BASIC PRINCIPLES 9

The reaction is also termed neutralisation, i. e., a reaction For example, carbon dioxide can be obtained by using anyone
between an acid and a base to form a salt and water molecule. of the following methods: '
I I
(a) by heating calcium carbonate,
FeCl 3 + 3HOH Fe(OHh + 3HCl
L..::....----l (b) by heating sodium bicarbonate,
Water
(c) by burning carbon in oxygen,
Reaction of above type is termed hydrolysis. (d) by reacting calcium carbonate with hydrochloric acid.
Whatever sample of carbon dioxide is taken, it is observed
1.8 LAWS OF CHEMICAL COMBINATION that carbon and oxygen are always combined in the ratio of
12 : 32 or 3 : 8.
In order to understand the composition of various compounds, it The converse of this law that when same elements combine in
is necessary to have a theory which accounts for both qualitative the same proportion, the same compound will be formed, is not
and quantitative observations during chemical changes. always true. For example, carbon, hydrogen and oxygen when
Observations of chemical reactions were most significant in the combine in the ratio of 12 : 3 : 8 may form either ethyl alcohol
development of a satisfactory theory of the nature of matter. (C2 H 5 0H) or dimethyl ether (CH 3 0CH 3 ) under different
These observations of chemical reactions are summarised in experimental conditions.
certain statements known as laws of chemical combination.
(iii) Law of multiple proportions: This law was put
(i) Law of conservation of mass: The law was fIrst stated forward by Dalton in 1808. According to this law, if two
by Lavoisier in 1774. It is also known as the law of elements combine to form more than one compound, then the
indestructibility of matter. According to this law, in all chemical different masses of one element which combine with a fIXed
changes, the total mass Qf a system remains constant or in a mass of the other element, bear a simple ratio to one another.
chemical change, mass is neither created nor destroyed. This ..Hydrogen and oxygen combine to form two compounds H 2 0
law ~as tested by Landolt. All chemical reactions follow this (water) and H 2 0 2 (hydrogen peroxide).
law. Thus, this law is the basis of all quantitative work in
In water, Hydrogen 2 parts Oxygen 16 parts
chemistry.
In hydrogen peroxide, Hydrogen 2 parts Oxygen 32 parts
Example: 1.70 g ofsilver nitrate dissolved in 100 g ofwater
is taken. 0.585 gofsodium chloride dissolved in 100g ofwater is The masses of oxygen which combine with same mass of
hydrogen in these two compounds bear a simple ratio 1 : 2.
added to it and chemical reaction occurs. 1.435 g of silver
chloride and 0.85 g of sodium nitrate are formed. Nitrogen forms fIve stable oxides.
Np Nitrogen 28 parts Oxygen 16 parts
Solution: Total masses before chemical change
Nitrogen 28 parts Oxygen 32 parts
Mass of AgN0 3 + Mass ofNaCI + Mass of water
Nitrogen 28 parts Oxygen 48 parts
=1.70g + 0.585g + 200.0g Nitrogen 28 parts Oxygen 64 parts
202.285 g N20 s Nitrogen 28 parts Oxygen 80 parts

Total masses after the chemical reaction, The masses of oxygen which combine with same mass of
nitrogen in the five compounds bear a ratio 16 : 32 : 48 : 64 : 80
= Mass of AgCl + Mass of NaN0 3 + Mass of water
or I : 2 : 3 : 4 : 5.
1.435 g + 0.85 g_+ 200.0 g (iv) Law of reciprocal proportions: This law was given
= 202.285g· by Richter in 1794. The law states that when definite mass of all
element A combines with two other elements Band C to form
Thus, in this chemical change,
two compounds and if Band C also combine to form a
Total masses of reactants = 'total masses of products compound, their combining masses are in same proportion or
bear a simple ratio to the masses of Band C which combine
This relationship holds good when reactants are completely
with a constant mass of A.
converted into products.

~(~)-----~-----
In case, the reacting materials are not completely consumed,
the relationship will be
Total masses of reactants = Total masses of products

r~ fR
+ Masses of unreacted reactants
(ii) Law of definite or constant proportions: This law
was presented by Proust in 1799 and may be stated as follows:
A chemical compound always contains the same element
combined together in fixed proportion by mass, i.e., a NaCI
chemical compound has a fIXed composition and it does not
For example, hydrogen combines with sodium and chlorine to
depend on the method of its preparation or the source from
form compounds NaH and HCI respectively.
which it has been obtained.
10 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

In NaH, Sodium 23 parts Hydrogen one part . Example 4. In an experiment, 2.4 g of iron oxide on
In HCI, Chlorine 35.5 parts Hydrogen one part reduction with hydrogen yield 1.68 g of iron. In another
Sodium and chlorine also combine to fonn NaCI in which 23
experiment, 2.9 gO/iron oxide give 2.03 g of iron on reduction
with hydrogen. Show that the above data illustrate the law of
parts of sodium and 35.5 parts of chlorine are present. These are
constant proportions.
the same parts which combine with one part of hydrogen in NaH
and HCl respectively. Solution:
In the first experiment

/
(~)
~. ~o
. The~ssoflrorioxide = 2.4 g
The mass ofJron after reduction 1.68g

S(B)~~O(c)
The mass of oxygen = Mass of iron oxide - Mass of iron
(2.4 - 1.68) O.72g
Ratio of oxygen and iron = 0.72: 1.68 1: 2.33
. 802 .
• In the second experiment
Hydrogen combines with sulphur and oxygen, to fonn The mass of iron oxide = 2.9 g
compounds H 2 S and H 20 respectively. The mass of iron afteueduction 2.03 g
In H2 S, Hydrogen 2 parts SUlphur 32 parts The triass of oxygen (2.9.;.. 2.03) ~ 0.87 g
In H 20, . Hydfogen 2 parts Oxygen 16 parts
Ratio of oxygen and iron = 0.87: 2.03 =1: 2.33
Thus, according to this law, sulphur should combine with Thus, the data illustrate the law of constant proportions, as in
oxygen in the ratio of 32 : 16 or a simple multiple of it. Actually, both the experiments the ratio of oxygen and iron is the same.
both combine to fonn S02 in the ratio of 32 : 32 or 1 : 1.
The law of reciprocal proportions is a special case of a more EXample S. Carbon combines with hydrogen to form three
general law, the law of equivalent masses, which can be stated compounds A, B and C. The percentages ofhydrogen in A, Band
as under: Care 25, 14.3 and 7.7 respectively. Which law of chemical
"In all chemical reactions, substances always react in the . combination is illustrated?
ratio oftheir equivalent masses." Solution:
(v) Law of gaseous volumes: This law was enunciated by
Gay-Lussac in 1808. According to this law, gases react with Compound % of Hydrogen % of Carbon
each other in the simple ratio of their volumes and if the product
is also in gaseous state, the volume of the product also bears a A 25.0 (100 - 25.0) = 75.0
simple ratio with the volumes of gaseous reactants when all B 14.3 (100 -14.3) = 85.7
volumes are measured under similar conditions of temperature
and pressure. . C 7.7 (100 - 7.7) = 92.3

ratio 1: 1:2 In Compound A

H2 + Cl z = 2HCI
I vol I vol Zvol 25 parts of hydrogen combine with 75 parts of carbon
2H z + O2 = 2H zO ratio 2: 1: 2 I part of hydrogen combines with 75/25
2vol I vol 2vol = 3 parts of carbon
2CO+ Oz 2002 ratio 2: 1 : 2 In Compound B
2 vol I vol 2 vol
14.3 parts of hydrogen combine with 85.7 parts of carbon
N2 + 3H2 = 2NH3 ratio 1:3:2 1 part of hydrogen combines with 85.7/14.3
I vol 3vol 2 vol
= 6.0parts of carbon
In Compound C
: :::::I_SOME SOLVED EXAMPLES\I:::::: 7.7 parts of hydrogen combine with 92.3 parts of carbon
Ex~mple 3. What mass of sodium chloride would be 1 part of hydrogen combines with 92.3/7.7
decomposed by 9.8 g of sulphuric ~id, if 12 g of sodium = 12.0 parts of carbon
bisulphate and 2.75 g of hydrogen chloride were produced in a Th1,lS, the masses of carbon in three compounds A, B and C,
reaction assuming that the law ofconservation of mass, is true? which combine with a fixed mass of hydrogen are in the ratio of
Solution: NaCI + H 2S0 4 = NaHS0 4 + HCl 3 : 6 : 12 or 1 : 2 : 4. This is a simple ratio. Hence, the data
According to law of conservation of mass, illustrate the law of multiple proportions:
Total masses of reactants = Total masses of products Example 6. Two compounds each containing only tin and
oxygen had the following composition:
Let the mass ofNaCI decomposed be x g, so Mass % of tin Mass % of oxygen
x + 9.8 12.0+ 2.75 Compound A 78.77 21.23
= 14.75 Compound B 88.12 11.88
x= 4.95g Show how this data illustrate the law ofmultiple proportions?
 BASIC PRINCIPLES 11

Solution: . Example 9. Carbon monoxide reacts oyith oxygen to form

In Compound A carbon dioxide according to the equation, 2CO + O 2 = 2COz.
21.23 parts of oxygen combine with 78.77parts of tin In an experiment, 400 mL of carbon monoxide and 180 mL of
oxygen were allowed to react, when 80% ofcarbon monoxide was
I part of oxygen combines with 78.77/21.23
transformed to carbon dioxide.
= 3.7 parts of tin All the volumes were measured under the same conditions of
In Compound B temperature and pressure. Find out ·the composition of the final .
11.88 parts of oxygen combine with 88.12 parts of tin mixture.
1 part of oxygen combines with 88.12111.88 Solution: 2CO + 02 = 2C0 2
2 vol I vol 2 vol
= 7.4 parts of tin
Thus, the mass of tin in compounds A and B which From the above equation, it is observed that volume of oxygen
combine with a fixed mass of oxygen are in the ratio 00.7: 7.4 required for the transformation of carbon monoxi& into carbon
or 1 : 2. This is a simple ratio. Hence, the data illustrate the law of dioxide is half the volume of carbon monoxide and the volume of
multiple proportions. carbon dioxide produced is same as that of carbon monoxide.
. Example 7. Illustrate the law of reciprocal proportions Volume of carbon monoxide transformed
from the following data: KCI contains 52.0% potassium, KJ = 80x 400 320mL
contains 23.(iO/o potassium and ICI contains 78.2% iodine. 100
Solution: In KCI: Potassium 52.0%, Hence, volume of oxygen required for transformation
Chlorine (100- 52)= 48% = ~ x 320= 160mL
2
In KI: Potassium 23.6%;
Volume of carbon dioxide produced
Iodine (l00 - 23.6) = 76.4%
= 320mL
23.6 parts of potassium combine with 76.4 ~arts of iodine
52.0 parts of potassium will combine with So, the composition of fmal mixture is
(76.4/23.6) x 52.0 168.3 parts of iodine. Carbon monoxide = (400 - 320)
The ratio of masses of chlorine and iodine which combines =80mL
with same mass ofpotassium = 48: 168.3 or I: 3.5 Carbon dioxide = 320 mL
In ICI: Iodine 78.2% and chlorine Oxygen = 180 - 160 = 20 mL
= (100-78.2) 21.8% Example 10. How much volume ofoxygen w(/l be required
for complete combustion of40mL of acetylene (C 2 H 2 ) and how
The ratio of chlorine and iodine in ICI = 21.8: 78.2 ::::: I: 3.5. much volume of carbon dioxide will be formed? All volumes are
Hence, the data illustrate the law of reciprocal proportions. measured at NTP.
Example 8. Zinc sulphate crystals contain 22.6% of zinc
and 43.9% of water. Assuming the law of constant proportions to
Solution: 2C 2 H2 + 5°2 - 4C0 2 +2H 2O
2 vol 5 vol 4 vol
be true, how much zinc should be used to produce 13.7 g ofzinc 40 mL
5 4
x40 mL x40 mL
sulphate and how much water will they contain? 2 2
40 mL 100 mL 80 mL
Solution: 100 g of zinc sulphate crystals are obtained from
So, for complete combustion of 40 mL of acetylene, 100 mL
= 22.6g zinc
of oxygen are required and 80 mL of carbon dioxide is formed.
I g of zinc sulphate crystals will be obtained from
=: 22.6/100g zinc 1'.91 DALTON'S ATOMIC THEORY
13.7 g of zinc sulphate crystals will be obtained from The concept that matter is composed of very small particles was
= 22.6 x 13.7 given by Indian and Greek philosophers. As early as 400 to 500
100 B.C. the Greek philosopher Democritus suggested that matter
= 3.0962gof zinc cannot be forever divided into smaller and smaller parts. The
ultimate particles were considered as indivisible. These particles
100 g of zinc sulphate crystals contain water were called atoms. The word atom has been derived from the
43.9g Greek word 'atomos' meaning 'indivisible'. These early ideas,
1g of zinc sulphate crystals contain water however, were not based on experiments but were mere
:::: 43.9f100g speculations. The existence of at'Oms was accepted by Boyle in
his bo'Ok 'The Sceptical Chymist' (1661) and by Newton in his
13.7 g of zinc sulphate crystals shall contain water b'Ooks ' Principia' and 'Opticks' (1704). The 'Old ideas were put 'On
43.9 3 a scientific scale by John Dalton in the years 1803 to 1808 in the
::::: 100 xl .7= 6.0143g
f'Orm of a theory known as Dalton's Atomic Theory which is a
12 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

landmark in the history of chemistry. The main points of Dalton's Some elements exist in more complex molecular forms. The
atomic theory are: molecule of phosphorus consists of four phosphorus atoms and
(i) Elements consist of minute, indivisible, indestructible the molecule of sulphur consists of eight sulphur atoms. Such
particles called atoms. molecules having more than two atoms are said to be polyatomic.
(ii) Atoms of an element are identical to each other. They A representation of the molecule of an element involves use of a
have the same mass and size. subscript to the right of the elemental symbol. The diatomic
(iii) Atoms of different elements differ in properties and have molecule of chlorine is represented as C1 2 , whereas molecules of
different masses and sizes. phosphorus and sulphur are represented as P4 and S8'
(iv) Compounds are formed when atoms of different elements respectively.
combine with each other in simple numerical ratios such The molecule is the smallest possible unit of a compound
as one-to-one, one-to-two, two-to-three and so on. which shows the properties of the compound. The molecules of
(v) Atoms cannot be created, destroyed or transformed into all compounds contain two or more different types of atoms.
atoms of other elements. These differ from the molecules of elements which contain only
(vi) The relative numbers and kind of atoms are always the one type of atoms.
same in a given compound. Thus, it becomes clear that atoms are the components of
The theory convincingly explained the various laws molecules and the molecules are components of elements or
of chemical combination, but the theory has undergone a compounds.
complete shake up with the modem concept of structure of atom. The formula is a group of symbols of elements which
However, the Daltonian atom still retains its significance as the represents one molecule of a substance. The formula of a
unit participating in chemical reactions. The following are the substance represents its chemical composition. Water consists of
modified views regarding Dalton's atomic theory: molecules containing two hydrogen atoms and one oxygen atom
(i) The atom is no longer supposed to be indivisible. The which are represented as H 2 0. The subscript to the right of the
atom is not a simple particle but a complex one. symbol for hydrogen indicates the number of hydrogen atoms
(ii) Atoms of the element may not necessarily possess the contained in a molecule. No subscript follows the symbol for
same mass but possess the same atomic number and oxygen which means, by convention, that only one atom of
show similar chemical properties (Discovery of oxygen is contained in the molecule.
isotopes). The subscripts representing the number of atoms contained in
(iii) Atoms of the different elements may possess the same a molecule of a compound are in no way related to the number of
mass but they always have different atomic numbers and atoms present in the molecule of a free element. Although both
differ in chemical properties (Discovery of isobars). hydrogen and oxygen are composed of diatomic molecules, a
(iv) Atoms of one element can be transmuted into atoms of water molecule contains only one atom of oxygen and two atoms
other element. (Discovery of artificial transmutation). of hydrogen. The two hydrogen atoms present in H2 0 are not
molecular hydrogen but rather two hydrogen atoms that have
(v) In certain organic compounds, like proteins, starch,
chemically combined with an oxygen atom.
cellulose, etc., the ratio in which atoms of different
elements combine cannot be regarded as simple. There For a chemical formula to be correct, it must contain two
are a number of compounds which do not follow the law pieces of information: (i) it must indicate the elements in the
of constant proRortions. Such compounds are called make up of the compound, and (ii) it must indicate the combining
non-stoichiometric compounds. ratio of atoms of these elements in the particular compound. The
first information is provided by including in the formula correct
chemical symbols for all the elements in the compound. The
1.~Oi ATOMS, MOLECULES AND FORMULAE second piece of information is provided by subscripts, i.e.,
An atom is the smallest particle of an element. The atom of numbers written to the right slightly below the chemical symbols
hydrogen is the smallest and the lightest. Atoms take part in of the elements.
chemical combination and remain as indivisible. All atoms do not Nitric acid is a combination of hydrogen, nitrogen and oxygen
occur free in nature. Avogadro introduced the idea of another giving a base formula RNO. These elements combine in the ratio
kind of particles called the molecules. A molecule is the smallest I : 1 : 3. Therefore, the correct formula for nitric acid is RN0 3 •
particle of an element or compound that can have a stable Some compounds are composed of ions rather than of
and independent existence. A molecule of an element consists molecules. Ions differ from atoms and molecules by being
of one or more atoms of the same element. Certain elements are electrically charged particles of matter. The charges may
capable of existence as single atoms and their atoms can be be positive or negative I and generally vary in magnitude.
regarded as molecules. A molecule of an element that consists of The positively charged idns are called cations and negatively
one atom only is called monoatomic molecule as in the case of charged ions are called: anions. Simple cations· and anions
inert gases. Oxygen is not stable in atomic form but is stable in come into existence by IQss and acceptance of an electron or
molecular form. A molecule of oxygen is diatomic in nature, i. e., electrons by neutral atoms respectively. Ions that consist of
its molecule consists of two oxygen atoms. Hydrogen, nitrogen, several atoms held together by chemical bonds similar to those
fluorine, chlorine, bromine, iodine are also diatomic like oxygen. involved in the molecules are called polyatomic ions or complex
 BASIC PRINCIPLES 13

ions. These complex ions differ from molecules in the sense that Common Name Chemical Name Chemical Formula
they bear a charge. Some of the common complex ions are:
Indian nitre Potassium nitrate KN0 3
N03" Nitrate PO~- Phosphate NH 4+ Ammonium
Limestone Calcium carbonate CaC0 3
so~- Sulphate CIO; Perchlorate PH4+ Phosphonium Lunar caustic Silver nitrate AgN03

SO~- Sulphite CO;- Carbonate MnO; Permanganate

Laughing gas Nitrous oxide Np
Litharge Lead monoxide PbO
When ions are present in a compound, the number of positive
Muratic acid Hydrochloric acid HCI
charges on a cation must balance with the negative charges on an
anion to produce electrically neutral matter. Since, the charge on Mohr's salt Ferrous ammonium FeS04(NH4)2S04
the anion may not always be equal to that on the cation, the sulphate ·6Hp
number of anions will not always be equal to the number of Milk of magnesia Magnesium hydroxide Mg(OH)2
cations. Microcosmic salt Sodium ammonium Na(NH 4 )HP04
Calcium nitrate consists of calcium and nitrate ions, Each hydrogen ortho-
calcium ion carries 2 units positive charge while each nitrate ion phosphate
carries I unit negative charge. Thus, to make net charge zero, two Marsh gas (Damp fire)Methane CH 4
nitrate ions will link with one calcium ion and the formula will be
Ca(N03 h, [Ca 2+ + 2NOi ]. Names and formulae of some Oleum Sulphuric acid H2~07
common chemical compounds are listed below: (Fuming)
Oxone Sodium peroxide Na 202
Common Name Chemical Name Chemical Formula
Plaster of Paris Calcium sulphate I
CaS0 4 ,-H2O
Alum Ammonium (NH4)2 S0 4' Al.2( S04)3 hemihydrate 2
aluminium sulphate '24H2O
Philosphers's wool Zinc oxide ZnO
Aspirin Acetyl salicylic acid C9H g04
Phosgene Carbonyl chloride COCl 2
Battery acid or Sulphuric acid H2SO 4
oil of vitriol Pearl ash Potassium carbonate K 2C0 3

Blue vitriol Copper sulphate CuS0 4,5HP Pyrene Carbon tetrachloride CCl 4

Baking soda Sodium bicarbonate NaHC0 3 Picric acid 2,4,6-Trinitrophenol C6H 2(OH)(N°2)3

Bleaching powder Calcium CaOCl 2 Quick lime Calcium oxide CaO

chlorohypochlorite Red lead (Minium) Lead tetroxide PbP4
BoraX Sodium tetraborate Na2BP7,I OH20 Sugar Sucrose C12H22011
Butter of tin Stannic chloride SnCI 4·5H2O Slaked lime (Milk of Calcium hydroxide Ca(OH)2
Caustic soda Sodium hydroxide NaOH lime)

Caustic potash Potassium hydroxide KOH Sid ammoniac Ammonium chloride NH4CI
Sugar oflead Lead acetate (CH 3COO)2Pb
Carbolic acid Phenol CJIsOH
Chile saltpetre Sonium nitrate NaN0 3 Sand Silicon dioxide Si02

Carborundum Silicon carbide SiC Table salt (Common Sodium chloride NaCI·
salt)
Corrosive sublimate Mercuric chloride HgCl2
TEL Tetra-ethyl lead Pb(C 2H s)4
Calomel Mercurous chloride Hg2Cl 2
Tear gas Chloropicrin CCl 3N02
Dry ice Carbon dioxide (solid) CO 2
Washing soda Sodium carbonate Na2C03,IOH20
Formalin Fonnaldehyde HCHO
(40% solution) Water glass Sodium silicate Na2Si03

Grain alcohol (Spirit) Ethyl alcohol . C2H sOH White vitriol Zinc sulphate ZnS°4 ,7H 2O

Green vitriol Ferrous sulphate FeS04,7H20

1~1~i! ATOMIC AND MOLECULAR MASS
Gypsum Calcium sulphate CaS04 ·2H20
One of the most important concepts derived from Dalton's atomic
Gammexane (BHC) Benzene hexachloride C6~C4;
theory is that of atomic mass, i. e" each element has a
llydrolith Calcium hydride CaH2 characteristic atomic mass. As atoms are very tiny particles, their
Hypo (Antichlor) Sodium thiosulphate Na2~03·5Hp
absolute masses are difficult to measure. However, it is possible
to determine the relative masses of different atoms if a small unit
14 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

of mass is taken as a standard. For this purpose, mass of one atom Thus,
of hydrogen was assumed as unity and was accepted as standard. 23
Tbe atomic mass of an element can be defined as tbe number lamu' 1.9924 x 10- = 1.66 X 10- 24 g or 1.66 x 10- 27 kg
wbich indicates how many times tbe mass of one atom of the 12 .
element is heavier in comparison to the mass of one atom: of
hydrogen.
A =Atomic mass of an element
Mass of one atom of the element
A == Atomic mass of an element =----------------------
lamu
Mass of one atom of the element
The atomic masses of some elements on the basis of carbon-l 2
Mass of one atom of hydrogen are given below:
In 1858, oxygen atom was adopted as a standard on account of Hydrogen 1.008 amu Iron 55.847 amu
the following reasons: Oxygen 16.00 amu Sodiuml 22.989 amu
(i) It is much easier to obtain compounds of elements with
Chlorine 35.453 amu Zinc 65.38 amu
oxygen than with hydrogen as oxygen is more reactive than
hydrogen. Magnesium 24.305 amu Silver 107.868 amu
(ii) The atomic masses of most of the elements. become Copper 63.546 amu
approximately whole numbers but with hydrogen as standard the
atomic masses of most of the elements are fractional. The actual mass of an atom of an element
The mass of one atom of natural oxygen was taken to be 16.0. The atomic mass of an element in amu x 1.66 x 10-24 g
Thus, atomic mass of an ~lement So, the actual mass of hydrogen atom
1.008 x 1.66 x 10-24 == 1.6736 X 10-24 g
element
Similarly, the actual mass of oxygen atom
16 th part of the mass of one atom of oxygen
16 x 1.66 x 10-24 = 2.656 X 10-23 g
Mass of one atom of the element x 16 It is clear from the above list of atomic masses that atomic
Mass of one atom of oxygen masses of a number of elements are not nearly whole numbers.
Actually, the above values are average relative masses. Most of
By accepting oxygen as a standard, the atomic mass of the elements occur in nature as a mixture of isotopes.
hydr6gen comes as 1.008, sodium 22.991 and sulphur 32.. 066. (Isotopes-the atoms. of the same element having different
In 1961, the International Union of Chemists selected a new atomic masses). With very few exceptions, however, elements
unit for expressing the atomic masses. They accepted the stable have constant mixtures of isotopes. Chlorine is found in nature as
isotope of carbon (12 C) with mass number of 12 as the standard. a mixture containing two isotopes CI-35 (34.969 amu) and CI-37
(36.966amu).These are found in the ratio of75.53% (CI-35) and
Atomic mass of an element can be defined as the number
24.47% (CI-37). Therefore, the average relative mass of chlorine
which indicates bow many times the mass of one atom of the
is calculated as:
element is heavier in comparison to 112 th part of the mass of
(34.969 x 0.7553)+ (3'6.966 x 0.2447)= 35.46 amu
one atom of carbon-ll (12 C).
Based on the average mass, the atomic mass of chlorine is
A == Atomic mass of an element 35.46 or 35.5 amu but it is never possible to have-an atom having
a relative mass 35.5 amu. It can have relative mass of about 35.0
Mass of one atom of the element or 37.0 amu depending on the particular isotope. Thus, average
~ th part of the mass of one atom of carbon -12 relative mass of any naturally occurring sample of chlorine is
35.46 or 35.5 amu as it is a mixture of two isotopes present in
= Mass of one atom of the element x 12 definite proportion. The same reasoning applies to all other
Mass of 'One atom of carbon -12 elements.
[The quantity' A' was formerly known as atomic weight. The average atomic masses of various elements are
However, this term is no longer used as the word 'weight' means det~ed by multiplying the atomic mass of each is.otope by its
gravitational force.] fractional abundance and adding the values thus obtained. The
Atomic mass unit: The quantity .
~
12
mass of an atom of fractional abundance is determined by· dividing percentage
abundance by hundred.
carbon-12 (12 C) is known as the atotinc~aSS unit and is
~ " m·X a+nxb
abbreviated as amu. The actual mass of one atom of carbon-l 2 is Average ISOtOPIC mass = - - - - - -
. m+n
1.9924 x 10-23 g or 1.99.24 x 10-26 kg.
here, a, b are atomic masses of isotopes in the ratio m : n.
*The term Dalton is used for one atomic mass unit, 1 Dalton = 1 amu.
 11. BASIC PRINCIPLES 15

..
Average IsotOPIC mass =-x X a+ X
b the mass of a molecule of a substance relative to the mass of an
100 100 atom of hydrogen as 1.008 or of oxygen taken as '16.00 or the
mass of one atom of carbon taken as 12. Molecular mass is a
here, x, y are percentage abundance of the two isotopes
number which indicates how many times one molecule of a
(y= 100-x).
substance is heavier in comparison to l~th of the mass of
Example 11. Boron has two isotopes boron-1O and boron-II
whose percentage abundances are 19.6% and 80.4% respectively. oxygen atom or l~ th of the mass of one atom of carbon-12.
What is the average atomic mass of boron?
M = Molecular mass
Solution: Mass of one molecule of the substance .
Contribution of boron-1O 10.0 X 0.196 = 1.96amu
12 th mass of one atom of carbon -12
Contribution of boron-II =ll.Ox 0.804 = 8.844amu
The mass of a molecule is equal to sum of the masses of the
Adding both =1.96+ 8.844 10.804amu atoms present in a molecule. One molecule of water consists of2
Thus, the average atomic mass of boron is 10.804 amu. atoms of hYQrogen and one atom of oxygen. Thus, molecular
. mass of water (2 x 1.008) + 16.06 18.016amu. One molecule
Example 12. Carbon occurs in nature as a mixture of of H 2 S04 (sulphuric acid) consists of2 atoms of hydrogen, one
carbon-12 and carbon-13. The average atomic mass ofcarbon is atom of sulphur and four atoms of oxygen. Thus, the molecular
12.011. What is the percentage abundance of carbon-12 in mass of sulphUric acid is
nature?
Solution: Let x be the percentage abundance of carbon-12;
=(2x 1.008)+ 32.00+ (4 x 16.00)
then (100 - x) will be the percentage abundance of carbon-I 3 . = 98.0160r 98.016amu

Therefore. 12x + 13(100- x) = 12.011 . Gram-molecular Mass or Gram Molecule

100 100' A quantity. of substance whose . mass. in grams is
or 12x+ 1300-13x= 1201.1 numerically equal to its molecular mass' is called gram-
or x= 98.9 molecular mass. In other words; molecular mass of a substance
expressed in grams is called gram-molecular mass ot gram
Abundance of carbon-12 is 98.9%. molecule. For exampl~, the molecular mass of chlorine is 71 and,
Gram-atomic Mass Or Gram Atum therefore, its gram-molecular mass or gram molecule is 71 g.
When numerical value of atomic mass of an element is Similarly, molecular mass of oxygen (02 ) is 32, i. e.,
expressed in grams, the value becomes gram-atomic mass or 2x 16= 32amu.
gram atom. The atomic mass of oxygen is 16 while gram-atomic Gram-molecular mass of oxygen 32 g
mass or gram atom of oxygen is 16 g. Similarly, the gram-atomic
masses of hydrogen, chlorine and nitrogen are 1.008 g, 35.5 g Molecular mass of nitric acid (HN0 3 ) is 63, i. e. ,
and 14.0 g respectively; Gram~atomic.mass orgram.atomof = 1+ 14 + 3 x 16 = 63 amu
every element consists of same number of atoms. This number Gram-molecular mass of nitric acid = 63 g
is called Avogadro's number. The value of Avogadro's number
is 6.02 x 1023 . . Gram-molecular mass should not be confused with the mass
Absolute mass of one oxygen atom of one molecule of the substance in grams. The mass of one
molecule of a substance is known as its actual mass. For
= 16amu =16x 1.66 x 10- g
24
example, the actual mass of one molecule of oxygen is equal to
Therefore, the mass of 6.02 x 10 23 atoms of oxygen will be . 32x 1.66 x 10-24 g, i~e., 5.32 x 10-23 g.
= 16x 1.66 x 10-24 x 6.02 X 1023 The number of gram molecules of a substance present in a
16g (gram-:atomic mass) given mass of a substance can be determined by the application
of following formula:
Thus, gram-atomic mass can be defined as the absolute
mass in grams of 6.02 x 10 23 atoms of any element. No. of gram molecules
Mass of a substance in grams
Number of gram atoms of any element can be calculated with
the help of the following formula: Molecular mass of the substance in grams
Mass of the element in grams . Molar mass in grams
No. of gram atoms Mass of smgle molecule - 23
Atomic mass of the element in grams 6.023 x 10
Molecular Mass = Molar mass in amu x 1.66x 10- 24 grams
Like an atom, a molecule of ~ substance is also a very small 'Example 13. Calculate the mass of 2.5 gram atoms of
particle possessing a mass of the order OflO-24 to 10-22 g. Similar oxygen.
to atomic mass, molecular mass is also expressed as a relative
Solution: We know that,
mass with respect to the mass of the standard substance which is Mass of the element in grams
an atom of hydrogen or an atom of oxygen or an atom of No. of gram atoms = - - - - - - - - - - - - -
carbon-12. The molecular mass of a substance may be defined as Atomic mass of the element in grams
16 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

So, Mass of oxygen = 2.5 x 32 80.0g (ii) Molecule: The smallest particle of a substance (element or
compound) which has free or independent existence and possesses
Example 14. Calculate the gram atoms in 2.3 g ofsodium. all characteristic properties of the substance. A molecule of an
element is composed of like atoms while a molecule of a compound
Solution: No. of gram atoms 2.3 == 0.1 contains fixed number of atoms of two or more different elements. A
23
[Atomic mass of sodium 23 g] molecule may be broken down into its constituent atoms but the
atom is indivisible during a chemical change.
Example 15. Calculate the mass of 1.5 gram molecule of Avogadro after making the above differentiation, presented a
sulphuric acid. hypothesis known as Avogadro hypothesis which can be stated
Solution: Molecular mass of as follows:
H 2 S04 =2x 1+ 32+ 4 x 16= 98.0amu "Under similar conditions of temperature and pressure,
Gram-molecular mass of H 2 S04 =98.0g equal volumes of aD gases contain equal number of molecules."
Mass of 1.5 gram molecule of H 2 S04 =98.0x 1.5=147.0g Avogadro hypothesis explains successfully the formation of
hydrogen chloride.
Example 16. Calculate the actual mass ofone molecule of Hydrogen + Chlorine = Hydrogen chloride
carbon dioxide (C0 2 ),
I vol I vol 2 vol
Solution: Molecular mass of CO 2 = 44 amu nmolecules nmolecules 2n molecules
larrru 1.66xlO- 24 g I molecule I molecule 2 molecules
24 I 1
So, The actual mass of CO 2 == 44 x 1.66 x 10- '2 molecule '2 molecule I molecule
23 I atom I atom I molecule
7.304 X 10- g
(Both hydrogen and chlorine are diatomic in nature.)
Thus, the hypothesis explains that the molecules of reacting
1i.~_~~ AVOGADRO'S HYPOTHESIS gases break up into constituent atoms during chemical change
According to Dalton's atomic theory, elements react with each which then combine to form new molecules of the product or
other in the simple ratio of their atoms. Gay-Lussac proposed that products.
gases combine in simple ratio of their volumes. In an attempt to
correlate Dalton's atomic theory with Gay-Lussac law of gaseous Applications of Avogadro's hypothesis
volumes, Berzelius stated that under similar conditions of 0) Atomicity*: Atomicity means number of atoms present
temperature and pressure, equal volume of all gases contain in one molecule of an elementary gas. Hydrogen, oxygen,
the same number of atoms. This hypothesis was subsequently nitrogen, chlorine, etc., are diatomic in nature. Noble gases are
found to be incorrect as it failed to interpret the experimental monoatomic while ozone is triatomic in nature. Avogadro's·
results and contradicted the very basic assumption of Dalton's hypothesis helps in determining the atomicity of elements.
atomic theory, i. e., an atom is indivisible. For example, the (ii) Relationship between molecular mass and vapour
formation of hydrogen chloride from hydrogen and chlorine density: The vapour density of any gas is the ratio of the
could not be explained on the basis of Berzelius hypothesis. densities of the· gas and hydrogen under similar conditions of
Hydrogen + Chlorine = Hydrogen chloride
temperature and pressure.
I vol I vol 2 vol Density of gas
n atoms natoms 2n compound atoms Vapour Density (V.D.)
Density of hydrogen
I atom I atom 2 compound atoms
Jiatom Ji atom I compound atom Mass of a certain volume of the gas
Mass of same volume of hydrogen at
i. e., for the formation of I compound atom of hydrogen the same temp. and pressure
chloride, ~ atom of hydrogen and ~ atom of chlorine are needed.
In other words, each atom of hydrogen and chlorine has been If n molecules are present in the given volume of a gas and
divided which is against Dalton's atomic. theory. Thus, the hydrogen under similar conditions of temperature and pressure,
hypothesis ofBerzelius was discarded. Mass of n molecules of gas
The Italian scientist, Amedeo Avogadro, in 1811, solved the V.D.=---------------------
Mass of n molecules of hydrogen
above problem by proposing two types of particles from which
whole of the matter is composed.of. Mass of I molecule of gas
(i) Atom: The smallest particle of an element that can take
part in chemical change but generally cannot exist freely as such. Mass of I molecule of hydrogen

*Atomicity can be ascertained with the values of ratio of two specific heats of gases ( ~; )

1.66 (Monoatomic), = lAO (Diatomic), Cp 1.33 (Polyatomic)

Cv · .. Cv
 BASIC PRIN.CIPLES 17
Molecular mass of gas apples or one gross books means 144 books or one gross oranges
Molecular mass of hydrogen means 144 oranges. In a similar way, for counting of atoms,
molecules, ions, etc., chemistS use the unit mole. The term mole
Mol. mass
was introduced by Ostwald in 1896. This is the Latin word
2 'moles' meaning heap or pile. A mole (mol) is defined as the
(since, moL mass of hydrogen = 2) number of atoms in 12.00 g of carbon-12. The number of atoms
Hence, 2x V.D.=Mol. mass in 12 g of carbon-12 has been found experimentally to be
6.02xlQ23, This number is also known as Avogadro's number
This fonnula can be used for the detennination of molecular
masses of volatile substances from vapour density. Vapour named in honour of Amedeo Avogadro (1776 - 1856). .
density is measured mainly by two methods: Thus, a mole contains 6.02 x 1023 units. These units can be
(a) Victor Meyer and (b) Duma's methods. atoms, molecules, ions, electrons or anything else.
. (iii) Gram-molecular volume: I g mole of any gas 1 mole of hydrogen atoms means 6.02 x 1023 hydrogen atoms.
occupies 22.4 litres or 22400 mL of volume at NTP or STP
1 mole of hydrogen molecules means 6.02 x 1023 hydrogen
conditions. 11<
The density of hydrogen at NTP is 0.00009 g mL- l . Thus, molecules.
0.00009 g of hydrogen will occupy volume at NTP 1 mL 1 mole of potassiiun ions means 6.02 x 1023 potassium ions.

1 g of hydrogen occupies volume at NTP = 1 mL 1 mole of electrons means 6.02 x 1023 electrons.
. 0:00009
The type of entity must be specified when the mole
Ig mole of hydrogen (2.0 16 g) occupies volume at NTP designation is used. A mole of oxygen atoms contains 6.02 x 10 23
i 2.016 = 22400 mL = 22.4 litre oxygen atoms. and a mole of oxygen molecules contains
0.00009 6.02 x 1023 oxygen molecules. Therefore, a mole of oxygen
According to Avogadro's hypothesis, equal volumes of molecules is equal to two moles of oxygen' atoms,
different gases contain same number of molecules under similar i. e. , 2 x 6.02 x 1023 oxygen atoms.
conditions of temperature and pressure. Thus, 22.4 litre or 22400 How much does one mole weigh? That depends on the nature
mL of any gas at NTP will contain one gram mole or its molecular of particles (units). The mass of one mole atoms of any element
mass in grams. is exactly equal to the atomic mass in grams (gram-atomic
Loschmidt number: Number of molecules in 1 em3 or 1 mL mass or gram atom) of that element.
of a gas at S.T.P. is known as Loschmidt number. For example, the atomic mass of aluminium is 27 amu. One
. 6.023x10 23 amu is equal to 1.6{i x 10-24 g. One mole of aluminium contains
LOSChmidt number = --.---
22400 6.02 x 1023 alJlIl1inium atoms.
= 2. 68x 1018 molecules mL-1 Mass of one atom aluminiUm = 27 x 1.66 x 10-24 g
(iv) Molecular formula: Avogadro's hypothesis helps in Mass of one mole aluminium =27 x 1.66 x 10-24 X 6.02 x 1023
[mding the molecular formulae of gases. Under similar =27g
conditions of temperature and pressure, 2 volumes of ozone after This is the atomic mass of aluminium in grams or it is one
decomposition give 3 volumes of oxygen. gram atomic mass or one gram atom of aluminium.
Deco~tion Similarly, the mass of 6.02 x 10 23 moleC"!Iles (1 mole) of a
Ozone ) Oxygen
2 vol 3 vol substance is equal to its molecular mass in grams c.r
gram-molecular mass Or gram molecule. For example,
2 molecules 3 molecules molecular mass of water is 18 amu. Thus, mass of one mole of
I molecu1e 3/2 molecules water will be 18 x 1.66 x 10-24 X 6.02 x 1023 , i. e., 18 g. This is
the molecular mass of water in grams or one gram-molecular
1 molecule 3 atoms
mass or one gram molecule.
Thus, the fonnula of ozone is ° 3, Mole concept is also applicable to ionic compounds which do
not contain molecules. In such cases, the fonnula of an ionic
1~t3 MOLE CONCEPT. compound represents the ratio between constituent ions. The
mass of 6.02 x 1023 fonnula units represents one mole of an ionic
For the counting of articles, the unit dozen or unit gross is compound.
commonly used irrespective of their nature. For example, one
dozen pencils means 12 pencils or one dozen apples means 12

... O°C or 273 K temperature and one atmosphere or 760 rom ofHg or 76 cm ofHg pressure are known as the standard conditions of temperature and
pressure (STP) or normal conditions oftemperature and pressure (NTP).
18 I G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

One mole ofBaC1 2 = 6.02 x 1023 BaCl 2 units Number of water mOlecules in one mole of water
23
= 208.2g BaCl 2 6.02 x 10
Molecular mass (fomrula mass) of BaCl 2
Number of molecules of water in 0.05 moles
:::: 6.02 x 1023 Ba 2+ ions + 2 x 6.02
X 10
23 CI- ions 0.05 x 6.02 x 1023
= 137.2+ 71.0 = 208.2g 3.01Ox 1022
One mole of a substance will have mass equal to formula As one molecule of water contains one oxygen atom,
mass of that substance expressed in grams. So, number of oxygen atoms in 3.010 x 1022 molecule of
It has been established by Avogadro's hypothesis that one water 3.010x 1022
gram-molecular mass of any gaseous substance occupies a
volume of 22.4 litres at NTP. One gram-molecular mass is Example 19. Calculate the mass of a single atom of
nothing but one mole of substance. Thus, one mole, sulphur and a single molecule ofcarbon dioxide.
i. e. , 6.02 x 1023 molecules of any gaseous substance occupies Solution:
22.4 litres as volume at NTP.
The following formulae satisfy the above discussion. Gram-atomic mass of sulphur = 32g
1 mole of a substance:::: 6.02 x 1023 particles ofthe substance Gram-atomic mass
Mass of one sulphur atom
Number of moles of a sub~tance 6.02 x 1023
Mass of substance in gram 32
::::---- 5.33 X 10-23 g
Mass of one mole of the substance in gram 23
6.02 x 10
No. of particles
Further, Number of moles = 23 Formula of carbon dioxide = CO 2
6.02 x 10
Molecular mass of CO 2 12+2xI6=44
Thus, Gram-molecular mass of CO 2 44 g
No. of particles = Mass of substance in gram Gram-molecular mass
6.02 x 10 23 Mass of one mole of the substance in gram Mass of one molecule of CO 2
6.02x
Mass of one atom of an element
Gram atom of an element = 4 4 = 7.308 x 10-23 g
6.02 x 1023
6.02 x 1023
Mass of one molecule of a substance Example 20. What is the mass of3.01 xl 022 molecules of
Gram-molecular mass of the substance ammonia?
=
6.02 x Solution: Gram-~olecular mass of ammonia 17 g
Number of molecules
:::: Vohune of gas in litres at NTP x 6.02 x 1023 Number of molecules in 17g (one mole)ofNH3 = 6.02 x 1023
22.4 Let the mass of 3.01x 1022 moieculesofNH3 be =xg

3.01 x 1022 x
,£ : :: ::: _SOME SOLVED eXAMPLES\ :::: : : ; So,
6.02x 1023 17
Example 17. A piece ofcopper weighs 0.635 g. How many 22
or x 17x 3.01x 10 =0.85g
atoms ofcopper does it contain? IeEE (Bihar) 19921
6.02x 1023
SolutiOn: Gram-atomic mass of copper:::: 63.5 g
Example21. From 200 mg of CO2 , 1021 molecules are
Number of moles in 0.635 g of copper 0.635 == 0.01
63.5 removed. How many moles ofCO2 are left?
23
Number of copper atoms in one mole = 6.02 x 10 Solution:
Number of copper atoms in 0.01 moles O.Olx 6.02 x 1023 Gram-molecular mass ofe02 :::: 44 g
6.02 x 1021
Example 18. How many molecules o/water-and oxygen Mass of 10 21 molecules of CO2 . 44 x 1021 = 0.073 g
6.02 x 1023
atoms are present in 0.9 g ofwater?
Mass of CO 2 left = (0.2 - 0.073) = 0.127 g
Solution: Gram-molecular mass of water :::: 18 g

Number of moles in 0.9g·of water = 0.9 = 0.05 Number of moles of CO 2 left 0.127 = 2.88 x 10- 3
18 44
 BASIC PRINCIPLES I 19

23
::Example 22. How many molecules and atoms of oxygen 6.02 x 10
are present.in 5,6 lUres of oxygen (0 2 )at NTP?
x 60x 60x 24 x 365
Solution: We know that, 22.4 litres of oxygen at NTP
contain 6.021< 1023 molecules of oxygen, = 19,089x 109 years = 1.9089x 1010 years

So, 5,6litres of oxygen at NTP contain .if

= 5,6 x 6,02 x 1023 molecules

22.4
23
1. Ill> mgof a compound on vaporisation in Victor Meyer's
1.505 x 10 molecules apparatus displaces 44.8 mL of air measured at STP. The
molecularmass of the compound is: [CEE (Kerala) 20041
1 molecule of oxygen contains 2 atoms of oxygen (a) 116 (b) 232 (c) 58 (d) 44,8 (e) 46.4
[Ans. (c)]
So, 1.505 x 1023 molecules of oxygen contain [Hint: Molar mass of compound
2 x 1.505 X 1023 atoms = Mass of22400 mL vapour at STP
23
3.01 X 10 atoms = OJ 16 x 22400 = 58]
44.8
Example 23. How many electrons are present in 1.6 g of
2. A gas has a vapour density 11.2. The volume occupied by I g
methane?
of the gas at NTP is: (JCECE 2004)
Solution: Gram-molecular mass of methane, (a) I L (b) 11.2 L (c)22.4 L (d) 4 L
(CH 4 ) 12+ 4 = 16g [Ans. (a)]
Number of moles in 1,6 g of methane [Hint: Molar mass 2 x 11,2 = 22.4 g
:::: 1.6 OJ 22.4
Volume of I g compound at STP ;=. - - = I L]
16 22.4
3. 3 g of hydrocarbon on combustion with 11.2 g of oxygen
Nunlber of molecules of metharie in 0.1 mole
produce 8,8 g of CO2 and 5.4 g of H 20. The data illustrate
:::: 0,1 x 6,02 x 1023 the law of:
6,02 x 10 22 (a) conservation of mass (b) multiple proportions
(c) constant proportions (d) reciprocal proportions
One molecule of methane has = 6 + 4 = Ioelectrons [Ans. (a)]
[Hint: L Masses of reactants L Masses of products
22
So, 6,02 x 10 molecules of methane have (3 + 11.2) g (8,8 + 5.4) g
Hence, law of conservation of mass is verified.]
10 x 6,02 X 1022 electrons
4. The maximum number of molecules is present in:
23
6,02 x 10 electrons [CBSE (PMT) 2004; Manipal (Medical) 20071
(a) 15 L ofIt2 gas at STP (b) ~ L ofN2 gas at STP
Example 24. The electric charge on the electron is
(c) 0,5 g ofH2 gas (d) 10 g of 02 gas
1.602 x 10- 19 coulomb, How much '~harge is present on 0,1 mole
[Ans. (a)]
ofCu 2+ ions? [Hint:
Solution: Charge on one mole of ele«trons 15
Number of molecules in 15 L H2 = - x N = 0.669 N
:::: 6,02 x 1023 x 1.602 X 10-19 coulomb 22.4
:::: 96500 coulomb :::: 1 faraday Number of molecules in 5 L No =_5_ x N = 0.223·N
- 22.4
Charge on one mole of Cu 2+ ions
= 2 x 96500 coulomb 2 faraday Number of molecules in 0,5 g H2 = 0.5 xN =: 0,25 N
2
Charge on 0.1 mole of Cu 2+ ions
Number of molecules in 10 g 0, =: .!Q x N = 0.312 N]
=O.lx 2 0.2faraday - 32
ExalJlple 25. How many years it would take to spend one 5. Insulin contains 3.4% sulphur. Then, the minimum molecular
mass of the insulin is about:
Avogadro snumber ofrupees at a rate oflO lakh of rupees in one
second? (MLNR 1990)
(a) 940 amu (b) 9400 amu
(c) 3600 amu (d) 970 amu
Solution: Number of rupees spent in one second:::: 10 6
[Ans. (a)]
Nunlber of rupees spent in one year [Hint: .: 3.4 g sulphur is present in 100 g insulin
:::: 106 X 60x 60x 24 x 365
:. 32 g sulphur will be present in 100 x 32 g insulin = 940
Avogadro's number of rupees will be spent 3,4

:. Molar mass of insulin is about 940 amu]

20 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

6. 25 g of MC1 4 contains 0.5 mol chlorine then its molecular active metals. A known mass of the active metal· is reacted with
mass is: (DPMT 2007) dilute mineral acid. Hydrogen gas thus evolved is measured
(a) 100g mol- 1 ,b) 200g mol- l under experimental conditions. The volume of hydrogen is then
reduced to NTP conditions. The mass of liberated hydrogen is
(c) 150g mol- l (d) 400g mol- l
determined using density of hydrogen (0.00009 at NTP).
[Ans. (b)]
[Hint: I mol of MCI 4 contaiils 4 mol of chlorine Equivalent IllIlSS Mass of element x 1.008
.,' 0.5 mol chlorine is present in 25 g of MCl 4 Mass of hydrogen

;;4 mol chlorine will be present in~ x 4, i.e.,200 g ofMCI 4.] Mass of element x 1.008
0.5
Volume in mL ot' hydrogen displaced atNTPx 0.00009

'1.'14 EQUIVALENT MASSES OR CHEMICAL Mass of element x 11200

EQUIVALENTS Volume in mL of hydrogen displaced at NTP

Equivalent mass of a substance (element or compound) is defmed (ii) Oxide formation method: A known mass of the
as the number of parts by mass of the substance which element is changed into oxide directly or indirectly. The mass of
combine or displace directly or indirectly 1.008 parts by mass oxide is noted.
of hydrogen or 8 parts by mass of oxygen or 35.5 parts by Mass of oxygen (Mass of oxide Mass of element)
mass of chlorine or 108 parts by mass of silver.
The equivalent mass is a pure number. When the equivalent Thus, the equivalent mass of the element
:-
mass of a substance is expressed in grams, it is called gram Mass of element x8
equivalent mass. For example, equivalent mass of sodium is 23, (Mass of oxide Mass of element)
hence, its gram equivalent mass is 23 g. • /
Mass of element x 8
The equivalent mass of a substance may ha~e different values
under different conditions. The -equivalent mass of an element Mass of oxygen
may vary with change of valency. For example, copper fornts two (iii) Chloride formation method: A known mass of the
oxides CUO and Cu 2 0. In CUO, 63.5 parts of copper combine element is changed into chloride directly or indirectly. The mass
with 16 parts of oxygen. Thus, equivalent mass of copper in this of the chloride is determined.
oxide is 6~5 31.75. In Cu 20' 2 x 63.5 parts ofcopper combine Mass of chlorine (Mass of chloride Mass of element)
with 16 parts of oxygen; thus, the equivalent mass of copper in Thus, the equivalent mass of the element
this oxide is:
Mass of element x 35.5
2 x 63.5 = 63.5
(Mass of chloride Mass of element)
2
Relation between atomic mass, equivalent mass and Mass of element x 35.5
valency: Suppose an element X combines with hydrogen to
. fom a compound, XH n , where n is the valency of the element X.
(iv) Metal to metal displacement method: A more active
n parts by mass of hydrogen combine with atomic mass of metal can displace less active metal from its salt's solution. For
elementX. example, when zinc is added to copper sulphate, copper is
1 part by m\lSS of hydrogen combines with precipitated. A known mass of active metal is added to the salt's
Atomic mass of element solution of less active metal. The precipitated metal after drying
n is accurately weighed. The masses of the displacing metal and the
.. Atomic mass of element. h . displaced metal bear the same ratio as their equivalent masses. If
B y ab ove defiImtlOn, IS t e eqUlva- El and E2 are the equivalent masses of two. elements and
n ml and m2 their respective masses, then,
lent mass of the element.
Atomic mass ml =~
Thus, Equivalent mass m2 E2
n
Knowing the equivalent mass of one metal, the equivalent
or Atomic mass Equivalent mass x Valency
mass of th.e other metal can be calculated.
Note: 'Detailed discussion'on equivalent masses of compounds (acids,
bases, salts, oxidising agents, reducing agents, etc.,) will be (v) Double decomposition method: This method is based
taken in chapter on volumetric analysis. on the following points:
The following methods are, used for the determination of (a) The mass of the compound reacted and the mass of
equivalent mass of elements. product formed are in the ratio of their equivalent masses.
(i) Hydrogen displacement method: This method is used (b) The equivalent mass of the compound (electrovalent) is
for those elements which can evolve hydrogen from acids, i.e., the sum of equivalent masses of its radicals.
 BASIC PRINCIPLES 21

(c) The equivalent mass of a radical is equalto the fonnula (a) 4 (b) 2 (c) 3 (d) 1.5
mass of the radical divided by its charge. [ADS. (d)]

AB+CD ~AD+CB , of metal = 16 = ~

[Hint: Equivalent mass n
ppt.
Mass of AB Equivalent mass of AB Where x atomic mass of metal
nvalency of metal
Mass of AD Equivalent mass of AD Molecular fonnula of metal oxide M 20 n
Eq. mass of A + Eq. mass of B Mass of metal oxide := 2 (16n) + 16(n) = 1.5]
Mass of metal ' 2 (l6n)
Eq. mass of A + Eq. mass of D \
Knowing the equivalent masses of Band D, equivalent mass 1.15' ME"rHODS FOR THE DETERMINATION
of A can be calculated. OF ATOMIC MASS
(i) Dulong and Petit's Law: According to this law, the
product of atomic mass and specific heat of a solid element is
7. An unknown element fonns an oxide. What will be the approximately equal to 6.4. The product of atomic mass and
equivalent mass of the element if the oxygen content is 20% specific heat is called atomic heat. Thus,
by mass: [JEE (W8) 2008] Atomic mass x Specific heat = 6.4
(a) 16 (b) 32 (c) 8 (d) 64
[ADS. (b)] or . mass (approximate
AtOlTIlC . ) = - - .6.4
---
. Mass of element . Specific heat
[Hint: Eqwvalent mass of element := x8
Mass of oxygen In above fonnula, the specific heat must he in cal/g unit.
80 The equivalent mass of the element is determined
= x 8 =32]
20 experimentally and the valency, which is always a whole number,
8. A metal M of equivalent mass E fonns an oxide of molecular can be obtained by dividing approximate atomic mass with the
fonnulaMxOy- The atomic mass of the metal is given by the equivalent mass and changing the value so obtained to the nearest
correct equation: [PMT (Kerala) 2008] whole number. In this way, exact atomic mass can be determined
(a) 2E(y/ x) (b).xyE by multiplying equivalent mass with valency.
(c) EI y (d) y/ E Example 26. A chloride of an element contains 49.5%
E x chlorine. The specific heat of the element is 0.056. Calculate the
(e)-x
2, y eqUivalent mass, valency and atomic mass of the element.
[Ans. (a)] Solution: Mass of chlorine in the metal chloride = 49.5%
[Hint: Let atomic mass of metal Mis' a'. Mass of metal (100- 49:5) 50:5
Mass of metal = a x x
. Mass of metal
Mass of oxygen = 16 X Y Eqmvalent mass of the metal '" x 35.5
. Mass of element Mass of chlorine
Eqwvalent mass of element = x8
Mass of oxygen 50.5 x 35.5 36.21
E=~x8 49.5
16y According to Dulong and Petit's law,

a 2E(~)] ·
ApproXlmate . mass 0 fth e metaI = -----.",....---
atOlTIlC
.
6.4
. Specifjc heat
9. The percentage of an element Mis 53 in its oxide of mlllecular 6.4 114.3
fonnula M 203' Its atomic mass is about: 0.056
[PET (Kerala) 2oo8J
(a) 45 (b) 9 (c) 18 (d) 38 Valency = Approximate atomic mass _ 114.3 3.1 '" 3
(e) 21 Equivalent mass 36.21
[Ans. (e)]
., . Mass of element Hence, exact atomic mass = 36.21 x 3 = 108.63
[HIDt: Eqwvalent mass of element = '. x 8
. Mass of oxygen Example 27. On dissolving 2.0 go/metal in sulphuric
53 " acid, 4.5lg o/the metal sulphate wasformed. The specific heat of
=-x8:::9
47 the metal is 0.051 cal g-I. What is the valency of the metal and
Atomic mass = Equivalent mass x Valency exact atomic mass?
=9x3 =27 amu.] Solution: Equivalent mass of SO~- radical
10. The equivaLent weight of a metal is double than that of
oxygen. How many times is the weight of its' oxide greater Ionic mass = 96 = 48
than the weight of metal? Valency 2
22 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Mass of metal sulphate = 4.51 g Examples of isomorphous compounds are:

Mass of metal 2.0g (a) K 2 S04 and K 2Cr0 4 (potassium sulphate and potassium
Mass of sulphate radical (4.51 ~ 2.0) == 2.51 g chromate)
2.51 g of sulphate combine with 2.0 g of metal. °
(b) ZnS04 ·7H2 and FeS04·7H 20 (zinc sulphate and
ferrous sulphate)
So, 48 g of sulphate will combine with (c) KCI04 and KMn0 4 (potassium perchlorate and
potassium permanganate)
=2. x 48 = 38.24 g metal
2.51 (d) K2 S04·Al2 (S04 ll' 24H2 0and K2S04·Cr2(S04k24H20
(potash alum and chrome alum).
Equivalent mass of metal = 38.24
The following conclusions have been deduced from the
According to Dulong and Petit'slaw, phenomenon of isomorphism:
(i) Masses of two elements that combine with same mass of
Approximate atomic mass 6.4 =~ = 112.5 other elements in their respective compounds are in the ratio of
Specific heat 0.057
their atomic masses. .
Approximate atomic mass
Valency = . Mass of one element (A) that combines
Equivalent mass with a certain mass of other elements Atomic mass of A
Mass of other element (B) that combines .Atomic mass of B
= 112.5 = 2.9"" 3 with the same mass of other elements
38.24
(ii) The valencies of the elements forming isomorphous
Exact atomic mass 38.24 x 3 114.72
compounds are the same.
(ii) Canniizaro's method: Atomic mass of an element Example 28. Potassium chromate is isomorphous to
may be defined as the smallest mass of the element present in the potassium sulphate (K 2 S0 4 ) and is found to contain 26.78%
molecular mass of anyone of its compounds. For this purpose, chromium. Calculate the atomic mass ofchromium (K = 39.10).
the following steps are followed: Solution: Since, the formula of potassium sulphate is
(a) Molecular masses of a number of compounds in which the K 2S04, so the formula of potassium chromate should be
element is present are determined. K 2Cr04 as it is isomorphous to K 2S0 4.
(b) Each compound is analysed. Mass of the element is If the atomic mass of chromium is A, then
determined in the molecular mass of each compound. . formula mass of potassium chromate should be
(c) The lowest mass ofthe element is taken its ato~c mass. = 2x 39.1+ A + 64 =(142.2+ A)
The following table shows the application of this method:
% of chromium A x 100
Vapour· Molecular % of carbon Mass of carbon In (142.2+ A)
Compound .density mass = by mass In one molecular mass 100A
(V,D.) lV.D. compound of the compound . So, 26.78
(142.2+ A)
75.0 x 16
Methane 8 16 75.0 = 12 g
100 100A = 26.78 (142.2+ A)
80.0 x 30
Ethane 15 30 80.0 =24 g or A 26: 78 x 142.2 52.00
100 73.22
Carbon 42.9 x.28
14 28 42.9 =12 g (iv) Atomic mass fmm vapour density of a chloride: The
monoxide 100
following steps are involved in this method:
Carbon 27.3 x44
22 44 27.3 = 12g (a) Vapour density of the chloride of the element is
dioxide 100
determined.
81.8 x 44
Propane 22 44 81.8 =36 g (b) Equivalent mass of the element is determined.
100
Let the valency ofthe element bex. The formula of its chloride
Least mass of carbon is 12 g.
Thus, the atomic mass of carbon is 12.
.
will be MCI x'
Molecular mass = Atomic mass of 1M + 35.Sx
(iii) The law of isomorphism: Isomorphous substances
form crystals which have same shape and size and can grow in = A + 35.Sx
the saturated solution of each other. They have a property of Atomic mass Equivalent mass x Valency
forming mixed crystals. Isomorphous substances have same
composition, i. e., they have same number of atoms arranged A =Exx
similarly. Molecular masS = E x x + 35.Sx
 BASIC PRINCIPLES 23
2 V.:O. x(E+ 35.5) Molecular formula mass
where, n
2V.D~ Empirical formula mass
x
E + 35.5 Molecular formula gives the following informations:
Knowing the value of valency, the atomic mass can be (i) Various elements present in the molecule.
determined. (ii) Number of atoms of various elements in the molecule.
Example 29. One gram ofa chloride was found to contain (iii) Mass ratio of the elements. present in the molecule.
0.835 g of chlorine. Its vapour density is 85. Calculate its The mass ratio of carbon and oxygen in CO 2 molecule is
molecular formula. 12 : 32 or 3 : 8.
Solution: Mass of metal chloride = Ig (iv) Molecular mass of the substance.
(v) The number written before the formula indicates the
Mass of chlorine = 0.835 g number of molecules, e.g., 2C0 2 means 2 molecules of
Mass of metal = (1- 0.835) = 0.165 g carbon dioxide.
. I f I 0.165x 35.5 (iii) Structural formula: It represents the way in which
E qillva ent mass 0 meta = - - - -
0.835 atoms of various elements present in the molecule are linked with
=7.01 one another. For example, ammonia is represented as:
2 V.D.
Va1ency 0 f the meta1= - - - H
E+ 35.5
2x 85 I
= N--H
7.01 + 35.5 I
4 H
Formula of the chloride = MC1 4
The formula indicates that three hydrogen atoms are linked to
Example 30. The oxide of an element contains 32.33 per one nitrogen atom by three single covalent bonds. '
cent of the element and the vapour density of its chloride is 79.
Calculate the atomic mass of the element.
1.17 PERCENTAGE COMPOSITION OF A
Solution: Mass of the element = 32.33 parts COMPOUND
Mass of oxygen (100- 32.33) = 67.67 parts Percentage composition of a compound is the relative mass of the
Equivalent mass of the element = 32.33 x 8 = 3.82 each of the constituent element in 100 parts of it. It is readily
67.67 calculated from the formula of the compound. Molecular mass of
2V.D. 2x 79 a compound is obtained from its formula by adding up the masses
VaIency 0 ftheeIement =4 of all the atoms of the constituent elements present in the
E + 35.5 3.82+ 35.5
molecule.
Hence, the atomic mass of the element = 3.82 x 4 Let the molecular mass of a compound be M and X be. the
mass of an element in the molecule.
= 15.28
Mass·of element
Percentage of element = x 100
1.16" TYPES OF FORMULAE M
X
As already stated in section 1.1 0, a formula is a group of symbols =-xI00
M
of the elements which represents one molecule of the substance.
Formula represents chemical composition of the substance. . Example 31. Calculate the percentage composition of
There are three kinds of formulae in the case of compounds. calcium nitrate.
(i) Empirical formula: .It represents the simplest relative Solution: The formula of calcium nitrate is Ca(N03 h.
whole number ratio of atoms of each element present in the
Thus, the formula mass or molecular mass
molecule of the substance. For example, CH is the empirical
formula of benzene in which ratio of the atoms of carbon and = At. mass of Ca + 2 x At. mass of N + 6 x At. mass of oxygen
hydrogen is 1 : 1. It also indicates that the ratio of carbon and 40 + 2 x 14 + 6 x 16
hydrogen is 12 : 1 by mass.
164
(ii) Molecular formula:· Molecular formula of a compound
is one which expresses as the actual number of atoIris of each % of Ca 40 x 100 = 24
element present in one molecule. C6H6 is the molecular formula 164
of benzene indicating that six carbon atoms and six hydrogen
%ofN= 28 x 100= 17
atoms are present in a molecule of benzene. Thus, 164
Molecular formula = n x Empirical formula %ofO 100-(24+17)=59
24 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Example 32. . Determine the percentage of water of

crystallisation, iron, sulphur and oxygen in pure ferrous sulphate
(FeS04·7H20 ). .
11. A gas mixture contains 50% helium and 50% methane by
Solution: The formula mass of ferrous sulphate volume. What is the percentage by mass of methane in the
mixture? IeEE (Kerala) 2004)
= At. mass of Fe -+ At. mass of S + 4 x At. mass ot:oxygen
(a) 19.97% (b) 20.05% (c) 50% (d) 75%
+7 x Mol. mass of J::i20
(e) 80.03%
= 56.0+ 32.0+ 4x 16.0+ 7x 18.0 [Ans. (e)]
[Hint: Molar and volume ratio will be same, i. e., 1 : L
= 278.0
:. Mass of 1 mole CH4 and He will be 16 and 4 g respectively.
So, % of water of crystallisation = 126 x 100 = 45.32 Mass of CH 4
278 Percentage by mass of CRt = x 100
Total mass
%of iron = 56 x 100=20.14 16
= x 100"" 80%]
278 20
12. The atomic composition of the entire universe is
%ofsulphur= 32 x 100=11.51
278 approximately iven in the table below:
64 of total no~ of atoms
%ofoxygen = - x ]00=23.02
278
H 93
(Oxygen present in water molecules is not taken into account.)
He 7
Example 33. It is found that 16.5 g of metal combine with
oxygen to form 35.60 g of metal oxide. Calculate the percentage Hydrogen atoms constitute what percentage of the universe
of metal and oxygen in the compound. by mass? .
Solution: (a) 77% (b) 23% (c) 37% (d) 73%
[Ans. (a)]
Mass of oxygen in oxide = (35.60- 16.50)= 19.10 g
[Hint: Mass of 93 'H' atoms = 93 amu
%of metal = 16.50 x 100= 46.3 Mass of7 'He' atoms = 28amu
35.60
93
. . 19.10 % Hydrogen by mass = x 100 = 77%]
%ofoxy.gen ~-- x tOO= 53.7 (93 + 28)
'., 35.60
13. Which pair of species has same percentage of carbon?
Exa~ple 34. Hydrogen. and o.xygim are combined in the
(a) CH 3COOH and C 6H l2 0 6
ratio 1: 16 . by mass in hydrogen peroxide. Calculate the
(b) CH 3 COOH and C 2 H 5 0H
percentagtt4Jf·hydrogen and oxygen in hydrogen peroxide.
(c) HCOOCH~ and Cl2H22011
Solution: 1i part<; of hydrogen peroxide contain hydrog~n
. = 1part . (d) C 6H I2 0 6 and (42H22011
[Ans. (a)]
100 parts of hydrogen peroxide contain hydrogen
1 [Hint: Percehtage of carbon in acetic acid = 24 x 100 = 40%
=~x tOO = 5.88 60
17
%of oxygen =(100- 5.88)=94.12 Percentage of carbon in CJil206 = I!:. x 100 = 40%]
180
Example 35. On analysis of an impure sample of sodium
14. Which of the following alkanes has 75% of carbon?
chloride, the percentage ofchlorine was found to be 45.5. What is
(a) Cz»t. (b) CH4 (c) C3Hg (d) C4 HlO
the percentage ofpure sodium chloride in the given sample? '.
[Ans. (b)]
Solution: The molecular mass of pure sodium chloride (NaCI)
= At. mass of Na + At. mass of chlorine [Hint: Percentage of carbon in methane:: 12 x 100 = 75%]
16
= (23 + 35.5) = 58.5
15. Which of the following two oxides of nitrogen have 30.5%
% of chlorine in pure NaCI nitrogen?
35.5 x 100= 60.6 (a) NO (b) NOz (c) N z0 4 (d) N 2 0 5
58.4 [Ans. (b) and (c)]
Thus, [Hint: Percentage of nitrogen in NO z = 14 x 100= 30.5%
% of purity of NaCI in the sample 46
:;:: 45.5 x tOO = 75 28
Percentage of nitrogen in N z0 4 - x 100 = 30.5%]
60.6 . 92
 I
BASIC PRINCIPLES 25

1~1)'; DETERMINATION OF EMPIRICAL AND The empirical fonnula is C 2 H 6 0.

MOLECULAR FORMULAE Empirical fonnula mass (2 x 12) + (6 x 1) + 16 = 46
The. following steps are followed to detennine ~he empirical Example 38. A compound of carbon, hydrogen and
fonnula of the compound: nitrogen contains these elements in the ratio 9: 1 : 3.5. Calculate
(i) The percentage composition of the compound is the empirical formula. If its molecular mass is 108, what is the
detennined by quantitative analysis. molecular formula?
(ii) The percentage of each element is divided by its atomic Solution:
mass. It gives atomic ratio of the elements present in the
Element Atomic Relative number Simplest
compound. Element
ratio mass of atoms ratio
(iii) The atomic ratio of each element is the divided by the
minimum value of atomic ratio as to get the simplest ratio 0.75 = 3
Carbon 9 12 9 =0.75
of the atoms of elements present in the compound. 12 0.25
(iv) If the simplest ratio is fractional, then values of simplest
I I =4
ratio of each element is multiplied by a smallest integer to Hydrogen - =I
I 0.25
get a simplest whole number for each of the element. ' .
(y) To -get the empirical fonnula, symbols of various 3.5 == 0.25 0.25 = I
Nitrogen 3.5 14
elements present are written side by side with their 14 0.25
respective whole number ratio as a subscript to the lower
The empirical fonnula = C 3 H4N
right hand comer of the symbol.
The molecular fennula of a substance may be determined Empirical formula mass =(3x 12)+ (4x 1)+ 14;::: 54
from the empirical fonnula if the molecular mass of the substance n;::: Mol. mass ;::: 108 =2
is known. The molecular fonnula is always a simple multiple of Emp.mass 54
empirical fonnula and the value of simple mUltiple is obtained by
dividing molecular mass with empirical fonnula mass. Thus, molecular fonnula of the compound
= 2 x Empirical fonnula
Example 36. Calculate the empirical formula for a
compound that contains 26.6% potassium, 35.4% chromium and = 2x C3 H4 N= C6 HgN2
38.1% oxygen.· Example 39. A carbon compound containing only carbon
[Given K 39.1; Cr= 52; 0 = 16] and oxygen has an approximate molecular mass of 290. On
Solution: analysis, it is found to contain 50% by mass of each element.
Relative Simplest What is the molecular formula of the compo":nd?
Per- Atomic Simplest Solution:
Element number of whole number
centage mass. ratio
atoms ratio Simplest
I ti num ber SI mpIest whole number
Element AtomIc Reave
Potassium 26.6 39.1
26.6
=0.68 0.68 =1 Ix2=2 percentage mass of atoms ratio ratio
39.1 0.68

Chromium 35.4 52.0 35.4 == 0.68 0.68 =1 1x2=2 4.166 = 1.33

52 0.68 Carbon 50.0 12 4.166 4
3.125

Oxygen 38.1 16.0 38.1 = 2.38 2.38 = 3.5 3.5 x 2= 7

' 16 3.125 = 1
16 0.68 Oxygen 50.0 3.125 3
3.125
Therefore, empirical fonnula is K 2 Cr2 0 7• The empirical fonnula ;::: C4 0 3
Example 37. A compound contains 34.8% oxygen, 52.2% Empirical formula mass = (4 X 12)+ (3 x 16) = 96
carbon and 13.0% hydrogen. What is the empiricalformula mass Molecular mass 290
ofthe compound?
Solution: n = Mol. mass -290 = 3 apprmamate
. 1
y
Emp. mass· 96
Atomic Relative number Simplest
Element Percentage. Molecular fonnula = n x Empirical fonnula
mass of atoms ratio
3 x C4 0 3 = C12 0 9
34.8 = 2.175 2.175 = 1
Oxygen 34.8 16 Example 40. A compound on analysis, was found to
16 2.175
have the follOWing composition: (i) Sodium = 14.31%,
Carbon 52.2 12 52.2 =4.35 4.35 = 2 (ii) Sulphur ;::: 9.97%, (iii) Oxygen = 69.50%, (iv) Hydrogen
12 2.175 ;::: 6.22%. Calculate the· molecular formula of the compound
13.0 13.0 = 6 assuming that whole of hydrogen in the compound is present as
Hydrogen 13.0 13.0
2.175 water ofcrystallisation. Molecular mass ofthe compound is 322.
26 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Solution: 17. A compound has an empirical formula c;H4o. An

Atomic Relative number independent analysis gave a value of 132.16 for its molecular
Element Percentage Simplest ratio mass. What is the correct molecular formula?
mass of atoms
[CET (Kerala) 2004[
0.622 = 2 (a) C4H40S (b) ~oHI2
Sodium 14.31 23 0.622
0.311 (c) <;03 (d) C6H I2 0 3
0.311 (e) C4 HgOs
Sulphur 9.97 32 0.311 =1
0.311 [Ans. (d)]
6.22 [Hint: Molecular formula = (C2H40)n
Hydrogen 6.22 1 6.22 = 20
0.311 n= Molecular mass = 132.16 = 3
4.34 =14 Empirical formula mass 44
Oxygen 69.50 16 4.34
0.311
Molecular formula = (C2H40)3 = CJI1203]
The empirical formula = Na2SH20014
18. An organic compound containing C and H has 92.30%
Empirical formula mass (2 x 23)+ 32+ (20 x 1)+ (14 x 16) carbon. Its empirical formula is:
=322 (a) CH (b) CH3
Molecular mass = 322 (c) CH2 (d) CH4
Molecular formula = Na 2SH 20 014 [Ans. (a)]
Whole of the hydrogen is present in the form of water. Thus, [Hint: Percentage of carbon =-12 x 100 = 92.30%
10 water molecules are present in the molecule.· . 13
So, molecular formula = Na 2SO 4 ·lOH2 ° 'Or'

Relative
Atomic Simplest
fUUSTRATfONSOF,OB]ECTlviQUESY}'ONS Element Percentage
mass
number of
ratio
atoms

Carbon . ~2.30 12 7.69

Hydrogen 7.70 7.70

Empirical formula = CH]

19. Two oxides of a metal contain 50% and 40% of metal M
respectively. If the formula of first oxide is MO, the formula
of 2nd oxide will be:
(a)M02
(c)M2 0
[Ans. (b)]
Hint:
% x Molecular mass = 43.86 x 146 = 4
Compound 1 Compound 2
° 100 Atomic mass 100 16
M 0 M 0
Molecular formula C1;H IO0 4
50% 50% 40% 60%
Molecular mass= 12 x 6 + 10 xl + 16 x 4::: 146
50 g 50 g 40t 60g
'Or'
50 60
Relative 1g -::lg 1g I.5g
Atomic Simplest 50 40
Element Percentage number of
mass ratio 2
, atoms
Carbon 49.30 12 4.10 1.5x2::3
Formula: MO M 20 3]
20. Two elements X and Y have atomic mass 75 and 16
Hydrogen 6.84 6.84 2:5 x 2 = 5
respectively. They combine to give a compound having
Oxygen 43.86 16 2.74 I x 2:: 2 75.8% X. The formula ofthe compound is:
(a) XY (b) X 2 Y
The empirical formula C3Hs02
(C)X 2Y2 (d)X 2Y3
2X73
n::: ";2 [Ans. (d)]
73
[Hint: Molecular mass of X 213 2 x 75 + 3 x 16 = 198
Molecular formula:: 2 X C3Hs02 C1;H IO0 4 ]
 BASIC PRINCIPLES 27
150
Percentage of' X = - x 100 = 75.80% Information Obtained from Chemical Equation
198
'Or' A balanced chemical equation provides the following
informations:
Relative
Atomic Simplest (i) What are the reactants and products involved in the
Element Percenta'ge number of
mass ratio
atoms chemical change?
X 75.80 75 1.01 Ix 2=2 (ii) The relative number of molecules of reactants and
y
products.
24.20 16 1.51 1.5 x 2 = 3
(iii) The relative number by parts of mass of reactants and
Formula = X 21'3] products.
21. The crystalline salt Na 2 S04 ·xH2 0 on heating loses 55.9% of (iv) Relative volumes of gaseous reactants and products.
its mass. The fonnula of crystalline salt is: For example, consider the following reaction:
(a) Na 2 S04 ·5H2 0 (b) Na 2 S04 ·7H2 0 CH 4(g)+ 20 2 (g)=C0 2 (g)+ 2H 20(g)
(c) Na 2 S04 ·2H2 0 (d) Na 2 S04 ·lOH2 0 This equation tells us that methane and oxygen are reactants
(e) Na 2 S04 ·6H2 0 [PMT (Kerala) 2007] and carbon dioxide and water are products~ One molecule of
JAns. (d)] methane reacts with two molecules of oxygen to produce one
[Hint: Molecular mass of Na 2S0 4 ·lOH20 molecule of coi' and two molecules of water or one mole of
=46+ 96+ 180= 322amu methane reacts with two moles of oxygen to produce one mole of
180 carbon dioxide and two moles of water or 16 g of methane reacts
% by mass ofHP = - x 100 = 55.9%]
. 322 with 64 g of oxygen to produce 44 g of CO 2 and 36 g of water.
This equation also tells that I vol. of methane reacts with 2 vol. of
oxygen to produce I vol. of CO 2 and 2 vol. of steam under
1.1.9; CHEMICAL EQUATION similar conditions of temperature and pressure.
A chemical equation is a symbolic representation of a chemical Limitations of Chemical Equation
change. A chemical equation fails to provide the following
The substances, in which the chemical change is brought, are informations:
called reactants and the substances which come into existence as (i) Actual concentration of the reactants taken and the actual
the result of chemical change are called products. The concentration of the products obtained.
relationship between reactants and products is represented in the (ii) Time taken for the completion of the chemical change.
form of a chemical equation. The symbols or formulae of the (iii) . Conditions applied for bringing the chemical change.
reactants are written on left hand side of equality (=) or ~ sign
(iv) Whether the reaction is reversible or irreversible.
and the symbols or formulae of products on right hand side. The
The following efforts have been made to make the chemical
symbols or formulae on both the sides are added by + sign. Such
equations more informative by introducing:
an equation is known as skeleton equation. The equation
(i) Experimental conditions: If a particular chemical
becomes balanced when total number of atoms of various
change occurs under certain temperature and pressure conditions,
elements are made equal on both the sides. Gases are always
these are mentioned above and below the (~ ) or ( = ) sign.
written in molecular form.
200 attn
KCl0 3 ~ KCl + 02 N2 + 3H 2 ~ 2NH3
450"C
This is the skeleton equation as it only represents reactant and
If the r.eaction occurs in presence of a catalyst, it is written
products involved in the chemical change but the following
.above the ( ~ ) or ( = ) sign.
equation is a balanced equation as the number of atoms of various
Pt
elements is equal on both sides. 2S0 2 + 02 ~ 2S03
2KCl0 3 = 2KCl + 302 (ii) Heat evolved or absorbed: Heat evolved or absorbed in
Reactant ~
a chemical change can be represented by adding or subtracting
The following notations are also used in chemical equations as the amount of heat on right hand side.
to provide more information about chemical change: N2 + 02 ~ 2NO- 43.2kcals
(i) Upper arrow (i) IS written immediately after the C + O 2 ~ CO 2 + 94.3 kcals
gaseous product. (iii) Reversible or irreversible nature: Reversible
(ii) Lower arrow (.!.) is written immediately after the reactions are shown by changing the sign of equality (=) or
insoluble substance (solid) which deposits from a solution. arrow (~ ) with sign of double arrow (, ').
(iii) Symbols, (s) for solid, (I) for liquid and (g) for gas are N 2(g)+ 3H2(g)~ 2NH 3(g)
also written to represent the physical state of the reactants
and products. Types of Chemical Equations
(iv) Symbol (aq.) is written for substances dissolved in water. Chemical equations are of two. types:
(v) Symbol (~) is written over an arrow or over an equality (i) Molecular equations
sign to represent heating. (ii) Ionic equations.
28 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Molecular equations are thoSe in which reactants and products Measures. This revised set of units is known as the
are represented in the form of molecules. International System of Units (abbreviated SI). Now the SI
BaCl 2 + Na 2 S0 4 BaS04 J, + 2NaCl units have been accepted by the scientists all over the world in all
2NaOH + H 2 S0 4 Na 2 S04 J, + 2H 2 0 branches of science, engineering and technology. '
Ionic equations are those in which reactants and products are The SI system have seven basic units. The various
written in ionic form. The molecular equation fundamental quantities that are expressed by these units along
BaCl 2 + Na2S04 BaS04 J, + 2NaCI with their symbols are tabulated below:
can be written in ionic form as:
Basic pbysical quantity Unit Symbol
Ba 2+ + 2CI- +2Na + + SO~- BaS04 J, + 2Na + + 2Cl-
Ba 2+ + SO~- = BaS04 J, Length Metre m
Note: Calculations based on chemical equations have been dealt in the Mass Kilogram kg
chapter 'Stoichiometry' in 'Inorganic Chemistry'.
Time Second s
1~20;' MEASUREMENT IN CHEMISTRY: Temperature Kelvin K
FUNDAMENTAL AN.D DERIVED UNITS Electric current Ampere amp orA
Chemistry is an experimental science. An experiment always Luminous intensity Candela cd
involves observation-of a· phenomenon under certain set of
conditions. The quantitative scientific observation generally Amount of substance Mole mol
requires the measurement of one or more physical quantities such
Sometimes, submultiples and multiples are used to reduce or
as mass, length, density, volume, pressure, temperature, etc.
enlarge the size of the different units. The names and symbols of
A physical quantity is expressed in terms of a number and a
sub-multiples and multiples are listed in the table given below.
unit. Without mentioning the unit, the number has no meaning. For
The name for the base unit for mass, the kilogram, already
example, the distance between two points is "four" has no meaning
contains a preftx. The names of other units of mass are obtained
unless a specific unit (inch, centimetre, metre, etc.,) is associated
by substituting other prefixes for prefix kilo. The names of no
with the number. The units of physical quantities depend on three
other base units contain prefixes.
basic units, i.e., units of mass, length and time. Since, these are
independent units and cannot be derived from any other units, they The use of SI system is slowly growing, however, older
are called fundamental units. It was soon realised that the three systems are still in use. Furthermore, the existence of older units
fundamental units cannot describe all the physical quantities such in scientific literature demands that one must be familiar with
as temperature, intensity of luminosity, electric current and the both old and new systems.'
amount of the .substance. Thus, seven units of measurement, Submultiples Multiples
namely mass, length, time, temperature, electric current, luminous
intensity and amount of substance are taken as basic units. All Prefix Symbol Sub-multiple Prefix Symbol Multiple
other units can be derived from them and are, therefore, called deci d 10- 1 deca da 10
derived units. The units of area, volume, force, work, density,
centi c 10-2 hecto h 102
velocity, energy, etc., are all derived units.
milli m 10-3 kilo k 103
SI Units of Measurement micro 10-6 mega M HI'
Jl
Various systems of units were in use prior to I91'l0. The 10-9
nano n giga G 109
common ones are the following:
pico P 10- 12 tera T 1012
(i) The English or FPS system: The system uses the foot,
the pound and the second for length, mass and time femto f 10- 15 peta P 1015

measurements respectively. It is not used now-a-days. atto a 10-18 exa E 10

18

(ii) MKS system: Here M stands for metre (a unit of zepto z 10-21 zeta Z 102 1
length), K for kilogram (a unit of mass) and S for second (a unit yocto 10-24 yotta Y 1024
of time). This is a decimal system. .
(iii) CGS system: Here the unit oflength is centimetre, the Greek Alphabets
unit of mass is gram and the unit of time is second. It is also a
Alpha A IX Nu N v
decimal system.
MKS system often known as metric system was very popular Beta B ~ Xi .:. ~
throughout the world, but the drawback with this system was that Gamma r y Omicron 0 '0
a number of different metric units for the same quantity were Delta 11 5 Pi n 11:
used in different parts of the world. In 1964, the National Bureau
of Standards adopted a slightly modified version of the metric
system, which had been officially recommended in 1960 by an
Epsilon
Zeta,
Eta
E
Z
H
,E Rho
Sigma
T\lu
P
L
p
cr
international body, General Conference of Weigbts and l'J 't 't
 BASIC PRINCIPLES 29
Theta e e UpsiloJ:l. r \) frequently by chemists. Certain other units which are not a part of
Iota I t Phi <I> .<\1 SI units are still retained for a limited period of time. The term
Kappa Chi atmosphere (atm), the unit of pressure, falls into this category.
K 1( X X
Few of the old units along with conversion factors are given
Lambda A A Psi lJ' lJf below:
Mu M j.t Omega Q ro Length: The interatomic distances are reported in units of
Numerical Prefix. angstrom (A), nanometre (nm) or picometre (pm).
lA= 10-8 cm= 10- 10 m
Prefix VaHue Prefix Value
Inm=10-7 cm=10-9 m=WA
Herni (II2) Deca 10
Mono Undeca 11 Ipm=W- IO cm 10- 12 m 10-2 A
Sesqui 1":
2
Dodeca 12 Inm 103 pm
DiorBi 2 Trideca 13 Mass: The basic unit of mass is generally taken as gram (g).
Tri 3 Tetradeca 14 The gram is 10-3 kg.
Tetra 4 Pentadeca 15 I kilogram (kg) = 103 g
, Penta 5 Hexadeca 16
Hexa 6 Heptadeca 17 1 milligram (mg)= 10-3 g
Hepta 7 Oetadeca 18 1 microgram (Ilg) =10-6 g
Oeta 8 Nonadeca 19 While dealing with atoms arid molecules, the term atomic
Nona 9 Eicosa 20 mass unit (amu) is used. One amu is taken exactly as ..!.. of the
. 12
SI Units for Some Common Derived Quantities mass ~f an atom of the carbon isotope, C : I2

(a) Area = length x breadth 1amu= 1.6605 x W~24 g = 1.6605 X 10-27 kg

= m x m = m 2 [square metre] Volume: The units of volume are reported as cubic
.(b) Volume length x breadth x height centimetre (cm 3) and cubic decimetre (dm 3). Cubic decimetre is
termed litre while cubic centimetre is termed miUilitre.
= mx mX m=m 3 [cubic metre]
llitre (lit or L) (Wcm)3 = WOOcm 3 = 10-3 m 3
· mass kg k -3
(c) Denslty=---=-= gm lmillilitre(mL)=(lcm)3 =lcm3 (cc) = 10-6 m 3
volume m 3
So, . I litre = 1000 mL
(d) Speed = distanc~ covered = ~etre = m S-I
time bme Temperature:. The celsius temperature scale which is not
. change in velocity ms- . -2·
I a part of SI system, is ·employed in scientific studies. This scale is
(e) Acceleration = = - - = ms based on the assignment of O°C to the normal freezing point of
time taken s .
water and 100°C to the normal boiling point of water. The celsius
(f) Force = mass x acceleration scale was formerly called the centigrade scale. .
The unit of temperature in SI system is Kelvin. A degree on
=kg x ms-2
the kelvin scale has the same magnitude as the degree on the
kg ms-2 (Newton, abbreviated as N) . celsius scale but zero on the kelvin scale is equal to -273.15"C.
The temperature (0 K) is often referred to as absolute zero.
(g) Pressure =force per unit area
So, K =: (OC+ 273.15)
k -2
g ms =kg m- I
or °c (K - 273.15)
m2
(Pascal-Pa) There is another important temperature scale known as
fahrenheit scale. In this scale, the normal freezing point of water
(h) Energy = force x distance travelled
is 32"F and normal boiling point is 212"F. Thus, 100"C equals
=kgms-2 xm 180"F. Both the scales are related by the following equations:
kg m 2 s-2 ( joule-J) oC=~ x (OF - 32)[since, 100 parts on celsius scale
9
Some Old Units Stili in Use = 180 parts on fahrenheit scale]
The use of some of the old units is still permitted. The 'litre' ,
for example, which is defined as I cubic decimetre is used
30 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

Pressure: There are three non-S1 units for pressure which Values of Some Useful Constants
are commonly used.
F\lndamental
(a) Atmosphere (atm) is defined as the pressure exerted by a Value In old units Value in SI units
constant
column of mercury of 760 mm or 76 cm height at O°C.
(b) Torr is defined as the pressure exerted by a I mm column 'Avogadro's
of mercury at O°C. number(N)
(c) Millimetre of mercury (mm Hg). Atomic mass . 1.6605 X 10-24 g 1.6605 X 10-27 g
These three units are related as: unit (amu)
Bohr radius (~) 0.52918A = 0.52918· 5.2918 X 10-11 m
latm=760torr 760mmHg 76cmHg 1.013 x 105 Pa
x 10-8 cm
Energy: Calorie has been used in the past as a unit of energy
Boltzmann 1.3807 X 10-23 JK- 1
measurement. The calorie was defmed as the. amount of heat constant (k)
required to raise the temperature of one gram of water from
14.5"C to 15.5°C. One calorie is defined as exactly equal to 4.184 Charge on e1ectron(-) 4.8029 x 10-10 esu (-) l.6021 X 10-19 coul
(e)
joules.
Ical 4.184 J or 11 = 0.2390 cal Charge to mass 1.7588 x 108 coul g-1
ratio el m of
1kcal = 1000 cal ==4.184 kJ electron
Conversion factors
I angstrom (A) = 10-8 cm=1O- IO m==IO-1 nm=10 2 pm
1 inch 2.54 cm.
39.37 inch = I metre
or

1 kg = 2.20 pounds (lb)

lcm =0.394 inch
1km=0.621mile
19 = 0.0353 ounce (0)
Electron rest
mass (me)
GaS constant
(R)
9.1091 x 10-28 g

1
0.0821lit atm deg- mol-

1.987
1

=2.0 cal deg-lmol- I

l
9.1091 X 10-31 kg

I
8.314 x 107 erg deg-1mor' 8.314 J K- mol-
.
1

..

1 pound (lb)=453.6g Molar volume 22.4 L mol- 1

1atomic IIm:ss unit (amu) == 1.6605 x 10- 24 g at NTP (Vm)
= 1.6605 x 1O- 27 kg Planck's constant 6.6252 x 10-27 erg sec 6.6252 X 10-34 J sec
(h)
= 1.492 x 10- 3 erg = 1.492 x 10- 10 J
Proton mass (mp) l.6726 x 10-24 g 1.6726 X 10-27 kg
3.564 X lOll cal =9.310 x 10 8 eV
Neutron mass 1.67495 x 10-24 g 1.67495 X 10-27 kg
=931.48 MeV (mn )
1atmosphere (atm) = 760 torr = 760 mmHg = 76cmHg Rydberg constant lQ9678 cm- 1
= 1.01325 x 10 5 Pa (R,,)

1calorie (cal)=4.1840x 10 7 erg =4.184 J Velocity of light 2.9979 x 1010 cm seC l 2.9979 X 108 m sec-I.
(c) in vacuum or 186281 miles sec-I
2.613 x 1019 eV
Faraday (F) 9.6487 x 104 C I equiv.
lcoulomb (coul) =2.9979 x 109 esu or 96500 C/equiv.
, 1curie (Ci) = 3.7 X 1010 disintegrations sec -I 0.8988xl0 1O N m 2C-2
lelectron volt (eV) 1.6021 x 1O-12 erg = L6021 x 10- 19 J or9x 109 N m 2e 2

= 3.827 xl 0-20 cal Derived SI Units

= 23.06 kcal mol- 1
Quantity with Symbol Unit (SI) Symbol
lerg 10-7 J = 2.389 x lO-scal 6.242 x 1011 eV
Velocity (v) metre per sec m
1electrostatic unit (esu) = 3.33564 x 10- 10 coul
Area (A) square metre
1faraday (F):= 9.6487 X 104 coul Volume (V) cubic metre
1dyne (dyne) = lO-s N Density (P) kilogram m- 3
1 joule = 107 erg = 0.2390 cal Acceleration (a) metre per sec2
llitre=1OOOcc 1000rnL=ldm3 Energy (E) joule (J) kgm 2
10- 3 m 3 Force (F) • newton (N) kgm
 BASIC PRINCIPLES 31

Power(W) watt0N) 243.4 has four significant figures.

Pressure (P) pascal (Pa) N m-2
. 24.123 has five significant figures.
(ii) A zero becomes significant figure if it appears between
Resistance (R) ohm(n) V A-I
two non-zero digits. For example,
Conduction (e) ohm-I, mho, siemens m-2 kg- I S3 A2 or n- I 5.03 has three significant figures.
Potential difference volt (V) kgm 2 S-3 A-I 5.604 has four significant figures.
4.004 has four significant figures.
Electrical charge coulomb (C) A-s (ampere-second)
(iii) Leading zeros or the zeros placed to the left of the
Frequency (v) hertz (Hz) cycle per sec number are never significant. For example,
Magnetic tesla (T) kg S-2 A-I =N A-I m- I 0.543 has three significant figures.
flux x density
0.045 has two significant figures.
Popular Units and their SI Equivalents 0.006 has one significant figure.
(iv) Trailing zeros or the zeros placed to the right of the
Physical quantity Unit with symbol .Equivalent in SI unit· number are significant. For example,
Mass 1 amu larnu = 1.6605 x 10-27 kg 433.0 has four significant figures.
19
433.00 has five significant figures.
Energy 1 electron volt (eV) 1.602 x 10- joule
343.000 has six significant figures.
Length I A 10- 10 m (10- 1 om) (v) In exponential notation, the numerical portion gives the
number of significant figures. For example,
Volume litre 10-3 m 3 =dm 3
1.32 x 10-2 has three significant figures.
Force dyne 10-5 N
1.32 x 10 4 has three significant figures.
Pressure I atmosphere 760 torr (760 mm Hg)
(vi) The non-significant figures in the measurements are
101325 pa or 105 pa rounded off.
1 bar 101325 pa or 105 pa (a) If the figure following the last number to be retained is
1 torr 133.322 N m-2 less than 5, all the unwanted figures are discarded and
the last number is left unchanged, e.g.,
Dipole moment debye,IO- 1S esu-cm 3.324 x 10-30 cm
5.6724 is 5.67 to three significant figures.
Magnetic flux density gauss (0) 10-4 T (b) If the figure following the last number to be retained
Area of nuclear I bam 10-28 m 2 is greater than 5, the last figure to be retained is
cross section increased by I unit and the unwanted figures are
discarded, e.g.,
Nuclear Diameter 1 fermi (1 femto) 10-15 m
8.6526 is 8.653 to four significant figures.
(c) If the figure following the last number to be retained
Significant Figures
is 5, the last figure is increased by 1 only in case it
There is always some degree of uncertainty in every scientific happens to be odd. In case of even number the last
measurement except in counting. The uncertainty in . figure remains unchanged.
measurement mainly depends upon two factors: 2.3524 is 2.4 to two significant figures. '
(i) Skill and accuracy of the observer,
7.4511 is 7.4 to two significant figures.
(ii) Limitation of the measuring scale.
To indicate the precision of a measurement, scientists use the Calculations Involving Significant Figures
term significant figures. The significant figures in a number are In most of the experiments, the observations of various
all certain digits plus one doubtful digit. The number of measurements are to be combined mathematically, i. e., added,
significant figures gives the information that except the digit at subtracted, multiplied or divided as to achieve the fmal result.
extreme right, all other digits are precise or reproducible. For Since, all the observations in measurements do not have the same
example, mass of an object is 11.24 g. This value indicates that precision, it is natural that the final result cannot be more prec~se
actual mass of the object lies between 11.23 g and 11.25 g. Thus, than the least precise measurement. The following two rules
one is sure of frrst three figures (1, 1 and 2) but the fourth figure should be followed to obtain th~ proper number of significant
is somewhat inexact. The total significant figures in this number figures in any calculation.
are four. Rule 1: The result of an addition or subtraction in the
The following rules are observed in counting the number of numbers having different precisions should be reported to the
significant figures in a given m;asured quantity: same number of decimal places as are present in the number
(i) All non-zero digits are significant. For example, having the least number of decimal places. The rule is illustrated
42.3 has three significant figures. by the following examples:
I
32 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

(a) 33.3 ~(has only one decimal place) posses~ed by the least precise term used in the calculation.
3.11 Examples are:
0.313 (a) 142.06
Sum 36.723 f-, (answer should be reported x 0.23 ~(two significant figures)
to one decimal place) 32.6738 ~ (answer should nave two

Correct answer = 36.7 significant figures)

Correct answer = 33
(b) 3.1421 (b) 51.028
0.241 x 1.31 ~(three significant figures)
0.09 ~(has 2 decimal places) 66.84668
Sum 3.473.1 ~ (answer should be reported to Correct answer = 66.8
2 decimal places) 0.90 = 0.2112676
(c)
Correct answer 3.47 4.26
(c) 62.831 ~(has 3 decimal places) Correct answer 0.21
- 24.5492 Note: (i) Same procedure is followed if an expression involves
multiplication as well as division. .
Difference 38.2818 ~ (answer should be reported
(ii) The presence of exact numbers in an expression does not
to 3 decimal places after affect the number of significant figures in the answer.
rounding off) Examples are:
Correct answer = 38.282 (a) 3.24 x 0.0866 = 0.055643 (b) 4.28 x 0.146 x 3 =44.84784
5.046 0.0418
Rule 2: The answer to a multiplication or division is
rounded off to the same number of significant figures as is Correct answer = 0.0556 Correct answer == 44.8

MISCELLANEOUS NUMERICAL EXAMPLES4'/////

."""" ..
Example 1. 0.44 g of a hydrocarbon on complete Solution: ~ccording to law of conservation of mass,
combustion with oxygen gave 1.8 g water and 0.88 g carbon Mass of lime + Mass of carbon dioxide = Mass of calcium
dioxide. Show that these results are in accordance with the law of carbonate
. conservation of mass.
560 g + Mass of CO 2 = 1000 g
Solution: A hydrocarbon is a compound which consists of
carbon and hydrogen only. It undergoes combustion forming Mass of CO 2 = 1000 - 560 = 440 g
carbon dioxide and water as products. Molecular mass of CO2 = 12 + 32 = 44 g (l mole)
Formula of carbon dioxide = CO 2 ;
Molecular mass 12 + 32 44 g No. of moles in 440 g of CO2 = 440 = 10
. 44
Formula of water H 2 0;
Molecular mass 2+ 16 = 18g 1 mole of CO 2 occupies volume at NTP = 22.4 litre
12 10 moles of CO 2 will occupy volume at NTP
Mass of carbon in 0.88 g of CO2 x 0.88 = 0.24 g
44
= 22.4 x 10·= 224 litre
Mass of hydrogen in 1.8 g of H 2 0 = 2 x 1.8 = 0.20 g Example 3. 10 mL of hydrogen combine with 5 mL of
18
Total masses of carbon and hydrogen in the products oxygen to yield water. When 200 mL of hydrogen at NTP are
0.24 + 0.20 = 0.44 g passed over heated CuO, the CuO loses 0.144 g of its mass. Do
these results correspond to the law ofconstant proportions?
This is equal to the mass of hydrocarbon before combustion.
Thus, the results are in accordance with the law of Solution: 1st Case:
conservation of mass. 2
Mass of 10 mL hytlrogen at NTP = - - x 10= 0.00089 g
Example 2. Calcium carbonate decomposes completely, 22400
on heating, into lime (CaO) and carbon dioxide (C0 2 ), 1 kg of 32
calcium carbonate is completely decomposed by heat, when 560 g Mass of5 mL of oxygen at NTP = - - x 5:::: 0.00714 g
22400
of lime are obtained. How much quantity of carbon dioxide in
grams, moles and litres at NTP is produced in the process?
 BASIC PRINCIPLES I 37

8.923 0.884 X 10- 16 cm 3

Mass of 22400 mLofHg vapouratNl})"" - - x 22400 .
,',,' 1000 ' .
Volume 0.884 x 10- 16
199:87 g' Mass of virus = = -----
Specific volume 0.75
Hence, molecular mass of Hg 199.87g
::: 1.178xlO- 16 g
.,. 6.4
(c) Approxunate atotnlc mass = - - - ,6.4 193.93 Molar maSs of vi~ = Mass of single virus x 6.023 x 1023 .
Sp. heat 0.033 g
= 1.178 X 10- 16 x 6.023 X 1023
Valency of Hg = 193.93:::: 2 (nearest whole number)
100 . = 7.095 xl0 7

So, accurate atomic mass Eq. mass x Valency Example 27. Weighing 31 04 carats (1 carat::: 200 mg), the
Cullinan diamond was the largest natural diamond ever found.
= 100 x 2= 200g How many carbon atoms were present in the stone?
.. Mol. mass 199.88_
A tOI11lClty ==--- 1 Solution: Mass of the stone
At. mass 200
= 3104 x 200= 620800mg::::: 620.8g
Hence, mercury molecules are monoatomic. Number of atoms of carbon
Example 23. How many grams of CaO are required to
Mass in gram x 6.023 x IOZ3
neutralise 852g ofP401O? (lIT 2005)
Gram-atomic mass
Solution: The reaction will be:
6CaO+ P4010 ~2Ca3(P04h 620.8 x 6.023 x 10 23 =3.12 X 1015
12
852 g P40 lO == 3 mol P40 lO Example 28. A cylinder of compressed gas contains
1 mole of P4010 neutralises 6 moles ofCaO. nitrogen and oxygen in the ratio 3: I by mole. If the cylinder is
known to contain 2.5 x 104 g ofoxygen, what is the total mass of
.. 3 moles OfP401O will neutralise 18 moles of.CaO.
the gas mixture?
Mass of CaO= 18;- 56=1008g
Solution: Number of moles of oxygen in the cylinder
. ' Example 24. If 1 grain is equal to 64.8 mg, how many\
Mass in 2.5 x 10 4
moles of aspirin (mol. wt: == 169) are present" in ,'tt 5 grain aspirin \
tablet? . \ Molecular mass in gram 32
Solution: Mass of aspirin in the tablet == 64.8 x 5 324 mg = 781.25
== 0.324g NumberofmoiesofN 2 3x781.25 2343.75
Number of moles = Mass = 0.324 Mass of nitrogen in the cylinder 2343.75 x 28
Molar mass 169
65625g
1.92 x 10-3
= 6.5625 x 104 g
Example 25. If the volume occupied in a crystal by a
Total mass of the gas in the cylinder
molecule of NaCI is 47 x 10-24 mL, calculate the volume of the
crystal weighing Ig. = 2.5 xl 04 + 6.5625 X 104 9.0625 X 104 g
Solution: Number of molecules of NaCI Example 29. Atmospheric air has 78% N z ;21% 02;
Mass x 6.023 x 1023 0.9 % Ar and 0.1 % CO 2 by volume. What is the molecular mass
Molar mass
_1_' x 6.023 x 1023 = 1.03x·1022
ofair in the atmosphere?
Solution:
.
Molecular mass of mixture
58.5
,L %ofeach M 1
- - - - x 0 ar mass
Volume of crystal = L03 x 1022 x 47 X 10-24 = 0.484 mL 100
,Example 26. A plant virus is found to consist of uniform 78 21 0.9 0.1
= x 28+-x 32+-x40+-,x44=28.964
cylindrical particles of 150 A in diameter and 5000 A long. The 100 100 100 100
specific volume ofthe virus is 0.75cm 3 /g. Ifthe virus is considered
Example 30. The famous toothpaste Forhans contains 0.76
to be a single particle, find its molecular mass. (lIT 1999)
g of sodium per gram of sodium monofluoroorthophosphate
Solntion: Volume of cylindrical virus = nrzl

3.14 x C~O x 10- 8

r x 500Qx 10-
8
(Na3P04F)in 100mL.
(a) How many fluorine atoms arepresent?
(b) How much fluorine in milligrams is present?
38 G.R.B.
-
PHYSICAL CHEMISTRY FOR COMPETITIONS

Solution:
Molar mass ofNa3P04F 3 x 23 + 31+ 16 x 4 + 19= 183
183 g Na3P04Fcontains 19 g fluorine
22. x gram of CaC03 was completely burnt in air. The mass of the
.. 0.76g Na3P04Fcontains 19 x 0.76g fluorine solid residue formed is 28 g. What is the value of' x' in gram?
183
(EAMCET 20(5)
:::: 0.0789 g= 78.9 mg fluorine (a) 44 (b) 200 (c) 150 (d) 50
Number of fluorine atoms [Ans. (d)]
_M_ass_in--=-_ _ x 6.023 xl 0 23 [Hint: CaC0 3(s) CaO(s) + CO 2 (g)
lOOg 56g
Gram-atomic mass
56 g residue 100 g CaC0 3
= 0.0789 x 6.023 x 1023
19 :. 28 g residue 50g CaC03]
= 2.5 x 1021 atoms 23. The mass of carbon anode consumed (giving only carbon
dioxide) in the production of270 kg of Al metal from bauxite
t~;txample31. An alloy of iron (54.7%), nickel (45%) and by Hall process is:
manganese (0.3%) has a density of 8.17 g/cm 3 . How many (a) 270 kg (b) 540 kg (c) 90 kg (d) 180 kg
iron atoms are there in a block of alloy measuring [Ans. (c)]
10cmx 20cmx 15 cm? [Hint: 3C + 2Al Z0 3 --,--74Al + 3C02
3x 12g 4X27=108g
Solution:
Volume of the block of alloy = lOx 20x 15cm3 .: 108 g Al is produced by consuming == 36 g carbon
:. 270 x 103 g AI will be produced by consuming
:::: 3000cm 3
,Mass of the block 3000 x 8.l7g ::::2451Og 36 x 270 x 103 g carbon
108
Mass of i,ron in the block
54.7 x 24510; 13406.97 g 90 x 103 g':" 90 kg carbon]
100
24. The equivalent mass of an element is 4. Its chloride has
Number of iron atoms in the block __M_a_ss__ x 6.023x 10 23
Atomic mass vapour density 59.25. Then the valency ofthe element is:
004 003 ~2 WI
13406.97 x 6.023 x 1023 [Ans. (b)]
56
[Hint: Molecular mass of MCI n == 59.25 x 2 = 118.5
:::: 1.442 X 1026
a + 35.5 x n= 118.5 ... (i) .
iiS'Example 32. An analysis of pyrex glass showed 12.9%
B 20 3 , 2.2% A12 0 3 , 3.8% Na20, 0.4% K 20 and remaining is Equivalent mass x n + 35.5 x n =118.5
Si0 2. What is the ratio ofsilicon to boron atoms in the glass ? 4n + 35.5n == 118.5 ... (ii)
'(BCECE 20(7)
n= 3]
Solution:
25. Sulphur trioxideis prepared by the fol1owing two reactions:
Percentage compositio~ ofB z 0 3 12.9%
S8(s) + 802 (g) ~ 8S02~g)
Percentage composition of
SiOz 100 - [12.9 + 2.2 + 3.8 + 0.4] 2S02(g) + 02(g ) ~ 2S03 (g)
How many grams of S03 are produced from 1 mole S8?
80.7%
(a) 1280 (b) 640
,. Mass . 12.9
NumberofmoiesofB 20 3 == =-=0.184 (c) 960 (d) 320
, Molar mass 70 [ADs. (b)]
Number of moles of boron atoms 2 x 0.184 [Hint: From the given reaction, it is clear that, I mole Sg will
give 8 moles of S03'
Number of moles ofSi0 2 Mass = 80.7 = 1.345 :. Mass of S03 formed will be = 80 x 8 = 640 g.]
Molar mass 60 26. Calculate the number of millilitresa.t STP of H2 S gas needed
Number of moles of silicon atoms 1.345 to precipitate cupric sulphide completely from 100 mL of a
Number of atoms of silicon NA x 1.345 :::: 7.3 solution containing 0.75 g ofCuCI 2 in 1 L.
(a) 21.4 (b) 14.2
Number of atoms of boron NA x 0.184 1
(c) 41.2 (d) 124
Where, NA = Avogadro's number [Ans. (d)]
 BASIC PRINCIPLES 39
[Hint: CuCl 2 + H2S ~ CuS + 2HCI the values of x and yare:
(a) 40,40
Number of moles of H2S = Number of moles of CuCl 2
(c) 30,30
=~=~.00557 [Ans. (a)]
134.5
[Hint: In RHS, there are 40 hydrogen atoms, hence only
Volume of H 2S =0.00557 x 22400 =124.8 mL J option (a) will be suitable.]
27. In the reaction,
As 2 SS +xHN0 3 ~5H2S04 + yN0 2 +2H 3 As0 4
+ 12H2 0

SUMMARY AND IMPORTANT POINTS TO REMEMBER

~_. -fa' 111,- • ... ... .... 111,_ ,lilt
1. Chemistry: Branch of physical science which deals properties ofa compound are altogether different from the
with the properties, composition and changes of matter. It properties of elements from which it has been constituted.
has several branches. Main branches are (i) organic 12. Mixture: A material containing two or more substances
(ii) inorganic (iii) physical and (iv) analytical.It is wide in its (elements or compounds) in any proportion, in which
scope and touches- almost every aspect of our lives. components do not lose their identity. Homogeneous mixtMre
2. Matter: It is anything which has mass and occupies space. has a single phase while heterogeneous has more than one
Matter exists in three physical states (i) solid (ii) liquid and phase. Mixture can be separated into components by
(iii) gas. It is chemically classified _into (a) elements physical methods.
(b) compounds and (c) mixtures. 13. Alloy: A homogeneous mixture of two or more
3. Energy: The capacity of doing work. It is of various forms. elements-metal and metal, ni.etal and non-metal or
One form can be converted into another but cannot be created non-metal and non-metal. They have unique properties.
c or de~troyed. The total amount of matter and en~:rgy available 14. Physical change: A temporary: change.mllo change in
in the universe is constant. The relationship between mass and chemical composition and mass. Physical properties alter. It
energy is given by Einstein equation, E = me 2 (where, can be reversed easily.
E energy, m mass, e velocity of light). 15. Chemical change: A permanent change, new substance is
4. Intensive properties: Do not depend on the quantity of formed which possesses different composition and
matter, e.g., colour, density, melting point, boiling point, etc. properties. It cannot be reversed easily. Chemical changes
5. Extensive properties: Depend on the quantity of matter, are of various types. The important ones are decomposition,
e.g., volume, mass, weight, etc. synthesis, substitution, addition, internal rearrangement,
6. Substance: A variety of matter, all samples of which have polymerisation, double decomposition, etc.
the same composition and properties. Pure substances are 16. Law of conservation of mass: (Lavoisier-:-1774) In a
divided into (i) elements and (ii) compounds. chemical change, mass is neither created nor destroyed. In
7. Element: A substance which cannot be decomposed into chemical reactions: ._.
anything more simpler by ordinary physical or chemical Total masses of reactants == Total masses of products.
means. 117 elements are known. 88 elements have been 17. Law of constant proportfons: (Proust-1799) A chemical
isolated from natural sources and remaining 29 have been compound always contains the same element combined
prepared by artificial means. Every element is represented by together in fixed proportion by mass.
a symbol which is a small abbreviation of its full and lengthy 18. Law of multiple proportions: (Dalton-l 808) When two
name. Oxygen is the most abundant element. Silicon, elements combine to form two or more compounds,_ the
aluminium, iron are second, third and fourth most abundant - different masses of one element which combine with a fixed
elements. Elements are classified as (i) metals mass of the other element, bear a simpl~ ratio to one another.
(ii) non-metals and (iii) metalloids. 19. Law of reciprocal proportions: (Richter-1794) When
8. Metals: Generally solids (Hg-exception). They have two different elements combine with the same mass of a
properties such as lustre, hardness, malleable, ductile, good third element, the ratio in which they do so will be the same
conductors of heat ·and electricity. Copper, zmc, iron, or simple multiple ifboth directly combine with each other.
aluminium are metals. In all chemical reactions, substances react in the ratio of their
9. Non-metals: Usually non-lustrous, brittle and poor equivalent mass~~.
conductors of electricity. Oxygen, carbon, nitrogen, 20. Law of gaseous volumes: (Gay-Lussac-1808) Gases
chlorine, helium, etc., are non-metals. react with each other in simple ratio of their volumes and if
10. Metalloids: Possess mixed properties of :metals and product is also in gaseous -state, its volume also bears a
non-metals both (e.g., As, 8b, 8n). simple ratio with the volumes of gaseous reactants under
11. Compound: Pure substance composed of two or more similar conditions of temperature and pressure.
different elements in a fixed proportion of mass. The
40 G.R.B. PHYSICAL CHEMISTRY FOR COMPETITIONS

21. Dalton's atomic theory: Every element is composed of 31. Gram molar volume: The volume occupied by one
small indivisible, indestructible particles called atoms. gram-molecular mass of any gas at NTP (O°C or 273 K and
Atoms of the same element are identical but differ in one atm or 76 em ofHg as pressure). Its value is 22.4 litre.
properties, mass and size. of atoms of other elements. Atoms 32. Vapour density:
of differe1Jt elements combine in. simple ratio to form Density of a gas
compounds. The relative number. and kind of atoms are V.D.
Density of hydrogen
always the same in a given compound. Atoms cannot be
Mass of a certain volume of a gas
created or destroyed.
Mass of same volume of hydrogen
22. Atom: The smallest particle of an element that takes part
in a chemical reaction. under same temperature and pressure
23 •. Molecule: The smallest· particle of· art element or 2Y.D.= Molecular mass
compound that can have a stable existence. 33. Mole: Amole(mol) isdefmed~sthenutnber of atoms in
24. For:mula: Gronp of symbols of elements which represents P,9 g of carbon-12 .. The number Qfatoms..is 6,07X10 23 •
one molecule of a substance. It represents also the chemical This number is called Avogadro's number.
composition. Mass of substan~e in gram
1
N 0.0f moes=--------------------~-------
25. Atomic
.
mass: Atomic mass of an1 element is the·ratio. of·
. Mass of one mole of the substance in gram
mass of one atom of an element.to 12th part of the mass of No. of particles
carbon.,,12. . i'
6.02 X 1023
Atomic mass of an element Volume of gas in Htres at NTP
= Mass of one atom of the element X 12 = 22.4
Mass of one atom of carbon-12 Mass of one atom of an element
26. Atomic mass unit (amu): ~th mass of carbon-12. It is Gram atom o'f an element
= 6.02 x
equal to 1.66 x 10-24 g.
Atomic mass of an el<~ment Mass of one molecule of a substance
Gram-molecular mass of a substance
Mass of one atom of the element
lamu 6.02 x 1023
34. Equivalent mass: The number of parts by mass of the
The actual mass of an atom of element Atomic mass in
substance which combine Or displace directly or indirectly
amu x 1.66 x 10-24 g. 1.008 parts by mass of hydrogen or 8 parts by mass of
The atomic masses of elements are actually average relative oxygen or 35.5 parts by mass of chlorine or 108 parts by
masses because elements occur as mixture of isotopes. mass of silver.
27. Gram-atomic mass or Gram atom: Atomic mass The equivalent mass of an element may vary with change of
expressed in grams. It is the absolute· mass· in grams of valency.
6.02 x 1023 atoms of any element. Eq. mass of an element
Mass of element in Mass of eleme;nt x 1.008
N 0.0f gram atoms = .
. Atomic mass of the element in grams Mass of hydrogen
Mass of element x 11200
28. Molecular mass: It indicates how many times one =-------------------------------
Volume in mL of hydrogen displaced at NTP
molecule of a substance is heavier in comparison to ~th of
. mass of one atom of carbon-12. Mass of a molecule is equal
= Mass of element x 8
Mass of oxygen
to sum of masses of the atoms present in a molecule.
29. Gram-molecular mass· or Gram molecule: Molecular Mass of element x 35.5
mass expressed in gram. It is the absolute mass in gram of Mass of chlorine
6.02 x 1023 molecules of any substance. . . m· E
35. Metal to metal displacement: _1 =_1
No. of gram molecules ~ E2
Mass of a substance in 36. Double decomposition: AB + CD ~ AD + CB
ppl.
. Molecular mass of the substance in gram
Mass of AB massofB
30. Avogadro's hypothesis: Under similar conditions of Mass of AD
temperature and pressure, equal volumes of all gases contain
same number of molecules. . Atomic mass of an element
:: Eq. mass of 1he element x Valency
 I
BASIC PRINCIPLES 41

37. Dulong and Petit's law: Molecular formula mass

n'=--------
Atomic mass (approximate) = ~A< ·J~;mpirical formula mass
. SpeCIfic heat >(iM Structimi1:'It represents the way in which atoms of
38. Cannizzaro's method: Atomic mass of an element is the various elements are linked with each other.
smallest mass of the element present in the molecular mass 42. Percentage of element:
of anyone of its compounds. Mass of element
Percentage of element = x 100
39. Law of isomorphism: Isomorphous compounds form Molecular mass .
crystals wnicnhave same size and shape and can grow in the 43. Chemical equation: It is a symbolic representation of a
saturated solution of each other. chemical change. The equation becomes balanced when total
Masses of two elements that combine with same mass of number of atoms of various elements are made equal on both
other elements in their respective compounds are in the ratio the sides of equation. Chemical equations are of two types
of their atomic masses. (i) molecular and (ii) ionic. Chemical equation is based on
40. Atomic mass from vapour density of a chloride: law of conservation of mass. '
2 V. D. of a volatile chloride 44. Unit: It is the primary standard chosen to measure any
V aIency 0 f an e1ement = - - - - - - - - - -
Eq. mass + 35.5 physical quantity.
41. Types offormulae:. The seven units of measurement, namely mass, length, time,
(i) Empirical: It represents the simplest relative whole temperature, electric current, luminous intensity and amount
number ratio ofatoms of each element present in the of substance are taken as basic units. All other units can be
molecule of a substance. derived from them and are, therefore, called derived units. SI
(il) Molecular: It represents the actual number of atoms of units are used these days in all branches of science.
each element present in one molecule of a substance. 45. Significant ,figure: It is the total number of certain digits .
plus one doubtfu,l digit. .
Molecular formula =n x Empirical formula.