SOME BASIC CONCEPTS OF CHEMISTRY 1

UNIT 1

SOME BASIC CONCEPTS OF CHEMISTRY

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Chemistry is the science of molecules and their

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

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hundred elements but of the infinite variety of molecules that
may be built from them ...
able to
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After studying this unit, you will be
Roald Hoffmann

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understand and appreciate the
role of chemistry in different
spheres of life;
explain the characteristics of
three states of matter;
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Chemistry deals with the composition, structure and

classify different substances into
elements, compounds and
properties of matter. These aspects can be best described
mixtures; and understood in terms of basic constituents of matter:
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define SI base units and list some atoms and molecules. That is why chemistry is called
commonly used prefixes; the science of atoms and molecules. Can we see, weigh
use scientific notations and and perceive these entities? Is it possible to count the
perform simple mathematical number of atoms and molecules in a given mass of matter
operations on numbers; and have a quantitative relationship between the mass and

differentiate between precision and number of these particles (atoms and molecules)? We will
accuracy;
like to answer some of these questions in this Unit. We
determine significant figures;
would further describe how physical properties of matter
convert physical quantities from
one system of units to another;
can be quantitatively described using numerical values
explain various laws of chemical
with suitable units.
combination; 1.1 IMPORTANCE OF CHEMISTRY
appreciate significance of atomic
mass, average atomic mass,
Science can be viewed as a continuing human effort to
molecular mass and formula systematize knowledge for describing and understanding
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mass; nature. For the sake of convenience science is sub-divided

describe the terms mole and into various disciplines: chemistry, physics, biology,
molar mass; geology etc. Chemistry is the branch of science that studies
calculate the mass per cent of the composition, properties and interaction of matter.
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different elements constituting a Chemists are interested in knowing how chemical

compound;
transformations occur. Chemistry plays a central role in
determine empirical formula and
science and is often intertwined with other branches of
molecular formula for a compound
from the given experimental data; science like physics, biology, geology etc. Chemistry also
perform the stoichiometric plays an important role in daily life.
calculations. Chemical principles are important in diverse areas, such
as: weather patterns, functioning of brain and operation
2 CHEMISTRY

of a computer. Chemical industries

manufacturing fertilizers, alkalis, acids, salts,
dyes, polymers, drugs, soaps, detergents,
metals, alloys and other inorganic and organic
chemicals, including new materials, contribute
in a big way to the national economy.
Chemistry plays an important role in meeting
human needs for food, health care products

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and other materials aimed at improving the
quality of life. This is exemplified by the large
scale production of a variety of fertilizers,

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improved varieties of pesticides and
insecticides. Similarly many life saving drugs Fig. 1.1 Arrangement of particles in solid, liquid
such as cisplatin and taxol, are effective in and gaseous state

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cancer therapy and AZT (Azidothymidine)
Everything around us, for example, book, pen,

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used for helping AIDS victims, have been
isolated from plant and animal sources or pencil, water, air, all living beings etc. are
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prepared by synthetic methods. composed of matter. You know that they have
mass and they occupy space.

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With a better understanding of chemical
principles it has now become possible to You are also aware that matter can exist in
design and synthesize new materials having three physical states viz. solid, liquid and gas.
specific magnetic, electric and optical The constituent particles of matter in these
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properties. This has lead to the production of three states can be represented as shown in
superconducting ceramics, conducting Fig. 1.1. In solids, these particles are held very
polymers, optical fibres and large scale close to each other in an orderly fashion and
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miniaturization of solid state devices. In recent there is not much freedom of movement. In
years chemistry has tackled with a fair degree liquids, the particles are close to each other
of success some of the pressing aspects of but they can move around. However, in gases,
environmental degradation. Safer alternatives the particles are far apart as compared to those
to environmentally hazardous refrigerants like present in solid or liquid states and their

CFCs (chlorofluorocarbons), responsible for movement is easy and fast. Because of such
ozone depletion in the stratosphere, have been arrangement of particles, different states of
successfully synthesised. However, many big matter exhibit the following characteristics:
environmental problems continue to be (i) Solids have definite volume and definite
matters of grave concern to the chemists. One shape.
such problem is the management of the Green
House gases like methane, carbon dioxide etc. (ii) Liquids have definite volume but not the
Understanding of bio-chemical processes, use definite shape. They take the shape of the
of enzymes for large-scale production of container in which they are placed.
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chemicals and synthesis of new exotic (iii) Gases have neither definite volume nor
materials are some of the intellectual challenges definite shape. They completely occupy the
for the future generation of chemists. A container in which they are placed.
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developing country like India needs talented These three states of matter are
and creative chemists for accepting such interconvertible by changing the conditions of
challenges. temperature and pressure.
1.2 NATURE OF MATTER he at
Solid




heat

liquid





Gas




cool




cool

You are already familiar with the term matter
from your earlier classes. Anything which has On heating a solid usually changes to a
mass and occupies space is called matter. liquid and the liquid on further heating
SOME BASIC CONCEPTS OF CHEMISTRY 3

changes to the gaseous ( or vapour) state. In composition is variable. Copper, silver, gold,
the reverse process, a gas on cooling liquifies water, glucose are some examples of pure
to the liquid and the liquid on further cooling substances. Glucose contains carbon,
freezes to the solid. hydrogen and oxygen in a fixed ratio and thus,
At the macroscopic or bulk level, matter like all other pure substances has a fixed
composition. Also, the constituents of pure
can be classified as mixtures or pure
substances cannot be separated by simple
substances. These can be further sub-divided
physical methods.
as shown in Fig. 1.2.

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Pure substances can be further classified
into elements and compounds. An element
consists of only one type of particles. These

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particles may be atoms or molecules. You may
be familiar with atoms and molecules from the
previous classes; however, you will be studying

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about them in detail in Unit 2. Sodium, copper,

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silver, hydrogen, oxygen etc. are some
examples of elements. They all contain atoms
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elements are different in nature. Some elements
such as sodium or copper, contain single
Fig. 1.2 Classification of matter atoms held together as their constituent
particles whereas in some others, two or more
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Many of the substances present around atoms combine to give molecules of the
you are mixtures. For example, sugar solution element. Thus, hydrogen, nitrogen and oxygen
in water, air, tea etc., are all mixtures. A mixture gases consist of molecules in which two atoms
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contains two or more substances present in it combine to give their respective molecules. This
(in any ratio) which are called its components. is illustrated in Fig. 1.3.
A mixture may be homogeneous or When two or more atoms of different
heterogeneous. In a homogeneous mixture, elements combine, the molecule of a
the components completely mix with each other

compound is obtained. The examples of some

and its composition is uniform throughout. compounds are water, ammonia, carbon
Sugar solution, and air are thus, the examples
of homogeneous mixtures. In contrast to this,
in heterogeneous mixtures, the composition
is not uniform throughout and sometimes the
different components can be observed. For
example, the mixtures of salt and sugar, grains
and pulses along with some dirt (often stone)
pieces, are heterogeneous mixtures. You can
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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
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using physical methods such as simple hand

picking, filtration, crystallisation, distillation
etc.
Pure substances have characteristics
different from the mixtures. They have fixed
composition, whereas mixtures may contain
the components in any ratio and their Fig. 1.3 A representation of atoms and molecules
4 CHEMISTRY

dioxide, sugar etc. The molecules of water and chemical properties are characteristic
carbon dioxide are represented in Fig 1.4. reactions of different substances; these include
acidity or basicity, combustibility etc.
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
Water molecule Carbon dioxide in which it is measured. For example length of

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(H2O) molecule (CO2 ) a room can be represented as 6 m; here 6 is
the number and m denotes metre the unit in

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Fig. 1.4 A depiction of molecules of water and which the length is measured.
carbon dioxide
Two different systems of measurement, i.e.
You have seen above that a water molecule the English System and the Metric System

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comprises two hydrogen atoms and one were being used in different parts of the world.

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oxygen atom. Similarly, a molecule of carbon The metric system which originated in France
dioxide contains two oxygen atoms combined in late eighteenth century, was more
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with one carbon atom. Thus, the atoms of convenient as it was based on the decimal

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different elements are present in a compound system. The need of a common standard
in a fixed and definite ratio and this ratio is system was being felt by the scientific
characteristic of a particular compound. Also, community. Such a system was established
the properties of a compound are different in 1960 and is discussed below in detail.
from those of its constituent elements. For
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example, hydrogen and oxygen are gases 1.3.1 The International System of Units
whereas the compound formed by their (SI)
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combination i.e., water is a liquid. It is The International System of Units (in French
interesting to note that hydrogen burns with Le Systeme International dUnits
a pop sound and oxygen is a supporter of abbreviated as SI) was established by the 11th
combustion, but water is used as a fire General Conference on Weights and Measures
extinguisher. (CGPM from Conference Generale des Poids

Moreover, the constituents of a compound et Measures). The CGPM is an inter

cannot be separated into simpler substances governmental treaty organization created by
by physical methods. They can be separated a diplomatic treaty known as Metre Convention
by chemical methods. which was signed in Paris in 1875.
The SI system has seven base units and
1.3 PROPERTIES OF MATTER AND
they are listed in Table 1.1. These units pertain
THEIR MEASUREMENT
to the seven fundamental scientific quantities.
Every substance has unique or characteristic The other physical quantities such as speed,
properties. These properties can be classified
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volume, density etc. can be derived from these

into two categories physical properties and quantities.
chemical properties.
The definitions of the SI base units are given
Physical properties are those properties
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in Table 1.2.
which can be measured or observed without
changing the identity or the composition of the The SI system allows the use of prefixes to
substance. Some examples of physical indicate the multiples or submultiples of a unit.
properties are colour, odour, melting point, These prefixes are listed in Table 1. 3.
boiling point, density etc. The measurement Let us now quickly go through some of the
or observation of chemical properties require quantities which you will be often using in this
a chemical change to occur. The examples of book.
SOME BASIC CONCEPTS OF CHEMISTRY 5

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
T ime t second s

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Electric current I ampere A
Thermodynamic T kelvin K

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temperature
Amount of substance n mole mol
Luminous intensity Iv candela cd

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Table 1.2 Definitions of SI Base Units
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Unit of length metre The metre is the length of the path travelled

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by light in vacuum during a time interval of
1/299 792 458 of a second.
Unit of mass kilogram The kilogram is the unit of mass; it is equal
to the mass of the international prototype
of the kilogram.
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Unit of time second The second is the duration of 9 192 631 770
periods of the radiation corresponding to the
transition between the two hyperfine levels
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of the ground state of the caesium-133 atom.

Unit of electric current ampere The ampere is that constant current which,
if maintained in two straight parallel
conductors of infinite length, of negligible
circular cross-section, and placed 1 metre

apart in vacuum, would produce between

these conductors a force equal to 2 107
newton per metre of length.
Unit of thermodynamic kelvin The kelvin, unit of thermodynamic
temperature temperature, is the fraction 1/273.16 of the
thermodynamic temperature of the triple
point of water.
Unit of amount of substance mole 1. The mole is the amount of substance of a
system which contains as many elementary
entities as there are atoms in 0.012
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kilogram of carbon-12; its symbol is mol.

2. When the mole is used, the elementary
entities must be specified and may be atoms,
molecules, ions, electrons, other particles,
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or specified groups of such particles.

Unit of luminous intensity candela The candela is the luminous intensity, in a
given direction, of a source that emits
monochromatic radiation of frequency
540 10 12 hertz and that has a radiant
intensity in that direction of 1/683 watt per
steradian.
6 CHEMISTRY

Table 1.3 Prefixes used in the SI System chemistry laboratories, smaller volumes are
used. Hence, volume is often denoted in cm3
Multiple Prefix Symbol
or dm3 units.
1024 yocto y
21
10 zepto z
18
10 atto a
15
10 femto f

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1012 pico p
109 nano n

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106
micro
3
10 milli m
2
10 centi c

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1
10 deci d

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10 deca da
102
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3
10 kilo k
6
10 mega M
9
10 giga G
12
10 tera T
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1015
peta P Fig. 1.5 Analytical balance
1018 exa E
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1021 zeta Z Maintaining the National

1024
yotta Y Standards of Measurement
The s ystem of units including unit
1.3.2 Mass and Weight definitions keeps on changing with time.

Whenever the accuracy of measurement

Mass of a substance is the amount of matter of a particular unit was enhanced
present in it while weight is the force exerted substantially by adopting new principles,
by gravity on an object. The mass of a member nations of metre treaty (signed in
substance is constant whereas its weight may 1875), agreed to change the formal
vary from one place to another due to change definition of that unit. Each modern
in gravity. You should be careful in using these industrialized country including India has
terms. a National Metrology Institute (NMI) which
maintains standards of measurements.
The mass of a substance can be determined
This responsibility has been given to the
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very accurately in the laboratory by using an National Physical Laboratory (NPL),

analytical balance (Fig. 1.5). New Delhi. This laboratory establishes
The SI unit of mass as given in Table 1.1 is experiments to realize the base units and
kilogram. However, its fraction gram derived units of measurement and
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(1 kg = 1000 g), is used in laboratories due to maintains National Standards of

Measurement. These standards are
the smaller amounts of chemicals used in
periodically inter -compared with
chemical reactions. standards maintained at other National
Volume Metrology Institutes in the world as well
Volume has the units of (length)3. So in SI as those established at the International
system, volume has units of m 3. But again, in Bureau of Standards in Paris.
SOME BASIC CONCEPTS OF CHEMISTRY 7

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

unit, is used for measurement of volume of
liquids.
1 L = 1000 mL , 1000 cm3 = 1 dm3
Fig. 1.6 helps to visualise these relations.
In the laboratory, volume of liquids or
solutions can be measured by graduated

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cylinder, burette, pipette etc. A volumetric flask
is used to prepare a known volume of a

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solution. These measuring devices are shown
in Fig. 1.7.
Density

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Fig. 1.6 Different units used to express volume

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Density of a substance is its amount of mass
per unit volume. So SI units of density can be
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obtained as follows:

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SI unit of mass
SI unit of density =
SI unit of volume
kg 3
= 3 or kg m
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m
This unit is quite large and a chemist often
expresses density in g cm3, where mass is
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expressed in gram and volume is expressed in

cm3.
Temperature

There are three common scales to measure Fig 1.7 Some volume measuring devices
temperature C (degree celsius), F (degree
fahrenheit) and K (kelvin). Here, K is the SI unit. o o
373.15 K 100 C Boiling point 212 F
The thermometers based on these scales are of water
shown in Fig. 1.8. Generally, the thermometer
with celsius scale are calibrated from 0 to 100
where these two temperatures are the freezing 310.15 K
o
37 C
Human body
temperature 98.6 F
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point and the boiling point of water 298.15 K

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25 C Room o
77 F
temperature
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respectively. The fahrenheit scale is

represented between 32 to 212. o Freezing point
273.15 K 0C 32 o F
of water
The temperatures on two scales are related
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to each other by the following relationship:

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F = (C) + 32
5 Kelvin Celsius Fahrenheit
The kelvin scale is related to celsius scale
as follows :
Fig. 1.8 Thermometers using different
K = C + 273.15
temperature scales
8 CHEMISTRY

It is interesting to note that temperature 1.4 UNCERTAINTY IN MEASUREMENT

below 0 C (i.e. negative values) are possible Many a times in the study of chemistry, one
in Celsius scale but in Kelvin scale, negative has to deal with experimental data as well as
temperature is not possible. theoretical calculations. There are meaningful
Reference Standard ways to handle the numbers conveniently and
After defining a unit of measurement such present the data realistically with certainty to
as the kilogram or the metre, scientists the extent possible. These ideas are discussed
agreed on reference standards that make below in detail.

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it possible to calibrate all measuring
devices. For getting reliable measurements, 1.4.1 Scientific Notation

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all devices such as metre sticks and As chemistry is the study of atoms and
analytical balances have been calibrated by
their manufacturers to give correct
molecules which have extremely low masses
readings. However, each of these devices and are present in extremely large numbers,
is standardised or calibrated against some a chemist has to deal with numbers as large

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reference. The mass standard is the as 602, 200,000,000,000,000,000,000 for the

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kilogram since 1889. It has been defined molecules of 2 g of hydrogen gas or as small
as the mass of platinum-iridium (Pt-Ir) as 0.00000000000000000000000166 g
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cylinder that is stored in an airtight jar at mass of a H atom. Similarly other constants

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International Bur eau of Weights and
such as Plancks constant, speed of light,
Measures in Sevres, France. Pt-Ir was
chosen for this standard because it is
charges on particles etc., involve numbers of
highly resistant to chemical attack and its the above magnitude.
mass will not change for an extremely long It may look funny for a moment to write or
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time. count numbers involving so many zeros but it

Scientists are in search of a new
offers a real challenge to do simple
standard for mass. This is being attempted
through accurate determination of mathematical operations of addition,
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Avogadr o constant. Work on this new subtraction, multiplication or division with

standard focuses on ways to measure such numbers. You can write any two
accurately the number of atoms in a well- numbers of the above type and try any one of
defined mass of sample. One such method, the operations you like to accept the challenge
which uses X-rays to determine the atomic

and then you will really appreciate the difficulty

density of a crystal of ultrapure silicon, has
an accuracy of about 1 part in 106 but has
in handling such numbers.
not yet been adopted to serve as a This problem is solved by using scientific
standard. There are other methods but notation for such numbers, i.e., exponential
none of them are presently adequate to notation in which any number can be
replace the Pt-Ir cylinder. No doubt, represented in the form N 10n where n is an
changes are expected within this decade. exponent having positive or negative values
The metre was originally defined as the
length between two marks on a Pt-Ir bar
and N is a number (called digit term) which
kept at a temperature of 0C (273.15 K). In varies between 1.000... and 9.999....
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1960 the length of the metre was defined Thus, we can write 232.508 as
as 1.65076373 106 times the wavelength 2.32508 102 in scientific notation. Note that
of light emitted by a krypton laser. while writing it, the decimal had to be moved
Although this was a cumbersome number,
to the left by two places and same is the
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it preserved the length of the metre at its

agreed value. The metre was redefined in exponent (2) of 10 in the scientific notation.
1983 by CGPM as the length of path Similarly, 0.00016 can be written as
travelled by light in vacuum during a time 1.6 104. Here the decimal has to be moved
interval of 1/299 792 458 of a second. four places to the right and ( 4) is the exponent
Similar to the length and the mass, there in the scientific notation.
are reference standards for other physical
quantities. Now, for performing mathematical
operations on numbers expressed in scientific
SOME BASIC CONCEPTS OF CHEMISTRY 9

notations, the following points are to be kept accuracy is the agreement of a particular value
in mind. to the true value of the result. For example, if
Multiplication and Division the true value for a result is 2.00 g and a
student A takes two measurements and
These two operations follow the same rules reports the results as 1.95 g and 1.93 g. These
which are there for exponential numbers, i.e. values are precise as they are close to each
(5.6 10 ) (6.9 10 ) = (5.6 6.9 ) (10 )
5 8 5+ 8
other but are not accurate. Another student
repeats the experiment and obtains 1.94 g and
= ( 5.6 6.9) 1013

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2.05 g as the results for two measurements.
= 38.64 1013 These observations are neither precise nor
accurate. When a third student repeats these

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= 3.864 1014 measurements and reports 2.01g and 1.99 g
(9.8 10 ) (2.5 10 ) = (9.8 2.5 ) (10 ( ) )
2 6 2+ 6 as the result. These values are both precise and
accurate. This can be more clearly understood
= ( 9.8 2.5) (10 )

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2 6
from the data given in Table 1.4

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Table 1.4 Data to Illustrate Precision and
= 24.50 108 Accuracy
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3
2.7 10
= ( 2.7 5.5 ) (10 3 4 ) = 0.4909 10 7 1 2 Average (g)
5.5 104 Student A 1.95 1.93 1.940
= 4.909 108 Student B 1.94 2.05 1.995
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Addition and Subtraction Student C 2.01 1.99 2.000

For these two operations, first the numbers are
written in such a way that they have same The uncertainty in the experimental or the
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exponent. After that, the coefficient are added calculated values is indicated by mentioning
or subtracted as the case may be. the number of significant figures. Significant
figures are meaningful digits which are known
Thus, for adding 6.65 104 and 8.95 103,
with certainty. The uncertainty is indicated by
6.65 104 + 0.895 104 exponent is made

writing the certain digits and the last uncertain

same for both the numbers. digit. Thus, if we write a result as 11.2 mL, we
Then, these numbers can be added as follows say the 11 is certain and 2 is uncertain and
(6.65 + 0.895) 104 = 7.545 104 the uncertainty would be +1 in the last digit.
Similarly, the subtraction of two numbers can Unless otherwise stated, an uncertainty of +1
be done as shown below : in the last digit is always understood.
2.5 102 4.8 103 There are certain rules for determining the
number of significant figures. These are stated
= (2.5 102) (0.48 102) below:
= (2.5 0.48) 102 = 2.02 102
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(1) All non-zero digits are significant. For

1.4.2 Significant Figures example in 285 cm, there are three
significant figures and in 0.25 mL, there
Every experimental measurement has some are two significant figures.
no

amount of uncertainty associated with it.

However, one would always like the results to (2) Zeros preceding to first non-zero digit are
be precise and accurate. Precision and not significant. Such zero indicates the
accuracy are often referred to while we talk position of decimal point.
about the measurement. Thus, 0.03 has one significant figure and
Precision refers to the closeness of various 0.0052 has two significant figures.
measurements for the same quantity. However, (3) Zeros between two non-zero digits are
10 CHEMISTRY

significant. Thus, 2.005 has four significant Since 2.5 has two significant figures, the
figures. result should not have more than two
(4) Zeros at the end or right of a number are significant figures, thus, it is 3.1.
significant provided they are on the right While limiting the result to the required
side of the decimal point. For example, number of significant figures as done in the
0.200 g has three significant figures. above mathematical operation, one has to keep
But, if otherwise, the terminal zeros are not in mind the following points for rounding off
significant if there is no decimal point. For the numbers

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example, 100 has only one significant 1. If the rightmost digit to be removed is more
figure, but 100. has three significant than 5, the preceding number is increased

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figures and 100.0 has four significant by one. for example, 1.386
figures. Such numbers are better If we have to remove 6, we have to round it
represented in scientific notation. We can
to 1.39
express the number 100 as 1102 for one

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significant figure, 1.010 2 for two 2. If the rightmost digit to be removed is less

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significant figures and 1.00102 for three than 5, the preceding number is not changed.
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significant figures. For example, 4.334 if 4 is to be removed,
then the result is rounded upto 4.33.

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(5) Counting numbers of objects, for example,
2 balls or 20 eggs, have infinite significant 3. If the rightmost digit to be removed is 5,
figures as these are exact numbers and can then the preceding number is not changed
be represented by writing infinite number if it is an even number but it is increased
of zeros after placing a decimal i.e., by one if it is an odd number. For example,
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2 = 2.000000 or 20 = 20.000000 if 6.35 is to be rounded by removing 5, we

In numbers written in scientific notation, have to increase 3 to 4 giving 6.4 as the
result. However, if 6.25 is to be rounded
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all digits are significant e.g., 4.01102 has three

significant figures, and 8.256 103 has four off it is rounded off to 6.2.
significant figures. 1.4.3 Dimensional Analysis
Addition and Subtraction of Significant Often while calculating, there is a need to

Figures convert units from one system to other. The

The result cannot have more digits to the right method used to accomplish this is called factor
of the decimal point than either of the original label method or unit factor method or
numbers. dimensional analysis. This is illustrated
12.11 below.
18.0 Example
1.012
A piece of metal is 3 inch (represented by in)
31.122
long. What is its length in cm?
Here, 18.0 has only one digit after the decimal
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We know that 1 in = 2.54 cm

point and the result should be reported only
up to one digit after the decimal point which From this equivalence, we can write
is 31.1. 1 in 2.54 cm
=1=
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Multiplication and Division of Significant 2.54 cm 1 in

Figures
1 in 2.54cm
In these operations, the result must be reported thus 2.54cm equals 1 and 1 in also
with no more significant figures as are there in
the measurement with the few significant equals 1. Both of these are called unit factors.
figures. If some number is multiplied by these unit
2.51.25 = 3.125 factors (i.e. 1), it will not be affected otherwise.
SOME BASIC CONCEPTS OF CHEMISTRY 11

Say, the 3 in given above is multiplied by i.e., 2 days = seconds

the unit factor. So,
The unit factors can be multiplied in
2.54cm series in one step only as follows:
3 in = 3 in 1 in = 3 2.54 cm = 7.62 cm
24 h 60 min 60 s
2 day
Now the unit factor by which multiplication 1day 1h 1 min
2.54 cm = 2 24 60 60 s

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is to be done is that unit factor ( in = 172800 s
1 in
the above case) which gives the desired units 1.5 LAWS OF CHEMICAL COMBINATIONS

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i.e., the numerator should have that part which The combination of elements
is required in the desired result. to form compounds is
It should also be noted in the above governed by the following five
basic laws.

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example that units can be handled just like

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other numerical part. It can be cancelled, 1.5.1 Law of Conservation
divided, multiplied, squared etc. Let us study of Mass
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one more example for it. It states that matter can Antoine Lavoisier

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Example neither be created nor (17431794)

A jug contains 2 L of milk. Calculate the volume destroyed.

of the milk in m3. This law was put forth by Antoine Lavoisier
Since 1 L = 1000 cm3 in 1789. He performed careful experimental
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and 1m = 100 cm which gives studies for combustion reactions for reaching
to the above conclusion. This law formed the
1m 100cm basis for several later developments in
=1 =
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100 cm 1m chemistry. Infact, this was the result of exact

measurement of masses of reactants and
To get m3 from the above unit factors, the
products, and carefully planned experiments
first unit factor is taken and it is cubed.
performed by Lavoisier.

3
1m 1 m3
= (1) = 1
3
6 3
1.5.2 Law of Definite Proportions
100 cm 10 cm This law was given by, a
Now 2 L = 21000 cm3 French chemist, Joseph
The above is multiplied by the unit factor Proust. He stated that a
given compound always
1m 3 2 m3
2 1000 cm 3 6 3
= 3
= 2 10 3 m 3 contains exactly the same
10 cm 10 proportion of elements by
Example weight.
Proust worked with two Joseph Proust
tt

How many seconds are there in 2 days? (17541826)

Here, we know 1 day = 24 hours (h) samples of cupric carbonate
one of which was of natural origin and the
1 day 24 h other was synthetic one. He found that the
no

or 24 h = 1 = 1d ay composition of elements present in it was same

for both the samples as shown below :
then 1h = 60 min
% of % of % of
1h 60 min copper oxygen carbon
or =1=
60 min 1h Natural Sample 51.35 9.74 38.91
so, for converting 2 days to seconds, Synthetic Sample 51.35 9.74 38.91
12 CHEMISTRY

Thus, irrespective of the source, a given Thus, 100 mL of hydrogen combine with
compound always contains same elements in 50 mL of oxygen to give 100 mL of water
the same proportion. The validity of this law vapour.
has been confirmed by various experiments. Hydrogen + Oxygen Water
It is sometimes also referred to as Law of
100 mL 50 mL 100 mL
definite composition.
Thus, the volumes of hydrogen and oxygen
1.5.3 Law of Multiple Proportions which combine together (i.e. 100 mL and

d
This law was proposed by Dalton in 1803. 50 mL) bear a simple ratio of 2:1.
According to this law, if two elements can Gay-Lussacs discovery of integer ratio in
combine to form more than one compound, the volume relationship is actually the law of

he
masses of one element that combine with a definite proportions by volume. The law of
fixed mass of the other element, are in the definite proportions, stated earlier, was with
ratio of small whole numbers. respect to mass. The Gay-Lussacs law was

pu T
For example, hydrogen combines with explained properly by the work of Avogadro

is
oxygen to form two compounds, namely, water in 1811.
and hydrogen peroxide.
re ER
Hydrogen + Oxygen Water
1.5.5 Avogadro Law

bl
In 1811, Avogadro proposed
2g 16g 18g that equal volumes of gases
Hydrogen + Oxygen Hydrogen Peroxide at the same temperature and
2g 32g 34g pressure should contain
be C

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

Avogadro made a distinction
which combine with a fixed mass of hydrogen
between atoms and
(2g) bear a simple ratio, i.e. 16:32 or 1: 2.
molecules which is quite Lorenzo Romano
o N

Amedeo Carlo
1.5.4 Gay Lussacs Law of Gaseous understandable in the Avogadro di
Volumes present times. If we consider Quareqa edi

This law was given by Gay again the reaction of hydrogen Carreto
(1776-1856)
and oxygen to produce water,

Lussac in 1808. He observed

that when gases combine or we see that two volumes of hydrogen combine
are produced in a chemical with one volume of oxygen to give two volumes
reaction they do so in a of water without leaving any unreacted oxygen.
simple ratio by volume Note that in the Fig. 1.9, each box contains
provided all gases are at equal number of molecules. In fact, Avogadro
same temperature and Joseph Louis could explain the above result by considering
Gay Lussac
pressure. the molecules to be polyatomic. If hydrogen
tt
no

Fig. 1.9 Two volumes of hydrogen react with One volume of oxygen to give Two volumes of water vapour
SOME BASIC CONCEPTS OF CHEMISTRY 13

and oxygen were considered as diatomic as understand what we mean by atomic and
recognised now, then the above results are molecular masses.
easily understandable. However, Dalton and
1.7.1 Atomic Mass
others believed at that time that atoms of the
same kind cannot combine and molecules of The atomic mass or the mass of an atom is
oxygen or hydrogen containing two atoms did actually very-very small because atoms are
not exist. Avogadros proposal was published extremely small. Today, we have sophisticated
in the French Journal de Physidue. In spite techniques e.g., mass spectrometry for
determining the atomic masses fairly

d
of being correct, it did not gain much support.
accurately. But, in the nineteenth century,
After about 50 years, in 1860, first scientists could determine mass of one atom

he
international conference on chemistry was held
relative to another by experimental means, as
in Karlsruhe, Germany to resolve various ideas.
has been mentioned earlier. Hydrogen, being
At the meeting, Stanislao Cannizaro presented
lightest atom was arbitrarily assigned a mass
a sketch of a course of chemical philosophy

pu T
of 1 (without any units) and other elements
which emphasised the importance of

is
were assigned masses relative to it. However,
Avogadros work.
the present system of atomic masses is based
re ER
1.6 DALTONS ATOMIC THEORY on carbon - 12 as the standard and has been

bl
agreed upon in 1961. Here, Carbon - 12 is one
Although the origin of idea
of the isotopes of carbon and can be
that matter is composed of
represented as 12C. In this system, 12C is
small indivisible particles
assigned a mass of exactly 12 atomic mass unit
called a-tomio (meaning
(amu) and masses of all other atoms are given
be C

indivisible), dates back to the

relative to this standard. One atomic mass
time of Democritus, a Greek
unit is defined as a mass exactly equal to one-
Philosopher (460 370 BC),
twelfth the mass of one carbon - 12 atom.
o N

it again started emerging as John Dalton

a result of several (17761884) And 1 amu = 1.660561024 g
experimental studies which led to the Laws Mass of an atom of hydrogen
mentioned above. = 1.67361024 g

In 1808, Dalton published A New System Thus, in terms of amu, the mass
of Chemical Philosophy in which he proposed 1.6736 1024 g
the following : of hydrogen atom =
1.66056 10 24 g
1. Matter consists of indivisible atoms.
= 1.0078 amu
2. All the atoms of a given element have
= 1.0080 amu
identical properties including identical
mass. Atoms of different elements differ in Similarly, the mass of oxygen - 16 (16O)
mass. atom would be 15.995 amu.
Today, amu has been replaced by u
3. Compounds are formed when atoms of
tt

which is known as unified mass.

different elements combine in a fixed ratio.
When we use atomic masses of elements in
4. Chemical reactions involve reorganisation calculations, we actually use average atomic
of atoms. These are neither created nor masses of elements which are explained
no

destroyed in a chemical reaction.

below.
Daltons theory could explain the laws of
chemical combination. 1.7.2 Average Atomic Mass
Many naturally occurring elements exist as
1.7 ATOMIC AND MOLECULAR MASSES more than one isotope. When we take into
After having some idea about the terms atoms account the existence of these isotopes and
and molecules, it is appropriate here to their relative abundance (per cent occurrence),
14 CHEMISTRY

the average atomic mass of that element can Solution

be computed. For example, carbon has the Molecular mass of glucose (C6H12O6)
following three isotopes with relative
= 6(12.011 u) + 12(1.008 u) +
abundances and masses as shown against 6(16.00 u)
each of them.
= (72.066 u) + (12.096 u) +
Isotope Relative Atomic (96.00 u)
Abundance Mass (amu)
= 180.162 u
(%)

d
12
C 98.892 12 1.7.4 Formula Mass

he
13
C 1.108 13.00335
Some substances such as sodium chloride do
14
C 2 1010 14.00317 not contain discrete molecules as their
From the above data, the average atomic constituent units. In such compounds, positive
(sodium) and negative (chloride) entities are

pu T
mass of carbon will come out to be :
arranged in a three-dimensional structure, as

is
(0.98892) (12 u) + ( 0.01108) (13.00335 u) + shown in Fig. 1.10.
(2 1012) (14.00317 u)
re ER
= 12.011 u

bl
Similarly, average atomic masses for other
elements can be calculated. In the periodic
table of elements, the atomic masses mentioned
for different elements actually represented their
be C

average atomic masses.

1.7.3 Molecular Mass

o N

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,
(CH4) = (12.011 u) + 4 (1.008 u) Fig. 1.10 Packing of Na+ and Cl ions in
sodium chloride
= 16.043 u
It may be noted that in sodium chloride,
tt

Similarly, molecular mass of water (H2O)

one Na+ is surrounded by six Cl and vice-versa.
= 2 atomic mass of hydrogen + 1 atomic
mass of oxygen The formula such as NaCl is used to
calculate the formula mass instead of
no

= 2 (1.008 u) + 16.00 u molecular mass as in the solid state sodium

= 18.02 u chloride does not exist as a single entity.

Problem 1.1 Thus, formula mass of sodium chloride =

atomic mass of sodium + atomic mass of
Calculate molecular mass of glucose chlorine
(C6H 12O 6) molecule.
= 23.0 u + 35.5 u = 58.5 u
SOME BASIC CONCEPTS OF CHEMISTRY 15

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 similar
magnitude is required.
Just as we denote one dozen for 12 items,

d
score for 20 items, gross for 144 items, we
use the idea of mole to count entities at the

he
microscopic level (i.e. atoms/molecules/
particles, electrons, ions, etc). Fig. 1.11 One mole of various substances
In SI system, mole (symbol, mol) was
1 mol of sodium chloride = 6.022 1023

pu T
introduced as seventh base quantity for the
formula units of sodium chloride

is
amount of a substance.
One mole is the amount of a substance Having defined the mole, it is easier to know
re ER
that contains as many particles or entities mass of one mole of the substance or the

bl
as there are atoms in exactly 12 g (or 0.012 constituent entities. The mass of one mole
kg) of the 12C isotope. It may be emphasised of a substance in grams is called its
that the mole of a substance always contain molar mass. The molar mass in grams is
the same number of entities, no matter what numerically equal to atomic/molecular/
formula mass in u.
be C

the substance may be. In order to determine

this number precisely, the mass of a carbon Molar mass of water = 18.02 g mol-1
12 atom was determined by a mass Molar mass of sodium chloride = 58.5 g mol-1
spectrometer and found to be equal to
o N

1.992648 1023 g. Knowing that one mole of 1.9 PERCENTAGE COMPOSITION

carbon weighs 12 g, the number of atoms in it So far, we were dealing with the number of
is equal to : entities present in a given sample. But many a
time, the information regarding the percentage

12 g/ mol 12C of a particular element present in a compound

1.992648 1023 g/12 Catom is required. Suppose an unknown or new
compound is given to you, the first question
= 6.0221367 1023 atoms/mol you would ask is: what is its formula or what
This number of entities in 1 mol is so are its constituents and in what ratio are they
important that it is given a separate name and present in the given compound? For known
symbol. It is known as Avogadro constant, compounds also, such information provides a
denoted by NA in honour of Amedeo Avogadro. check whether the given sample contains the
To really appreciate largeness of this number, same percentage of elements as is present in a
tt

let us write it with all the zeroes without using pure sample. In other words, one can check
any powers of ten. the purity of a given sample by analysing this
602213670000000000000000 data.
no

Hence, so many entities (atoms, molecules Let us understand it by taking the example
or any other particle) constitute one mole of a of water (H2O). Since water contains hydrogen
particular substance. and oxygen, the percentage composition of both
these elements can be calculated as follows :
We can, therefore, say that 1 mol of
hydrogen atoms = 6.0221023 atoms Mass % of an element =
1 mol of water molecules = 6.0221023 mass of that element in the compound 100
water molecules molar mass of the compound
16 CHEMISTRY

Molar mass of water = 18.02 g Problem 1.2

21.008 A compound contains 4.07 % hydrogen,
Mass % of hydrogen = 100
18.02 24.27 % carbon and 71.65 % chlorine.
= 11.18 Its molar mass is 98.96 g. What are its
empirical and molecular formulas ?
16.00
Mass % of oxygen = 100 Solution
18.02
Step 1. Conversion of mass per cent
= 88.79

d
to grams.
Let us take one more example. What is the
Since we are having mass per cent, it is
percentage of carbon, hydrogen and oxygen

he
convenient to use 100 g of the compound
in ethanol? as the starting material. Thus, in the
Molecular formula of ethanol is : C2H5OH 100 g sample of the above compound,
Molar mass of ethanol is : (212.01 + 61.008 4.07g hydrogen is present, 24.27g

pu T
+ 16.00) g carbon is present and 71.65 g chlorine is

is
present.
= 46.068 g
re ER
Mass per cent of carbon Step 2. Convert into number moles of
each element

bl
24.02g
= 46.068g 100 = 52.14% Divide the masses obtained above by
respective atomic masses of various
Mass per cent of hydrogen elements.
be C

6.048 g 4.07 g
= 100 = 13.13% Moles of hydrogen = = 4.04
46.068 g 1.008 g
o N

Mass per cent of oxygen 24.27 g

16.00 g Moles of carbon = = 2.021
12.01g
= 46.068g 100 = 34.73%
71.65g
After understanding the calculation of per

Moles of chlorine = 35.453 g = 2.021

cent of mass, let us now see what information
can be obtained from the per cent composition Step 3. Divide the mole value obtained
data. above by the smallest number
Since 2.021 is smallest value, division by
1.9.1 Empirical Formula for Molecular it gives a ratio of 2:1:1 for H:C:Cl .
Formula In case the ratios are not whole numbers,
An empirical formula represents the simplest then they may be converted into whole
whole number ratio of various atoms present number by multiplying by the suitable
tt

in a compound whereas the molecular coefficient.

formula shows the exact number of different Step 4. Write empirical formula by
types of atoms present in a molecule of a mentioning the numbers after writing
compound. the symbols of respective elements.
no

If the mass per cent of various elements CH2Cl is, thus, the empirical formula of
present in a compound is known, its the above compound.
empirical formula can be determined. Step 5. Writing molecular formula
Molecular formula can further be obtained if (a) Determine empirical formula mass
the molar mass is known. The following Add the atomic masses of various atoms
example illustrates this sequence. present in the empirical formula.
SOME BASIC CONCEPTS OF CHEMISTRY 17

For CH2Cl, empirical formula mass is 1.10 STOICHIOMETRY AND

12.01 + 2 1.008 + 35.453 STOICHIOMETRIC CALCULATIONS
= 49.48 g The word stoichiometry is derived from two
Greek words - stoicheion (meaning element)
(b) Divide Molar mass by empirical and metron (meaning measure).
formula mass Stoichiometry, thus, deals with the calculation
of masses (sometimes volumes also) of the
Molar mass 98.96 g reactants and the products involved in a
=

d
Empirical formula mass 49.48 g chemical reaction. Before understanding how
to calculate the amounts of reactants required
= 2 = (n) or the products produced in a chemical

he
(c) Multiply empirical formula by n reaction, let us study what information is
obtained above to get the molecular available from the balanced chemical equation
formula of a given reaction. Let us consider the
combustion of methane. A balanced equation

pu T
Empirical formula = CH2Cl, n = 2. Hence
for this reaction is as given below :

is
molecular formula is C2H4Cl2.
CH4 (g) + 2O2 (g) CO2 (g) + 2 H2O (g)
re ER
Balancing a chemical equation

bl
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 err or. Let us take the reactions of a few metals and non-metals with
oxygen to give oxides
be C

4 Fe(s) + 3O2 (g) 2Fe2 O3 (s) (a) balanced equation

2 Mg(s) + O 2(g) 2MgO(s) (b) balanced equation
P4 (s) + O2 (g) P4O 10(s) (c) unbalanced equation
o N

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

P4 (s) + 5O2 (g) P4O 10(s) balanced equation

Now let us take combustion of propane, C3 H8 . This equation can be balanced in steps.
Step 1 Write down the correct for mulas of reactants and products. Here propane and oxygen are
reactants, and carbon dioxide and water are products.
C3H 8(g) + O2 (g) CO2 (g) +H2O(l) unbalanced equation
Step 2 Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore,
three CO2 molecules are required on the right side.
C3H 8 (g) + O2 (g) 3CO2 (g) + H2 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
tt

required for eight hydrogen atoms on the right side.

C3H 8 (g) +O2 (g) 3CO2 (g)+4H 2O (l)
Step 4 Balance the number of O atoms: There are ten oxygen atoms on the right side (3 2 = 6 in
CO2 and 4 1= 4 in water). Therefore, five O2 molecules are needed to supply the required ten
no

oxygen atoms.
C3H 8 (g) +5O2 (g) 3CO2 (g) + 4H2 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 ten 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.
18 CHEMISTRY

Here, methane and dioxygen are called CH 4 (g) gives 2 mol of H2O (g).
reactants and carbon dioxide and water are
2 mol of water (H2O) = 2 (2+16)
called products. Note that all the reactants and
= 2 18 = 36 g
the products are gases in the above reaction
and this has been indicated by letter (g) in the 18 g H2O
brackets next to its formula. Similarly, in the 1 mol H2O = 18 g H2O 1mol H O = 1
2
case of solids and liquids, (s) and (l) are written
respectively.
18 g H2O

d
The coefficients 2 for O2 and H2O are called Hence 2 mol H2O
1 mol H2O
stoichiometric coefficients. Similarly the

he
coefficient for CH4 and CO2 is one in each case. = 2 18 g H 2O = 36 g H2O
They represent the number of molecules (and Problem 1.4
moles as well) taking part in the reaction or
formed in the reaction. How many moles of methane are required

pu T
to produce 22 g CO2 (g) after combustion?
Thus, according to the above chemical

is
reaction, Solution

re ER
One mole of CH4(g) reacts with two moles According to the chemical equation,

bl
of O2(g) to give one mole of CO2(g) and CH4 ( g ) + 2O2 ( g ) CO2 ( g ) + 2H2O ( g )
two moles of H 2O(g)
44g CO 2 (g) is obtained from 16 g CH4 (g).
One molecule of CH 4(g) reacts with
[ 1 mol CO2(g) is obtained from 1 mol of
2 molecules of O2(g) to give one molecule
be C

CH4(g)]
of CO2(g) and 2 molecules of H2O(g)
mole of CO 2 (g)
22.7 L of CH4(g) reacts with 45.4 L of O2 (g)
1 mol CO2 (g)
o N

to give 22.7 L of CO2 (g) and 45.4 L of H2O(g)

= 22 g CO 2 (g)
16 g of CH4 (g) reacts with 232 g of O2 (g) 44 g CO2 (g)
to give 44 g of CO2 (g) and 218 g of = 0.5 mol CO 2 (g)
H 2O (g).
Hence, 0.5 mol CO2 (g) would be obtained

From these relationships, the given data from 0.5 mol CH4 (g) or 0.5 mol of CH4 (g)
can be interconverted as follows : would be required to produce 22 g
mass moles no. of molecules CO 2 (g).

Mass
= Density 1.10.1 Limiting Reagent
Volume
Many a time, the reactions are carried out
when the reactants are not present in the
Problem 1.3
amounts as required by a balanced chemical
Calculate the amount of water (g)
tt

reaction. In such situations, one reactant is in

produced by the combustion of 16 g of excess over the other. The reactant which is
methane. present in the lesser amount gets consumed
Solution after sometime and after that no further
no

reaction takes place whatever be the amount

The balanced equation for combustion of
methane is : of the other reactant present. Hence, the
reactant which gets consumed, limits the
CH4 ( g ) + 2O2 ( g ) CO2 ( g ) + 2H2O ( g ) amount of product formed and is, therefore,
called the limiting reagent.
(i)16 g of CH4 corresponds to one mole.
In performing stoichiometric calculations,
(ii) From the above equation, 1 mol of this aspect is also to be kept in mind.
SOME BASIC CONCEPTS OF CHEMISTRY 19

Problem 1.5 17.0 g NH3 (g)

3.30103 mol NH3 (g)
50.0 kg of N2 (g) and 10.0 kg of H 2 (g) are 1 mol NH3 (g)
mixed to produce NH3 (g). Calculate the
NH3 (g) formed. Identify the limiting = 3.3010317 g NH 3 (g)
reagent in the production of NH3 in this = 56.1103 g NH3
situation. = 56.1 kg NH3
Solution

d
A balanced equation for the above reaction 1.10.2 Reactions in Solutions
is written as follows : A majority of reactions in the laboratories are

he
Calculation of moles : carried out in solutions. Therefore, it is
important to understand as how the amount
N 2 ( g ) + 3H2 ( g ) 2 NH3 ( g ) of substance is expressed when it is present in
moles of N2 the form of a solution. The concentration of a

pu T
solution or the amount of substance present

is
1000 g N2 1 mol N2 in its given volume can be expressed in any of
= 50.0 kg N2
re ER 1 kg N2 28.0 g N2 the following ways.

bl
= 17.86102 mol 1. Mass per cent or weight per cent (w/w %)
moles of H2 2. Mole fraction
1000 g H2 1mol H2 3. Molarity
= 10.00 kg H2
1 kg H2 2.016 g H2 4. Molality
be C

= 4.96103 mol Let us now study each one of them in detail.

According to the above equation, 1 mol 1. Mass per cent
o N

N2 (g) requires 3 mol H2 (g), for the reaction. It is obtained by using the following relation:
Hence, for 17.86102 mol of N2, the moles
of H2 (g) required would be Massof solute
Massper cent = 100
Massof solution
3 mol H2 (g)

17.86102 mol N2
1mol N2 (g)
Problem 1.6
= 5.36 103 mol H2
A solution is prepared by adding 2 g of a
But we have only 4.96103 mol H2. Hence, substance A to 18 g of water. Calculate
dihydrogen is the limiting reagent in this the mass per cent of the solute.
case. So NH3(g) would be formed only from
that amount of available dihydrogen i.e.,
Solution
4.96 103 mol
Since 3 mol H2(g) gives 2 mol NH3(g) Massof A
tt

Mass per centof A = 100

Massof solution
2 mol NH3 (g)
4.96103 mol H2 (g)
3mol H2 (g) 2g
= 100
no

= 3.3010 mol NH3 (g)

3
2 gof A +18 gof water
3.30103 mol NH3 (g) is obtained.
2g
If they are to be converted to grams, it is = 100
done as follows : 20 g
1 mol NH3 (g) = 17.0 g NH 3 (g) = 10 %
20 CHEMISTRY

2. Mole Fraction Thus, 200 mL of 1M NaOH are taken and

It is the ratio of number of moles of a particular enough water is added to dilute it to make it
component to the total number of moles of the 1 litre.
solution. If a substance A dissolves in In fact for such calculations, a general
substance B and their number of moles are formula, M1 V1 = M 2 V2 where M and V are
n A and nB respectively; then the mole fractions molarity and volume respectively can be used.
of A and B are given as In this case, M1 is equal to 0.2; V 1 = 1000 mL
and, M 2 = 1.0; V 2 is to be calculated.

d
Mole fraction of A Substituting the values in the formula:
No.of moles of A 0.2 M 1000 mL = 1.0 M V2

he
=
No.of moles of solution 0.2M 1000mL
nA V2 = = 200 mL
= 1.0M
n A + nB

pu T
Note that the number of moles of solute

is
(NaOH) was 0.2 in 200 mL and it has remained
Mole fraction of B the same, i.e., 0.2 even after dilution ( in 1000
re ER
No.of moles of B
mL) as we have changed just the amount of

bl
= solvent (i.e. water) and have not done anything
No.of moles of solution with respect to NaOH. But keep in mind the
nB concentration.
=
n A + nB Problem 1.7
be C

3. Molarity Calculate the molarity of NaOH in the

It is the most widely used unit and is denoted solution prepared by dissolving its 4 g in
enough water to form 250 mL of the
o N

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

the solute in 1 litre of the solution. Thus, solution.
Solution
No. of moles of solute
Molarity (M) = Since molarity (M)
Volume of solution in litres

Suppose we have 1 M solution of a No. of moles of solute

=
substance, say NaOH and we want to prepare Volume of solution in litres
a 0.2 M solution from it. Mass of NaOH/Molar mass of NaOH
1 M NaOH means 1 mol of NaOH present =
0.250 L
in 1 litre of the solution. For 0.2 M solution we
require 0.2 moles of NaOH in 1 litre solution. 4 g / 40 g 0.1 mol
= =
Hence, we have to take 0.2 moles of NaOH 0.250 L 0.250 L
and make the solution to 1 litre. = 0.4 mol L 1
tt

Now how much volume of concentrated = 0.4 M

(1M) NaOH solution be taken which contains Note that molarity of a solution depends
0.2 moles of NaOH can be calculated as upon temperature because volume of a
no

follows: solution is temperature dependent.

If 1 mol is present in 1 L or 1000 mL
4. Molality
then 0.2 mol is present in
It is defined as the number of moles of solute
1000 mL present in 1 kg of solvent. It is denoted by m.
0.2 mol
1 mol No. of moles of solute
= 200 mL Thus, Molality (m) =
Mass of solvent in kg
SOME BASIC CONCEPTS OF CHEMISTRY 21

Problem 1.8
The density of 3 M solution of NaCl is No. of moles of solute
Molality =
1.25 g mL1. Calculate molality of the Mass of solvent in kg
solution.
3 mol
Solution =
1.0745 kg
M = 3 mol L1
= 2.79 m
Mass of NaCl

d
in 1 L solution = 3 58.5 = 175.5 g Often in a chemistry laboratory, a solution
of a desired concentration is prepared by
Mass of

he
diluting a solution of known higher
1L solution = 1000 1.25 = 1250 g concentration. The solution of higher
1
(since density = 1.25 g mL ) concentration is also known as stock
Mass of solution. Note that molality of a solution

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water in solution = 1250 175.5 does not change with temperature since

is
= 1074.5 mass remains unaffected with
g temperature.
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bl SUMMARY
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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
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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 out of which 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
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quantities is very important. The measurements of quantities in chemistry are spread

over a wide range of 1031 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
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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 Lussacs Law of Gaseous Volumes and Avogadro Law. All
these laws led to the Daltons atomic theory which states that atoms are building
blocks of matter. The atomic mass of an element is expressed relative to 12C isotope of
22 CHEMISTRY

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 23). This is known as 1 mol of

d
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

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

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

is
expressed in number of ways, e.g., mass per cent, mole fraction, molarity and molality.
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bl
EXERCISES
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1.1 Calculate the molar mass of the following :

(i) H2O (ii) CO 2 (iii) CH4
1.2 Calculate the mass per cent of different elements present in sodium sulphate
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(Na2 SO4).
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 bur nt 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 (CH3COONa) required to make 500 mL of
0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol1.
1.6 Calculate the concentration of nitric acid in moles per litre in a sample which
has a density, 1.41 g mL1 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 (CuSO4) ?
1.8 Determine the molecular formula of an oxide of iron in which the mass per cent
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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
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35
Cl 75.77 34.9689
37
Cl 24.23 36.9659
1.10 In three moles of ethane (C2H 6), calculate the following :
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
SOME BASIC CONCEPTS OF CHEMISTRY 23

(iii) Number of molecules of ethane.

1.11 What is the concentration of sugar (C12H 22O 11) in mol L1 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 L1, 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 m2

d
If mass of air at sea level is 1034 g cm2 , calculate the pressure in pascal.
1.14 What is the SI unit of mass? How is it defined?

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1.15 Match the following prefixes with their multiples:
Prefixes Multiples
(i) micro 106

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(ii) deca 109

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(iii) mega 106
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(iv) giga
(v) femto 10
1015

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1.16 What do you mean by significant figures ?
1.17 A sample of drinking water was found to be severely contaminated with chloroform,
CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15
ppm (by mass).
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(i) Express this in percent by mass.

(ii) Determine the molality of chloroform in the water sample.
1.18 Express the following in the scientific notation:
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(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
(iv) 126,000
(v) 500.0
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(vi) 2.0034
1.20 Round up the following upto three significant figures:
(i) 34.216
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(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
24 CHEMISTRY

(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

d
(ii) 1 mg = ...................... kg = ...................... ng
(iii) 1 mL = ...................... L = ...................... dm3

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1.22 If the speed of light is 3.0 108 m s1 , calculate the distance covered by light in
2.00 ns.
1.23 In a reaction
A + B2 AB2

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

is
(i) 300 atoms of A + 200 molecules of B
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(ii) 2 mol A + 3 mol B

bl
(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
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to the following chemical equation:

N2 (g) + H 2 (g) 2NH3 (g)
(i) Calculate the mass of ammonia produced if 2.00 103 g dinitrogen reacts
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with 1.00 103 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 Na2CO3 and 0.50 M Na 2CO3 different?

1.26 If ten 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
1.28 Which one of the following will have largest number of atoms?
(i) 1 g Au (s)
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(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
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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 12C atom in g ?
1.31 How many significant figures should be present in the answer of the following
calculations?
0.02856 298.15 0.112
(i) (ii) 5 5.364
0.5785
SOME BASIC CONCEPTS OF CHEMISTRY 25

(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
36 1
Ar 35.96755 g mol 0.33 7%
38
Ar 37.96272 g mol1 0.06 3%
40 1
Ar 39.9624 g mol 99.60 0%

d
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.

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

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formula.

is
1.35 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to
the reaction, CaCO3 (s) + 2 HCl (aq) CaCl2 (aq) + CO2 (g) + H2O(l)
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What mass of CaCO 3 is required to react completely with 25 mL of 0.75 M HCl ?

bl
1.36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with
aqueous hydrochloric acid according to the reaction
4 HCl (aq) + MnO2(s) 2H2 O (l) + MnCl2 (aq) + Cl2 (g)
How many grams of HCl react with 5.0 g of manganese dioxide?
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