Importance and Scope of Chemistry

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2 Importance of Chemistry Importance and Scope of Chemistry Chemistry plays an essential role in daily life. It has helped us to meet all our requirements for a better living. Without the knowledge of Chemistry, our life would have been very dull and dreary. Some important applications of Chemistry are given below: o Food industry o Health and sanitation o Safer alternatives for environment hazards o Applications in industry o Nuclear energy Branches of Chemistry Organic Chemistry Organic compounds or carbon and its compounds Chemistry Inorganic Chemistry Physical Chemistry Elements other than carbon and its compounds Explanation of fundamental principles Analytical Chemistry Qualitative and quantitative analyses of chemical substances Nature of Matter Everything which is around us and the presence of which can be felt with the help of our five senses sight, touch, smell, hearing and taste is called matter. Matter is anything which has mass and occupies space. 2

3 Classification of Matter Matter is classified in two ways physical classification and chemical classification. Classification of Matter Physical Classification Chemical Classification Solids Pure Substances Mixtures Liquids Elements Compounds Homogeneous Mixtures Gases Organic Compounds Heterogeneous Mixtures Inorganic Compounds Physical Classification Matter can exist in three physical states solid, liquid and gas under ordinary conditions of temperature and pressure. The three states of matter are interconvertible by changing the conditions of temperature and pressure. Heat Cool Heat Cool Solid Liquid Gas On heating, a solid usually changes to a liquid, and the liquid on further heating changes to the gaseous or vapour state. In the reverse process, a gas on cooling liquefies to the liquid and the liquid on further cooling freezes to the solid. Chemical Classification At the macroscopic or bulk level, matter can be classified as pure substances or mixtures. These can be further sub-divided as follows: 3

4 Pure Substances Pure substances have fixed composition. They can be further classified into elements and compounds. Element It is the simplest form of matter. An element consists of only one type of particle. These particles may be atoms or molecules. The smallest unit of an element is known as atom. Two or more atoms combine to give molecules of the element. The total number of known elements is 118; 98 elements occur naturally and 20 are formed by artificial transmutation. Examples: Na, K, Mg, Al, Si, P, C, F, Br Compound A compound is formed by chemical combination of two or more atoms of different elements in a fixed ratio. It is a non-elemental pure compound. Examples: H 2 O, CO 2, C 6 H 12 O 6 Mixture A mixture contains two or more substances present in it in any ratio which are called its components. Chemical identity of pure components remains maintained in mixtures. A mixture may be homogeneous or heterogeneous. o Homogeneous mixtures In a homogeneous mixture, the components completely mix with each other and its composition is uniform throughout. Examples: Sugar solution, air o Heterogeneous mixtures In heterogeneous mixtures, the composition is not uniform throughout and sometimes different components can be observed. Examples: Mixture of salt and sugar, grains or pulses with dirt and stone pieces The components of a mixture can be separated by using physical methods such as simple handpicking, filtration, crystallisation and distillation. Properties of Matter and their Measurement The quantities which we come across during our scientific studies are called physical quantities. For example, we buy vegetables in kilograms, milk in litres and cloth in metres. However, during scientific studies, in addition to the measurements of mass, volume and length, we come across the measurement of several other quantities such as temperature, pressure, concentration, force, work and density. The International System of Units (SI) The International System of Units was established by the 11 th Measurements in General Conference on Weights and 4

5 Fundamental Units These units can neither be derived from one another nor be further resolved into any other units. SI has seven fundamental units (base units): Physical Quantity Symbol for Quantity Name of the Unit Symbol of the Unit Time t second S Mass m kilogram kg Length l metre m Thermodynamic Temperature T kelvin K Electric current I ampere A Luminous intensity I ν candela Cd Amount of substance n mole Mol Derived Units These units are the function of more than one fundamental unit. Quantity with Symbol Unit (SI) Symbol Velocity (v) metre per sec ms 1 Area (A) square metre m 2 Volume (V) cubic metre m 3 Density (r) kilogram m 3 kg m 3 Energy (E) Joule (J) kg m 2 s 2 Force (F) Newton (N) kg ms 2 Frequency (n) Hertz cycle per sec Pressure (P) Pascal (Pa) nm 2 Electrical charge Coulomb (C) a s (ampere second) Subsidiary Units (Prefix used in SI) Quite often we require units which may be multiples or fractions of the base units. SI recommends multiples such as 10 3, 10 6 and 10 9 and fractions such as 10 3, 10 6 and The powers are multiples of 3. These are indicated by special prefixes. 5

6 Mass and Weight Mass of a substance is the amount of matter present in it, while weight is the force exerted by gravity on an object. The mass of a substance is constant, whereas its weight may vary from one place to another because of a change in gravity. The mass of a substance can be determined very accurately in the laboratory by using an analytical balance. The SI unit of mass is kilogram. However, its fraction gram (1 kg = 1000 g) is used in laboratories because of the smaller amount of chemicals used in chemical reactions. Volume Volume has the units of (length) 3. So, in SI, volume has units of m 3. However, in chemistry laboratories, smaller volumes are used. Hence, the volume is often denoted in cm 3 or dm 3. A common unit, litre (L), which is not an SI unit, is used for measurement of volume of liquids. 1 L = 1000 ml, 1000 cm 3 = 1 dm 3 In the laboratory, the volume of liquids or solutions can be measured by a graduated cylinder, burette or pipette. A volumetric flask is used to prepare a known volume of a solution. Density Density of a substance is its amount of mass per unit volume. So, the SI unit of density can be obtained as follows: SI unit of mass SI unit of density SI unit of volume kg 3 or kg m m 3 This unit is quite large and a chemist often expresses density in g cm 3, where mass is expressed in gram and volume is expressed in cm

7 Temperature The three common scales to measure temperature are C (degree Celsius), F (degree Fahrenheit) and K (Kelvin). Here, K is the SI unit. A thermometer with a Celsius scale is generally calibrated from 0 to 100, where these two temperatures are the freezing point and the boiling point of water, respectively. The Fahrenheit scale is represented from 32 to 212. The relationship of the temperature of the two scales is 9 F ( C) 32 5 The relationship between the Kelvin scale and the Celsius scale is K C Note that temperature below 0 C (i.e. negative values) is possible in the Celsius scale, but negative temperature is not possible in the Kelvin scale. Uncertainty in Measurement In the study of Chemistry, both experimental data and theoretical calculations need to be dealt with simultaneously. There are interesting ways to handle the numbers conveniently and present the data realistically with certainty to the extent possible. These ideas are discussed below. Scientific Notation Chemistry is the study of atoms and molecules which have extremely low masses and are present in extremely large numbers. This offers a real challenge for performing simple mathematical operations of addition, subtraction, multiplication or division for such numbers. This problem is solved by using the scientific notations for such numbers called exponential notation in which any number can be represented in the form of N 10 n. where N = digit term n = exponent with positive or negative value Rules for exponential numbers 1) In scientific notation, we can write as In this case, the decimal had to be moved to the left by two places and the same is the exponent 2 of 10 in the scientific notation. 2) Similarly, can be written as In this case, the decimal has to be moved four places to the right and - 4 is the exponent in the scientific notation. 7

8 Mathematical Operations of Scientific Notation Multiplication and Division Follow the same rules for exponential number. Example: ( ) ( ) = ( ) (10 [3 + ( 7)] ) = The result cannot have more digits to the right of the decimal point than either of the original numbers ( ) / ( ) = (7.0/8.0) (10 [3 ( 7)] ) = = Addition and Subtraction Numbers are written in a way such that they have the same exponent and thereafter coefficients are added or subtracted. ( ) + ( ) = ( ) + ( ) = ( ) 10 3 = Similarly, the subtraction of two numbers can be done: = ( ) ( ) = ( ) 10 2 = Significant figures Every experimental measurement has some kind of uncertainty associated with it. One would always like the results to be precise and accurate. Precision and Accuracy Precision: Closeness of outcomes of various measurements taken for the same quantity. Accuracy: Agreement of the experimental value to the true value. Example: If the true value for a result is 7.00 g, and a student X takes two measurements and reports the results as 6.95 g and 6.93 g, then these values are said to be precise as they are close to each other but are not accurate. Another student Y repeats the experiment and obtains 6.94 g and 7.05 g as the results for two measurements. These values are neither precise nor accurate. A third student Z repeats these measurements and reports 7.01 g and 6.99 g as the results. These values are both precise and accurate. 8

9 Rules for determining the number of significant figures: 1) All non-zero digits are significant. Example: In 395 cm, there are three significant figures, and in 0.75 ml, there are two significant figures. 2) Zeroes preceding the first non-zero digit are not significant. Such zeros indicate the position of the decimal point. 3) Zeroes between two non-zero digits are significant. Example: has four significant figures. 4) Zeroes at the end of a number are significant when they are on the right side of the decimal point. Example: g has three significant numbers. But, the terminal zeros are not significant if there is no decimal point. In case of 100, it has only one significant figure has four significant figures. These numbers are better represented in scientific notations. We can express the number 100 as for one significant figure for two significant figures and for three significant figures. 5) Counting numbers of objects having infinite significant figures. Example: 2 balls or 20 eggs have infinite significant figures as these are exact numbers and can be represented by writting infinite number of zeros after placing a decimal, i.e. 2 = or ) In numbers written in scientific notation, all digits are significant. Example: has three significant figures, and has four singinificant figures. 9

10 Rules for limiting the result of mathematical operations: 1) If the rightmost digit to be removed is more than 5, then the preceding number is increased by one. Example: The given number is If we have to remove 6, then we have to round it to ) If the rightmost digit to be removed is less than 5, then the preceding number is not changed. Example: The given number is If 4 is to be removed, then the result is rounded to ) If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number, but is increased by one if it is an odd number. Example: If 6.35 is to be rounded by removing 5, then we have to increase 3 to 4, giving 6.4 as the result. However, if 6.25 is to be rounded off, then it is rounded off to 6.2. Dimensional Analysis The derived unit is first expressed in dimension and each fundamental quantity such as mass, length and time are converted to the other system of the desired unit to work out the conversion factor. Original quantity Conversion factor = Equivalent quantity (former unit) (final unit) Atomic Mass The atomic mass of an element is the average relative mass of its atoms as compared with an atom of carbon-12. Earlier, hydrogen was taken as a standard for measuring atomic masses. Scientists compared the mass of an atom with the mass of hydrogen to determine its relative atomic mass. Hydrogen, being the smallest atom, its mass was taken as 1 unit. Example: If one atom of nitrogen weighs as much as 14 atoms of hydrogen, then the atomic mass of nitrogen is 14 units. Later, carbon-12 isotope was chosen as a standard for measuring the atomic masses. In this system, carbon-12 is assigned a mass of exactly 12 atomic mass unit (amu) and masses of all other atoms are given relative to this standard. Recently, the unit of atomic mass amu was replaced by u, meaning unified mass. A sophisticated technique such as mass spectrometry is used for determining the atomic masses. Definition of atomic mass unit (amu): One atomic mass unit is a mass unit equal to exactly one-twelfth 1 th the mass of one 12 atom of carbon

11 Example: The mass of the hydrogen atom, in terms of amu, can be computed as 1 amu = g Mass of an atom of hydrogen = g g Thus, in terms of amu, the mass of hydrogen atom = g amu amu Similarly, the mass of oxygen-16 ( 16 O) atom would be amu. Average Atomic Mass Average atomic mass is the sum of the product of the relative abundance and the atomic masses of the isotopes. When we take into account the existence of these isotopes and their relative abundance (percent occurrence), the average atomic mass of the element can be computed. Example: Carbon has three isotopes with relative abundances and atomic masses as shown in the table. Isotope Relative Abundance (%) Atomic Mass (amu) 12 C C C From the above data, we can calculate the average atomic mass of carbon. ( ) (12 u) + ( ) ( u) + ( ) ( u) = u Molecular Mass Molecular mass is the sum of atomic masses of the elements present in a molecule. It is obtained by multiplying the atomic mass of each element by the number of its atoms and adding them together. Example: Molecular mass of water which contains two hydrogen atoms and one oxygen atom can be obtained as follows: Molecular mass of water (H 2 O) = 2 atomic mass of hydrogen + 1 atomic mass of oxygen = 2 (1.008 u) + 1 (16.00 u) = Gram Molecular Mass The molecular mass of a substance expressed in grams is called its gram molecular mass. Molecular mass of H 2 SO 4 = 98.0 u Gram molecular mass of H 2 SO 4 (or one gram molecule of H 2 SO 4 ) = 98.0 g 11

12 Formula Mass Mass of a molecule of an ionic compound. It is the sum of atomic masses of all the elements present in one formula unit of a compound. It is used for measuring the molecular mass of entities which do not exist in the solid form. In sodium chloride, one Na + is surrounded by six Cl and vice versa. Hence, in such cases, the formula mass is used to calculate the mass instead of the molecular mass. Just like the molecular mass, it is calculated by adding the atomic masses of the atoms present in one formula unit. Thus, formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine = 23.0 u u = 58.5 u The elements chemically combine with each other to form compounds. The process of chemically combining of elements to form compounds is governed by five basic laws. Law of Conservation of Mass Avogadro's Law Laws of Chemical Combination Law of Definite Proportions Gay- Lussac's Law of Gaseous Volumes Law of Multiple Proportions Law of Conservation of Mass This law was put forth by Antoine Lavoisier in He performed careful experimental studies on combustion reactions. Matter can neither be created nor destroyed in a chemical reaction. 12

13 This law was the result of exact measurements for the masses of reactants and products. According to the law of conservation of mass, sum of the masses of reactants is always equal to the sum of the masses of products formed during the reaction. Example: Burning of carbon Mass of Carbon Mass of Oxygen Mass of Carbon dioxide Law of Definite Proportions (Law of Definite Compositions) This law was proposed by the French chemist Joseph Proust. A compound always contains exactly the same proportions of elements by weight. The law was experimentally proved by following examples: 1. Naturally occurring pure sample of copper carbonate or cupric carbonate contains 51.35% of copper, 38.91% of carbon and 9.74% of oxygen. Later, a pure sample of cupric carbonate was synthesized in laboratory and was found that the percentage by weight of copper, carbon and oxygen were exactly identical to that of sample of naturally occurring sample of cupric carbonate % Copper Naturally occuring and synthesized sample of Cupric Carbonate 38.91% Carbon 9.74% Oxygen 13

14 2. This law was further supported by different samples of pure water. It was observed that irrespective of sources of water it contained 88.81% oxygen and 11.19% hydrogen by weight. Pure water 88.81% oxygen 11.19% hydrogen 3. The pure samples of sugar obtained 51.4% oxygen, 42.1 % carbon and 6.5 % hydrogen by weight. Sugar 51.4% Oxygen 42.1 % Carbon 6.5 % Hydrogen Law of Multiple Proportions The law was proposed by British scientist John Dalton, in If two elements can combine to form more than one compound, then the masses of one element which combines with the fixed mass of the other element are in the ratio of small whole numbers. Example: 1. Carbon reacts chemically with oxygen forming two compounds carbon monoxide and carbon dioxide. Carbon monoxide is a poisonous combustible gas. While carbon dioxide is non-poisonous, non combustible gas. On analysis it was found that, 1g of carbon reacts with 1.33 g of oxygen to form carbon monoxide and 1 g of carbon reacts with 2.66 g of oxygen to yield carbon dioxide. Hence the ratio of weights of oxygen to that of carbon for two compounds are, 14

15 Therefore the ratio of masses of oxygen that combine with the same mass of carbon is 2 : Hydrogen reacts with oxygen to form two compounds water and hydrogen peroxide. On analysis it is found that water (H 2 O) contains 2g of hydrogen and 16g of oxygen. Hydrogen 2g Oxygen 16g Water 18g 3. Hydrogen peroxide (H 2 O 2 ) contains 2g of hydrogen and 32g of oxygen. Hydrogen 2g Oxygen 32g Hydrogen Peroxide 34g Hence the masses of oxygen 16g and 32g which combine with a fixed mass of hydrogen (2g) bear a simple ratio, i.e., 16:32 or 1:2. Gay-Lussac s Law of Gaseous Volumes (Law of Combining Volumes) This law was given by Gay-Lussac in He investigated a large number of chemical reactions occurring in gases. As a result of his experiments, Gay-Lussac found that there exists a definite relationship among the volumes of gaseous reactants and products. 15

16 When gases react together, they always do so in volumes which bear a simple ratio to one another and to the volumes of the products, if these are also gases, provided all measurements of volumes are performed under similar conditions of temperature and pressure. Consider the following illustrations. When gases react together to produce gaseous products, the volumes of reactants and products bear a simple whole number ratio with each other, provided volumes are measured at same temperature and pressure. Example 1: Under identical conditions of temperature and pressure equal volumes, say 1 litre i.e., 1L of each hydrogen and chlorine gases react together to produce double the volume, 2L of hydrogen chloride gas. Hydrogen 1L (1Vol) Chlorine 1L (1 Vol) Hydrogen chloride 2L (2 Vol) Thus, the ratio of volumes is 1 : 1 : 2. Example 2: Under identical conditions of temperature and pressure 2l of hydrogen gas reacts with 1L of oxygen gas to produce 2L of steam (water vapour). Hydrogen 2L (2 Vol) Oxygen 1L (1 Vol) Water vapour 2L (2 Vol) Thus, the ratio of volumes is 2 : 1 : 2. Example 3: Under identical conditions of temperature and pressure 1L of nitrogen gas reacts with 3L of hydrogen to produce 2L of ammonia gas. Nitrogen 1L (1 Vol) Hydrogen 3L (3 Vol) Ammonia 2L (2 Vol) 16

17 Here the ratio of volumes is 1 : 3 : 2. Conclusion: All examples cited above indicate that during gaseous reaction, gaseous reactants and products bear a simple ratio of whole number of volumes with each other. Gay-Lussac s law was explained in detail by Avogadro in Avogadro s Law In 1811, Amedeo Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules. Equal volumes of gases at the same temperature and pressure contain equal number of molecules. The distinction between atoms and molecules was made quite understandable by Avogadro. Let us again consider the reaction of hydrogen and oxygen to produce water. We see that two volumes of hydrogen combine with one volume of oxygen to give two volumes of water without leaving any unreacted oxygen. In the figure, each box contains equal number of molecules. In fact, Avogadro could explain the above result by considering the molecules to be polyatomic; for example, the diatomic molecules such as hydrogen and oxygen. Avogadro s proposal was published in the French Journal de Physique. Despite being correct, it did not gain much support. However, Dalton and others believed at that time that atoms of the same kind cannot combine and molecules of oxygen or hydrogen containing two atoms do not exist. Dalton s Atomic Theory John Dalton, a British chemist, provided a basic theory about the nature of matter. In 1808, he presented his atomic theory, which was a turning point in the study of matter. An atom is the smallest indivisible particle of matter. It is similar to a hard, solid sphere. Dalton s theory did not propose anything about the charges in an atom as he considered the atom to be indivisible. Therefore, his theory was not able to explain the behaviour and properties of matter. However, his theory could explain the laws of chemical combination. Postulates of Dalton's Atomic Theory 1. Matter consists of indivisible atoms. 2. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass. 3. Compounds are formed when atoms of different elements combine in a fixed ratio. 17

18 4. Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction. We know that a dozen is 12 items, a century is 100 items and a gross is 144 items. As we use the terms such as dozen, century and gross to express certain quantities of substances, a mole is a term used to describe a collection of particles, i.e. atoms, molecules or ions. Definition of Mole One mole is the amount of a substance which contains as many particles or entities as there are atoms in exactly 12 g (or kg) of the 12 C isotope. The mass of one mole of a substance in grams is called its molar mass. The atomic mass expressed in grams is the gram atomic mass. The molecular mass expressed in grams is the gram molecular mass. Avogadro experimentally found that one mole of any substance always contained particles. This number is called the Avogadro s constant denoted by N A. 1 mole of a substance is equal to particles, i.e. atoms, molecules or ions, of the substance. 1 mole (of anything) = Percentage Composition The mass percentage of each constituent element present in any compound is called its percentage composition. It can be calculated by using the given formula: Mass of element in 1 molecule of the compound Mass % of the element = 100 Molar mass of the compound Empirical Formula for Molecular Formula An empirical formula represents the simplest whole number ratio of various atoms present in a compound. The molecular formula expresses the exact number of different types of atoms present in a molecule of a compound. Relation between Empirical Formula and Molecular Formula The molecular formula of a compound is a simple whole number multiple of its empirical formula. Mathematically, Molecular formula = n Empirical formula where n is any integer such as 1, 2, 3,

19 When n = 1, Molecular Formula = Empirical Formula When n = 2, Molecular Formula = 2 Empirical Formula and so on. The value of 'n' can be obtained from the relation: n Molecular mass Emperical formula mass Calculation of the Empirical Formula The following steps are involved in the calculation of the empirical formula: Step 1: Conversion of mass percent to grams Step 2: Calculate the number of moles of each element Moles of an element Mass of element in grams Atomic mass of the element The moles of different elements thus calculated represent the relative number of moles. Step 3: Calculate the simplest molar ratio Step 4: Write the empirical formula Step 5: Write the molecular formula Balancing a Chemical Equation Balancing of a chemical equation means making the number of atoms of each element equal on both sides of the equation. The following methods are generally used for balancing of chemical equations: (1) Hit and trial method or trial and error method (2) Partial equation method (3) Oxidation number method (4) Ion-electron method The first two methods are briefly described below: Hit and Trial Method The simplest method to balance a chemical equation is by the hit and trial method also called the trial and error method. The method involves the following steps: Step 1: Write the correct formulae of the reactants and the products with plus signs in between with an arrow pointing from the reactants to the products. This is called the skeleton equation. 19

20 Step 2: Select the largest formula from the skeleton equation and equalise the number of atoms of each of its constituent elements on both sides of the chemical equation by suitable multiplications. Or Balance the atoms of that element which occurs at the minimum number of places on both sides of the chemical equation first. Atoms which occur at a maximum number of places are balanced last of all. Step 3: When an elementary gas (diatomic) such as hydrogen, oxygen, chlorine or nitrogen appears as a reactant or a product, the equation is balanced more easily by keeping the elementary gas in the atomic state. The balanced atomic equation is then made molecular by multiplying the whole equation by two. Step 4: Verify that the number of atoms of each element is balanced in the final equation. Step 5: The chemical equation can be made more informative by mentioning the physical states of the reactants and the products. Thus, the gaseous, liquid, aqueous and solid states of the reactants and products are represented by the notations (g), (l), (aq) and (s), respectively. Note: Subscripts in the formulae of reactants and products cannot be changed for the sake of convenience to balance an equation. An integer placed in front of chemical formula multiples every atom of that formula by that integer. Thus, 2 NH 3 is 2 N atoms and 6 H atoms. 3 H 2 O is 6 H atoms and 3 O atoms. Partial Equation Method The hit and trial method is useful only for balancing simple chemical reactions. For a reaction in which the same element is repeated in several compounds, the partial equation method is more helpful. The method involves the following steps: Step 1: The chemical reaction represented by the equation is supposed to proceed in two or more steps. Step 2: The skeleton equations representing each step are written and then balanced by the hit and trial method. These equations are known as partial equations. Step 3: If necessary, the partial equations are multiplied by suitable integers so as to cancel those intermediate products which do not occur in the final equation. Step 4: The partial equations are added to get the final balanced equation. Stoichiometry The word stoichiometry is derived from two Greek words stoicheion (meaning element) and metron (meaning measure). Stoichiometry is the amount of reactants required or the products produced in a chemical reaction. It deals with the calculation of masses and volumes of the reactants and products involved in a chemical reaction. Example: In a given reaction, CH 4(g) + 2O 2(g) CO 2(g) + 2H 2 O (g) 20

21 The coefficient 2 for O 2 and H 2 O is called stoichiometric coefficient. Similarly, the coefficient for CH 4 and CO 2 is one in each case. They represent the number of molecules and moles taking part in the reaction or formed in the reaction. Thus, according to the above chemical reaction, One mole of CH 4(g) reacts with two moles of O 2(g) to give one mole of CO 2(g) and two moles of H 2 O (g). One molecule of CH 4(g) reacts with 2 molecules of O 2(g) to give one molecule of CO 2(g) and 2 molecules of H 2 O (g) 22.7 L of CH 4(g) reacts with 45.4 L of O 2 (g) to give 22.7 L of CO 2 (g) and 45.4 L of H 2 O (g). 16 g of CH 4 (g) reacts with 2 32 g of O 2 (g) to give 44 g of CO 2 (g) and 2 18 g of H 2 O (g). From these relationships, the given data can be interconverted as follows: Mass Moles Number of molecules Limiting Reagent Quite often, one of the reactants is present in a larger amount than the other as required according to the balanced equation. In such situations, one reactant is in excess over the other. The reactant which is present in the lesser amount gets consumed after sometime, and thereafter, no further reaction takes place whatever be the amount of the other reactant present. Hence, the reactant which gets consumed completely limits the amount of product formed and is therefore called the limiting reagent. The reactant which is not consumed completely in the reaction is called excess reactant which is left unreacted. Reactions in Solutions Most reactions are carried out in solutions in the laboratories. Therefore, it is important to understand how the amount of substance is expressed when it is present in the form of a solution. The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways: 1. Mass percent or weight percent (w/w %) 2. Mole fraction 3. Molarity 4. Molality 21

22 Mass Percent It is obtained by using the relation Mass of solute Mass percent 100 Atomic of solution Mole Fraction It is the ratio of the number of moles of a particular component to the total number of moles of the solution. If a substance A dissolves in substance B and their number of moles is n A and n B, respectively; then the mole fractions of A and B are Mole fraction of A Mole fraction of B No. of moles of A No. of moles of solution n n A A n A n n B B No. of moles of B No. of moles of solution n B Molarity It is the most widely used unit and is denoted by M. It is the number of moles of the solute in 1 litre of the solution. Molarity(M) No. of moles of solute Volume of solution in litres Molality It is the number of moles of solute present in 1 kg of solvent. It is denoted by m. No. of moles of solute Molality (m) Mass of solvent in kg 22