Chapter 1 Some basic concepts of Chemistry (atoms molecules and chemical arithmetic) by TEACHING CARE online tuition and coaching classes

Chapter 1 Some basic concepts of Chemistry (atoms molecules and chemical arithmetic) by TEACHING CARE online tuition and coaching classes


Chemistry is basically an experimental science. In it we study physical and chemical properties of substance and measure it upto possibility. The results of measurement can we reported in two steps : (a) Arithmetic number, (b) Unit of measurement.

Every experimental measurement vary slightly from one another and involves some error or uncertainty depending upon the skill of person making the measurements and measuring instrument. The closeness of the set of values obtained from identical measurement called precision and a related term, refers to the closeness of a single measurement to its true value called accuracy.

In the measured value of a physical quantity, the digits about the correctness of which we are surplus the last digit which is doubtful, are called the significant figures. Number of significant figures in a physical quantity depends upon the least count of the instrument used for its measurement.

  • Common rules for counting significant figures : Following are some of the common rules for counting significant figures in a given expression:

Rule 1. All non zero digits are significant.


Example :

x = 1234

has four significant figures. Again

x = 189

has only three significant figures.


Rule 2. All zeros occurring between two non zero digits are significant.


Example :

x = 1007

has four significant figures. Again

x = 1.0809

has five significant figures.


Rule 3. In a number less than one, all zeros to the right of decimal point and to the left of a non zero digit are not significant.


Example :

x = 0.0084

has only two significant digits. Again,

x = 1.0084 has five significant figures. This is on


account of rule 2.

Rule 4. All zeros on the right of the last non zero digit in the decimal part are significant.


Example :

x = 0.00800

has three significant figures 8, 0, 0. The zeros before 8 are not significant again 1.00


has three significant figures.

Rule 5. All zeros on the right of the non zero digit are not significant.


Example :

x = 1000

has only one significant figure. Again

x = 378000 has three significant figures.


Rule 6. All zeros on the right of the last non zero digit become significant, when they come from a measurement. Example : Suppose distance between two stations is measured to be 3050 m. It has four significant figures.

The same distance can be expressed as 3.050 km or 3.050 ´105 cm . In all these expressions, number of significant figures continues to be four. Thus we conclude that change in the units of measurement of a quantity does not change the number of significant figures. By changing the position of the decimal point, the number of significant

digits in the results does not change. Larger the number of significant figures obtained in a measurement, greater is

the accuracy of the measurement. The reverse is also true.

  • Rounding off : While rounding off measurements, we use the following rules by convention:

Rule 1. If the digit to be dropped is less than 5, then the preceding digit is left unchanged.


Example :

x = 7.82

is rounded off to 7.8, again

x = 3.94

is rounded off to 3.9.


Rule 2. If the digit to be dropped is more than 5, then the preceding digit is raised by one. Example : x = 6.87 is rounded off to 6.9, again x = 12.78 is rounded off to 12.8.

Rule 3. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one. Example : x = 16.351 is rounded off to 16.4, again x = 6.758 is rounded off to 6.8.




Rule 4. If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even. Example : x = 3.250 becomes 3.2 on rounding off, again x = 12.650 becomes 12.6 on rounding off.

Rule 5. If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd. Example : x = 3.750 is rounded off to 3.8. again x = 16.150 is rounded off to 16.2.

(3)  Significant figure in calculation

  • Addition and subtraction : In addition and subtraction the following points should be remembered :
  • Every quantity should be changed into same
  • If a quantity is expressed in the power of 10, then all the quantities should be changed into power of
  • The result obtained after addition or subtraction, the number of figure should be equal to that of least, after decimal

(ii)  Multiplication and division

  • The number of significant figures will be same if any number is multiplied by a
  • The product or division of two significant figures, will contain the significant figures equal to that of

The chosen standard of measurement of a quantity which has essentially the same nature as that of the quantity is called the unit of the quantity. Following are the important types of system for unit,

  • G.S. System :    Length (centimetre), Mass (gram), Time (second)
  • K.S. System :    Length (metre), Mass (kilogram), Time (second)
  • P.S. System :    Length (foot), Mass (pound), Time (second)
  • I. System : The 11th general conference of weights and measures (October 1960) adopted International system of units, popularly known as the SI units. The SI has seven basic units from which all other units are derived called derived units. The standard prefixes which helps to reduce the basic units are now widely used.

Dimensional analysis : The seven basic quantities lead to a number of derived quantities such as pressure, volume, force, density, speed etc. The units for such quantities can be obtained by defining the derived quantity in terms of the base quantities using the base units. For example, speed (velocity) is expressed in distance/time.


So the unit is

ms -2 .

m / s

or ms -1 . The unit of force (mass ´ acceleration) is kg ms -2


Seven basic SI units

and the unit for acceleration is



Length Mass Time Temperature Electric Current Luminous Intensity Amount of substance
metre (m) Kilogram (kg) Second (s) Kelvin (K) Ampere (A) Candela (Cd) Mole (mol)




Derived Units


Physical quantity Unit Symbol
Area square metre m2
Volume cubic metre m3
Velocity metre per second ms–1
Acceleration metre per second square ms–2
Density kilogram per cubic metre kg m–3
Molar mass kilogram per mole kg mol–1
Molar volume cubic metre per mole m3 mol–1
Molar concentration mole per cubic metre mol m–3
Force newton (N) kg m s–2
Pressure pascal (Pa) N m–2
Energy work joule (J) kg m2 s–2, Nm

Standard prefixes use to reduce the basic units


Multiple Prefix Symbol Submultiple Prefix Symbol
1024 yotta Y 10–1 deci d
1021 zetta Z 10–2 centi c
1018 exa E 10–3 milli m
1015 peta P 10–6 micro m
1012 tera T 10–9 nano n
109 giga G 10–12 pico p
106 mega M 10–15 femto f
103 kilo k 10–18 atto a
102 hecto h 10–21 zeto z
101 deca da 10–24 yocto y




Conversion factors


1 m = 39.37 inch 1 cal = 4.184 J 1 e.s.u. = 3.3356 ´ 10–10 C 1 mole of a gas = 22.4 L at STP
1 inch = 2.54 cm 1 eV = 1.602 ´ 10–19 J 1 dyne = 10–5 N 1 mole a substance = N0 molecules
1 litre = 1000 mL 1 eV/atom =96.5 kJ mol–1 1 atm = 101325 Pa 1 g atom = N0 atoms
1 gallon (US) = 3.79 L 1 amu = 931.5016 MeV 1 bar = 1 ´ 105 N m–2 t (oF) = 9/5 t (oC) + 32
1 lb = 453.59237 g 1 kilo watt hour = 3600 kJ 1 litre atm = 101.3 J 1 g cm–3 = 1000 kg m–3
1 newton =1 kg m s–2 1 horse power = 746 watt 1 year = 3.1536 ´ 107 s 1Å = 10–10 m
1 J = 1 Nm =1 kg m2 s–2 1 joule = 107 erg 1 debye (D) = 1 ´ 10 –18 esu cm 1nm = 10–9 m




Matter is the physical material of the universe which occupies space and has mass e.g., water, sugar, metals, plants etc. Matter can be classified as,



Everything that has mass and occupies space

Physical classification                                                                         Chemical classification


Solids          Liquids          Gases                                   MIXTURES                                                 PURE SUBSTANCES


  • Variable composition
  • Components retain their characteristic properties
  • May be separated into pure components by physical methods
  • Mixtures of different composition may have widely different properties
  • Fixed composition
  • Cannot be separated into simpler substances by physical methods
  • Properties do not vary




Homogeneous              Hetrogeneous                  Elements                     Compounds


  • Have same composition throughout
  • Components are indistin-guishable for example : a gaseous mixture or a liquid solution
  • Do not have same composition throughout
  • Components are distin- guishable for example carbon and sulphur mixture (gun powder)
  • Can not be decom- posed into simpler substances by chemical changes
  • Can be decomposed into simpler substances by chemical changes, always at constant composition



Inorganic          Organic



Metals            Metalloids            Non-metals


Each component of a mixture retains its own properties and thus a mixture can be separated into its components by taking advantages of the difference in their physical and chemical properties. The different methods which are employed to separate the constituents from a mixture to purify an impure sample of a substance are,

  • Sedimentation : It is also called gravity It is used for a mixture in which one component is a liquid and the other is insoluble solid heavier than the liquid. Example : Sand dispersed in water.
  • Filtration : It is used for a mixture containing two components one of which is soluble in a particular solvent and the other is not. Example : (i) A mixture of salt and paper using water as solvent (ii) A mixture of sand and sulphur using carbon disulphide as (iii) A mixture of glass powder and sugar, using water as a solvent in which sugar dissolves but glass does not. (iv) A mixture of sand and sulphur, using carbon disulphide as the solvent in which sulphur dissolves but sand does not.
  • Sublimation : It is used for a mixture containing a solid component, which sublimes on heating from


non-volatile solids. Example : A mixture of sand + naphthalene or powdered glass +


/ camphor / iodine


etc. can be separated by the method of sublimation because substances like naphthalene, etc. form sublimates whereas sand, glass etc. do not.

NH 4 Cl , iodine, camphor




  • Evaporation : It is used for a mixture in which one component is a non–volatile soluble salt and other is a Example : Sodium chloride dissolved in sea–water.
  • Crystallization : It is a most common method for a mixture containing solid components and based upon the differences in the solubilities of the components of the mixture into a For separation, a suitable solvent is first selected. It is of two types :
    • Simple crystallization : It is applicable when there is a large difference in the solubilities of different components of a
    • Fractional crystallization : It is applicable when there is a small difference in the solubilities of different


components of a mixture in the same solvent. Example :


and KCl . Here

K 2 Cr2 O7

is less soluble in


water and hence crystallizes first. A series of repeated crystallization separates the two components in pure form.

  • Distillation : It is the most important method for purifying the liquids. It involves the conversion of a liquid to its vapours on heating (vaporisation) and then cooling the vapours back into the liquid (condensation). It can be used to separate, (i) A solution of a solid in a e.g., aqueous copper sulphate solution. (ii) A solution of two liquids whose boiling points are different. Several methods of distillation are employed.
    • Simple distillation : It is used only for such liquids which boil without decomposition at atmospheric pressure and contain non–volatile impurities. Example : (a) Pure water from saline water. (b) Benzene from
    • Fractional distillation : It is used for the separation and purification of a mixture of two or more miscible liquids having different boiling The liquid having low boiling point vaporises first, gets condensed and is collected in the receiver. The temperature is then raised to the boiling point of second so that the second liquid vaporises and is collected in other receiver. If two liquids present in a mixture have their boiling points closer to each other, a fractionating column is used. Example : (a) Crude petroleum is separated into many useful products such as lubricating oil, diesel oil, kerosene and petrol by fractional distillation. (b) A mixture of acetone and methyl alcohol.
    • Vacuum distillation or distillation under reduced pressure : It is used for such liquids which decompose on heating to their boiling points. At reduced pressure, the boiling point of liquid is also reduced. Example : Glycerol is distilled under pressure as it decomposes on heating to its boiling point.
    • Steam distillation : It is used for liquids which are partially miscible with water, volatile in e.g., aniline, oils etc. are purified by steam distillation. The principle involved is of reduced pressure distillation. If Pw and


Pl are vapour pressures of water and liquid at distillation temperature, then

Pw + Pl = P = 76 cm

i.e., atmospheric


pressure. Thus, a liquid boils at relatively low temperature than its boiling point in presence of steam.

Example :    Some solids   like   naphthalene,   o-nitrophenol   which   are   steam   volatile   can   be   purified.

Nitrobenzene, chlorobenzene, essential oils are also extracted or separated by this process

  • Solvent extraction : It is used for the separation of a compound from its solution by shaking with a suitable The extraction follows Nernst distribution law. The solvent used must be insoluble with other phase in which compound is present as well as the compound should be more soluble in solvent. The extraction becomes more efficient if the given extracting liquid is used for more number of extractions with smaller amounts than using it once with all extracting liquid. Example : (i) Aqueous solution of benzoic acid by using benzene. (ii) Aqueous solution of Iodine by using chloroform or carbon tetrachloride. (iii) Flavour of tea from the tea leaves by boiling with water.




  • Magnetic separation : It is used for a mixture in which one component is magnetic while the other is non–magnetic. Example : iron ore from the non–magnetic
  • Chromatography : It is based on the differences in the rates at which different components of a mixture are absorbed on a suitable solvent. It is used in separation, isolation, purification and identification of a substance. It was proposed by a Russian botanist
  • Atmolysis : It is used for separation of the mixture of gases or It is based on the difference in


their rates of diffusion through a porous substance. Example : their hexa–fluorides.

H 2 , SO2 , CH4

and O , U 235 & U 238 in the form of


  • Electrophoresis : It is based upon the differences in the electrical mobility of the different components of the
  • Ultracentrifugation : It is based upon the difference in sedimentation velocity of the components in a centrifugal

Various chemical reactions take place according to the certain laws, known as the Laws of chemical combination. These are as follows,

  • Law of conservation of mass : It was proposed by Lavoisier and verified by Landolt. According to this law, Matter is neither created nor destroyed in the course of chemical reaction though it may change from one form to other. The total mass of materials after a chemical reaction is same as the total mass before


Example : A reaction between


solution and KI solution.


AgNO3 (aq)

  • KI(aq)

¾¾®  AgI + NaNO3 (aq)  (yellow ppt.)


Mass of

AgNO3 (aq)

  • Mass of KI(aq)

= Mass of the ppt. of AgI

  • Mass of NaNO3 (aq)


According to the modified statement of the law, The total sum of mass and energy of the system remains constant.

  • Law of constant or definite proportion : It was proposed by Proust. According to this law, A pure chemical compound always contains the same elements combined together in the fixed ratio of their weights whatever its methods of preparation may


Example :


can be formed by either of the following processes:


  • By heating CaCO3  :

Ca CO3

¾¾D ®

Ca O + CO2


  • By heating

NaHCO3    :

2 NaHCO3

¾¾D ®  Na   CO   + H  O  +  CO


2         3          2                  2


is collected separately as a product of each reaction and the analysis of


of each collection reveals


that it has the combination ratio of carbon and oxygen as 12 : 32 by weight.

  • Law of multiple proportion : It was proposed by Dalton and verified by Berzelius. According to this law, When two elements A and B combine to form more than one chemical compounds then different weights of A, which combine with a fixed weight of B, are in proportion of simple whole




Example : Nitrogen forms as many as five stable oxides. The analysis of these oxides (N 2 O, NO, N 2 O3 , N 2 O4



N 2 O5 ) reveals that for 28 gm. nitrogen, the weight of oxygen that combines is in the ratio 16 : 32 : 48 : 64 :


80 i.e., 1 : 2 : 3 : 4 : 5 in

N 2 O, NO, N 2 O3 , N 2 O4  and

N 2 O5 respectively.


  • Law of equivalent proportion or law of reciprocal proportion : It was proposed by Ritcher. According to this law, The weights of the two or more elements which separately react with same weight of a third element are also the weights of these elements which react with each other or in simple multiple of


Example : Formation of

H 2 S, H 2 O

and SO2

can be done as follows,


  • Hydrogen combines with sulphur forming hydrogen sulphide; 2 of hydrogen reacts with 32gm of sulphur. (ii) Hydrogen combines oxygen forming water; 2 gm. of hydrogen reacts with 16 gm. of oxygen. (iii) Sulphur combines with oxygen forming sulphur dioxide; 32 gm. of sulphur reacts with 32 gm. of oxygen i.e., in the ratio 32 : 32. This ratio is double of the ratio weights of these elements which combine with 2 gm. of hydrogen. i.e., 32/16 : 32/32 = 2 : 1

Law of Reciprocal proportion can be used to obtain equivalent weights of elements. As elements always combine in terms of their equivalent weights.

  • Gay-Lussac’s Law: It was proposed by Gay–Lussac and is applicable only for gases. According to this law, When gases combine, they do so in volumes, which bear a simple ratio to each other and also to the product

formed provided all gases are measured under similar conditions. The Gay-Lussac’s law, was based on experimental observation.


Example : (i) Reaction between hydrogen and oxygen.

H 2 (g )

  • 1 O



2 (g )

¾¾® HO




One volume of

H2 reacts with half volume of O2 to give one volume of

H 2 O .


(ii) Reaction between nitrogen and hydrogen.

N 2 (g )

+ 3 H 2 (g )

® 2 NH 3 (g )


One volume of

N2 reacts with three volumes of

H 2 to give two volumes of

NH 3 .



  • Atomic hypothesis : Keeping in view various law of chemical combinations, a theoretical proof for the validity of different laws was given by John Dalton in the form of hypothesis called Dalton’s atomic hypothesis. Postulates of Dalton’s hypothesis is as followes,
    • Each element is composed of extremely small particles called atoms which can take part in chemical
    • All atoms of a given element are identical e., atoms of a particular element are all alike but differ from atoms of other element.
    • Atoms of different elements possess different properties (including different masses).
    • Atoms are indestructible e., atoms are neither created nor destroyed in chemical reactions.
    • Atoms of elements take part to form molecules e., compounds are formed when atoms of more than one element combine.
    • In a given compound, the relative number and kinds of atoms are




  • Modern atomic hypothesis : The main modifications made in Dalton’s hypothesis as a result of new discoveries about atoms are,
  • Atom is no longer considered to be
  • Atoms of the same element may have different atomic e.g., isotopes of oxygen O16 , O17 , and O18 .
  • Atoms of different element may have same atomic e.g., isobars Ca 40 and Ar 40 .
  • Atom is no longer In many nuclear reactions, a certain mass of the nucleus is converted into energy in the form of a, b and g rays.
  • Atoms may not always combine in simple whole number ratios. g., in sucrose (C12 H 22 O11 ) , the elements carbon, hydrogen and oxygen are present in the ratio of 12 : 22 : 11 and the ratio is not a simple whole number ratio.
  • Berzelius hypothesis : “Equal volumes of all gases contain equal number of atoms under same conditions of temperature and pressure”. When applied to law of combining volumes, this hypothesis predicts that atoms are divisible and hence it is contrary to Dalton’s
  • Avogadro’s hypothesis : “Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of ” Avogadro hypothesis has been found to explain as follows :
  • Provides a method to determine the atomic weight of gaseous
  • Provides a relationship between vapour density (V.D.) and molecular masses of substances. Vapour density =Volume of definite amount of a gas

Volume of same amount of hydrogen

or   Vapour denstiy =    Mass of ‘n’ molecule of a gas

Mass of ‘n’ molecule of hydrogen



Vapour density =



Mass of 1 molecule of a gas Mass of 1 molecule of hydrogen


Molecular mass = 2 ´ vapour density




  • It helps in the determination of mass of fixed volume of a particular



Vapour density =

Mass of 1 molecule of gas =

Mass of 1 molecule of H 2

Mass of 1 ml of gas =

Mass of 1 ml of H 2

Mass of 1 ml of gas



( 1 ml

H 2 = 0.0000897 gm.) at NTP


\      Mass of 1 ml gas = V.D. ´ 0.0000897 gm.


  • It also helps in the determination of molar volume at T.P.

 V.D. ´ 0.0000897 gm. gas has volume = 1 ml


\ 2 ´ V.D.(i.e., molecular mass) gm. has volume =



1 ´ 2 ´ V.D.



ml = 22400 ml


V.D. ´ 0.0000897

\ Molar mass of a gas or its 1 mole occupies 22.4 L volume at S.T.P.




Note : ®           22.4 litres of any gas at S.T.P. weigh equal to the molecular weight of the gas expressed in grams. This is called Gram-molecular volume (G.M.V.) law.

  • It helps in determination of molecular formulae of gases and is very useful in gas By knowing the molecular volumes of reactants and products of reaction, molecular composition can be determined easily.


  • Atomic Mass : It is the average relative mass of atom of element as compared with an atom of carbon – 12 isotope taken as

Average atomic mass : If an elements exists in two isotopes having atomic masses ‘a’ and ‘b’ in the ratio m :

n, then average atomic mass = (m ´ a+ (n ´ b) . Since the atomic mass is a ratio, it has no units and is expressed in amu,

m + n


1 amu = 1.66 ´ 10 -24 g . One atomic mass unit (amu) is equal to

th of the mass of an atom of carbon-12 isotope.



Gram atomic mass (GAM) : Atomic mass of an element expressed in grams is called Gram atomic mass or gram atom or mole atom.

  • Number of gram atoms or mole atoms = Mass of an element


  • Mass of an element in = Number of gm. atom ´ GAM
  • Number of atoms in 1 GAM = 02 ´ 10 23


\     Number of atoms in a given substance = No. of GAM ´ 6.02 ´ 10 23 =

Mass   ´ 6.02 ´ 1023



  • Number of atoms in 1 gm of element =

6.02´10 23


Atomic mass


  • Mass of one atom of the element (in ) = GAM   

6.02 ´10 23

Methods of determination of atomic mass

  • Dulong and Pettit’s method : According to Dulong and Pettit’s law Atomic mass ´ Specific heat = 4 (approx.)


Atomic mass (approx. ) =



Specific heat (in cals.)


This law is applicable to solid elements only except Be, B, C and Si.

Atomic mass = Equivalent mass ´ Valency

  • Vapour density method : It is suitable for elements whose chlorides are


Valency of the element = Molecular mass of chloride =

Equivalent mass of chloride

2 ´ Vapour density of chloride Equivalent mass of metal + 35.5




  • Specific heat method : It is suitable only for The two types of specific heats of gases are CP (at

constant pressure) and Cv     (at constant volume). Their ratio is known as g whose value is constant (1.66 for

monoatomic, 1.40 for diatomic and 1.33 for triatomic gases).

  • Volatile chloride method : Different steps are given below

Step I. The element (M) whose atomic mass is to be determined is converted into its volatile chloride whose vapour density is determined by Victor Meyer method.

Thus, Molecular mass of the chloride = 2 ´ V. D.

Step II. Equivalent mass of the element (M) of valency X is determined as usual.

Atomic mass of the element = Equivalent mass of the element ´ X = Z ´ X Step III. The formula of the volatile chloride is derived as below,

M X                              Cl1





Step IV. Molecular mass of the chloride = (Z ´ X)

Cl X


(35.5 ´ X) = X(Z + 35.5)



From (I) and (IV)

X (Z + 35.5) = 2´ Vapour density or

X =   2 ´ V.D.

Z + 35.5


Thus, Atomic mass of the element = Z ´ X

  • Isomorphism method : It is based on law of isomorphism which states that compounds having identical crystal structure have similar constitution and chemical

Example : K 2 SO4 , K 2 CrO 4 and K 2 SeO4 (valency of S, Cr, Se = 6),

ZnSO4 .7 H 2 O, MgSO4 .7 H 2 O, FeSO4 .7 H 2 O (valency of Zn, Mg, Fe = 2).

Applications of isomorphism

  • The valencies of two elements forming isomorphism salts are essentially Thus if valency of one of the elements is known that of other will be the same.
  • Masses of two elements, that combine with the same mass of other elements in their respective isomorphous compounds, are in the ratio of their atomic masses e.,


Mass of element A that combines with a certain mass of other elements Mass of element B that combines with the same mass of other elements

= Atomic mass of A

Atomic mass of B


  • By knowing the percentage of two elements of their isomorphous compound and atomic mass of one element, atomic mass of other element can be
    • Molecular mass : Molecular mass of a molecule, of an element or a compound may be defined as a

number which indicates how many times heavier is a molecule of that element or compound as compared with

1 of the mass of an atom of carbon–12. Molecular mass is a ratio and hence has no units. It is expressed in a.m.u.





Molecular mass = Mass of one molecule of the substance
1 / 12 ´ Mass of one atom of C – 12

Actual mass of one molecule = Molecular mass ´1.66 ´10 -24 gm.

Molecular mass of a substances is the additive property and can be calculated by adding the atomic masses present in one molecule.

Gram molecular mass (GMM) and Gram molar volume : Molecular mass of an element or compound when expressed in gm. is called its gram molecular mass, gram molecule or mole molecule.

Number of gm molecules or mole molecules = Mass of substances


Mass of substances in gm = Number of gm. molecules ´ GMM

Volume occupied by one mole of any gas at STP is called Gram molar volume. The value of gram molar volume is 22.4 litres. Volume of 1 mole of any gas at STP = 22.4 litres

Expression for mass and density

Mass of 11.2L of any gas at STP = V.D. of that gas in gm.



Density of a gas at NTP =

Mol. mass in gm.

22400 ml


Important generalisations

Number of atoms in a substance = Number of GMM ´ 6.02 ´ 1023 ´ Atomaticity

Number of electrons in given substance = Number of GMM ´ 6.02 ´ 1023 ´ Number of electrons

Methods of determination of molecular mass : following methods are used to determine molecular mass

  • Diffusion method (For gases) : The ratio of rates of diffusion of two gases is inversely proportional to the square root of their molecular
  • Vapour density method (For gases only) : Mass of a fixed volume of the vapour is compared with the mass of the same volume of hydrogen under same conditions. The ratio of these masses is called Vapour density or Relative density.



Note : ®           Vapour density of a substance increases due to molecular association (e.g.,



and decreases due to dissociation (e.g.,

NH 4 Cl,

PCl5 , etc.)


  • Victor Meyer method (For volatile liquids or solids) : It is based on Dalton’s law of partial pressure and Avogadro’s hypothesis (gram molar volume).

22400 ml of vapours of a substance = Molecular mass of that substance

  • Colligative property method (For non-volatile solids) : Discussed in colligative properties of

Average atomic mass and molecular mass




A (Average atomic mass) = å Ai Xi

å X total

; M (Average molecular mass) =

å Mi Xi

å X total



A1 , A2 , A3 ,…. are atomic mass of species 1, 2, 3,…. etc. with % ratio as X 1 , X 2 , X3………………….. etc. Similar terms


are for molecular masses.


  • Equivalent mass : The number of parts by mass of a substance that combines with or displaces 1.008 parts by mass of hydrogen or 8.0 parts of oxygen or 35.5 parts of chlorine is called its equivalent mass (EM). On the other hand quantity of a substance in grams numerically equal to its equivalent mass is called its gram equivalent mass (GEM) or gram equivalent.

Expressions for equivalent mass (EM)

  • EM of an element = Atomic mass


  • EM of an acid = Molecularmass


(Basicity of acid is the number of replaceable hydrogen atoms in one molecule of the acid).

  • EM of a base = Molecular mass


(Acidity of a base is the number of replaceable– OH groups in one molecule of the base).

  • EM of a salt = Formula mass            

Total positive or negative charge



  • EM of a reducing agent =

Formula mass


Number of electrons lost per molecule or Total change in oxidation number


Example : Eq. mass of oxalic acid (C2 H 2 O4 )

2     4

C  O  2-   ¾¾®  2 CO   + 2 e


(O.N. of

C = + 3)

(O.N. of

C = + 4)


\ Number of electrons lost = 2

or Change in O.N. per C atom (from + 3 to + 4) = 1

\ Total change in O.N. = 1 ´ 2 = 2



\ Equivalent mass of C2

H 2 O4

= 90 = 45



  • EM of an oxidising agent = Formula mass                                

Number of electrons gained per molecule or Total change in O.N.

Equivalent mass of common oxidising agent changes with the medium of the reaction.


Example : (a) Equivalent mass of

KMnO4 in acidic medium




4                                                                       2

MnO  –   + 8 H  + + 5 e –   ¾¾® Mn2 +   +  4 H  O + 5 e

Change in O.N. of Mn from + 7 to + 2 = + 5



\ Equivalent mass of


= 158 = 31.60



  • Equivalent mass of KMnO4

in neutral medium


4                                                        2             2

MnO  –  + 4 H +  + 3 e –   ¾¾® MnO   + 2 H  O  or    KMn O   ¾¾® Mn O

4                           2



\ Equivalent mass of


= 158 = 52.67



  • Equivalent mass of KMnO4

in basic medium



– + e – ¾¾® MnO 2-

or  KMnO4

¾¾® K 2




Equivalent mass of KMnO4


  • EM of radicals =Formula mass of radical  

Number of units of charge

= 158 = 158




Example : Equivalent mass of SO 2- = 96


= 48


Methods of determination of equivalent mass

  • Hydrogen displacement method : The mass of metal which displaces 11200 ml of hydrogen at NTP from an acid, alkali or alcohol is the equivalent mass of the
    • Equivalent mass of metal = Mass of metal      ´008 = W ´ 1.008g


Mass of H 2 displaced

  • Equivalent mass of metal = Mass of metal           

Vol.(ml) of H 2 displaced at STP



´11200 =

W ´11200



This method is useful for metals which can displace hydrogen from acids or can combine with hydrogen

(Mg, Zn, Na, Ca etc.)

  • Oxide formation method : The mass of the element which combines with 8 grams of oxygen is the equivalent mass of the
    • Equivalent mass of metal = Mass of metal´ 8

Mass of oxygen

  • Equivalent mass of metal = Mass of metal        ´ 5600

Vol. of O2 at S.T.P. in ml

  • Chloride formation method : The mass of an element which reacts with 5 gm. of chlorine is the equivalent mass of that element.


  • Equivalent mass of metal =

Mass of metal Mass of chlorine

´ 35.5


  • Equivalent mass of metal

=          Mass of metal

Vol. of Cl 2 (in ml.) at STP





  • Neutralisation method : (For acids and bases).

Equivalent mass of acid (or base) =  W          , Where W = Mass of acid or base in gm., V = Vol. of base or

V ´ N

acid in litre required for neutralisation and N is Normality of base or acid

  • Metal displacement method : It is based on the fact that one equivalent of a more electropositive metal displaces one gm equivalent of a less electropositive metal from its salt solution.


Mass of metal added Mass of metal displaced

= Eq. mass of metal added     ;

Eq. mass of metal displaced

W1  = E1

W2         E2


  • Electrolytic method : The quantity of substance that reacts at electrode when 1 faraday of electricity is passed is equal to its gram equivalent

Gram equivalent mass = Electrochemical equivalent ´ 96500

W1  = E1
W2 E2


The ratio of masses of two metals deposited by the same quantity of electricity will be in the ratio of their equivalent masses.

  • Double decomposition method : AB  +  CD ¾¾® AD  ¯ +  CB

Mass of compound ABEq. mass of A + Eq. mass of B

Mass of compound AD   Eq. mass of A + Eq. mass of D


or    Mass of salt taken (W1 )    =

Mass of ppt. obtained (W2 )

Eq. mass of salt (E1 ) Eq. mass of salt in ppt.(E2 )


  • Conversion method : When one compound of a metal is converted to another compound of the same metal, then


Mass of compound I (W1 ) = E + Eq. mass of radical I

(E = Eq. mass of the metal)


Mass of compound II (W2 )      E + Eq. mass of radical II

(ix)  Volatile chloride method



Valency of metal =

2´ V.D. of Chloride     =

Eq. mass of metal chloride

2´ V.D. E + 35.5

E = 2´ V.D. of Chloride – 35.5 Valency


  • Silver salt method (For organic acids)

Equivalent Mass of acid = 108 ´ Mass of silver salt

Mass of Ag metal


– 107



Molecular mass of acid = Equivalent mass of acid ´ Basicity


Example: 1        Oxygen contains 90% O16 and 10% O18 . Its atomic mass is                                                             [KCET 1998]

(a) 17.4                        (b) 16.2                              (c) 16.5                        (d) 17


Solution: (b)       Average atomic mass of oxygen = 90 ´ 16 + 10 ´ 18


= 16.20


Example: 2        The total number of electrons present in 1.6 gm. of methane is                      [IIT 1976; Roorkee 1985; CPMT  1987, 92]






6.02 ´ 1023


6.02 ´ 10 22

6.02 ´ 1021

4.02 ´ 1020


Solution: (a)       Number of GMM of methane in 1.6 gm. of methane


=       Mass of methane       =  1.6

= 0.1


Mol. mass of methane           16

Number of electrons in 1 molecule of methane (CH4 ) = 6 + 4 = 10 Hence, Number of electrons in 1.6 gm. of methane

= Number of GMM ´ 6.02 ´ 1023 ´ No. of electrons


Example: 3

= 0.1 ´ 6.02 ´ 1023 ´ 10 = 6.02 ´ 1023

KClO3 on heating decomposes to KCl and O2 . The volume of O2 at STP liberated by 0.1 mole KClO3 is [BIT 1991]

  • 36 L (b) 3.36 L                                    (c) 2.36 L                            (d) none of these


Solution: (b)       On heating KClO3

dissociates as:


2 KClO3

2 moles

¾¾D ®  2 KCl + 3 O2

3 ´ 22.4 L at STP


   2 moles of KClO3 on heating produces = 67. 2 L of O2 at STP



 0.1 mole of

KClO3 on heating produces = 67.2 ´ 0.1 L = 3.36 L of O2 at STP



Example: 4         At S.T.P. the density of CCl 4

vapour in g/L will be nearest to                                                  [CBSE PMT 1988]


(a) 6.84                          (b) 3.42                       (c) 10.26                         (d) 4.57


Solution: (a)       1 mole of CCl 4

vapour = 12 + 4 ´ 35.5

= 154 gm = 22.4 L at S.T.P.


\ Density = 154 gmL1 = 6.875 gmL1


Example:5          4.4 g of an unknown gas occupies 2.24 litres of volume at NTP. The gas may be                                   [MP PMT 1995]

  • Carbon dioxide (b) Carbon monoxide           (c) Oxygen                   (d) Sulphur dioxide

Solution: (a)       Mass of 2.24 litres gas = 4.4 gm.

 Mass of 22.4 litres gas = 44 gm.

Here, out of four given gases, the molecular mass of only carbon dioxide is 44 gm.

Example: 6        One gm. of a metal carbonate gave 0.56 gm. of its oxide on heating. The equivalent mass of the metal is (a) 40  (b) 30                                 (c)  20                                 (d) 10


Mass of metal carbonate


E + Eq. mass of CO3 2

1.0     E + 30


Solution: (c)

Mass of metal oxide

=                                        2-

E + Eq. mass of O2

or           =

0.56     E + 8


or (E + 8) ´ 1.0 = (E + 30)(0.56)           (Where E is the equivalent mass of the metal)

\            E = 20

The mole (abbreviated as mol) is the SI base unit for a amount of a chemical species. It is always associated with a chemical formula and refers to Avogadro’s number ( 6.022 ´ 1023 ) of particles represented by the formula. It


is designated as

N0 . Thus, 12 molecules of HCl is a dozen, 144 molecules of HCl is a gross and

6.022 ´ 1023


molecules of HCl is a mole.




1 mole of a substance = 6.022 ´ 1023 species

The molar mass of a substance is the mass in grams of 1 mole of that substance.



Mole of a substance =

mass in grams molar mass


Under STP conditions when temperature is 273K and pressure is 1 atm, volume of one mole of an ideal gas is 22.4L



Example: 7 The number of gram molecules of oxygen in 6.02 ´ 1024 CO molecules is                                               [IIT 1990]




Solution: (b)

  • 10 gm molecules (b) 5 gm molecules        (c) 1 gm molecules           (d) 5 gm molecules

6.02 ´ 1023 molecules = 1 mole of CO

\ 6.02 ´ 1024 CO molecules = 10 moles of CO

= 10 gms of Oxygen atom =5 gm molecules of O2


Example: 8        1 c.c of


at NTP contains :                                                                                        [CBSE PMT 1988]






1.8 ´ 10 22 atoms                                                (b)


  6.02 ´ 10 23 molecules





1.32 ´ 10 23 electrons                                            (d) All the above



Solution: (d) 22400 c.c. = 6.02 ´ 10 23 molecules

6.02 ´ 10 23


1 c.c. of N 2O =


3 ´ 6.02 ´ 10 23






= 6.02 ´ 10 23


atoms (Since N 2 O has three atoms)


  • 22 electron (Because number of electrons in N 2 O are 22)


Example: 9 The mass of carbon present in 0.5 mole of K4 [Fe(CN)6 ] is




1.8 g

18 g

3.6 g

36 g


Solution: (d) 1 mole of K4 [Fe(CN)6 ] = 6 gm atoms of carbon

0.5 mole of K4 [Fe(CN)6 ] = 3 gm atoms of carbon

= 3 ´ 12 = 36 g

Example: 10      The number of moles of oxygen in one litre of air containing 21% oxygen by volume under standard conditions is                                                                                                                   [CBSE PMT 1995]

(a) 0.186 mole              (b) 0.21 mole                      (c) 0.0093 mole                   (d) 2.10 mole

Solution: (c)        100 ml of air at STP contains 21 ml of O2.

\ 1000 ml of air at STP contains 210 ml of O2.





\ No. of moles of O2 =


Vol. of O2 in litres under


STP conditions

= 210 / 1000 = 21  

= 0.0093 mole


22.4 litre

22.4       2240


Example: 11      The number of moles of BaCO3

which contains 1.5 moles of oxygen atoms is                            [EAMCET   1991]


(a) 0.5                          (b) 1                                  (c)  3                                  (d) 6.02 ´ 1023


Solution: (a)        1 mole of

BaCO3 contains 3 moles of oxygen atoms.


\ 1 mole (0.5) of


BaCO3 contains 1.5 moles of oxygen atoms.



  • Percentage composition of a compound : Percentage composition of the compound is the relative mass of each of the constituent element in 100 parts of If the molecular mass of a compound is M and B is the mass of an element in the molecule, then
  • Empirical formula : The chemical formula that gives the simplest whole number ratio between the atoms of various elements present in one molecule of the compound is known as empirical formula (simplest formula).


For example, the empirical formula of glucose is CH 2 O

1 : 2 : 1 in a molecule of glucose.

which shows that C, H and O are present in the ratio


Empirical formula mass of a compound is equal to the sum of atomic masses of all atoms present in the empirical formula of the compound.

Calculation of the empirical formula : The empirical formula of a chemical compound can be deduced by knowledge of the, (i) Percentage composition of different elements. (ii) Atomic masses of the elements.

The following steps are involved in the calculation of the empirical formula,

Step I.       Calculate the relative number of atoms or atomic ratio.



Atomic ratio =

Percentage of an element Atomic mass of the same element


Step II.      Calculate the simplest atomic ratio.

Step III.    Calculate the simple whole number ratio.

Step IV. Write the empirical formula.

  • Molecular formula : The chemical formula that gives the actual number of atoms of various elements present in one molecule of the For example, the molecular formula of glucose is C6 H12 O6 .

Relation between empirical and molecular formula : The molecular formula of a compound is a simple whole number multiple of its empirical formula.

Molecular formula = n ´Empirical formula ;         Where n is any integer such as 1, 2, 3….etc.




The value of ‘n’ can be obtained from the following relation:

n =      Molecular mass      .

Empirical formula mass




The molecular mass of a volatile compound can be determined by Victor Meyer’s method or by employing

Molecular mass = 2 ´ Vapour density

the relation,                                                                                   .

Calculation of the molecular formula : The molecular formula of a compound can be deduced from its :

(i) Empirical formula, (ii) Molecular mass

The determination of molecular formula involves the following steps:

Setp I.  Calculation of empirical formula from the percentage composition.

Setp II. Calculation of empirical formula mass.

Setp III. Calculation of the value of ‘n’.

Setp IV. Calculation of molecular formula by multiplying the empirical formula of the compound by ‘n’.


Example:12        The oxide of a metal contains 40% by mass of oxygen. The percentage of chlorine in the chloride of the metal is [BIT Ranchi 1997]

(a) 84.7                        (b) 74.7                              (c) 64.7                              (d) 44.7


Solution: (b)     % of oxygen =    8   ´ 100 = 40

m + 8


40(m + 8)= 800


m + 8 = 20


m =12


\ % of chlorine =    35.5   ´100 =

m + 35.5



12 + 35.5

´ 100 = 74.7 (Where m is the atomic mass of metal)


Example: 13 The empirical formula of an organic compound containing carbon and hydrogen is

CH 2 . The mass of one


litre of this organic gas is exactly equal to that of one litre of

N 2 . Therefore, the molecular formula of the


organic gas is                                                                                                                     [EAMCET 1985]



CH 4


CH 6


C6 H12


C4 H 8


Solution: (a)      Molar mass of 1L of gas = mass of 1L N 2

\ Molecular masses will be equal i.e., molecular mass of the gas = 28, hence formula is C2 H 4

Example: 14      A sample of pure compound is found to have Na = 0.0887 mole, O = 0.132 mole, C = 2.65 ´ 1022 atoms. The empirical formula of the compound is                                                                                        [CPMT 1997]



Na2 CO3


Na3 O2 C5




Solution: (a)      

Na0.0887 00.132 C2.65 ´1022

6.02 ´ 1023 atoms of C = 1 mole of C

  • NaCO


\ 2.65 ´ 1022 atoms of  C = 1´ 2.65 ´ 1022

6.02 ´ 1023

mole =    2.65    = 0.044

6.02 ´ 10




Element Relative number of moles Simplest ratio of moles
 Na 0.0887 0.0887 = 2
 O 0.132 0.132 = 3
 C 0.044 0.044 = 1




Thus, the empirical formula of the compound is Na2CO3 .

Example: 15      An organic compound containing C, H and N gave the following on analysis: C = 40%, H = 13.3% and

N = 46.67%. Its empirical formula would be                                                                    [CBSE    PMT 1999, 2002]


  • CHN (b)

Solution: (c)      Calculation of empirical formula

C2H2 N




C2H7 N



Element Symbol Percentage of element At. mass elements of Relative number of atoms

= Percentage

At. mass

Simplest atomic ratio Simplest whole number atomic ratio
  C 40 12 40 = 3.33 3.33 = 1 1
Carbon       12 3.33  
Hydroge H 13.3 1 13.3 = 13.3 13.3 = 4 4
n       1 3.33  
  N 46.67 14 46.67 = 3.33 3.33 = 1 1
Nitrogen       14 3.3  

Thus, the empirical formula is CH4 N .

Example: 16      An organic substance containing C, H and O gave the following percentage composition :

C = 40.687%, H = 5.085% and O = 54.228%. The vapour density of the compound is 59. The molecular formula of the compound is



Solution: (a)


C4 H6O4

C4 H6O2

C4 H4O2

  • None of these



Element Symbol Percentage of element Atomic mass of element Relative number of atoms

= Percentage

At. mass

Simplest atomic ratio Simplest whole number atomic ratio
Carbon C 40.687 12 40.687 = 3.390


  3.390 = 1 2
Hydroge n H



5.085 1 5.085 = 5.085


5.085 = 1.5


Oxygen 54.228 16 54.228 = 3.389 3.389 = 1 2
      16 3.389  


\   Empirical formula is C2H3O2

\ Empirical formula mass of C2H3O2 = 59 Also, Molecular mass = 2 ´ Vapour density



= 2 ´ 59 = 118


\     n =       Molecular mass        118 = 2

Empirical formula mass         59

Now, Molecular formula = n ´ (Empirical formula) = 2 ´ (C2H3O2 ) = C4 H6O4



\            Molecular formula is C4 H6O4 .


  • Chemical equations : Chemical reactions are represented in a concise way by chemical equations. A chemical equation represents an actual chemical change taking place in terms of the symbols and the formulae of the reactants and e.g., Methane burns in oxygen to produce carbon dioxide and water. The chemical

reaction can be represented as: CH 4 + 2O2 ¾¾® CO2 + 2H 2 O



Essentials of a chemical equation

  • It must represent an actual chemical re



  • It must be balanced e., the total number of atoms of various substances involved on both sides of the equation must be equal.
  • It should be The elementary gases like hydrogen, oxygen etc. must be written in the molecular form as H2, O2 etc.

Information conveyed by a chemical equation : A Chemical equation conveys both qualitative and quantitative information.

  • Qualitative information : Qualitatively a chemical equation conveys the names of the reactants and products taking part in the
  • Quantitative information : Quantitatively a chemical equation conveys the following information :
  • It conveys the actual number of reactants and product species (atoms or molecules) taking part in the
  • It tells the relative masses of the reactants and products participating in the
  • It conveys the relative number of reactant and product
  • It also conveys the volumes of the gaseous reactants and products if Example : Reaction between CaCO3 and aqueous HCl.


CaCO3  +

100 gm.

1 mole


73 gm.

2 mole

¾¾® CaCl2

111 gm.

1 mole

  • H 2O +

18 gm.

1 mole


44 gm.

  • mole

22.4 litres at STP


Note : ® All chemical equations are written under STP conditions provided no other conditions are mentioned.

Limitations of a chemical equation and their removal : Although a chemical equation conveys a number of informations, it suffers from certain limitations or drawbacks. The major limitations and the steps taken for their removal are given below:

  • Physical states of the reactants and products : The chemical equation fails to convey the physical states of the reactants and products. These are specified by the use of letters ‘s’(for solids), ‘l’(for liquids), ‘g’(for gases) and ‘aq’(for aqueous solutions).


Example : CaCO3 (s)

  • 2HCl(aq)

¾¾® CaCl2 (s)

  • H 2 O(l )
  • CO2 (g )




  • Conditions of temperature, pressure and catalyst : These conditions are normally not mentioned in the These can be expressed on the arrow head which separates the reactants from the products.


Example :

N2(g ) +

3 H2(g )

¾¾Fe,7¾23¾K  ® 2 NH


3(g )


  • Speed of reaction : The speed of a particular reaction whether slow or fast can be mentioned by writing the word slow or fast on the arrow


Example :

NO2 (g )

F2 (g )

¾slow®  NO

F(g )

  • F(g )




NO2 (g )

  • F(g )



F(g )


  • Heat change accompanying the reaction : The heat evolved or absorbed in a chemical reaction can be written on the product The S.I. unit of heat is kJ.


Example : CH 4( g )  + 2O2( g )  ¾¾® CO2 ( g )

+  2 H 2 O(l )  + 393.5 KJ

(Heat is evolved)




H 2 (g )

I 2 (g )

¾¾® 2 HI(g )

– 53.9 KJ

(Heat is absorbed)



  • Reversible nature of a reaction : Certain chemical reactions proceed both in the forward and backward The reversible nature of the reaction can be indicated by two arrows pointing in the opposite direction (⇌).


Example : H 2 (g )

  • Cl2 (g )




  • HCl(g )


  • Formation of precipitate and evolution of a gas : Formation of a precipitate in the chemical reaction can be indicated by writing the word or by an arrow pointing downwards.


Ag NO3 (aq)

  • NaCl(aq)

¾¾® AgCl ¯ + NaNO3 (aq)




The evolution of a gas is expressed by an arrow which points upwards.



  • 2 HCl(aq)

¾¾® Mg Cl2 (aq)

H 2 (g ) ­


  • Balancing of chemical equations : A correct chemical equation must be in accordance with the law of conservation of mass i.e., the number of atoms of each kind in the reactants must be equal to the number of atoms of same kind in the products. Balancing of a chemical equation means to equalise the atoms of different elements or compounds which are involved in

Let us consider a chemical reaction which occurs due to passing of steam over red hot iron forming iron oxide and hydrogen gas. It can be represented as:


Skeleton equation :

Fe (s)

  • H 2 O (v)

¾¾® FeO4 (s)

H 2 (g )




Balanced equation :

  • Fe(s)

+ 4 H 2 O(v)

¾¾®  Fe3 O4 (s)

+ 4 H 2 (g )


The balancing of equations is done by the following methods:

  • Hit and Trial method, (ii) Partial Equation method




(iii) Oxidation-Number method,          (iv)  Ion-Electron method

The first two methods are discussed here, while the remaining two methods will be taken up for discussion in redox reactions.

  • Hit and Trial method : This method involves the following steps:
  • Write the symbols and formulae of the reactants and products in the form of skeleton
  • If an elementary gas like H2, O2 or N2 appears on either side of the equation, write the same in the atomic form.
  • Select the formula containing maximum number of atoms and start the process of
  • In case the above method is not convenient, then start balancing the atoms which appear minimum number of
  • Balance the atoms of elementary gases in the
  • When the balancing is complete, convert the equation into molecular


Let us balance the skeleton equation

Mg 3 N 2

+   H 2 O ¾¾®  Mg (OH)2

  • NH 3


The balancing is done in the following steps


Step I. Balance the Mg atoms

Mg 3 N 2

+ HO ¾¾® 3 Mg (OH)2

  • NH 3


Step II. Balance the N atoms

Mg 3 N 2

+ HO ¾¾® 3 Mg(OH)2

  • 2 NH 3


Step III. Balance the O atoms Mg 3 N 2 + 6 H 2 O ¾¾® 3 Mg (OH)2 + 2 NH 3

The hydrogen atoms are already balanced. Hence, final balanced equation is


Mg 3 N 2 +

6HO ¾¾® 3 Mg (OH)2

  • 2 NH 3


  • Partial equation method : Chemical equations which involve a large number of reactants and products can not be balanced easily by the hit and trial method. In partial equation method, the overall reaction is assumed to take place into two or more simpler reactions known as partial equations. The balancing of the equation involves the following steps:
  • Split the chemical equation into two or more simpler equations or partial
  • Balance each partial equation separately by hit and trial
  • Multiply the partial equations with suitable coefficient if necessary so as to cancel out the final substances which do not appear in the final
  • Finally, add up the partial equations to get the final equation.


Let us balance the skeleton equation

NaOH + Cl2 ¾¾® NaCl +


  • H 2 O


This reaction is supposed to take place in the following steps: The probable partial equations for the above reaction are:


Na OH + Cl2 ¾¾® Na Cl +

Na ClO + H 2 O  and

Na Cl O ¾¾® Na Cl O3

+ NaCl


Balance the partial chemical equations separately by hit and trial method as


2 Na OH  + Cl2

¾¾®  NaCl  +

Na ClO + H 2 O


3 Na ClO3

¾¾® Na ClO3

+ 2 NaCl


Multiply the first partial equation by 3 in order to cancel out NaClO which does not appear in the final equation. Finally add the two partial equations to get the final equation.




2 NaOH + Cl2 ¾¾® NaCl + NaClO + H 2 O] ´ 3

3 NaClO ¾¾® NaClO3 + 2 NaCl

6 NaOH + 3 Cl2 ¾¾® NaClO3  + 5 NaCl + 3 H 2 O


Calculation based on chemical equations is known as chemical stoichiometry. Stoichiometry can be broadly classified into two groups: (1) Gravimetric analysis (Stoichiometry-I), (2) Volumetric analysis (Stoichiometry-II)

  • Gravimetric analysis (Stoichiometry-I) : With the help of chemical equation, we can calculate the weights of various substances reacting and weight of substances For example,

MgCO3 ¾¾® MgO + CO2  ­

This equation implies :


  • 1 mol of

MgCO3  gives 1 mol of MgO  and 1 mol of CO2 .


  • 84 g of

MgCO3 (Mol. wt. of

MgCO3 ) gives 40 g of MgO and 44 g of CO2 .


Hence, chemical equation provide us information regarding :

  • Molar ratio of reactants and
  • Mass ratio between reactants and
  • Volume ratio between gaseous reactant and

The calculation based upon chemical equation (Stoichiometry–I) are based upon three types namely :

(a) Mass-mass relationship (b) Mass-volume relationship (c) Volume-volume relationship

  • Volumetric analysis (Stoichiometry-II) : It is a method which involves quantitative determination of the amount of any substance present in a solution through volume measurements. For the analysis a standard solution is required. (A solution which contains a known weight of the solute present in known volume of the solution is known as standard )

To determine the strength of unknown solution with the help of known (standard) solution is known as titration. Different types of titrations are possible which are summerised as follows :

  • Redox titrations : To determine the strength of oxidising agents or reducing agents by titration with the help of standard solution of reducing agents or oxidising


KCr2 O7 + 4 H 2 SO4 ® KSO4 + Cr2 (SO4 )3 + 4 H 2 O + 3[O] [2FeSO4 + H 2 SO4 + O ® Fe 2 (SO4 )3 + H 2 O] ´ 3

6FeSO4  + KCr2 O7  + 7H 2 SO4  ® 3Fe(SO4 )3  + KSO4  + Cr2 (SO4 )3 7H 2 O

2KMnO4 + 3H 2 SO4 ® KSO4 + 2MnSO4 + 3H 2 O + 5[O] [2FeSO4 + H 2 SO4 + O ® Fe2 (SO4 )3 + H 2 O] ´ 5

10FeSO4  + 2KMnO4  + 8H 2 SO4  ® 5Fe2 (SO4 )3  + KSO4  + 2MnSO4  + 8H 2 O


Similarly with

H 2 C2 O4


2KMnO4  + 3H 2 SO4  + 5H 2 C2 O4  ® KSO4  + 2MnSO4  + 8H 2 O + 10CO2





  • Acid-base titrations : To determine the strength of acid or base with the help of standard solution of base or




NaOH + HCl ® NaCl + H 2 O


NaOH + CH3 COOH ® CH3 COONa + H 2 O



  • Iodiometric titrations : To determine the reducing agents with the help of standard iodine solution is known as


For example:

As 2 O3

Reducing agent

+ 2I 2  + 2H 2 O ® As 2 O3  + 4 HI

Na2 S2 O3 + I 2  ® Na2 S4 O6  + 2NaI


  • Iodometric titrations : To determine the oxidising agent indirectly by titration of liberated I 2

of standard hypo solution is known as iodometric titrations.

with the help


Examples: Oxidising agents such as

KMnO4 , KCr2 O7 , CuSO4 , ferric salts, etc. are reduced quantitatively


when treated with large excess of KI in acidic or neutral medium and liberate equivalent amount of

2CuSO4 + 4 KI ® Cu2 I 2  + 2K2 SO4  + I 2

Kr2 Cr2 O7 + 7H 2 SO4  + 6KI ® Cr2 (SO4 )3  + 4 KSO4  + 7H 2 O + 3I 2

I 2 .


This I 2

is estimated with hypo


I 2 + 2Na2 S2 O3 ® Na2 S4 O6 + 2NaI

  • Precipitation titrations : To determine the anions like CN -, AsO3- , PO3- , X

etc, by precipitating with


3            4

AgNO3 provides examples of precipitation titrations.

NaCl + AgNO3  ® AgCl ¯ + NaNO3 ;            KSCN + AgNO3  ® AsSCN  ¯ +KNO3

End point and equivalence point : The point at which titration is stopped is known as end point, while the point at which the acid and base (or oxidising and reducing agents) have been added in equivalent quantities is known as equivalence point. Since the purpose of the indicator is to stop the titration close to the point at which the reacting substances were added in equivalent quantities, it is important that the equivalent point and the end point be as close as possible.

Normal solution : A solution containing one gram equivalent weight of the solute dissolved per litre is called a normal solution; e.g. when 40 g of NaOH are present in one litre of NaOH solution, the solution is known as normal (N) solution of NaOH. Similarly, a solution containing a fraction of gram equivalent weight of the solute dissolved per litre is known as subnormal solution. For example, a solution of NaOH containing 20 g (1/2 of g eq. wt.) of NaOH dissolved per litre is a sub-normal solution. It is written as N/2 or 0.5 N solution.

Formula used in solving numerical problems on volumetric analysis

  • Strength of solution = Amount of substance in g litre -1
  • Strength of solution = Amount of substance in g moles litre -1
  • Strength of solution = Normality ´ wt. of the solute
  • Molarity =Moles of solute

Volume in litre


Number of moles = Wt. in gm = M ´ V

Volume in litres

at NTP (only for gases)


Mol. wt.

(in l )



  • Number of millimoles = in gm ´1000 = Molarity ´ Volume in ml. mol. wt.



  • Number of equivalents = in gm= x ´ No. of moles ´ Normality ´ Volume in litre

Eq. wt.

  • Number of milliequivalents (meq.) = in gm ´1000 = normality ´ Volume in ml.

Eq. wt.


  • Normality

= x ´ No. of millimoles = x ´ Molarity =

Strength in gmlitre -1


Eq. wt.



x = Mol. wt. , x* = valency or change in oxi. Number.

Eq. wt.


  • Normality formula,

N1 V1 = N 2 V2


  • % by weight =

Wt. of solvent ´ 100 Wt. of solution


  • % by volume =


  • % by strength =

Wt. of solvent Vol. of solution Vol. of solvent Vol. of solution

´ 100


´ 100


  • Specific gravity = of solution

Vol. of solution

= Wt. of 1 ml. of solution


  • Formality =

Wt. of ionic solute Formula Wt. of solute ´ Vinl


  • Wt. = V.D ´ 2 (For gases only)

In many situations, an excess of one or more substance is available for chemical reaction. Some of these

excess substances will therefore be left over when the reaction is complete; the reaction stops immediately as soon as one of the reactant is totally consumed.

The substance that is totally consumed in a reaction is called limiting reagent because it determines or limits the amount of product. The other reactant present in excess are called as excess reagents.

Let us consider a chemical reaction which is initiated by passing a spark through a reaction vessel containing 10 mole of H2 and 7 mole of O2.

2 H 2 (g) + O2 (g) ¾¾® 2 HO (v)


Moles before reaction 10 7 0
Moles after reaction 0 2 10

The reaction stops only after consumption of 5 moles of O2 as no further amount of H2 is left to react with unreacted O2. Thus H2 is a limiting reagent in this reaction



Example: 17      When a solution containing 4.77 gm. of NaCl is added to a solution of 5.77 gm. of AgNO3, the weight of precipitated AgCl is                                                                                                                  [IIT 1978]

(a) 11.70 gm.                             (b) 9.70 gm.                                (c) 4.86 gm.                                (d) 2.86 gm.


Solution: (c)


  • NaCl



  • NaNO3


Moles before mixing



108 + 14 + 48

4.77               0                    0



= 0.0339               = 0.0815    (Here

AgNO3 is limiting reactant, thus)



Moles after mixing             0        0.0815 – 0.0339             0.0339            0.0339

= 0.0476

\   Moles of AgCl formed =    0.0339


\  Mass of AgCl formed = Mol. mass ´ No. of moles

= 143.5 ´ 0.0339

= 4.864 gm.


Example: 18      The volume of oxygen at STP required to completely burn 30 ml of acetylene at STP is                       [Orissa JEE  1997]

(a) 100 ml                                   (b) 75 ml                              (c) 50 ml                                      (d) 25 ml

Solution: (b)     The balanced chemical equation for the reaction can be written as:


CH 2

1 Vol.


+   5 / 2 O2 ¾¾® 2 CO2 + H 2O

5 / 2 Vol.

5 / 2 ml


30 ml      30 ´ 5 / 2 = 75 ml


Hence, volume of the oxygen at STP required to burn 30 ml of acetylene at STP = 75 ml.

Example: 19      What is the volume (in litres) of oxygen at STP required for complete combustion of 32 g of CH4

(a) 44.8                        (b) 89.6                              (c) 22.4                        (d) 179.2


[EAMCET 2001]


Solution: (b)     According to Avogadro’s hypothesis, volume occupied by one mole of any gas at STP is 22.4 litres.


CH4 (g)            +

2 O2(g)


CO2(g)      +

2 H 2 O (l)


  • mole 2 moles
  • moles 4 moles

2´16 gm. = 32 gm.               4 ´ 22.4 litres = 89.6 litres

Example: 20 A metal oxide has the formula Z 2O3 . It can be reduced by hydrogen to give free metal and water. 0.1596 g

of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

[CBSE PMT 1989]

(a) 27.9                          (b) 159.6                      (c) 79.8                          (d) 55.8


Solution: (d)   Valency of metal in

Z 2 O3  = 3


Z 2 O3  + 3H 2  ¾¾® 2Z + 3H 2 O


0.1596 gm of


react with H 2 = 6mg = 0.006 gm


1 gm of H 2 react with Z

2 O3

= 0.1596 = 26.6 gm



Equivalent wt. of Atomic weight of


ZO3 = 26.6 = equivalent wt. of Z +

Z = 18.6 ´ 3 = 55.8


equivalent wt. of O = E + 8 = 26.6 or E = 18.6