Preview Modern Approach to Chemical Calculations R C Mukherjee Ramendra Chandra Mukerjee IIT JEE Physical Chemistry Bharati Bhawan not Mukherji Mukerji by R C Mukherjee Ramendra Chandra Mukerjee Mukherji Muke (2019)

Preview Modern Approach to Chemical Calculations R C Mukherjee Ramendra Chandra Mukerjee IIT JEE Physical Chemistry Bharati Bhawan not Mukherji Mukerji by R C Mukherjee Ramendra Chandra Mukerjee Mukherji Muke (2019) Preview Modern Approach to Chemical Calculations R C Mukherjee Ramendra Chandra Mukerjee IIT JEE Physical Chemistry Bharati Bhawan not Mukherji Mukerji by R C Mukherjee Ramendra Chandra Mukerjee Mukherji Muke (2019) Preview Modern Approach to Chemical Calculations R C Mukherjee Ramendra Chandra Mukerjee IIT JEE Physical Chemistry Bharati Bhawan not Mukherji Mukerji by R C Mukherjee Ramendra Chandra Mukerjee Mukherji Muke (2019)

(An Autonomous College of Allahabad University)

The book continues the tradition of providing a firm foundation in chemicalcalculations I have included quite a good number of problems with hintswherever necessary The challenge to me has always been to present a solidunderstanding of the basic facts and principles of chemistry I am fullyconfident that the present book will help the students prepare for theJEEandother engineering and medical entrance examinations.

The design of question papers, particularly of theJEE, changes from time

to time Remember that chemistry has not changed I feel strongly that if youunderstand the subject well, you can solve any question in any form I havealso noticed that questions are now being asked from all chapters includingthe mole concept, stoichiometric calculations, etc Previously, it used to beselective Another important fact is that the questions are sometimes toodifficult for students of +2 level In that case you may consult my book

‘Modern Approach to Physical Chemistry, volumes I and II’ for an advancedknowledge of physical chemistry Basically, read every chapter very carefullytill you have a vivid picture of the subject in your mind Finally, follow SwamiVivekananda’s saying,

Arise, Awake and Stop Nottill the goal is achieved

Ramendra C Mukerjee

PREFACE TO THE EIGHTH EDITION

I take immense pleasure in presenting you the eighth edition of this book Onthe basis of helpful suggestions given by learned teachers and students fromall over the country, and also in view of the recent changes in the syllabus and

in the examination pattern of IIT-JEE, all the chapters have been criticallyreviewed Some changes in the presentation of the text and the necessaryadditions, both in the text and in the problems, have been carried out Therevision chapters 21 and 22 have been updated by including the questionsasked in IIT-JEE of the recent past years My goal in this revision is to providestudents with the best possible tool for learning numerical chemistry, by

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concepts Here are some of the new features:

The presentation of thermodynamics has been reorganized Theconsistent presentation helps to consolidate the concepts of thermodynamics

As the central concepts of chemical change are best understood in thesequence of chemical thermodynamics (Is the forward or the reverse reactionfavoured?), chemical thermodynamics is now placed as Chapter 14, followed

by chemical equilibrium (Chapter 15), ionic equilibrium (Chapter 16—Howfar the reaction would go?) and then by chemical kinetics (Chapter 17—Howfast the reaction would go?)

Though the oxidation and reduction processes have been discussed inchapters 6, 7 and 18, a new chapter (19) on the concepts of oxidation numberand its application to balance redox reactions has been added Chapter 20 onsolid state has been extended by including the liquid state Chapters 21 and 22contain only problems for the purpose of revision

Many of my colleagues have helped me write this book, to all of whom I

am sincerely thankful I especially thank Mr D Kumar, former Head,Chemistry Department of our college, with whom I have discussed manyproblems, and Dr G K Verma, C M P Degree College, for offering me manyuseful suggestions I am greatly indebted to Prof S S Shukla, LamarUniversity, Texas, USA, for going through the manuscript of the first editionand making invaluable suggestions I wish to express my appreciation to allthe students and teachers who have been kind enough to write letters withhelpful suggestions for the improvement of the book Special thanks are due

to my own students for pointing out a number of errors Finally, I thank allthe members of the Bharati Bhawan family who were involved in theproduction of this edition

I believe this edition will continue to prove useful to the students andinteresting to the teachers

Allahabad 211 006

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Dear Students,

This is a textbook on chemical calculations primarily intended forstudents who are preparing for the entrance tests of IIT and otherengineering colleges This text is equally useful to the students ofIntermediate(+2) and BSc classes of Indian universities

Chemistry is just one of the subjects you have to study and, therefore,you have limited time for each subject Keeping in mind your limitedtime and requirements, I have tried my best to write this book in such away that it fulfils your requirements The special features of the book are:

1 This book has been written in the language of the mole, as the concept

of mole is a basic tool in understanding chemistry Try to have fun withthe mole; I did

2 Sufficient fundamental principles have been provided before introducingproblems These principles, in the form of rules or equations, have beenreferred to throughout the book while solving problems This will helpyou in following each step of the solutions of these problems

3 A wide variety of problems have been selected; at the same timerepetitions of similar types of problems have been avoided This allowed

me to give maximum coverage in minimum space

4 Mixed system of units including SI units have been used in this text inaccordance with the latest trend of the entrance tests of IIT and most ofthe other engineering colleges Moreover, a selected list of basic andderived SI units and conversions of selected nonsystem units to SI unitshave been provided at the beginning of the book to help you solveproblems using different systems of units

Finally, I would like to suggest two important points so that you can takethe maximum advantage from this book

 Read Chapter 1 and Chapter 14 very carefully as both these chaptershave wide applications in other chapters

 Do not jump directly to solving the problems until you havethoroughly gone through the text provided at the beginning of eachchapter

Ramendra C Mukerjee

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Amount of substance mole mol

Derived Units:

Quantity Unit Conversion factor to SI

millimetre of Hg

or torr

105 Pa101325  105 Pa133322 Pa

Energy, work, amount

160219 10 19 J

41868 J41868 J

10 7 J

‰

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Quantity Symbol Value

Planck constant h  66262 10 27 erg  s

00821 L  atm deg 1 mole 1

1987 cal deg 1 mole 1Molar volume of gas at V 22414 L mole 1

NTP 22414  10 3 m3 mole 1 SIVelocity of light c 2998  1010 cm s 1

2998  108 m s 1SI

Electron mass m e 9109  10 28 g

9109  10 31 kg SIElectron charge e 4803  10 10 esu

1602  10 19 coulomb SIFaraday constant F 96485coulomb eq 1

96484coulomb mol 1 SI

Acceleration due to gravity g 980665 cm s 2

980665 m s 2 SI

Atomic mass unit amu 166  10  27 kg SI

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Principle of Atom Conservation POAC

1 Elementary Problems Based on

19 Oxidation Number and Balancing of Redox Reactions 712

Appendix

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ELEMENTARY PROBLEMS BASED ON

DEFINITION OF MOLE: THE MOLE CONCEPT

The mole concept is an essential tool for the study of the fundamentals ofchemical calculations This concept is simple but its application in problemsrequires a thorough practice

Mole in Latin means heap or mass or pile A mole of atoms is a collection

of atoms whose total weight is the number of grams equal to the atomicweight As equal numbers of moles of different elements contain equalnumbers of atoms, it is convenient to express amounts of the elements interms of moles Just as a dozen means twelve objects, a score means twentyobjects, chemists have defined a mole as a ‘definite number’ of particles, viz.,atoms, molecules, ions or electrons, etc This ‘definite number’ is called theAvogadro constant, equal to 60221023, in honour of Amedeo Avogadro.However, for many years scientists have made use of the concept of a molewithout knowing the value of the Avogadro constant Thus, a mole ofhydrogen atoms or a mole of hydrogen molecules or a mole of hydrogen ions

or a mole of electrons means the Avogadro constant of hydrogen atoms,hydrogen molecules, hydrogen ions or electrons respectively

The value of the Avogadro constant depends on the atomic-weight scale

At present the mole is defined as the amount of a substance containing as many atoms, molecules, ions, electrons or other elementary entities as there are carbon atoms in exactly 12 g of 12 C The value of the Avogadro constant

was changed by a very small amount in 1961 when the basis of theatomic-weight scale was changed from the naturally occurring mixture ofoxygen isotopes at 16 amu to 12C , which put oxygen at 15.9994 amu

It is quite interesting and surprising to know that the mole is such a bignumber that it will take 1016 years to count just one mole at the rate of onecount per second, and the world population would be only of the order of

10–14 mole in chemical terminology

1

In modern practice, a gram-molecule and a gram-atom are termed as a mole of molecules and a mole of atoms respectively, e.g., 1 gram-molecule of oxygen and 1 gram-atom of oxygen are expressed as 1 mole of O2 and 1 mole

of O respectively Similarly, gram-equivalent shall be written as equivalent.

The number of moles of a substance can be calculated by various means.The choice of method depends on the data provided

These rules are very important and have been frequently appliedthroughout the book In the beginning, the rules have been mentionedwherever they are applied in solving problems Before discussing the moleconcept in further detail, the understanding of a balanced chemical equation

is necessary

Significance of Chemical Equations

A chemical equation describes the chemical process both qualitatively andquantitatively The stoichiometric coefficients in the chemical equation givethe quantitative information of the chemical process These coefficientsrepresent the relative number of molecules or moles of the reactants andproducts, e.g.,

RULES IN BRIEF

The following are the definitions of ‘mole’ represented in the form of

equations:

(1) Number of moles of molecules  molecular weight weight in g

(2) Number of moles of atoms  atomic weight weight in g

(3) Number of moles of gases  standard volume at molar NTP volume

(Standard molar volume is the volume occupied by 1 mole of any gas at NTP , which is equal to 22 4 litres.)

(4) Number of moles of atoms / molecules / ions / electrons

 no of atoms / molecules Avogadro constant / ions / electrons

(5) Number of moles of solute  molarity  volume of solution

2 KClO3 s  2KCl s + 3O2 g

2 molecules 2 molecules 3 molecules

or 2 N molecules 2 N molecules 3 N molecules

(if N is Av const.)

Again, Avogadro’s principle states that under the same conditions oftemperature and pressure, equal volumes of gases contain the same number

of molecules Thus, for homogeneous gaseous reactions, the stoichiometriccoefficients of the chemical equation also signify the relative volumes of eachreactant and product under the same conditions of temperature and pressure,e.g.,

H2 g + I2 g  2 HI g

1 molecule 1 molecule 2 molecules

(T & p constant)

(T & V constant)

The coefficients, however, do not represent the relative weights of

reactants and products, which is due to the fact that 1 mole is a fixed number

of particles but not a fixed weight It is to be noted that the coefficients do

not even represent the relative number of equivalents (eq.) or equivalents (m.e.) of reactants and products In any reaction the number ofequivalents or milli-equivalents of each reactant and product is the same, e.g.,

milli-2 KClO3  2 KCl + 3 O2

no of equi-  no of equi-  no of

equi-valents of valents of valents of

or no of m.e of  no of m.e  no of m.e

The principle of conservation of mass, expressed in the concepts of atomictheory, means the conservation of atoms And if atoms are conserved, moles

of atoms shall also be conserved This is known as the principle of atomconservation This principle is in fact the basis of the mole concept

In order to solve problems of nearly all stoichiometric calculations, let usfirst see how this principle works Choose an example,

KClO3 s  KCl s  O2 g

Apply the principle of atom conservation (POAC) for K atoms

Moles of K atoms in reactant = moles of K atoms in products

or moles of K atoms in KClO3 = moles of K atoms in KCl.

Now, since 1 molecule of KClO3 contains 1 atom of K

or 1 mole of KClO3 contains 1 mole of K, similarly, 1 mole of KClcontains 1 mole of K

Thus, moles of K atoms in KClO3  1  moles of KClO3

and moles of K atoms in KCl  1  moles of KCl

           moles of KClO3  moles of KCl

or wt of KClO3 in g

mol wt of KClO3  wt of KCl in g mol wt of KCl  (Rule 1)

The above equation gives the weight relationship between KClO 3 and KCl which is important in stoichiometric calculations.

Again, applying the principle of atom conservation for O atoms,moles of O in KClO3 = moles of O in O2

But since 1 mole of KClO3 contains 3 moles of O and 1 mole of O2 contains

2 moles of O,

thus, moles of O in KClO3 3 moles of KClO3

moles of O in O2  2  moles of O2

 3  moles of KClO3  2  moles of O2

or 3  mol.wt of wt ofKClO KClO3

3 2  standardvol. of O molar2 at NTP vol.  (Rules 1 and 3)

The above equation thus gives the weight–volume relationship of reactants and products.

From the above discussion we see that the procedure to use the mole method is first to set up an equation based on the principle of atom conservation and then apply rules 1 to 6 to it.

Advantages of the Mole Method over other Methods

The advantages of the Mole Method are as under:

(1) Balancing of chemical equations is not required in the majority ofproblems as the method of balancing the chemical equation is based on theprinciple of atom conservation

(2) Number of reactions and their sequence, leading from reactants toproducts, need not be given

(3) It is a general method, applicable in solving many types of problems

as may be seen in different chapters

[Note: The students should carefully note that POAC should be applied for onlythose atoms which remain conserved in a chemical reaction.]

Let us first solve some elementary problems based on the definition ofmole

Elementary Chemical Calculations

Ex 1 Calculate the weight of 6022 1023 molecules of CaCO3.

 6022  1023

6022  1023  1

Weight of CaCO3  no of moles  molecular wt (Rule 1)

 1  100  100 g

Ex 2 Calculate the weight of 12044 1023 atoms of carbon.

6022  1023  2 (Rule 4)

Wt of C atoms  no of moles  at wt (Rule 2)

 2  12  24 g

Ex 3 What will be the number of oxygen atoms in 1 mole of O2?

the oxygen molecule is diatomic, i.e., 1 molecule contains 2 atoms, the

no of oxygen atoms in 1 mole of O2 is equal to 2  6022  1023

Ex 4 A piece of Cu weighs 0 635 g How many atoms of Cu does it contain?

 0635

635  001.

No of atoms of Cu  no of moles of atoms  Av const (Rule 4)

 001  6022  1023  6022  1021

Ex 5 Calculate the number of molecules in 11 2 litres of SO2 gas at NTP

std molar volume litres (Rule 3)

 112224  05.

No of molecules of SO2  no of molesAv const (Rule 4)

 05  6022  1023

 3011  1023

Ex 6 One atom of an element X weighs 6644  1023 g Calculate the number of gram-atoms in 40 kg of it.

Solution : Wt of 1 mole of atoms of X  wt of 1 atom  Av const.

 6644  10 236022  1023

 40 g

Thus the at wt of X = 40.

No of moles (or gram-atoms) of X  weight in gat wt. (Rule 2)  40  100040  1000

Ex 7 From 200 mg of CO2 , 1021 molecules are removed How many moles of CO2are left?

 02

44  000454.

No of moles removed  1021

6022  1023 000166 (Rule 4)

No of moles of CO2 left  000454 – 000166 000288

Ex 8 What will be the volume occupied by 1 mole atom of a (i) monoatomic gas,

and (ii) diatomic gas at NTP?

1 mole atom occupies 224 litres at NTP, and for diatomic gases,

1 mole atom occupies 112 litres at NTP, as 1 mole of O2 contains 2 moles

of O.

Ex 9 Calculate the volume of 20 g of hydrogen gas at NTP

Volume of the gas at NTP no of moles  224

 10  224  224 litres

Ex 10 What volume shall be occupied by 6022  1023 molecules of any gas at NTP?

gas at NTP occupies a volume of 224 litres

Ex 11 Calculate the number of atoms present in 5 6 litres of a (i) monoatomic, and (ii) diatomic gas at NTP

 no of molecules ofthegas 14  6022  1023 (Rule 4)  15  1023.

Now, if the gas is monoatomic, the no of atoms of the gas

 no of molecules = 15  1023

And if the gas is diatomic,

no of atoms  2  no of molecules  2  15  1023

 30  1023

Ex 12 Calculate the number of sulphate ions in 100 mL of 0001M H2SO4 solution.

 0001  01  00001

Now,

1 molecule of H2SO4 contains 1 SO42  ion

 1 mole of H2SO4 contains 1 mole of SO42 

 00001 mole of H2SO4 contains 00001 mole of SO4 2 

 number of sulphate ions  moles of ions  Av const

 00001  6022  1023  6022  1019 (Rule 4)

Ex 13 How many atoms are there in 100 amu of He?

or weight of one 12C atom = 12 amu (at wt of C 12 amu)

Similarly, as the atomic weight of He is 4 amu,

weight of one He atom = 4 amu

Thus, the number of atoms in 100 amu of He  1004  25

Ex 14 If a mole were to contain 1  1024 particles, what would be the mass of (i) one mole of oxygen, and (ii) a single oxygen molecule?

 mass of a single O2 molecule  no of moleculesmass of 1 molein 1 mole

 32

1 1024  32  1023g

Ex 15 The density of O2 at NTP is 1429 g/L Calculate standard molar volume

of the gas.

Solution : Standard molar volume is the volume occupied by 1 mole of thegas at NTP Now,

1429 g of O2 at NTP occupies a volume of 1 litre

 32 g (i.e., 1 mole) of O2 occupies a volume  142932 L

 2239 litres

Ex 16 The measured density of He at NTP is 01784 g/L What is the weight of

1 mole of it?

 standard molar volume (litres)  01784 224 g

 4 g

[Note: Weight of 1 mole is the molecular weight in g.]

Ex 17 A metal M of atomic weight 54 94 has a density of 742 g/cc Calculate the apparent volume occupied by one atom of the metal.

or 1 mole of H2SO4 contains 2 moles of H

or 5 moles of H2SO4 contain 10 moles of H

1 mole of H2SO4 contains 1 mole of S

or 5 moles of H2SO4 contain 5 moles of S

and again,

1 mole of H2SO4 contains 4 moles of O

or 5 moles of H2SO4 contain 20 moles of O.

Ex 19 Calculate the number of oxygen atoms and its weight in 50 g of CaCO3.

(mol wt of CaCO3  100)

Now,

1 molecule of CaCO3 contains 3 atoms of O

or 1 mole of CaCO3 contains 3 moles of O

or 05 mole of CaCO3 contains 15 moles of O.

No of atoms of O 15  6022  1023 (Rule 4)

 9033  1023

Wt of atoms of O  no of moles  at wt of O (Rule 2)

 15  16  24 g

Ex 20 Calculate the number of atoms of each element present in 122 5 g of

KClO3

1225  1.

(mol wt of KClO3  1225

From the formula KClO3 , we know that 1 mole of KClO3 contains 1 mole

of K atoms, 1 mole of Cl atoms and 3 moles of O atoms.

Ex 21 Calculate the total number of electrons present in 1 6 g of CH4

No of molecules in 16 g of CH4  01  6022  1023 (Rule 4)  6022 1022 molecules

B 1 molecule of CH4 has 10 electrons

 6022 1022 molecules of CH4 has 10  6022  1022 electrons

 6022  1023 electrons

Ex 22 Find the charge of 1 g-ion of N3  in coulombs.

(charge on an electron 1602 1019 coulombs)

 6022 10 23 ions (1 g-ion or 1 mole) carry a charge of

3  1602  1019  6022  1023 coulombs

 2894  105 coulombs

Ex 23 Find the charge of 27 g of Al3  ions in coulombs.

same magnitude of charge as that on an electron

No of moles of Al3  ions wt in gat wt. (Rule 1)

 2727 1

No of Al3  ions in 27 g no.ofmoles Av const (Rule 4)

 1  6022  1023.Charge of 27 g of Al3  ions 3  charge of a proton  no of Al3+ ions

 3  1602  1019  6022 1023

 2894 105 coulombs

Ex 24 Equal masses of oxygen, hydrogen and methane are taken in a container in

identical conditions Find the ratio of the volumes of the gases.

Therefore, O2 : H2 : CH4

Weight – X X X

No of moles – 32X X2 16X (Rule 1)Volume ratio – 32X : X2 : (Avogadro’s principle)16XHence, O2 : H2 : CH4 1 : 16 : 2

Ex 25 If the components of air are N2, 78%; O2, 21%; Ar, 09% and CO2, 01%

by volume, what would be the molecular weight of air?

(Avogadro’s principle)

 mol wt of air  78  28  21  32  09  40  01  44

78  21  09  01 (wt in g per mole)

 28964

N2 28, O2  32, Ar 40 and CO2 44

Ex 26 The atomic weights of two elements (A and B) are 20 and 40 respectively.

If x g of A contains y atoms, how many atoms are present in 2x g of B?

Solution : Number of moles of A  20x  (Rule 2)

Number of atoms of A 20x  N. (Rule 4)

(N is the Av const.)

Ex 27 Oxygen is present in a 1-litre flask at a pressure of 76  10  10 mmHg at

0C Calculate the number of oxygen molecules in the flask

 076  10  10cm

Ex 28 The density of mercury is 13 6 g/cc Calculate approximately the diameter

of an atom of mercury, assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom.

of one Hg atom

 volume occupied by 1 Hg atom  X3 cc

and mass of one Hg atom  136  X3 g

Mass of one Hg atom  Av const.at wt.  200

6022 1023 g

(B mass of 1 mole of atoms is the atomic weight in g, and 1 mole containsthe Av const of atoms)

[Li  6939, F  18998 1Å 10  8cm.]

of LiF molecules present in 1 mole As 1 mole of LiF weighs 25937 g

LiF 6939  18998,

volume of 1 mole  wt of 1 mole

wt per cc density 

25937265  978 cc

Since this volume is supposed to be of a cube,

the length of each edge of the cube  3978

 004375

 molarity of H2SO4  004375

0125  035 M

Ex 31 It is found that in 11 2 litres of any gaseous compound of phosphorus at

NTP, there is never less than 155 g of P Also, this volume of the vapour of phosphorus itself at NTP weighs 62 g What should be the atomic weight and molecular weight of phosphorus?

of P atoms, and also 224 litres (at NTP) of the gaseous compound contains

31 g of phosphorus, therefore, 1 mole of phosphorus weighs 31 g, i.e., 31

is the atomic weight of phosphorus Similarly, the molecular weight ofphosphorus is 124 as 1 mole of its vapour weighs 124 g

Ex 32 A polystyrene, having the formula Br3C6H3C3H8n, was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air If it was found to contain 1046% bromine by weight, find the value of n.

 no of moles of Br in 100 g of polystyrene  1046799  01309

From the formula of polystyrene, we have,

no of moles of Br  3  moles of Br3C6H3C3H8n

or 01309  3  mol wt.wt. 3147  44n3  100 

Ex 33 It has been estimated that 93% of all atoms in the entire universe are

hydrogen and that the vast majority of those remaining are helium Based on only these two elements, estimate the mass percentage composition of the universe.

7 atoms are of He, that is, the number of moles of H and He atoms, out

of 100 moles, are 93 and 7 respectively

Mass of H  93  1  93 g

Mass of He  7  4  28 g

 mass percentage of H  93  2893  100  76.86%

 mass percentage of He = 23.14%

Ex 34 The molecular weight of haemoglobin is about 65,000 g/mol Haemoglobin

contains 0.35% Fe by mass How many iron atoms are there in a haemoglobin molecule?

Solution : 100 g of haemoglobin contain 0.35 g of iron

or, 65000100 mole of haemoglobin contains 0.3556 mole of Fe

 1 mole of haemoglobin contains 0.3556  65000100 moles of Fe

= 4.06 moles of Fe

Thus one molecule of haemoglobin contains four iron atoms

Ex 35 At room temperature, the density of water is 1.0 g/mL and the density of

ethanol is 0.789 g/mL What volume of ethanol contains the same number of molecules as are present in 175 mL of water?

as are present in 175 mL of H2O be v mL As given,

moles of C2H5OH in v mL  moles of H2O in 175 mL

Now, mol wt of Cwt of C2H5OH

2H5OH 

wt of H2Omol wt.ofH2O

46 1.0  17518 

 v  566.82 mL

Ex 36 A sample of ethane has the same mass as 10.0 million molecules of methane.

How many C2H6 molecules does the sample contain?

mass of C2H6  mass of 107 molecules of CH4

Av const.n  mol wt of C2H6  Av c10o7nst.  mol wt of CH4

mass of W in w g of CaWO4  mass of W in 569 g of FeWO4

Moles of W in CaWO4  at wt of W

= moles of W in FeWO4  at wt of W

As both CaWO4 and FeWO4 contain 1 atom of W each,

 moles of CaWO4  at wt of W = moles of FeWO4  at wt of W

moles ofmolesofAO4

2  0.752.0 moles of A/4

moles of O/2 

0.752.0 moles of AmolesofO34

Thus, the empirical formula of the product is A3O4 Further, as 2 moles

of O2 give 1 mole of A3O4 (for gases, pressure  mole at constanttemperature and volume), A3O4 is also the molecular formula of the product

2  moles of O2  y  moles of AxOy

2  2  y 1; y4

In the following chapters, we shall apply the principle of atom conservation (POAC) along with the said rules in tackling the various problems encountered in chemical practice.

PROBLEMS

(Answers bracketed with questions)

1 Find the number of atoms in 48 g of ozone at NTP (18066  10 24 )

2 What is the ratio of the volumes occupied by 1 mole of O2 and 1 mole of O3 in

4 The vapour density of a gas is 112 Calculate the volume occupied by 112 g of the gas at NTP

5 Calculate the number of oxygen atoms in 02 mole of Na2CO310H2O 156  10 24

6 Calculate the number of moles of CuSO4 contained in 100 mL of 1 M CuSO4solution Also, find the number of SO4 2  ions in it 01 mole, 0.6022  10 23 

7 Find the total number of nucleons present in 12 g of 12

C atoms 12  6022  10 23 

8 Find (i) the total number of neutrons, and (ii) the total mass of neutrons in 7 mg

of 14 C (Assume that the mass of a neutron = mass of a hydrogen atom)

[Hint: 1 14 C atom contains 8 neutrons.] 24088  1020, 0004 g

9 How many moles are there in 1 metre3 of any gas at NTP ? (1 m 3 = 10 3 litres)

(446 moles)

10 3 g of a salt of molecular weight 30 is dissolved in 250 g of water Calculate the

11 Calculate the volume occupied by 525 g of nitrogen at 26C and 742 cm of

15 The measured density at NTP of a gaseous sample of a compound was found to

be 178 g/L What is the weight of 1 mole of the gaseous sample? (399 g)

16 If the concentration of a solution is 2 M calculate the number of millimoles present

in 2 litres of the solution.

17 How many moles of oxygen are contained in one litre of air if its volume content

18 How many atoms do mercury vapour molecules consist of if the density of mercury

vapour relative to air is 692? Hg  200 The average mass of air is 29 g/mole.

(One)

19 Calculate the total number of atoms in 05 mole of K2 Cr2 O7.

[Hint: Follow Example 19] 331  10 24



20 What is the volume of 6 g of hydrogen at 1 atm and 0C? (672 litres)

21 What is the density of oxygen gas at NTP ?

[Hint: See Example 14] (1429 g/L)

22 Calculate the total number of electrons present in 18 mL of water.

10  6022  10 23 

23 Calculate the number of electrons, protons and neutrons in 1 mole of 16 O  2 ions.

10  6022  10 23 , 8  6022  10 23 , 8  6022  10 23 

24 Find the mass of the nitrogen contained in 1 kg of (i) KNO3 , (ii) NH4NO3 , and (iii) (NH4)2 HPO4 [(i) 1385 g (ii) 350 g and (iii) 212 g]

25 Find the mass of each element in 784 g of FeSO4 NH4 2 SO4 6H2O What will

26 The density of solid AgCl is 556 g/cc The solid is made up of a cubic array of alternate Ag and Cl  ions at a spacing of 2773 Å between centres From these

27 Three atoms of magnesium combine with 2 atoms of nitrogen What will be the

weight of magnesium which combines with 186 g of nitrogen?

28 600 mL of a mixture of O3 and O2 weighs 1 g at NTP Calculate the volume of

29 The vapour density (hydrogen = 1) of a mixture consisting of NO2 and N2O4 is 383 at 267C Calculate the number of moles of NO2 in 100 g of the mixture [Hint: Wt of NO2  x g.

 obs mol wt (wt./mole)  total moleswt in g  100

30 A nugget of gold and quartz weighs 100 g Sp gr of gold, quartz and the nugget

are 193, 26 and 64 respectively Calculate the weight of gold in the nugget.

31 The nucleus of an atom of X is supposed to be a sphere with a radius of

5  10  13 cm Find the density of the matter in the atomic nucleus if the atomic weight of X is 19.

32 Copper forms two oxides For the same amount of copper, twice as much oxygen

was used to form the first oxide than to form the second one What is the ratio

of the valencies of copper in the first and second oxides?

[Hint: Assume that the oxides are Cu2Ox and Cu2Oy and apply Rule 6] 2 : 1

33 105 mL of pure water 4C is saturated with NH3 gas, producing a solution of density 09 g/mL If this solution contains 30% of NH3 by weight, calculate its volume.

[Hint: Density total volumetotal mass 105 H2O V45NH3 ]

16667 mL

34 How many iron atoms are present in a stainless steel ball bearing having a radius

of 01 inch (1 inch = 254 cm)? The stainless steel contains 856% Fe by weight and

35 How many litres of liquid CCl4 (d  15 g/cc) must be measured out to contain

36 A sample of potato starch was ground in a ball mill to give a starchlike molecule

of lower molecular weight The product analysed 0086% phosphorus If each molecule is assumed to contain one atom of phosphorus, what is the molecular

37 The dot at the end of this sentence has a mass of about one microgram Assuming

that the black stuff is carbon, calculate the approximate number of atoms of carbon needed to make such a dot (1 microgram  1  10 6 g 5  10 16atoms

38 To what volume must 50 mL of 350 M H2SO4 be diluted in order to make 2 M H2SO4?

39 Sulphur molecules exist under various conditions as S8, S6, S4, S2 and S.

(a) Is the mass of one mole of each of these molecules the same?

(b) Is the number of molecules in one mole of each of these molecules the same? (c) Is the mass of sulphur in one mole of each of these molecules the same? (d) Is the number of atoms of sulphur in one mole of each of these molecules the

40 Two minerals that contain Cu are CuFeS2 and Cu2S What mass of Cu2S would contain the same mass of Cu as is contained in 125 lb of CuFeS2? (54.2 lb)

41 What is the maximum number of moles of CO2 that could be obtained from the

42 What mass of NaCl would contain the same total number of ions as 245 g of

43 An unknown sample weighing 1.5 g was found to contain only Mn and S The

sample was completely reacted with oxygen and it produced 1.22 g of Mn(II) oxide and 1.38 g of SO3 What is the simplest formula for this compound? (MnS)

44 The two sources of Zn, that is, ZnSO4 and ZnCH3COO2  2H2O, can be purchased

at the same price per kilogram of compound Which is the most economical source

45 How many moles of H2O form when 25.0 mL of 0.10 M HNO3 solution is

46 Which would be larger: an atomic mass unit based on the current standard or one

based on the mass of a Be-9 atom set at exactly 9 amu? (latter)

47 The enzyme carbonic anhydrase catalyses the hydration of CO2 This reaction: CO2  H2O  H2CO3, is involved in the transfer of CO2 from tissues to the lungs via the bloodstream One enzyme molecule hydrates 10 6 molecules of CO2 per second How many kg of CO2 are hydrated in one hour in one litre by

48 An oxybromo compound, KBrOx , where x is unknown, is analysed and found to

49 Radium disintegrates at an average rate of 2.24  10 13 -particles per minute Each

-particle takes up two electrons from the air and becomes a neutral helium atom After 420 days, helium gas collected was 0.5 mL, measured at 27°C and 750 mmHg.

50 If the value of Avogadro number is 6023  10 23 mol 1 and the value of Boltzmann constant is 1380  10 23 JK 1 , then the number of significant digits in the calculated value of the universal gas constant is (IIT 2014 Adv.) (4)

[Hint: Since k and N, both have four significant figures, the value of R is also

rounded off up to four significant figures.]

51 Three moles of B2H6 are completely reacted with methanol The number of moles

of boron containing product formed is (IIT 2015 Adv.) (2)

Objective Problems

1 The density of chlorine relative to air is

(c) found only experimentally (d) 4

2 A gaseous oxide contains 304% of nitrogen, one molecule of which contains one nitrogen atom The density of the oxide relative to oxygen is

3 The mass of an oxygen atom is half that of a sulphur atom Can we decide on

this basis that the density of sulphur vapour relative to oxygen is 2?

4 Density of air is 0001293 g/cc Its vapour density is

5 56 litres of oxygen at NTP is equivalent to

(a) 1 mole (b) 12 mole (c) 14 mole (d) 18 mole

6 224 litres of water vapour at NTP , when condensed to water, occupies an approximate volume of

7 Which of the following has the highest mass?

(c) 10 mL of water (d) 3011  10 23 atoms of oxygen

8 6022  10 22 molecules of N2 at NTP will occupy a volume of

(a) 224 litres (b) 224 litres (c) 602 litres (d) 602 mL

9 How many grams are contained in 1 gram-atom of Na?

10 The weight of 350 mL of a diatomic gas at 0C and 2 atm pressure is 1 g The wt.

of one atom is

11 The number of atoms present in 16 g of oxygen is

(a) 602  10 115 (b) 301  10 23 (c) 301  10 115 (d) 602  10 23

12 1 mole of a compound contains 1 mole of C and 2 moles of O The molecular

weight of the compound is

15 2 moles of H atoms at NTP occupy a volume of

(a) 112 litres (b) 448 litres (c) 2 litres (d) 224 litres

16 No of electrons in 18 mL of H2O (l) is

(a) 602  10 23 (b) 3011  10 23 (c) 06022  10 23 (d) 6022  10 23

17 Molecular weight of a gas, 112 litres of which at NTP weighs 14 g, is

18 The weight of 1 mole of a gas of density 01784 g/L at NTP is

19 Number of HCl molecules present in 10 mL of 01 N HCl solution is

(a) 6022  10 23 (b) 6022  10 22 (c) 6022  10 21 (d) 6022  10 20

20 Number of atoms in 12 g of 12 6C is

21 5 moles of a gas in a closed vessel was heated from 300 K to 600 K The pressure

of the gas doubled The number of moles of the gas will be

22 Which of the following contains the greatest number of oxygen atoms?

23 If the atomic weight of carbon were set at 24 amu, the value of the Avogadro

constant would be

(a) 6022  10 23 (b) 12044  10 23 (c) 3011  10 23 (d) none of these

24 If 32 g of O2 contain 6022  10 23 molecules at NTP then 32 g of S, under the same conditions, will contain,

(a) 6022  10 23 S (b) 3011  10 23 S (c) 12044  10 23 S (d) 1  1023S

25 How many moles of electrons weigh one kilogram?

(IIT 2013 Main)

27 The ratio of masses of oxygen and nitrogen of a particular gaseous mixture is 1:4.

The ratio of number of molecules is

(IIT 2014 Main)

28 The most abundant amounts by mass in the body of a healthy human adult are

oxygen (61.4%), carbon (22.96%) hydrogen (10.0%) and nitrogen (2.6%) The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is

29 1 g of a carbonate M2CO3 on treatment with excess HCl produces 0.01186 mole

of CO2 The molar mass of M2O3 is g mol  is

PROBLEMS BASED ON EQUATIONS: STOICHIOMETRY

The word ‘stoichiometry’ is derived from the Greek words stoicheion, which means element, and metrein, which means to measure The numerals used to

balance a chemical equation are known as stoichiometric coefficients Thesenumbers are essential for solving problems based on chemical equations.Hence, such problems are also called stoichiometric calculations The molemethod is very useful in such calculations

For stoichiometric calculations, the mole relationships between differentreactants and products are required, as from them, the mass–mass,mass–volume and volume–volume relationships between different reactantsand products can be obtained

For a given balanced equation say,

2 KClO3  2KCl  3O2

we can get such relationships directly from the stoichiometric coefficients,e.g.,

3  moles of KClO3  2  moles of O2

2  moles of KClO3  2  moles of KCl

and, 3  moles of KCl  2  moles of O2

For balanced chemical equations, one can also apply the Factor-Label Method (Ex 38 and 39)

The above equations can also be obtained from an unbalanced equation say

KClO3  KCl  O2

by applying POAC for different atoms as explained in Chapter 1

Another important method used for solving problems based on chemicalequations is the equivalent (or milli-equivalent) method. This method is based

on the fact that for the different amounts of reactants and products involved

in the reaction, the number of equivalents (or milli-equivalents) of eachreactant and each product are equal

For the reaction,

VO  Fe2O3  FeO  V2O5

eq of VO = eq of Fe2O3 = eq of FeO = eq of V2O5

or, m.e of VO = m.e of Fe2O3 = m.e of FeO = m.e of V2O5

In this method too, balancing of chemical equations is not required. Thismethod is generally applied in volumetric stoichiometric calculations

22

Concept of Limiting Reagent

In single-reactant reactions, the calculations are carried out with only thatamount of the reactant which has converted to the product This is done inall the methods mentioned above

In the reactions where more than one reactant is involved, one has tofirst identify the limiting reactant, i.e., the reactant which is completely consumed.

All calculations are to be carried out with the amount of the limiting reactantonly Now the question is how to identify the limiting reactant? The procedure issimple as illustrated below:

Initially 5 moles

A  12 moles2 B  0 moles4 C

If A is the limiting reactant: moles of C produced = 20

If B is the limiting reactant: moles of C produced = 24

The reactant producing the least number of moles of the product is the limiting reactant and hence A is the limiting reactant Thus,

Initially 

Finally 

5 molesA

However, in the following examples we shall mainly follow the POAC method, just for the practice of this modern concept.

[Note: In problems involving complicated reactions in which their balanced

chemical equations are given, one should prefer to apply the mole methodrather than the m.e method as followed in Example 50 in Ch 7.]

EXAMPLES

Ex 1 What amount of CaO will be produced by 1 g of calcium?

Applying POAC for Ca atoms as Ca atoms are conserved,

moles of Ca atoms in the reactant

 moles of Ca atoms in the product

 moles of Ca atoms in CaO  1  moles of CaO(B 1 mole of CaO contains 1 mole of Ca atoms)

wt of Ca

at wt of Ca  mol.wt of CaO wt of CaO

(Rule 2) (Rule 1)

401 wt of CaO56 ; wt of CaO  5640  14 g.

Ex 2 What weight of oxygen will react with 1 g of calcium? (Ca  40)

Since all the atoms of Ca have changed into CaO, the amount of Ca inCaO is 1 g Now from the formula of CaO, we have,

moles of Ca moles of O (Rule 6)

Applying POAC for Ca atoms,

moles of Ca in the reactant  moles of Ca in CaO

1

40 Rule 2 1 moles of CaO(B 1 mole of CaO contains 1 mole of Ca atom)

 moles of CaO  401   (1)

Again applying POAC for oxygen atoms,

moles of O in O2  moles of O in CaO

2  moles of O2  1  moles of CaO  (2)(B 1 moles O2 contains 2 moles of O and 1 mole of CaO contains 1 mole

[Note: The chemical equation of the above given problem is simple, i.e., easy to

balance But in complicated reactions (Ex 5, 6, etc.) where the balancing

is not very easy, the student can apply POAC without balancing theequation This is where the mole method has its importance.]

Ex 3(a) Calculate the volume occupied by 1 mole of He, H and O atoms at NTP.

But hydrogen and oxygen being diatomic gases, 1 mole of their respectiveatoms will occupy a volume of 112 litres at NTP.

(b) What volume of oxygen (NTP) will be required to react with 1 g of Ca?

we have,

moles of Ca  moles of O (atoms)

wt of at wt.Cavol occupied vol of by O1 mole of O atoms at atoms at NTP NTP

(i) weight of oxygen produced,

(ii) weight of KClO3 originally taken, and

(iii) weight of KCl produced

(K = 39, Cl  355 and O  16)

Solution : (i) Mole of oxygen  22400448 002 (Rule 3)

Wt of oxygen  002  32  064 g (Rule 1)(ii) KClO3  KCl  O2

Applying POAC for O atoms,

moles of O atoms in KClO3  moles of O atoms in O2

3  moles of KClO3  2  moles of O2

(1 mole of KClO3 contains 3 moles of O and 1 mole of O2 contains 2 moles

of O)

3  mol.wt wt of KClOof KClO3

3 2 vol at NTP litres

224 (Rule 1) (Rule 3)

3  wt of KClO3

1225  2 

0448224 

Wt of KClO3  1634 g

(iii) Again applying POAC for K atoms,

moles of K atoms in KClO3  moles of K atoms in KCl

or 1  moles of KClO3 1  moles of KCl

(1 mole of KClO3 contains 1 mole of K and 1 mole of KCl contains 1 mole

of K)

1  mol.wt of KClO wt of KClO3

3  1  mol wt of wt. of KClKCl 1634

1225wt of KCl745 

Wt of KCl  09937 g

Ex 5 276 g of K2CO3 was treated by a series of reagents so as to convert all of its carbon to K2Zn3 [FeCN6]2 Calculate the weight of the product.

Since C atoms are conserved, applying POAC for C atoms,

moles of C in K2CO3  moles of C in K2Zn3 [FeCN6]2

1  moles of K2CO3  12  moles of K2Zn3 [FeCN6]2

(B 1 mole of K2CO3 contains 1 mole of C

and 1 mole of K2Zn3 [FeCN6]2 contains 12 moles of C)

mol wt.wt of Kof K2CO2CO3 3 12  mol.wt of the product wt of product

wt of K2Zn3[FeCN6]2  276

138  69812  116 g

[mol wt of K2CO3 = 138 and mol wt of K2Zn3[FeCN6]2  698]

Ex 6 In a gravimetric determination of P, an aqueous solution of dihydrogen

phosphate ion H2PO4  is treated with a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate, MgNH4PO4 6H2O

This is heated and decomposed to magnesium pyrophosphate, Mg2P2O7 , which

is weighed A solution of H2PO4  yielded 1054 g of Mg2P2O7 What weight of

NaH2PO4 was present originally?

Na  23, H  1, P  31, O 16, Mg 24

heated Mg2P2O7

Since P atoms are conserved, applying POAC for P atoms,

moles of P in NaH2PO4 moles of P in Mg2P2O7

1  moles of NaH2PO4  2  moles of Mg2P2O7

(B 1 mole of NaH2PO4 contains 1 mole of P and 1 mole of Mg2P2O7contains 2 moles of P)

mol wt of NaHwt. of NaH2PO4

[Note: In Ex 5 and Ex 6 the students should note that if in any reaction, a

particular atom is conserved, the principle of atom conservation (POAC)with respect to that atom can be applied regardless of the number of steps

of the reaction and their sequence.]

Ex 7 What weight of AgCl will be precipitated when a solution containing

477 g of NaCl is added to a solution of 577 g of AgNO3?

Na 23, Cl  355, Ag 108, N  14 and O  16

No of moles of NaCl  477

moles of Ag in AgNO3  moles of Ag in AgCl

1  moles of AgNO3  1  moles of AgCl

wt of AgNO3

mol wt of AgNO3 mol wt wt of AgClof AgCl

wt of AgCl  003394  1435  487g

Ex 8 What is the number of moles of Fe OH3 s that can be produced by allowing

1 mole of Fe2S3 , 2 moles of H2O and 3 moles of O2 to react?

2Fe2S3s  6H2O l  3O2g  4Fe OH3s  6Ss

mole ratio in the reaction, (i.e., 2 : 6 : 3) it is clear that H2O is the limiting reagent as H2O shall be fully consumed in the reaction Thus,

moles of FeOH3 produced by 2 moles of H2O  46  2  134

Ex 9 Equal weights of phosphorus and oxygen are heated in a closed vessel

producing P2O3 and P2O5 in a 1 : 1 mole ratio If the limiting component is exhausted, find which component and also what fraction of it is left over.

Solution : As 1 mole each of P2O3 and P2O5 is formed, the mole ratio of thereacting P and O atoms is 4 : 8 or 1 : 2.

As the given weights of phosphorus and oxygen are same, say w g, their

moles of atoms are w

31 and 16w respectively.

 2 moles of O combine with 1 mole of P

 16w moles of O combine with w

Ex 10 In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde

containing manganese(II) acetate (catalyst) under pressure at 60°C.

2CH3CHO  O2  2CH3COOH

In a laboratory test of this reaction, 20 g of CH3CHO and 10 g of O2 were put into a reaction vessel.

(a) How many grams of CH3COOH can be produced?

(b) How many grams of the excess reactant remain after the reaction is complete?

(a) CH3CHO is the limiting reagent and hence

amount of CH3COOH produced  04545  60 g

 2727 g.(b) Amount of O2 left  00853  32

stoichiometric coefficients of the reactants would be 075 : 2 or 3 : 8 for

A4 and O2 Thus,