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MOLE CONCEPT, MATTER, , PHYSICAL, , CHEMICAL, Colloidal, state, , Liquid, , Solid, , size between definite shape definite volume, but no, 10Å - 2000Å, & volume, definite shape, , Gas, , Plasma, , no definite subs present in, ionic form is, shape, & volume under extreme, condition, , Pure Substance, , fixed ratio of, masses, of constituents, , Elements, , Compounds, , cannot be decomposed made of two or more, further into simpler, elements in fixed, substances, ratio of by mass, , Ionic, , Metals, , Metalloids, Non, Metals, , Covalent, , Mixture, , Homogeneous, , same composition, throughout &, component are, indistinguishable., eg. gases, liquid, solution, alloys,, atmospheric, air nearly, homogeneous, , Heterogeneous, , composition is, not uniform., eg. sand & water,

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1., , (1), , CL ASSIFICATION OF UNIVERSE :, (1), , Matter, , (2), , Energy, , M AT T E R, The thing which occupy space and having mass which is feel by our five senses is called as matter., , (I), , , , 2 types, , (I), , Physical classification, , (II), , Chemical classification, , Physical Classification :, It is based on physical state under ordinary conditions of temperature and pressure, matter is classified into, the following three types :, , (a), , (a), , Solid, , (b), , Liquid, , (c), , 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., , (b), , Liquid :, A substance is said to be liquid if it possesses a definite volume but not definite shape. They take up the shape, of the vessel in which they are put., e.g. water, milk, oil, mercury, alcohol etc., , (c), , Gas :, A substance is said to be gas if it neither possesses a definite volume nor a definite shape. This is because they, fill up the whole vessel in which they are put., e.g. hydrogen(H 2), oxygen(O 2), carbon dioxide(CO 2), etc.', , (II), , (A), , Chemical Classification :, , , 2 Type s, , (A), , Pure, , (B), , Mixture, , Substa nce, , Pure Substa nce :, A material containing only one type of substance. Pure Substance can not be seperated into simpler substance by physical method., e.g. :, , Element, , =, , Na, Mg, Ca .................. etc., , Compound, , =, , HCl, H2O, CO2, HNO3 .......... etc., , , , 2 types, , (a), , Element, , (b), , Compound, , (a), , Element : The pure substance containing only one kind of atoms ., 3 types (depend on physical and chemical property), (a'), , Metal, , (b'), , Non-metal, , (c'), , Metalloids

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(b), , Compound :, It is defined as pure substance containing more than one kind of atoms which are combined together, in a fixed ratio by weight and which can be decomposed into simpler substance by the suitable, chemical method. The properties of a compound are differnt from those of its components., e.g. :, , H2 O , HCl, HNO3 .......... etc., 2 : 16, 1 : 8 by wt., , 2 types, , , , (B), , (a'), , Organic Compound, , (b'), , Inorganic Compound, , Mixture :, A material which contain more than one type of substances and which is mixed any ratio by wt.is called as, mixture., , , The property of the mixture is the property of its components, , , , The mixture is seperated by simple physical method., , , , 2 types, , (a'), , Homogeneous mixture, , (b'), , Hetrogeneous mixture, , (a'), , Homogeneous mixture :, The mixture, in which all the components are present in uniform is called as homogeneous mixture., e.g. :, , (b'), , Water + Salt, Water + Sugar, Water + alcohol,, , Hetrogeneous mixture :, The mixture in which all the components are present in nonuniform is called as Hetrogeneous, mixture., e.g. :, , , , Water + Sand, Water + Oil,, , INTRODUCTION :, There are a large number of objects around us which we can see and feel., Any t h i ng t hat occupie s space a nd has mass is cal led mat ter., It was John Dalton who firstly developed a theory on the structure of matter, later on which is known, as Dalton's atomic theory., , , , 1., , DALTON'S ATOMIC THEORY :, , 1., , Matter is made up of very small undivisible particle called atoms., , 2., , All the atoms of a given element is identical in all respect i.e. mass, shape, size, etc., , 3., , Atoms cannot be created nor destroyed by any chemical process., , 4., , Atoms of different elements are different in nature., , 2., , THE L AW OF CHEMICAL COMBINATION :, , Atoi ne Lavoisier, John Dalton and other scientists formulate certain law concerning the composition, of matter and chemical reactions. These laws are known as the law of chemical combination.

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3., , , , 4., , THE L AW OF CONSERVATION OF M ASS :, It is given by Lavoisier., In a chemical change total mass remains conserved., i.e. mass before reaction is always equal to mass after reaction., Example :, 1, H 2 (g) +, O (g) , H 2O (), 2 2, 1, Before reaction initialy 1 mole, mole, 2, After the reaction, 0, 0, 1 mole, 1, mass before reaction = mass of 1 mole H 2 (g) +, mole O 2 (g), 2, = 2 + 16 = 18 g, mass after reaction = mass of 1 mole water = 18 g, L AW OF CONSTANT OR DEFINITE PROPORTION :, It is given by Proust., All chemical compounds are found to have constant composition irrespective of their method of preparation or sources., , , , Example :, In water (H 2O), Hydrogen and Oxygen combine in 2 : 1 molar ratio, the ratio remains constant whether, it is tap water, river water or sea water or produced by any chemical reaction., , Ex., Sol., , 1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated in, steam gave 2.50 g of its oxide. Show that these results illustrate the law of constant proportion., In the first sample of the oxide,, wt. of metal = 1.80 g,, wt. of oxygen = (3.0 – 1.80) g = 1.2 g, , wt.of metal, 1.80g, =, = 1.5, wt.of oxygen, 1.2g, In the second sample of the oxide,, wt. of metal = 1.50 g,, wt. of oxygen = (2.50 – 1.50) g = 1 g, , , , , wt.of metal, 1.50g, =, = 1.5, wt.of oxygen, 1g, , Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen are fixed. Hence,, the results follows the law of constant proportion., 5., , THE L AW OF MULTIPLE PROPORTION :, It is given by Dalton., When one element combines with the other element to form two or more different compounds, the mass of, one elements, which combines with a constant mass of the other, bear a simple ratio to one another., , Note :, , Simple ratio here means the ratio between small natural numbers, such as 1 : 1, 1 : 2, 1 : 3, Later on, this simple ratio becomes the valency and then oxidation state of the element., Example : Carbon and Oxygen when combine, can form two oxides viz CO (carbonmonoxide), CO 2, (Carbondioxides), In CO, 12 g carbon combined with 16 g of oxygen., In CO 2 , 12 g carbon combined with 32 g of oxygen., Thus, we can see the mass of oxygen which combine with a constant mass of carbon (12 g) bear simple, ratio of 16 : 32 or 1 : 2.

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6., , LAW OF RECIPROCAL PROPORTION (OR L AW OF EQUIVALENT W T.) :, , , It is given by Richter., , Statement :, The ratio of the weights of two elements A and B which combine seperately with a fixed weight of the third, element C is either the same or simple ratio of the weights in which A and B combine directly with each, other., e.g., , , , , C O2, , O, , H2 O, 2 : 16, 1:8, , , , , , 12 : 32, 3: 8, , , , , , , , , , , , , , , CH4, 12 : 4, 3:1, C  H, , 3:1, , , Special Note : This law is also called as law of equivalent wt. due to each element combined in their, equivalent wt. ratio., E, , , , M w At.wt., V.F., , For ions, V.F. = Total no. of positive charge, or, , V.F. = Total no. of negative charge, , , , EXA MPLE BASED ON LAW OF RECIPROCAL PROPORTION, , Ex.1, , Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains 88.90% of oxygen and, 11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of nitrogen. Show that, these data illustrate the law of reciprocal proportions., , Sol., , In NH3, 17.65g of H combine with N = 82.35g, , , 1 g of H combine with, , N=, , 82.35, g, 17.65, , = 4.67 g, , In H2O, 11.10 g of H combine with O = 88.90 g, , 88.90, g = 8.01 g, 11.10, , , , 1 g of H combine with O =, , , , Ratio of the weights of N and O which combine with fixed weight (=1g) of H, = 4.67 : 8.01, , = 1 : 1.7, , In N2O3, ratio of weights of N and O which combine with each other = 36.85 : 63.15, = 1 : 1.7, Thus the two ratios are the same. Hence it illustrates the law of reciprocal proportions.

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, , Other Method, N, N2, , O3, , N, , H3, , 36.85 : 63.15, , 82.53 : 17.65, , 36.85 63.15, :, 36.85 36.85, , 82.53, 17.65, :, 17.65, 17.65, , : 1.7, , 4.67, , O, , :, , H, H2, , O, , 11.1 : 88.90, , N, , 11.1, 88.90, :, 11.1, 11.1, , O, , 4.67 : 8.01, , 4.67 8.01, :, 4.67 4.67, , : 8.01, , : 1.7, Thus the two ratios are the same. Hence it illustrates the law of reciprocal proportions., 7., , REL ATIVE ATOMIC M ASS :, One of the most important concept come out from Dalton's atomic theory was that of relative atomic mass, or relative atomic weight. This is done by expressing mass of one atom with respect to a fixed standard., Dalton used hydrogen as the standard (H = 1). Later on oxygen (O = 16) replaced hydrogen as the reference. Therefore relative atomic mass is given as, , Mass of one atom of an element Mass of one atom of an element, =, 1, mass of one hydrogen atoms,  mass of one Oxygen atom, 16, The present standard unit which was adopted internationally in 1961, is based on the mass of one carbon12 atom., Relative atomic mass (R.A.M.) =, , , , 8., , ATOMIC M ASS UNIT (OR A MU) :,  1 , The atomic mass unit (amu) is equal to one twelvth ,  the mass of one atom of carbon-12 isotope.,  12 , , , , , 1 amu =, , 1, × mass of one C-12 atom, 12, , = 1.66 × 10 –24 g or 1.66 × 10 –27 kg, , One amu is also called one Dalton (Da)., Now the relative atomic mass is given as, mass of one atom of the element, 1,  mass of one C  12 atom, 12, Atomic mass, R.A.M. =, 1 amu, Atomic mass = R.A.M. × 1 amu, , Relative atomic mass =, , Relative molecular mass =, , , , mass of one molecule of the substance, 1,  mass of one C  12 atom, 12, , Molecular mass = Relative molecular mass × 1 amu

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9., , MOLE :, A mole is the amount of a substance that contains as many entities (atoms, molecules or, ot her par t icle s) as t here are atoms exactly i n 0.012 kg (or 12 g) of t he carbon - 12 isotope., From mass spectrometer we found that there are 6.023 × 1023 atoms are present in 12 g of C-12 isotope., The number of entities in 1 mol is so important that it is given a separate name and symbol known as, Avogadro constant denoted by N A ., i.e. on the whole we can say that 1 mole is the collection of 6.02 × 1023 entities. Here entities may represent, atoms, ions, molecules or even pens, chair, paper etc., 1 mole of atom is also termed as 1 g – atom, 1 mole of ions is also termed as 1 g – ion, 1 mole of molecule is also termed as 1 g – molecule, Methods of Calculations of mole :, (a), (b), , Given no., NA, Given wt., If weight of a given species is given, then no. of moles =, (for atoms),, Atomic wt., If no. of some species is given, then no. of moles =, , Given wt., (for molecules), Molecular wt., If volume of a gas is given along with its temperature (T) and pressure (P)., or =, , (c), , PV, RT, where R = 0.0821 lit-atm/mol-K (when P is in atmosphere and V is in litre), use n =, , 1 mole of any gas at STP occupies 22.4 litre., Ex., , Sol., , Chlorophyll the green colouring material of plants contains 3.68 % of magnesium by mass. Calculate the, number of magnesium atom in 5.00 g of the complex., , 3.68, × 5.00 = 0.184 g, 100, , Mass of magnesium in 5.0 g of complex =, Atomic mass of magnesium = 24, 24 g of magnesium contain = 6.023 × 10, , 23, , atoms, , 6.023  10 23, 21, × 0.184 = 4.617 × 10 atom, 24, 21, Therefore, 5.00 g of the given complex would contain 4.617 × 10 atoms of magnesium., , , , 10., , 0.184 g of magnesium would contain =, , GR A M ATOMIC M ASS :, The atomic mass of an element expressed in gram is called gram atomic mass of the element., OR, It is also defined as mass of 6.02 × 10 23 atoms., OR, It is also defined as the mass of one mole atoms., For example for oxygen atom :, Atomic mass of 'O' atom = mass of one 'O' atom = 16 amu, gram atomic mass = mass of 6.02 × 10 23 'O' atoms, = 16 amu × 6.02 × 10 23, (, , , , = 16 × 1.66 × 10 –24 g × 6.02 × 10 23 = 16 g, 1.66 × 10 –24 × 6.02 × 10 23 ~ 1), , Now see the table given below and understand the definition given before.

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Element, , R.A.M., , Atomic mass, , (Relative Atomic Mass), N, , 14, , He, , 4, , C, , 12, , Gram Atomic mass/weight, , (mass of one atom), 14 amu, 4 amu, 12 amu, , 14 gm, 4 gm, 12 gm, , Average atomic weight =  % of isotope s X molar mass of isotope s., 11., , GRAM MOLECULAR MASS :, The molecular mass of a substance expressed in gram is called the gram-molecular mass of the substance., OR, 23, It is also defined as mass of 6.02 × 10 molecules, OR, It is also defined as the mass of one mole molecules., For example for 'O 2 ' molecule :, Molecular mass of 'O 2 ' molecule =, =, =, =, gram molecular mass, =, , mass of one 'O 2 ' molecule, 2 × mass of one 'O' atom, 2 × 16 amu, 32 amu, mass of 6.02 × 10 23 'O 2 ' molecules, , = 32 amu × 6.02 × 10 23, = 32 × 1.66 × 10 –24 g × 6.02 × 10 23, = 32 gm, Average molecule wt. =, 12., ,  ni Mi, where ni = no. of moles of compound, mi = molecular mass of compound,  ni, , GAY-LUSSAC'S L AW OF COMBINING VOLUME :, According to him elements combine in a simple ratio of atoms, gases combine in a simple ratio of their, volumes provided all measurements should be done in the same temperature and pressure, H 2 (g) + Cl 2 (g)  2HCl, 1 vol, 1 vol, 2 vol, , 13., , AVOGA DRO'S H YPOTHESIS :, Equal volume of all gases have equal number of molecules (not atoms) at same temperature and pressures conditions., S.T.P. (Standard Temperature and Pressure), At S.T.P. condition :, temperature = 0°C or 273 K, pressure = 1 atm = 760 mm of Hg, and volume of one mole of gas at STP is found to be experimentaly equal to 22.4 litres which is known, as molar volume., , 14., , Y-M AP :, , Number, , , ×N, N, , A, , A, , Mole, , lt, .4 t, 2, × 2 2.4l, 2, , ,  mol.wt × mol.wt,  At.wt × At.wt, Mass, , Volume at STP

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15., , PERCENTAGE COMPOSITION AND MOLECUL AR FORMUL A :, Here we are going to find out the percentage of each element in the compound by knowing the molecular, formula of compound., We known that according to law of definite proportions any sample of a pure compound always possess, constant ratio with their combining elements., , , , Example :, Every molecule of ammonia always has formula NH 3 irrespective of method of preparation or sources., i.e. 1 mole of ammonia always contains 1 mol of N and 3 mole of H. In other wards 17 g of NH3 always, contains 14 g of N and 3 g of H. Now find out % of each element in the compound., , 16., , Mass % of N in NH 3 =, , Mass of N in 1 mol NH 3, 14 g, Mass of 1 mol of NH 3 × 100 = 17 × 100 = 82.35 %, , Mass % of H in NH 3 =, , Mass of H is1 mol NH 3, 3, Mass of 1 mol of NH 3 × 100 = 17 × 100 = 17.65%, , EMPIRICAL AND MOLECULAR FORMULA :, We have just seen that knowing the molecular formula of the compound we can calculate percentage, composition of the elements. Conversely if we know the percentage composition of the elements initially,, we can calculate the relative number of atoms of each element in the molecules of the compound. This, gives as the empirical formula of the compound. Further if the molecular mass is known then the molecular, formula can be easily determined., Thus, the empirical formula of a compound is a chemical formula showing the relative number of atoms, in the simplest ratio, the molecular formula gives the actual number of atoms of each element in a molecule., i.e., , Empirical formula : Formula depicting constituent atom in their simplest ratio., Molecular formula : Formula depicting actual number of atom in onemolecule of the compound., , The molecular formula is generally an integral multiple of the empirical formula., i.e., , molecular formula = empirical formula × n, , where n =, Ex., , molecular formula mass, empirical formula mass, , An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition., C = 40.687 % ; H = 5.085 % and O = 54.228 %, The molecular weight of the compound is 118. Calculate the molecular formula of the compound., , Sol., , Step-1, To calculate the empirical formula of the compound., Element, , Symbol, , Pe r c e n t a g e, of, element, , At. mass, of, element, , Relative no., , Simplest, , Simplest, , of atoms =, , atomic, , no. atomic, , Percentage, At.mass, , ratio, , ratio, , Carbon, , C, , 40.687, , 12, , 40.687, =3.390, 12, , 3.390, 1, 3.389, , 2, , Hydrogen, , H, , 5.085, , 1, , 5.085, =5.035, 1, , 5.085, =1.5, 3.389, , 3, , Oxygen, , O, , 54.228, , 16, , 54.228, =3.389, 16, , 3.389, =1, 3.389, , , , Empirical formula is C 2H 3O 2, , 2, , whole

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, , Step - 2, To calculate the empirical formula mass., The empirical formula of the compound is C 2H 3 O 2 ., , , Empirical formula mass, = (2 × 12) + (3 × 1) + (2 × 16) = 59., , , , Step - 3, To calculate the value of 'n', n =, , , , molecular mass, 118, =, = 2, Empirical formula mass, 59, , Step - 4, To calculate the molecular formula of the salt, Molecular formula, , = n × (Empirical formula), = 2 × C 2H 3O 2 = C 4H 6O 4, , Thus the molecular formula is C 4 H 6O 4 ., 17., , DENSITY :, It is of two type., 1., Absolute density, For liquid and solids, , 2., , Relative density, , mass, volume, , Absolute density =, , Relative density or specific gravity =, , density of the subs tan ce, density of water at 4 C, , For gases :, , PM, RT, where P is pressure of gas, M = mol. wt. of gas, R is the gas constant, T is the temperature., Absolute density (mass / volume) =, , 18., , REL ATIVE DENSIT Y OR VAPOUR DENSIT Y :, Vapour density is defined as the density of the gas with respect to hydrogen gas at the same temperature, and pressure., Vapour density =, , V.D. =, 19., , M gas, MH2, , =, , d gas, dH2, , M gas, 2, , , , PM gas / RT, PM H 2 / RT, ,  M gas  2 V.D., , CHEMICAL EQUATION :, All chemical reaction are represented by chemical equations by using formule of reactant and products., Qualitatively a chemical equation simply describes what the reactants and products are. However, a balanced chemical equation gives us a lot of quantitative information mainly the molar ratio in which reactants, combine and the molar ratio in which products are formed., Example :, When potassium chlorate (KClO 3) is heated it gives potassium chloride (KCl) and oxygen (O 2)., , , KClO 3  KCl + O 2 (unbalanced chemical equation), , , 2KClO 3  2 KCl + 3 O 2 (balanced chemical equation)

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, , Remember a balanced chemical equation is one which contains an equal number atoms of each element, on both sides of equation., , 20., , GR AVIMETRIC ANALYSIS :, Once we get a balanced chemical equation then we can interpret a chemical equation by following ways, 1., , Mass - mass analysis, , 2., , Mass - volume analysis, , 3., , Mole - mole analysis, , 4., , Vol - Vol analysis (separately discussed as eudiometry or gas analysis), , Now you can understand the above analysis by following example, 1., , Mass - mass analysis :, Consider the reaction, , 2 KClO 3  2KCl + 3O 2, , mass-mass ratio : 2 × 122.5 :, or, Ex., , Mass of KClO 3, 2  122.5, =, Mass of KCl, 2  74.5, , According to stoichiometry of the reaction, , 2 × 74.5 : 3 × 32, , , Mass of KClO 3, 2  122.5, =, Mass of O 2, 3  32, , Calculate the weight of iron which will be converted into its oxide by the action of 36 g of steam., (Given : 3Fe + 4H2O  Fe3O4 + H2), , Sol., , Mole ratio of reaction suggests,, , 3, Mole of Fe, =, Mole of H 2 O 4,  Mole of Fe =, =, , mol of H2O, , 3 36, 3, , =, 4 18, 2, , wt. of Fe =, Ex., , 3, ×, 4, , 3, × 56 = 84 g, 2, –, , In a gravimetric determination of P of an aqueous solution of dihydrogen phosphate in H2PO4 is treated with, a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate,, Mg(NH4)PO4.6H2O. This is heated and decomposed to magnesium pyrophosphate, Mg2P2O7. A solution, –, , of H 2PO 4 yielded 2.054 g of (Mg 2P 2 O 7) which is weighed. What weight of NaH 2PO 4 was present, originally?, Sol., , NaH 2PO 4 + Mg, , 2+, , +, , , + NH 4  Mg(NH 4)PO 4.6H 2O ,  Mg 2P 2O 7, , As P atoms are conserved, applying POAC for P atoms, moles of P in NaH2PO4 = Moles of P in Mg2P2O7, , , , Ex., , 1 ×, , Moles of NaH2PO4 = 2 × Moles of Mg2P2O 7, , W NaH 2 PO 4, M NaH 2 PO 4, , = 2 ×, , WMg2 P2 O 7, M Mg2 P2 O 7, , , , W NaH 2 PO 4, 120, , = 2 ×, , 2.054, 222, , W NaH 2 PO 4 = 2.22 g, , A solid mixture weighing 5.00 g containing lead nitrate and sodium nitrate was heated below 600°C until, the mass of the residue was constant. If the loss of mass is 30 %, find the mass of lead nitrate and, sodium nitrate in mixture., (At. wt. of Pb = 207, Na = 23, N = 14, O = 16)

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Sol., , 2Pb(NO3)2, 2 × 331, 2NaNO3, , , ,  2PbO +, 2 × 223, , 4NO2  + O2 , , 2NaNO2 + O2 , , , , , , 2 × 85, , 2 × 69, , Let, wt. of Pb(NO3)2 in mixture = x, wt. of NaNO3 = (5 – x) g, 662 g of Pb(NO3)2 will give residue = 446, , , 446,  (x) = 0.674x g, 662, , xg of Pb(NO3)3 will give residue =, , 170 g of NaNO3 give residue = 138 g, 138,  (5 – x) = 0.812 × (5 – x), 170, 30 , , Actual wt. of residue obtained =  5  5 , = 3.5 g, 100 , , , , , (5 – x), g NaNO3 will give residue =, , , , 0.674x + 0.812 × (5 – x) = 3.5  0.138 x = 0.56, , , , x = 4.05 g = wt. of Pb(NO3)2, , , , wt. of NaNO3 in the mixture = (5 – 4.05) = 0.95 g, , Ex., , 3.0 g an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride, solution when 1.74 g of BaSO 4 were obtained as dry precipitate. Calculate the percentage purity of, sample., , Sol., , Na2SO4 + BaCl2  BaSO4, 142g, , +, , 2NaCl, , 223g, , 223 g of BaSO4 are produced from 142 g of Na2SO4,  1.74 g of BaSO4 would be produced by =, , 142, × 1.74 = 1.06 g of Na2SO4, 233, , 1.06,  100 = 35.33 %, 3.0, Mass - volume analysis :, , % purity of Na2SO 4 =, 2., , Now again consider decomposition of KClO 3, 2 KClO 3  , mass volume ratio :, , 2KCl, , +, , 2 × 122.5 g : 2 × 74.5 g :, , 3 O2, 3 × 22.4 L at STP, , we can use two relation for volume of oxygen, , 2  122.5 g, Mass of KClO 3, =, 3  22.4 L, volume of O 2 at STP, , .......(i), , 2  74.5 g, Mass of KCl, =, ......(ii), 3  22.4 L, volume of O 2 at STP, How much marble of 90.5 % purity would be required to prepare 10 litres of CO2 at STP when the, marble is acted upon by dilute HCl ?, and, , Ex., Sol., , CaCO3, 100 g, , +, , 2HCl, , , , CaCl2 +, , H2O, , +, , CO2, , 22.4litre, , 22.4 L of CO2 at STP will be obtained from 100 g of CaCO3,  10 L of CO2 at STP will be obtained from pure CaCO3 =,  Impure marble required =, , 100, ×, 90.5, , 44.64 = 49.326 g, , 100,  10 = 44.64 g, 22.4

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3., , Mole - mole analysis :, This analysis is very much important for quantitative analysis point of view., Now consider again the decomposition of KClO 3 ., 2 KClO 3   2 KCl + 3 O 2, In very first step of mole-mole analysis you should read the balanced chemical equation like 2 moles KClO 3, on decomposition gives you 2 moles KCl and 3 moles O 2 and from the stoichiometry of reaction, we can write, , Moles of KClO 3, Moles of KCl, Moles of O 2, =, =, 2, 2, 3, Now for any general balance chemical equation like, a A + b B  c C + d D, you can write., , Moles of A reacted Moles of B reacted Moles of C reacted Moles of D reacted, =, =, =, a, b, c, d, 21., , PRINCIPLE OF ATOM CONSERVATION (POAC) :, Infact POAC is nothing but the conservation of mass, expressed before in the concepts of atomic theory., And if atoms are conserved, moles of atoms shall also be conserved., The principle is fruitful for the students when they don't get the idea of balanced chemical equation in, the problem., This principle can be under stand by the following example., Consider t he decomposit ion of KClO 3 (s)  KCl (s) + O 2 (g) (unbala nced chemical react ion), Apply the principle of atom conservation (POAC) for K atoms., Moles of K atoms in reactant = moles of K atoms in products, or moles of K atoms in KClO 3 = moles of K atoms in KCl, Now, since 1 molecule of KClO 3 contains 1 atom of K, or 1 mole of KClO 3 contains 1 mole of K, similarly 1 mole of KCl contains 1 mole of K, Thus, moles of K atoms in KClO 3 = 1 × moles of KClO 3, and moles of K atoms in KCl = 1 × moles of KCl, , , , , , moles of KClO 3 = moles of KCl, , or, , w t. o f K C l in g, w t . o f K C l O 3 in g, =, m ol. wt. of K Cl, m ol. wt. of K ClO3, , The above equation gives the mass-mass relationship between KClO 3 and KCl which is important in, stoichiometric calculations., Again, applying the principle of atom conservation for O atoms,, moles of O in KClO 3 = 3 × moles of KClO 3, moles of O in O 2 = 2 × moles of O 2, , , , , , 3 × moles of KClO 3 = 2 × moles of O 2, , or, , 3 ×, , wt.of KClO 3, vol.of O 2 at NTP, = 2 ×, mol. wt.of KClO 3, standard molar vol.(22.4 lt), , The above equations thus gives the mass-volume relationship of reactants and products.

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Ex., , 27.6 g K 2CO 3 was treated by a series of reagents so as to convert all of its carbon to K 2Zn 3 [Fe(CN) 6] 2., Calculate the weight of the product., [mol. wt. of K 2 CO 3 = 138 and mol. wt. of K 2 Zn 3 [Fe(CN) 6] 2 = 698], , Sol., , Here we have not knowledge about series of chemical reactions but we known about initial reactant and, final product accordingly, Several, , K 2CO 3 , K 2Zn 3 [Fe(CN) 6] 2, Steps, Since C atoms are conserved, applying POAC for C atoms,, moles of C in K 2 CO 3 = moles of C in K 2Zn 3 [Fe(CN) 6] 2, 1 × moles of K 2CO 3 = 12 × moles of K 2Zn 3 [Fe(CN) 6 ] 2, (, , 1 mole of K 2CO 3 contains 1 moles of C), , wt.of K 2 CO 3, wt.of the product, mol. wt.of K 2 CO 3 = 12 × mol. wt.of product, , Ex., , 27.6, 698, ×, = 11.6 g, 138, 12, A sample of 3 g containing Na2CO3 and NaHCO3 loses 0.248 g when heated to 300°C, the tempera, ture at which NaHCO3 decomposes to Na2CO3, CO2 and H2O. What is the percentage of Na2CO3 in, the given mixture?, , Sol., , The loss in weight is due to removal of CO2 and H2O which escape out on heating., , wt. of K 2 Zn 3 [Fe(CN) 6 ] 2 =, , wt. of Na2CO3 in the product = 3.00 – 0.248 = 2.752 g, Let wt. of Na2CO3 in the mixture be x g,  wt. of NaHCO3 = (3.00 – x) g, Since Na2CO 3 in the products contains x g of unchanged reactant Na2CO 3 and rest produced from, NaHCO 3 ., The wt. of Na2CO3 produced by NaHCO3 = (2.752 – x)g, NaHCO 3  Na2CO 3 + (H 2O + CO 2) , (3.0 – x), , (2.752 – x), , Applying POAC for Na atom, 1 × moles of NaHCO3 = 2 × moles of Na2CO3 , , (3  x), (2.752  x ), = 2x, 84, 106, ,  x = 2.3244 g,  % of Na2CO 3 =, 22., , 2.3244, × 100 = 77.48 %, 3, , LIMITING REAGENT :, The reactant which consumed first into the reaction, When we are dealing with balance chemical equation then if number of moles of reactants are not in the, ratio of stoichiometric coefficient of balanced chemical equation, then there should be one reactant which, should be limiting reactant., , Ex., , Three mole of Na 2CO 3 is reacted with 6 moles of HCl solution. Find the volume of CO 2 gas produced at, STP. The reaction is, Na 2 CO 3 + 2 HCl  2 NaCl + CO 2 + H 2 O, , Sol., , From the reaction :, , Na 2 CO 3 + 2 HCl  2 NaCl + CO 2 + H 2 O, , gives moles, , 3 mol, , given mole ratio, , 1, , :, , Stoichiometric coefficient ratio, , 1, , : 2, , 6 mol, 2

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, , See here given moles of reactant are in stoichiometric coefficient ratio therefore non reactant left over., Now use Mole-mole analysis to calculate volume of CO 2 produced at STP, , Moles of Na 2 CO 3, Mole of CO 2 produced, =, 1, 1, Moles of CO 2 produced = 3, volume of CO 2 produced at STP = 3 × 22.4 L = 67.2 L, Ex., , 6 moles of Na2CO3 is reacted with 4 moles of HCl solution. Find the volume of CO2 gas produced at STP., The reaction is, Na 2 CO 3 + 2 HCl  2 NaCl + CO 2 + H 2 O, , Sol., , , , From the reaction :, , Na 2 CO 3 + 2 HCl  2 NaCl + CO 2 + H 2 O, , gives moles of reactant, , 6, , :, , 4, , given molar ratio, , 3, , :, , 2, , Stoichiometric coefficient ratio, , 1, , :, , 2, , See here given number of moles of reactants are not in stoichiometric coefficient ratio. Therefore there, should be one reactant which consumed first and becomes limiting reagent., But the question is how to find which reactant is limiting, it is not very difficult you can easily find it. According, to the following method., , 23., , HOW TO FIND LIMITING REAGENT :, Step : I, Divided the given moles of reactant by the respective stoichiometric coefficient of that reactant., Step : II, See for which reactant this division come out to be minimum. The reactant having minimum value is limiting, reagent for you., Step : III, Now once you find limiting reagent then your focus should be on limiting reagent, From, Step I & II, Na 2 CO 3, HCl, 6, 4, 6,  2 (division in minimum), 1, 2, , HCl is limit i ng reagent, From Step III, Mole of HCl, Mole of CO 2 produced, From, =, 2, 1, , mole of CO 2 produced = 2 moles, , volume of CO 2 produced at S.T.P. = 2 × 22.4 = 44.8 L, , Ex., , Calculate the weight of FeO from 4 g VO and 5.75 g of Fe2O3. Also report the limiting reactant., VO + Fe2O3  FeO + V2O 5, , Sol., , Balanced equation, , 2VO, , Moles before reaction, , 4, 67, , 5.75, 160, , 0.05970, , 0.03590, , =, Moles after reaction, , +, , (0.05970–0.0359), , 3Fe2O3   , , 0, , 6FeO, , +, , 0, , 6, ,  3  0.0359 , , , , V2O5, 0, , 1, ,  3  0.0359 , , , , As 2 moles of VO react with 3 moles of Fe2O3, , , 0.05970 g moles of VO =, , 3, × 0.05970 = 0.08955 moles of Fe2O 3, 2

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Moles of Fe 2O 3 available = 0.0359 only, Hence, Fe2O3 is the limiting reagent., , 6, × 0.0359, 3, , Moles of FeO formed =, , , Weight of FeO formed = 0.0359 × 2 × 72 = 5.17 g, ,  n FeO, 6,    n = 6  n Fe O, , , 2 3, FeO, 3,  n Fe2 O 3 3 , 6,  n Fe2 O 3  M Fe2 O 3, 3, +, A mixture of KBr, NaBr weighing 0.56 g was treated with aqueous solution of Ag and the bromide, ion was recovered as 0.97 g of pure AgBr. What was the weight of KBr in the sample ?, WFeO =, Ex., Sol., , KBr, , +, , a g, , NaBr, , +, , Ag, , +, , AgBr, ,  , , (0.56 – a)g, , 0.97 g, , Applying POAC for Br atoms,, Moles of Br in KBr + Moles of Br in NaBr = Moles of Br in AgBr, or 1 × Moles of KBr + 1 × Moles of NaBr = 1 × Moles of AgBr, , , , a, (0.56  a) 0.97, , =, (MKBr = 199, MNaBr = 103, MAgBr = 188), 119, 103, 188, a = 0.2124 g, , Percentage of KBr in the sample =, Ex., , 0.2124,  100 = 37.93, 0.560, , The reaction, 2C + O2  2CO, is carried out by taking 24 g of carbon and 128 g of O2., Find out :, (i), , Which reactant is left in excess ?, , (ii), , How much of it is left ?, , (iii), , How many moles of CO are formed ?, , (iv), , How many grams of other reactant should be taken so that nothing is left at the end, of reaction ?, , Sol., , 2C, Mole before reaction, Mole after reaction, , +, , 24, 12, 0, , O2, , , , 128, 32, 3, , 2CO, , 2, , , , Mole ratio of C : O2 : CO :: 2 : 1 : 2, , (i), , O2 is left in excess., , (ii), , 3 moles of O2 or 96 g of O2 is left., , (iii), , 2 moles of CO or 56 g of CO is formed., , (iv), , To use O2 completely, total 8 moles of carbon or 96 g of carbon is needed., , PERCENTAGE YIELD : The percentage yield of product =, , actual yield, × 100, the theoretical maximum yield

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24., , SOLUTIONS :, A mixture of two or more substances can be a solution. We can also say that a solution is a homogeneous, mixture of two or more substances 'Homogeneous' means 'uniform throughout'. Thus a homogeneous mixture,, i.e., a solution, will have uniform composition throughout., , 25., , CONCENTR ATION TERMS :, The following concentration terms are used to expressed the concentration of a solution. These are :, 1., , Molarity (M), , 2., , Molality (m), , 3., , Mole fraction (x), , 4., , % calculation, , 5., , ppm, , , , Remember that all of these concentration terms are related to one another. By knowing one concentration, term you can also find the other concentration terms. Let us discuss all of them one by one., , 1., , Molarity (M) : The number of moles of a solute dissolved in 1 L (1000 ml) of the solution, is known as the molarity of the solution., , number of moles, volume of solution in litre, Let a solution is prepared by dissolving w g of solute of mol. wt. M in V mL water., i.e., Molarity of solution =, , 2., , , , Number of moles of solute dissolved =, , , , V mL water have, , , , 1000 mL water have, , w, M, , w, mole of solute, M, , w  1000, , M  V (in mL ), ,   Molarity (M) =, , w  1000, (Mol. wt of solute)  V (in mL ), , Molality (m) : The number of moles of solute dissolved in 1000 g (1 kg) of a solvent is known, as the molality of the solution., i.e.,, , molality =, , number of moles of solute, × 100, mass of solvent in gram, , Let y g of a solute is dissolved in x g of a solvent. The molecular mass of the solute is m. Then, y/m mole of the solute are dissolved in x g of the solvent. Hence, , y, × 1000, mx, Mole fract ion (x) : T he rat io of number of mole s of the solute or solvent pre sent i n the solut ion, a nd t he total number of mole s pre sent i n t he solut ion is know n as t he mole fract ion of substa nce s concerned., Molality =, , 3., , Let number of moles of solute in solution = n, Number of moles of solvent in solution = N, , , n, n N, , N, , also, x1 + x2 = 1, n N, % Calculation : The concentration of a solution may also expressed in terms of percentage in the following, way., , , 4., , Mole fraction of solute (x 1) =, , Mole fraction of solvent (x 2) =, , (i) % weight by weight (w/w) : It is given as mass of solute present in per 100 g of solution., , mass of solute in g, × 100, mass of solution in g, [X % by mass means 100 g solution contains X g solute ;  (100 – X) g solvent], i.e., , % w/w =

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(ii) % weight by volume (w/v) : It is given as mass of solute present in per 100 mL of solution., i.e., , % w/v =, , mass of solute in g, × 100, volume of solution in mL, , w, [X %   means 100 mL solution contains X g solute], V, , (iii) % volume by volume (V/V) : It is given as volume of solute present in per 100 mL solution., i.e., , % V/V =, , Volume of solute, × 100, Volume of solution in mL, Mass of solute, Mass of solute, × 10 6  , × 10 6, Mass of solution, Mass of solvent, , 5., , Parts per million (ppm) :, , 26., , EUDIOMETRY OR GAS ANALYSIS :, Gaseous reactions are carried out in a special type of tube known as eudiometer tube. The tube is graduated, in millimeters for volume measurement. The reacting gases taken in the eudiometer tube are exploded, by sparks. T he volume s of the product s of a gase s are determi ned by absorbing them i n suitable, reagents,, , Eg., , Solvent, , gas (es) absorb, , KOH, , CO 2 , SO 2 , Cl 2, , Ammonical Cu 2 Cl 2, , CO, , Turpentine oil, , O3, , Alkaline pyrogallol, , O2, , Water, , NH 3 , HCl, , CuSO 4, , H 2O, , Eudiometry is mainly bases on Avogadro's law i.e. V  n at the same temperature and pressure., , The mole concept may be applied in solving the problems, keeping in mind that in a gaseous reaction, the relative volumes (measured under identical conditions) of each reactant and product represent their, relative numbers of moles., eg :, , , , A(g), , +, , B(g), , , , C(g), , +, , D(g), , a volumes, , b volumes, , c volumes, , d volumes, , a moles, , b moles, , c moles, , d moles, , Generally, explosions are carried out at STP and H 2O is assumed to be in liquid state, means its volume, is negligible as compared to product gases., Burni ng of hydrocarbon :, , 1., , Hydrocarbon containing carbon and hydrogen only., , y, y, , C x H y (g) +  x   O 2 (g)  xCO 2 (g) +, H O (), 2 2, 4, , 2., , Ex., Sol., , Hydrocarbon containing carbon and hydrogen and oxygen., , y, y z, , C x H yO z (g) +  x    O 2 (g)  xCO 2 (g) +, H O (), 2 2, 2 2, , 3, What volume of oxygen at STP is required to effect complete combustion of 400 cm of acetylene and, what would be the volume of carbon dioxide formed?, 2C 2H 2 + 5O 2  4CO 2 + 2H 2O, 2 volume of C2H2 require O2 for complete combustion = 5 vol., 3, ,  400 cm of C2H2 will require O2 for complete combustion =, , 5,  400, 2

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3, , = 1000 cm at STP, 2 volume of C2H2 produce CO2 = 4 volume, , 4,  400 = 800 cm3, 2, 3, Thus, volume of CO2 produced = 800 cm at STP., 3, ,  400 cm of C2H2 at STP will produce CO2 =, , Ex., , A gaseous hydrocarbon requires 6 times its own volume of O2 for complete oxidation and produces, 4 times its volume of CO2. What is its formula ?, , Sol., , The balanced equation for combustion, , y, y, , CxHy +  x   O2  xCO2 +, HO, 2 2, 4, , y, , 1 volume  x   volume, 4, , ,  x +, , y, = 6 (by equation), 4, , or 4x + y = 24, , .......(1), , Again x = 4 since evolved CO2 is 4 times that of hydrocarbon, , Ex., , 16 + y = 24 or y = 8  formula of hydrocarbon C4H8, , A 30 c.c. mixture of CO, CH4 and He gases is exploded by an electric discharge at room temperature, with excess of oxygen. The decrease in volume is found to be 13 c.c. A further contraction of 14 c.c., occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous, mixture in terms of volume percentage., , Sol., , Let the volume of CO be 'a' c.c. and CH4 be 'b' c.c, , , Volume of He = (30 – a – b), , on explosion with oxygen, , 1, O2 (g)  CO2 (g), 2, CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (), CO (g) +, , 'a' c.c. of CO give 'a' c.c. of CO2 and 'b' c.c. of CH4 gives 'b' c.c. of CO2., Therefore the volume decrease is due to the consumption of O2. O2 consumed for 'a' c.c. of CO is, c.c. and O2 consumed for 'b' c.c. of CH4 is '2b' c.c., , a, + 2b = 13, 2, The further contraction occurs because of the absorption of CO2 by KOH, a + b = 14, , , , , b = 4 c.c., , , , a = 10 c.c., , , , Percentage composition of CO =, , 10, × 100 = 33.33 %, 30, 4, Percentage composition of CH 4 =, × 100 = 13.33 %, 30, (30  10  4), Percentage composition of He =, × 100 = 53.33 %, 30, , a, 2