Page 1 :
CLASS XI- CHEMISTRY NOTES, CHAPTER-1, , SOME BASIC CONCEPTS OF CHEMISTRY, MATTER, Matter is defined as any thing that occupies space possesses mass and the, presence of which can be felt by any one or more of our five senses. Matter can, exist in 3 physical states viz. solid, liquid, gas. Solid - a substance is said to be, solid if it possesses a definite volume and a definite shape, e.g., sugar, iron,, gold, wood etc. Liquid- A substance is said to be liquid, if it possesses a definite, volume but no definite shape. They take up the shape of the vessel in which, they are put, e.g., water, milk, oil, mercury, alcohol etc. Gas- a substance is, said to be gaseous if it neither possesses definite volume nor a definite shape., This is because they fill up the whole vessel in which they are put, e.g.,, hydrogen, oxygen etc. The three states are inter convertible by changing the, conditions of temperature and pressure as follows, , CLASSIFICATION OF MATTER AT MACROSCOPIC LEVELL, At the macroscopic or bulk level, matter can be classified as, (a) mixtures (b) pure substances, These can be further sub-divided as shown below
Page 2 :
Classification of matter :, a) Mixtures : Amixture contains two or more substances present in it (in, any ratio) which are called its components. A mixture may be, homogeneous or heterogeneous, Homogeneous mixture- in homogeneous mixture the components, completely mix with each other and its composition is uniform throughout, i.e it consist of only one phase. Sugar solution and air are thus, the, examples of homogeneous mixtures., Heterogeneous mixtures- In heterogeneous mixture the composition is not, uniform throughout and sometimes the different phases can be observed., For example, grains and pulses along with some dirt (often stone) pieces,, are heterogenous mixtures, b)Substances :- A material containing only one substance is called a pure, substance, In chemistry, a substance is a form of matter that has constant chemical, composition and characteristic properties. It cannot be separated into, components by physical separation methods, i.e. without breaking chemical, bonds. They can be solids, liquids or gases
Page 3 :
Pure substances can be further classified into elements and compounds., Element- An element is defined as a pure substance that contains only one, kind of particles. Depending upon the physical and chemical properties, the, elements are further subdivided into three classes, namely (1) Metals (2), Nonmetals and (3) Metalloids., Compound- A compound is a pure substance containing two or more than, Pure two elements combined together in a fixed proportion by mass., Further, the properties of a compound are completely different from those of, its constituent elements. Moreover, the constituents of a compound cannot, be separated into simpler substances by physical methods. They can be, separated by chemical methods., , PROPERTIES OF MATTER, Every substance has unique or characteristic properties. These properties can, be classified into two categories – physical properties and chemical properties., Physical Properties : Physical properties are those properties which can be, measured or observed without changing the identity or the composition of the, substance. Some examples of physical properties are color, odor, melting point,, boiling point, density etc., Chemical properties: Chemical properties are those in which a chemical, change in the substance occurs. The examples of chemical properties are, characteristic reactions of different substances; these include acidity or, basicity, combustibility etc., , MEASUREMENT, Physical quantities All such quantities which we come across during our, scientific studies are called Physical quantities. Evidently, the measurement of, any physical quantity consists of two parts, (1) The number,, , and, , (2) The unit, , A unit is defined as the standard of reference chosen to measure any physical, quantity, S.I. UNITS, The International System of Units (in French Le SystemeInternational d’Unités, – abbreviated as SI) was established by the 11th General Conference on
Page 4 :
Weights and Measures (CGPM from Conference Generale des Poids at, Measures). The CGPM is an inter governmental treaty organization created by a, diplomatic treaty known as Meter Convention which was signed in Paris in, 1875. The SI system has seven base units and they are listed in table given, below, These units pertain to the seven fundamental scientific quantities. The other, physical quantities such as speed, volume, density etc. can be derived from, these quantities. The definitions of the SI base units are given below :, Definitions of SI Base Units, Unit of length, , metre, , Unit of mass, , Kilogram, , Unit of time, , second, , Unit of electric current, , ampere, , Unit, of, temperature, , kelvin, , thermodynamic, , Unit of amount of substance, , mole, , Unit of luminous intensity, , candela, , The metre is the length of the, path travelled by light in vacuum, during a time interval of 1/299, 792 458 of a second, The kilogram is the unit of mass;, it is equal to the mass of the, internationl prototype of the, kilogram, The second is the duration of 9, 192 631 770 periods of the, radiation corresponding to the, transition between the two, hyperfine levels of the ground, state of the caesium-133 atom., The ampere is that constant, current which, if maintained in, two straight parallel conductors, of infinite length, of negligible, circular cross-section, and placed, 1 metre apart in vacuum, would, produce, between, these, conductors a force equal to 2 ×, 10–7 newton per metre of length, The, kelvin,, unit, of, thermodynamic,, is, the, temperature fraction 1/273. 16, of, the, thermodynamic, temperature of the triple point of, water., When the mole is used, the, elementary entities must be, specified and may be atoms,, molecules, ions, electrons, other, particles, or specified groups of, such particles., The candela is the luminous, intensity, in a given direction, of, a, source, that, emits, monochromatic, radiation, of, frequency 540 × 1012 hertz and, that has a radiant intensity in, that direction of 1/683 watt per, steradian
Page 5 :
SOME IMPORTANT TERMS:, 1.MASS & WEIGHT:, Mass: Mass of a substance is the amount of matter present in it., The mass of a substance is constant. The mass of a substance can be, determined accurately in the laboratory by using an analytical, balance. SI unit of mass is kilogram., Weight: It is the force exerted by gravity on an object. Weight of substance may, vary from one place to another due to change in gravity., 2.Volume: Volume means the space occupied by matter. It has the units of, (length)3. In SI units, volume is expressed in metre3 (m3). However, a popular, unit of measuring volume, particularly in liquids is litre (L) but it is not in SI, units or an S.I. unit., Mathematically,, 1L = 1000 mL = 1000 cm3 = 1dm3., Volume of liquids can be measured by different devices like burette, pipette,, cylinder, measuring flask etc. All of them have been calibrated., 3.Temperature: There are three scales in which temperature can be measured., These are known as Celsius scale (°C), Fahrenheit scale (°F) and Kelvin scale, (K)., -> Thermometres with Celsius scale are calibrated from 0°C to 100°C., -> Thermometres with Fahrenheit scale are calibrated from 32°F to 212°F., -> Kelvin’scale of temperature is S.I. scale and is very common these, days.Temperature on this scale is shown by the sign K., The temperature on two scales are related to each other by the relationship, , 4.Density: Density of a substance is its amount of mass per unit volume. So,, SI unit of density can be obtained as follows:, , This unit is quite large and a chemist often expresses density in g cm3 where, mass is expressed in gram and volume is expressed in cm3.
Page 6 :
Uncertainty in Measurements, All scientific measurements involve certain degree of error or uncertainty. The, errors which arise depend upon two factors., (i) Skill and accuracy of the worker ii) Limitations of measuring instruments., • Scientific Notation, It is an exponential notation in which any number can be represented in the, form N x 10n where n is an exponent having positive or negative values and N, can vary between 1 to 10. Thus, 232.508 can be written as 2.32508 x 102 in, scientific notation., Now let us see how calculations are carried out with numbers expressed in, scientific notation., (i) Calculation involving multiplication and division, , (ii) Calculation involving addition and subtraction: For these two operations,, the first numbers are written in such a way that they have the same exponent., After that, the coefficients are added or subtracted as the case may be. For, example,, , Significant Figures, Significant figures are meaningful digits which are known with certainty. There, are certain rules for determining the number of significant figures. These are, stated below:, 1. All non-zero digits are significant. For example, in 285 cm, there are three, significant figures and in 0.25 mL, there are two significant figures., 2. Zeros preceding to first non-zero digit are not significant. Such zeros, indicates the position of decimal point., For example, 0.03 has one significant figure and 0.0052 has two significant, figures., 3. Zeros between two non-zero digits are significant. Thus, 2.005 has four
Page 7 :
significant figures., 4. Zeros at the end or right of a number are significant provided they are on the, right side of the decimal point. For example, 0.200 g has three significant, figures., 5. Counting numbers of objects. For example, 2 balls or 20 eggs have infinite, significant figures as these are exact numbers and can be represented by, writing infinite number of zeros after placing a decimal., i.e., 2 = 2.000000, or 20 = 20.000000, • Addition and Subtraction of Significant Figures, In addition or subtraction of the numbers having different precisions, the final, result should be reported to the same number of decimal places as in the term, having the least number of decimal places., For example, let us carry out the addition of three numbers 3.52, 2.3 and 6.24,, having different precisions or different number of decimal places., , The final result has two decimal places but the answer has to be reported only, up to one decimal place, i.e., the answer would be 12.0., Subtraction of numbers can be done in the same way as the addition., , The final result has four decimal places. But it has to be reported only up to, two decimal places, i.e., the answer would be 11.36., • Multiplication and Division of Significant Figures, In the multiplication or division, the final result should be reported upto the same, number of significant figures as present in the least precise number., Multiplication of Numbers: 2.2120 x 0.011 = 0.024332, According to the rule the final result = 0.024, Division of Numbers: 4.2211÷3.76 = 1.12263, The correct answer = 1.12, , LAWS OF CHEMICAL COMBINATIONS, The combination of elements to form compounds is governed by the following, five basic laws., (i) Law of Conservation of Mass, (ii) Law of Definite Proportions
Page 8 :
(iii) Law of Multiple Proportions, (iv) Law of Gaseous Volume (Gay Lussac’s Law), (v) Avogadro’s Law, (i) Law of Conservation of Mass, The law was established by a French chemist, A. Lavoisier. The law states:, In all physical and chemical changes, the total mass of the reactants is equal to, that of the products., In other words, matter can neither be created nor destroyed., The following experiments illustrate the truth of this law., (a) When matter undergoes a physical change., , It is found that there is no change in weight though a physical change has, taken place., (b) When matter undergoes a chemical change., For example, decomposition of mercuric oxide., , During the above decomposition reaction, matter is neither gained nor lost., (ii) Law of Definite Proportions, According to this law:, A pure chemical compound always consists of the same elements combined, together in a fixed proportion by weight., For example, Carbon dioxide may be formed in a number of ways i.e.,, , (iii) Law of Multiple Proportions, If two elements combine to form two or more compounds, the weight of one of, the elements which combines with a fixed weight of the other in these
Page 9 :
compounds, bears simple whole number ratio by weight., , For example,, (iv) Gay Lussac’s Law of Gaseous Volumes, The law states that, under similar conditions of temperature and pressure,, whenever gases combine, they do so in volumes which bear simple whole, number ratio with each other and also with the gaseous products. The law may, be illustrated by the following examples., (a) Combination between hydrogen and chlorine:, , (b) Combination between nitrogen and hydrogen: The two gases lead to the, formation of ammonia gas under suitable conditions. The chemical equation is, , (v) Avogadro’s Law: Avogadro proposed that, equal volumes of gases at the, same temperature and pressure should contain equal number of molecules., For example,, If we 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., • DALTONS ATOMIC THEORY, In 1808, Dalton published ‘A New System of Chemical Philosophy’ in which he, proposed the following:, 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., 4. Chemical reactions involve reorganisation of atoms. These are neither, created nor destroyed in a chemical reaction.
Page 10 :
1) Atomic Mass : It is defined as the average relative mass of an atom of an, element as compared to the mass of an atom of carbon – 12 taken as 12., Atomic mass is represented by ‘u’ (unified mass)., 1u = 1.66056 × 10–24 g, . It is also known as unified mass., , 2) Molecular mass : It is algebraic the sum of the atomic mass of the elements, present in the molecule. 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 number of its atoms and adding them together, For example : Molecular mass of CH4 = (1 × 12) + (4 × 1) = 16 u, 3) Formula Mass:, Ionic compounds such as NaCl, KNO3, Na2C03 etc. do not consist of molecules, i.e., single entities but exist “as ions closely packed together in a three, dimensional space, In such cases, the formula is used to calculate the formula mass instead of, molecular mass. Thus, formula mass of NaCl = Atomic mass of sodium +, atomic mass of chlorine = 23.0 u + 35.5 u = 58.5 u, 4) Empirical Formula :, The formula of the compound which gives the simplest whole number ratio of, the atoms of yarious elements present in one molecule of the compound., For example, the formula of hydrogen peroxide is H202. In order to express its, empirical formula, we have to take out a common factor 2. The simplest whole, number ratio of the atoms is 1:1 and the empirical formula is HO. Similarly,, the formula of glucose is C6H1206. In order to get the simplest whole number of, the atoms,, Common factor = 6, The ratio is = 1 : 2 : 1 The empirical formula of glucose = CH20, In case n is 1, Molecular formula of a compound = Empirical formula of the, compound., 5) Molecular Formula :, The formula of a compound which gives the actual ratio of the atoms of various
Page 11 :
elements present in one molecule of the compound., For example, molecular formula of hydrogen peroxide = H202and Glucose =, C6H1206, Molecular formula = n x Empirical formula, Where n is the common factor and also called multiplying factor. The value of n, may be 1, 2, 3, 4, 5, 6 etc., 6) Mole (n) : It is amount of a substance that contains as many particles or, entities as the number of atoms in exactly 12 grams of pure C-12., 1 mole of a substance = Molar mass of substance = Avogadro’s Number, e.g., 1 mole of CH4 = 16g of CH4 = 6.023 × 1023 molecules of CH4 = 22.4L at STP, 𝑤, , n=𝑀=, , V L(at STP), 22.4L, , =, , x paticles, NA, , = MV/1000, , 7) Avogadro Number : It is the amount of atoms or molecules present in, one mole of a substance., Avogadro number (NA) = 6.023 × 1023 mol–1, 8) Molar Mass : The mass of one mole of a substance in grams is called its, molar mass. For example :, Molar mass of CH4 = (1 × 12) + (4 × 1) = 16g mol–1 of chemical units = 22.4L, volume at STP of gaseous substance, 9) Molar Volume (Vm) : It is volume occupied by one mole of gas at STP., Molar volume of a gas = 22.4L at STP (273 K, 1atm) or 22.7L at STP (273K, 1, bar), Calculating Molar Volume: PV = NRT, ., , 10) Percentage Composition, One can check the purity of a given sample by analysing this data. Let us understand, by taking the example of water (H20). Since water contains hydrogen and oxygen, the, percentage composition of both these elements can be calculated as follows:
Page 12 :
•Stoichiometry and Stoichiometric Calculations, , The word ‘stoichiometry’ is derived from two Greek words—Stoicheion (meaning, element) and metron (meaning measure). Stoichiometry, thus deals with the, calculation of masses (sometimes volume also) of the reactants and the, products involved in a chemical reaction. Let us consider the combustion of, methane. A balanced equation for this reaction is as given below:, , Limiting Reactant/Reagent, Sometimes, in alchemical equation, the reactants present are not the amount as, required according to the balanced equation. The amount of products formed then, depends upon the reactant which has reacted completely. This reactant which reacts, completely in the reaction is called the limiting reactant or limiting reagent. The, reactant which is not consumed completely in the reaction is called excess reactant., Reactions in Solutions, When the reactions are carried out in solutions, 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
Page 13 :
1. Mass percent: It is obtained by using the following relation:, , 2. Mole fraction: It is the ratio of number of moles of a particular component to the, total number of moles of the solution. For a solution containing n2 moles of the solute, dissolved in n1 moles of the solvent,, , 3. Molarity: It is defined as the number of moles of solute in 1 litre of the solution., , 4. Molality: It is defined as the number of moles of solute present in 1 kg of solvent. It, is denoted by m., , .
