Atoms & Molecules

Atoms, Molecules and Chemical Arithmetic MODULE - 1
Some Basic Concepts of

Chemistry
1
Notes
ATOMS, MOLECULES AND

CHEMICAL ARITHMETIC
Chemistry is the study of matter and the changes it undergoes. Chemistry is

often called the central science, because a basic knowledge of chemistry is
essential for the study of biology, physics, geology, ecology, and many other
subjects.
Although chemistry is an ancient science, its modern foundation was laid in the
nineteenth century, when intellectual and technological advances enabled
scientists to break down substances into ever smaller components and consequently
to explain many of their physical and chemical characteristics.
Chemistry plays a pivotal role in many areas of science and technology e.g.
in health, medicine, energy and environment, food, agriculture and new
materails.
As you are aware, atoms and molecules are so small that we cannot see them
with our naked eyes or even with the help of a microscope. Any sample of matter
which can be studied consists of extremely large number of atoms or molecules.
In chemical reactions, atoms or molecules combine with one another in a definite
number ratio. Therefore, it would be pertinent if we could specify the total
number of atoms or molecules in a given sample of a substance. We use many
number units in our daily life. For example, we express the number of bananas
or eggs in terms of ‘dozen’. In chemistry we use a number unit called mole
which is very large.
With the help of mole concept it is possible to take a desired number of atoms/
molecules by weighing. Now, in order to study chemical compounds and reactions
in the laboratory, it is necessary to have adequate knowledge of the quantitative
relationship among the amounts of the reacting substances that take part and
CHEMISTRY 1
MODULE - 1 Atoms, Molecules and Chemical Arithmetic

Chemistry products formed in the chemical reaction. This relationship is knows as
stoichiometry. Stoichiometry (derived from the Greek Stoicheion = element and
metron = measure) is the term we use to refer to all the quntatitative aspects of
chemical compounds and reactions. In the present lesson, you will see how
chemical formulae are determined and how chemical equations prove useful in
predicting the proper amounts of the reactants that must be mixed to carry out a
Notes complete reaction. In other words we can take reactants for a reaction in such a
way that none of the reacting substances is in excess. This aspect is very vital in
chemistry and has wide application in industries.
OBJECTIVES
After reading this lesson you will be able to :
z explain the scope of chemistry;
z explain the atomic theory of matter;
z state the laws of chemical combinaton;
z explain Dalton’s atomic theory;
z define the terms element, atoms and molecules.
z state the need of SI units;
z list base SI units;
z explain the relationship between mass and number of particles;
z define Avogadro’s constant and state its significance;
z calculate the molar mass of different elements and compounds;
z define molar volume of gases at STP.
z define empirical and molecular formulae;
z differentiate between empirical and molecular formulae;
z calculate precentage by mass of an element in a compound and also work out
empirical formula from the percentage composition;
z establish relationship between mole, mass and volume;
z calculate the amount of substances consumed or formed in a chemical reaction
using a balanced equation and mole concept, and
z explain the role of limiting reagent in limiting the amount of the products
formed.
2 CHEMISTRY
1.1 SCOPE OF CHEMISTRY Chemistry
Chemistry plays an important role in all aspects of our life. Let us discuss role
of chemistry in some such areas.
1.1.1 Health and Medicine

Three major advances in this century have enabled us to prevent and treat Notes
diseases. Public health measures establishing sanitation systems to protect vast
numbers of people from infectious diseases; surgery with anesthesia, enabling
physicians to cure potentially fatal conditions, such as an inflamed appendix; and
the introduction of vaccines and antibiotics that made it possible to prevent
diseases spread by microbes. Gene therapy promises to be the fourth revolution
in medicine. (A gene is the basic unit of inheritance.) Several thousand known
conditions, including cystic fibrosis and hemophilia, are carried by inborn
damage to a single gene. Many other ailments, such as cancer, heart disease,
AIDS, and arthritis, result to an extent from impairment of one or more genes
involved in the body’s defenses. In gene therapy, a selected healthy gene is
delivered to a patient’s cell to cure or ease such disorders. To carry out such
a procedure, a doctor must have a sound knowledge of the chemical properties
of the molecular components involved.
Chemists in the pharmaceutical industry are researching potent drugs with few
or no side effects to treat cancer, AIDS, and many other diseases as well as
drugs to increase the number of successful organ transplants. On a broader scale,
improved understanding of the mechanism of ageing will lead to a longer and
healthier lifespan for the world’s population.
11.2 Energy and the Environment

Energy is a by-product of many chemical processes, and as the demand for
energy continues to increase, both in technologically advanced countries like the
United States and in developing ones like India. Chemists are actively trying
to find new energy sources. Currently the major sources of energy are fossil
fuels (coal, petroleum, and natural gas). The estimated reserves of these fuels
will last us another 50-100 years at the present rate of consumption, so it is
urgent that we find alternatives.
Solar energy promises to be a viable source of energy for the future. Every year
earth’s surface receives about 10 times as much energy from sunlight as is
contained in all of the known reserves of coal, oil, natural gas, and uranium
combined. But much of this energy is “wasted” because it is reflected back into
space. For the past thirty years, intense research efforts have shown that solar
energy can be harnessed effectively in two ways. One is the conversion of
CHEMISTRY 3

Chemistry sunlight directly to electricity using devices called photovoltaic cells. The other
is to use sunlight to obtain hydrogen from water. The hydrogen can then be
fed into a fuel cell to generate electricity. Although our understanding of the
scientific process of converting solar energy to electricity has advanced, the
technology has not yet improved to the point where we can produce electricity
on a large scale at an economically acceptable cost. By 2050, however, it has
Notes been predicted that solar energy will supply over 50 percent of our power needs.
Another potential source of energy is nuclear fission, but because of environmental

concerns about the radioactive wastes from fission processes, the future of the
nuclear industry is uncertain. Chemists can help to devise better ways to dispose
of nuclear waste. Nuclear fusion, the process that occurs in the sun and other
stars, generates huge amounts of energy without producing much dangerous
radioactive waste. In another 50 years, nuclear fusion will likely be a significant
source of energy.
Energy production and energy utilization are closely tied to the quality of our
environment. A major disadvantage of burning fossil fuels is that they give off
carbon dioxide, which is a greenhouse gas (that is, it promotes the heating of
Earth’s atmosphere), along with sulfur dioxide and nitrogen oxides, which result
in acid rain and smog. Harnessing solar energy has no such detrimental effects
on the environment. By using fuel-efficient automobiles and more effective
catalytic converters, we should be able to drastically reduce harmful auto
emissions and improve the air quality in areas with heavy traffic. In addition,
electric cars, powered by durable, long-lasting batteries, should be more
prevalent in the next century, and their use will help to minimize air pollution.
1.1.3 Materials and Technology

Chemical research and development in the twentieth century have provided us
with new materials that have profoundly improved the quality of our lives and
helped to advance technology in countless ways. A few examples are polymers
(including rubber and nylon), ceramics (such as cookware), liquid crystals (like
those in electronic displays), adhesives, and coatings (for example, latex paint).
What is in store for the near future? One likely possibility is room-temperature
superconductors. Electricity is carried by copper cables, which are not perfect
conductors. Consequently, about 20 percent of electrical energy is lost in the
form of heat between the power station and our homes. This is a tremendous
waste. Superconductors are materials that have no electrical resistance and can
therefore conduct electricity with no energy loss.
4 CHEMISTRY
1.1.4 Food and Agriculture Chemistry
How can the world’s rapidly increasing population be fed? In poor countries,
agricultural activities occupy about 80 percent of the workforce and half of an
average family budget is spent on foodstuffs. This is a tremendous drain on a
nation’s resources. The factors that affect agricultural production are the
richness of the soil, insects and diseases that damage crops, and weeds that
compete for nutrients. Besides irrigation, farmers rely on fertilizers and Notes
pesticides to increase crop yield.
1.2 PARTICULATE NATURE OF MATTER

Chemistry deals with study of structure and composition of matter. Since ancient
time people have been wondering about nature of matter. Suppose we take a
piece of rock and start breaking it into smaller and smaller particles can this
process go on far ever resulting in smaller and smaller particles or would it come
to stop when such particles are formed which can no longer to broken into still
smaller particles? Many people including Greek philosophers Plato and Aristotle
believed that matter is continuous and the process of subdivision of matter can
go on.
On the other hand, many people believed that the process of subdivision of mater
can be repeated only a limited nuimber of times till such particles are obtained
which cannot be further subdivided. They believed that mattr is composed of
large number of very tiny particles and thus has particle naturew. The smallest
indivisible particles of matter were given the name ‘atom’ from the Greek word
“atoms” meaning ‘indivisible’. It is generally agreed that the Greek philosopher
Leucippus and his student Democritus were the first to propose this idea, about
440 B.C.. However, Maharshi Kanad had propounded the atomic concept of
matter earlier (500 BC) and had named the smallest particle of matter as
"PARMANU".
1.3 LAWS OF CHEMICAL COMBINATIONS

There was tremendous progress in Chemical Sciences after 18th century. It arose
out of an interest in the nature of heat and the way things burn. Major progress
was made through the careful use of chemical balance to determine the change
in mass that occurs in chemical reactions. The great French Chemist Antoine
Lavoisier used the balance to study chemical reactions. He heated mercury in
a sealed flask that contained air. After several days, a red substance mercury
(II) oxide was produced. The gas remaining in the flask was reduced in mass.
The remaining gas was neither able to support combustion nor life. The
remaining gas in the flask was identified as nitrogen. The gas which combined
with mercury was oxygen. Further he carefully performed the experiment by
CHEMISTRY 5

Chemistry taking a weighed quantity of mercury (II) oxide. After strong heating, he found
that mercury (II) oxide, red in colour, was decomposed into mercury and
oxygen. He weighed both mercury and oxygen and found that their combined
mass was equal to that of the mercury (II) oxide taken. Lavoisier finally came
to the conclusion that in every chemical reaction, total masses of all the
reactants is equal to the masses of all the products. This law is known as the
law of conservation of mass.
Notes
There was rapid progress in science after chemists began accurate determination
of masses of reactants and products. French chemist Claude Berthollet and
Joseph Proust worked on the ratio (by mass) of two elements which combine
to form a compound. Through a careful work, Proust demonstrated the
fundamental law of definite or constant proportions in 1808. In a given
chemical compound, the proportions by mass of the elements that compose
it are fixed, independent of the origin of the compound or its mode of
preparation.
In pure water, for instance, the ratio of mass of hydrogen to the mass of oxygen
is always 1:8 irrespective of the source of water. In other words, pure water
contains 11.11% of hydrogen and 88.89% of oxygen by mass whether water
is obtained from well, river or from a pond. Thus, if 9.0 g of water are
decomposed, 1.0 g of hydrogen and 8.0 g of oxygen are always obtained.
Furthermore, if 3.0 g of hydrogen are mixed with 8.0 g of oxygen and the mixture
is ignited, 9.0 g of water are formed and 2.0 g of hydrogen remains unreacted.
Similarly sodium chloride contains 60.66% of chlorine and 39.34% of sodium
by mass whether we obtained it from salt mines or by crytallising it from water
of ocean or inland salt seas or synthesizing it from its elements sodium and
chlorine. Of course, the key word in this sentence is ‘pure’. Reproducible
experimental results are highlights of scientific thoughts. In fact modern science
is based on experimental findings. Reproducible results indirectly hint for a
truth which is hidden. Scientists always worked for findings this truth and in
this manner many theories and laws were discovered. This search for truth plays
an important role in the development of science.
The Dalton’s atomic theory not only explained the laws of conservations of mass
and law of constant proportions but also predicted the new ones. He deduced
the law of multiple proportions on the basis of his theory. The law states that
when two elements form more than one compound, the masses of one
element in these compound for a fixed mass of the other element are in
the ratio of small whole numbers. For example, carbon and oxygen form two
compounds: carbon monoxide and carbon dioxide. Carbon monoxide contains
1.3321 g of oxygen for each 1.0000 g of carbon, whereas carbon dioxide
contains 2.6642 g of oxygen for 1.0000 g of carbon. In other words, carbon
dioxide contains twice the mass of oxygen as is contained in carbon monoxide
6 CHEMISTRY
(2.6642 g = 2 × 1.3321 g) for a given mass of carbon. Atomic theory explains Chemistry
this by saying that carbon dioxide contains twice as many oxygen atoms for a
given number of carbon atoms as does carbon monoxide. The deduction of law
of multiple proportions from atomic theory was important in convincing
chemists of the validity of the theory.
1.4 DALTON’S ATOMIC THEORY Notes

As we learnt earlier, Lavosier laid the experimental foundation of modern
chemistry. But the British chemist John Dalton (1766–1844) provided the basic
theory; all matter – whether element, compound, or mixture –is composed of
small particles called atoms. The postulates, or basic assumptions of Dalton's
theory are presented below in this section.
1.4.1 Postulates of Dalton's Atomic Theory

The English scientist John Dalton was by no means the first person to propose
the existence of atoms, as we have seen in the previous section, such ideas date
back to classical times. Dalton’s major contribution was to arrange those ideas
in proper order and give evidence for the existence of atoms. He showed that
the mass relationship expressed by Lavoisier and Proust (in the form of law of
conservation of mass and law of constant proportions) could be interpreted most
suitably by postulating the existence of atoms of the various elements.
In 1803, Dalton published a new system of chemical philosophy in which the

following statements comprise the atomic theory of matter:
1. Matter consists of indivisible atoms.
2. All the atoms of a given chemical element are identical in mass and in all
other properties.
3. Different chemical elements have different kinds of atoms and in particular
such atoms have different masses.
4. Atoms are indestructible and retain their identity in chemical reactions.
5. The formation of a compound from its elements occurs through the
combination of atoms of unlike elements in small whole number ratio.
Dalton’s fourth postulate is clearly related to the law of conservation of mass.
Every atom of an element has a definite mass. Also in a chemical reaction there
is rearrangement of atoms. Therefore after the reaction, mass of the product
should remain the same. The fifth postulate is an attempt to explain the law of
definite proportions. A compound is a type of matter containing the atoms of
CHEMISTRY 7

Chemistry two or more elements in small whole number ratio. Because the atoms have
definite mass, the compound must have the elements in definite proportions by
mass.
The Dalton’s atomic theory not only explained the laws of conservations of mass
and law of constant proportions but also predicted the new ones. He deduced
the law of multiple proportions on the basis of his theory. The law states that
Notes when two elements form more than one compound, the masses of one
element in these compound for a fixed mass of the other element are in
the ratio of small whole numbers. For example, carbon and oxygen form two
compounds: Carbon monoxide and carbon dioxide. Carbon monoxide contains
1.3321 g of oxygen for each 1.000g of carbon, whereas carbon dioxide contains
2.6642 g of oxygen for 1.0000 g of carbon. In other words, carbon dioxide
contains twice the mass of oxygen as is contained in carbon monoxide (2.6642
g = 2 × 1.3321 g) for a given mass of carbon. Atomic theory explains this by
saying that carbon dioxide contains twice as many oxygen atoms for a given
number of carbon atoms as does carbon monoxide. The deduction of law of
multiple proportions from atomic theory was important in convincing chemists
of the validity of the theory.
1.4.2 What is an Atom?

As you have just seen in the previous section that an atom is the smallest particle
of an element that retains its (elements) chemical properties. An atom of one
element is different in size and mass from the atoms of the other elements. These
atoms were considered ‘indivisible’ by Indian and Greek ‘Philosophers’ in the
beginning and the name ‘atom’ was given as mentioned earlier. Today, we know
that atoms are not indivisible. They can be broken down into still smaller particles
although they lose their chemical identity in this process. But inspite of all these
developments atom still remains a building block of matter.
1.4.3 Molecules
A molecule is an aggregate of at least two atoms in a definite arrangement
held together by chemical forces (also called chemical bonds). It is smallest
particle of matter, an element or a compound, which can exist independently.
A molecule may contain atoms of the same element or atoms of two or more
elements joined in a fixed ratio, in accordance with the law of definite
proportions stated. Thus, a molecule is not necessarily a compound, which, by
definition, is made up of two or more elements. Hydrogen gas, for example,
is a pure element, but it consists of molecules made up of two H atoms each.
Water, on the other hand, is a molecular compound that contains hydrogen and
8 CHEMISTRY
oxygen in a ratio of two H atoms and one O atom. Like atoms, molecules are Chemistry
electrically neutral.
The hydrogen molecule, symbolized as H2, is called a diatomic molecule
because it contains only two atoms. Other elements that normally exist as
diatomic molecules are nitrogen (N2) and oxygen (O2), as well as the Group
17 elements-fluorine (F2), chlorine (Cl2), bromine (Br2), and iodine (I2). Of
course, a diatomic molecule can contain atoms of different elements. Examples Notes
are hydrogen chloride (HCl) and carbon monoxide (CO).
The vast majority of molecules contain more than two atoms. They can be atoms
of the same element, as in ozone (O3), which is made up of three atoms of
oxygen, or they can be combinations of two or more different elements.
Molecules containing more than two atoms are called polyatomic molecules.
Like ozone, water (H2O) and ammonia (NH3) are polyatomic molecules.
1.4.4 Elements
Substances can be either elements or compounds. An element is a substance
that cannot be separated into simpler substances by chemical means. To date,
118 elements have been positively identified. Eighty-three of them occur
naturally on Earth. The others have been created by scientists via nuclear
processes.
For convenience, chemists use symbols of one or two, letters to represent the
elements. The first letter of a symbol is always capitalized, but the following
letter is not. For example, Co is the symbol for the element cobalt, whereas CO
is the formula for the carbon monoxide molecule. Table 1.l shows the names
and symbols of some of the more common elements; a complete list of the
elements and their symbols appears inside the front cover of this book. The
symbols of some elements are derived from their Latin names for example, Au
from auram (gold), Fe from ferrurn. (iron), and Na from natrium (sodium) while
most of them come from their English names.
Table 1.1: Some Common Elements and Their Symbols
Name Symbol Name Symbol Name Symbol
Aluminium Al Fluorine F Oxygen 0

Arsenic As Gold Au Phosphorus P
Barium Ba Hydrogen H Platinum Pt
Bismuth Bi Iodine I Potassium K
Bromine Br Iron Fe Silicon Si
CHEMISTRY 9

Chemistry Calcium Ca Lead Pb Silver Ag
Carbon C Magnesium Mg Sodium Na
Chlorine Cl Manganese Mn Sulfur S
Chromium Cr Mercury Hg Tin Sn
Cobalt Co Nickel Ni Tungsten W
Notes Copper Cu Nitrogen N Zinc Zn
Chemists use chemical formulas to express the composition of molecules and

ionic compounds in terms of chemical symbols. By composition we mean not
only the elements present but also the ratios in which the atoms are combined.
INTEXT QUESTIONS 1.1

1. Chemistry plays a vital role in many areas of science and technology. What
are those areas?
2. Who proposed the particulate nature of matter?
3. What is law of conservation of mass?
4. What is an atom?
5. What is a molecule?
6. Why is the symbol of sodium Na?
7. How is an element different from a compound?
1.5 SI UNITS (REVISITED)

Measurement is needed in every walk of life. As you know that for every
measurement a ‘unit’ or a ‘reference standard’ is required. In different countries,
different systems of units gradually developed. This created difficulties whenever
people of one country had to deal with those of another country. Since scientists
had to often use each other’s data, they faced a lot of difficulties. For a practical
use, data had to be first converted into local units and then only it could be used.
In 1960, the ‘General Conference of Weights and Measures’, the international
authority on units proposed a new system which was based upon the metric system.
This system is called the ‘International System of Units’ which is abbreviated as
SI units from its French name, Le Système Internationale d’Unitès. You have
learned about SI units in your earlier classes also and know that they are based
upon seven base units corresponding to seven base physical quantities. Units
needed for various other physical quantities can be derived from these base SI
units. The seven base SI units are listed in Table 1.2.
10 CHEMISTRY
Table 1.2: SI Base Units Chemistry
Physical Quantity Name of SI Unit Symbol for SI unit
Length Metre m
Mass Kilogram kg
Time Second s
Electrical current Ampere A Notes
Temperature Kelvin K
Amount of substance Mole mol
Luminous intensity Candela cd
For measuring very large or very small quantities, multiples or sub-multiples of
these units are used. Each one of them is denoted by a symbol which is prefixed
to the symbol of the unit. For example, to measure long distances we use the unit
kilometre which is a multiple of metre, the base unit of length. Here kilo is the
prefix used for the multiple 103. Its symbol is k which is prefixed to the symbol of
metre, m . Thus the symbol of kilometre is km and
1 km = 1.0 × 103 m = 1000 m
Similarly, for measuring small lengths we use centimetre (cm) and millimetre
(mm) where
1 cm = 1.0 × 10–2 m = 0.01 m
1 mm = 1.0 × 10–3 m = 0.001 m
Some prefixes used with SI units are listed in Table 1.3.
Table 1.3: Some prefixes used with SI units
Prefix Symbol Meaning Example
Tera T 1012 1 terametre (Tm) = 1.0 ×1012 m

Giga G 109 1 gigametre (Gm) = 1.0 × 109 m
Mega M 106 1 megametre (Mm) = 1.0 × 106 m
Kilo k 103 1 kilometre (km) = 1.0 × 103 m
Hecta h 102 1 hectametre (hm) = 1.0 × 102 m
Deca da 101 1 decametre (dam) = 1.0 × 101 m
Deci d 10-1 1 decimetre (dm) = 1.0 × 10–1 m
Centi c 10–2 1 centimetre (cm) = 1.0 × 10–2 m
Milli m 10–3 1 millimetre (mm) = 1.0 × 10–3 m
Micro µ 10–6 1 micrometre (µm) = 1.0 × 10–6 m
Nano n 10–9 1 nanometre (nm) = 1 × 10–9 m
Pico p 10–12 1 picometre (pm) = 1 × 10–12m
Before proceeding further try to answer the following questions:
CHEMISTRY 11

Chemistry
1. Name the SI Unit of mass
2. What symbol will represent 1.0 × 10–6 g ?
3. Name the prefixes used for (i) 102 and (ii) 10–9
Notes (i) ..................................................................................................................
(ii) .................................................................................................................
4. What do the following symbols represent?
(i) Ms (ii) ms
(i) ..................................................................................................................
(ii) .................................................................................................................
1.6 RELATIONSHIP BETWEEN MASS AND NUMBER

OF PARTICLES
Suppose you want to purchase 500 screws. How, do you think, the shopkeeper
would give you the desired quantity? By counting the screws individually? No,
he would give the screws by weight because it will take a lot of time to count
them. If each screw weighs 0.8 g, he would weigh 400 g screws because it is the
mass of 500 screws (0.8 × 500 = 400 g). You will be surprised to note that the
Reserve Bank of India gives the desired number of coins by weight and not by
counting.This process of counting by weighing becomes more and more labour
saving as the number of items to be counted becomes large. We can carry out the
reverse process also. Suppose we take 5000 very tiny springs (used in watches)
and weigh them. If the mass of these springs is found to be 1.5 g, we can conclude
that mass of each spring is 1.5 ÷ 5000 = 3 × 10–4 g.
Thus, we see that mass and number of identical objects or particles are inter-
related. Since atoms and molecules are extremely tiny particles it is impossible to
weigh or count them individually. Therefore we need a relationship between the
mass and number of atoms and molecules (particles). Such a relationship is
provided by ‘mole concept’.
1.7 MOLE – A NUMBER UNIT

Mass of an atom or a molecule is an important property. However, while discussing
the quantitative aspects of a chemical reaction, the number of reacting atoms or
molecules is more significant than their masses.ACTIVITY 1.1
12 CHEMISTRY
It is observed experimently that iron and sulphur do not react with each other in Chemistry
a simple mass ratio. When taken in 1:1 ratio by mass (Fe:S), some sulphur is left
unreacted and when taken in 2:1 ratio by mass (Fe:S) some iron is left unreacted.
Let us now write the chemical equation of this reaction
Fe + S → FeS
From the above chemical equation, it is clear that 1 atom of iron reacts with 1 Notes
atom of sulphur to form 1 molecule of iron (II) sulphide (FeS). It means that if
we had taken equal number of atoms of iron and sulphur, both of them would
have reacted completely. Thus we may conclude that substances react in a simple
ratio by number of atoms or molecules.
From the above discussion it is clear that the number of atoms or molecules of a
substance is more relevant than their masses. In order to express their number we
need a number unit. One commonly used number unit is ‘dozen’, which, as you
know, means a collection of 12. Other number units that we use are ‘score’ (20)
and ‘gross’(144 or 12 dozens). These units are useful in dealing with small numbers
only. The atoms and molecules are so small that even in the minute sample of any
substance, their number is extremely large. For example, a tiny dust particle
contains about 1016 molecules. In chemistry such large numbers are commonly
represented by a unit known as mole. Its symbol is ‘mol’ and it is defined as.
A mole is the amount of a substance that contains as many elementary entities
(atoms, molecules or other particles) as there are atoms in exactly 0.012 kg
or 12 g of the carbon-12 isotope.
The term mole has been derived from the Latin word ‘moles’ which means
a ‘heap’ or a ‘pile’. It was first used by the famous chemist Wilhelm Ostwald
more than a hundred years ago.
Here you should remember that one mole always contains the same number of
entities, no matter what the substance is. Thus mole is a number unit for dealing
with elementary entities such as atoms, molecules, formula units, electrons etc.,
just as dozen is a number unit for dealing with bananas or oranges. In the next
section you will learn more about this number.
1.8 AVOGADRO’S CONSTANT

In the previous section we have learned that a mole of a substance is that amount
which contains as many elementary entities as there are atoms in exactly 0.012
kilogram or 12 gram of the carbon-12 isotope. This definition gives us a method
by which we can find out the amount of a substance (in moles) if we know the
number of elementary entities present in it or vice versa. Now the question arises
CHEMISTRY 13

Chemistry how many atoms are there in exactly 12 g of carbon-12. This number is determined
experimentally and its currently accepted value is 6.022045 × 1023. Thus 1 mol =
6.022045 × 1023 entities or particles, or atoms or molecules.
For all practical purposes this number is . rounded off to 6.022 × 1023.
The basic idea of such a number was first conceived by an Italian scientist
Notes Amedeo Avogadro. But, he never determined this number. It was
determinned later and is known as Avogadro’s constant in his honour.
This number was earlier known as Avogadro’s number. This number alongwith
the unit, that is, 6.022 × 1023 mol–1 is known as Avogadro constant. It is
represented by the symbol NA. Here you should be clear that mathematically a
number does not have a unit. Avogadro’s number 6.022 × 1023 will not have any
unit but Avogradro’s constant will have unit of mol–1. Thus Avogradro’s constant,
NA = 6.022 × 1023 mol–1.
Significance of Avogadro’s Constant

You know that 0.012 kg or 12 g of carbon –12 contains its one mole of carbon
atoms. A mole may be defined as the amount of a substance that contains 6.022 ×
1023 elementary entities like atoms, molecules or other particles. When we say
one mole of carbon –12, we mean 6.022 × 1023 atoms of carbon –12 whose mass
is exactly 12 g. This mass is called the molar mass of carbon-12. The molar mass
is defined as the mass ( in grams) of 1 mole of a substance. Similarly, a mole of
any substance would contain 6.022 × 1023 particles or elementary entities. The
nature of elementary entity, however,depends upon the nature of the substance as
given below :
S.No. Type of Substance Elementary Entity

1. Elements like Na, K, Cu which Atom
exist in atomic form
2. Elements like O, N, H, which Molecule
exist in molecular form (O2, N2, H2)
3. Molecular compounds like NH3, H2O, CH4 Molecule
4. Ions like Na+, Cu2+, Ag+, Cl–, O2– Ion
5. Ionic compounds like NaCl, NaNO3, K2SO4 Formula unit
Formula unit of a compound contains as many atoms or ions of different types

as is given by its chemical formula. The concept is applicable to all types of
compounds. The following examples would clarify the concept.
14 CHEMISTRY
Formula Atoms/ions present in one formula unit Chemistry
H2O Two atoms of H and one atom of O

NH3 One atom of N and three atoms of H
NaCl One Na+ ion and one Cl– ion
NaNO3 One Na+ ion and one NO3– ion
Notes
K2SO4 Two K+ ions and one SO2—
4
ion
Ba3(PO4)2 Three Ba2+ ions and two PO3—

4
ions
Now, let us take the examples of different types of substances and correlate their
amounts and the number of elementary entities in them.
1 mole C = 6.022 × 1023 C atoms
1 mole O2 = 6.022 × 1023 O2 molecules
1 mole H2O = 6.022 × 1023 H2O molecules
1 mole NaCl = 6.022 × 1023 formula units of NaCl
1 mole Ba2+ ions = 6.022 × 1023 Ba2+ ions

We may choose to take amounts other than one mole and correlate them with
number of particles present with the help of relation :
Number of elementary entities = number of moles × Avogadro’s constant
1 mole O2 = 1 × (6.022 × 1023) = 6.022 × 1023 molecules of O2
23
0.5 mole O2 = 0.5 × (6.022 × 1023) = 3.011 × 10 molecules of O2
22
0.1 mole O2 = 0.1 × (6.022 × 1023) = 6.022 ×10 molecules of O2

1. A sample of nitrogen gas consists of 4.22 × 1023 molecules of nitrogen.
How many moles of nitrogen gas are there?
2. In a metallic piece of magnesium, 8.46 × 1024 atoms are present. Calculate
the amount of magnesium in moles.
3. Calculate the number of Cl2 molecules and Cl atoms in 0.25 mol of Cl2 gas.
CHEMISTRY 15

Chemistry 1.9 MOLE, MASS AND NUMBER RELATIONSHIPS
You know that 1 mol = 6.022 × 1023 elementary entities
and Molar mass = Mass of 1 mole of substance
= Mass of 6.022 × 1023 elementary entities.
As discussed earlier the elementary entity can be an atom, a molecule, an ion or a
Notes formula unit. As far as mole – number relationship is concerned it is clear that one
mole of any substance would contain 6.022 × 1023 particles (elementary entities).
For obtaining the molar mass, i.e., mole-mass relationship we have to use atomic
mass scale.
1.9.1 Atomic Mass Unit

By inernational agreement, a unit of mass to specify the atomic and molecular
masses has been defined. This unit is called atomic mass unit and its symbol is
‘amu’. The mass of one C-12 atom, is taken as exactly 12 amu. Thus, C-12 atom
serves as the standard. The Atomic mass unit is defined as a mass exactly equal
to the 1/12th of the mass of one carbon-12 atom.
Mass of one C -12 atom
1 amu =
12
Atomic mass unit is also called unified atomic mass unit whose symbol is ‘u’.
Another name of atomic mass unit is dalton (symbol Da). The latter is mainly
used in biological sciences.
1.9.2 Relative Atomic and Molecular Masses

You are aware that atomic mass scale is a relative scale with C-12 atom (also
written as 12C) chosen as the standard. Its mass is taken as exactly 12. Relative
masses of atoms and molecules are the number of times each atom or molecules
1
is heavier than th of the mass of one C-12 atom. Often, we deal with elements
12
and compounds containing isotopes of different elements. Therefore, we prefer
to use average masses of atoms and molecules. Thus
Average mass of 1 atom of the element

Relative atomic mass =
1
th of the mass of one C -12 atom
12
Average mass of 1 molecule of the substance
and Relative molecular mass = 1
th of the mass of one C -12 atom
12
Experiments show that one O-16 atom is 1.333 times as heavy as one C-12 atom.
Thus
16 CHEMISTRY
Relative atomic mass of O-16 = 1.333 × 12 = 15.996 ~ 16.0 Chemistry
The relative atomic masses of all elements have been determined in a similar
manner. Relative molecular masses can also be determined experimentally in a
similar manner . In case we know the molecular formula of a molecule, we can
calculate its relative molecular mass by adding the relative atomic masses of all
its constituent atoms. Let us calculate the relative molecular mass of water, H2O.
Relative molecular mass of water, H2O = (2 × relative atomic mass of H) + Notes
(relative atomic mass of O)
= (2 × 1) + (16) = 2 + 16 = 18
The relative atomic and molecular masses are just numbers and dimensionless,
unit-less quantities.
1.9.3 Atomic, Molecular and Formula Masses

From the definition of atomic mass unit, we can calculate the atomic masses. Let
us again take the example of oxygen-16 whose relative atomic mass is 16. By
definition:
mass of one O -16 atom
Relative atomic mass of O-16 = 16 =
1
th the mass of one C -12 atom
12
1
Since 1 amu = th the mass of one C-12 atom
12
mass of one O -16 atom
∴ 16 =
1 amu
Mass of one O-16 atom = 16 amu
Or Atomic mass of O-16 = 16 amu.
From this example we can see that numerical value of the relative atomic mass
and atomic mass is the same. Only, the former has no unit while the latter has the
unit amu.
Molecular and formula masses can be obtained by adding the atomic or ionic
masses of all the constituent atoms or ions of the molecule or formula unit
respectively. Let us understand these calculations with the help of following
examples.
Example 1.1 : Calculate the molecular mass of ammonia, NH3.
Solution : One molecule of NH3 consists of one N atom and three H atoms.
Molecular mass of NH3 = (Atomic mass of N) + 3 (Atomic mass of H)
CHEMISTRY 17

Chemistry = [14 + (3 × 1)] amu
= 17 amu
Example 1.2 : Calculate the formula mass of sodium chloride (NaCl).
Solution : One formula unit of sodium chloride consists of one Na+ ion and one
Cl– ion.
Notes Formula mass of NaCl = (Ionic mass of Na+ ) + (Ionic mass of Cl–)
= 23 amu + 35.5 amu
= 58.5 amu.
You would have noticed in the above example that ionic mass of Na+ ion has
been taken as 23 amu which is the same as the atomic mass of Na atom. Since
loss or gain of few electrons does not change the mass significantly, therefore
atomic masses are used as ionic masses. Similarly we have taken ionic mass of
Cl– as 35.5 amu which is the same as the atomic mass of Cl–.
1.9.4 Molar Masses

We know that molar mass is the mass of 1 mol of the substance. Also, 1 mol of
any substance is the collection of its 6.022 × 1023 elementary entities. Thus
Molar mass = Mass of 6.022 × 1023 elementary entities.
(i) Molar mass of an element

You know that the relative atomic mass of carbon–12 is 12. A 12g sample of it
would contain 6.022 × 1023 atoms. Hence the molar mass of C-12 is 12 g mol–
1. For getting the molar masses of other elements we can use their relative atomic
masses.
Since the relative atomic mass of oxygen -16 is 16, a 16 g sample of it would
contain 6.022 × 1023 oxygen atoms and would constitute its one mole. Thus, the
molar mass of O–16 is 16 g mol–1. Relative atomic masses of some common
elements have been listed in Table 1.4
Table 1.4 : Relative atomic masses of some elements
(upto 1st place of decimal)
Element Relative Element Relative
Atomic Mass Atomic Mass
Hydrogen, H 1.0 Phosphorus, P 31.0
Carbon, C 12.0 Sulphur, S 32.1
Nitrogen, N 14.0 Chlorine, Cl 35.5
Oxygen, O 16.0 Potassium, K 39.1
Sodium, Na 23.0 Iron, Fe 55.9
18 CHEMISTRY
(ii) Molar mass of a molecular substance Chemistry
The elementary entity in case of a molecular substance is the molecule. Hence,
molar mass of such a substance would be the mass of its 6.022 × 1023 molecules,
which can be obtained from its relative molecular mass or by multiplying the
molar mass of each element by the number of its moles present in one mole of the
substance and then adding them.
Notes
Let us take the example of water, H2O. Its relative molecular mass is 18. Therefore,
18 g of it would contain 6.022 × 1023 molecules. Hence, its molar mass is 18 g
mol–1 . Alternately we can calculate it as :
Molar mass of water, H2O = (2 × molar mass of H) + (molar mass of O)
= (2 × 1 g mol–1) + (16 g mol–1)
= 18 g mol–1
Table 1.5 lists molecular masses and molar masses of some substances.
Table 1.5 : Molecular masses and molar masses of some substances
Element or Compound Molecular mass / amu Molar mass / (g mol–1)

O2 32.0 32.0
P4 124.0 124.0
S8 256.8 256.8
H2O 18.0 18.0
NH3 17.0 17.0
HCl 36.5 36.5
CH2Cl2 85.0 85.0
(iii) Molar masses of ionic compounds

Molar mass of an ionic compound is the mass of its 6.022 × 1023 formula units.
It can be obtained by adding the molar masses of ions present in the formula unit
of the substance. In case of NaCl it is calculated as
Molar mass of NaCl = molar mass of Na+ + molar mass of Cl–
= (23 g mol–1) + (35.5 g mol–1)
= 58.5 g mol–1
Let us take some more examples of ionic compounds and calculate their molar
masses.
CHEMISTRY 19

Chemistry Example 1.3 : Calculate the molar mass of
(i) K2SO4 (ii) Ba3(PO4)2
Solution :
(i) Molar mass of K 2 SO 4 = (2 × molar mass of K+) + (molar mass of SO2–
4
)
= (2 × molar mass of K+) +

Notes (molar mass of S + 4 × molar mass of O)
= [(2 × 39.1) + (32.1 + 4 × 16)] g mol–1]
= (78.2 + 32.1 + 64) g mol–1 = 174.3 g mol–1
(ii) Molar mass of Ba3(PO4)2 = (3 × molar mass of Ba2+) +
2 (molar mass of PO3–
4
)
= (3 × molar mass of Ba2+) +

2 (molar mass of P + 4 × molar mass of O)
= [(3 × 137.3) + 2 (31.0 + 4 × 16.0)] g mol–1
= (411.9 + 190.0) g mol–1 = 601.9 g mol–1
Now you have learned about the mole, mass and number relationships for all
types of substances. The following examples would illustrate the usefulness of
these relationships.
Example 1.4 : Find out the mass of carbon -12 that would contain 1.0 × 1019
carbon-12 atoms.
Solution : Mass of 6.022 × 1023 carbon-12 atoms = 12 g
12 × 1 × 1019
Mass of 1.0 × 1019 carbon-12 atoms = g
6.022 × 1023
= 1.99 × 10–4 g
Example 1.5 : How many molecules are present in 100 g sample of NH3?
Solution : Molar mass of NH3 = (14 + 3) g mol–1 = 17 g mol–1
∴ 17 g sample of NH3 contains 6.022 × 1023 molecules
6.022 × 1023 molecule
Therefore, 100 g sample of NH3 would contain × 100g
17g
= 35.42 × 1023 molecules
= 3.542 × 1024 molecules
20 CHEMISTRY
Example 1.6 : Molar mass of O is 16 g mol–1. What is the mass of one atom and Chemistry
one molecule of oxygen?
Solution : Mass of 1 mol or 6.022 × 1023 atoms of O = 16 g
16g
∴ Mass of 1atom of O =
6.022 × 10 23
= 2.66 × 10–23 g Notes
Since a molecule of oxygen contains two atoms (O2),

its mass = 2 × 2.66 × 10–23 g = 5.32 × 10–23 g.

1. Calculate the molar mass of hydrogen chloride, HCl.
2. Calculate the molar mass of argon atoms, given that the mass of single atom
is 6.634 × 10–26 kg.
3. Calculate the mass of 1.0 mol of potassium nitrate, KNO3 (atomic masses :
K = 39 amu; N = 14 amu, O = 16 amu).
4. The formula of sodium phosphate is Na3PO4. What is the mass of 0.146
mol of Na3PO4? (atomic masses : Na = 23.0 amu, P = 31.0 amu; O = 16.0
amu).
1.10 MASS, MOLAR MASS AND NUMBER OF MOLES

Mass, molar mass and number of moles of a substance are inter-related quantities.
We know that :
Molar mass (M) = Mass of one mole of the substance.
Molar mass of water is 18 g mol–1. If we have 18 g of water, we have 1mol of it.
Suppose we have 36 g water (18 × 2), we have 2 mol of it. In general in a sample
of water of mass (n × 18) g, the number of moles of water would be n. We may
generalize the relation as
mass of the substance
Number of moles (amount) of a substance = molar mass of the substance
m
n =
M
or m =n×M
These relations are useful in calculations involving moles of substances.
CHEMISTRY 21

Chemistry Example 1.7 : In a reaction, 0.5 mol of aluminium is required. Calculate the
amount of aluminium required in grams? (atomic mass of Al = 27 amu)
Solution : Molar mass of Al = 27 g mol–1
Required mass = no. of moles × molar mass
= (0.5 mol) × (27 g mol–1)
Notes = 13.5 g
1.11 MOLAR VOLUME, Vm

Molar volume is the volume of one mole of a substance. It depends upon
temperature and pressure. It is related to the density, by the relation.
Molar mass
Molar volume =
Density
In case of gases, we use their volumes at standard temperature and pressure

(STP). For this purpose 0 0C or 273 K temperature is taken as the standard
temperature and 1bar pressure is taken as the standard pressure. At STP, the
molar volume of an ideal gas is 22.7 litre*. You will study that gases do not
behave ideally and therefore their molar volume is not exactly 22.7 L. However,
it is very close to 22.7 L and for all practical purposes we take the molar volume
of all gases at STP as 22.7 L mol–1.

1. How many moles of Cu atoms are present in 3.05 g of copper (Relative
atomic mass of Cu = 63.5).
2. A piece of gold has a mass of 12.6 g. How many moles of gold are present
in it? (Relative atomic mass of Au = 197)
3. In a combustion reaction of an organic compound, 2.5 mol of CO2 were
produced. What volume would it occupy at STP (273K, 1bar) ?
1.12 MOLCULAR AND EMPIRICAL FORMULAE

In your previous classes, you have studied how to write chemical formula of a
sustance. For example, water is represented by H2O, carbon dioxide is represented
* Earlier 1 atmosphere pressure was taken as the standard pressure and at STP (273K, 1atm) the molar
volume of an ideal gas was taken as 22.4 L mol–1. The difference in the value is due to the change in the
standard pressure (1bar) which is slightly less than 1atm.
22 CHEMISTRY
by CO2, methane is represented by CH4, dinitrogen penta oxide is represented by Chemistry
N2O5, and so on. You are aware, formula for a molecule uses a symbol and subscript
number to indicate the number of each kind of atoms present in the molcule
(subscript 1 is always omitted). Such a formula is called molecular formula as it
represents a molecule of a substance. A molecule of water consists of two hydrogen
atoms and one oxygen atom. So its molecular formula is written as H2O. Thus a
molecular formula shows the actual number of atoms of different elements
Notes
in a molecule of a compound.
There is another kind of formula, the empirical formul of a compound, which
gives only relative number of atoms of different elements. These numbers are
expressed as the simplest ratio. For example, empirical formula of glucose, which
consists of carbon, hydrogen and oxygen in the ratio of 1:2:1 is CH2O (empirical
formulae are also called simplest formulae). Molecular formula of a substance is
always an integral multiple of its empirical formula (i.e. molecular formula = Xn
where X is empirical formula and n is an integer). For example molecular formula
of glucose is C6H12O6 which is 6 × its empirical formula. Thus, while empirical
formula gives only a ratio of atoms, the molecular formula gives the actual number
of atoms of each element in an individual molecule. In some cases the ratio of
atoms shown in a molecular formula cannot be reduced to smaller integers. In
such cases molecular and empirical formulae are the same, for example, sucrose
C12H22O11 which is popularly known as cane-sugar. In case of certain elements, a
molecule consists of several atoms for example P4, S8, etc. In such cases, empirical
formula will be symbol of the element only.
As you know, common salt, which is chemically called sodium chloride is
represented as NaCl. This salt is ionic in nature and does not exist in molecular
form. Therefore, NaCl is its empirical formula which shows that sodium and chlorine
atoms are present in NaCl in the ratio of 1:1. Similar is the case with all ionic substanes.
KCl, NaNO3, MgO are examples of empirical formulae as these are all ionic
compounds. Table 1.6 provides a few more examples.
Table 1.6: Molecular and Empirical Formulae
Substance Molecular formula Empirical formula

Ammonia NH3 NH3
Carbon dioxide CO2 CO2
Ethane C2H6 CH3
Fructose C6H12O6 CH2O
Sulphur S8 S
Benzene C6H6 CH
Sodium chloride — NaCl
Calcium oxide — CaO
CHEMISTRY 23

Chemistry 1.13 CHEMICAL COMPOSITION AND FORMULAE
How much carbon is present in one kilogram of methane whose molecular formula
is CH4? How much nitrogen is present in one kilogram of ammonia, NH3? If we
have prepared a substance that is made of 58.8% carbon, 28.4% oxygen, 8.28%
nitrogen and 6.56% hydrogen, what is its empirical formula? You have studied
atomic masses, formulae, and the mole concept. Can we solve the problem
Notes using these basic concepts? The answer is ‘yes’. Atomic masses, formulae and the
mole concept are the basic tools needed to solve such problems. What is percentage
composition? Let us take up this aspect in a little detail and try to understand.
1.13.1 Percentage Composition

If we know the formula of a compound, we can find out how much of each of the
elements is present in a given quantity of the compound. Aluminium is obtained
from its oxide. Al2O3 (which is found as the ore, bauxite). From the formula we
can calculate how much aluminium can be obtained, at least in prinicple, from a
given amount of aluminium oxide. Calculation is done by making use of the idea
of percentage composition
Percentage mass of an element in a compound
mass of element in one molecular formula or
in one empirical formula
= × 100
molecular mass or empirical formula
mass of compound
Mass of element in 1mol of compound

= × 100
Molar mass of compound
Let us calculate percentage composition of aluminium oxide, Al2O3
Mass of aluminium in 1 mol Al2 O3
Pecentage of aluminium = × 100
Molar mass of Al2 O3
Molar mass of Al2O3= (2 × 27.0) g + (3 × 16.0) g = 102.0 g
Since 1 mol of Al2O3 contains 2 mol of Al atoms, the mass of Al is 2 × 27.0 g =
54.0 g Al
54.0 g
Percentage of Aluminium = 102.0 g × 100 = 52.9 %
We can calculate percentage of oxygen in the same way. One mole of Al2O3
contains 3 mole of O atoms, that is, 3 × 16.0 g oxygen therefore
3 × 16.0 g
Percentage of oxygen = 102.0 g × 100 = 47.1%
24 CHEMISTRY
Example 1.8: Butanoic acid, has the formula C4H8O2. What is the elemental Chemistry
analysis of butanoic acid?
Solution : Molecular formula of the butanoic acid is C4H8O2.
In one mole of butanoic acid there are 4 mol of carbon atoms, 8 mol of hydrogen
atoms and 2 mol of oxygen atoms. Thus, 1 molar mass of butanoic acid will be
equal to the sum of 4 × molar mass of carbon atoms, 8 × molar mass of hydrogen
atoms, and 2 × molar mass of oxygen atoms. Notes
Molar mass of butanoic acid = 4 × 12.0 g + 8 × 1.0 g + 2 × 16.0 g = 88.0 g
48.0 g
Percentage of C by mass = 88.0 g × 100 = 54.5%
8.0 g
Percentage of H by mass= 88.0 g × 100 = 9.1%
32.0 g
Percentage of O by mass = 88.0 g × 100 = 36.4%
The percentage of O in butanoic acid can also be calculated as follows :

Percentage of O by mass = 100 – (Percentage of C by mass +
Percentage of H by mass)
= 100 – (54.5 + 9.1) = 36.4%
1.14 DETERMINATION OF EMPIRICAL FORMULAE –

FORMULA STOICHIOMETRY
We have just seen that if we know the formula of a compound we can calculate
the percentage composition. Now the question arises, can we determine the
formula of the compound if we know the percentage composition of a compound.
The answer will be ‘yes’, but this formula will not be molecular formula; instead
it would be empirical formula as it would give simplest ratio of different atoms
present in a compound. Normally we determine the percentage composition of
different elements present in an unknown compound and determine its formula.
Let us take a simple example of water. Water consists of 11.11% hydrogen and
88.89% oxygen by mass. From the data, we can determine empirical formula of
water. Now if we assume that we have a 100.00 g sample of water, then the
percentage composition tells us that 100.0 g of water contains 11.11 g of hydrogen
atoms and 88.89 g of oxygen atoms.
From the atomic mass table, we find that 1 mol of hydrogn atoms has a mass of
1.0g, and 1 mol of oxygen atoms has a mass of 16.0 g. Now we can write unit
conversion factors so that the mass of hydrogen can be converted to moles of H
atoms and the mass of oxygen can be converted to moles of O atoms.
CHEMISTRY 25

Chemistry Mass of H 11.11 g
Moles of H = =
Molar mass of H 1.0 g mol−1
Similarly,
Mass of O 88.89 g
Moles of O = = = 5.55 mol
Molar mass of O 16.0 g mol−1
Notes
Thus in water, the ratio of moles of hydrogen atoms to moles of oxygen atoms is
11.11 : 5.55.
Since a mole of one element contains the same number of atoms as a mole of
another element, the ratio of moles of atoms in a compound is also the ratio
of the number of atoms. Therefore, the ratio of hydrogen atoms to oxygen
atoms is 11.11:5.55. Now by dividing each by the smaller of the two numbers we
can convert both numbers to integers
11.11 5.55
= 2 and =1
5.55 5.55
Thus ratio hydrogen and oxygen atoms in water is 2 : 1 and empirical formula of
water is H2O.

1. For the compound Fe3O4, calculate percentage of Fe and O.
2. State percent composition for each of the following:
(a) C in SrCO3 (b) SO3 in H2SO4
3. What are the empirical formulae of substances having the following
molecular formulae?
H2O2, C6H12, Li2CO3, C2H4O2, S8, H2O, B2H6, O3, S3O9, N2O3
4. A compound is composed of atoms of only two elements, carbon and
oxygen. If the compound contain 53.1% carbon, what is its empirical
formula.
1.15 CHEMICAL EQUATION AND REACTION

STOICHIOMETRY
You have studied that a reaction can be represented in the form of a chemical
equation. A balanced chemical equation carries a wealth of information
qualitative as well as quantitative. Let us consider the following equation and
learn what all information it carries.
4Fe(s) + 3O2(g) → 2Fe2O3(s) ...(1.1)
26 CHEMISTRY
1. Qualitative Information Chemistry
Qualitatively the equation (2.1) tells that iron reacts with oxygen to form iron
oxide.
2. Quantitative Information
Quantitatively a balanced chemical equation specifies numerical relationship
among the quantities of its reactants and products. These relationships can be Notes
expressed in terms of :
(i) Microscopic quantities, namely, atoms, molecules and formula units.
(ii) Macroscopic quantities, namely, moles, masses and volumes (in case of
gaseous substances) of reactants and products.
Now let us again take the reaction (1.1) given earlier and get the quantitative
information out of it.
1.15.1 Microscopic Quantitative Information

The reaction (1.1)
4Fe(s) + 3O2(g) → 2Fe2O3(s) ...(1.2)
tells that 4 atoms of iron react with 3 molecules of oxygen to form 2 formula
units of iron oxide. Often this information is written below each reactant and
product for ready reference as shown below:
4Fe(s) + 3O2(g) → 2Fe2O3(s) ...(1.2a)
4 atoms of Fe 3 molecules of O2 2 formula units of Fe2O3
1.15.2 Macroscopic Quantitative Information

The microscopic quantitative information discussed in the previous section can
be converted into macroscopic information with the help of mole concept which
you have learnt in unit 1.
(a) Mole Relationships

We know that Avogadro number of elementary entities like atoms, molecules,
ions or formula units of a substance constitute one mole of it. Let us multiply the
number of atoms, molecules and formula masses obtained in the previous section
(Eq.2.1a) by Avogadro’s constant, NA
4 Fe(s) + 3O2(g) → 2Fe2O3(s) ..(1.3)
4 atoms of Fe 3 molecules of O2 2 formula units of Fe2O3
4 × NA atoms of Fe 3 × NA molecules of O2 2 × NA formula units of Fe2O3
4 mol of Fe 3 mol of O2 2 mol of Fe2O3
CHEMISTRY 27

Chemistry We may rewrite the above equation as
4Fe(s) + 3O2(g) → 2Fe2O3(s) ...(1.3a)
4 mol of Fe 3 mol of O2 2 mol of Fe2O3
The above equation (1.3a) gives us the mole relationship between reactants and
products. Here 4 mol of Fe react with 3 mol of O2 and produce 2 mol of Fe2O3.
(b) Mass Relationships
Notes
The mole relationships which you have learnt in the previous section, can be
converted into mass relationship by using the fact that mass of one mole of any
substance is equal to its molar mass which can be calculated from its formula
with the help of relative atomic masses of its constituent elements.
In the reaction that we are discussing, the relative atomic masses of iron and
oxygen are 55.8 and 16.0 respectively. Therefore
(i) molar mass of Fe = 55.8 g mol–1
(ii) molar mass of O2 = 2 × 16.0 = 32 g mol–1
(iii) molar mass of Fe2 O3 = (2 × 55.8 + 3 × 16.0) g mol–1
= 159.6 g mol–1
Using these molar masses we can convert the mole relationship given by equa-
tion 2.1b into mass relationship as given below :
4Fe(s) + 3O2(g) → 2Fe2O3(s)
4 mol Fe 3 mol O2 2 mol Fe2O3
(4 × 55.8) g Fe (3 × 32) g O2 (2 × 159.6) g Fe2O3
223.2 g Fe 96 g O2 319.2 g Fe2O3
Thus 223.2 g iron would react with 96 g oxygen and produce 319.2 g iron oxide,
We may rewrite the above equation as
4Fe(s) + 3O2(g) → 2Fe2O3(s) (1.3b)
223.2 g Fe 96 g O2 319.2 g Fe2O3
(c) Volume Relationships

We know that one mole of any gas occupies a volume of 22.7 L* at STP (stan-
dard temperature and pressure, 0oC and 1 bar pressure). We can use this informa-
tion to arrive at volume relationships between gaseous substances. The reaction
that we are considering involves only one gaseous substance, O2. We may rewrite
the equation (2.1b) as
4Fe(s) + 3O2(g) → 2Fe2O3(s) (2.1b)
4 mol 3 mol 2 mol.
(3 × 22.7) L at STP
68.1 L at STP
* Earlier, the standard pressure was taken as 1 atmosphere and the volume of one mole of gas at STP
was taken as 22.4 L.
28 CHEMISTRY
Thus 4 mol of iron would react with 68.1 L of oxygen at STP to produce 2 mol Chemistry
of iron oxide. (The volume relationship becomes more useful for reactions in-
volving 2 or more gaseous substances).
We can express microscopic as well macroscopic quantitative relationships in-
volved in the above reaction as shown below:
4Fe(s) + 3O2(s) → 2Fe2O3(s)
4 atoms 3 molecules 2 formula units Notes
4 mol 3 mol 2 mol
223.2 g 96 g 319.2 g
– 68.1 L at STP –
We may use even mixed relations. For example, we may say 4 mol of iron would
react with 68.1 L (at STP) of oxygen to produce 319.2 g of iron oxide.
Let us understand these relationships with two more examples.
(a) Let us work out the mole, mass and volume relationships for the reaction
involved in manufacture of ammonia by Haber’s process.
Microscopic relationship N2(g) + 3H2(g) 2NH3(g) ...(2.2)
Microscopic relationships 1 Molecule 3 Molecules 2 Molecules
(i) Moles 1 mol 3 mol 2 mol

(ii) Mass 28 g (3 × 2.0) = 6.0 g (2 × 17.0) = 34 g
(iii) Volume 1 × 22.7 L (3 × 22.7) (2 × 22.7)
= 22.7 L = 68.1 L = 45.4 L
or 1 vol 3 vol 2 vol
(b) Let us take one more reaction, the combustion reaction of butane and work
out the different types of relationships. The reaction is :
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
2 molecules 13 molecules 8 molecules 10 molecules
2 mol 13 mol 8 mol 10 mol
2 × (4 × 12 +10 × 1) g (13 × 32) g 8 × (12 + 2 × 16) g 10 × (2 ×1+16) g
116 g 416 g 352 g 180 g
2 × 22.7 = 45.4 L 13 × 22.7 = 295.1 L 8 × 22.7 = 181.6 L 10 × 22.7 = 227 L
2 vol 13 vol 8 vol 10 vol
Now let us use the mole, mass and volume relationships to make some calcula-
tions.
Example 1.9 : In the manufacture of ammonia by Haber process, nitrogen reacts
with hydrogen at high temprature and high pressure in the presence of a catalyst
and gives ammonia.
N2(g) + 3H2(g) 2NH3(g)
How much hydrogen would be needed to produce one metric ton of ammonia?
CHEMISTRY 29

Chemistry Solution : We should first find out the mass relationships for the reaction.
N2(g) + 3H2(g) 2NH3(g)
1 mol 3 mol 2 mol
1 × 28g = 28 g 3 × 2g = 6.0 g 2 × 17g = 34 g
We know that :
1 metric ton = 1000 kg = 103 kg = 106 g
Notes
From the mass relationship 34 g NH3 requires 6.0 g H2 for its manufacture.
6.0 × 106
∴ 10 g NH3 would require
6
g = 1.76 × 105g of H2.
34
Thus 1 metric ton of ammonia will be obtained by using 1.176 × 105g of Hydrogen.
Example 1.10: In a rocket motor fuelled by butane, C4H10, how many kg of O2

should be provided with each kg of butane to provide for complete combustion?
Solution : The combustion reaction of butane is
2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)
2 mol 13 mol
2 × 58 = 116 g 13 × 32 = 416 g
Thus, to completely burn 116 g butane, oxygen required is 416g.

Therefore, to completely burn 1 kg (1000 g) butane, oxygen required will be
416 × 1000
= g O2
116
= 3586 g O2
= 3.586 kg O2 ≈ 3.59 kg O2
Example 1.11: When lead sulphide; PbS and lead oxide, PbO, are heated to-
gether the products are lead metal and sulphur dioxide, SO2 ,
heat
PbS(s) + 2PbO(s) 3Pb (1) + SO2(g)
If 14.0 g of lead oxide reacts according to the above equation, how many (a)
moles of lead (b) grams of lead, (c) atoms of lead and (d) grams of sulphur
dioxide are formed?
(Atomic mass : Pb = 207.0, S = 32.1 ; O =16.0)
Solution : For each part of the question we will use the balanced equation
heat
PbS(s) + 2PbO(s) 3Pb (1) + SO2(g)
1mol 2mol 3 mol 1mol
30 CHEMISTRY
Now formula mass of PbO = (207.0 + 16.0) = 223.0 amu Chemistry
Thus, one mole of lead oxide formula units have a mass of 223.0 g. Therefore,
14.0 g PbO
14.0 g of PbO is 223.0 g mol –1 PbO = 6.28 × 10–2 mol PbO
(a) The balanced equation shows that 2 mol of PbO form 3 mol of Pb. There-
Notes
fore, 6.28 × 10-2 mol of PbO form
3 mol Pb
6.28 × 10-2 mol PbO × 2 mol PbO = 9.42 × 10-2 mol Pb
(b) The atomic mass of Pb is 207.0 ; this tells us that one mol of lead has a mass
207.0 g. Thus, 9.42 × 10-2 mol of Pb has a mass of
207.0 g Pb
9.42 × 10-2 mol Pb × = 19.5 g Pb
1mol Pb
(c) 9.42 × 10-2 mol of Pb is

9.42 × 10-2 mol of Pb × 6.022 × 1023 atoms mol–1 = 5.67 × 1022 Pb atoms
(d) The balanced equation shows that 2 mol of PbO form 1 mol of SO2.
Therefore, 6.28 × 10-2 mol of PbO formula unit forms
1mol SO 2
6.28 × 10-2 mol PbO × 2 mol PbO
= 3.14 × 10-2 mol SO2

Now the relative molecular mass of SO2 = 32.1 + 2(16.0) = 64.1
Molar mass of SO2 = 64.1 g mol–1
Therefore, 3.14 × 10-2 mol of SO2 molecules have a mass of 3.14 × 10–2 mol × 64.1
g mol–1 = 2.01 g

1. How many grams of NH3 can be made according to the reaction
N2(g) + 3H2(g) ⎯⎯→ 2NH3 (g)
from (a) 0.207 mol of N2 (b) 22.6 g of H2
CHEMISTRY 31

Chemistry 2. In reaction
C2H4(g) + 3O2(g) ⎯⎯→ 2CO2(g) + 2H2O( l )
How many (a) moles of O2 are consumed and (b) moles of H2O are formed
when 4.16 X 10-2 mol of C2H4 react?
Notes 1.16 LIMITING REAGENT

We generally find that substances which react with each other are not present in
exactly the same proportionin a reaction mixture as stated by a balanced chemi-
cal equation. For example, if 2 mol each of hydrogen and oxygen are mixed and
a spark is passed through the mixture, water is formed, according to the equation
2H2 + O2 ⎯⎯→ 2H2O
2 mol 1mol 2 mol
Here, 2 mol of hydrogen react with only 1 mol of oxygen, and 1 mol of oxygen
therefore remains unreacted. In this example hydrogen is said to be the limiting
reagent or reactant because its amount becomes zero and the reaction therefore
stops before the other reactant; that is, the oxygen is used up completly. The
amount of hydrogen present initially limits the amount of product that is formed.
Example 1.12: 3 mol of sulphur dioxide SO2 is mixed with 2 mol of oxygen O2,
and after reaction is over sulphur trioxide, SO3 is obtained.
(i) Which is the limiting reagent?
(ii) What is the maximum amount of SO3 that can be formed?
Solution : (i) We must first write the balanced equation
2SO2 + O2 → 2SO3
According to the above equation
(a) 2 mol of SO3 can be formal from 2 mol of SO2.
∴ Amount of SO that can be formed from 3 mol of SO .
3 2
2mol SO3
= (3 mol SO2) × = 3 mol SO3
2mol SO 2
(b) 2 mol of SO3 can be formed from 1 mol of O2. Therefore, the amount of
SO3 that can be formed from 2 mol of O2.
2mol SO3
= (2 mol O2) × = 4 mol SO3
1mol O 2
According to the definition, the limiting reactant is that reactant which gives the
smallest amount. In this case SO2 is the limiting reactant.
32 CHEMISTRY
(ii) The maximum amount of product that can be obtained is the amount Chemistry
formed by the limiting reagent. Thus a the maximum amount of SO3 that
can be obtained is 3 mol.
Example 1.13: 2.3 g of sodium metal is introduced into a 2L flask filled with
chlorine gas at STP (273 K, 1bar). After the reaction is over, find :
(i) What is the limiting reagent in this reaction? Notes
(ii) How many moles of sodium chloride are formed?
(iii) Which substance is left unconsumed at the end of the reaction? Find out
its mass in grams.
(iv) What percentage of the substance present in excess is converted into
sodium chloride?
(Given : Na = 23, Cl = 35.5)
Solution :
2 Na (s) + Cl2(g) 2NaCl (s)
2mol 1 mol 2 mol
or 22.7 L at STP
2.3g
(i) Moles of sodium introduced = = 0.1 mol
23g mol –1
From the above equation, it is clear that 2 mol NaCl is formed from 2 mol Na
2 × 0.1
Therefore 0.1 mol Na can produce = = 0.1 mol NaCl
2
Molar volume at STP = 22.7 L
2L
Therefore moles of chlorine in 2 L volume at STP = = 0.088 mol
22.7 Lmol –1
From equation : 1 mol Cl2 can produce 2 mol NaCl
Therefore 0.088 mol Cl2 can produce 2 × 0.088 = 0.176 mol NaCl.
Since sodium produces less amount of NaCl, it is the limiting reagent.
(ii) Sodium being the limiting reagent, as calculated in (i), the moles of NaCl
produced = 0.1 mol
(iii) From above equation, 2 mol NaCl is produced from 1 mol Cl2
1 × 0.1
Therefore 0.1 mol NaCl is produced from = 0.05 mol Cl2
2
Initial moles of Cl2 = 0.088 mol
CHEMISTRY 33

Chemistry Moles of Cl2 left unconsumed = (0.088 – 0.05) mol = 0.038 mol
Therefore, mass of Cl2 left unconsumed = 0.038 g × 71.0 g mol–1 = 2.698 g
(because molar mass of Cl2 = 2 × 35.5 = 71.0 g mol–1)
(iv) Moles of Cl2 consumed = 0.05 mol out of 0.088 mol
0.05
Notes ∴ Percent of Cl2 consumed and converted into NaCl = × 100 = 56.8 %
0.088
Example 1.14: 2.0 g mixture of MgCO3 and CaCO3 are heated till no further loss
of weight takes place. The residue weighs 1.04 g. Find the percentage compo-
sition of the mixture. (Mg = 24, Ca = 40, C = 12, O = 16)
Solution : Mixture of MgCO3 and CaCO3 taken = 2.0 g
Let the mass of MgCO3 be = x g
Therefore the mass of CaCO3 = (2.0 – x) g
The decomposition reactions are
MgCO3 (s) → MgO(s) + CO2(g) (i)
(24+12+48) g (24+16) g
84 g 40 g (Residue)
CaCO3 (s) → CaO(s) + CO2(g) (ii)
(40 + 12 + 48) g (40 + 16) g
100 g 56 g (Residue)
From the equation (i)

84 g MgCO3 leaves a residue = 40 g
40 x
x g MgCO3 will leave residue = g
84
From the equation (ii)
100 g CaCO3 leaves a residue = 56 g
56 × (2.0 – x)
(2.0 – x) g CaCO3 will leave residue = g
100
40 x 56 × (2.0 – x )
Total mass of the residue = + = 1.04 g (given)
84 100
40 × 100x + 84 × 56 × 2 – 84 × 56x = 84 × 100 × 1.04
4000x + 9408 – 4704x = 8736
9408 – 8736 = (4704 – 4000)x
672 = 704x
34 CHEMISTRY
672 Chemistry
Therefore, mass of MgCO3 in the mixture = x = = 0.96 g
704
0.96
Therefore, percentage of MgCO3 = × 100 = 48 %
2.0
and percentage of CaCO3 = 100 – 48 = 52 %

Notes
WHAT YOU HAVE LEARNT

z Chemistry plays an important role in many aspects of our life like health and
medicine, energy and environment, materials and technology, food and
agriculture.
z Matter has particulate matter.
z According to the law of conservation of mass, in any chemical reaction, the
total mass of all the reactants is equal to the total mass of all the products.
z According to the law of definite proportion, in a chemical compound, the
proportions by mass of the elements that compose it are fixed and
independent of the origin of the compound or its mode of preparation.
z According to the law of multiple proportions when two elements form two
or more compounds, the masses of one element that combine with a fixed
mass of the other element are in the ratio of small whole numbers.
z John dalton gave the atomic theory in which he proposed that it is the
smallest indivisible particle of matter. Atoms of the same element are all
identicle while atoms of different elements differ. Atoms of different elements
combine in a simple whole number ratio to form a molecule.
z An atom is the smallest particle of an element that retains its chemical
properties.
z A molecule is the smallest particle of matter which can exist independently.
z An element is a substance that cannot be separated into simipler substnaces
by chemical means.
z Mole is the amount of a substance which contains as may elementary entities
as there are atoms present in 0.012 kg or 12 g of C-12. Thus mole denotes a
number.
z The number of elementary entities present in one mole of a substance is

6.022 × 1023.
CHEMISTRY 35

Chemistry z The mass of one mole of a substance is called its molar mass It is numerically
equal to relative atomic mass or relative molecular mass expressed in grams
per mole (g mol–1) or kilogram per mole (kg mol–1).
z Molar volume is the volume occupied by one mole of a substance. One mole
of an ideal gas at standard pressure and temperature, STP (273 K and 1 bar)
occupies 22.7 litres.
Notes
z In ionic substances, molar mass is numerically equal to the formula mass of
the compound expressed in grams.
z If the molar mass of a substance is known, then the amount of a substance
present in a sample having a definite mass can be calculated. If M is the molar
mass, then, the amount of substance n, present in a sample of mass m is
m
expressed as n = .
M
z A chemical formula is used not only to represent the name of a compound
but also to indicate its composition in terms of (i) relative number of atoms
and (ii) relative number of moles of atoms.
z A molecular formula of a substance shows(i) the number of atoms of
different elements in one molecule.(ii) the number of moles of atoms of
different elements in one mole of molecule.
z An empirical formula shows only a ratio of (i) number of atoms, and (ii)
moles of atoms in a compound.
z Molecular formula is always an integral multiple of the empirical formula.
z The empirical formula of a compound can be determined from its chemical
analysis.
z In order to determine a compound’s molecular formula, molecular mass also
must be known.
z Stoichiometry is the quantitative study of the composition of chemical com-
pounds (compound or formula stoichiometry) and of the substances con-
sumed and formed in chemical reactions (reaction or equation stoichiom-
etry).
z Chemical equations specify not only the identities of substances consumed
and formed in a reaction, but also the relative quantities of these substances
in terms of (a) atoms, molecules, and formula units and (b) moles of these
entities.
z A balanced chemical equation demonstrates that all the atoms present in the
reactants are accounted for in the product; atoms are neither created nor
destroyed in a reaction.
36 CHEMISTRY
z The stoichiometric ratios among the moles of reactants shown in a balanced Chemistry
equation are useful for determining which substance is entirely consumed
and which substance(s) is (are) left over.
TERMINAL EXERCISE
Notes
1. How many atoms are present in a piece of iron that has a mass of 65.0 g/
(atomic mass; Fe = 55.9 amu).
2. A piece of phosphorus has a mass of 99.2 g. How many moles of phosphorus,
P4 are present in it? (atomic mass, P = 31.0 amu)
3. Mass of 8.46 × 1024 atoms of fluorine is 266.95 g. Calculate the atomic
mass of fluorine.
4. A sample of magnesium consists of 1.92 × 1022 Mg atoms. What is the mass
of the sample in grams? (atomic mass = 24.3 amu)
5. Calculate the molar mass in g mol–1 for each of the following:
(i) Sodium hydroxide, NaOH
(ii) Copper Sulphate CuSO4 .5H2O.
(iii) Sodium Carbonate, Na2CO3 .10H2O
6. For 150 gram sample of phosphorus trichloride ( PCl3 ), calculate each of
the following:
(i) Mass of one PCl3 molecule.
(ii) The number of moles of PCl3 and Cl in the sample.

(iii)The number of grams of Cl atoms in the sample.
(iv)The number of molecules of PCl3 in the sample.
7. Find out the mass of carbon-12, that would contain 1 × 1019 atoms.
8. How many atoms are present in 100 g sample of C-12 atom?
9. How many moles of CaCO3 would weigh 5 g?
10. If you require 1.0 × 1023 molecules of nitrogen for the reaction N2 + 3H2
→ 2NH3.
(i) What is the mass (in grams) of N2 required?
(ii) How many moles of NH3 would be formed in the above reaction from
1.0 × 1023 molecules of N2?
(iii) What volume would NH3 gas formed in (ii) occupy at STP?
CHEMISTRY 37

Chemistry 11. Write empirical formulae of the following compounds:
CO, Na2SO3, C4H10, H2O2, KC1
12. The empirical formula of glucose is CH2O which has a formula mass of 30
amu. If the molecular mass of glucose is 180 amu. Determine the molecular
formula of glucose
13. What is ratio of masses of oxygen that are combined with 1.0 gram of nitro-
Notes gen in the compound NO and N2O3?
14. A compound containing sulphur and oxygen on analysis reveals that it con-
tains 50.1% sulphur and 49.9% oxygen by mass. What is the simplest for-
mula of the compound?
15. Hydrocarbons are organic compound composed of hydrogen and carbon.
A, 0.1647 g sample of a pure hydrocarbon on burning in a combustion tube
produced 0.5694 g of CO2 and 0.0845 g of H2O. Determine the percentage
of these elements in the hydrocarbon.
16. On combustion 2.4 g of a compound of carbon, hydrogen and oxygen gave
3.52 g of CO2 and 1.44 g of H2O. The molecular mass of the compound was
found to be 60.0 amu.
(a) What are the masses of carbon, hydrogen and oxygen in 2.4 g of the
compound?
(b) What are the empirical and molecular formulae of the compound?
17. (i) What mass of oxygen is required to react completely with 24 g of CH4
in the following reaction?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(1)
(ii) How much mass of CH4 would react with 96 g of oxygen.
18. In the reaction H2 + C12 → 2HC1
How many grams of chlorine, C12 are needed to react completely with
0.245 g of hydrogen, H2, to give hydrogen chloride, HC1? How much HC1
is formed?
19. 3.65 g of H2 and 26.7 g of O2 are mixed and reacted. How many grams of
H2O are formed?
20. Caustic soda NaOH can be commercially prepared by the reaction of Na2CO3
with slaked line, Ca(OH)2. How many grams of NaOH can be obtained by
treating 2.0 kg of Na2CO3 with Ca(OH)2?
21. A portable hydrogen generator utilizes the reaction
CaH2 + H2O → Ca(OH)2 + 2H2
How many grams of H2 can be produced by a 100 g cartridge of CaH2?
38 CHEMISTRY
22. The reaction 2Al + 3MnO → Al2O3 + 3Mn proceeds till the limiting Chemistry
substance is consumed. A mixture of 220 g Al and 400 g MnO was heated to
initiate the reaction. Which initial substance remained in excess and by how
much ? ( Al = 27, Mn = 55).
23. KClO4 may be prepared by means of following series of reactions
Cl2 + 2KOH → KCl + KClO + H2O
Notes
3KClO → 2KCl + KClO3
4KClO3 → 3KClO4 + KCl
How much Cl2 is needed to prepare 400 g KClO4 by the above sequence?
(K = 39, Cl = 35.5, O = 16, H = 1)
24. 2.0 g of a mixture of Na2CO3 and NaHCO3 was heated when its weight
reduced to 1.876 g. Determine the percentage composition of the mixture.
25. Calculate the weight of 60 % sulphuric acid required to decompose 150 g of
chalk (calcium carbonate). Given Ca = 40, C = 12, O = 16, S = 32)
ANSWERS TO INTEXT QUESTIONS
1.1
1. Health, medicine, energy, food, agriculture etc.
2. Leucippus and his student Democritus
3. In every chemical reaction total masses of all the reactants is equal to the
masses of all the products.
4. An atom is extremely small particles of matter that retains its identity during
chemical reaction.
5. Molecule is an aggregate of at least two atoms in a definite arragement held
togethrer its chemical forces.
6. It is derived from the Latin name of sodium i.e. Natrium
7. An elements comprises of atoms of one type only while a compound
comprises atoms of two or more types combined in a simple but fixed ratio.
1.2
1. Kilogram
2. μg
3. (i) h (ii) n
CHEMISTRY 39

Chemistry 4. (i) Megasecond, 106 s
(ii) millisecond, 10–3 s.
1.3
4.22 × 1023 molecules
1. Moles of N2 gas = 6.022 × 1023 molecules mol –1 = 0.70 mol
Notes
8.46 × 1024 atoms
2. Amount of magnesium (moles) = 6.022 × 1023 atoms mol –1 = 14.05 mol
3. No. of Cl2 molecules in 0.25 mol Cl2 = 0.25 × 6.022 × 1023 molecules
= 1.5055 × 1023 molecules
Since each Cl2 molecule has 2 Cl atoms, the number of Cl atoms = 2 × 1.5055
× 1023 = 3.011 × 1023 atoms.
1.4
1. Molar mass of hydrogen chloride = molar mass of HCl
= 1 mol of H + 1 mol of Cl
= 1.0 g mol–1 + 35.5 g mol–1
= 36.5 g mol–1
2. Molar mass of argon atoms = mass of 1 mol of argon
= mass of 6.022 × 1023 atoms of argon.
= 6.634 × 10–26 kg × 6.022 × 1023 mol–1
= 39.95 × 10–3 kg mol–1
= 39.95 g mol–1
3. Molar mass of KNO3 = mass of 1 mol of K + mass of 1 mol of N + mass of 3
mol of O.
Since molar mass of an element is numerically equal to its atomic mass but
has the units of g mol–1 in place of amu = 39.1 g + 14.0 g + 3 × 16.0 g
∴ Molar mass of KNO3 = 39.1 g + 14.0 g + 48.0 g = 101.1 g mol–1
4. Mass of 1 mol of Na3PO4 = 3 × (mass of 1mol of Na) + mass of 1 mol
of P + 4 × (mass of 1mol of oxygen)
= 3 (23.0 g) + 31.0 g + 4(16.0) g
= 69.0 g + 31.0 g + 64.0 g = 164.0 g
∴ Mass of 0.146 mol of Na3PO4 = 0.146 × 164.0 g = 23.94 g
40 CHEMISTRY
1.5 Chemistry
3.05 g
1. Moles of Cu atoms in 3.05 g copper = 63.5 g mol –1 = 0.048 mol
12.6 g
2. Moles of gold, Au = 197 g mol –1 = 0.064 mol
Notes
3. Molar volume of any gas at STP (298 K, 1 bar) = 22.7 L
∴ Volume occupied by 2.5 mol CO2 at STP = 2.5 × 22.7 L = 56.75 L
1.6
1. Molar mass of Fe3O4 = 3 × 56.0 + 4 × 16.0
= (168.0 + 64.0) = 232.0 g mol–1
168.0
Percentage of Fe = × 100 = 72.41%
232.0
64.0
Percentage of O = × 100 = 27.59%
232.0
2. (a) Molar mass of SrCO3 = 87.6 + 12.0 + 48.0 = 147.6 g mol–1
12.0
Percentage of carbon C in SrCO3 = × 100 = 8.13%
147.6
(b) Molar mass of H2SO4 = 2.0 + 32.1 + 64.0 = 98.1 g mol–1
Molar mass of SO3 = 32.1 + 48.0 = 80.1 g mol–1
80.1× 100
Percentage of SO3 in H2SO4= = 81.65%
98.1
3. Substance Empirical formula
H2O2 HO
C6 H12 CH2
Li2CO3 Li2CO3
C2H4O2 CH2O
S8 S
H2O H2O
B2 H 6 BH3
O3 O3
S3O9 SO3
N2O3 N2O3
CHEMISTRY 41

Chemistry 4. Percentage of carbon = 53.1%
Percentage of Oxygen = 46.9%
Suppose we take 100 g of the substance then moles of carbon
53.1
= g = 4.43 mol
12.0
Notes
46.0
mole of oxygen = = 2.93 mol
16.0
4.43 2.93
molar ratio of C and O = :
2.93 2.93
= 1.50 : 1 or 3 : 2
Empirical formula of the compound is C3O2
1.7
1. In equation
N2(g) + 3H2(g) → 2NH3(g)
1 mol 3 mol 2 mol
0.207 mol of N2 gives 0.414 mol of NH3

0.414 mol of NH3 = 0.414 mol × 17.0 g mol–1 = 7.038 g of NH3
22.6
22.6 g of hydrogen = = 11.3 mol of hydrogen
2.0
2
11.3 mol of hydrogen will give × 11.3 mol of NH3 = 7.53 mol
3
Therefore, mass of NH3 = 7.53 mol × 17.0 g mol–1 = 128.01 g

2. C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)
1 mol 3 mol 2 mol 2 mol
(a) 4.16 × 10-2 mol of C2H4 will consume 3 × 4.16 × 10-2 mol of oxygen
= 12.48 × 10-2 = 1.248 × 10-1 mol of O2
(b) moles of H2O formed = 2 × 4.16 × 10-2 mol
= 8.32 × 10-2 mol of H2O
42 CHEMISTRY

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Atoms, Molecules and Chemical Arithmetic MODULE - 1

Some Basic Concepts of

ATOMS, MOLECULES AND

Chemistry is the study of matter and the changes it undergoes. Chemistry is

Some Basic Concepts of

1.1.1 Health and Medicine

11.2 Energy and the Environment

Some Basic Concepts of

Another potential source of energy is nuclear fission, but because of environmental

1.1.3 Materials and Technology

1.2 PARTICULATE NATURE OF MATTER

1.3 LAWS OF CHEMICAL COMBINATIONS

Some Basic Concepts of

1.4 DALTON’S ATOMIC THEORY Notes

1.4.1 Postulates of Dalton's Atomic Theory

In 1803, Dalton published a new system of chemical philosophy in which the

Some Basic Concepts of

1.4.2 What is an Atom?

Name Symbol Name Symbol Name Symbol

Aluminium Al Fluorine F Oxygen 0

Some Basic Concepts of

Chemists use chemical formulas to express the composition of molecules and

INTEXT QUESTIONS 1.1

1.5 SI UNITS (REVISITED)

Physical Quantity Name of SI Unit Symbol for SI unit

Tera T 1012 1 terametre (Tm) = 1.0 ×1012 m

Before proceeding further try to answer the following questions:

Some Basic Concepts of

1.6 RELATIONSHIP BETWEEN MASS AND NUMBER

1.7 MOLE – A NUMBER UNIT

1.8 AVOGADRO’S CONSTANT

Some Basic Concepts of

Significance of Avogadro’s Constant

S.No. Type of Substance Elementary Entity

Formula unit of a compound contains as many atoms or ions of different types

H2O Two atoms of H and one atom of O

Ba3(PO4)2 Three Ba2+ ions and two PO3—

1 mole C = 6.022 × 1023 C atoms

1 mole O2 = 6.022 × 1023 O2 molecules

1 mole H2O = 6.022 × 1023 H2O molecules

1 mole NaCl = 6.022 × 1023 formula units of NaCl

1 mole Ba2+ ions = 6.022 × 1023 Ba2+ ions

INTEXT QUESTIONS 1.3

Some Basic Concepts of

1.9.1 Atomic Mass Unit

1.9.2 Relative Atomic and Molecular Masses

Average mass of 1 atom of the element

1.9.3 Atomic, Molecular and Formula Masses

Some Basic Concepts of

1.9.4 Molar Masses

(i) Molar mass of an element

Element or Compound Molecular mass / amu Molar mass / (g mol–1)

(iii) Molar masses of ionic compounds

Some Basic Concepts of

= (2 × molar mass of K+) +

= (3 × molar mass of Ba2+) +

Since a molecule of oxygen contains two atoms (O2),

INTEXT QUESTIONS 1.4

1.10 MASS, MOLAR MASS AND NUMBER OF MOLES

Some Basic Concepts of