FYJC Chemistry Textbook PDF

Standard XI I Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune 154.00 The Coordination Committee formed by GR No...

Standard XI I Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune 154.00 The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 20.6.2019 and it has been decided to implement it from academic year 2019-20. CHEMISTRY Standard XI Download DIKSHA App on your smartphone. If you scan the Q.R. Code on this page of your textbook, you will be able to access full text. If you scan the Q.R. Code provided, you will be able to access audio-visual study material relevant to each lesson, provided as teaching and learning aids. 2019 Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. First Edition : © Maharashtra State Bureau of Textbook Production and 2019 Curriculum Research, Pune - 411 004. Reprint : 2020 The Maharashtra State Bureau of Textbook Production and Curriculum Research reserves all rights relating to the book. No part of this book should be reproduced without the written permission of the Director, Maharashtra State Bureau of Textbook Production and Curriculum Research, ‘Balbharati’, Senapati Bapat Marg, Pune 411004. Subject Committee Illustration Shri. Pradeep Ghodke Dr. Chandrashekhar V. Murumkar, Chairman Shri. Shubham Chavan Dr. Sushama Dilip Joag, Member Cover Shri. Vivekanand S. Patil Dr. Shridhar Pandurang Gejji, Member Typesetting Dr. Satyawati Sudhir Joshi, Member DTP Section, Textbook Bureau, Pune Dr. Rajashree Vikas Kashalkar, Member Dr. Laxman Shamrao Patil, Member Co-ordination Shri. Rajesh Vamanrao Roman, Member Shri. Rajiv Arun Patole Special Officer for Chemistry Shri. Rajiv Arun Patole, Member Secretary Paper Study group 70 GSM Creamwove Dr. Anjali Deepak Ruikar Print Order Shri. Sachin Ashok Bartakke Printer Smt. Archana Sanjeev Kale Smt. Pushpalata Babanrao Gangarde Smt. Archana Dipak Harimkar Production Shri. Vishnu Rustumrao Deshmukh Shri Sachchitanand Aphale Chief Production Officer Shri. Sharad Ajabrao Mankar Shri Liladhar Atram Shri. Ritesh Vijay Bijewar Production Officer Shri. Rupesh Dinkar Thakur Shri. Milind Chandrakant Gaikwad Publisher Shri. Gajanan Shivajirao Suryawanshi Shri Vivek Uttam Gosavi, Controller Maharashtra State Textbook Smt. Prachi Ravindra Sathe Bureau, Prabhadevi, Chief Co-ordinator Mumbai - 400 025  The Constitution of India Preamble WE, THE PEOPLE OF INDIA, having solemnly resolved to constitute India into a SOVEREIGN SOCIALIST SECULAR DEMOCRATIC REPUBLIC and to secure to all its citizens: JUSTICE, social, economic and political; LIBERTY of thought, expression, belief, faith and worship; EQUALITY of status and    of opportunity; and to promote among them all FRATERNITY assuring the dignity of the individual and the unity and integrity of the Nation; IN OUR CONSTITUENT ASSEMBLY this twenty-sixth day of November, 1949, do HEREBY ADOPT, ENACT AND GIVE TO OURSELVES THIS CONSTITUTION. NATIONAL ANTHEM  Preface Dear Students, We welcome you all to std. XI. For the first time, you are being introduced to the subject of chemistry discipline. You have already been acquainted with some of the concepts of chemistry from standard five onwards, especially in the subject of general science up to standard eight and science and technology for standard nine and ten. Chemistry is very broad subject that covers many aspects of our everyday experience. This textbook aims to create awareness and to understand certain essential aspects by the national curriculum framework (NCF) which was formulated in 2005, followed by the state curriculum framework (SCF) in 2010. Based on these two framework, reconstruction of the curriculum and prepartion of a revised syllabus has been done and designed now. The subject chemistry is the study of substances, their properties, structures and transformation. The world is full of chemical substances and we need chemicals for many useful purposes. Our body is a huge chemical factory. Keeping this in mind, the textbook is written in organized manner. You can learn a very basic principles, understand facts and put them into practice by learning in the classroom and laboratory. The textbook is presented in a simple language with relevant diagrams, graphs, tables, photographs. This will help you to understant various terminology, concepts with more clarity. All the illustrations are in color form. The new syllabus focuses on the basic principles, concepts, laws based on precise observations, their applications in everyday life and ability to solve different types of problems. The general teaching - learing objectives of the revised syllabus are further determined on the basis of the ‘Principle of constructivism’ i.e. self learning. The curriculum and syllabus is designed to make the students to think independently. The student are encouraged to read, study more through the additional information given in the colored boxes. Activities have been introduced in each chapter. These activities will not only help to understand the content knowledge on your own efforts. QR code have been introduced for gaining the additional information, abstracts of chapters and practice questions/ activities. The efforts taken to prepare the text book will help the students think about more than just the content of the chemical concepts. Teachers, parents as well as those aspiring condidates preparing for the competitive examinations will also be benefited. We look forward to a positive response from the teachers and students. Our best wishes to all ! (Dr. Sunil Magar) Director Pune Maharashtra State Bureau of Date : 20 June 2019 Textbook Production and Bharatiya Saur : 30 Jyeshtha 1941 Curriculum Research, Pune 4  - For Teachers - Dear Teachers, We are happy to introduce the revised textbook Each unit is structured in a definite manner. of chemistry for std. XI. This book is a sincere It starts from the basic concepts of general attempt to follow the maxims of teaching as well chemistry required for each branch of as develop a ‘constructivist’ approach to enhance chemistry. Application of this knowledge will the quality of learning. The demand for more help students to understand further chapters in activity based, experiential and innovative learning each unit. opportunities is the need of the time. The present Each chapter provides solved problems on curriculum has been restructured so as to bridge the each and every concept and various laws. The credibility gap that exists in the experience in the solved problems are put into boxes. Teachers outside world. Guidelines provided below will help should expalin each step of the problem and to enrich the teaching - lerning process and achieve give them pratice. the desired learning outcomes. Ask the students about the related information, To begin with, get familiar with the textbook backgroud about the chapter. You are provided, yourself. for this with the different boxes like ‘Can You The present book has been prepared for Recall’, ‘Do you know?’ construtivism and activity based learning. Encourage the students to collect related Teachers must skillfully plan and organize the information by providing them the websites. activities provided in each chapter to develop Teaching- learning interactions, processes and interest as well as to stimulate the thought participation of all students are necessary and process among the students. so is your active guidance. Always teach with proper planning. Do not use the content of the boxes titles ‘Do Use teaching aids as required for the proper you know’? for evaluation. understanding of the subject. Exercises include parameters such as co- Do not finish the chapter in short. relation, critical thinking, analytical reasoning Follow the order of the chapters strictly as etc. Evaluation pattern should be based on the listed in the contents because the units are given paramerters. Equal weightage should introduced in a graded manner to facilitate be assigned to all the topics. Use different knowledge building. combinations of questions. Remember Try this Can you recall? Can you tell? Front Page : The photograph depicts transmission electron micrograph (TEM) of a few layer Use your brain power Graphene (left). The electron diffraction pattern (hexagonal arrangement of spots corresponds to the hexagonal symmetry of the structure of Just think Activity : Graphene (Right). Picture Credit : Prof. Dr. M. A. More, Department of Physics, Savitribai Phule Pune Universiy, Pune 411007. Do you know ? Exercises Observe and Discuss Find out Internet my friend DISCLAIMER Note : All attempts have been made to contact copy right/s (©) but we have not heard from them. We will be pleased to acknowledge the copy right holder (s) in our next edition if we learn from them.  Competency Statements - Standard XI Area/ Unit/ After studying the contents in Textbook students..... Lesson Understand the SI unit of important fundamental scientific quantities. Explain various fundamental laws of chemical combination, which are applied in day-to-day life. Relate basic concepts of number of moles and molecules. Differentiate between quantitative and qualitative analysis. Develop accuracy, precision, concentration ability in taking accurate reading. General Calculate empirical formula and molecular formula of compounds. chemistry Obtain information about different techniques to purify substance as well as separation of miscible solids and liquids. Gain the information about various theories, principles, put up by eminent Scientists leading to atomic stucture. Classify elements isotopes, isobars and isotones. Understand the duel nature of electron. Application of concept of quantum number in writing electronic configuration of various elements. Inculcate social and scientific awareness by gaining knowledge of oxidation-reduction concept. Evaluate oxidation number of elements and balance the redox reaction by different methods. Categorize oxidizing and reducing agents with their applications. Classify elements based on electronic configuration. Understand co-relation between the various properties like atomic size, valency, oxidation state, ionization enthalpy and electronegativity in a group and in a period. Recognize isoelectronic species. Inorganic Compare the trends in physical and chemical properties in group I and chemistry group II. Understand the diagonal relationship. Gain the knowledge of hydrogen from periodic table. Develop interest in systematic study of elements present in Group 13, Group 14 and group15. Learn anomalous behaviors of boron, carbon and nitrogen. Draw the structures of some compounds of boron, carbon and nitrogen. Elaborate information about various theories to explain nature of bonding in formation of molecules. Inculcate skill to draw Lewis structure of molecules. Assign the structures of various compounds with respect to geometry, bond angle and types of bond.  Generate environmental awareness by compiling concepts of adsorption phenomenon. Learn science behind the fact about colloids in day to day life. Interpret nature, difference and relation of equilibrium constant. Design the suitable conditions to get more yield of the desired product. Differentiate nuclear reactions with ordinary chemical reaction. Physical Acquire knowledge of natural radioactivity and related terms like chemistry nuclear transmutation, nuclear fission, nuclear fusion. Clarify the beneficial and harmful effects of radioactivity. State the applications of radioactive elements like carbon dating, nuclear reactor, generation of electricity and medicinal uses. Develop mathematical skills in finding radioactive decay constant, half life period and nuclear binding energy. Interpret the structure and functional group of organic compounds. IUPAC nomenclature of organic compounds. Understand the influence of electronic displacement and reactivity in organic molecules. Organic Draw the formulae of various isomers of organic compounds. chemistry Illustrate different methods of preparation and chemical properties of hydrocarbons. Infer importance of hydrocarbon. Gain information of medicinal properties of some chemical compounds and chemistry behind food quality and cleasing action. CONTENTS Sr. No Title Page No 1 Some Basic Concepts of Chemistry 1 - 12 2 Introduction to Analytical Chemistry 13 - 26 3 Basic Analytical Techniques 27 - 34 4 Structure of Atom 35 - 54 5 Chemical Bonding 55 - 80 6 Redox Reactions 81- 92 7 Modern Periodic Table 93 - 109 8 Elements of Group 1 and 2 110 - 122 9 Elements of Group 13, 14 and 15 123 - 134 10 States of Matter 135 - 159 11 Adsorption and Colloids 160 - 173 12 Chemical Equilibrium 174 - 189 13 Nuclear Chemistry and Radioactivity 190 - 203 14 Basic Principles of Organic Chemistry 204 - 232 15 Hydrocarbons 233 - 260 16 Chemistry in Everyday Life 261 - 270  1. Some Basic Concepts of Chemistry 1.1 Introduction : Chemistry is the study of 1.2.1. Matter : You have learnt earlier that matter, its physical and chemical properties matter occupies space and has mass. Matter and the physical and chemical changes it can be further classified into pure substances undergoes under different conditions. and mixtures on the basis of chemical composition. Matter pure substances mixtures elements compounds homogeneous heterogeneous metals nonmetals metalloids Chemistry is a central science. Its Let us understand first what are pure knowledge is required in the studies of physics, substances and mixtures. biological sciences, applied sciences, and earth 1.2.2 Pure substances versus mixtures : and space sciences. The scope of chemistry is Pure substances have a definite chemical in every aspect of life, for example, the air we composition. They always have the same breathe, the food we eat, the fluids we drink, properties regardless of their origin. Mixtures our clothing, transportation and fuel supplies, have no definite chemical composition and and so on. hence no definite properties. Though it is an ancient science, due to development and advancement in science Examples of pure substances : Pure metal, and technology, chemistry has developed as distilled water, etc. modern science. Technological development Examples of mixtures : Paint (mixture of oils, in sophisticated instruments expanded our pigment, additive), concrete (a mixture of knowledge of chemistry which, now, has been sand, cement, water) used in applied sciences such as medicine, dentistry, engineering, agriculture and in daily home use products. Can you tell? 1.2 Nature of Chemistry : Chemistry is Which are mixtures and pure substances traditionally classified further into five branches : organic, inorganic, physical, bio from the following ? and analytical. Organic chemistry is the study i. sea water ii. gasoline iii. skin iv. a rusty of the properties and reactions of compounds nail v. a page of the textbook. vi. diamond of carbon. Inorganic chemistry is the study of all substances which are not organic. Physical Pure substances are further divided into chemistry is the study of principles underlying elements and compounds. Elements are chemistry. It deals with the studies of properties pure substances which can not be broken of matter. It is study of atoms, molecules, and fundamental concepts related to electrons, down into simpler substances by ordinary energies and dynamics therein. It provides chemical changes. Elements are further basic framework for all the other branches of classified as metals, nonmetals and metalloids. chemistry. 1 1. Metals : apart as compared to those in liquid and solid i. have a lustre (a shiny appearance). state. ii. conduct heat and electricity. Three states of matter are interconvertible iii. can be drawn into wire (are ductile). by changing the conditions of temperature and iv. can be hammered into thin sheets (are pressure. malleable). Examples : gold, silver, copper, iron. Mercury Can you tell? is a liquid metal at room temperature. Classify the following as element and 2. Non-metals : compound. i. have no lustre. (exception : diamond, iodine) i. mercuric oxide ii. helium gas iii. water iv. ii. are poor conductors of heat and electricity. table salt v. iodine vi. mercury vii. oxygen (exception : graphite) viii. nitrogen iii. can not be hammered into sheets or drawn into wire, because they are brittle. 1.3 Properties of matter and their Examples : Iodine, nitrogen, carbon, etc. measurement : 3.Metalloids : Some elements have properties intermediate between metals and non-metals and are called metalloids or semi-metals. Examples include arsenic, silicon and germanium. Compounds are the pure substances which can be broken down into simpler substances by ordinary chemical changes. In a compound, two or three elements are Fig : 1.1 Burning of magnesium wire combined in a fixed proportion. Mixture contains two or more substances Different kinds of matter have in no fixed proportions and may be separated characteristic properties, which can be by physical methods. Mixtures are further classified into two categories as physical divided into homogeneous and heterogeneous. properties and chemical properties. Solutions are homogeneous mixtures, because Physical properties are those which can the molecules of constituent solute and solvent be measured or observed without changing are uniformly mixed throughout its bulk. the chemical composition of the substance. In heterogeneous mixtures the molecules Colour, odour, melting point, boiling point, of the constituents are not uniformly mixed density, etc. are physical properties. Chemical throughout the bulk. For example : Suspension properties are the properties where substances of an insoluble solid in a liquid. undergo a chemical change and thereby 1.2.3 States of matter : You are also aware exhibit change in chemical composition. For that matter exists in three different states example, coal burns in air to produce carbon namely gas, liquid and solid. You are going to dioxide or magnesium wire burns in air in the learn about these states in unit 3 (chapter 10). presence of oxygen to form magnesium oxide. In solids, constituent atoms or molecules (Fig. 1.1) (particles) are tightly held in perfect order and 1.3.1 Measurement of properties : Many therefore solids possess definite shape and properties of matter are quantitative in nature. volume. Liquids contain particles close to When you measure something, you are each other and they can move around within comparing it with some standard. The standard the liquid. While in gases, the particles are far quantity is reproducible and unchanging. 2  Many properties of matter such as mass, SI units : length, area, pressure, volume, time, etc. In 1960, the general conference of weights are quantitative in nature. Any quantitative and measure, proposed revised metric system, measurement is expressed by a number called International System of units, that is, followed by units in which it is measured. SI units. For example, length of class room can be The metric system which originated in represented as 10 m. Here 10 is the number and France in late eighteenth century, was more 'm' denotes metre-the unit in which the length convenient as it was based on the decimal is measured. system. Later, based on a common standard The standards are chosen orbitrarily system, the International System of Units (SI with some universally acceped criteria. ''The units) was established. arbitrarily decided and universally accepted The SI system has seven base units as standards are called units.'' listed in Table 1.1. These are fundamental There are several systems in which units scientific quantities. Other units like speed, are expressed such as CGS (centimetre for volume, density, etc. can be derived from these length, gram for mass and second for time), quantities. FPS (foot, pound, second) and MKS (metre, kilogram, second) systems, etc. Table 1.1 SI Fundamental units Base Physical Quantity Symbol for Quantity Name of SI Unit Symbol for SI Unit Length l metre m Mass m kilogram kg Time t second s Electric current I ampere A Thermodynamic temperature T Kelvin K Amount of substance n mole mol Luminous intensity Iv candela cd 1.3.2 Physical properties i. Mass and weight : We know that matter ii. Length : In chemistry we come across has mass. So mass is an inherent property of 'length' while expressing properties such as matter. It is the measure of the quantity of the atomic radius, bond length, wavelenght of matter a body contains. The mass of a body electromagnetic radiation, and so on. These does not vary as its position changes. On the quantities are very small therefore fractional other hand, the weight of a body is result of the units of the SI unit of length are used for mass and gravitational attraction. The weight example, nanometre (nm), picometre (pm). of a body varies because the gravitational Here 1nm = 10-9 m, 1 pm = 10-12 m. attraction of the earth for a body varies with iii. Volume : It is the amount of space occupied the distance from the centre of the earth. by a three dimensional object. It does not Hence, the mass of a body is more depend on shape. For measurement of volume fundamental property than its weight. of liquids and gases, a common unit, litre (L) The basic unit of mass in the SI system is which is not an SI unit is used. the kilogram as given in Table 1.1. However, 1 L = 1 dm3 = 1000 mL = 1000 cm3 a fractional quantity 'gram' is used for 1000 cm3 = 10 cm × 10 cm × 10 cm of volume weighing small quantities of chemicals in the SI unit of volume is expressed as (metre)3 laboratories. Therefore, in terms of grams it is or m3. defined (1kg = 1000 g = 103 g ) 3  Volume : 1000 cm3 ; Fahrenheit) and K (Kelvin). Here K is the SI 1000 mL; 1dm3 ; unit. Figure 1.4 shows the thermometers based 1L. on these scales. Generally, the thermometer with celsius scale are calibrated from 0 0C to 100 0C where these two temperatures are respectively the freezing point and the boiling point of water Volume : 1 cm3 at atmospheric pressure. These are represented 1 mL 1 cm on fahrenheit scale as 320 F to 2120 F. 10 cm = 1dm 1 cm Water boils Fig. 1.2 : Litre and SI unit of volume 373.15 – 370 100 – 100 212 – – 210 – 200 at sea level 90 – 190 Different kinds of glassware are used to 360 350 80 – 180 – 200 measure the volume of liquids and solutions. 340 70 60 – 160 – 150 – 140 For example, graduated cylinder, burette, 330 320 50 – 130 – 120 Body temperature pipette, etc. A volumetric flask is used to 310 40 30 – 100 – 90 98.6º F, 37º C 300 – 80 prepare a known volume of a solution. Figure 290 20 – 70 – 60 1.3 shows the types of apparatus used in 273.15 – 280 0– 10 0 32 – – 50 – 40 Water freezes 270 – 30 at sea level laboratory for measuring volume of liquids. 260 -10 – 20 – 10 -20 –0 250 – 0 - 10 Calibration 0– -273.15 – -459.67 – Absolute Zero — all molecular mark indicates mL motion STOPS 25-ml volume 0 mL 1 100 2 Kelvin Celsius Fahernheit 3 Calibration 90 80 4 mark indicates 250-ml volume Fig 1.4 : Thermometers of different 70 60 44 45 temperature scale 50 46 47 The temperatures on two scales are related to 40 Value 48 30 (stopcock) 49 50 each other by the following relationship : 20 controls the 9 0 10 liquid flow 0 F= ( C) + 32 5 100-mL 25-mL pipette 50-mL burette 250-mL graduated cylinder volumetric flask The Kelvin scale is related to Celsius scale as Fig. 1.3 : Volumetric glass apparatus follows : iv. Density : Density of a substance is its K = 0C + 273.15 mass per unit volume. It is determined in 1.4 Laws of Chemical Combination : The the laboratory by measuring both the mass elements combine with each other and form and the volume of a sample. The density is compounds. This process is governed by five calculated by dividing mass by volume. It is basic laws discovered before the knowledge of the characteristic property of a substance. molecular formulae. So SI unit of density can be obtained as follows: 1.4.1 Law of conservation of mass : Antoine SI unit of mass Lavoisier (1743-1794) a SI unit of density = French scientist is often SI unit of volume kg referred to as the father = or kg m-3 of modern chemistry. m3 g He carefully performed CGS units it is or g mL-1 or g cm-3 many combustion mL experiments, namely, v. Temperature : Temperature is a measure of burning of phosphorus and mercury, both in the hotness or coldness of an object. There are the presence of air. Both resulted in an increase three common scales to measure temperature, in weight. After several experiments he found namely 0C (degree Celsius), 0F (degree that the weight gained by the phosphorus was 4 exactly the same as the weight lost by the air. hydrogen peroxide. He observed that, Hydrogen + Oxygen Water Total mass of reactants 2g 16 g 18 g = Total mass of products Hydrogen+ Oxygen Hydrogen Peroxide When hydrogen gas burns and combines 2g 32 g 34 g with oxygen to yield water, the mass of the Here, it is found that, the two masses of water formed is equal to the mass of the oxygen i.e. 16 g and 32 g which combine with hydrogen and oxygen consumed. Thus, the a fixed mass of hydrogen (2g) are in the ratio law of conservation of mass states that 'mass of small whole numbers, i. e. 16:32 or 1:2. can neither be created nor destroyed.' ii. Nitrogen and oxygen combine to form two 1.4.2 Law of Definite Proportions : compounds, nitric oxide and nitrogen dioxide. French chemist, Joseph Proust performed Nitrogen + Oxygen Nitric Oxide 14 g 16 g 30 g experiments on two samples of cupric Nitrogen + Oxygen Nitrogen Dioxide carbonate. One of the samples was natural in 14 g 32 g 46 g origin and the other was a synthetic one. He Here, you find that the two masses of oxygen found that the composition of elements present i.e. 16 g and 32 g when combine with a fixed in it was same for both the samples as shown mass of Nitrogen (14 g) are in the ratio of below : small whole numbers i.e. 16:32 or 1:2. Cupric % of % of % of carbon (Similar examples such as CO and CO2 Carbonate copper oxygen (1:2 ratio), SO2 and SO3 (2:3 ratio), can be Natural 51.35 38.91 9.74 found.) sample 1.4.4 Gay Lussac Law of Gaseous Volume Synthetic 51.35 38.91 9.74 : This law was put forth by Gay Lussac sample in 1808. The law states that when gases This led Joseph Proust to state the law of combine or are produced in a chemical definite proportion as follows : reaction they do so in a 'A given compound always contains exactly simple ratio by volume, the same proportion of elements by weight.' provided all gases are at Irrespective of the source, a given compound same temperature and always contains same elements in the same pressure. proportion. The validity of this law has been Illustration : i. Under confirmed by various experiments. This law the same conditions of is sometimes referred to as Law of definite temperature and pressure, composition. 100 mL of hydrogen 1.4.3 Law of multiple proportions : combines with 50 mL of oxygen to give 100 This law was proposed by John Dalton in mL of water vapour. Hydrogen (g) + Oxygen (g) Water(g) 1803. It has been observed that two or more 100 mL 50 mL 100 mL elements may form more than one compound. (2 vol) (1 vol) (2 vol) Law of multiple proportions summarizes Thus, the volumes of hydrogen gas and oxygen many experiments on such compounds. When gas which combine together i.e. 100 mL and two elements A and B form more than one 50 mL producing two volumes of water vapour compounds, the masses of element B that which amounts to 100 mL bear a simple ratio combine with a given mass of A are always of 2:1:2 in the ratio of small whole numbers. For ii. Under the same condition of temperature example, i. Hydrogen combines with oxygen and pressure, to form two compounds, namely water and 5 1 L of nitrogen gas combines with 3 L of Therefore, 2 molecules of hydrogen hydrogen gas to produce 2 L of ammonia gas. gas combine with 1 molecule of oxygen to Nitrogen (g) + Hydrogen (g) Ammonia(g) give 2 molecules of water vapour. Avogadro 1L 3L 2L could explain the above result by considering (1 vol) (3 vol) (2 vol) the molecules to be polyatomic. If hydrogen Thus, the volume of nitrogen gas and hydrogen and oxygen were considered as diatomic, as gas which combine together i.e. 1 L and 3 L recognized now, then the above results are and volume of ammonia gas produced i. e. 2 L easily understandable. bear a simple ratio of 1:3:2. Remember Remember Avogadro made a distinction between atoms Gay Lussac's discovery of integer ratio in and molecules, which is quite understandable volume relationship is actually the law of in the present time. definite proportion by gaseous volumes. Can you tell? If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how 1 volume of 1 volume of 1 volume of 2 volume of many volumes of water vapour would be hydrogen hydrogen oxygen water vapour produced? Fig. 1.5 : two volume of hydrogen react with one volume of oxygen to give two volumes of 1.5 Avogadro Law : In water vapour 1811, Avogadro proposed that equal volumes of 1.6 Dalton's Atomic Theory : In 1808, all gases at the same Dalton published ''A New System of chemical temperature and pressure philosophy'' in which he proposed the contain equal number of following features, which later became famous molecules. as Dalton's atomic theory. If we consider the 1. Matter consists of tiny, indivisible particles reaction of hydrogen and oxygen to produce called atoms. water vapour. 2. All the atoms of a given elements have Hydrogen (g) + Oxygen (g) Water (g) identical properties including mass. Atoms 100 mL 50 mL 100 mL of different elements differ in mass. (2 vol) (1 vol) (2 vol) 3. Compounds are formed when atoms of (Gay Lussac Law) different elements combine in a fixed ratio. 2n molecules n molecules 2n molecules 4. Chemical reactions involve only the (Avogadro law) reorganization of atoms. Atoms are neither 2 molecules 1 molecule 2 molecules created nor destroyed in a chemical reaction. We see that 2 volumes of hydrogen Dalton's theory could explain all the laws of combine with 1 volume of oxygen to give 2 chemical combination. volumes of water vapour, without leaving any unreacted oxygen. According to Avogadro Can you recall? law, if 1 volume contains n molecules, then What is an atom and a moleule ? 2n molecules of hydrogen combine with n What is the order of magnitude of mass of molecules of oxygen to give 2n molecules of one atom ? What are isotopes? water. 6 1.7 Atomic and molecular masses : You 1.7.2 Average Atomic Mass : Many naturally know about the terms atoms and molecules. occuring elements exist as mixture of more Thus it is appropriate here to understand what than one isotope. Isotopes have different we mean by atomic and molecular masses. atomic masses. The atomic mass of such an 1.7.1 Atomic Mass : Every element has a element is the weighted average of atomic characteristic atomic mass. Atomic mass is the masses of its isotopes (taking into account the mass of an atom. It is actually very very small. atomic masses of isotopes and their relative For example, the mass of one hydrogen atom abundance i.e. percent occurrance). This is is 1.6736 × 10-24 g. This is very small quantity called average atomic mass of an element. and not easy to measure. For example, carbon has the following three In the present system, mass of an atom isotopes with relative abundances and atomic is determined relative to the mass of a carbon masses as shown against each of them. - 12 atom as the standard and this has been Isotope Atomic Relative agreed upon in 1961 by IU PAC. In this system, mass (u) Abundance (%) an atom of carbon-12 is assigned a mass of 12 C 12.00000 98.892 exactly 12.00000 atomic mass unit (amu) and 13 C 13.00335 1.108 all other atoms of other elements are given a 14 C 14.00317 2 × 10-10 relative atomic mass, to that of carbon - 12. From the above data, the average atomic The atomic masses are expressed in amu. mass of carbon One amu is defined as a mass exactly equal = (12 u) (98.892/100) + (13.00335 u) to one twelth of the mass of one carbon-12 (1.108/100) + (14.00317) (2 × 10-10/100) atom. Later on the exact value of atomic mass = 12.011 u unit in grams was experimentally established. Similary, average atomic masses for other 1 1 amu = 12 × mass of one C-12 elements can be calculated. 1 Remember = 12 × 1.992648 × 10-23 g In the periodic table of elements, the = 1.66056 × 10-24 g atomic masses mentioned for different Recently, amu has been replaced by elements are actually their average unified mass unit called dalton (symbol 'u' or atomic masses. 'Da'), 'u' means unified mass. For practical purpose, the average atomic mass is rounded off to the nearest Problem 1.1 : Mass of an atom of oxygen in whole number when it differs from it by gram is 26.56896 × 10-24 g. What is the atomic a very small fraction. mass of oxygen in u ? Solution : Mass of an atom of oxygen in gram Element Isotopes Average Rounded is 26.56896 × 10-24 g, and atomic off 1.66056 × 10-24 g = 1 u mass atomic ∴ 26.56896 × 10-24 g = ? mass Carbon 12 C, 13C, 14C 12.011 u 12.0 u 26.56896 × 10-24 g = = 16.0 u 1.66056 × 10-24 g/u Nitrogen 14 N, 15N 14.007 u 14.0 u Oxygen 16 O, 17O, 18O 15.999 u 16.0 u Similarly mass of an atom of hydrogen Chlorine 35 Cl, 37Cl 35.453 u 35.5 u = 1.0080 u Bromine 79 Br, 81Br 79.904 u 79.9 u 7  Problem 1.2 : Calculate the average atomic mass of neon using the following data : Isotope Atomic mass Natural Abundance 20 Ne 19.9924 u 90.92% 21 Ne 20.9940 u 0.26 % 22 Ne 21.9914 u 8.82 % Solution : Average atomic mass of Neon (Ne) Atomic mass of 20Ne × % + Atomic mass of 21Ne × % + Atomic mass of 22Ne × % = 100 (19.9924u)(90.92) + (20.9940u)(0.26) + (21.9914u)(8.82) = = 20.1707 u 100 1.7.3 Molecular Mass : Molecular mass of a In sodium chloride crystal, one Na⊕ ion substance is the sum of average atomic masses is surrounded by six Cl ions, all at the same of all the atoms of elements which constitute distance from it and vice versa. Therefore, the molecule. Molecular mass of a substance NaCl is the formula used to represent is the mass of one molecule of that substance sodium chloride, though it is not a molecule. relative to the mass of one carbon-12 atom. Similarly, a term 'formula mass' is used for It is obtained by multiplying average atomic such ionic compounds, instead of molecular mass of each element by the number of its mass. The formula mass of a substance is atoms and adding them together. the sum of atomic masses of the atoms For example, the molecular mass of carbon present in the formula. dioxide (CO2) is = 1(average atomic mass of C) Problem 1.4 : Find the formula mass of + 2 (average atomic mass of O) i. NaCl ii. Cu (NO3)2 = 1 (12.0 u) + 2 (16.0 u) = 44.0 u i. Formula mass of NaCl Some more examples of calculations of = average atomic mass of Na molecular mass. + average atomic mass of Cl i. H2O = 2 × 1 u + 16 u = 18 u = 23.0 u + 35.5u = 58.5 u ii. C6H5Cl = (6 × 12 u) + (5 × 1 u) + (35.5 u) ii. Formula mass of Cu(NO3)2 = 112.5 u = average atomic mass of Cu + 2 × (average iii. H2SO4 =(2 × 1 u) + (32 u) +(4 × 16 u) = 98 u atomic mass of nitrogen + average atomic mass of three oxygen) = (63.5) + 2(14 + 3 x 16) = 187.5 u Problem 1.3 : Find the mass of 1 molecule of oxygen (O2) in amu (u) and in grams. Solution : Molecular mass of O2 = 2 × 16 u Try this ∴mass of 1 molecule = 32 u ∴mass of 1 molecule of O2 Find the formula mass of CaSO4 = 32.0 × 1.66056 × 10-24 g If atomic mass of Ca = 40.1 u, = 53.1379 × 10-24 g S =32.1 u and O = 16.0 u 1.7.4 Formula Mass 1.8 Mole concept and molar mass Some substances such as sodium chloride do not contain discrete molecules as the Can you recall? constituent units. In such compounds, cationic (sodium) and anionic (chloride) entities are 1. One dozen means how many items ? arranged in a three dimensional structure. 2. One gross means how many items ? 8 Mole : Expressing large count of objects is Molar Mass : The mass of one mole of a made easy by using quantitative adjectives substance (element/compound) in grams is such as dozen, gross. You know that even called its molar mass. The molar mass of a small amount of any substance contains any element in grams is numerically equal to very large number of atoms or molecules. We atomic mass of that element in u. use a quantitative adjective 'mole' to express Element Atomic Molar mass the large number of submicroscopic entities mass (u) (g mol-1) like atoms, ions, electrons, etc. present in a H 1.0 u 1.0 g mol-1 substance. C 12.0 u 12.0 g mol-1 Definiton : One mole is the amount of a O 16.0 u 16.0 g mol-1 substance that contains as many entities or Simillary molar mass of any substance, particles as there are atoms in exactly 12 g existing as polyatomic molecule, in grams is (or 0.012 kg) of the carbon -12 isotope. numerically equal to its molecular mass or Let us calculate the number of atoms in formula mass in u. 12.0000 g of Carbon-12 isotopes. Mass of Polyatomic Molecular/ Molar mass one carbon-12 atom (determined by mass substance formula mass (u) (g mol-1) spectrometer) = 1.992648 × 10-23 g, O2 32.0 u 32.0 g mol-1 Mass of one mole carbon atom = 12 g H 2O 18.0 u 18.0 g mol-1 ∴Number of atoms in 12 g of carbon -12 NaCl 58.5 u 58.5 g mol-1 12g/mol Molar mass of O atoms = 1.992648 × 10-23 g/atom = 6.022 × 1023atom/mol × 16 u/atom × 1.66056 × 10-24 g/u = 16.0 g/mol = 6.0221367 × 1023 atom/mol Problem 1.5 : Calculate the number of moles Thus one mole is the amount of a substance and molecules of urea present in 5.6 g of urea. that contains 6.0221367 × 1023 particles/ Solution : Mass of urea = 5.6 g entities (such as atoms, molecules or ions). Molecular mass of urea, NH2CONH2 Note that the name of the unit is mole and = 2 (average atomic mass of N) + 4 (average the symbol for the unit is mol. atomic mass of H) + 1(average atomic mass of C) + 1(average atomic mass of O) = 2 × 14 u +1 × 12 u +4 ×1 u + 1 × 16 u Remember = 60 u ∴molar mass of urea = 60 g mol-1 The number 6.0221367 × 1023 is known as Number of moles Avogadro's Constant 'NA' in the honour of Amedo Avogadro. mass of urea in g = In SI system, mole (Symbol mol) was molar mass of urea in g mol-1 introduced as seventh base quantity for the 5.6 g = = 0.0933 mol amount of a substance. 60 g mol-1 Number of molecules = Number of moles × Avogadro's constant Example : Number of molecules of urea 1 mole of oxygen atoms = 6.0221367 × 1023 = 0.0933 × 6.022 × 1023 molecules/mol atoms of oxygen = 0.5618 × 1023 molecules 1 mole of water molecules = 6.0221367 × 1023 = 5.618 × 1022 molecules molecules of water Ans : Number of moles = 0.0933 mol 1 mole of sodium chloride = 6.0221367 × 1023 Number of molecules of urea formula units of NaCl = 5.618 × 1022 molecules 9  Number of moles of a gas (n) = Problem 1.6 : Calculate the number of atoms in each of the following Volume of the gas at STP Molar volume of gas i. 52 moles of Argon (Ar) ii. 52 u of Helium (He) Thus iii. 52 g of Helium (He) Number of moles of a gas (n) = Solution : Volume of the gas at STP i. 52 moles of Argon 22.4 dm3mol-1 1 mole Argon atoms = 6.022 × 1023 atoms of Ar Number of molecules = number of moles × ∴52 moles of Ar 6.022 × 1023 molecules mol-1 6.022 × 1023 (Note : IUPAC has recently changed the = 52 moles × atoms standard pressure to 1 bar. Under these new 1mol STP conditions the molar volume of a gas is = 313.144 × 1023 atoms of Argon 22.71 L mol-1) ii. 52 u of Helium Atomic mass of He = mass of 1 atom of Problem 1.7 : Calculate the number of He = 4.0u moles and molecules of ammonia (NH3) ∴4.0 u = 1 He ∴52 u = ? 1atom gas in a volume 67.2 dm3 of it measured = 52 u × 4.0 u = 13 at STP. atoms of He Solution : iii. 52 g of He Volume of NH3 at STP = 67.2 dm3 Mass of 1 mole of He = 4.0 g molar volume of a gas = 22.4 dm3 mol-1 Number of moles of He Number of moles (n) mass of He Volume of the gas at STP = mass of 1mole of He = Molar volume of gas 52 g 67.2 dm3 = 4.0 g mol-1 = 13 mol Number of moles of NH3 = 22.4 dm3 mol-1 Number of atoms of He = 3.0 mol = Number of moles × 6.022 × 1023 Number of molecules = Number of moles × = 13 mol × 6.022 × 1023 atoms/mol 6.022 × 1023 molecules mol-1 = 78.286 × 1023 atoms of He. Number of molecules of NH3 = 3.0 mol × 6.022 × 1023 molecules mol-1 1.9 Moles and gases : Many substances exist = 18.066 × 1023 molecules as gases. If we want to find the number of moles of gas, we can do this more conveniently by measuring the volume rather than mass of the Try this gas. Chemists have deduced from Avogardro law that ''One mole of any gas occupies a Calculate the volume in dm3 occupied volume of 22.4 dm3 at standard temperature by 60.0 g of ethane at STP. (00C) and pressure (1 atm) (STP). The volume of 22.4 dm3 at STP is known as molar volume of a gas. 10  Exercises 1. Choose the most correct option. 2. Answer the following questions. A. A sample of pure water, whatever the A. State and explain Avogadro's law. source always contains by B. Point out the difference between 12 g of mass of oxygen and 11.1 % by mass of carbon and 12 u of carbon hydrogen. C How many grams does an atom of a. 88.9 b. 18 c. 80 d. 16 hydrogen weigh ? B. Which of the following compounds can D. Calculate the molecular mass of the NOT demonstrate the law of multiple following in u. proportions ? a. NH3 b. CH3COOH c. C2H5OH a. NO, NO2 b. CO, CO2 E. How many particles are present in 1 c. H2O, H2O2 d. Na2S, NaF mole of a substance ? C. Which of the following temperature F. What is the SI unit of amount of a will read the same value on celsius and substance ? Fahrenheit scales. G. What is meant by molar volume of a a. - 400 b. + 400 gas ? c. -80 0 d. -200 H. State and explain the law of conservation D. SI unit of the quantity electric current is of mass. I. State the law of multiple proportions. a. Volt b. Ampere 3. Give one example of each c. Candela d. Newton A. homogeneous mixture E. In the reaction N2 + 3H2 2NH3, B. heterogeneous mixture the ratio by volume of N2, H2 and NH3 C. element D. compound is 1 : 3 : 2 This illustrates the law of 4. Solve problems : a. definite proportion A. What is the ratio of molecules in 1 mole b. reciprocal proportion of NH3 and 1 mole of HNO3. c. multiple proportion (Ans. : 1:1) d. gaseous volumes F. Which of the following has maximum B. Calculate number of moles of hydrogen number of molecules ? in 0.448 litre of hydrogen gas at STP a. 7 g N2 b. 2 g H2 (Ans. : 0.02 mol) c. 8 g O2 d. 20 g NO2 C. The mass of an atom of hydrogen is G. How many g of H2O are present in 0.25 1.008 u. What is the mass of 18 atoms mol of it ? of hydrogen. (18.144 u) a. 4.5 b. 18 D. Calculate the number of atom in each c. 0.25 d. 5.4 of the following (Given : Atomic mass H. The number of molecules in 22.4 cm3 of of I = 127 u). nitrogen gas at STP is a. 254 u of iodine (I) a. 6.022 x 1020 b. 254 g of iodine (I) b. 6.022 x 1023 (Ans. : 2 atoms, 1.2044 x 1024 atoms) c. 22.4 x 1020 E. A student used a carbon pencil to write d. 22.4 x 1023 his homework. The mass of this was I. Which of the following has the largest found to be 5 mg. With the help of this number of atoms ? calculate. a. 1g Au (s) b. 1g Na (s) a. The number of moles of carbon in his c. 1g Li (s) d. 1g Cl2 (g) homework writing. (Ans : 4.16 x 10-4) 11  b. The number of carbon atoms in 12 N. Calculate the number of moles of mg of his homework writting magnesium oxide, MgO in i. 80 g and (Ans : 6.022 x 1020) ii. 10 g of the compound. (Average F. Arjun purchased 250 g of glucose atomic masses of Mg = 24 and O = 16) (C6H12O6) for Rs 40. Find the cost of (Ans. i. 2 mol ii. 0.25 mol) glucose per mole. O. What is volume of carbon dioxide, CO2 (Ans : Rs 28.8) occupying by i. 5 moles and ii. 0.5 mole G. The natural isotopic abundance of 10B of CO2 gas measured at STP. is 19.60% and 11B is 80.40 %. The exact (Ans. i. 112 dm3 ii. 11.2dm3) isotopic masses are 10.13 and 11.009 P. Calculate the mass of potassium respectively. Calculate the average chlorate required to liberate 6.72 dm3 of atomic mass of boron oxygen at STP. Molar mass of KClO3 is (Ans. :10.81) 122.5 g mol-1. H. Convert the following degree Celsius (Ans. 24.5 g) temperature to degree Fahrenheit. Q. Calculate the number of atoms of a. 40 0C b. 30 0C hydrogen present in 5.6 g of urea, (Ans. : A. 104 0F, B. 86 0F ) (NH2)2CO. Also calculate the number I. Calculate the number of moles and of atoms of N, C and O. molecules of acetic acid present in 22 g (Ans. : No. of atoms of H = 2.24 x 1023, of it. N =1.124 x 1023 and C = 0.562 x 1023, O (Ans. : 0.3666 mol, 2.2076 x 1023 = 0.562 x 1023) molecules ) R. Calculate the mass of sulfur dioxide J. 24 g of carbon reacts with some oxygen produced by burning 16 g of sulfur in to make 88 grams of carbon dioxide. excess of oxygen in contact process. Find out how much oxygen must have (Average atomic mass : S = 32 u, been used. O = 16 u) (Ans. : 64.0 ) (Ans. 32 g) K. Calculate number of atoms is each of 5. Explain the following. (Average atomic mass : A. The need of the term average atomic N = 14 u, S = 32 u) mass. a. 0.4 mole of nitrogen B. Molar mass. b. 1.6 g of sulfur C. Mole concept. (Ans. : A. 2.4088 x 1023 , D. Formula mass with an example. B. 3.011 x 1022 atom ) E. Molar volume of gas. L. 2.0 g of a metal burnt in oxygen gave F. Types of matter (on the basis of chemical 3.2 g of its oxide. 1.42 g of the same composition). metal heated in steam gave 2.27 of its oxide. Which law is verified by these data ? Activity : M. In two moles of acetaldehyde (CH3CHO) calculate the following Collect information of various scientists a. Number of moles of carbon and prepare charts of their contribution in b. Number of moles of hydrogen chemistry. c. Number of moles of oxygen d. Number of molecules of acetaldehyde (Ans. : A. 4 mol, B. 8 mol, C. 2 mol, D. 12.044 x 1023 molecules ) 12  2. Introduction to analytical chemistry 2.1 Intrduction : Analytical chemistry 2.2 Analysis : Analysis is carried out on a facilitates investigation of chemical small sample of the material to be tested, and composition of substances. It uses the not on the entire bulk. When the amount of instruments and methods to separate, identify a solid or liquid sample is a few grams, the and quantify the matter under study. The analysis is called semi-microanalysis. It is analysis thus provides chemical or physical of two types : qualitative and quantitative. information about a sample. Analysis may Classical qualitative analysis methods include be qualitative or quantitative. Qualitative separations such as precipitation, extraction analysis is concerned with the detection of the and distillation. Identification may be based presence or absence of elements in compounds on differences in colour, odour, melting and mixture of compounds. Quantitative point, boiling point, and reactivity. Classical analysis deals with the determination of the quantitative methods consist of volumetric relative proportions of elements in compounds analysis, gravimetric analysis, etc. and mixture compounds. 2.2.1 Chemical methods of qualitative analysis : Chemical analysis of a sample is Remember carried out mainly in two stages : by the dry The branch of chemistry which deals method in which the sample under test is not with the study of seperation, identification, dissolved and by the wet method in which the qualitative and quantitative determination sample under test is first dissolved and then of the compositions of different substances, analyzed to determine its composition. The dry is called analytical chemistry. method is usually used as preliminary tests in Importance of analytical chemistry : the qualititative analysis. The course of analytical chemistry The semi-micro qualitative analysis is extends the knowledge acquired by the carried out using apparatus such as : test tubes, students in studying general, inorganic beakers, evaporating dish, crucible, spot plate, and organic chemistry. Chemical analysis watch glass, wire guaze, water bath, burner, is one of the most important methods of blow pipe, pair of tongues, centrifuge, etc. monitoring the composition of raw materials, The qualitative analysis of organic and intermediates and finished products, and also inorganic compounds involves different types the composition of air in streets and premises of tests. The majority of organic compounds of industrial plants. In agriculture, chemical are composed of a relatively small number analysis is used to determine the compostion of elements. The most important ones are : of soils and fertilizers; in medicine, to carbon, hydrogen, oxygen, nitrogen, sulphur, determine the composition of medicinal halogen, phosphorous. Elementary qualitative preparations.Analytical chemistry has analysis is concerned with the detection of the applications in forensic science, engineering presence of these elements. The identification and industry. Industrial process as a whole and of an organic compound involves tests such as the production of new kinds of materials are detection of functional group, determinition closely associated with analytical chemistry. of melting/ boiling point, etc. The qualitative Analytical chemistry consists of classical, wet analysis of simple inorganic compounds chemical methods and modern instumental involves detection and confirmation of cationic methods. 13 and anionic species (basic and acidic radical) values as 6.022 x 1023 and 1.66 x 10-24 g. The in them. number 123.546 becomes 1.23546 x 102, in 2.2.2 Chemical methods of quantitative scientific notation. Note that while writing it, analysis : Quantitative analysis of organic we have moved the decimal to the left by two compounds involves methods such as (i) places and same is the exponent (2) of 10 in determination of percentage constituent the scientific notation. Similarly, 0.00015 can element, (ii) concentrations of a known be written as 1.5 x 10-4. compound in the given sample, etc. Quantitative analysis of simple inorganic Problem 2.1 : For adding 5.55 x 104 and compounds involves methods based on (i) 6.95 x 103, first the exponent is made decomposition reaction (gravimetric analysis), equal. Thus and (ii) the progress of reaction between two 5.55 x 104 + 0.695 x 104. Then these solutions till its completion (titrametric or numbers can be added as follows : volumetric analysis), etc. The quantitative (5.55 + 0.695) x 104 = 6.245 x 104 analytical methods involve measurement of quantities such as mass and volume, by Problem 2.2 : The subtraction of two means of some equipment/ apparatus such as numbers can be done as shown below: weighing machine, burette. 3.5 x 10-2 - 5.8 x 10-3 2.3 Mathematical operation and error = (3.5 x 10-2) - (0.58 x 10-2) analysis : The accuracy of measurement is of = (3.5 - 0.58) x 10-2 a great concern in analytical chemistry. Also = 2.92 x 10-2 there can be intrinsic errors in the analytical Here the decimal has to be moved four measurement. The numerical data, obtained places to the right and (-4) is the exponent experimentally, are treated mathematically to in the scientific notation. Now let us perform reach some quantitative conclusion. Therefore, mathematical operations on numbers an anlytical chemist has to know how to report expressed in scientific notation. the quantitative analytical data, indicating the extent of the accuracy of measurement, Problem 2.3 : ( 5.6 x 105) x (6.9 x 108) perform the mathematical operation and = (5.6 x 6.9) (105+8) properly express the quatitative error in the = (5.6 x 6.9) x 1013 result. In the following subsection we will = 38.64 x 1013 consider these aspects related to measurments = 3.864 x 1014 and calculation. 2.3.2 Scientific notation (exponential Problem 2.4 : (9.8 x 10-2) x (2.5 x 10-6) notation) : A chemist has to deal with numbers = (9.8 x 2.5) (10-2 + (-6)) as large as 602,200,000,000, 000, 000, 000, 000 = (9.8 x 2.5) x (10-2-6) for the molecules of 2 g of hydrogen gas or as = 24.50 x 10-8 small as 0.00000000000000000000000166g. = 2.45 x 10-7 that is, mass of a H atom. To avoid the writing Addition and subtraction : To perform of so many zeros in mathematical operations, addition operation, first the numbers are scientific notations i.e. exponential notations written in such a way that they have the same are used. Here, any number can be represented exponent. The coefficients are then added. into a form N x 10n where 'n' is an exponent (Problems 2.1 and 2.2) having positive or negative values and N can Multiplication : The rule for the multiplication vary 1 < N < 10. Thus, we can write the above of exponential numbers can be well explained from the solved problems 2.3 and 2.4. 14 2.3.2 Precision and accuracy of measurement Multiple readings of the same quantity Aim of any measurement is to get the are noted to minimize the error. If the actual value called true value or accepted readings match closely, they are said to value of a quantity. Nearness of the measured have high precision. High percision implies value to the true value is called the accuracy of reproducibility of the readings. High precision measurement. Larger the accuracy smaller the is a prerequisite for high accuracy. Precision error. Accuracy depends upon the sensitivity is expressed in terms of deviation. An absolute or least count (the smallest quantity that can deviation is the modulus of the difference be measured) of the measuring quuipment. between an observed value and the arithmetic consider, for example, a burette reading of mean for the set of several measurements made 10.2 mL. For all the three situations in the in the same way. It is a measure of absolute Fig. 2.1 the reading would be noted as 10.2 error in the repeated observation. mL It means that there is an uncertainty about Absolute deviation = Observed value - Mean the digit appearing after the decimal point in Arithmetic mean of all the absolute the reading 10.2 mL. This is because the least deviations is called the mean absolute count of the burette is 0.1 mL. The meaning deviation in the measurements. The ratio of of the reading 10.2 mL is that the true value mean absolute deviation to its arithmentic of the reading lies between 10.1 mL and 10.3 means is called relative deviation. mL. This is indicated by writing 10.2 ± 0.1 Relative deviation mL. Here, the burette reading has an error of ± = Mean absolute deviation x 100 % 0.1mL (Fig. 2.1). Mean Errors may be expressed as absolute or relative error. Problem 2.5 : In laboratory experiment, Absolute error = Observed value - True value 10 g potassium chlorate sample on decomposition gives following data ; The Relative error is generally a more useful sample contains 3.8 g of oxygen and the quantity than absolute error. Relative error is actual mass of oxygen in the quantity of the ratio of an absolute error to the true value. potassium chlorate is 3.92 g. Calculate It is expressed as a percentage. absolute error and relative error. Absolute error Solution : The observed is 3.8 g and Relative error = x 100 % accepted value is 3.92 g True value There can be error in a measurement due Absolute error = Observed value - True value to a number of reasons including inefficiency = 3.8 - 3.92 = - 0.12 g of the person doing measurement. The negative sign indicates that your experimental result is lower than the true value. Absolute error The relative error = x 100% True value -0.12 = x 100% 3.92 = -3.06 % Fig. 2.1 : Three possibilities of a burette reading 10.2 mL 15  Problem 2.6 : The three identical samples of 2.3.4 Rules for deciding significant figures : potassium chlorate are decomposed. The mass 1. All non zero digits are significant; e. g. of oxygen is determined to be 3.87 g, 3.95 g and 127.34 g contains five significant figures 3.89 g for the set. Calculate absolute deviation which are 1, 2, 7, 3 and 4. and relative deviation. 2. All zeros between two non zero digits are significant e. g. 120.007 m contains six Solution : significant figures. 3.87 + 3.95 + 3.89 mean = = 3.90 3. Zeroes on the left of the first non zero digit 3 are not significant. Such a zero indicates the Average deviation of a set of position of the decimal point. For example, measurements 0.025 has two significant figures, 0.005 has Sample Mass of oxygen Deviation one significant figure. 1 3.87g 0.03g 4. Zeroes at the end of a number are significant 2 3.95g 0.05g if they are on the right side of the decimal point. 3 3.89g 0.01g Terminal zeros are not significant if there is no Mean 0.03 decimal point. (This is beacause the least count absolute of an instrument contains decimal point) For deviation example 0.400 g has three singnificant figures. Absolute deviation The measurements here indicates that it is made = Observed value - Mean on a weighing machine having least count of ∴ Mean absolute deviation = ± 0.03 g. 0.001 g. Significant figures are also indicated The relative deviation, in scientific notation by means of decimal = Mean absolute deviation x 100 % point. For example, the measurment 400 g has Mean one significant figure. The measurement 4.0 × 0.03 102 g has two significant figures, wheras the = x 100% = 0.8% 3.9 measurment 4.00 × 102 g has three significant figures. The zeros after the decimal points in 2.3.3 Significant Figures : Uncertainty these cases indicates that the least counts of in measured value leads to uncertainty in the weighing machines are 1 g, 0.1 g and 0.01 calculated result. Uncertainty in a value g, respectively. is indicated by mentioning the number of 5. In numbers written is scientific notation, all significant figures in that value. Consider, the digits are significant. For example, 2.035×102 column reading 10.2 ± 0.1 mL recorded on a has four significant figures, and 3.25 × 10-5 has burette having the least count of 0.1 mL. Here three significant figures. it is said that the last digit ‘2’ in the reading is uncertain, its uncertainty is ±0.1 mL. On the other hand, the figure ‘10’ is certain. Problem 2.7 : How many significant figures The significant figures in a measurement are present in the following measurements ? or result are the number of digits known a. 4.065 m b. 0.32 g c. 57.98 cm3 with certainty plus one uncertain digit. In a d. 0.02 s e. 4.0 x 10-4 km scientific experiment a result is obtained by f. 604.0820 kg g. 307.100 x 10-5 cm doing calculation in which values of a number of quantities measured with equipment of Ans. : a. 4 b. 2 c. 4 d. 1 different least counts are used. e. 2 f. 7 g. 6 Following rules are to be followed during such calculation. 16  In general, a quantity measured with 2.4 Determination of molecular formula : an instrument of smaller least count will Molecular formula of a compound have more significant figures and will be is the formula which indicates the actual more accurate than when measured with an number of atoms of the constituent elements instrument of larger least count. in a molecule. It can be obtained from the 2.3.5 Calculations with significant figures : experimentally determined values of percent When performing calculations with elemental composition and molar mass of that measured quantities the rule is that the compound. accuracy of the final result is limited to the 2.4.1 Percent composition and empirical accuracy of the least accurate measurement. formula : Compounds are formed by In other words, the final result can not be chemical combination of different elements. more accurate than the least accurate number Quantitative determination of the constituent involved in the calculation. element by suitable methods provides the Rounding off : The final result of a calculation percent elemental composition of a compound. often contains figures that are not significant. If the percent total is not 100, the difference When this occurs the final result is rounded is considered as percent oxygen. From the the off. The following rules are used to round off per cent composition, the ratio of the atoms a number to the required number of significant of the constituent elements in the molecule is figures : calculated. The simplest ratio of atoms of the If the digit following the last digit to constituent elements in a molecule is called be kept is less than five, the last digit is left the empirical formula of that compound. unchanged. Molecular formula can be obtained from e.g. 46.32 rounded off to two significant the empirical formula if the molar mass is figures is 46. known. The molar mass of the substance under If the digit following the last digit to be examination is determined by some convenient kept is five or more, the last digit to be kept is method. The following example illustrates this increased by one. e.g. 52.87 rounded to three sequence. significant figures is 52.9. Problem 2.9 : A compound contains 4.07 Problem 2.8 : Round off each of the % hydrogen, 24.27% carbon and 71.65 % following to the number of significant digits chlorine by mass. Its molar mass is 98.96 indicated : g. What is its empirical formula ? Atomic a. 1.223 to two digits b. 12.56 to three masses of hydrogen, carbon and chlorine are digits c. 122.17 to four digits d. 231.5 to 1.008, 12.000 and 35.453 u, respectively three digits. Solution : Ans. : i. 1. 2; the third digit is less than 5, so Step I : Check whether the sum of all the we drop it all the others to its right. percentages is 100. ii. 12.6 ; the fourth digit is greater than 5, so 4.07 + 24.27 + 71.65 = 99.99 ≈100 we drop it and add 1 to the third digit. Therefore no need to consider presence of iii. 122.2 ; the fifth digit is greater than 5, so oxygen atom in the molecule. we do it and add 1 to the fourth digit. Step II : Conversion of mass percent to iv. 232; the fourth digit is 5, so we drop it grams. Since we are having mass percent, it and add 1 to the third digit. is convenient to use 100 g of the compound as the starting material. Thus in the 100