Chemistry and Solution

CHEMISTRY HIGHER SECONDARY – FIRST YEAR VOLUME – I REVISED BASED ON THE RECOMMENDATIONS OF THE TEXT BOOK DEVELOPMENT COMMITTEE A Publication Under Government of Tamilnadu Distribution of Free Textbook Programme (NOT FOR SALE) Untouchability is a sin Untouchability is a crime Untouchability is inhuman TAMILNADU TEXTBOOK CORPORATION College Road, Chennai – 600 006 © Government of Tamilnadu First Edition – 2004 Revised Edition – 2007 CHAIRPERSON & AUTHOR Dr. V. BALASUBRAMANIAN Professor of Chemistry (Retd. ) Presidency College, (Autonomous), Chennai – 600 005. REVIEWERS AUTHORS Dr. S.

P. MEENAKSHISUNDRAM Professor of Chemistry, Annamalai University, Annamalai Nagar 608 002. Dr. R. RAMESH Senior Lecturer in Chemistry, Bharathidasan University Trichirapalli 620 024. Mrs. T. VIJAYARAGINI P. G. Teacher in Chemistry, SBOA Mat. Higher Secondary School Chennai – 600 101. Dr. S. MERLIN STEPHEN, P. G. Teacher in Chemistry CSI Bain Mat. Hr. Sec. School Kilpauk, Chennai – 600 010. Dr. K. SATHYANARAYANAN, P. G. Teacher in Chemistry, Stanes Anglo Indian Hr. Sec. School, Coimbatore – 18. Dr. M. RAJALAKSHMI P. G. Teacher in Chemistry, Chettinad Vidyashram Chennai – 600 028.

We will write a custom essay sample on
Chemistry and Solution
or any similar topic only for you
Order now

Dr. M. KRISHNAMURTHI Professor of Chemistry Presidency College (Autonomous) Chennai – 600 005. Dr. M. KANDASWAMY Professor and Head Department of Inorganic Chemistry University of Madras Chennai – 600 025. Dr. M. PALANICHAMY Professor of Chemistry Anna University Chennai – 600 025. DR. J. SANTHANALAKSHMI Professor of Physical Chemistry University of Madras Chennai – 600 025. Mr. V. JAISANKAR, Lecturer in Chemistry L. N. Government Arts College, Ponneri – 601 204. Price : Rs. This book has been prepared by the Directorate of School Education on behalf of the Government of Tamilnadu.

This book has been printed on 60 G. S. M paper Printed by Offset at : (ii) PREFACE Where has chemistry come from ? Throughout the history of the human race, people have struggled to make sense of the world around them. Through the branch of science we call chemistry we have gained an understanding of the matter which makes up our world and of the interactions between particles on which it depends. The ancient Greek philosophers had their own ideas of the nature of matter, proposing atoms as the smallest indivisible particles.

However, although these ideas seems to fit with modern models of matter, so many other Ancient Greek ideas were wrong that chemistry cannot truly be said to have started there. Alchemy was a mixture of scientific investigation and mystical quest, with strands of philosophy from Greece, China, Egypt and Arabia mixed in. The main aims of alchemy that emerged with time were the quest for the elixir of life (the drinking of which would endue the alchemist with immortality), and the search for the philosopher’s stone, which would turn base metals into gold.

Improbable as these ideas might seem today, the alchemists continued their quests for around 2000 years and achieved some remarkable successes, even if the elixir of life and the philosopher’s stone never appeared. Towards the end of the eighteenth century, pioneering work by Antoine and Marie Lavoisier and by John Dalton on the chemistry of air and the atomic nature of matter paved the way for modern chemistry. During the nineteenth century chemists worked steadily towards an understanding of the relationships between the different chemical elements and the way they react together.

A great body of work was built up from careful observation and experimentation until the relationship which we now represent as the periodic table emerged. This brought order to the chemical world, and from then on chemists have never looked back. Modern society looks to chemists to produce, amongst many things, healing drugs, pesticides and fertilisers to ensure better crops and chemicals for the many synthetic materials produced in the twenty-first century. It also looks for an academic understanding of how matter works and how the environment might be protected from the source of pollutants.

Fortunately, chemistry holds many of the answers ! (iii) Following the progressing trend in chemistry, it enters into other branches of chemistry and answers for all those miracles that are found in all living organisms. The present book is written after following the revised syllabus, keeping in view with the expectations of National Council of Educational Research & Training (NCERT). The questions that are given in each and every chapter can be taken only as model questions. A lot of self evaluation questions, like, choose the best answer, fill up the blanks and very short answer type questions are given in all chapters.

While preparing for the examination, students should not restrict themselves, only to the questions/problems given in the self evaluation. They must be prepared to answer the questions and problems from the entire text. Learning objectives may create an awareness to understand each and every chapter. Sufficient reference books are suggested so as to enable the students to acquire more informations about the concepts of chemistry. Dr. V. BALASUBRAMANIAN Chairperson Syllabus Revision Committee (Chemistry) & XI Std Chemistry Text Book Writing Committee (iv) Syllabus : Higher Secondary – First Year Chemistry

INORGANIC CHEMISTRY Unit I – Chemical Calculations Significant figures – SI units – Dimensions – Writing number in scientific notation – Conversion of scientific notation to decimal notation – Factor label method – Calculations using densities and specific gravities – Calculation of formula weight – Understanding Avogadro’s number – Mole concept-mole fraction of the solvent and solute – Conversion of grams into moles and moles into grams Calculation of empirical formula from quantitative analysis and percentage composition – Calculation of molecular formula from empirical formula – Laws of chemical combination nd Dalton’s atomic theory – Laws of multiple proportion and law of reciprocal proportion – Postulates of Dalton’s atomic theory and limitations – Stoichiometric equations – Balancing chemical equation in its molecular form – Oxidation reduction-Oxidation number – Balancing Redox equation using oxidation number – Calculations based on equations. Mass/Mass relationship Methods of expressing concentration of solution – Calculations on principle of volumetric analysis – Determination of equivalent mass of an element Determination of equivalent mass by oxide, chloride and hydrogen displacement method – Calculation of equivalent mass of an element and compounds Determination of molar mass of a volatile solute using Avogadro’s hypothesis.

Unit 2 – Environmental Chemistry Environment – Pollution and pollutants – Types of pollution – Types of pollutants – Causes for pollution – Effects of pollution – General methods of prevention of environmental pollution. Unit 3 – General Introduction to Metallurgy Ores and minerals – Sources from earth, living system and in sea Purification of ores-Oxide ores sulphide ores magnetic and non magnetic ores Metallurgical process – Roasting-oxidation – Smelting-reduction – Bessemerisation – Purification of metals-electrolytic and vapour phase refining – Mineral wealth of India. v) Unit 4 – Atomic Structure – I Brief introduction of history of structure of atom – Defects of Rutherford’s model and Niels Bohr’s model of an atom – Sommerfeld’s extension of atomic structure – Electronic configuration and quantum numbers – Orbitals-shapes of s, p and d orbitals. – Quantum designation of electron – Pauli’s exclusion principle – Hund’s rule of maximum multiplicity – Aufbau principle – Stability of orbitals Classification of elements based on electronic configuration.

Unit 5 – Periodic Classification – I Brief history of periodic classification – IUPAC periodic table and IUPAC nomenclature of elements with atomic number greater than 100 – Electronic configuration and periodic table – Periodicity of properties Anomalous periodic properties of elements. Unit 6 – Group-1s Block elements Isotopes of hydrogen – Nature and application – Ortho and para hydrogen – Heavy water – Hydrogen peroxide – Liquid hydrogen as a fuel – Alkali metals – General characteristics – Chemical properties – Basic nature of oxides and hydroxides – Extraction of lithium and sodium – Properties and uses.

Unit 7 – Group – 2s – Block elements General characteristics – Magnesium – Compounds of alkaline earth metals. Unit 8 -p- Block elements General characteristics of p-block elements – Group-13. Boron Group Important ores of Boron – Isolation of Born-Properties – Compounds of BoronBorax, Boranes, diboranes, Borazole-preparation. roperties – Uses of Boron and its compounds – Carbon group – Group -14 – Allotropes of carbon Structural difference of graphite and diamond – General physical and chemical properties of oxides, carbides, halides and sulphides of carbon group – Nitrogen – Group-15 – Fixation of nitrogen – natural and industrial – HNO3-Ostwald process – Uses of nitrogen and its compounds – Oxygen – Group-16 – Importance of molecular oxygen-cell fuel – Difference between nascent oxygen and molecular oxygen – Oxides classification, acidic basic, amphoteric, neutral and peroxide Ozone preparation, property and structure – Factors affecting ozone layer. vi) Physical Chemistry Unit 9 – Solid State – I Classification of solids-amorphous, crystalline – Unit cell – Miller indices Types of lattices belong to cubic system. Unit 10 – Gaseous State Four important measurable properties of gases – Gas laws and ideal gas equation – Calculation of gas constant ‘‘R” – Dalton’s law of partial pressure Graham’s law of diffusion – Causes for deviation of real gases from ideal behaviour – Vanderwaal’s equation of state – Critical phenomena – Joule-Thomson effect and inversion temperature – Liquefaction of gases – Methods of Liquefaction of gases.

Unit 11 – Chemical Bonding Elementary theories on chemical bonding – Kossel-Lewis approach – Octet rule – Types of bonds – Ionic bond – Lattice energy and calculation of lattice energy using Born-Haber cycle – Properties of electrovalent compounds Covalent bond – Lewis structure of Covalent bond – Properties of covalent compounds – Fajan’s rules – Polarity of Covalent bonds – VSEPR Model Covalent bond through valence bond approach – Concept of resonance Coordinate covalent bond.

Unit 12 – Colligative Properties Concept of colligative properties and its scope – Lowering of vapour pressure – Raoul’s law – Ostwald – Walker method – Depression of freezing point of dilute solution – Beckmann method – Elevation of boiling point of dilute solution – Cotrell’s method – Osmotic pressure – Laws of Osmotic pressure Berkley-Hartley’s method – Abnormal colligative properties Van’t Hoff factor and degree of dissociation.

Unit 13 – Thermodynamics – I Thermodynamics – Scope – Terminology used in thermodynamics Thermodynamic properties – nature – Zeroth law of thermodynamics – Internal energy – Enthalpy – Relation between ‘‘H and “E – Mathematical form of First law – Enthalpy of transition – Enthalpy of formation – Enthalpy of combustion (vii) Enthalpy of neutralisation – Various sources of energy-Non-conventional energy resources.

Unit 14 – Chemical Equilibrium – I Scope of chemical equilibrium – Reversible and irreversible reactions Nature of chemical equilibrium – Equilibrium in physical process – Equilibrium in chemical process – Law of chemical equilibrium and equilibrium constant Homogeneous equilibria – Heterogeneous equilibria. Unit 15 – Chemical Kinetics – I Scope – Rate of chemical reactions – Rate law and rate determining step Calculation of reaction rate from the rate law – Order and molecularity of the reactions – Calculation of exponents of a rate law – Classification of rates based on order of the reactions.

ORGANIC CHEMISTRY Unit 16 – Basic Concepts of Organic Chemistry Catenation – Classification of organic compounds – Functional groups Nomenclature – Isomerism – Types of organic reactions – Fission of bonds Electrophiles and nucleophiles – Carbonium ion Carbanion – Free radicals Electron displacement in covalent bond. Unit 17 – Purification of Organic compounds Characteristics of organic compounds – Crystallisation – Fractional Crystallisation – Sublimation – Distillation – Fractional distillation – Steam distillation – Chromotography.

Unit 18 – Detection and Estimation of Elements Detection of carbon and hydrogen – Detection of Nitrogen – Detection of halogens – Detection of sulphur – Estimation of carbon and hydrogen – Estimation of Nitrogen – Estimation of sulphur – Estimation of halogens. Unit 19 – Hydrocarbons Classification of Hydrocarbons – IUPAC nomenclature – Sources of alkanes – General methods of preparation of alkanes – Physical properties (viii) Chemical properties – Conformations of alkanes – Alkenes – IUPAC nomenclature of alkenes – General methods of preparation – Physical properties – Chemical properties – Uses – Alkynes – IUPAC Nomenclature of lkynes – General methods of preparation – Physical properties – Chemical properties – Uses. Unit 20 – Aromatic Hydrocarbons Aromatic Hydrocarbons – IUPAC nomenclature of aromatic hydrocarbons – Structure of Benzene – Orientation of substituents on the benzene ring Commercial preparation of benzene – General methods of preparation of Benzene and its homologues – Physical properties – Chemical properties – Uses Carcinogenic and toxic nature.

Unit 21 – Organic Halogen Compounds Classification of organic hydrogen compounds – IUPAC nomenclature of alkyl halides – General methods of preparation – Properties – Nucleophilic substitution reactions – Elimination reactions – Uses – Aryl halide – General methods of preparation – Properties – Uses – Aralkyl halides – Comparison arylhalides and aralkyl halides – Grignard reagents – Preparation – Synthetic uses. (ix) CHEMISTRY PRACTICALS FOR STD XI I. II. Knowledge of using Burette, Pipette and use of logarithms is to be demonstrated. Preparation of Compounds. 1.

Copper Sulphate Crystals from amorphous copper sulphate solutions 2. Preparation of Mohr’s Salt 3. Preparation of Aspirin 4. Preparation of Iodoform 5. Preparation of tetrammine copper (II) sulphate III. Identification of one cation and one anion from the following. (Insoluble salt should not be given) Cation : Pb++, Cu++, Al++, Mn2+, Zn++, Ca++, Ba++, Mg++, NH4+. Anions : Borate, Sulphide, Sulphate, Carbonate, Nitrate, Chloride, Bromide. IV. V. Determination of Melting point of a low melting solid. Acidimetry Vs Alkalimetry 1. 2. 3. 4. Preparation of Standard solution of Oxalic acid and Sodium Carbonate solution.

Titration of HCl Vs NaOH Titration of HCl Vs Na2CO3 Titration of Oxalic acid Vs NaOH (x) CONTENTS UNIT NO. Inorganic Chemistry 1. 2. 3. 4. 5. 6. 7. Chemical Calculations General Introduction to Metallurgy Atomic Structure – I Periodic Classification – I Group 1 s-Block elements Group 2 s-Block elements p-Block elements PAGE NO. 1 43 57 76 110 133 146 Physical Chemistry 8. 9. Solid state – I Gaseous state – I 175 194 (xi) INORGANIC CHEMISTRY 1. CHEMICAL CALCULATION OBJECTIVES * * * * * * * * Know the method of finding formula weight of different compounds. Recognise the value of Avogadro number and its significance.

Learn about the mole concept and the conversions of grams to moles. Know about the empirical and molecular formula and understand the method of arriving molecular formula from empirical formula. Understand the stoichiometric equation. Know about balancing the equation in its molecular form. Understand the concept of reduction and oxidation. Know about the method of balancing redox equation using oxidation number. 1. 1 Formula Weight (FW) or Formula Mass The formula weight of a substance is the sum of the atomic weights of all atoms in a formula unit of the compound, whether molecular or not.

Sodium chloride, NaCl, has a formula weight of 58. 44 amu (22. 99 amu from Na plus 35. 45 amu from Cl). NaCl is ionic, so strictly speaking the expression “molecular weight of NaCl” has no meaning. On the other hand, the molecular weight and the formula weight calculated from the molecular formula of a substance are identical. Solved Problem Calculate the formula weight of each of the following to three significant figures, using a table of atomic weight (AW): (a) chloroform CHCl3 (b) Iron (III) sulfate Fe2 (SO4)3. Solution a. 1 x AW of C = 12. 0 amu 1 x AW of H = 1. 0 amu 3 x AW of Cl = 3 x 35. 5 = 106. 4 amu Formula weight of CHCl3 = 119. 4 amu 1 The answer rounded to three significant figures is 119 amu. b. Iron(III)Sulfate 2 x Atomic weight of Fe = 2 x 55. 8 = 111. 6 amu 3 x Atomic weight of S = 3 x 32. 1 = 96. 3 amu = 399. 9 amu 3 x 4 Atomic weight of O =12×16= 192. 0 amu Formula weight of Fe2 (SO4)3 Problems for Practice Calculate the formula weights of the following compounds a. NO2 b. glucose (C6H12O6) f. PCl3 c. NaOH g. K2 CO3 d. Mg(OH)2 e. methanol (CH3 OH) The answer rounded to three significant figures is 4. 00 x 102 amu. 1. 2 Avogadro’s Number (NA)

The number of atoms in a 12-g sample of carbon – 12 is called Avogadro’s number (to which we give the symbol NA). Recent measurements of this number give the value 6. 0221367 x 1023, which is 6. 023 x 1023. A mole of a substance contains Avogadro’s number of molecules. A dozen eggs equals 12 eggs, a gross of pencils equals 144 pencils and a mole of ethanol equals 6. 023 x 1023 ethanol molecules. Significance The molecular mass of SO2 is 64 g mol-1. 64 g of SO2 contains 6. 023 x 1023 molecules of SO2. 2. 24 x 10-2m3 of SO2 at S. T. P. contains 6. 023 x 1023 molecules of SO2. Similarly the molecular mass of CO2 is 44 g mol-1. 4g of CO2 contains 6. 023 x 1023 molecules of CO2. 2. 24 x 10-2m3 of CO2 at S. T. P contains 6. 023 x 1023 molecules of CO2. 1. 3 Mole concept While carrying out reaction we are often interested in knowing the number of atoms and molecules. Some times, we have to take the atoms or molecules of different reactants in a definite ratio. 2 Eg. Consider the following reaction 2 H2 + O2 > 2H2O In this reaction one molecule of oxygen reacts with two molecules of hydrogen. So it would be desirable to take the molecules of H2 and oxygen in the ratio 2:1, so that the reactants are completely consumed during the reaction.

But atoms and molecules are so small in size that is not possible to count them individually. In order to overcome these difficulties, the concept of mole was introduced. According to this concept number of particles of the substance is related to the mass of the substance. Definition The mole may be defined as the amount of the substance that contains as many specified elementary particles as the number of atoms in 12g of carbon – 12 isotope. (i. e) one mole of an atom consists of Avogadro number of particles. One mole One mole of oxygen molecule One mole of oxygen atom One mole of ethanol = = = = 6. 23 x 1023 particles 6. 023 x 1023 oxygen molecules 6. 023 x 1023 oxygen atoms 6. 023 x 1023 ethanol molecules In using the term mole for ionic substances, we mean the number of formula units of the substance. For example, a mole of sodium carbonate, Na2CO3 is a quantity containing 6. 023 x 1023 Na2CO3 units. But each formula unit of Na2CO3 contains 2 x 6. 023 x 1023 Na+ ions and one CO32ions and 1 x 6. 023 x 1023 CO32- ions. When using the term mole, it is important to specify the formula of the unit to avoid any misunderstanding. Eg. A mole of oxygen atom (with the formula O) contains 6. 23 x 1023 Oxygen atoms. A mole of oxygen molecule (formula O2) contains 6. 023 x 1023 O2 molecules (i. e) 2 x 6. 023 x 1023 oxygen. Molar mass The molar mass of a substance is the mass of one mole of the substance. The mass and moles can be related by means of the formula. 3 Mass Molar mass = ____ mole Eg. Carbon has a molar mass of exactly 12g/mol. Problems Solved Problems 1. What is the mass in grams of a chlorine atom, Cl? 2. What is the mass in grams of a hydrogen chloride, HCl? Solution 1. The atomic weight of Cl is 35. 5 amu, so the molar mass of Cl is 35. 5 g/mol. Dividing 35. 5 g (per mole) by 6. 23 x 1023 gives the mass of one atom. 35. 5 g Mass of a Cl atom = __________ 6. 023 x 1023 = 5. 90 x 10-23 g 2. The molecular weight of HCl equal to the atomic weight of H, plus the atomic weight of Cl, (ie) (1. 01 + 35. 5) amu = 36. 5 amu. Therefore 1 mol of HCl contains 36. 5 g HCl 36. 5 g _________ Mass of an HCl molecule = 6. 02 x1023 = 6. 06×10-23g Problems For Practice 1. What is the mass in grams of a calcium atom, Ca? 2. What is mass in grams of an ethanol molecule, C2H5OH? 3. Calcualte the mass (in grams) of each of the following species. a. Na atom b. S atom c. CH3Cl molecule d.

Na2SO3 formula unit 1. 3. 1 Mole Calculations To find the mass of one mole of substance, there are two important things to know. 4 i. How much does a given number of moles of a substance weigh? ii. How many moles of a given formula unit does a given mass of substance contain. Both of them can be known by using dimensional analysis. To illustrate, consider the conversion of grams of ethanol, C2H5OH, to moles of ethanol. The molar mass of ethanol is 46. 1 g/mol, So, we write 1 mol C2H5OH = 46. 1 g of C2 H5OH Thus, the factor converting grams of ethanol to moles of ethanol is 1mol C2H5OH/46. 1g C2H5OH.

To covert moles of ethanol to grams of ethanol, we simply convert the conversion factor (46. 1 g C2H5OH/1 mol C2H5OH). Again, suppose you are going to prepare acetic acid from 10. 0g of ethanol, C2H5OH. How many moles of C2H5OH is this? you convert 10. 0g C2H5OH to moles C2H5OH by multiplying by the appropriate conversion factor. 1 mol C2H5OH 10. 0g C H OH x _____________ 2 5 46. 1 g C2H5OH = 0. 217 mol C2H5 OH 1. 3. 2 Converting Moles of Substances to Grams Solved Problems 1. ZnI2, can be prepared by the direct combination of elements. A chemist determines from the amounts of elements that 0. 654 mol ZnI2 can be formed. Solution The molar mass of ZnI2 is 319 g/mol. (The formula weight is 319 amu, which is obtained by summing the atomic weight in the formula) Thus 319 g ZnI2 0. 0654 mol ZnI x __________ 2 1 mol ZnI2 = 20. 9 gm ZnI2 5 Problems for Practice 1. H2O2 is a colourless liquid. A concentrated solution of it is used as a source of oxygen for Rocket propellant fuels. Dilute aqueous solutions are used as a bleach. Analysis of a solution shows that it contains 0. 909 mol H2O2 in 1. 00 L of solution. What is the mass of H2O2 in this volume of solution?. 2.

Boric acid, H3BO3 is a mild antiseptic and is often used as an eye wash. A sample contains 0. 543 mol H3BO3. What is the mass of boric acid in the sample?. 3. CS2 is a colourless, highly inflammable liquid used in the manufacture of rayon and cellophane. A sample contains 0. 0205 mol CS2. Calculate the mass of CS2 in the sample. Converting Grams of Substances to Moles In the preparation of lead(II)chromate PbCrO4, 45. 6 g of lead(II)chromate is obtained as a precipitate. How many moles of PbCrO4 is this? The molar mass of PbCrO4 is 323 g/mol (i. e) 1 mol PbCrO4 = 323 g PbCrO4 Therefore, 45. g PbCrO4 x 1 mol. PbCrO4 ___________________________ 323 g PbCrO4 = 0. 141 mol PbCrO4 Problems for Practice 1. Nitric acid, HNO3 is a colourless, corrosive liquid used in the manufacture of Nitrogen fertilizers and explosives. In an experiment to develop new explosives for mining operations, a 28. 5 g sample of HNO3 was poured into a beaker. How many moles of HNO3 are there in this sample of HNO3? 2. a. c. e. g. Obtain the moles of substances in the following. 3. 43 g of C b. 7. 05 g Br2 d. 35. 4 g Li2 CO3 76g C4 H10 2. 57 g As f. 7. 83 g P4 h. 153 g Al2 (SO4)3 41. 4 g N2H4 6

Calculation of the Number of Molecules in a Given Mass Solved Problem How many molecules are there in a 3. 46 g sample of hydrogen chloride, HCl? Note: The number of molecules in a sample is related to moles of compound (1 mol HCl = 6. 023 x 1023 HCl molecules). Therefore if you first convert grams HCl to moles, then you can convert moles to number of molecules). Solution 3. 46g HClx 1 mol HCl 36. 5g HCl x 6. 023 x 10 23 HClmolecul es 1 mol HCl = 5. 71 x 1022 HCl molecules Problems for Practice 1. How many molecules are there in 56mg HCN? 2. a. b. c. Calculate the following Number of molecules in 43g NH3 Number of atoms in 32. g Br2 Number of atoms in 7. 46 g Li 1. 4 Calculation of Empirical Formula from Quantitative Analysis and Percentage composition Empirical Formula “An empirical formula (or) simplest formula for a compound is the formula of a substance written with the smallest integer subscripts”. For most ionic substances, the empirical formula is the formula of the compound. This is often not the case for molecular substances. For example, the formula of sodium peroxide, an ionic compound of Na+ and O22-, is Na2O2. Its empirical formula is NaO. Thus empirical formula tells you the ratio of numbers of atoms in the compound.

Steps for writing the Empirical formula The percentage of the elements in the compound is determined by 7 suitable methods and from the data collected, the empirical formula is determined by the following steps. i. Divide the percentage of each element by its atomic mass. This will give the relative number of moles of various elements present in the compound. ii. Divide the quotients obtained in the above step by the smallest of them so as to get a simple ratio of moles of various elements. iii. Multiply the figures, so obtained by a suitable integer of necessary in order to obtain whole number ratio. v. Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound. Solved Problem A compound has the following composition Mg = 9. 76%,S = 13. 01%, 0 = 26. 01, H2O = 51. 22, what is its empirical formula? [Mg = 24, S = 32, O = 16, H = 1] Solution Element Magnesium Relative No. of Simple ratio moles moles 9. 76 0. 406 _____ = 1 9. 76 ____ = 0. 406 24 0. 406 0. 406 13. 01 _____ = 1 13. 01 _____ = 0. 406 % 32 26. 01 26. 01 _____ = 1. 625 16 51. 22 51. 2 _____ = 2. 846 18 0. 406 1. 625 _____ = 4 0. 406 2. 846 _____ = 7 0. 406 7 Simplest whole No. ratio 1 Sulphur 1 Oxygen 4 Water Hence the empirical formula is Mg SO4. 7H2O. Problems for Practice 8 1. A substance on analysis, gave the following percentage composition, Na = 43. 4%, C = 11. 3%, 0 = 43. 3% calculate its empirical formula [Na = 23, C = 12, O = 16]. Ans:- Na2CO3 2. What is the simplest formula of the compound which has the following percentage composition: Carbon 80%, hydrogen 20%. Ans:- CH3 3. A compound on analysis gave the following percentage composition: C – 54. 54%, H = 9. 9%, 0 = 36. 36% Ans:- C2H4O 1. 4. 1 Molecular Formula from Empirical Formula The molecular formula of a compound is a multiple of its empirical formula. Example The molecular formula of acetylene, C2H2 is equivalent to (CH)2, and the molecular formula of benzene, C6H6 is equivalent to (CH)6. Therefore, the molecular weight is some multiple of the empirical formula weight, which is obtained by summing the atomic Weights from the empirical formula. For any molecular compound. Molecular Weight = n x empirical formula weight. Where `n’ is the whole number of empirical formula units in the molecule.

The molecular formula can be obtained by multiplying the subscripts of the empirical formula by `n’ which can be calculated by the following equation Molecular Weight _____________________ n = Empirical formula Weight Steps for writing the molecular formula i. Calculate the empirical formula ii. Find out the empirical formula mass by adding the atomic mass of all the atoms present in the empirical formula of the compound. iii. Divide the molecular mass (determined experimentally by some 9 suitable method) by the empirical formula mass and find out the value of n which is a whole number. iv.

Multiply the empirical formula of the compound with n, so as to find out the molecular formula of the compound. Solved Problem 1. A compound on analysis gave the following percentage composition C = 54. 54%, H, 9. 09% 0 = 36. 36. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound. Solution Calculation of empirical formula Element % Relative No. of moles 54. 54 _____ = 4. 53 12 9. 09 _____ = 9. 09 1 36. 36 _____ = 2. 27 16 Simple ratio moles 4. 53 ____ = 2 2. 27 9. 09 ____ = 4 2. 27 2. 27 ____ = 1 2. 27 Simplest whole No. ratio 2 C 54. 54 H 9. 09 4 O 36. 6 1 Empirical formula is C2 H4 O. Calculation of Molecular formula Empirical formula mass = 12 x 2 + 1 x 4 + 16 x 1 = 44 Molecular mass = 2 x Vapour density = 2 x 44 = 88 n Molecular mass 88 = ____________________ = ____ = 2 Empirical Formula mass 44 Molecular formula = Empirical formula x n 10 = C2 H4 O x 2 = C4 H8 O2 2. A compound on analysis gave the following percentage composition: Na=14. 31% S = 9. 97%, H = 6. 22%, O = 69. 5%, calcualte the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation.

Molecular mass of the compound is 322 [Na = 23, S = 32, H = 1, 0 = 16]. Solution :- Calculation of empirical formula Element % Relative No. of Simple ratio Simplest whole moles moles No. ratio 0. 62 ___ = 2 0. 31 0. 31 ___ = 1 0. 31 6. 22 ___ = 20 0. 31 4. 34 ___ = 14 0. 31 2 Na S H 14. 31 14. 31 ____ = 0. 62 23 19. 97 9. 97 ____ = 0. 31 32 6. 22 6. 22 ___ = 6. 22 1 69. 5 ___ = 4. 34 16 1 20 O 69. 5 14 The empirical formula is Na2 SH20O14 Calculation of Molecular formula Empirical formula mass = (23 x 2) + 32 + (20 x 1) + (16 x 14) = 322 n Molecular mass ___________________ = Empirical formula mass 322 ____ = 1 = 322

Hence molecular formula = Na2 SH20 O14 11 Since all hydrogens are present as H2O in the compound, it means 20 hydrogen atoms must have combined. It means 20 hydrogen atoms must have combined with 10 atoms of oxygen to form 10 molecules of water of crystallisation. The remaining (14 – 10 = 4) atoms of oxygen should be present with the rest of the compound. Hence, molecular formula = Na2SO4. 10H2O. Problems for Practice 1. An organic compound was found to have contained carbon = 40. 65%, hydrogen = 8. 55% and Nitrogen = 23. 7%. Its vapour – density was found to be 29. 5.

What is the molecular formula of the compound? Ans:- C2H5NO 2. A compound contains 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75. Calculate the empirical and molecular formula. Ans:- C2H3O3, C4H6O6 3. An acid of molecular mass 104 contains 34. 6% carbon, 3. 85% hydrogen and the rest is oxygen. Calcualte the molecualr formula of the acid. 4. What is the simplest formula of the compound which has the following percentage composition: carbon 80%, Hydrogen 20%, If the molecular mass is 30, calcualte its molecular formula. 1. 5 Stoichiometry Equations

Stoichiometry Stoichiometry is the calculation of the quantities of reactants and products involved in the chemical reaction. It is the study of the relationship between the number of mole of the reactants and products of a chemical reaction. A stoichiometric equation is a short scientific representation of a chemical reaction. Rules for writing stoichiometric equations i. In order to write the stoichiometric equation correctly, we must know the reacting substances, all the products formed and their chemical formula. ii. The formulae of the reactant must be written on the left side of 12 rrow with a positive sign between them. iii. The formulae of the products formed are written on the right side of the arrow mark. If there is more than one product, a positive sign is placed between them. The equation thus obtained is called skeleton equation. For example, the Chemical reaction between Barium chloride and sodium sulphate producing BaSO4 and NaCl is represented by the equation as BaCl2 + Na2SO4 > BaSO4 + NaCl This skeleton equation itself is a balanced one. But in many cases the skeleton equation is not a balanced one. For example, the decomposition of Lead Nitrate giving Lead oxide, NO2 and oxygen.

The skeletal equation for this reaction is Pb(NO3)2 > PbO + NO2 + O2 iv. In the skeleton equation, the numbers and kinds of particles present on both sides of the arrow are not equal. v. During balancing the equation, the formulae of substances should not be altered, but the number of molecules with it only be suitably changed. vi. Important conditions such as temperature, pressure, catalyst etc. , may be noted above (or) below the arrow of the equation. vii. An upward arrow (^) is placed on the right side of the formula of a gaseous product and a downward arrow (v) on the right side of the formulae of a precipitated product. viii.

All the reactants and products should be written as molecules including the elements like hydrogen, oxygen, nitrogen, fluorine chlorine, bromine and iodine as H2, O2, N2, F2, Cl2, Br2 and I2. 1. 5. 1 Balancing chemical equation in its molecular form A chemical equation is called balanced equation only when the numbers and kinds of molecules present on both sides are equal. The several steps involved in balancing chemical equation are discussed below. Example 1 Hydrogen combines with bromine giving HBr 13 H2 + Br2 > HBr This is the skeletal equation. The number of atoms of hydrogen on the left side is two but on the right side it is one.

So the number of molecules of HBr is to be multiplied by two. Then the equation becomes H2 + Br2 > Example 2 Potassium permanganate reacts with HCl to give KCl and other products. The skeletal equation is KMnO4 + HCl > KCl + MnCl2 + H2O + Cl2 If an element is present only one substance in the left hand side of the equation and if the same element is present only one of the substances in the right side, it may be taken up first while balancing the equation. According to the above rule, the balancing of the equation may be started with respect to K, Mn, O (or) H but not with Cl. There are L. H. S. R. H.

S K=1 Mn = 1 O =4 1 1 1 2HBr This is the balanced (or) stoichiometric equation. So the equation becomes KMnO4 + HCl > KCl + MnCl2 + 4H2O + Cl2 Now there are eight hydrogen atoms on the right side of the equation, we must write 8 HCl. KMnO4 + 8HCl > KCl + MnCl2 + 4H2O + Cl2 Of the eight chlorine atoms on the left, one is disposed of in KCl and two in MnCl2 leaving five free chlorine atoms. Therefore, the above equation becomes KMnO4+8HCl > KCl+MnCl2+4H2O+5/2 Cl2 Equations are written with whole number coefficient and so the 14 equation is multiplied through out by 2 to become 2KMnO4+16 HCl>2KCl+2 MnCl2+8H2O+5Cl2 1. 5. Redox reactions [Reduction – oxidation] In our daily life we come across process like fading of the colour of the clothes, burning of the combustible substances such as cooking gas, wood, coal, rusting of iron articles, etc. All such processes fall in the category of specific type of chemical reactions called reduction – oxidation (or) redox reactions. A large number of industrial processes like, electroplating, extraction of metals like aluminium and sodium, manufactures of caustic soda, etc. , are also based upon the redox reactions. Redox reactions also form the basis of electrochemical and electrolytic cells.

According to the classical concept, oxidation and reduction may be defined as, Oxidation is a process of addition of oxygen (or) removal of hydrogen Reduction is a process of removal of oxygen (or) addition of hydrogen. Example Reaction of Cl2 and H2S Oxidation H2S + Cl2 > Reduction In the above reaction, hydrogen is being removed from hydrogen sulphide (H2S) and is being added to chlorine (Cl2). Thus, H2S is oxidised and Cl2 is reduced. Electronic concept of oxidation and Reduction According to electronic concept, oxidation is a process in which an atom taking part in chemical reaction loses one or more electrons.

The loss of electrons results in the increase of positive charge (or) decrease of negative of the species. For example. Fe2+> Fe3+ + e- [Increase of positive charge] Cu> Cu2+ + 2e- [Increse of positive charge] 15 2HCl + S The species which undergo the loss of electrons during the reactions are called reducing agents or reductants. Fe2+ and Cu are reducing agents in the above example. Reduction Reduction is a process in which an atom (or) a group of atoms taking part in chemical reaction gains one (or) more electrons. The gain of electrons result in the decrease of positive charge (or) increase of negative charge of the species.

For example, Fe3+ + e- >Fe2+ [Decrease of positive charges] Zn2+ + 2e- > Zn [Decrease of positive charges] The species which undergo gain of electrons during the reactions are called oxidising agents (or) oxidants. In the above reaction, Fe3+ and Zn2+ are oxidising agents. Oxidation Number (or) Oxidation State Oxidation number of the element is defined as the residual charge which its atom has (or) appears to have when all other atoms from the molecule are removed as ions. Atoms can have positive, zero or negative values of oxidation numbers depending upon their state of combination.

General Rules for assigning Oxidation Number to an atom The following rules are employed for determining oxidation number of the atoms. 1. The oxidation number of the element in the free (or) elementary state is always Zero. Oxidation number of Helium in He = 0 Oxidation number of chlorine in Cl2 = 0 2. The oxidation number of the element in monoatomic ion is equal to the charge on the ion. 3. The oxidaton number of fluorine is always – 1 in all its compounds. 4. Hydrogen is assigned oxidation number +1 in all its compounds except in metal hydrides. In metal hydrides like NaH, MgH2, CaH2, LiH, 16 etc. , the oxidation number of hydrogen is -1. . Oxygen is assigned oxidation number -2 in most of its compounds, however in peroxides like H2O2, BaO2, Na2O2, etc its oxidation number is 1. Similarly the exception also occurs in compounds of Fluorine and oxygen like OF2 and O2F2 in which the oxidation number of oxygen is +2 and +1 respectively. 6. The oxidation numbers of all the atoms in neutral molecule is Zero. In case of polyatomic ion the sum of oxidation numbers of all its atoms is equal to the charge on the ion. 7. In binary compounds of metal and non-metal the metal atom has positive oxidation number while the non-metal atom has negative oxidation number.

Example. Oxidation number of K in KI is +1 but oxidation number of I is – I. 8. In binary compounds of non-metals, the more electronegative atom has negative oxidation number, but less electronegative atom has positive oxidation number. Example : Oxidation number of Cl in ClF3 is positive (+3) while that in ICl is negative (-1). Problem Calculate the oxidation number of underlined elements in the following species. CO2, Cr2O72-, Pb3O4, PO43Solution 1. C in CO2. Let oxidation number of C be x. Oxidation number of each O atom = -2. Sum of oxidation number of all atoms = x+2 (-2) ? x – 4.

As it is neutral molecule, the sum must be equal to zero. ? x-4 = 0 (or) x = + 4 2. Cr in Cr2O72-. Let oxidation number of Cr = x. Oxidation number of each oxygen atom =-2. Sum of oxidation number of all atoms 2x + 7(-2) = 2x – 14 Sum of oxidation number must be equal to the charge on the ion. Thus, 2x – 14 = -2 17 2x = +12 x x = 12/2 = 6 Problems for Practice Calculate the oxidation number of underlined elements in the following species. a. MnSO4 Oxidation “A chemical process in which oxidation number of the element increases”. Reduction “A chemical process in which oxidation number of the element decreases”. Eg.

Reaction between H2S and Br2 to give HBr and Sulphur. Decrease of Oxidation Number (Br) +1 -2 H2S 0 Br2 > +1 -1 2HBr 0 S b. S2O3 c. HNO3 d. K2MnO4 e. NH4+ Oxidation and Reduction in Terms of Oxidation Number + + Increase of oxidation Number (S) In the above reaction, the oxidation number of bromine decreases from 0 to -1, thus it is reduced. The oxidation number of S increases from -2 to 0. Hence H2S is oxidised. Under the concept of oxidation number, oxidising and reducing agent can be defined as follows. i. Oxidising agent is a substance which undergoes decrease in the oxidation number of one of its elements. 18 ii.

Reducing agent is a substance which undergoes increase in the oxidation number of one of its elements. In the above reaction H2S is reducing agent while Br2 is oxidising agent. Solved Problem Identify the oxidising agent, reducing agent, substance oxidised and substance reduced in the following reactions. MnO2 + 4HCl > MnCl2 + Cl2 + 2H2O Solution Increase of Oxidation Number +4 MnO2 + -1 4HCl > +2 MnCl2 + 0 Cl2 +1 -2 + 2H2 O Decrease of oxidation Number As it is clear, manganese decrease its oxidation number from +4 to +2. Hence, MnO2 gets reduced and it is an oxidising agent. Chlorine atom in HCl increases its oxidation number from -1 to 0.

Thus, HCl gets oxidised and it is reducing agent. Balancing Redox reaction by Oxidation Number Method The various steps involved in the balancing of redox equations according to this method are : 1. Write skeleton equation and indicate oxidation number of each element and thus identify the elements undergoing change in oxidation number. 2. Determine the increase and decrease of oxidation number per atom. Multiply the increase (or) decrease of oxidation number of atoms undergoing the change. 3. Equalise the increase in oxidation number and decrease in oxidation number on the reactant side by multiplying the respective formulae with 19 uitable integers. 4. Balance the equation with respect to all atoms other than O and H atoms. 5. Balance oxygen by adding equal number of water molecules to the side falling short of oxygen atoms. 6. H atoms are balanced depending upon the medium in same way as followed in ion electron method. Let us balance the following equations by oxidation number method. MnO2 + ClStep 1 MnO2 + Cl- > Mn2+ + Cl2 + H2O Step 2 O. N. Decreases by 2 per Mn 2+ > Mn + Cl2 + H2O in acidic medium 4+ + MnO2 -1 Cl +2 Mn + Cl2 + H2O O. N. increases by 1 per Cl Step 3 Equalise the increase / decrease in O.

N by multiply MnO2 by 1 and Cl-1 by 2. MnO2 + 2 Cl- >Mn2+ + Cl2 + H2O Step 4 Balance other atoms except H and O. Here they are all balanced. Step 5 Balance O atoms by adding H2O molecules to the side falling short of oxygen atoms. MnO2 + 2Cl- > Mn2+ + Cl2 + H2O + H2O 20 Step 6 Balance H atoms by adding H+ ions to the side falling short of H atoms MnO2 + 2Cl- + 4H+ > Mn2+ + Cl2 + 2H2O Problems for practice Balance the following equations 1. Mg + NO3- > Mg2+ + N2O + H2O (in acidic medium) 3+ 2. Cr + Na2O2 > CrO4- + Na+ 3. S2- + NO3- > NO + S 4. FeS + O2 > Fe2O3 + SO2 (molecular form) 5. Cl2 + OH- > Cl- + ClO3- + H2O . 6 Calculations based on chemical equations Stoichiometric problems are solved readily with reference to the equation describing the chemical change. From a stoichiometric chemical equation, we know how many molecules of reactant react and how many molecules of products are formed. When the molecular mass of the substances are inserted, the equation indicates how many parts by mass of reactants react to produce how many parts by mass of products. The parts by mass are usually in kg. So it is possible to calculate desired mass of the product for a known mass of the reactant or vice versa. . 6. 1 Mass / Mass Relationship Example 1 Calculate the mass of CO2 that would be obtained by completely dissolving 10kg of pure CaCO3 in HCl. CaCO3 + 2 HCl > CaCl2 + H2O + CO2 100 kg x 10-3 44 kg x 10-3 100 kg of CaCO3 produces 44 x 10-3 kg of CO2 44 x 10-3 21 10 kg of CaCO3 produces = ________ x 10 100 x 10-3 = 4. 4 kg of CO2 Example 2 Calculate the mass of oxygen obtained by complete decomposition of 10kg of pure potassium chlorate (Atomic mass K=39, O=16 and Cl = 35. 5) 2KClO3 > 2 KCl + 3O2 Molecular mass of KClO3 = 39+35. 5+48=122. 5 Molecular Mass of O2 = 16 + 16 = 32.

According to the Stoichiometric equation written above (2 x 122. 5) x 10 kg of KClO3 on heating gives (3 x 32) x 10-3 kg of oxygen. -3 10kg of KClO3 gives = 2 x 122. 5 x 10-3 3 x 32 x 10-3 ____________ x 10 = 3. 92 kg of O2 Example 3 Calculate the mass of lime that can be prepared by heating 200 kg of limestone that is 90% pure CaCO3 CaCO3 > CaO + CO2 100 kg x 10-3 56 kg x 10-3 90 200 kg of 90% pure CaCO = 200 x ___ 3 100 = 180 kg pure CaCO3 100×10-3 kg of pure CaCO3 on heating gives 56 x 10-3 kg of Ca0 56 x 10-3 x 180 ____________ 180 kg of CaCO = 3 gives on heating = 100 x 10-3 100. 8 kg CaO 1. Methods of Expressing the concentration of solution 22 The concentration of a solution refers to the amount of solute present in the given quantity of solution or solvent. The concentration of a solution may be expressed quantitatively in any of the following ways. 1. Strength The Strength of a solution is defined as the amount of the solute in grams, present in one litre of the solution. It is expressed in g L-1. Strength Mass of solute in grams = _______________________ Volume of solution in litres X x 1000 = _______ V If X gram of solute is present in V cm3 of a given solution then Strength 2.

Molarity (M) Molarity of a solution is defined as the number of gram-moles of solute dissolved in 1 litre of a solution Molarity No. of moles of solute = ______________________ Volume of Solution in litres X __________ x = Mol. mass 1000 _____ V If `X’ grams of the solute is present in V cm3 of a given solution, then, Molarity Molarity is represented by the symbol M. Molarity can also be calculated from the strength as follows Molarity Strength in grams per litre _______________________ = Molecular mass of the solute 23 Example A 0. 1M solution of Sugar, C12H22O11 (mol. mass = 342), means that 34. g of sugar is present in one litre (1000 cm3) of the solution. 3. Normality Normality of a solution is defined as the number of gram equivalents of the solute dissolved per litre of the given solution. Number of gram-equivalents of solute Normality = ______________________________ Volume of Solution in litre If X grams of the solute is present in V cm3 of a given solution, then, Normality = X ________ x Eq. mass 1000 mL _______ V Normality is represented by the symbol N. Normality can also be calculated from strength as follows Strength in grams per litre ______________________ Normality = Eq. ass of the solute Example A 0. 1N (or decinormal) solution of H2SO4 (Eq. mass = 49), means that 4. 9 g of H2SO4 is present in one litre (1000 cm3) of the solution. 4. Molality (m) Molality of a solution is defined as the number of gram-moles of solute dissolved in 1000 grams (or 1 kg) of a Solvent. Mathematically, Molality Number of moles of solute = _______________________ Mass of solvent in kilograms “If X grams of the solute is dissolved in b grams of the solvent”, then 24 Molality X 1000g ________ x _____ = Mol. mass bg Molality is represented by the symbol ‘m’. Example A 0. 1m Solution of glucose C6H12O6 (Mol. ass = 180), means that 18g of glucose is present in 1000g (or one kilogram) of water. 5. Mole Fraction Mole fraction is the ratio of number of moles of one component (Solute or Solvent) to the total number of moles of all the components (Solute and Solvent) present in the Solution. It is denoted by X. Let us suppose that a solution contains 2 components A&B and suppose that nA moles of A and nB moles of B are present in the solution. Then, nA Mole fraction of A, XA = _______ nA + nB ….. (1) nB ________ …… (2) Mole fraction of B, XB = nA + nB Adding 1 & 2 we get, nA XA + XB = nA+nB + nA+nB nB = nA+nB nA+nB

Thus, sum of the two mole fractions is one. Therefore, if mole fraction of one component in a binary solution is known, that of the other can be calculated. Solved Problems 1. 4. 5g of urea (molar mass = 60g mol-1) are dissolved in water and solution is made to 100 ml in a volumetric flask. Calculate the molarity of solution. 25 Solution Mass of Urea = 4. 5g Mass 4. 5g _________ = _______ Moles of Urea = Molar Mass 60gmol-1 = 0. 075mol Volume of Solution 100 ____ L = 1000 ml = 1000 = 0. 1L Molarity of Solution Mass of Solute in g _____________________ = Volume of Solution in litres 0. 075 = _____ mol = 0. 5M 0. 12 2. Calculate the normality of solution containing 3. 15g of hydrated oxalic acid (H2C2O4. 2H2O) in 250ml of solution (Mol. mass = 126). Solution Mass of oxalic acid = 3. 15g Mol. mass Equivalent mass of oxalic acid = ________ Basicity 126 = ?? = 63 g equiv-1 2 Mass of solute Equivalents of oxalic acid = ____________ Eq. Mass 26 3. 15g -1 ???? = 0. 05 equiv 63 g equiv-1 250 Volume of solution = 250ml = ?? L = 0. 25L 1000 Equivalent of Solute __________________ Normality = Volume of solution inL 0. 05 equiv = 0. 2N = ??? 0. 25 L 3. Calculate the molality of an aqueous solution containing 3. g of urea (mol. mass=60) in 250g of water. Solution Mass of solute = 3. 0g Mass of solute Moles of solute = ____________ Molar Mass 3. 0 g ________ = 0. 05 mol = 60 g mol-1 = 250g 250 = ?? = 0. 25 kg 1000 Moles of Solute __________________ = Mass of Solvent in kg. 0. 05 Mol ________ = 0. 2m = 0. 25 kg 27 Mass of Solvent Molality of Solution Problems for practice 1. Calculate the volume of 14. 3m NH3, solution needed to prepare 1L of 0. 1M solution. Ans:-6. 77 mL 2. How would you make up 425 mL of 0. 150M HNO3 from 68. 0% HNO3? The density of 68. 0% HNO3 is1. 41g/mL. Ans: 4. 25 mL 3.

Calculate the molarity of a solution obtained by mixing 100 mL of 0. 3 M H2SO4 and 200 mL of 1. 5M H2SO4 Ans:1. 1M 4. Calculate the molality of a solution by dissolving 0. 850g of ammonia (NH3) in 100g of water. Ans:0. 5m 1. 8 Calculations based on Principle of Volumetric Analysis 1. 8. 1 Volumetric Analysis An important method for determining the amount of a particular substance is based on measuring the volume of reactant solution. Suppose substance A in solution reacts with substance B. If you know the volume and concentration of a solution of B that just reacts with substance A in a sample, you can determine the amount of A.

Titration is a procedure for determining the amount of substance A by adding a carefully measured volume of a solution of A with known concentration of B untill the reaction of A and B is just completed. Volumetric analysis is a method of analysis based on titrations. Law “Equal volume of equinormal solutions exactly neutralise the other solution having same concentration and volume”. V1 N1 = V2N2 V1, V2 N1, N2 Volume of solutions. Strength of solutions. 28 Solved problem Calculating the volume of reactant solution needed 1. What volume of 6M HCl and 2M HCl should be mixed to get one litre of 3M HCl?

Solution Suppose the volume of 6M HCl required to obtain 1L of 3M HCl = XL Volume of 2M HCl required = (1-x)L Applying the molarity equation + M2V2 = M3V3 M1V1 6MHCl+ 2MHCl= 3MHCl 6x+2(1-x) = 3×1 6x+2-2x = 3 4x = 1 x = 0. 25L hence, volume of 6M HCl required Volume of 2M HCl required = 0. 25L = 0. 75L 2. How much volume of 10M HCl should be diluted with water to prepare 2. 00L of 5M HCl. Solution N1V1 10xV1 ? V1 = N2V2 = 5 x 2. 00 5 x 2. 00 = _______ 10 = 1. 00L Problems for Practice 1. NiSO4 reacts with Na3PO4 to give a yellow green precipitate of Ni3(PO4)2 and a solution of Na2SO4. 29 10N HCl = 5N HCl NiSO4(aq) + 2Na3PO4(aq) > Ni3(PO4)2 (s) + 3Na2SO4(aq) How many mL of 0. 375M NiSO4 will react with 45. 7 mL of 0. 265M Na3PO4? 2. What volume of 0. 250M HNO3 reacts with 42. 4 mL of 0. 150M Na2CO3 in the following reaction? 2HNO3(aq) + Na2CO3(aq) > 2NaNO3(aq) + H2O(aq)+CO2(g) 3. A flask contains 53. 1 mL of 0. 150M Ca(OH)2 solution. How many mL of 0. 350M Na2CO3 are required to react completely with Ca(OH)2 in the following reaction. Na2CO3(aq) + Ca(OH)2(aq) > CaCO3(aq) + 2NaOH(aq) 1. 8. 2 1. 2. 3. 4. Determination of equivalent masses of elements Hydrogen displacement method Oxide method Chloride method Metal displacement method

Equivalent masses can be determined by the following methods: Hydrogen displacement method This method is used to determine the equivalent mass of those metals such as magnesium, zinc and aluminium which react with dilute acids and readily displace hydrogen. Mg + 2HCl Zn + H2SO4 2Al + 6HCl MgCl2 + H2 ZnSO4 + H2 2AlCl3 + 3H2 From the mass of the metal and the volume of hydrogen displaced, the equivalent mass of the metal can be calculated. Problem 1 0. 548 g of the metal reacts with dilute acid and liberates 0. 0198 g of hydrogen at S. T. P. Calculate the equivalent mass of the metal. . 548 g of the metal displaces 0. 0198 g of hydrogen 30 The mass of the metal which will displace 1. 008 g of hydrogen = 1. 008 x 0. 548 ___________ g of metal 0. 0198 The equivalent mass of the metal = 27. 90 g equiv-1 Oxide Method This method is employed to determine the equivalent mass of those elements which readily form their oxides eg. magnesium, copper etc. Oxide of an element can be formed by direct method or by indirect method. Magnesium forms its oxide directly on heating 2Mg + O2 > 2 MgO In the case of copper, its oxide is obtained in an indirect manner i. e. copper s dissolved in concentrated nitric acid and the copper(II) nitrate formed after evaporation is strongly heated to give copper (II) oxide. Cu + 4HNO3 > Cu(NO3)2 + 2H2O + 2 NO2 2Cu(NO3)2 > 2CuO + 4 NO2 + O2 Calculations Mass of the element taken Mass of oxygen = w1 g = (w2-w1) g Mass of the oxide of the element = w2 g (w2 – w1) g of oxygen has combined with w1 g of the metal. w1 ______ x 8 ? 8 g of oxygen will combine with w2 – w1 This value represents the equivalent mass of the metal. Problem 2 0. 635 g of a metal gives on oxidation 0. 795g g of its oxide. Calculate the equivalent mass of the metal. 31 Mass of the metal oxide = 0. 95g Mass of the metal = 0. 635g Mass of oxygen = 0. 795 – 0. 635 = 0. 16g 0. 16 g of oxygen has combined with 0. 635 g of a metal ? 8 g of oxygen will combine with 8 x 0. 635 ???? = 31. 75 of the metal 0. 16 Equivalent mass of the metal = 31. 75g equiv-1 Chloride Method The equivalent mass of those elements which readily form their chlorides can be determined by chloride method. For example, a known mass of pure silver is dissolved completely in dilute nitric acid. The resulting silver nitrate solution is treated with pure hydrochloric acid when silver chloride is precipitated. It is then filtered, dried and weighed.

Thus from the masses of the metal and its chloride, the equivalent mass of the metal can be determined as follows : Calculations Mass of the metal = w1 g Mass of the metal chloride = w2 g Mass of chlorine = (w2 – w2) g (w2 – w1) g of chlorine has combined with w1 of the metal 35. 46 g of chlorine will combine with 35. 46 x w1 _________ g of the metal (w2 – w1) This value gives the equivalent mass of the metal. Uses of volumetric laws If the volume of the acid is represented as V1, the normality of the acid as N1, volume of base as V2 the normality of the base as N2, then from the law of volumetric analysis it follows that V1 x N1 = V2 x N2 32

All volumetric estimations are based on the above relationship. If any three quantities are known, the fourth one can readily be calculated using the above expression. 1. 8. 3 Equivalent mass of acid, base, salt, oxidising agent and reducing agent Acids contain one or more replaceble hydrogen atoms. The number of replaceble hydrogen atoms present in a molecule of the acid is referred to its basicity. Equivalent mass of an acid is the number of parts by mass of the acid which contains 1. 008 part by mass of replaceble hydrogen atom. Equivalent= mass of an acid molar mass of the acid __________________________ No. f replaceble hydrogen atom (or) = molar mass of the acid ___________________ basicity of the acid For example, the basicity of sulphuric acid is 2. Equivalent mass of H2SO4 Molar mass of H2SO4 = ????????? 2 98 = ___ = 49 2 2. Equivalent mass of the base Equivalent mass of a base is the number of parts by mass of the base which contains one replaceable hydroxyl ion or which completely neutralises one gram equivalent of an acid. The number of hydroxyl ions present in one mole of a base is known as the acidity of the base. Sodium hydroxide, potassium hydroxide, ammonium hydroxide are examples of monoacidic bases.

Calcium hydroxide is a diacidic base 33 In general, molar mass of the base __________________ equivalent mass of a base = acidity of the base Equivalent mass of KOH = 56 /1 = 56 3. Equivalent mass of a salt Equivalent mass of a salt is a number of parts by mass of the salt that is produced by the neutralisation of one equivalent of an acid by a base. In the case of salt like potassium chloride, the salt formed by the neutralisation of one equivalent of an acid by a base. KOH + HCl > KCl + H2O Therefore, the equivalent mass of the salt is equal to its molar mass. 4.

Equivalent mass of an oxidising agent The equivalent mass of an oxidising agent is the number of parts by mass which can furnish 8 parts by mass of oxygen for oxidation either directly or indirectly. For example, potassium permanganate is an oxidising agent. In acid medium potassium permanganate reacts as follows 2 KMnO4 + 3 H2SO4 > K2SO4 + 2 MnSO4 + 3 H2O + 5 [O] 316 80 80 parts by mass of oxygen are made available from 316 parts by mass of KMnO4 8 parts by mass of oxygen will be furnished by 316 x 8 ______ = 3. 16 80 Equivalent mass of KMnO4 = 31. 6g equiv-1 5.

Equivalent mass of a reducing agent The equivalent mass of a reducing agent is the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or with one equivalent of any oxidising agent. 34 (i) Ferrous sulphate reacts with an oxidising agent in acid medium according to the equation 2 FeSO4 + H2SO4 + (O) > Fe2 (SO4)3 + H2O 2 x 152g 16g 16 parts by mass of oxygen oxidised 304 parts by mass of ferrous sulphate 8 parts by mass of oxygen will oxidise 304/16 x 8 parts by mass of ferrous sulphate. The equivalent mass of ferrous sulphate (anhydrous) is 152.

The equivalent mass of crystalline ferrous sulphate, FeSO4 7H2O is 152 + 126 = 278 126 is the mass corresponding to 7 moles of water. (ii) In acid medium, oxalic acid is oxidised according to the equation (COOH)2 + (O) > 2 CO2 + H2O 16 Parts by mass of oxygen oxidised 90 parts by mass of anhydrous oxalic acid. 8 parts by mass of oxygen will oxidise 90/16 x 8 = 45 parts by mass of anhydrous oxalic acid. ? Equivalent mass of anhydrous oxalic acid = 45 g equiv -1 But equivalent mass of crystalline oxalic acid, (COOH)2. 2H2O=126/2 = 63 g equiv-1. 1. 8. Determination of Molecular Mass Victor-Meyer’s Method Principle In this method a known mass of a volatile liquid or solid is converted into its vapour by heating in a Victor-Meyer’s tube. The vapour displaces its own volume of air. The volume of air displaced by the vapour is measured at the experimental temperature and pressure. The volume of the vapour at s. t. p is then calculated. From this the mass of 2. 24 x 10-2m3 of the vapour at S. T. P. is calculated. This value represents the molecular mass of the substance. 35 1. Sample tube 2. Liquid of boiling point higher than that of the volatile substance 3. Victor-Meyer tube 4.

Outer jacket Figure 1. 1 Determination of molecular mass by victor-Meyers method The apparatus consists of an inner Victor-Meyer tube, the lower end of which is in the form of a bulb. The upper end of the tube has a side tube which leads to a trough of water. The Victor-Meyer tube is surrounded by an outer jacket. In the outer jacket is placed a liquid which boils at a temperature at least 30 K higher than the boiling point of the volatile substance under study. A small quantity of glass wool or asbestos fiber covers the bottom of the Victor-Meyer tube to prevent breakage when the bottle containing the substance is dropped in.

Procedure The liquid in the outer jacket is allowed to boil and when no more air escapes from the side tube, a graduated tube filled with water is inverted over the side tube dipping in a trough full of water. A small quantity of the substance is exactly weighed in a small stoppered bottle and quickly dropped in the heated Victor-Meyer tube and corked immediately. The bottle falls on the asbestos pad and its content suddenly changes into vapour, blow out the stopper and displace an equal volume of air which collects in the graduated tube.

The volume of air in the graduated tube is measured by taking it out by closing its mouth with the thumb and dipping 36 it in a jar full of water. When the water levels outside and inside the tube are the same, the volume of air displaced is noted. The atmospheric pressure and laboratory temperature are noted. Calculations Mass of the volatile substance = Volume of air displaced Laboratory temperature wg = Volume of vapour = V1 m3 = T1 K Let the atomospheric pressure be P Pressure of dry vapour = Atomospheric pressure – aqueous tension at. T1 K Let the aqueous tension be p Nm-2 at that temperature.

Pressure of dry vapour Standard temperature Standard pressure = P1 = T0 = P0 = [P-p] = 273 K = 1. 013 x 105 Nm-2 Let the volume of the vapour at standard temperature and pressure be V0 m3 From the gas equation, it follows P0V0 ____ T0 V0 P1 V1 = _____ T1 P1V1 = _____ x T1 3 T0 ____ P0 The mass of V0 m of vapour at s. t. p is w g. The mass of 2. 24 x 10-2 m3 of the vapour at s. t. p. is 2. 24×10-2xW __________ V0 The value thus calculatd gives the molecular mass 37 Molecular mass = 2 x vapour density Molecular mass _____________ Vapour density= 2 Problem In the determination of molecular mass by Victor-Meyer’s Method 0. 90 g of a volatile liquid displaced 1. 696 x 10-4m3 of moist air at 303 K and at 1 x 105 Nm-2 pressure. Aqueous tension at 303 K is 4. 242 x 103 Nm-2. Calculate the molecular mass and vapour density of the compound. Mass of the organic compound Volume of Vapour Volume of air displaced = 0. 79 g = V1=1. 696×10-4m3 = Volume of vapour. P1 = (atmospheric pressure – aqueous tension) = (1. 0 x 105) – (4. 242 x 103) = 0. 958 x 105 Nm-2 T1 = 303 K Lab Values V1 = 1. 696 x 10-4 Nm-2 P1 = 0. 958 x 10-5 x 105 Nm-2 T1 = 303 K P0V0 P1V1 _____ = _____ T1 T0 V0 P1V1T0 _______ = P0T1 Values at S. T. P Vo = ? Po=1. 013×105 Nm-2 T0 = 273 K 73 V0 = ?????????? x ____ 303 1. 013 x 105 38 0. 958 x 105 x 1. 696 x 10-4 V0 = 1. 445 x 10-4 m3 The mass of 1. 445 x 10-4m3 of vapour at S. T. P = 0. 79 g. The mass of 2. 24 x 10-2m3 of vapour at S. T. P is 2. 24 x 10-2 x 0. 79 ______________ = 1. 445 x 10-4 The molecular mass of the substance = 122. 46 molecular mass _____________ Vapour density of = the compound 2 122. 46 = ______ = 61. 23 2 Questions A. Choose the best answer : 1. The volume occupied by 16g of oxygen at S. T. P. a) 22. 4 L b) 44. 8 L c) 11. 2 L d) 5. 6 L 2. Avogadaro’s number represents the number of atoms in b) 320g of S a) 12g of C12 c) 32g of Oxygen d) 12. g of iodine. 3. The value of gram molecular volume of ozone at S. T. P is a) 22. 4 L b) 2. 24 L c) 11. 2 L d) 67. 2 L 4. The number of atoms present in 0. 5 gram- atoms of Nitrogen is same as the atoms in a) 12g of C b) 32g of S c) 8g of the oxygen d) 24g of magnesium. 5. The number of gram-atoms of oxygen in 128g of oxygen is a) 4 b) 8 c) 128 d) 8×6. 02×1023 6. The total number of moles present in 111 g of CaCl2 is a) One mole b) Two moles c) Three moles d) Four moles 7. Which of the following weighs the most? a) One gram-atom of nitrogen b) One mole of water c) One mole of Sodium d) One molecule of H2SO4 39 . Which of the following contains same number of carbon atoms as are in 6. 0g of carbon (C-12)? a) 6. 0g ethane b) 8. 0g methane c) 21. 0g Propane d) 28. 0g CO 9. Which of the following contains maximum number of atoms? a) 2. 0g hydrogen b) 2. 0g oxygen c) 2. 0g nitrogen d) 2. 0g methane 10. Which one among the following is the standard for atomic mass? c) 146C d) 168O a) H b) 126C 11. Which of the following pair of species have same number of atoms under similar conditions ? a) 1L each of SO2 and CO2 b) 2L each of O3 and O2 c) 1L each of NH3 and Cl2 d) 1L each of NH3 and 2L of SO2 12. 2. g of oxygen contains number of atoms same as in a) 4 g of S b) 7 g of nitrogen d) 12. 3 g of Na c) 0. 5 g of H2 13. The number of gm-molecules of oxygen in 6. 02 x 1024 CO molecules is a) 1 gm-molecule b) 0. 5 gm-molecule c) 5 gm-molecule d) 10 gm-molecule 14. Hydrogen phosphate of certain metal has a formula MHPO4, the formula of metal chloride is c) MCl2 d) MCl4 a) MCl b) MCl3 15. A compound contains 50% of X (atomic mass 10) and 50% Y (at. mass 20). Which formulate pertain to above date ? c) X4Y3 d) (X2)3 Y3 a) XY b) X2Y 16. Which of the following compound has / have percentage of carbon same as that in ethylene (C2H4) ? ) propene b) Cyclohexane c) Ethyne d) Benzene 17. 5L of 0. 1 M solution of sodium Carbonate contains b) 106 g of Na2CO3 a) 53 g of Na2CO3 d) 5 x 102 millimoles of Na2CO3 c) 10. 6 of Na2CO3 B. Fill in the blanks atoms. 1. One mole of a triatomic gas contains 2. One mole of Sulphuric acid contains Oxygen atoms. oxy


Hi there, would you like to get such a paper? How about receiving a customized one? Check it out