Introduction

A dynamic environment of matter and material surrounds man. Since his birth he has been exposed to these changing surroundings. His curiosity to understand his surroundings and the events occurring around him, has led him to systematically enquire and collect information through experiments and observations. This knowledge is referred to as science. Science may be defined as systematized knowledge gained by mankind through observations and experimentation. Science has been further classified into different branches due to its enormous expansion and diversified fields. Chemistry is one of its most important disciplines and refers to the study of the materials that make up the universe.Importance of Chemistry

The branch of science, which deals with the study of matter, its composition, its properties and the changes that it undergoes in composition as well as in energy during various processes, is defined as chemistry. It has been further divided into different branches depending upon specialized fields of study. The major branches of chemistry are: Branches of Chemistry Organic Chemistry This branch deals with study of carbon compounds especially hydrocarbons and their derivatives. Inorganic Chemistry This branch deals with the study of compounds of all other elements except carbon. It largely concerns itself with the study of minerals found in the Earth's crust. Physical Chemistry The explanation of fundamental principles governing various chemical phenomena is the main concern of this branch. It is basically concerned with laws and theories of the different branches of chemistry. Industrial Chemistry The chemistry involved in industrial processes is studied under this branch. Analytical Chemistry This branch deals with the qualitative and quantitative analysis of various substances. Biochemistry This branch deals with the chemical changes going on in the bodies of living organisms; plants and animals. Nuclear Chemistry Nuclear reactions, such as nuclear fission, nuclear fusion, transmutation processes etc. are studied under this branch. In addition, with new developments emerging in recent years, more branches of chemistry are being added. These include, pharmaceutical chemistry, geochemistry, agricultural chemistry, medicinal chemistry, solid-state chemistry, etc. As chemistry concerns itself with matter and materials of the universe, it obviously plays a very significant role in almost all walks of life. Modern daily life involves the use of many chemical products and many chemical changes, although we are unaware of them. Major contributions of chemistry to the different walks of life in modern world are as given below: Chemistry in Medicine The eradication of a number of diseases and major contributions to improving human health, have been made by

modern chemical discoveries. Some important contributions in this field are: 

 Life saving drugs and prophylactics, i.e., disease preventing serum and vaccines 

 Anaesthetics and antiseptics 

 Disinfectants and germicides Chemistry in industry Chemistry plays an important role in most industrial processes. Important examples are manufacture of: 

 Synthetic fibre like rayon, nylon, dacron, orlon etc. 

 Plastics like bakelite, teflon, polythene etc. 

 Paints, varnishes, enamels, dyes etc. 

 Cement, glass and ceramics and extraction of metals like silver, magnesium, gold, iron etc. Chemistry in Agriculture Significant contributions towards securing enough food grains and protecting crops have been made by the use of chemistry. These are: 

 Chemical fertilizers like urea, ammonium sulphate, calcium nitrate etc., for the better production of crops. 

 Insecticides such as DDT, BHC, chlorinated hydrocarbons, aldrin, malathion, dieldrin, furodan etc., for the protection of crops from insects and to help in the safe storage of foodgrains. 

 Preservatives like sodium benzoate, sodium metabisulphite and salicylic acid for better preservation of food and check its wastage. Chemistry in Life Styles In every walk of life, chemistry has been the pioneer contributor towards enabling the comfort, convenience and pleasures of man. Chemistry has helped us in getting a number of amenities. Some examples are: 

 Domestic requirements such as better fabrics, soaps, shampoos, cosmetics, dyes, oils, flavouring essences, paper etc. 

 Air conditioning, refrigeration and domestic gadgets: To keep people cool and food fresh in all seasons, we make use of air conditioning. Chemical substances like liquid ammonia, liquid sulphur dioxide, CFC etc., are used in these machines. All consumer electronic gadgets based on Printed Circuit Board technology (PCB'S) use specialized chemical coatings for their functioning. Chemistry and Energy Resources The energy resources of the world are mainly petroleum, coal, wood, nuclear fuels. To conserve the fast depleting reserves of coal and petroleum, attempts are being made by chemists to utilize alternate energy fuels. Nuclear energy, solar energy, ocean energy (hydrogen from seawater), which may provide inexhaustible sources, are being studied under chemistry for future energy needs. Dark Side of Chemistry Chemistry has also been misused by unscrupulous people, endangering the society. It has given the menace of harmful drugs and their bulk processing, like LSD, cocaine, brown sugar, angel dust, etc. that cripple the society. Chemistry has also developed explosives such as TNT, RDX and other deadly chemical weapons of destruction - the atomic and hydrogen bombs. Terrorists and aggressive societies have used these with devastating effects.

Physical Quantities and their Measurement

The development of chemistry is based upon the systematic approach of the scientists in carrying out different experiments. Scientific theories, inferences and generalizations are made on the observations recorded. The quantitative information involves the measurement of one or more quantities. These measurements are often used in calculations to obtain other related quantities. The ability of a chemist to measure properties as accurately as possible depends upon his skill and the degree of accuracy of the instrument. Chemistry is largely an experimental science. It deals with physical things that can be measured. A number of devices are used in laboratories to make simple measurements such as, a metre rod measures length; a physical balance measures mass; burette, pipette, graduated cylinder etc. measure volume; thermometer measures temperature and so on. Any measured physical quantity consists of two parts: 

 The numerical value 

The unit For example, if one measures the length of a wall as 24.8 m, then 24.8 represents the numerical value and 'm' (which stands for metre) represents the unit.

S.I. Units

The value of any physical quantity is always expressed as its numerical value as well as a unit. One often hears the weight expressed as grams or kilograms; the distance as foot, yard or mile; the volume as litres or gallons. The use of a number of units cause confusion and complications. Hence, the need for a uniform system of measurement was felt as a necessity. In 1971, the French Academy of Science devised the metric system of measurement based on the decimal system. In October 1960, The XIth General Conference of Weights and Measures (Conference Generale des Poids et Mesures) adopted an International System of Units. This system is known as SI system of units after the French word, Systeme' International d'Unites. The SI system has seven basic units from which all other units are derived. The Seven Basic Units 

 Note: Although the SI unit of temperature is Kelvin, Celsius scale (oC) is commonly used. K=oC + 273.15. Similarly Angstrom (Å) is the commonly used unit but nanometer or picometer is the SI unit.1nm = 10-9m, 1pm = 10-12m.  

Prefixes in SI system The SI units of some of the physical quantities are either too small or too large. To change the order of magnitude, the metric system, which is a decimal system, relates this in powers of ten. These powers are indicated by prefixes. Standard prefixes for reducing or enlarging the size of any unit 

 Some rules on the use of SI units 

 While writing a unit, only its singular form is used. 

The abbreviation does not have a full stop. 

The sign of degree (o) is omitted when Kelvin scale is used. 

Unit combinations should be indicated by means of either a dot or leaving space in between.      Metre Kelvin = m.K or m K 

Words and symbols should not be used in mixed forms. J per mole should be either written as joule per mole or J mole-1. 

 A unit with a prefix is a power for the complete unit. cm3 means (centimetre)3 and not centi (metre)3

  Exponents also operate on prefixes. 1cm2 = (10-2 m)2 = 10-4 m2

Derived Units

The units of all physical quantities can be derived from the seven basic units. These units are called derived units because they can be derived from the basic units algebraically by multiplication and division. It is frequently necessary to convert one set of units to another.

                                  Some of the Common derived units 

Uncertainty in Measurement

When dealing with objects that can be counted, one can always get an exact answer. But while measuring the weight of a person, the volume of milk in a container, the length of a cloth etc. it is not possible to measure them exactly. The difference in the two situations arises because objects are measured by a discrete (distinct) variable (there can be three or four balls and nothing in between) while quantities like weight, volume etc. are measured by a continuous variable (weight can be 42kg or 43kg or anything in between). Measurement of continuous variable can only be as precise as the skill of the observer and the choice of the measuring apparatus. The limit of reproducibility of any measurement is known as its precision. A measurement may be absolutely precise or uncertain depending upon the instrument or the person making the measurement. The degree of conformity of a measure to its true value is termed as its accuracy. 

Skill and Accuracy of the Observer Suppose three students A, B and C measure the volume of the liquid in a cylinder. They report the following values: Result of A: 23.4 mL; Result of B= 23.5 mL; Result of C= 23.6 mL If the correct volume of the liquid is 23.5 mL, it means that the student B has measured the volume of the liquid correctly while the students A and C have made some error. 

 

Limitation of the Measuring Instrument Limitation of the measuring instrument is an important factor. This leads to uncertainty in measurement. In order to understand this, let us assume that we want to measure the volume of the liquid with a graduated cylinder. If the volume of the liquid is more than 23 mL and less than 24 mL, it lies between 23 and 24 mL. With the scale on the measuring cylinder, one cannot have a correct reading in the first decimal place. We can guess

that it is approximately in the middle of 23 and 24 and therefore, the volume of the liquid may be reported as 23.5 mL. Here 2 and 3 are certain digits while 5 is an uncertain digit. Therefore, different persons measuring the volume of the

liquid might report volumes that differ by  0.1 or  0.2 mL. So, every scientific measurement has some degree of uncertainty. This is mainly due to the skill of the observer and limitation of the instrument. This uncertainty of measurement is expressed in figures called significant figures.

Significant Figures

The reliability of a measurement is indicated by the number of digits used to represent it. To express it more accurately we express it with digits that are known with certainty. These are called as Significant figures. They contain all the certain digits plus one doubtful digit in a number. If the volume of the liquid is reported to be 23.5mL, it means that digits 2 and 3 are certain while 5 is uncertain. Therefore, the number 23.5 has three significant figures. The number of significant figures conveys that except the last digits all others are correct. The number of digits depends on the precision of the scale used. 

Rules for Determining the Number of Significant Figures 

 All non-zero digits are significant. For example, 6.9 has two significant figures, while 2.16 has three significant figures. The decimal place does not determine the number of significant figures. 

 A zero becomes significant in case it comes in between non zero numbers. For example, 2.003 has four significant figures, 4.02 has three significant figures. 

 Zeros at the beginning of a number are not significant. For example, 0.002 has one significant figure while 0.0045 has two significant figures. 

 All zeros placed to the right of a number are significant. For example, 16.0 has three significant figures, while 16.00 has four significant figures. Zeros at the end of a number without decimal point are ambiguous. 

In exponential notations, the numerical portion represents the number of significant figures. For example, 0.00045 is expressed as 4.5 x 10-4 in terms of scientific notations. The number of significant figures in this number is 2, while in Avogadro's number (6.023 x 1023) it is four. 

 The decimal point does not count towards the number of significant figures. For example, the number 345601 has six significant figures but can be written in different ways, as 345.601 or 0.345601 or 3.45601 all having same number of significant figures.

Problems - Significant Figures

1. State the number of significant figures in each of the following: (i) 208.91 (ii) 0.00456 (iii) 453 (iv) 2.945 x 104 (v) 0.346 Solution (i) 208.91 has five significant figures. (ii) 0.00456 has three significant figures. (iii) 453 has three significant figures. (iv) 2.945 x 104 has four significant figures. (v) 0.346 has three significant figures. 2. Express 0.00000345 in scientific notation and calculate the number of significant figures. Solution 

0.00000345 can be expressed in scientific notation as 3.45 x 105. The significant figures are three. 

Calculations Involving Significant Figures Rule I When addition or subtraction is to be carried out in the numbers having different precisions, the final result should be reported up-to the same number of decimal places as are present in the term having the least number of decimal places. 

 Addition of numbers In the addition of 42.2 + 4.22 + 0.422, all the three numbers have three significant figures, but 42.2 has the least number of decimal place, namely one. The answer is therefore limited to one decimal place. 

 Subtraction of numbers The subtraction of numbers is carried out in the same way as addition. In the difference of 5.2748 and 5.2722 the answer (0.0026) has four decimal places. The numbers have five significant figures but the answer has only two significant figures. Rule II In the multiplication or division of numbers, the final result should be reported up to the same number of significant figures as are present in the term with the least number of significant figures. 

 Multiplication of numbers In the multiplication of 5.1028 (five significant figures) with 1.30 (three significant figures) the value is 6.63364, but the correct answer is 6.63. 

 Division of numbers In the division of 5.2765 (five significant figures) by 1.25 (three significant figures) the result is 4.2212 but the correct answer is 4.22. 

Retention of Significant Figures - Rounding off Figures The rounding off procedure is applied to retain the required number of significant figures. 

 If the digit coming after the desired number of significant figures happens to be more than 5, the preceding significant figure is increased by one, 4.317 is rounded off to 4.32. 

 If the digit involved is less than 5, it is neglected and the preceding significant figure remains unchanged, 4.312 is rounded off to 4.31. 

 If the digit happens to be 5, the last mentioned or preceding significant figure is increased by one only in case it happens to be odd. In case of even figure, the preceding digit remains unchanged. 8.375 is rounded off to 8.38 while 8.365 is rounded off to 8.36. The rounding off is done only in the final answer when the problem involves more than one step. Problems 3. Express the results of the following calculations to the appropriate number of significant figures. 

 Solution

 

 = 0.05608 = 0.0561 The number of significant figures in the term 3.24 is 3, therefore the result should have 3 significant figures. Therefore, the correct answer is 0.0561. The number after zero is 8 and therefore it is rounded off to one. 

 = 0.2615 x 10-4 = 0.3 x 10-4

 The answer should have one significant figure because 0.5 has one significant figure. 4. Calculate to proper significant figures: (i) 12.6 x 11.2 (ii) 172.8/15 Solution (i) 12.6 x 11.2 = 141.12 Correct answer = 141 (up to three significant figures) (ii) 172.8/15 = 11.52 Correct answer = 12 (up to two significant figures) 

Dimensional Analysis It is often necessary to convert one set of units to another. This can be done in a systematic way by the method called dimensional analysis or the conversion factor method. Conversion factor is equal to the ratio of the magnitudes of the same quantity in the two units. To obtain a conversion factor the magnitude of the physical quantity in the unit to be converted is placed in the denominator, while that in the desired unit is placed in the numerator. For example, to convert 25.6 metres to centimetres, we know that 

 100 cm/1m is called the conversion factor because it converts a quantity expressed in one unit to a quantity expressed in another unit. Multiply the given quantity by the conversion factor and retain the units of the physical quantity and that of the conversion factor in such a way that all units cancel out leaving behind only the required units. 

 Similarly to convert 10 minutes to seconds, one proceeds as follows: 1min = 60 s 

 10 min = 10 min x 1 

 

 The advantage of this method is that if the equation is written correctly, all the units except the desired one get cancelled. If the conversion factor is written incorrectly, then the desired unit is not obtained. To convert the units of a physical quantity involving many terms, it is convenient to place the units of each physical quantity directly in the steps of the calculations along with the number, cancel out the common units and finally convert the units by using proper conversion factor. Problems 5. Express each of the following in SI units: (1) 93 million miles (distance between earth and sun) (2) 5 feet and 2 inches (height of an Indian female) (3) 100 miles per hour (speed of Rajdhani Express) (4) 14 pounds per square inch (atmospheric pressure) (5) 0.4 Å (bond length of hydrogen molecule) Solution (1) 93 million miles = 93 x 106 miles 1 mile = 1.60934 km = 1.60934 x 103 m 

 

 

 = 1.5 x 1011 m (2) 5 feet and 2 inch = (5 x 12 + 2) inch = 62 inch 1 inch = 2.54 x 10-2 m 

 then, 5 feet and 2 inch = 62 inch 

 

 1 mile = 1.60934 x 103 m, 1 min 60 x 60 s 

 

 

 1 pound force = 4.448 N, 1 inch = 2.54 x 10-2 m 

 

 = 9.66 x 104 N m-2

 

 

 

 = 0.74 x 10-10 m 7.4 x 10-11 m 6. The density of ethanol is 0.79 g cm-3 at 25o C. What is its value in SI units? Solution 1 cm3 of ethanol weighs 0.79 g or 0.79 x 10-3 kg 1 m3 i.e., 106 cm3 weighs = 0.79 x 10-3 x 106 kg = 0.79 x 103 kg 0.79 g cm-3 = 0.79 x 103 kg m-3

Classification of Matter

All around us are substances that are made of matter. Matter has mass and occupies some space. Different kinds of substances are made up of different matter. Matter can be thus classified in many ways. Two basic classifications of matter are based on its physical and chemical properties. The physical classification of matter shows its physical appearance that describe the shape, hardness, softness, melting and boiling points. The chemical classification of matter shows the ability of matter to change its composition to form new substances. 

Physical Classification of Matter Based on its physical characteristics, matter is divided into solids, liquids and gases. Solids The particles are closely packed and bound by strong inter-particle attraction, which makes solids rigid and geometrical. This gives them definite shape and definite volume, for e.g., rock, wood, steel girders etc. Liquids The particles are loosely packed and are bound to each other with forces weaker than those of solids. This gives them definite volume but not definite shape. Thus liquids are mobile and take the shape of the container in which they are placed. Examples: Water, oil, milk etc. Gases The particles are separated by much greater distances, almost 10 to 100 times the size of the particles. The operating forces of attraction are virtually non-existent in gases, resulting in loosely packed particles, which are free to move in any direction. Thus gases possess neither definite volume and shape and occupy the whole volume of the vessel in which they are placed e.g., air in balloons, oxygen in cylinders. The three states of matter are inter-convertible. This is achieved by heating and cooling, which affect the kinetic energy of the particles. Heating increases the kinetic energy of matter: as a result, solids when heated become liquids and liquids become gas, when heated. Comparison of the properties of solids, liquids and gases 

 

  

Chemical Classification of Matter Based on its broad chemical composition, matter is classified as elements, compounds and mixtures. Elements An element is the simplest form of matter that cannot be split into simpler substances or built from simpler substances by any ordinary chemical or physical method. There are 110 elements known to us, out of which 92 are naturally occurring while the rest have been prepared artificially. Elements are further classified into metals, non-metals and metalloids. Metals All elements except hydrogen, which form positive ions by losing electrons during chemical reactions are called metals. Thus metals are electropositive elements. They are characterized by bright lustre, hardness, ability to resonate sound and are excellent conductors of heat and electricity. Metals are solids under normal conditions except for Mercury. They are ductile (can be drawn into wire) and malleable (can be beaten into very thin sheets). 

Non-metals Elements that tend to gain electrons to form anions during chemical reactions are called non-metals. These are electronegative elements. They are non-lustrous, brittle and poor conductors of heat and electricity (except graphite). Non-metals can be gaseous, liquids or solids. Metalloids Elements that behave like both metals and non-metals are called metalloids. Common metal, non-metals and metalloids 

 

Compounds A compound is a pure substance made up of two or more elements combined in a definite proportion by mass, which could be split by suitable chemical methods. Characteristics of compound 

 Compounds always contain a definite proportion of the same elements by mass. Water as a compound, always contains hydrogen and oxygen in the ratio 1:8 by mass. 

 The properties of compounds are totally different from the elements from which they are formed. For example, while water is normally used for extinguishing fire, its elements are not. Hydrogen is combustible and oxygen is a supporter of combustion. 

 Compounds are homogeneous. 

 During the formation of a compound, energy in the form of heat, light or electricity is either evolved or absorbed. Coal when burnt, gives heat and light energies. Compounds are broadly classified into inorganic and organic compounds. Inorganic compounds are those, which are obtained from non-living sources such as minerals. For example, common salt, marble and limestone. Organic compounds are those, which occur in living sources such as plants and animals. They all contain carbon. Common organic compounds are oils, wax, fats etc.

Mixtures

A mixture is a combination of two or more elements or compounds in any proportion so that the components do not lose their identity. Air is an example of a mixture (several gases mix to form air). Mixtures are of two types, homogeneous and heterogeneous. 

 Homogeneous mixtures have the same composition throughout the sample. The components of such mixtures cannot

be seen under a powerful microscope. They are also called solutions. Examples of homogeneous mixtures are air, sea-water, gasoline, brass etc. 

 Heterogeneous mixtures consist of two or more parts (phases), which have different compositions. These mixtures have visible boundaries of separation between the different constituents and can be seen with the naked eye e.g., sand and salt, chalk powder in water etc. Classification of matter can be summarized as follows: 

 

 

Separation of Mixtures The separation of mixtures into its constituents in a pure state is an important process in chemistry. The constituents of any mixture can be separated on the basis of their differences in their physical and chemical properties e.g., particle size, solubility, effect of heat, acidity or basicity etc. Some of the methods for separation of mixtures are: 

Sedimentation or Decantation Aim To separate the mixture of coarse particles of a solid from a liquid e.g., muddy river water. Experiment Aim To separate the mixture of coarse particles of a solid from a liquid e.g., muddy river water. Principle The coarse particles of the solid being heavier than the liquid (usually water), settle down due to gravity. The clear upper layer of the liquid is then gently poured out into another container. Settling down of the coarse particles due to the effect of gravity is called sedimentation. The mechanical transfer of the clear upper liquid without disturbing the settled solid particles is called decantation. Process The mixture is taken in a container and allowed to stand for sometime. The solid particles settle down with time. Settling down of the particles leaves the upper layers of the liquid clearer. Bigger particles settle down faster than the finer particles. Sometimes the sedimentation process is hastened by adding a small quantity of alum. Al3+ ions in alum cause the coagulation of the fine particles by undergoing hydrolysis to Al(OH)3 that has a strong tendency for

adsorption. 

 

 Fig: 1.1 - Separation of the coarse particles of a solid from a liquid by sedimentation and decantation 

Filtration Aim To separate the insoluble solid component of a mixture from the liquid completely i.e. separating the precipitate (solid phase) from any solution. Experiment Aim To separate the insoluble solid component of a mixture from the liquid completely i.e. separating the precipitate (solid phase) from any solution. Principle The solvent molecules and the molecules/ions present in the solution can pass through the porous membranes while the suspended particles cannot and are retained on the porous membrane. Process The solution containing the suspended impurities is made to pass through the porous membrane such as filter paper, filter cloth etc. The solvent or solution containing dissolved substances passes through the porous membrane, which is called filtrate. The insoluble solid suspended particles that remain on the porous membrane is termed residue. When the suspended impurities are very fine, a small amount of alum added to the suspension makes filtration faster. 

            Fig:1.2 - Separation by filtration 

Evaporation Aim To separate a non-volatile soluble salt from a liquid or recover the soluble solid solute from the solution. The solvent is lost into the surroundings. Experiment Aim To separate a non-volatile soluble salt from a liquid or recover the soluble solid solute from the solution. The solvent is lost into the surroundings. Principle Liquids evaporate at all temperatures. Evaporation becomes faster at higher temperatures. Process The solution containing the mixture is taken in a china dish and heated gently. Gradually the solvent evaporates and the solution containing the dissolved solute becomes thicker. The semi-solid mass left on the china dish is slowly heated to dryness. 

 Fig: 1.3 - Evaporation of a solution Crystallization This method is used to separate a solid compound in pure and geometrical form. A nearly saturated solution of an impure substance is prepared in a hot solvent. The prepared solution is quickly filtered and the filtrate is then allowed to cool slowly in a china dish. The resulting pure crystals that form are removed with the help of a spatula. They are dried by pressing them between the folds of filter papers and finally put into a dessicator. Sublimation This method is used to separate volatile solids, from a non-volatile solid. The mixture is taken in a china dish covered with a perforated filter paper on which an inverted glass funnel is placed to collect the vapours. Upon heating, the substance vapourizes and gets deposited on the walls of the funnel. The non-volatile substances are left in the dish. Distillation Distillation is used for separating the constituents of a liquid mixture, which differ in their boiling points. Depending upon the difference in the boiling points of the constituents, different types of distillation like fractional distillation, steam distillation etc. are employed. 

Magnetic Separation Method Aim To separate a magnetic component from a mixture containing non- magnetic components. Experiment Aim To separate a magnetic component from a mixture containing non-magnetic components. Principle The magnetic component of the mixture is separated with the help of the magnetic attraction. Process A magnet is moved over the mixture containing the magnetic substance e.g., iron filings. These get attracted to the magnet. The process is repeated until the magnetic material is completely separated from the mixture. 

 Fig: 1.4 - Separation of a magnetic substance by a magnet 

Gravity method Aim To separate mixtures in which components have different densities. Principle Particles with higher density settle to the bottom while the lighter particles are separated through various processes such as winnowing (grain separation) and washing (panning of gold in the river bed). Process Riverbed sand containing fine particles of gold is repeatedly washed in a pan with flowing water. Gold particles settle to the bottom of the pan because of higher density while lighter sand particles are washed over the edge of the pan. 

Solvent Extraction Method Organic compounds, which are easily soluble in organic solvents but insoluble or immiscible with water forming two separate layers can be easily separated. The aqueous solution of the mixture is mixed with a small quantity of organic solvent in a separating funnel. The separating funnel is stoppered and shaken strongly. It is allowed to stand for some time. The organic liquid and water form separate layers, which are collected by opening the stop-cock. The aqueous layer is again transferred to the separating funnel. The process is repeated with more organic solvent. 

Chromatography This technique is based on the differential adsorption of various components of a mixture on a suitable adsorbent called the stationary phase while the liquid in which the substance is dissolved is called the mobile or moving phase. Depending on the nature of the two phases there are various types of chromatography. In the method of adsorption chromatography used in column chromatography the adsorbent alumina is packed in a column, which acts as a stationary phase. The mixture is dissolved in a suitable solvent and the solution is poured on top of this column. The mixture moves down and the different components of the mixture get adsorbed in different strengths on the alumina surface. The adsorbed components are then eluted out by the mobile phase (solvent).

Laws of Chemical Combination

The study of reactions is one of the important areas of Chemistry. There are many aspects of a reaction like the rate of reaction, energy absorbed or evolved, mechanism of a reaction that can be evaluated only with weight and volume relationships. A study of reactions is possible by understanding the concept of atoms and molecules and the definite laws and patterns regulating them. 

Law of Conservation of Mass This law was put forth by the French chemist A. Lavoisier (1774). It states that in 'any chemical reaction, the total mass of the system before and after the reaction is the same although its matter can undergo a physical change'. If 'a' grams of A and 'b' grams of B react to give 'c' grams of C and 'd' grams of D, then, Total mass of the reactants = Total mass of the products (a + b) grams = (c + d) grams Therefore, matter is neither created nor destroyed as a result of any chemical or physical change. This law can be verified by a simple experiment. A solution of sodium chloride and silver nitrate are taken separately in the two limbs of an 'H' shaped tube. The tube is sealed and weighed precisely. The two reactants are made to react by inverting the tube. The following reaction takes place. 

 The whole tube is kept undisturbed for sometime so that the reaction is complete. When the tube is weighed again it is observed that: Weight before the reaction = Weight after the reaction This verifies the law of conservation of mass. 

Law of Definite Proportions or Law of Constant Composition This law was proposed by Louis Proust in 1799, which states that: 'A chemical compound always consists of the same elements combined together in the same ratio, irrespective of the method of preparation or the source from where it is taken'. For example, water is always found to contain only two elements hydrogen and oxygen combined in a fixed ratio by mass (1:8) irrespective of its source. Similarly carbon dioxide is always found to contain carbon and oxygen in the mass ratio of 3:8. This law can be verified by studying the electrolysis of different samples of water into hydrogen and oxygen when it is found that the ratio of hydrogen and oxygen is 2:1 (by volume) and 1:8 (by mass). However, this law does not hold good when the compound is obtained by using different isotopes of the combining elements and when the compounds are non stoichiometric. 

Law of Multiple Proportions Proposed by Dalton in 1803, this law states that: 'When two elements combine to form two or more compounds, then the different masses of one element, which combine with a fixed mass of the other, bear a simple ratio to one another'. This law is evident when one element combines with another to give more than one compound of different stoichiometry e.g., nitrogen reacts with oxygen to form nitrous oxide (N2O) and nitric oxide (NO). This law does not hold good when different isotopes are used to obtain different compounds. This law can be verified by heating 1.00 g each of the oxide of copper CuO and Cu2O in a current of hydrogen. Both the oxides react with hydrogen producing metallic copper. From the weight of the copper obtained, the respective weights of oxygen in the two compounds are determined. From this, the different weights of oxygen, which combine

with the same weight of copper in the two compounds are calculated. Since these weights are found to bear a simple whole number ratio, the law is verified. 

Law of Reciprocal Proportions or Law of Equivalent Proportions This law was proposed by Ritcher in 1792. It states that 'when two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so will be the same or some simple multiple of the mass ratio in which they combine with each other'. For example, oxygen and sulphur react with copper to give copper oxide and copper sulphide respectively. Suphur and oxygen also react with each other to give SO2. Then, In CuS, Cu:S = 63.5:32 In CuO, Cu:O = 63.5:16 

 S:O = 32:16 S:O = 2:1 Now in SO2

 S:O = 32:32 S:O = 1:1 Thus the ratio between the two ratios is 

 

Gay Lussac's Law of Gaseous Volumes This law put forward by Gay Lussac states that 'whenever gases react together under constant conditions of temperature and pressure, the volumes of the reacting gases as well as products (gases) bear a simple whole number ratio between themselves'. For example, in the reaction, 

 The ratio of volumes of hydrogen, chlorine, and hydrogen chloride is 1:1:2 (a simple ratio).

Dalton's Atomic Theory

John Dalton provided a simple theory of matter to provide theoretical justification to the laws of chemical combinations in 1805. The basic postulates of the theory are: 

 All substances are made up of tiny, indivisible particles called atoms. 

 Atoms of the same element are identical in shape, size, mass and other properties. 

 Each element is composed of its own kind of atoms. Atoms of different elements are different in all respects. 

 Atom is the smallest unit that takes part in chemical combinations. 

 Atoms combine with each other in simple whole number ratios to form compound atoms called molecules. 

 Atoms cannot be created, divided or destroyed during any chemical or physical change. The main achievement of Dalton's theory was the derivation of the laws of chemical combination. But later discoveries found that it could not explain: 

 the law of gaseous volumes 

 why atoms of different elements have different masses, sizes, valencies etc. 

 why atoms of different elements combine with each other to form molecules. 

 the nature of forces that bind together atoms in a molecule 

 the fundamental particles that make the element or the compound

Modern Atomic Theory

In view of new discoveries about atoms, Dalton's atomic theory was modified as given below: 

 Atom is no longer indivisible as it has a complex structure of sub atomic particles (electrons, protons and neutrons). 

 Atoms of the same element may posses different relative masses (isotopes). 

 Certain atoms of different elements have same relative masses but their chemical properties are entirely different (isobars). 

 Although the atom is composed of sub-atomic particles, it still is the smallest unit that takes part in a chemical reaction. 

 The ratio in which the different atoms combine may be fixed and integral but may not always be simple. 

 Atom of one element may be changed into atoms of another element (transmutation). 

 The mass of an atom can be changed into energy (E = mc2) shows that atom is no longer indestructible.

Avogadro's Hypothesis

In trying to correlate Dalton's atomic theory with Gay Lussac's law of gaseous volume, Berzelius put forward the hypothesis that equal volumes of all gases contain equal number of atoms under similar conditions of temperature. However, it was found that even a fraction of an atom could be involved in certain chemical reactions. This went against Dalton's atomic theory. Therefore, Avogadro postulated the existence of molecules along with atoms as two kinds of ultimate particles. According to Avogadro's postulates, 

 The smallest particle of an element, which may or may not have independent existence and takes part in a chemical reaction is an atom. 

 The smallest particle of a substance (element or compound) capable of independent existence is called molecule. Statement Avogadro's law states that 'under similar conditions of temperature and pressure, equal volumes of all gases contain equal number of molecules'. For example, if equal volumes of different gases like hydrogen, oxygen and chlorine are held in separate containers under similar conditions of temperature and pressure, then each container will contain the same number of molecules. The total mass and size of each gas in the container may differ from each other. 

 Fig:1.5 - Different gases under same conditions of pressure and temperature Experimental evidences show that one mole of all gases under similar conditions of temperature and pressure occupy the same volume called molar volume (Vm). Molar volume of all gases under normal pressures (1 atm) and normal temperature (273K) have a value of 22.4 litre and contains 6.023 x 1023 molecules. The number is called the Avogadro's number or Avogadro's constant and is denoted by NA. Avogadro's law is very useful in chemistry and has some important applications as in the deduction of the atomicity of the elementary gases, the derivation of relationship between molar mass and vapour density and determination of the molecular formula of a gaseous compound.

Problems

 7. 1.375 g of cupric oxide on reduction in hydrogen gas gives 1.098 g of copper. In another experiment, 1.179 g of metallic copper produced 1.476 g of copper oxide. Show that these results illustrate the law of constant proportions. Solution Mass of copper oxide taken in experiment 1. = 1.375 g Mass of copper obtained = 1.098 g 

 Mass of copper oxide produced in experiment 2 = 1.476 g Mass of copper used = 1.179 g 

 Since the percentage of copper in the two samples of copper oxide is the same, the law of definite proportion is verified.  8. Carbon and oxygen are known to form two compounds. The carbon content in one of these is 42.9% while in the other it is 27.3%. Show that this data is in agreement with the law of multiple proportions. Solution For the first compound, mass % of C = 42.9 Mass % of O = 57.1 

Thus, 42.9 g of C reacts with 57.1 g of oxygen 

 =1.33 g of oxygen For the second compound, mass % of C = 27.3 Mass % of O = 72.7 Thus, 27.3 g of C reacts with 72.7 g of oxygen 

 = 2.66 g of oxygen The ratio of oxygen masses, which combine with 1 g of C is 1.33:2.66 or 1:2 This is a simple ratio and supports law of multiple proportions.  9. Hydrogen sulphide contains 94.11% sulphur. Sulphur dioxide contains 50% oxygen. Water contains 11.11% hydrogen. Show that the results are in agreement with the law of reciprocal proportions. Solution In H2S, 100-94.11 = 5.89 g of hydrogen combines with 94.11 g of sulphur. So, 1g of hydrogen combines with  

 of sulphur. In H2O, 100-11.11 = 88.89 g of oxygen combines with 11.11 g of hydrogen. So, 1g of hydrogen combines with  

 of oxygen. Ratio of the masses of sulphur (H2S) and oxygen (H2O) 

 Ratio of the masses of sulphur and oxygen (SO2) 

 The first ratio is double of the second, which is a simple ratio. This illustrates the law of reciprocal proportions.

Atoms and Molecules

The smallest particle of an element, which may or may not have independent existence is called an atom, while the smallest particle of a substance which is capable of independent existence is called a molecule. Molecules are classified as homoatomic and heteroatomic. Homoatomic molecules are made up of the atoms of the same element and have different atomicity (number of atoms in a molecule of an element) like monoatomic, diatomic, triatomic and

polyatomic. 

Atomic Mass Unit An atom is a very minute particle and so the actual mass of an atom is extremely small. It is physically not possible to measure such small masses easily. In 1811, Avogadro suggested that atomic and molecular masses can be expressed on an atomic mass scale relative to that of a standard reference atom. The standard reference atom chosen by IUPAC in 1961 is carbon-12 isotope designated as 12C6 or simply as12C. The 12C isotope has been assigned an atomic mass of 12.000000 atomic mass unit. Hence, one atomic mass unit (amu) may be defined as follows: The mass equal to 1/12th of the mass of a 12C atom. It is abbreviated as amu or mau

 

, Absolute mass of 12C atom is 1.9924 x 10-23g 

Atomic and Molecular Masses

Atomic Mass It is known that the atomic mass of an element/compound is not the mass of one atom or molecule of the said element/compound because they occur as a mixture of isotopes. As the atomic and molecular masses are expressed on a relative scale based on a mass of12C atom, all atomic and molecular masses are in fact the weighted average of the mass of these isotopes. Thus, atomic mass of an element is defined as the average relative mass of an atom of an element as compared to the mass of an atom of carbon (12C ) taken as 12. 

 For example, chlorine occurs in nature in the form of two isotopes with atomic mass 35 and 37 in the ratio of 3:1. Therefore, 

 Thus, an atom of chlorine is 35.5 times heavier that 1/12th of the mass of a 12C atom. 

Gram Atomic Mass The quantity of an element whose mass in grams is numerically equal to its atomic mass. In simple terms, atomic mass of an element expressed in grams is the gram atomic mass or gram atom. For example, the atomic mass of oxygen = 16 amu Therefore gram atomic mass of oxygen = 16 g The gram atom should not be confused with the mass of one atom of the element in grams, which is the actual mass of the atom. 

Molecular Mass Molecular mass of a substance is defined as the average relative mass of its molecule as compared to the mass of an

atom of 12C taken as 12. It expresses as to how many times the molecule of a substance is heavier than 1/12th of the mass of an atom of carbon. For example, a molecule of carbon dioxide is 44 times heavier than 1/12th of the mass of an atom of carbon. Therefore the molecular mass of CO2 is 44 amu. 

Gram Molecular Mass A quantity of substance whose mass in grams is numerically equal to its molecular mass is called gram molecular mass. In simple terms, molecular mass of a substance expressed in grams is called gram molecular mass. For e.g., the molecular mass of oxygen = 32 amu Therefore, gram molecular mass of oxygen = 32 g The gram molecular mass should not be confused with the mass of one molecule of the substance in grams, which is the actual mass of the molecule.

Mole Concept

In chemistry, the term mole represents a pile or mass of atoms, molecules, ions or electrons. Just as common man measures quantity in terms of kilograms or dozens, a chemical scientist deals with a 'mole' of atoms, molecules, ions or electrons. Mole is defined as the amount of a substance, which contains the same number of chemical units (atoms, molecules, ions or electrons) as there are atoms in exactly 12 grams of pure carbon-12. A mole represents a collection of 6.023 x 1023 ( Avogadro's number) chemical units. Thus a mole represents the quantity of material, which contains one Avogadro's number of chemical units of any substance. The unit of mole is denoted as 'mol'. For example One mole of hydrogen atoms = 6.023 x 1023 atoms of hydrogen One mole of hydrogen molecules = 6.023 x 1023 molecule of hydrogen One mole of electrons = 6.023 x 1023 electrons One mole of sodium ions (Na+) = 6.023 x 1023 Na+ ions It can be thus concluded that, One mole of atoms = 6.023 x 1023 atoms = Gram atomic mass of the element. One mole of molecules = 6.023 x 1023 molecules = Gram molecular mass. 

 

Molar Volume The volume occupied by one mole of any substance is called its molar volume. It is denoted by Vm. Molar volume of the substance depends on temperature and pressure. One mole of all gaseous substances at 273 K and 1 atm pressure occupies a volume equal to 22.4 litre or 22,400 mL. The unit of molar volume is litre per mol or millilitre per mol. The various relationships of mole can be summarized as follows: 

                                 Fig: 1.6 - Mole relationships Problems 10. Calculate the average atomic weight of silicon having 92.2% of Si-28 isotope of relative mass 27.98 amu, 4.7% of Si-29 isotope of relative mass 28.98 amu and 3.1% of Si-30 isotope of relative mass 29.97 amu. Solution Data given: 92.20% of silicon of mass 27.98 amu 4.7% of silicon of mass 28.98 amu 3.1 % of silicon of mass 29.97 amu Average atomic weight 

 = 25.80 + 1.36 + 0.93 = 28.09 amu. Average atomic weight of silicon is 28.09 amu.  11. One million silver atoms weigh 1.79 x 1016 g. Calculate the gram atomic mass of silver. Solution Number of silver atoms = 1 million = 1 x 106

 Mass of one million Ag atoms = 1.79 x 1016 g 

 = 107.8 g

 Gram atomic mass of silver is equal to the mass of 6.023 x 1023 atoms of silver. So, the gram atomic mass of silver is 107.8 g  12. Chlorophyll, the green colouring matter of plants responsible for photosynthesis, contains 2.68% of magnesium by weight. Calculate the number of Mg atoms in 2.00 g of chlorophyll. Solution 100 g of chlorophyll contains 2.68 g of Mg 

 1 mole of Mg = 24 g of Mg = 6.023 x 1023 atom of Mg 24 g of Mg contains, 6.023 x 1023 atom of Mg 

 

 =1.34 x 1021 Mg atoms  13. A sample of gaseous substance weighing 0.5 g occupies a volume of 1.12 litre under NTP conditions. Calculate the molar mass of the substance. Solution 1 mole of any gaseous substance at NTP occupies 22.4 L. 1.12 L of gaseous substance = 0.5 g 

 The molar mass of the substance therefore is 10 g/mol.

Chemical Formulae

Every chemical substance is known by a specific name. But many a times these names are cumbersome, confusing, and do not provide information about its chemical composition. To overcome this, each chemical compound is represented by a chemical formula that gives its composition (constituent elements present) and the number of elements of each type present. There are two types of chemical formula. They are: 

Molecular Formula The formula that gives the symbolic representation of the actual number of atoms of various elements present in one molecule of the compound is called the molecular formula. Discrete molecules can be described by this formula. As it represents one molecule of the substance giving the names and number of atoms of the various elements present, it denotes the molecular mass of the substance. For example, the molecular formula of water is H2O, which means that one molecule of water contains two atoms of

hydrogen and one atom of oxygen. This also represents the molar mass, which is the sum of the gram atomic mass of all the atoms. Gram atomic mass of 2 hydrogen atoms = 2 x 1.008 g Gram atomic mass of oxygen atom = 16 g Total molecular mass = 18 g 

Empirical Formula Empirical formula is defined as the simplest formula of the substance, which gives the relative number of atoms of each element present in the molecule of that substance. Substances which have no discrete molecules such as ionic and network covalent compounds are described by this empirical formula. This is also called as stoichiometric formula. It gives the simplest whole number ratio between the number of atoms of all the elements present in the compound. For example, in the compound benzene, C6H6 there are six carbon atoms and six hydrogen atoms. The lowest whole number ratio between them is 1:1 (6:6 can be simplified to 1:1). Therefore, the empirical formula of benzene having molecular formula of C6H6 is CH. Empirical formula mass or formula mass is equal to the sum of atomic masses of all the atoms present in the empirical formula. The empirical formula of benzene is CH. So, The formula mass of CH is (12+1) = 13 amu or 13 g/mol. Relationship between empirical and molecular formulae The two formulas are related as Molecular formula = n x empirical formula where 'n' may have whole number values 1,2,3… . The value of 'n' can be obtained by the following relationship 

 For example, the molecular mass of benzene is 78 and its empirical formula is CH and therefore, its empirical formula mass is 13. 

 Therefore, the molecular formula of benzene is 6 x (CH) = C6H6. Determination of the empirical formula of a compound The empirical formula of a compound is determined from the percentage composition of different elements and atomic masses of the elements. The various steps involved in determining the empirical formula are: 

 The percentage of each element is divided by its atomic mass. This gives the relative number of atoms of various elements in the molecule of the compound. 

 The result obtained in the above step is divided by the smallest value to get the simplest ratio of various elements. 

 The values obtained are made to the nearest whole number ratio (multiplied if necessary by a suitable integer to make the values whole numbers). 

 The symbols of various elements are written side by side and the numerical value at the right hand lower corner of each symbol is inserted.

Determination of the Molecular Formula of a Compound

 The empirical formula of the compound is written. 

 The empirical formula mass is calculated by adding the atomic masses of all atoms present in the empirical formula. 

 The molecular mass or molar mass is obtained from experiment or from the vapour density relationship:     molecular mass = 2 x vapour density 

  Obtain the molecular formula from: Molecular formula = n x empirical formula

 

Percentage Composition of a Compound The mass percentage of each constituent element present in any compound is called its percentage composition. This may be obtained from the molecular formula of the compound by using the relationship: 

 This can also be determined by a suitable chemical method of chemical analysis of the compound by using the same relationship. Problems 10. Determine the percentage composition of potassium nitrate, (KNO3) (Atomic masses in amu are, K = 39, N= 14, O=16) Solution Formula mass of KNO3 = (39 amu + 14 amu + 3 x 16 amu) = 101 amu Then, 

 

 

 The percentage composition of KNO3 is K=38.6%, N=13.9%, O=47.5%. 11. A compound contains 75% carbon and 25% hydrogen. Determine its empirical formula. The molecular mass of this compound is 16 amu. Determine its molecular formula also. The atomic masses are: C = 12 amu, H = 1 amu. Solution The above results are written as follows: 

 So, The empirical formulae of the compound = C1H4 or CH4

 The empirical formula mass = (1 x 12) + (4 x 1) = 12 + 4 = 16 amu Molecular mass (given) = 16 amu 

 Therefore, Molecular formula = 1 x Empirical formula = 1 x CH4 = CH4.

Stoichiometry of Chemical Reactions

A chemical reaction can be represented by a chemical equation, which may be defined as a chemical change in terms of symbols and formulae of the substances involved in the reaction. The substances that react among themselves to bring about the chemical changes are known as reactants, where as the substances that are produced as a result of the chemical change, are known as products. 

Essentials of a Chemical Equation A chemical equation should satisfy the following conditions: 

 It should represent a true chemical reaction. 

 It should be molecular in nature. The ionic reactions must be represented by ionic equations. 

 It should be a balanced equation i.e. the number of atoms of each elements on both sides should be equal. 

Writing of a Chemical Equation Steps followed to write a chemical equation are: 

 The reactants and products of the reaction are identified and written down. 

 The formulae or symbols of the reactants are written on the left hand side with a '+' sign in between them. 

 The formulae or symbols of the products are written on the right hand side with a '+' sign in between them. 

 The two sides (reactants and products) are separated either by a sign of equality (=) or that of an arrow ( ) pointing towards the products. 

 For a reversible reaction a sign ( ) is used in place of an arrow. 

 The number of atoms of each element are counted on both the sides. If they are equal, then it is termed as balanced chemical equation. If they are not equal, the balancing of the equation is done by adjusting the coefficients before the symbols and formulae of the reactants and products. In qualitative terms, a chemical equation conveys: 

 The names of various reactants and the products.

 In quantitative terms it represents: 

 The relative number of atoms and molecules (reactant and product species) taking part in the reaction. 

 The relative number of moles of the reactants and products. 

 The relative masses of the reactants and products. 

 The relative volumes of gaseous reactants and products. Thus the chemical equation gives us the quantitative relationship between the reactants and products or the stoichiometry of the reaction. 

Making a Chemical Equation more Informative 

 The physical states of the reactants and products are specified. We use 's' for solids, 'l' for liquids and 'g' for gases. Sometimes 'aq' is used for aqueous - to represent that the given substance has been dissolved in excess of water. 

  The strength of acid or base used in the reaction is prefixed by the words 'conc'. for concentrated and 'dil'. for

diluted. 

  The conditions of the reaction such as catalyst, temperature, pressure etc. are written on the arrow between the

reactants and products. 

 The above equation indicates that the reaction has been carried out in the presence of a catalyst Fe/Mo at 723 K at 456000 mm of Hg pressure. 

 Heat changes (absorption and evolution) in the reaction may be expressed in the equation as 

  The distinction between slow and fast reactions can be made by writing the words slow and fast on the arrow head.

 

 The evolution of a gas in the reaction can be indicated by an arrow pointing upward ( ) while the formation of a

precipitate can be indicated by an arrow pointing downwards ( ) or by writing the word 'ppt'. 

 The reversible nature of the reaction is indicated by a double headed arrow, indicating that the reaction occurs in the forward as well as backward direction. 

Balancing of Chemical Equation

A chemical equation should be balanced so as to satisfy the requirements of the law of conservation of mass, as no matter is destroyed or created during a chemical reaction. 

The two popular methods of balancing are the hit and trial method and partial equation method. 

Hit and Trial Method or Inspection Method 

 The symbols and formulae of the reactant and products are written as a skeletal equation. 

 Any elementary gas (O2, H2, N2 etc.) appearing on either side of the skeletal equation, is written in the atomic state. 

 The formula containing the maximum number of atoms is selected to begin the process of balancing. If this method is not convenient, then balancing of the atoms begin with atoms, which appear minimum number of times. 

 Atoms of elementary gases are balanced at the last. 

When the balancing is complete, the equation is converted to the molecular form. Problem 12. Balance the chemical equation, 

 By hit and trial method. Solution (i) Skeletal equation 

 (ii) Elementary gas in atomic form 

 (iii) Starting with KMnO4: K and Mn are balanced (iv) Balancing O atoms 

 (v) Balancing H atoms 

 (vi) Balancing Cl atoms 

 The equation is now balanced in the atomic form. (vii) To make it molecular 

 So the balanced molecular equation is 

 

Partial Equation Method When equations contain many reactants and products they cannot be balanced by the hit and trial method. They are then balanced by the partial equation method. In this method the overall reaction is assumed to take place through two or more simpler reactions, which can be represented by partial equations. The steps involved are: Steps and problem 

 The given chemical equation is split into two or more partial equations. 

 Each partial equation is separately balanced by the hit and trial method. 

 These balanced partial equations are multiplied with suitable coefficients in order to exactly cancel out those common substances which do not appear in the overall chemical equations. 

 The balanced partial equations so obtained, are added to arrive at the balanced chemical equation. Problem 13. Balance the equation, 

 by partial equation method. Solution The given skeleton equation can be split into two partial equations. Partial eq.1 

 Partial eq.2 

 The two partial equations are balanced by hit and trial method. Balanced partial eq.1 

 Balanced partial eq.2 

 NaClO does not appear in the overall equation and so to cancel it, the partial balanced equation 1 is multiplied by 3 and the two equations are added to get the overall balanced equation. 

 

Limiting Reagent

 In a desired chemical reaction some reactants may be present in lesser or greater proportions than the stoichiometry as indicated by the balanced chemical equation. The reactant, which is completely used up first as per the stoichiometry, limits the amount of product that can be formed and does not allow the reaction to go further. This is the limiting reagent. The excess reactants are left behind as unconsumed reagents, being limited by the limiting reagent. In such cases the desired reaction does not go to 100% completion.

Stoichiometric Calculations Involving Reactions/Solutions

Most chemical reactions are carried out in solutions. It is therefore important to know how to express the concentration of the substance in its solution. Most common ways of expressing concentration of any solute in a solution are: 

 Mass Percent is the mass of the solute in grams per 100 grams of the solution. 

 A 5 % solution of sodium chloride means that 5 g of NaCl is present in 100g of the solution. 

 Volume percent is the number of units of volume of the solute per 100 units of the volume of solution. 

 A 5 % (v/v) solution of ethyl alcohol contains 5 cm3 of alcohol in 100 cm3 of the solution. 

Molarity and Molarity Equation 

 Molarity of the solution is defined as the number of moles of solute dissolved per litre (dm3) of the solution. It is denoted by the symbol M. Measurements in Molarity can change with the change in temperature because solutions expand or contract accordingly. 

 The Molarity of the solution can also be expressed in terms of mass and molar mass 

 In terms of weight, molarity of the substance can be expressed as: 

 The units of molarity are moles per liter (mol L-1). Molarity equation To calculate the volume of a definite solution required to prepare solution of other molarity, the following equation is used: M1V1 = M2V2, where M1= initial molarity, M2= molarity of the new solution, V1= initial volume and V2= volume of the new solution. The molarity equation is commonly used to calculate the molarity of solutions after dilution.

Normality

 Normality is defined as the number of gram equivalents of solute present per litre (dm3) of the solution at any given temperature. It is expressed as N.  

 The Normality of the solution can also be expressed in terms of mass and equivalent mass 

 In terms of weight, Normality of the substance can be expressed as: 

 Measurements in normality can change with the change in temperature because solutions expand or contract accordingly.

Normality Equation

To calculate the volume of a definite solution required to prepare solutions of other normality, the following equation is used: N1V1=N2V2, where N1=initial normality and N2=normality of the new solution, a nd V1= initial volume and V2= volume of the new solution. The normality equation is commonly used to calculate the normality of solutions after dilution. The above equation is also called as dilution formula because it helps in calculating the volume of the solvent required in diluting a concentrated solution. The same equation holds good for calculations involving molarity (M).

Relationship between Normality and Molarity of Solutions

Normality and Molarity are related as: 

 For acids Normality = Molarity x Basicity where basicity is the number of H+ ions a molecule of an acid can give. For bases Normality = Molarity x Acidity where acidity is the number of OH- ions a molecule of a base can give. Problems

 14. Commercially available concentrated hydrochloric acid contains 38% HCl by mass. (a) What is the molarity of this solution? The density is 1.19 g/mL. (b) What volume of concentrated HCl is required to make 1.00 L of 0.10 M HCl? Solution (a) 38% solution, means 38 g of HCl in 100 g of solution. Then, Mass of the solution= 100 g 

 Molar mass of HCl = 36.5 g mol-1

 

 

 (b) The molarity of conc. HCl sample = 12.38 mol/L Molarity of HCl solution to be prepared = 0.10 mol/L Volume of HCl solution to be prepared =1.00 L= 1000mL Then, using molarity equation, M1V1 = M2V2

 

 Thus, to obtain 1.0 L of 0.10 M HCl, one should dissolve 8.08 mL of concentrated HCl to make up the volume to 1.0L. 15. Concentrated nitric acid used as a laboratory reagent is usually 69% by mass of nitric acid. Calculate the volume of the solution which contained 23 g HNO3. Density of the concentrated HNO3 solution is 1.41 g cm-3. Solution Let, mass of conc. HNO3 sample = 100g Mass of HNO3 in 100 g of sample = 69 g Mass of water in 100 g of sample = 31 g Density of conc. HNO3 sample = 1.41 g cm-3

 

 

= 70.92 cm3

 Thus, 69 g of HNO3 is contained in 70.92 cm3 of conc. HNO3

 

 

 Thus 23.6 cm3 concentrated HNO3 sample contained 23 g of HNO3.

Other Methods of Expressing Concentration of Solution

 Molality is defined as the number of moles of solute dissolved per 1000 g (1 kg) of solvent. Molality is expressed as 'm'. 

  Mole Fraction is the ratio of number of moles of one component to the total number of moles (solute and solvents)

present in the solution. It is expressed as 'x'. 

 

 Mole fraction of the solute + Mole fraction of solvent = 1