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Some Basic Concept Of Chemistry

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Published in: Chemistry
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Chemistry Concepts

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  1. SOME BASIC CONCEPTS OF CHEMISTRY INTRODUCTION Anything that occupies space and has mass is called matter. Ancient Indian and greek philospher's believed that the wide variety of object around us are made from combination of five basic elements: Earth, Fire , Water , Air and Sky. The Indian philosopher Kanad (600 BC) was of the view that matter was composed of very small, indivisible particle called "parmanus". Ancient Greek philosopher also believed that all matter was composed of tiny building blocks which were hard and indivisible. The Greek philosopher Domocritus named these building blocks as atoms, All these people had their philosphical views about matter, these views experimental test. It was John Dalton who firstly developed a theory on the structure of matter, known as Dalton's atomic theory. DALTON'S ATOMIC THEORY: Matter is made up of very small indivisible particle called atoms. All the atoms of a element are identical in all respect i.e. mass, shape, size , different elements are different in nature. Atoms cannot be created or destroyed by any chemical process. Classification of matter meaning indivisible. were never put to latter on which was etc. and atoms of On the basis of physical behaviour Solids Gases Liquids On the basis of chemical behaviour Pure substances Mixtures Compound Element The combination of elements to form compounds is governed by the following five basic laws. LAW OF CHEMICAL COMBINATION : (a) Law of Conservation of Mass : It states that matter can neither be created nor destroyed. For any balanced chemical reaction mass of reactant is equal to mass of products. [Exception of law is nuclear reactions where einstein equation is applicable.] H 2 + —02 1-420 2 2gm + 16gm 18gm Example 1: Solution: When 20 g of NaHC03 is heated, 12.62 g of Na2C03 and 5.24g of C02 is produced. How many grams of H20 is produced? Total mass of NaHC03 heated = 20 gms; Total mass Na2C03 produced = 12.62 gms Total mass of C02 produced = 5.24 gms . Mass of 1420 produced = 20—12.62—5.24 = 2.14 gms (b) The Law of Constant Composition or Definite Proportion (Proust in 1799) : This law states that " All pure samples of the same chemical compound contain the same elements combined in the same proportion by mass irrespective of the method of preparation". Example : Different samples of carbon dioxide contain carbon and oxygen in the ratio of 3 : 8 by mass. Similarly in water ratio of weight of hydrogen to oxygen is 1 : 8.
  2. Example 2: Solution: When 50 g of ammonia is heated it gives 41.18 g of Nitrogen. When 10 g. of Nitrogen is combined with required amount of hydrogen it produces 12.14g ammonia. Show that the given data follows the law of constant compositions. If 50g of Ammonia gives 41.18g of Nitrogen, then the percentage of Nitrogen in ammonia 41.18 x 100 = 82.36%. is 50 If 10g of Nitrogen gives 12.14 g of Ammonia then percentage of Nitrogen in ammonia is 10 x 1 00 = 82.37% . 12.14 (c) The Law of Multiple Proportion (Dalton) This law states that :when two elements A and B combine together to from more than one compound, then several, masses of A which separately combine with a fixed mass of B, are in a simple ratio. Example : co 12:16 ratio and 16 : 32 i.e. C02 12 : 32 (d) The Law of Reciprocal Proportions (Richer in 1792 - 94) This law states that "when two elements combines separately with third element and form different types of molecules their combining ratio is directly reciprocated if they combine directly. Example : C combines with O to form C02 and with H to form CH4. In C02 12 g of C reacts with 32g of O, whereas in CH4 12 g of reacts with 4g of H. Therefore when O combines with H, they should combine in the ratio of 32 : 4 (i.e. 8 : 1) or in simple multiple of it. The same is found to be true in H20 molecules. The ratio of weight of H and O in 1-120 is 1 : 8. (e) The Law of Gaseous Volume (Gay Lussac in 1808) This law states that "when gas combine, they do so in volume which bear a simple ratio to each other and also to the product formed provided all gases are measured under similar conditions. Or in other words volume of reacting gases and product gases have a simple numerical ratio to one another. Example H2(g) 1 unit vol. (f) The Avogadro Law C12(g) 1 unit vol 2HCl(g) 2 unit vol. ratio = 1 This law states that "equal voume of all gaseous under similar conditions of temperature and pressure contain equal number of molecules". 21-12 2 vol. 02 1 vol. 21-120 2 vol. ratio : 2 It provides a relationship between vapour density and molecular mass of substances. 2 x vapour density (VD) = molecular mass of gas. ATOMIC AND MOLECULAR MASSES Atomic Mass : Average mass of an atom Atomic mass 1/12>< Mass of an atom of C 2 Gram Atomic Mass : The atomic mass of an element expressed in grams is called gram atomic mass of that element. It is also defined as mass of 6.022 x 1023 atoms. It is also defined as the mass of one mole atoms. It is also defined as the mass of 1 gram atom of the element.
  3. Atomic mass unit (amu) : = —L the mass of a carbon - Atomic mass Unit (amu) 12 lamu = 1.660539 x 10 24 gm. Mass of one c162 atom = 12 a.m.u 12 1 1 a.m.u = mass of one c162 atoms NO X 12 NO 12 atom Nowadays amu has been replaced by 'u' which is known as unified mass. Molecular mass : Molecular mass of a molecule, of an element or a compound may be defined as a number which 1 indicates how many times heavier is a molecule of that element or compound as compared with — of 12 the mass of an atom of carbon—12. Molecular mass is a ratio and hence has no units. It is expressed in a.m.u. Mass of one molecule of the substance Molecular mass 1/12 x Mass of one atom of C-12 Actual mass of one molecule = Mol. mass xl .66 x 10 24 gm. Gram Molecular Mass : The molecular mass of a substance expressed in gram is called the gram-molecular mass of the substance. It is also defined as mass of 6.022 x 1023 molecules . It is also defined as the mass of 1 mole molecules. It is also defined as the mass of 1 gram molecule. Average atomic mass and molecular mass Ä (Average atomic mass) = total (Average molecular mass) = total Where A A A ..... are atomic mass of species 1, 2, 3, terms are for molecular masses. .. etc. with % ratio as . etc. Similar Equivalent concept : w = no. of equivalent Equivalent mass (E) . w n x no. of moles m no. of equivalent of A =no. of equivalent of B = no. of equivalent of C weight (grams) Number of moles of a species = Atomic or molecular mass (g / mole) Volume occupied by gas at NT P Number of moles of a gas = w Volume occupied by 1 mole of the gas at NT P . Number of equivalents of solute = n x number of moles of solute
  4. Ml / n v (in litre) Ml V (in litre) . Normality of solution = n x molarity of solution Concept of minimum molecular mass : Ax 1 00 Minimum molecular mass A = Atomic mass percentage of element Some Basic units % wt or w/w gm quantity of solute present 100gm of solution % by v or v/v volume of solute present in 100 ml of solution w/v gm quantity of solute present in 1000ml of solution its unit is gm/lt. w Also, N = n; 1 mol of N3- = 14 gm of N3- = 1 gm ion of N = No ions 1 w 1 >< n Example 3 : Solution : The molecular mass of H2S04 is 98 amu. Calculate the number of moles of each element in 294 g of H2S04. Gram molecular mass of H2S04 = 98 gm 294 Moles of H2S04 = 3 moles H2S04 one molecule one mole . 3 mole MOLE CONCEPT 98 2 atoms 2 x NA atoms 2 mole 6 mole s one atom 1 x NA atoms one mole 3 mole O 4 atoms 4 x NA atoms 4 mole 12 mole Mole means heap or collection of things. It is basic measuring unit of chemical substance. One mole of any chemical substance contains fixed no. of entity i.e. particles and known as Avogadro 23 number (No or Na) equal to 6.023 x 10 N14 For element : 7 1 mol of N = 14 gm of N = 1 gm atom = No atoms For molecule/compound N2, N02 etc. 1 mol of N2 = 28gm of N2 = 1 gm molecule of N2 = No molecule = 2 No atoms For ions N3-, CF no.of moles = n gmquantity m Atomic mass/molecular mass 1 mol of any gas contains fixed volume i.e: 22.4 It at NT P (OOC & 1 bar) weight (grams) Number of moles of a species = w Atomic or molecular mass (g/ mole) M Volume occupied by gas at NT P Number of moles of a gas = Avogadro's hypothesis : Volume occupied by 1 mole of the gas at NT P Equal volume of the gases have equal number of molecules ( not atoms) at same temperature and pressure condition. S.T.P. : (Standard Temperature and Pressure) At S.T. P. condition Tempereture = OOC or 273 K pressure = 1 atm = 760 mm of Hg
  5. and volume of one mole of gas at STP is found to be experimently equal to 22.4 litres which is known as molar volume. Avogadro number = 6.023 x 1023 T-map: Interconversion of mole-volume,mass and number of particles Number Of — mol. wt — At. wt Empirical and Molecular Formula : Mole x mol. wt x At. wt Mass Volume at STP + 22.4 It Dividing % by atomic mass gives molar ratio from which empirical formula in obtained. molecular mass n empirical formula mass Molecular formula = (Empirical formula)n (iii) Molecular mass = mass of 22.41 of gas or vapour at S.T.P Example 4 : A substance, on analysis, gave the following percentage composition: Na = 43. 4%, 11.3%, O = 45.3%. Calculate its empirical formula. [Na = 23, C = 12, O = 16] Solution: Element Sodium Carbon Oxygen SYMBOL Na c O % age 43.4 11.3 45.3 Atomic Mass 23 12 16 Relative number moles 43.4 = 1 .88 23 11 •3 =o.94 12 45.3 - 2.83 16 Simplest Simple ratio of of moles 1.88 2 0.94 0.94 2.83 -3 0.94 whole no. ratio 2 1 3 Therefore, the empirical formula is Na2C03. STOICHIOMETRY 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 volumes also) of the reactants and the products involved in a chemical reaction. Before understanding how to calculate the amounts of reactants required or those produced in a chemical reaction, let us study what information is available from the balanced chemical equation of a given reaction. Let us consider the combustion of methane. A balanced equation for this reaction is as given below: CH4 (g) + 202 (g) C02 (g) + 2 H20(g) The above balance reaction gives the following information: a) b) c) For every 1 mole of CH4, 2 mole of 02 will be required to produce 1 mole of C02 and 2 moles of 1-120. this signifies Mole — Mole relation For every 16 gms of CH4 , 64 gms of 02 will be required to produce 44gms of C02 and 36 gms of H20 this signifies Mass — Mass relation Ratio of moles of C02 : 1-120 at any time = 1 : 2
  6. d) e) There will be no change in total mass of all reactants and products at any time for any chemical reaction. For the above reaction only, there will be no change in total number of moles of all reactants and products. In order to solve the problems based on chemical calculations the following steps, in general,are quite helpful. (i) Write the balanced chemical equation. (ii) Write the atomic mass/molecular mass/moles/molar volumes of the species involved in calculations. (iii) Calculate the result by applying unitary method. INTERPRETATION OF BALANCED CHEMICAL EQUATIONS Once we get a balanced chemical equation then we can interpret a chemical equation by following ways Mass — mass analysis Mass — volume analysis Volume — volume analysis Now you can understand the above analysis by following example Mass — mass analysis Consider the reaction 2KC103 2KCl +302 According to stoichiometry of the reaction Mass — mass ratio : 2 x 122.5 : 2 x 74.5 : 3 x 32 Or Solution Mass of KC103 Mass of KCI Mass of KC103 Mass of 02 2 x 74.5 Example 5 : 367.5 gram KC103( M = 122.5) is heated. How many gram KCI and oxygen is produced. Balanced chemical equation for heating of KC103 is 2KC103 2KCl + 302 Mass — mass ratio: 2 x 122.5 gm Mass of KC103 Mass of KCI = 3 X 74.5 = 223.5 gm Mass of KC103 Mass of 02 w = 144 gm oxygen Mass — volume analysis : 2 x 74.5 gm: 3 x 32 gm 367.5 w 367.5 w Now again consider decomposition of KC103 2KC103 2KCl + 302 mass volume ratio 2 x 122.5 gm : 2 x 74.5 gm we can use two relation for volume of oxygen. Mass of KC103 volume of 02 at STP 3 x 22.41t 122.5 74.5 3 x 32 : 3 x 22.4 It at S.T.P. Mass of KCI and 2 x 74.5 3 x 22.4 It Example 6 : Solution: Calculate the volume of 02 and volume of air needed for combustion of 1 kg carbon at STP. C02 volume of 02 at STP C + 02
  7. 12 g C requires 02 = 22.4 litre of 02 = 1 mole of O 1000 g C requires 02 litre 32 g of 02 var -5XVO 12 = 1866.67 litre 02 5 x 1866.67 = 9333.35 litre Volume — Volume Relationship : It relates the volume of gaseous species ( reactants or product ) with the volume of another gaseous species ( reactant or product ) involved in a chemical reaction. Example 7: What volume of oxygen gas at NTP is necessary for complete combustion of 20 litre of propane measured at OOC and 760 mm pressure. Solution: The balanced equation is +502 = 3C02 +41420 Ivol 5 vol Ilitre 51itre 1 litre of propane requires = 5 litre of oxygen 20 litre of propane will require = 5 x 20 = 100 litre of oxygen at 760 mm pressure and OOC. CONCENTRATION TERMS Molarity (M) : Number of moles present in one It of solution (mol/lt) w Molarity x 1000 m x V(ml) Example 8: 149 gm of potassium chloride (KCI) is dissolved in 10 Lt of an aqueous solution. Determine the molarity of the solution (K = 39, Cl = 35.5) Solution : Molecular mass of KCI = 39 + 35.5 = 74.5 gm 149gm Moles of KCI = 2 74.5gm 2 Molarity of the solution = — 0.2M 10 Normality (N) : Number of equivalent (w/E) present in one It of solution w Normality x 1000 Ex v(ml) Molality (m) : Number of moles of solute present in 1000 gm of solvent known as molality w Molality mx W gmquantityof solvent Molarity(M) and Molality(m) for Pure Substances: Water : Let the sample of water has 1000 ml Mass of water = 1000 gm [density of water = lgm/mL.] 1000 Moles of water mol 18 1000 Molarity 18 55.55M and molality = 1 1000 mol 18 1 kg 55.55 m 2. Pure ethanol : d gm/ml (density of ethanol) (C2H50H) let volume of ethanol taken be 1000 ml.
  8. :.wtof ethanol in 1000 ml = gm 1 OOOd Mol of ethanol = 46 IOOOd IOOOd Molarity 46 mol 1000 46 & molality of ethanol = kg 1000 46 Parts per million (ppm) —Amount of solute ( in g ) with 106 g solvent Parts per billion ( ppb) •amount of solute ( in g ) with 109 g solvent Example 9 : Solution : 255 gm of an aqueous solution contains 5 gm of urea. What is the concentration of the solution in terms of molality. (Mol. wt. of urea = 60) Mass of urea = 5 gm Molecular mass of urea = 60 5 Number of moles of urea = — 0.083 60 Mass of solvent = (255 — 5) = 250 gm Number of moles of solute Molality of the solution = x 1000 Mass of solvent in gram 0.083 x 1000 = 0.332 250 Mole Fraction (X) The ratio of number of moles of the solute or solvent present in the solution and the total number of moles present in the solution is known as the mole fraction of substance concerned. Let number of moles of solute in solution Number of moles of solvent in solution = N n Mole fraction of solute (Xi ) Mole fraction of solvent (X2) Also Xl + = 1 Mole fraction is a pure number. It will remain independent of temperature changes. Formality (F) : Number of gram formula weight of a solute dissolve per litre of the solution. mass of solute (g) x 1 formula mass of solute Volume of solution (L) PERCENTAGE CONCENTRATION The concentration of a solution may also be expressed in terms of percentage in the following way. % weight by weight (w/w): it is given as mass of solute present in per 100 gm of solution. i.e. mass of solute in gm x 100 mass of solution in gm % weight by volume (w/v) : It is given as mass of solute present in per 100 ml of solution
  9. i.e Example 10 : Solution i.e. % volume by volume (VIV) : It is given as volume of solute present in per 100 ml solution. x 100-1.96 mass of solute in gm x 100 volume of solution in ml Volume of solute in ml x 100 Volume of solution in ml 0.5 g of a substance is dissolved in 25 g of a solvent. Calculate the percentage amount of the substance in the solution. Mass of substance = 0.5 g Mass of solvent = 25 g 0.5 . Percentage of the substance (w/w) 0.5+25

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