743–1794)

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15SOME BASIC CONCEPTS OF CHEMISTRY

1.5.2 Law of Definite Proportions

This law was given by, a

French chemist, Joseph

Proust. He stated that a given

compound always contains

exactly the same proportion

of elements by weight.

Proust worked with two

samples of cupric carbonate

— one of which was of natural

origin and the other was synthetic. He found

that the composition of elements present in it

was same for both the samples as shown below:

produced in a chemical

reaction they do so in a

simple ratio by volume,

provided all gases are at

the same temperature and

pressure.

Thus, 100 mL of hydrogen

combine with 50 mL of oxygen

to give 100 mL of water

vapour.

Hydrogen + Oxygen → Water

100 mL 50 mL 100 mL

Thus, the volumes of hydrogen and oxygen

which combine (i.e., 100 mL and

50 mL) bear a simple ratio of 2:1.

Gay Lussac’s discovery of integer ratio in

volume relationship is actually the law of

definite proportions by volume. The law of

definite proportions, stated earlier, was with

respect to mass. The Gay Lussac’s law was

explained properly by the work of Avogadro

in 1811.

1.5.5 Avogadro’s Law

In 1811, Avogadro proposed that equal

volumes of all gases at the same temperature

and pressure should contain equal number

of molecules. Avogadro made a distinction

between atoms and molecules which is quite

understandable in present times. If we

consider again the reaction of hydrogen and

oxygen to produce water, we see that two

volumes of hydrogen combine with one volume

of oxygen to give two volumes of water without

leaving any unreacted oxygen.

Note that in the Fig. 1.9 (Page 16) each box

contains equal number of

molecules. In fact, Avogadro

could explain the above result

by considering the molecules

to be polyatomic. If

hydrogen and oxygen were

considered as diatomic as

recognised now, then the

above results are easily

understandable. However,

Dalton and others believed at

that time that atoms of the

Joseph Proust

(1754–1826)

Joseph Louis

Gay Lussac

Lorenzo Romano

Amedeo Carlo

Avogadro di

Quareqa edi

Carreto

(1776–1856)

% of % of % of

copper carbon oxygen

Natural Sample 51.35 9.74 38.91

Synthetic Sample 51.35 9.74 38.91

Thus, he concluded that irrespective of the

source, a given compound always contains

same elements combined together in the same

proportion by mass. The validity of this law

has been confirmed by various experiments.

It is sometimes also referred to as Law of

Definite Composition.

1.5.3 Law of Multiple Proportions

This law was proposed by Dalton in 1803.

According to this law, if two elements can

combine to form more than one compound, the

masses of one element that combine with a

fixed mass of the other element, are in the

ratio of small whole numbers.

For example, hydrogen combines with

oxygen to form two compounds, namely, water

and hydrogen peroxide.

Hydrogen + Oxygen

→ Water

2g 16g 18g

Hydrogen + Oxygen → Hydrogen Peroxide

2g 32g 34g

Here, the masses of oxygen (i.e., 16 g and 32 g),

which combine with a fixed mass of hydrogen

(2g) bear a simple ratio, i.e., 16:32 or 1: 2.

1.5.4 Gay Lussac’s Law of Gaseous

Volumes

This law was given by Gay Lussac in 1808. He

observed that when gases combine or are

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16 CHEMISTRY

same kind cannot combine and molecules of

oxygen or hydrogen containing two atoms did

not exist. Avogadro’s proposal was published

in the French Journal de Physique. In spite of

being correct, it did not gain much support.

After about 50 years, in 1860, the first

international conference on chemistry was held

in Karlsruhe, Germany, to resolve various

ideas. At the meeting, Stanislao Cannizaro

presented a sketch of a course of chemical

philosophy, which emphasised on the

importance of Avogadro’s work.

1.6 DALTON’S ATOMIC THEORY

Although the origin of the idea that matter is

composed of small indivisible particles called ‘a-

tomio’ (meaning, indivisible), dates back to the

time of Democritus, a Greek

Philosopher (460–370 BC), it

again started emerging as a

result of several experimental

studies which led to the laws

mentioned above.

In 1808, Dalton

published ‘A New System of

Chemical Philosophy’, in

which he proposed the following :

1. Matter consists of indivisible atoms.

2. All atoms of a given element have identical

properties, including identical mass. Atoms

of different elements differ in mass.

3. Compounds are formed when atoms of

different elements combine in a fixed ratio.

4. Chemical reactions involve reorganisation

of atoms. These are neither created nor

destroyed in a chemical reaction.

Dalton’s theory could explain the laws of

chemical combination. However, it could not

explain the laws of gaseous volumes. It could

not provide the reason for combining of

atoms, which was answered later by other

scientists.

1.7 ATOMIC AND MOLECULAR MASSES

After having some idea about the terms atoms

and molecules, it is appropriate here to

understand what do we mean by atomic and

molecular masses.

1.7.1 Atomic Mass

The atomic mass or the mass of an atom is

actually very-very small because atoms are

extremely small. Today, we have

sophisticated techniques e.g., mass

spectrometry for determining the atomic

masses fairly accurately. But in the

nineteenth century, scientists could

determine the mass of one atom relative to

another by experimental means, as has been

mentioned earlier. Hydrogen, being the

lightest atom was arbitrarily assigned a mass

of 1 (without any units) and other elements

were assigned masses relative to it. However,

the present system of atomic masses is based

on carbon-12 as the standard and has been

agreed upon in 1961. Here, Carbon-12 is one

of the isotopes of carbon and can be

represented as

C. In this system,

C is

assigned a mass of exactly 12 atomic mass

unit (amu) and masses of all other atoms are

given relative to this standard. One atomic

mass unit is defined as a mass exactly equal

John Dalton

(1776–1884)

Fig. 1.9 Two volumes of hydrogen react with one volume of oxygen to give two volumes of water vapour

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17SOME BASIC CONCEPTS OF CHEMISTRY

to one-twelfth of the mass of one carbon - 12

atom.

And 1 amu = 1.66056×10

–24

Mass of an atom of hydrogen

= 1.6736×10

–24

Thus, in terms of amu, the mass

of hydrogen atom =

= 1.0078 amu

= 1.0080 amu

Similarly, the mass of oxygen - 16 (

atom would be 15.995 amu.

At present, ‘amu’ has been replaced by ‘u’,

which is known as unified mass.

When we use atomic masses of elements

in calculations, we actually use average

atomic masses

of elements, which are

explained below.

1.7.2 Average Atomic Mass

Many naturally occurring elements exist as

more than one isotope. When we take into

account the existence of these isotopes and

their relative abundance (per cent occurrence),

the average atomic mass of that element can

be computed. For example, carbon has the

following three isotopes with relative

abundances and masses as shown against

1.7.3 Molecular Mass

Molecular mass is the sum of atomic masses

of the elements present in a molecule. It is

obtained by multiplying the atomic mass of

each element by the number of its atoms and

adding them together. For example, molecular

mass of methane, which contains one carbon

atom and four hydrogen atoms, can be

obtained as follows:

Molecular mass of methane,

(CH

) = (12.011 u) + 4 (1.008 u)

= 16.043 u

Similarly, molecular mass of water (H

= 2 × atomic mass of hydrogen + 1 × atomic

mass of oxygen

= 2 (1.008 u) + 16.00 u

= 18.02 u

1.7.4 Formula Mass

Some substances, such as sodium chloride,

do not contain discrete molecules as their

constituent units. In such compounds, positive

(sodium ion) and negative (chloride ion) entities

are arranged in a three-dimensional structure,

as shown in Fig. 1.10.

Isotope Relative Atomic

Abundance Mass (amu)

(%)

C 98.892 12

C 1.108 13.00335

C 2 ×10

–10

14.00317

each of them.

From the above data, the average atomic

mass of carbon will come out to be:

(0.98892) (12 u) + (0.01108) (13.00335 u) +

(2 × 10

–12

) (14.00317 u) = 12.011 u

Similarly, average atomic masses for other

elements can be calculated. In the periodic

table of elements, the atomic masses

mentioned for different elements actually

represent their average atomic masses.

It may be noted that in sodium chloride,

one Na

ion is surrounded by six Cl

–

ion and

vice-versa.

The formula, such as NaCl, is used to

calculate the formula mass instead of molecular

mass as in the solid state sodium chloride does

not exist as a single entity.

Fig. 1.10 Packing of Na

and Cl

–

ions

in sodium chloride

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18 CHEMISTRY

This number of entities in 1 mol is so

important that it is given a separate name and

symbol. It is known as ‘Avogadro constant’,

or Avogadro number denoted by N

in honour

of Amedeo Avogadro. To appreciate the

largeness of this number, let us write it with

all zeroes without using any powers of ten.

602213670000000000000000

Hence, so many entities (atoms, molecules or

any other particle) constitute one mole of a

particular substance.

We can, therefore, say that 1 mol of hydrogen

atoms = 6.022×10

atoms

1 mol of water molecules = 6.022×10

water

molecules

1 mol of sodium chloride = 6.022 × 10

formula units of sodium chloride

Having defined the mole, it is easier to know

the mass of one mole of a substance or the

constituent entities. The mass of one mole

of a substance in grams is called its

molar mass. The molar mass in grams is

numerically equal to atomic/molecular/

formula mass in u.

Molar mass of water = 18.02 g mol

-1

Molar mass of sodium chloride = 58.5 g mol

-1

1.9 PERCENTAGE COMPOSITION

So far, we were dealing with the number of

entities present in a given sample. But many a

time, information regarding the percentage of

a particular element present in a compound is

required. Suppose, an unknown or new

compound is given to you, the first question

Thus, the formula mass of sodium chloride is

atomic mass of sodium + atomic mass of chlorine

= 23.0 u + 35.5 u = 58.5 u

Fig. 1.11 One mole of various substances

Problem 1.1

Calculate the molecular mass of glucose

) molecule.

Solution

Molecular mass of glucose (C

)

= 6(12.011 u) + 12(1.008 u) +

6(16.00 u)

= (72.066 u) + (12.096 u) +

(96.00 u)

= 180.162 u

1.8 MOLE CONCEPT AND MOLAR MASSES

Atoms and molecules are extremely small in

size and their numbers in even a small amount

of any substance is really very large. To handle

such large numbers, a unit of convenient

magnitude is required.

Just as we denote one dozen for 12 items,

score for 20 items, gross for 144 items, we

use the idea of mole to count entities at the

microscopic level (i.e., atoms, molecules,

particles, electrons, ions, etc).

In SI system, mole (symbol, mol) was

introduced as seventh base quantity for the

amount of a substance.

The mole, symbol mol, is the SI unit of

amount of substance. One mole contains

exactly 6.02214076

elementary entities.

This number is the fixed numerical value of

the Avogadro constant, N

, when expressed in

the unit mol

–1

and is called the Avogadro

number. The amount of substance, symbol n,

of a system is a measure of the number of

specified elementary entities. An elementary

entity may be an atom, a molecule, an ion, an

electron, any other particle or specified group

of particles. It may be emphasised that the

mole of a substance always contains the same

number of entities, no matter what the

substance may be. In order to determine this

number precisely, the mass of a carbon–12

atom was determined by a mass spectrometer

and found to be equal to 1.992648 × 10

–23

Knowing that one mole of carbon weighs 12 g,

the number of atoms in it is equal to:

1 992648 10

23 12

g mol C

g C atom

. /×

−

= ×6 0221367 10

. atoms/mol

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19SOME BASIC CONCEPTS OF CHEMISTRY

you would ask is: what is its formula or what

are its constituents and in what ratio are they

present in the given compound? For known

compounds also, such information provides a

check whether the given sample contains the

same percentage of elements as present in a

pure sample. In other words, one can check

the purity of a given sample by analysing this

data.

Let us understand it by taking the example

of water (H

O). Since water contains hydrogen

and oxygen, the percentage composition of

both these elements can be calculated as

follows:

Mass % of an element =

mass of that element in the compound 100

molar mass of th

ee compound

Molar mass of water = 18.02 g

Mass % of hydrogen =

= 11.18

Mass % of oxygen =

16.00

18.02

100×

= 88.79

Let us take one more example. What is the

percentage of carbon, hydrogen and oxygen

in ethanol?

Molecular formula of ethanol is: C

Molar mass of ethanol is:

(2×12.01 + 6×1.008 + 16.00) g = 46.068 g

Mass per cent of carbon

24.02 g

46.068 g

100×

= 52.14%

Mass per cent of hydrogen

6.048 g

46.068 g

100×

= 13.13%

Mass per cent of oxygen

16.00 g

46.068 g

100×

= 34.73%

After understanding the calculation of

per cent of mass, let us now see what

information can be obtained from the

per cent composition data.

1.9.1 Empirical Formula for Molecular

Formula

An empirical formula represents the simplest

whole number ratio of various atoms present in

a compound, whereas, the molecular formula

shows the exact number of different types of

atoms present in a molecule of a compound.

If the mass per cent of various elements

present in a compound is known, its empirical

formula can be determined. Molecular formula

can further be obtained if the molar mass is

known. The following example illustrates this

sequence.

Problem 1.2

A compound contains 4.07% hydrogen,

24.27% carbon and 71.65% chlorine. Its

molar mass is 98.96 g. What are its

empirical and molecular formulas?

Solution

Step 1. Conversion of mass per cent

to grams

Since we are having mass per cent, it is

convenient to use 100 g of the compound

as the starting material. Thus, in the

100 g sample of the above compound,

4.07g hydrogen, 24.27g carbon and

71.65g chlorine are present.

Step 2. Convert into number moles of

each element

Divide the masses obtained above by

respective atomic masses of various

elements. This gives the number of moles

of constituent elements in the compound

Moles of hydrogen =

4.07 g

1.008 g

= 4.04

Moles of carbon =

24.27 g

g12 01

2 021

Moles of chlorine =

71.65 g

g35 453

2 021

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20 CHEMISTRY

Step 3. Divide each of the mole values

obtained above by the smallest number

amongst them

Since 2.021 is smallest value, division by

it gives a ratio of 2:1:1 for H:C:Cl .

In case the ratios are not whole numbers, then

they may be converted into whole number by

multiplying by the suitable coefficient.

Step 4. Write down the empirical formula

by mentioning the numbers after writing

the symbols of respective elements

Cl is, thus, the empirical formula of

the above compound.

Step 5. Writing molecular formula

(a) Determine empirical formula mass by

adding the atomic masses of various

atoms present in the empirical formula.

For CH

Cl, empirical formula mass is

12.01 + (2 × 1.008) + 35.453

= 49.48 g

(b) Divide Molar mass by empirical

formula mass

= 2 = (n)

obtained above to get the molecular

formula

Empirical formula = CH

Cl, n = 2. Hence

molecular formula is C

available from the balanced chemical equation

of a given reaction. Let us consider the

combustion of methane. A balanced equation

for this reaction is as given below:

(g) + 2O

(g) → CO

(g) + 2 H

O (g)

Here, methane and dioxygen are called

reactants and carbon dioxide and water are

called products. Note that all the reactants and

the products are gases in the above reaction

and this has been indicated by letter (g) in the

brackets next to its formula. Similarly, in case

of solids and liquids, (s) and (l) are written

respectively.

The coefficients 2 for O

and H

O are called

stoichiometric coefficients. Similarly the

coefficient for CH

and CO

is one in each case.

They represent the number of molecules (and

moles as well) taking part in the reaction or

formed in the reaction.

Thus, according to the above chemical

reaction,

• One mole of CH

(g) reacts with two moles

of O

(g) to give one mole of CO

(g) and

two moles of H

O(g)

• One molecule of CH

(g) reacts with

2 molecules of O

(g) to give one molecule

of CO

(g) and 2 molecules of H

O(g)

• 22.7 L of CH

(g) reacts with 45.4 L of O

(g)

to give 22.7 L of CO

(g) and 45.4 L of H

O(g)

• 16 g of CH

(g) reacts with 2×32 g of O

(g) to

give 44 g of CO

(g) and 2×18 g of H

O (g).

From these relationships, the given data can

be interconverted as follows:

Mass

Volume

Density=

1.10.1 Limiting Reagent

Many a time, reactions are carried out with the

amounts of reactants that are different than

the amounts as required by a balanced

chemical reaction. In such situations, one

reactant is in more amount than the amount

required by balanced chemical reaction. The

reactant which is present in the least amount

1.10 STOICHIOMETRY AND

STOICHIOMETRIC CALCULATIONS

The word ‘stoichiometry’ is derived from two

Greek words — stoicheion (meaning, element)

and metron (meaning, measure).

Stoichiometry, thus, deals with the calculation

of masses (sometimes volumes also) of the

reactants and the products involved in a

chemical reaction. Before understanding how

to calculate the amounts of reactants required

or the products produced in a chemical

reaction, let us study what information is

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21SOME BASIC CONCEPTS OF CHEMISTRY

gets consumed after sometime and after that

further reaction does not take place whatever

be the amount of the other reactant. Hence,

the reactant, which gets consumed first, limits

the amount of product formed and is, therefore,

called the limiting reagent.

In performing stoichiometric calculations,

this aspect is also to be kept in mind.

1.10.2 Reactions in Solutions

A majority of reactions in the laboratories are

carried out in solutions. Therefore, it is

important to understand as how the amount

of substance is expressed when it is present in

the solution. The concentration of a solution

or the amount of substance present in its

given volume can be expressed in any of the

following ways.

1. Mass per cent or weight per cent (w/w %)

2. Mole fraction

3. Molarity

4. Molality

Let us now study each one of them in detail.

Balancing a chemical equation

According to the law of conservation of mass, a balanced chemical equation has the same

number of atoms of each element on both sides of the equation. Many chemical equations can

be balanced by trial and error. Let us take the reactions of a few metals and non-metals with

oxygen to give oxides

4 Fe(s) + 3O

(g) → 2Fe

(s) (a) balanced equation

2 Mg(s) + O

(g) → 2MgO(s) (b) balanced equation

(s) + O

(g) → P

(s) (c) unbalanced equation

Equations (a) and (b) are balanced, since there are same number of metal and oxygen atoms on

each side of the equations. However equation (c) is not balanced. In this equation, phosphorus

atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on

the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the

right side of the equation.

(s) + 5O

(g) → P

(s) balanced equation

Now, let us take combustion of propane, C

. This equation can be balanced in steps.

Step 1 Write down the correct formulas of reactants and products. Here, propane and oxygen

are reactants, and carbon dioxide and water are products.

(g) + O

(g) → CO

(g) + H

O(l) unbalanced equation

Step 2 Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore,

three CO

molecules are required on the right side.

(g) + O

(g) → 3CO

(g) + H

O (l)

Step 3 Balance the number of H atoms: on the left there are 8 hydrogen atoms in the reactants

however, each molecule of water has two hydrogen atoms, so four molecules of water will be

required for eight hydrogen atoms on the right side.

(g) +O

(g)

→ 3CO

(g)+4H

O (l)

Step 4 Balance the number of O atoms: There are 10 oxygen atoms on the right side (3 × 2 = 6 in

and 4 × 1= 4 in water). Therefore, five O

molecules are needed to supply the required 10

and 4 × 1= 4 in water). Therefore, five O

molecules are needed to supply the required 10

oxygen atoms.

(g) +5O

(g)

→ 3CO

(g) + 4H

O (l)

Step 5 Verify that the number of atoms of each element is balanced in the final equation. The

equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side.

All equations that have correct formulas for all reactants and products can be balanced. Always

remember that subscripts in formulas of reactants and products cannot be changed to balance

an equation.

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22 CHEMISTRY

Problem 1.3

Calculate the amount of water (g)

produced by the combustion of 16 g

of methane.

Solution

The balanced equation for the combustion

of methane is :

(g) + 2O

(g)

→

(g) + 2H

O (g)

(i)16 g of CH

corresponds to one mole.

(ii) From the above equation, 1 mol of

(g) gives 2 mol of H

O (g).

2 mol of water (H

O) = 2 × (2+16)

= 2 × 18 = 36 g

1 mol H

O = 18 g H

⇒

18 g H O

1mol H O

= 1

Hence, 2 mol H

O ×

18 g H O

1mol H O

= 2 × 18 g H

O = 36 g H

Problem 1.4

How many moles of methane are required

to produce 22g CO

(g) after combustion?

Solution

According to the chemical equation,

(g) + 2O

(g)

→

(g) + 2H

O (g)

44g CO

(g) is obtained from 16 g CH

(g).

[

∴

1 mol CO

(g) is obtained from 1 mol of

(g)]

Number of moles of CO

(g)

= 22 g CO

(g) ×

mol CO g

g CO g

( )

= 0.5 mol CO

(g)

Hence, 0.5 mol CO

(g) would be

obtained from 0.5 mol CH

(g) or 0.5

mol of CH

(g) would be required to

produce 22 g CO

(g).

Problem 1.5

50.0 kg of N

(g) and 10.0 kg of H

(g) are

mixed to produce NH

(g). Calculate the

amount of NH

(g) formed. Identify the

limiting reagent in the production of NH

in this situation.

Solution

A balanced equation for the above reaction

is written as follows :

Calculation of moles :

Number of moles of N

50 0

1000

28 0

kg N

g N

kg N

mol N

g N

× ×

= 17.86×10

mol

Number of moles of H

10 00

1000

2 016

kg H

g H

kg H

mol H

g H

× ×

= 4.96×10

mol

According to the above equation, 1 mol

(g) requires 3 mol H

(g), for the reaction.

Hence, for 17.86×10

mol of N

, the moles

of H

(g) required would be

17 86 10

. × ×

( )

mol N

mol H g

mol N g

= 5.36 ×10

mol H

But we have only 4.96×10

mol H

. Hence,

dihydrogen is the limiting reagent in this

case. So, NH

(g) would be formed only

from that amount of available dihydrogen

i.e., 4.96 × 10

mol

Since 3 mol H

(g) gives 2 mol NH

(g)

4.96×10

mol H

(g) ×

mol NH g

mol H g

( )

= 3.30×10

mol NH

(g)

3.30×10

mol NH

(g) is obtained.

If they are to be converted to grams, it is

done as follows :

1 mol NH

(g) = 17.0 g NH

(g)

3.30×10

mol NH

(g) ×

17 0

. g NH g

mol NH g

( )

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23SOME BASIC CONCEPTS OF CHEMISTRY

= 3.30×10

×17 g NH

(g)

= 56.1×10

g NH

= 56.1 kg NH

3. Molarity

It is the most widely used unit and is denoted

by M. It is defined as the number of moles of

the solute in 1 litre of the solution. Thus,

Molarity (M)

No. of moles of solute

of solution in litresVolume

Suppose, we have 1 M solution of a

substance, say NaOH, and we want to prepare

a 0.2 M solution from it.

1 M NaOH means 1 mol of NaOH present

in 1 litre of the solution. For 0.2 M solution,

we require 0.2 moles of NaOH dissolved in

1 litre solution.

Hence, for making 0.2M solution from 1M

solution, we have to take that volume of 1M NaOH

solution, which contains 0.2 mol of NaOH and

dilute the solution with water to 1 litre.

Now, how much volume of concentrated

(1M) NaOH solution be taken, which contains

0.2 moles of NaOH can be calculated as follows:

If 1 mol is present in 1L or 1000 mL

solution

then, 0.2 mol is present in

1000

0 2

mol

mol solution× .

= 200 mL solution

Thus, 200 mL of 1M NaOH are taken and

enough water is added to dilute it to make it 1 litre.

In fact for such calculations, a general

formula, M

× V

= M

× V

where M and V are

molarity and volume, respectively, can be used.

In this case, M

is equal to 0.2M; V

= 1000 mL

and, M

= 1.0M; V

is to be calculated.

Substituting the values in the formula:

0.2 M × 1000 mL = 1.0 M × V

Note that the number of moles of solute

(NaOH) was 0.2 in 200 mL and it has remained

the same, i.e., 0.2 even after dilution ( in 1000

mL) as we have changed just the amount of

solvent (i.e., water) and have not done anything

with respect to NaOH. But keep in mind the

concentration.

1. Mass per cent

It is obtained by using the following relation:

Problem 1.6

A solution is prepared by adding 2 g of a

substance A to 18 g of water. Calculate

the mass per cent of the solute.

Solution

2. Mole Fraction

It is the ratio of number of moles of a particular

component to the total number of moles of the

solution. If a substance ‘A’ dissolves in

substance ‘B’ and their number of moles are

and n

, respectively, then the mole fractions

of A and B are given as:

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24 CHEMISTRY

Problem 1.7

Calculate the molarity of NaOH in the

solution prepared by dissolving its 4 g in

enough water to form 250 mL of the solution.

Solution

Since molarity (M)

Note that molarity of a solution depends

upon temperature because volume of a

solution is temperature dependent.

SUMMARY

Chemistry, as we understand it today is not a very old discipline. People in ancient

India, already had the knowledge of many scientific phenomenon much before the

advent of modern science. They applied the knowledge in various walks of life.

The study of chemistry is very important as its domain encompasses every sphere

of life. Chemists study the properties and structure of substances and the changes

undergone by them. All substances contain matter, which can exist in three states

– solid, liquid or gas. The constituent particles are held in different ways in these

states of matter and they exhibit their characteristic properties. Matter can also be

classified into elements, compounds or mixtures. An element contains particles of

only one type, which may be atoms or molecules. The compounds are formed where

atoms of two or more elements combine in a fixed ratio to each other. Mixtures occur

widely and many of the substances present around us are mixtures.

When the properties of a substance are studied, measurement is inherent. The

quantification of properties requires a system of measurement and units in which

the quantities are to be expressed. Many systems of measurement exist, of which

Problem 1.8

The density of 3 M solution of NaCl is

1.25 g mL

–1

. Calculate the molality of the

solution.

Solution

M = 3 mol L

–1

Mass of NaCl

in 1 L solution = 3 × 58.5 = 175.5 g

Mass of

1L solution = 1000 × 1.25 = 1250 g

(since density = 1.25 g mL

–1

)

Mass of water in solution = 1250 –75.5

= 1074.5 g

Molality

No. of moles of solute

of solvent in kgMass

3 mol

1 0745. kg

= 2.79 m

Often in a chemistry laboratory, a solution

of a desired concentration is prepared by

diluting a solution of known higher

concentration. The solution of higher

concentration is also known as stock

solution. Note that the molality of a solution

does not change with temperature since

mass remains unaffected with temperature.

4. Molality

It is defined as the number of moles of solute

present in 1 kg of solvent. It is denoted by m.

Thus, Molality (m)

No. of moles of solute

Mass of solvent in kg

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25SOME BASIC CONCEPTS OF CHEMISTRY

the English and the Metric Systems are widely used. The scientific community, however,

has agreed to have a uniform and common system throughout the world, which is

abbreviated as SI units (International System of Units).

Since measurements involve recording of data, which are always associated with a

certain amount of uncertainty, the proper handling of data obtained by measuring the

quantities is very important. The measurements of quantities in chemistry are spread

over a wide range of 10

–31

to 10

+23

. Hence, a convenient system of expressing the numbers

in scientific notation is used. The uncertainty is taken care of by specifying the number

of significant figures, in which the observations are reported. The dimensional analysis

helps to express the measured quantities in different systems of units. Hence, it is possible

to interconvert the results from one system of units to another.

The combination of different atoms is governed by basic laws of chemical combination

— these being the Law of Conservation of Mass, Law of Definite Proportions, Law of

Multiple Proportions, Gay Lussac’s Law of Gaseous Volumes and Avogadro Law. All

these laws led to the Dalton’s atomic theory, which states that atoms are building

blocks of matter. The

atomic mass of an element is expressed relative to

C isotope of

carbon, which has an exact value of 12u. Usually, the atomic mass used for an element is

the average atomic mass obtained by taking into account the natural abundance of

different isotopes of that element. The molecular mass of a molecule is obtained by

taking sum of the atomic masses of different atoms present in a molecule. The molecular

formula can be calculated by determining the mass per cent of different elements present

in a compound and its molecular mass.

The number of atoms, molecules or any other particles present in a given system are

expressed in the terms of Avogadro constant (6.022 × 10

). This is known as 1 mol of

the respective particles or entities.

Chemical reactions represent the chemical changes undergone by different elements

and compounds. A balanced chemical equation provides a lot of information. The

coefficients indicate the molar ratios and the respective number of particles taking part

in a particular reaction. The quantitative study of the reactants required or the products

formed is called stoichiometry. Using stoichiometric calculations, the amount of one or

more reactant(s) required to produce a particular amount of product can be determined

and vice-versa. The amount of substance present in a given volume of a solution is

expressed in number of ways, e.g., mass per cent, mole fraction, molarity and molality.

EXERCISES

1.1 Calculate the molar mass of the following:

(i) H

O (ii) CO

(iii) CH

1.2 Calculate the mass per cent of different elements present in sodium sulphate

(Na

1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and

30.1% dioxygen by mass.

1.4 Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

1.5 Calculate the mass of sodium acetate (CH

COONa) required to make 500 mL of

0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol

–1

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26 CHEMISTRY

1.6 Calculate the concentration of nitric acid in moles per litre in a sample which

has a density, 1.41 g mL

–1

and the mass per cent of nitric acid in it being 69%.

1.7 How much copper can be obtained from 100 g of copper sulphate (CuSO

1.8 Determine the molecular formula of an oxide of iron, in which the mass per cent

of iron and oxygen are 69.9 and 30.1, respectively.

1.9 Calculate the atomic mass (average) of chlorine using the following data:

% Natural Abundance Molar Mass

Cl 75.77 34.9689

Cl 24.23 36.9659

1.10 In three moles of ethane (C

), calculate the following:

(i) Number of moles of carbon atoms.

(ii) Number of moles of hydrogen atoms.

(iii) Number of molecules of ethane.

1.11 What is the concentration of sugar (C

) in mol L

–1

if its 20 g are dissolved in

enough water to make a final volume up to 2L?

1.12 If the density of methanol is 0.793 kg L

–1

, what is its volume needed for making

2.5 L of its 0.25 M solution?

1.13 Pressure is determined as force per unit area of the surface. The SI unit of

pressure, pascal is as shown below:

1Pa = 1N m

–2

If mass of air at sea level is 1034 g cm

–2

, calculate the pressure in pascal.

1.14 What is the SI unit of mass? How is it defined?

1.15 Match the following prefixes with their multiples:

Prefixes Multiples

(i) micro 10

(ii) deca 10

(iii) mega 10

–6

(iv) giga 10

–15

(v) femto 10

1.16 What do you mean by significant figures?

1.17 A sample of drinking water was found to be severely contaminated with chloroform,

CHCl

, supposed to be carcinogenic in nature. The level of contamination was 15

ppm (by mass).

(i) Express this in per cent by mass.

(ii) Determine the molality of chloroform in the water sample.

1.18 Express the following in the scientific notation:

(i) 0.0048

(ii) 234,000

(iii) 8008

(iv) 500.0

(v) 6.0012

1.19 How many significant figures are present in the following?

(i) 0.0025

(ii) 208

(iii) 5005

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27SOME BASIC CONCEPTS OF CHEMISTRY

(iv) 126,000

(v) 500.0

(vi) 2.0034

1.20 Round up the following upto three significant figures:

(i) 34.216

(ii) 10.4107

(iii) 0.04597

(iv) 2808

1.21 The following data are obtained when dinitrogen and dioxygen react together to

form different compounds:

Mass of dinitrogen Mass of dioxygen

(i) 14 g 16 g

(ii) 14 g 32 g

(iii) 28 g 32 g

(iv) 28 g 80 g

(a) Which law of chemical combination is obeyed by the above experimental data?

Give its statement.

(b) Fill in the blanks in the following conversions:

(i) 1 km = ...................... mm = ...................... pm

(ii) 1 mg = ...................... kg = ...................... ng

(iii) 1 mL = ...................... L = ...................... dm

1.22 If the speed of light is 3.0 × 10

m s

–1

, calculate the distance covered by light in

2.00 ns.

1.23 In a reaction

A + B

d AB

Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

1.24 Dinitrogen and dihydrogen react with each other to produce ammonia according

to the following chemical equation:

(g) + H

(g)

d 2NH

(g)

(i) Calculate the mass of ammonia produced if 2.00 × 10

g dinitrogen reacts

with 1.00 ×10

g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?

1.25 How are 0.50 mol Na

and 0.50 M Na

different?

1.26 If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how

many volumes of water vapour would be produced?

1.27 Convert the following into basic units:

(i) 28.7 pm

(ii) 15.15 pm

(iii) 25365 mg

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28 CHEMISTRY

1.28 Which one of the following will have the largest number of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1 g of Cl

(g)

1.29 Calculate the molarity of a solution of ethanol in water, in which the mole fraction of

ethanol is 0.040 (assume the density of water to be one).

1.30 What will be the mass of one

C atom in g?

1.31 How many significant figures should be present in the answer of the following

calculations?

(i)

0.02856 298.15 0.112× ×

0 5785.

(ii) 5 × 5.364

(iii) 0.0125 + 0.7864 + 0.0215

1.32 Use the data given in the following table to calculate the molar mass of naturally

occuring argon isotopes:

Isotope Isotopic molar mass Abundance

Ar 35.96755 g mol

–1

0.337%

Ar 37.96272 g mol

–1

0.063%

Ar 39.9624 g mol

–1

99.600%

1.33 Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii)

52 u of He (iii) 52 g of He.

1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it

in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A

volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g.

Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular

formula.

1.35 Calcium carbonate reacts with aqueous HCl to give CaCl

and CO

according to the

reaction, CaCO

(s) + 2 HCl (aq) → CaCl

(aq) + CO

(g) + H

O(l)

What mass of CaCO

is required to react completely with 25 mL of 0.75 M HCl?

1.36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO

) with

aqueous hydrochloric acid according to the reaction

4 HCl (aq) + MnO

(s) → 2H

O (l) + MnCl

(aq) + Cl

(g)

How many grams of HCl react with 5.0 g of manganese dioxide?

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