Page 1 : Atoms and Molecules, , reactants. For example, 12 g carbon combines with 32, g of oxygen to form 44 g of carbon dioxide:, C, , O2, , CO2, 12 g, 32 g, 44 g, , The idea of tiniest unit of matter, viz. "anu" and, "parmanu", was put forth by Maharishi Kanad in Vedic, period in our country. Another Indian philosopher,, Pakudha Katyayama, elaborated this doctrine and said, that these particles normally exist in combined form,, which gives us various forms of matter., Democritus, a Greek philosopher, also proposed that, matter is made up of extremely small particles, "the, atoms". The name comes from Greek word 'atomos', means "indivisible"., However, all these theories proposed by the Indian, and Greek philosophers were based on abstract, thinking and not on experiments. For about 2000, years, the atomic theory remained a mere speculation, till John Dalton, an English school teacher,, propounded his theory on atoms. Dalton's theory was, based on the chemical laws of combination known at, that time. These laws of chemical combination were, given by A.L. Lavoisier and Joseph L. Proust and were, based upon extensive study of chemical reactions., , To verify the law of conservation of mass i.e. mass of, the reactants is same as the mass of the products in, a chemical reaction., • Prepare 5% aqueous solutions of barium chloride, and sodium sulphate in separate test tubes., • Take a small amount of sodium sulphate solution in, a conical flask and some solution of barium chloride in, an ignition tube (a small tube made of glass)., • Hang the ignition tube in the flask carefully; see that, the solutions do not get mixed. Put a cock on the, flask., • Weigh the flask with its content carefully., • Now, tilt the flask so that the solutions mix with, each other., • Weigh again., • Observations: As soon as the two solutions mix, a, chemical reaction takes place to form new products., Barium chloride + Sodium sulphate  Barium, sulphate + Sodium chloride, BaCl2  Na2 SO4 ,  BaSO4  2 NaCl, , Laws of Chemical Combination, By studying the quantitative measurements of many, reactions it was observed that the chemical reactions, taking place between various substances are, governed by certain laws. These laws are called the, 'laws of chemical combination'. These laws formed, the basis of Dalton's atomic theory. There are two, main laws of chemical combination which are as, follows:, (1) Law of Conservation of Mass, (2) Law of Constant Proportion (also known as the law, of definite composition), , ( white ppt .), , We put a cork on the mouth of the flask to avoid any, exchange of matter between the flask and its, surrounding., It is observed that the mass of the flask and its, contents remain the same even after the chemical, reaction has taken place., • Conclusion: The activity shows that in chemical, reaction, mass of the reactants is the same as the, mass of the products. This proves the law of, conservation of mass., , Law of Conservation of Mass, This law deals with the relation between the mass of, the reactants and the products during the chemical, changes. It was stated by a French chemist, A., Lavoisier in 1774. The law of conservation of mass, states that during any chemical change, the total mass, of the products is equal to the total mass of the, reactants. In other words, the law of conservation of, mass means that mass can neither be created nor, destroyed during any chemical combination. This law, is also known as law of indestructibility of matter., From a large number of experiments, it was, concluded that although substances may undergo, chemical changes, the total mass of the products of, the reaction is exactly equal to the total mass of the, , 1., , Sol.:, , 2, , 10.0 g of CaCO3 on heating gave 4.4 g of CO2, and 5.6 g CaO. shows that these observations, are in agreement with the law of conservation, of mass., The above reaction takes place according to, the following equation:, , CaO3( s ) , CaO( s )  CO2( g ) 

Page 2 : Mass of reactant (CaCO3 )  10 g, Mass, of, products, (CaO  CO2 )  (5.6  4.4) g  10 g, Since, mass of the reactants is equal to the, mass of the products, therefore, these, observations are in agreement with the law of, conservation of mass., , • Observation: It is observed that 90 g of water on, decomposition gives 10 g of hydrogen gas and80 g of, oxygen gas,, Mass of hydrogen gas obtained 10 g 1, Thus,, , , Mass of oxygen gas formed, 80 g 8, From the above two experiments we observe that, water formed in the laboratory had hydrogen and, oxygen gases combined in the ratio of 1 : 8 by mass., At the same time, water obtained fromany source, also had hydrogen and oxygen gases combined in the, ratio of 1 : 8 by mass., The above results are in accordance with the law of, constant composition., , Law of Constant Proportion, This law states that in a pure chemical substance the, elements are always present in definite proportions, by mass., This law is also known as the law of definite, proportions. This means that, whatever, maybe the, source from which a compound is obtained, it is, always made up of the same elements in the same, proportion by mass., , 2., , Experiment I (Production of water), • Collect the water formed by burning 10 g hydrogen, gas completely in 80 g of oxygen., • Record the mass of the water so obtained., • Observation: The mass of water so obtained is, found out to be 90 g., , Ratio in which hydrogen and oxygen combine, 10 1,  , 80 8, , Sol.:, , , , , , Weight of copper oxide obtained, by heating, 4.32 g of metallic copper with nitric acid and, subsequent ignition, was 5.40 g. In another, experiment 2.30 g of copper oxide oil, reduction yielded 1.84 g of copper. Show, that these findings are in. accordance with the, law of constant proportion., Experiment I, Weight of copper = 4.32 g, Weight of copper oxide = 5.40 g, Weight of oxygen = Weight of copper oxide, Weight of copper,  5.40  4.32  1.08 g, Ratio of copper arid oxygen, Mass of copper 4.32 4, , , Mass of oxygen 1.08 1, ratio is 4:1, , Experiment II, Weight of copper = 1.84 g, Weight of copper oxide = 2.30 g, , Weight of oxygen = Weight of copper oxide, Weight of copper,  2.30  1.84  0.46g, Ratio of copper and oxygen, Mass of copper 1.84 4, , , Mass of oxygen 0.46 1,  ratio is 4:1, As the ratio of copper and oxygen (by mass) is same in, the two experiments, law of constant proportion is, verified., , Thus, we see that hydrogen and oxygen combine in, the ratio of 1:8., Hydrogen and oxygen combine in the ratio of 1:8 by, mass to form water., Experiment II (Decomposition of water), • Take water from any source (say river, pond, lake,, tap water, well, etc.) and distill it to remove unwanted, salts., • Now take 90 g of pure water and pass electricity, through it to cause its decomposition. Collect the, gases so obtained., 3

Page 3 : 3., , Sol.:, , - Atoms of the same element are identical in all, respects, i.e., size, shape, mass and properties., - Atoms of different elements have different sizes and, masses and also possess different properties., - Atoms of the same or different elements combine, together to form molecules or compounds., - When atoms of different elements combine together, to form compounds, they do so in a simple whole, number ratio such as 1 : 1, 2 : 1, 2 : 3 etc., - Atoms of two different elements may combine in, different ratios to form more than one compound. For, example, carbon and oxygen may combine to form, carbon monoxide (CO) and carbon dioxide (CO2) in, which the ratios of the combining atoms (C and O) are, 1 : 1and 1 : 2 respectively., - The number and lend of atoms in a given compound, is always fixed., - Atom is the smallest particle that takes place in a, chemical reaction. In other words, whole atoms rather, than fractions of atoms take part in a chemical, reaction., - An atom can neither be created nor destroyed, i.e.,, atom is indestructible., , Copper oxide was prepared by two different, methods. In one case, 1.75 g of the metal, gave 2.19 g of oxide. In the second case, 1.14, g of the metal gage 1.43 g of the oxide. Show, that the given data illustrate the law of, constant proportions., In case I, mass of copper = 1.75 g, Mass of copper oxide = 2.19 g, % of copper in the oxide, Mass of copper, ,  100, Mass of copper oxide, , 1.75, 100  79.9%, 2.19,  % of oxygen  100  79.9  20.1%, In case II, mass of copper = 1.14 g, Mass of copper oxide = 1.43 g, 1.14, 100  79.7%, % of copper in the oxide, 1.43,  % oxygen  100  79.7  20.3%, Thus, copper oxide prepared by any of the given, methods contains copper and oxygen in the same, proportion by mass (within the experimental error.), Hence, it proves the law of constant proportions., , , (1), , 4., , Calculate the mass of carbon present in 2 g of, carbon dioxide., Sol.: Carbon dioxide (CO2 ) contains C and O in the, fixed proportion by mass, which is 12 : 32, i.e.,, 3 : 8. This means that 3 g of carbon combine with 8 g, of oxygen to form 11 g of CO2 . In other words,, 11 g of CO2 contains C  3 g, 3,  2 g of CO2 will contain C   2  0.545 g., 11, , (2), , (3), , Dalton's Atomic Theory, John Dalton provided the basic theory about the, nature of matter. He took the name atoms for the, smallest particle of matter as given by the Greek., Based on the laws of chemical combination, he, proposed a model of atom known as Dalton atomic, theory., The main postulates of the Dalton's atomic theory, are:, - All matter, whether an element, a compound or a, mixture is made up of extremely small particles called, atoms (i.e., same name was used for the smallest, indivisible particles as used by Greek philosophers)., , (4), , 5., , 4, , Drawbacks of Dalton's Atomic Theory, Atom is no longer considered as the smallest, indivisible particles. This is because recent, studies have shown that atom is made up of, still smaller particles called electrons, protons, and neutrons., Atoms of the same element may have, different masses. For example, there are two, types of atoms of chlorine with masses 35 and, 37 (on the atomic scale, called 'mass, numbers'). Such atoms of the same element, with different mass numbers are called, isotopes., Atoms of different elements may have same, masses. For example, atoms of potassium and, calcium are known with same mass number, (40). Such atoms of different elements with, same mass numbers are known as isobars., Substances made up of the same kind of, atoms may have different properties. For, example, charcoal, graphite and diamond are, all made up of carbon atoms but have, different physical properties., , Dalton postulated that “Atoms of same, element are same in all respects.” How was, this statement proved wrong?

Page 4 : Sol.:, , The discovery of isotopes i.e., atoms of same, element having same atomic number but, different mass number, proved the above, statement wrong., , the basis of some important property of the element., However, as more and more elements were, discovered, an International Committee was set up,, called International Union of Pure and Applied, Chemistry (IUPAC),which approved the names of the, different elements., The symbol of an element is the "first letter" or "first, letter and another letter" of the English name or, Latin name of the element. However, in all cases, the, first letter is always capital and another letter (if, added) is always a small letter., , What is an Atom?, The smallest unit of an element, which may or may, not exist independently, but always takes part in a, chemical reaction, is called an atom., Atoms are the building blocks of matter. They are, smaller than anything you can imagine or compare, with. More than a million atoms when stacked upon, one another would make a layer as thick as the sheet, of paper of your science book., The size of an atom is indicated by its radius which is, called 'atomic radius'. It is measured in 'nanometers'., One nanometer is one billionth part of a metre., 1 nanometre (nm)  109 metre (m) or 1 metre (m),  109 nanometre (nm), The table below gives the relative size of the radius of, an atom of hydrogen with respect toother objects., Species, Hydrogen atom, Molecule of water, Molecule of haemoglobin, Grain of sand, An ant, Watermelon, , Symbols of elements derived from English names, English name of the, Symbol, element, 1. Argon, Ar, 2. Arsenic, As, 3. Aluminium, Al, 4. Boron, B, 5. Barium, Ba, 6. Bromine, Br, 7. Carbon, C, 8. Calcium, Ca, 9. Chlorine, Cl, 10. Chromium, Cr, 11. Cobalt, Co, 12. Fluorine, F, 13. Hydrogen, H, 14. Helium, He, 15. Iodine, I, 16. Lithium, Li, 17. Magnesium, Mg, 18. Manganese, Mn, 19. Nitrogen, N, 20. Neon, Ne, 21. Nickel, Ni, 22. Oxygen, O, 23. Phosphorus, P, 24. Platinum, Ft, 25. Sulphur, S, 26. Uranium, U, 27. Zinc, Zn, , Atomic radius, 1010 m, 109 m, 108 m, 104 m, 102 m, 101 m, , Symbols Used to Represent Atoms of Different, Elements, In case of elements, symbol means a short method of, representing the full name of an element. Dalton was, the first scientist to suggest specific symbols in terms, of figures for different elements known at that time, (which were limited in number). In fact, the symbol, used by him also represented the quantity of the, element, i.e., one atom of the element. A few of these, symbols, as proposed by Dalton are given below:, , Symbols derived from Latin names, English name of Latin name of the Symbol, the element, element, 1. Antimony, Stibium, Sb, 2.Copper, Cuprum, Cu, 3. Gold, Aurum, Au, 4. Iron, Ferrum, Fe, 5.Lead, Plumbum, Pb, , Different elements have been named either after the, name of the place where they were first discovered or, after the name of the scientist who discovered it or on, 5

Page 5 : 6. Mercury, 7. Potassium, 8. Silver, 9.Sodium, 10. Tin, , Hydragyrum, Kalium, Argentum, Natrium, Stannum, , Hg, K, Ag, Na, Sn, , 23. Platinum, 24. Gold, 25. Lead, 26. Uranium, , It is the quantity in grams which is numerically equal, to the atomic mass of an element on a.m.u. scale., For example, 1 gram atom of hydrogen = 1.008 g,, 1 gram atom of sodium = 23 g, The relationship between gram atoms, mass and, atomic mass of a substance is,, Mass in grams, Number or gram atoms , Gram atomic mass, , Atoms are extremely small particles. Hence, the actual, masses of the atoms are so small that, it is difficult to, determine the actual masses of individual atoms., In 1961, IUPAC recommended the use of an isotope of, carbon with mass number 12 as the standard, reference for measuring atomic masses. It is called, carbon-twelve (C-12) and is represented as 12 C. The, atomic mass (or earlier called as "atomic weight") is, now defined as follows:, The atomic mass of an element is the relative mass, of its atoms as compared with the mass of an atom, of carbon-12 isotope taken as 12 units., 1, th of, Atomic mass unit (amu) may be defined as, 12, the mass of an atom of carbon-12 isotopeon the, 1, atomic scale, i.e., 1 amu  th of mass of C-12, 12, isotope., , 6., , Sol.:, , Atomic masses of some common elements (in amu or u)., , Symbol, H, He, Li, B, C, N, O, F, Ne, Na, Mg, Al, P, S, Cl, Ar, K, Ca, Fe, Cu, Zn, Ag, , 195, 197, 207, 238, , Gram Atomic Mass, , Atomic Mass, , Element, 1. Hydrogen, 2. Helium, 3. Lithium, 4. Boron, 5. Carbon, 6. Nitrogen, 7. Oxygen, 8. Fluorine, 9. Neon, 10. Sodium, 11. Magnesium, 12. Aluminium, 13. Phosphorus, 14. Sulphur, 15. Chlorine, 16. Argon, 17. Potassium, 18. Calcium, 19. Iron, 20. Copper, 21. Zinc, 22. Silver, , Pt, Au, Pb, U, , Atomic Mass, 1, 4, 7, 11, 12, 14, 16, 19, 20, 23, 24, 27, 31, 32, 35.5, 40, 39, 40, 56, 63.5, 65, 108, , Calculate the number of gram atoms in, (i) 640 g of sulphur, (ii) 360 g of magnesium, (i) Atomic mass of sulphur =32 amu,  One gram atom of sulphur =32g, Now, 32 g of stilphiir = 1 grain atom of sulphur, 1,  640 g of sulphur   640  20 g sulphur, 32, (ii) Atomic mass of magnesium = 24 amu,  One gram atom of magnesium = 24 g, Now 24 g of magnesium = 1 g atom, 1,  360 g   360  15 g, 24, , 7., , An atom of copper has a mass of, 1.0625  1022. How, many atoms of copper are, there in 16 g?, Soln.: In 1.0625 1022 g of copper, number of atoms, =1,  in 16 g of copper, number of atoms, 1, , 16  1.505 1023, 22, 1.0625 10, How do Atoms Exist?, The atoms of only a few elements called noble gases, (such as helium, neon, argon, etc) are chemically un, reactive and exist in free state or single atoms. Atoms, of most of the elements are chemically reactive and, do not exist in Free State or single atom. Atoms, usually exist in two ways:, (i) in the form of molecules., (ii) in the form of ions., 6

Page 6 : Molecules, , means that two atoms of oxygen combine together to, , A molecule is a group of two or more atoms which are, held together strongly by some kind of attractive, forces. Such an attractive force holding the atoms, together is called a chemical bond., We may also define a molecule as follows:, A molecule is the smallest particle of an element or a, compound which can exist freely under ordinary, conditions and shows all the properties of that, substance., , form a molecule. The formula of oxygen molecule is, O2 ., , Atomicity, The number of atoms present in one molecule of an, element is called its atomicity., The atomicity of an element is indicated by writing, the number as a subscript on the right hand side, bottom of the symbol., For example, H 2 shows that the atomicity of, hydrogen is 2. P4 shows that the atomicity of, phosphorus is 4, He shows that the atomicity of, helium is 1., On the basis of their atomicities, the elements may be, classified as monoatomic, diatomic, triatomic, tetra, atomic, etc., , Molecules of an Element, A molecule of an element consists of the same type of, atoms bonded together. For example, a molecule of, oxygen is formed when 2 atoms of oxygen combine, together., , Oxygen, , atom, , alone, , cannot, , exist, , independently. It exists as a diatomic molecule. This, S. No, 1., , Atomicity, 1, , 2, , 2, , 3., 4., 5., , 3, 4, More than 4, , Name of the Class, Monoatomic, , Diatomic, , Triatomic, Tetratomic, Polyatomic, , Examples, (i) Noble gas : Helium (He), Argon (Ar),, Neon (Ne), Krypton (Kr)., (ii)Metals (Exists as large clusters), Examples : Sodium (Na), Magnesium, (Mg), Aluminium (Al)., (iii) Carbon, (i) Hydrogen (H2) (ii) Oxygen (O2),, (iii) Chlorine (Cl2), (iv) Fluorine (F2),, (v) Nitrogen(N2), Ozone (O3), Phosphorus (P4), Sulphur (S8), Fullerenes (C60), , Molecules of Compounds, , proportion. The molecules of compounds may also be, diatomic, triatomic, tetra-atomic and polyatomic in, nature depending upon the number of the atoms, linked or combined by chemical bonds. For example,, , In the molecules of compounds, the atoms of, different elements are combined or bonded together, by chemical bonds. These are present in definite, proportion by mass according to law of constant, Compound, Hydrogen chloride (HCl), Water (H2O), Ammonia (NH3), Carbon dioxide (CO2), , Combining elements, Hydrogen, Chlorine, Hydrogen, oxygen, Hydrogen, nitrogen, Carbon, oxygen, , The molecules of elements are homoatomic in nature, which means that the atoms present in them are the, same. The molecules of the compounds are, heteroatomic in nature in the sense that different, , Nature, Diatomic, Triatomic, Tetra-atomic, Triatomic, , Ratio by mass, 1 : 35.5, 1:8, 14:3, 3:8, , atoms are present in them. For example, a molecule, of nitrogen element is homoatomic in nature (N2). A, molecule of methane (CH4) which is a compound is, heteroatomic in nature., 7

Page 7 : Ions, , Information Conveyed by a Chemical Formula, , An ion may be defined as an atom or group of atoms, having positive or negative charge., • The ion which has one or more positive charges is, called Cation. At the same time, the ion carrying one, or more negative charges is known as Anions., • A few positively charged ions or cations are:, Na+(sodium ion), 1C (potassium ion), Ca+ (calcium ion),, Al 3 (aluminium ion)., • A few negatively charged ions or anions are:, Cl  (chloride ion), S 2 (sulphide ion), (OH  ), , By looking at the chemical formula of a substance, we, can gather the following information’s:, 1. Name of the substance., 2. Name of various elements present in that, substance., 3. Chemical formula of a substance represents one, molecule of that substance., 4. Relative number of atoms of various elements, present in one molecule of that element or, compound., 5. Relative masses of various elements in the, compound., 6. It represents one mole of that substance., 7. We can calculate the gram molecular mass of that, substance., For example, the chemical formula of CO2 conveys, the following informations:, 1. The substance is carbon dioxide., 2. Carbon dioxide is composed of the elements;, carbon and oxygen., 3. In a molecule of carbon dioxide, carbon and, oxygen atoms combine together in the ratio1 : 2, by number., 4. The ratio of mass of carbon and oxygen is 3 : 8., 5. It represents one mole of carbon dioxide, molecules., 6. The molecular mass of carbon dioxide is 44 amu., , (hydroxide ion), ( SO42 ) (sulphate ion)., Molecular Formula of a Compound, The symbolic representation of a molecule of a, compound, is called its molecular formula., The molecule of a compound contains two or more, than two types of elements. If a molecule of a, compound contains two different kinds of elements, only, it is called binary compound., For example, water ( H 2 O ), sodium chloride ( NaCl ),, iron sulphide (FeS), etc. are binary compounds., To calculate the ratio of number of atoms in a, molecule of a compound., The ratio of number of atoms in a molecule of a, compound can be calculated, if we know the ratio of, their masses and the atomic masses of the elements,, constituting the molecule of the compound., , 8., , Writing Chemical Formula of Compounds, Chemical formula of a molecular compound, represents the actual number of atoms of different, elements present in one molecule of the compound,, e.g., chemical formula of water is H2O, that of, ammonia is NH3, while that of carbon dioxide is CO2, and so on., Chemical formula of an ionic compound simply, represents the ratio of the cations and anions present, in the structure of the compound, e.g., the formula, Na  Cl  simply represents that sodium chloride, contains Na  and Cl  ions in the ratio of 1 : 1, (though the actual crystal of sodium chloride consists, of very large but equal number of Na  ions and Cl , ions). Similarly, the formula Mg 2 Cl2 simply tells that, , The ratio of masses of carbon and oxygen is, 3:8 and their atomic masses are 12 and 16, respectively, in molecule of carbon dioxide., Calculate the ration of atoms of carbon and, oxygen in the molecule of carbon dioxide., , Sol.:, Element, , Ratio, by, Mass, , Atomic, Mass (u), , Mass ratio , Atomic Mass, , Simple, ratio, , C, 3, 12, 1, 3  12  0.25, 8  16  0.50, O, 8, 16, 2,  Ratio of atoms of carbon to oxygen in carbon, dioxide = 1 : 2, , it contains Mg 2 and Cl  ions in the ratio of 1 : 2 and, so on., , 8

Page 8 : Valency of an Element, Valency of an element is defined as the combining, capacity of the element. It is equal to the number of, hydrogen atoms or number of chlorine atoms or, double the number of oxygen atoms with which one, atom of the element combines., In addition to the atoms, the ions which are charged, species, also have some valencies. Positive ions or, cations have positive valencies. Negative ions or, anions have negative valencies. The valencies of the, poly-valent ions are expressed by enclosing them in, bracket and putting the positive or negative signs, outside it. Let us write the valencies of some, commonly used positive and negative ions., , Sodium Na , , Magnesium, Mg 2, , Iron Fe3, , Silver Ag , , Zinc Zn 2 , , Gold Au 3, , Copper Cu , Gold Au , Ammonium, ( NH 4 ), , Cobalt Co 2 , Copper Cu 2, Iron Fe2, , Some elements have shown more than one valencies., In such cases, Roman Numerals are used to denote, the valencies. These are put in bracket. For example,, copper (I) and copper (II);similarly, iron (II) and iron, (III)., Valencies of Negative Ions, , Valencies of Positive Ions, Positive ions may be monovalent, bivalent, trivalent,, tetravalent etc. depending upon the charge present, on them. These are listed in the following table., List of some common positive ions (Cations), Monovalent, Bivalent, Trivalent, 2, , Hydrogen H Barium Ba Aluminium Al 3, Potassium K  Calcium Ca 2  Chromium Cr 3, , Valencies of Negative Ions, Like positive ions, negative ions may also be, monovalent, bivalent, trivalent, etc. in nature. These, are also listed;, , List of some common negative ions (Anions), Monovalent, Bivalent, Trivalent, , 2, S, Chloride, Sulphide, Nitride, Cl, N 3, Oxide, O 2, Bromide, Phosphide, Br , P3, Iodide, I, Carbonate (CO3 )2, Phosphate ( PO4 )3, Hydroxide, , (OH ), , Sulphate ( SO4 )2, , Borate ( BO3 )3, , Nitrate, , ( NO3 ), , Sulphite (SO3 )2, , Arsenate ( AsO3 )3, , Nitrite, , ( NO2 ), , Manganate (MnO4 )2, , Bicarbonate, , ( HCO3 ) Oxalate (C2O4 )2, , Cyanide, , (CN ), , Chromate (CrO4 )2, , Permanganate ( MnO4 ) Dichromate (Cr2O7 )2, Chlorate, , (ClO3 ), , Formulae of Ionic Compounds, , Step I: Write the symbol of the cation showing the, charge on it. Write the symbol of the anion showing, the charge on it, on the right hand side of the cation., Step II: If a compound contains polyatomic ions, the, formula of the ion is enclosed within brackets before, crisscrossing the valencies., Step III: Divide the valency number by common, factor, if any, to get simple ratio. Now ignore the (+), and (-) symbols., , One of the most important points to remember while, writing the formula of a chemical compound is that it, is always electrically neutral. In other words, the, positive and negative valencies of the ions present in, the chemical compound add up to zero. To write a, formula, follow the steps given below. This method of, writing formula is called the crisscross method., 9

Page 9 : Step IV: Now, write the valency of the cation at the, bottom right of the anion and the charge number of, the anion at the bottom of the cation. Thus, the, symbol of cation is subscribed with the charge, number of the anion and the anion is subscribed with, the charge number of the cation. This is called the, criss-crossing of valencies., , Step V: If these subscripts are 1, these are not written, in the final stoichiometric formulae., Examples:, , Compound, Calcium chloride, , Symbols with Valency, Ca 2 Cl , , Magnesium sulphate, , Mg 2 SO42 or Mg1SO41, (Dividing by C.F .  2 ), , MgSO4, , Aluminium sulphate, , Al 3 SO4 2, , Al2 ( SO4 )3, , Aluminium phosphate, , Al 3 PO43 or Al1PO41, (Dividing by C.F .  3 ), , AlPO4, , Ammounium phosphate, , NH 43 PO43, , ( NH 4 )3 PO4, , Potassium dichromate, , K 1 Cr2O7 2, , K 2Cr2O7, , Ca 2 HCO31, , Ca ( HCO3 ) 2, , Calcium bicarbonate, , Shifting Valency number, , Formula, CaCl2, , Formulae of Molecular Compounds, , 3. Formula of ammonia, , While writing the chemical formulae of the molecular, compounds, we write the constituent elements and, their valencies as shown below. Then, we crossover, the valencies of the combining atoms., , Formula of ammonia is NH3., 4. Formula of carbon tetrachloride, , 1. Formula of hydrogen chloride, , Formula of carbon tetrachloride is CCl4 ., Formula of hydrogen chloride is HCl., , Molecular Formulae of Important Acids and Bases, , 2. Formula of hydrogen sulphide, , Molecular Formulae of Important Acids, Acid, Molecular, Acid, Molecular, Formula, Formula, Hydrochloric HCl, Acetic acid CH 2COOH, , Formula of hydrogen sulphide is H2S., 10

Page 11 : nitrate, Zinc nitrate, , nitrate, , chloride, Magnesium, chloride, Zinc chloride, , Zn( NO3 ) 2, , Molecular Formulae of Chlorides, Chloride, Molecular Chloride Molecular, Formula, Formula, Ammonium, Iron, (II) FeCl2, NH 4 Cl, chloride, chloride, Sodium, Iron, (III) FeCl3, NaCl, chloride, chloride, Potassium, Lead, (II) PbCl2, KCl, chloride, chloride, Calcium, Tin, (II) SnCl2, CaCl2, , Aluminium, chloride, , MgCl2, ZnCl2, AlCl3, , Molecular Formulae of Sulphates, Sulphate, Molecular, Sulphate, Formula, Ammonium sulphate, ( NH 4 ) 2 SO4 Iron (11) sulphate, Sodium sulphate, Iron (III) sulphate, Na2 SO4, Potassium sulphate, Lead (II) sulphate, K 2 SO4, Calcium sulphate, Tin (II) sulphate, CaSO4, Magnesium sulphate, Copper (II) sulphate, MgSO4, Aluminium sulphate, Mercury (II) sulphate, Al2 ( SO4 )3, Zinc sulphate, Silver (I) sulphate, ZnSO4, , Molecular mass of a substance (element or, compound) is the average relative mass of its, molecules as compared with that of an atom of C-12, isotope taken as 12. In other wards, molecular mass, of a substance represent the number of times the, molecule of that substance is heavier than 1/12th of, the mass of an atom of C-12 isotope., Calculation of Molecular Mass, , In case of the compounds formed by ions (ionic, compounds) formula of the compound does not, represent its molecule, but only represents the ratio, of different ions in the compounds., This is called formula unit of the ionic compound., Formula mass of an ionic compounds is obtained by, adding atomic masses of all the atoms in a formula, unit of the compound., , As molecules are made up of two or more atoms of, the same or different elements, and each atom has a, definite atomic mass, therefore, molecular mass of a, molecule of a substance can be calculated by adding, atomic masses of all the atoms present in one, molecule of the substance., , mass, , Miolecular, Formula, FeSO4, Fe2 ( SO4 )3, PbSO4, SnSO4, CuSO4, HgSO4, Ag 2 SO4, , Similarly, molecular mass of CO2  atomic mass of, C  2 (atomic mass of oxygen),  12  2(16)  12  32  44 a.m.u., Other examples, Molecular mass of H 2  2  atomic mass of hydrogen,  2(1)  2 a.m.u., Molecular mass of NH 3  atomic mass of nitrogen + 3, (atomic mass of hydrogen),  14  3(1)  17 a.m.u., Formula Mass, , Molecular Mass, , For example,, Molecular mass of H 2O  2 (atomic, hydrogen) + atomic mass of oxygen,  2(1)  16  18 a.m.u., , chloride, Copper (II) CuCl2, chloride, Mercury, HgCl2, (II) chloride, Silver, (I) AgCl, chloride, , of, , 12

Page 12 : For Example,, Formula mass of calcium oxide (CaO)  atomic mass, of calcium + atomic mass of oxygen, , Sol.:, ,  40  16  56 u, , Gram Atomic Mass and Gram Molecular Mass, , Therefore, the valency of calcium (i.e., metal), is 2., (b) Na2 O has been formed by, , Atomic mass expressed in grams is called gram atomic, mass of that element., For example,, Atomic mass of hydrogen = 1.0 u, Gram atomic mass of hydrogen = 1.0 g, Atomic mass of oxygen = 16.0 u, Gram atomic mass of oxygen = 16.0 g, , So, the valency of sodium (i.e., metal) is 1., (c) Fe2 O3 has been formed by, , The amount of an element having mass equal to gram, atomic mass is called one "gram atom" (or g atom) of, the element., For example,, 1 g atom of hydrogen = 1.0 g, 1 g atom of oxygen = 16.0 g, , Therefore, the valency of iron (i.e., metal) is 3., 11., , Molecular mass expressed in grams is called gram, molecular mass of that substance., For example, Molecular mass of H 2  2.0 u, Gram molecular mass of H 2  2.0 g, Molecular mass of O2  32.0 u, Gram molecular mass of O2  32.0 g, , Sol.:, , Find the molecular mass of the following:, (a) O2, (b) Cl2, C, H, (c) 2 6, (d) PCl3, [Atomic masses: O  16, C  12,, Cl  35.5, H  1, P  31 ], O2  2 (atomic mass of oxygen), Sol.:,  2(16)  32u, (b) Cl2  2 (atomic mass of chlorine),  2(35.5)  71u, (c) C2 H 6  2 (atomic mass of C) +6 (atomic mass of, hydrogen),  2(12)  6(1)  24  6  30u, (d) PCl3  atomic mass of phosphorus + 3 (atomic, mass of chlorine),  31  3(35.5), , Sol.:, , 10., , Write the formula of compounds formed by:, (a) Ag  and S 2, (b), (c), (a), (b), (c), , Write the formula for the following:, (a) Ferric chloride, (b) Ferrous sulphate, (c) Cupric iodide, (d) Cupric oxide, (e) Ammonium hydroxide, (f) Calcium phosphate, (a) FeCl3, (b) FeSO4, (c) Cul2, (d) CuO, (d) NH 4OH, (f) Ca3 ( PO4 )2, , 12., , The amount of the substance having mass equal to its, gram molecular mass is called one "gram molecule", (or g molecule) of the substance. Thus,, 1 g molecule of H 2  2.0 g, 1 g molecule of O2  32.0 g, , 9., , (b) Na2 O, (c) Fe2 O3, (a) CaCl2 has been formed by, , Cr 3 and O 2 , Al 3 and SO42, Ag 2 S, Cr2 O3, Al2 ( SO4 )3, ,  31  106.5  137.5u, , What is the valency of the metal in the, following formula?, (a) CaCl2, , 13., Find the formula mass of the following:, (a) ZnO, (b) Na2 O, (c) CuSO4 .5H 2O, [Atomic masses: Zn  65, O  16,, 13

Page 13 : Na  23, Cu  63.5, H  1 ], Sol.: (a) ZnO  atomic mass of Zn + atomic mass of, oxygen,  65  16  81 u, (b) Na2O  2 (atomic mass of sodium) + atomic mass, of oxygen,  2(23)  16  46  16  62 u, (c) CuSO4 .5H 2O  atomic mass of Cu  atomic mass, of S  4 (atomic mass of O) + 5 (molecular mass of, H 2O ),  63.5  32  4(16)  5(18),  249.5 u, , 14., , Sol.:, , Calculate the mass percentage composition of, copper in copper sulphate., [Cu  64, S  32, O  16], Unit formula mass of CuSO4,  1(Cu)  1( S )  4(O)  1(64)  1(32)  4(16),  64  32  64  160 amu., ,  Mass % age composition of copper, 64 100, ,  40%, 160, 15., 4.9 g of sulphuric acid contains 0.1 g of, hydrogen, 1.6 g of sulphur and rest oxygen., Calculate the mass percentage composition of, all the elements of sulphuric acid., Soln.: Mass of sulphuric acid (W) = 4.9 g, Mass of hydrogen (1 )  0.1 g, Mass of sulphur (2 )  1.6 g,  Mass percentage of hydrogen, w, 0.1,  1 100 , 100  2.04%, W, 4.9, Mass percentage of sulphur, w, 1.6,  2  100 ,  100  32.65%, W, 4.9,  Mass percentage of oxygen,  100  (2.04  32.65)  65.31%, 16., Calculate the mass of 0.72 gram molecule of, carbon dioxide (CO2 )., Sol.: Molecular mass of CO2  Atomic mass of, Atomic, mass, of, C  2, O  (12 u  2 16 u)  44 u Gram molecular, mass of CO2  44 g 1 gram molecule, , Mass Percentage Composition of an Element in a, Compound, The percentage composition of an element in a, compound can be determined, if we know the, molecular mass or unit formula mass and the atomic, weight of the element., If M is the molecular mass (or unit formula mass) of a, compound and A, the atomic mass of its atoms for, some particular element then, Mass percentage composition of the element =, Sum of atomic masses of the atoms of element, A,  100  100, Molecular mass, M, , However, if molecular mass and atomic mass are not, given, instead given the mass of compound and mass, of element in it, then percentage composition can be, calculated as follows:, Let W be the mass of a compound which contain  as, the mass of an element.,  Mass percentage composition or the element =, Mass element, w,  100   100, Mass of compound, W, , CO2  44 g, 0.72 gram molecule of CO2, , , , , , , To convert atomic mass into gram atomic mass, and to convert molecular mass in gram molecular, mass, we have to simply replace 'u' by 'g'., The masses of hydrogen and oxygen in water can, be determined by electrolyzing water. The, volumes of hydrogen and oxygen obtained during, electrolysis are measured, and their masses, calculated., , (44),  (0.72)  31.68 g, (1), , Mole Concept, Quite commonly, we use different units for counting, such as dozen for 12 articles, score for 20 articles and, gross for 144 articles irrespective of their nature., In a similar way, chemists use the unit 'mole' for, counting atoms, molecules ions, etc.. A mole is a, collection, of 6.022 1023 particles. Thus:, A mole represents 6.022 1023 particles., , 14

Page 14 : The number 6.022 1023 is called Avogadro number, and is symbolized as N A or N 0 . In other words, a, mole is an Avogadro number of particles. For example, 1 mole of hydrogen atoms  6.022  1023 hydrogen, atoms., 1 mole of hydrogen molecules  6.022  1023, hydrogen molecules., 1 mole of sodium ions  6.022  1023 sodium ions., 1 mole of electrons  6.022  1023 electrons., The amount of substance that contains the same, number of entities (atoms, molecules, ions or other, particles), as the number of atoms present in 0.012 kg, (or 12 g) of the carbon-12 isotope., , 19., Sol.:, , Mole in Terms of mass, A mole of atoms is defined as that amount of the, substance (element) which has mass equal to gram, atomic mass (i.e., atomic mass expressed in grams). In, other words, it is equal to one gram atom of the, element., A mole of molecules is defined as that amount of the, substance (element or compound) which has mass, equal to gram molecular mass (i.e., molecular mass, expressed in grams).In other words, it is equal to one, gram molecule of the substance. For example,, If the number of chemical entities of a substance is, known, then the number of moles of that substance, can be calculated as follows,, Number of moles , , 17., Sol.:, , 18., Sol.:, , 44, 1021  0.073 g, 23, 6.022 10, Mass of CO2 left  (0.2  0.073)  0.127 g, 0.127 g, Moles of CO2 left , 44 g, 3, So, 2.8 10 moles of CO2 are left., How many electrons are present in 1.6 g of, methane?, Gram-molecular mass of methane (CH4),  12  4  16 g, Number of moles in 1.6 g of methane, 1.6, ,  0.1, 16, Number of molecules of methane in 0.1 mole,  0.1 6.022  1023,  6.022 1022, One molecule of methane has  6  4  10, electrons, So, 6.022 1022 molecules of methane have,  10  6.022 1022 electrons,  6.022  1023 electrons, Calculate the number of moles in, 3.0115  1023 atoms of calcium,, 1 mole of any substance contains 6.022 1023, particles., , 6.022 1023 atoms of calcium = 1, mole, 1, 1 atom of calcium , moles, 6.022  1023, 3.0115  1023 atoms of calcium, 1, ,  3.0115 1023, 23, 6.022 10,  0.5 mole, Calculate the mass of, (a) 0.2 mol molecules of oxygen, (b) one atom of aluminium, (c) 3.0 mol of Cl ion, (d) 2.5 mol of NaCl, (Atomic masses:, O  16 u, Cl  35.5 u, Na  23 u. Al  27 u ), (a) Molecular mass of oxygen (O2 )  32 u, , , 20., Sol.:, , Number of chemical entities of a subs tan ce ( N ), N, or n , 6.022  1023 ( Avogadro ' s Number , N A ), NA, , 21., , What is the number of molecules in 0.25, moles of oxygen?, We know that:, 1 mole of oxygen contains,  6.022  1023 molecules, So, 0.25 mole of oxygen contains,  6.022 1023  0.25  1.505 1023 molecules, Thus, 0.25 mole of oxygen contains, 1.505 1023 molecules., From 200 me, of CO2 , 1021 molecules are, removed. How many moles of CO2 are left?, Gram-molecular mass of CO2  44 g, , Sol.:, ,  Molar mass of O2  32 g mol 1, Now, 1 .mole of O2 molecule = 32 g,  0.2 mole of O2 molecule  32  0.2  6.4 g, (b) Atomic mass ( Al )  27 u,  Molar mass  27 g mol 1, , Mass of 1021 molecules of CO2, , Now 6.022  1023 atoms of aluminium weigh, 15

Page 15 :  27 g,  1 atom of aluminium will weigh, 27, ,  4.48  1023 g, 23, 6.022 10, (c) Formula mass of Cl ion  35.5 u, , Molar mass of iron ( Fe)  56 g mol 1, Molar mass of magnesium (Mg )  24 g mol 1, Molar mass of H 2  2.0 g mol 1, Molar mass of O2  32.0 g mol 1, Molar mass of H 2O  18.0 g mol 1, ,  Molar mass of Cl ions  35.5 g mol 1, Number of moles = 3 (given), m( Mass), Now, n , M ( Molar mass),  m  n  M  3  35.5  106.5 g, (d) Formula mass of NaCl  23  35.5  58.5 u, , Molar mass of NH3  17.0 g mol 1, Molar mass is usually represented by the symbol ‘M’., Thus, M ( H 2O)  18 g mol 1 and so on., The volume occupied by one mole of any gas at STP is, always same and equal to 22400 mL or 22.4 L or, 22.4 dm3 . This volume is called molar volume or gram, molecular volume (G.M.V.)., ,  Molar mass of NaCl (M )  58.5 g mol 1, Number of moles = 2.5 (given), m, Now, n , M, , Mole concept for Ionic Compounds, ,  m  n M,  2.5  58.5, , Ionic compounds (like NaCl , KNO3 , etc.) do not, consist of molecules. They consist of a cluster of ions., In case of ionic compounds, a mole is defined as, follows:, (1), A mole of an ionic compound is that amount, of the substance which has mass equal to, gram formula unit mass, i.e., formula unit, mass of the ionic compound expressed in, grams., (2), A mole is defined as that amount of the, substance which contains Avogadro's number, of formula units., e.g., 1 mole of sodium chloride ( NaCl ) , Gram formula unit mass of NaCl,  23  35.5 g  58.5 g, ,  146.25 g, , Mole in Terms of Volume, In case of gaseous substances, it is found that, Avogadro's number of molecules (i.e., one mole of, molecules) of any gas under standard conditions of, temperature and pressure, i.e., STP conditions (0°C, and one atmospheric pressure) occupy the same, volume which is equal to 22400 mL or 22.4 litres., Hence,, A mole of a gaseous substance is defined as that, amount of the substance which has volume equal to, 22400 mL at STP conditions., Thus, 1 mole of CO2 gas = 22400 mL of CO2 at STP, 1 mole of NH3 gas = 22400 mL of NH3 at STF, 1 mole of SO2 gas = 22400 mL of SO2 at STP, , = Avogadro's number (6.022 1023 ) of formula units, of NaCl,  6.022 1023 Na  ions  6.022 1023 Cl ions, Similarly,, 1, mole, of, potassium, nitrate, ( KNO3 )  39  14  3  16 g,  101 g, , Molar mass and Molar volume, The mass of 1 mole of the substance (i.e., Avogadro's, number of particles) is called molar mass of that, substance. If the substance is atomic, its molar mass is, equal to gram atomic mass. If the substance is, molecular, its molar mass is equal to gram molecular, mass. As it is mass of one mole of the substance, its, units are gram per mole ( g mol 1 ) or kilogram per, mole (kg mol 1 )., Thus,, Molar, mass, 1, , of, ,  12 g mol or 0.012 kg mol, , C-12, ,  6.022  1023 formula units of KNO,, ,  6.022 1023 K ions  6.022 1023 NO3ions,  6.022 1023 K atoms  6.022 1023 N atoms, 3  6.022 1023 O atoms, Combining all the definitions of mole, we arrive at the, following results:, 1 mole of atoms = Gram atomic mass  6.022  1023, atoms, 1 mole of molecules = Gram molecular mass,  6.022  1023 molecules, = 22.4 L at STP if the substance is a gas., , isotope, , 1, , 16

Page 16 : 1 mole of an ionic compound = Gram formula unit, mass,  6.022  1023 formula units., Relationship between mole, Avogadro, , number and mass., , 1 mole of carbon, atoms, , 12 g of carbon, , atoms of C, , 1 mole of hydrogen, atoms, , 1 g of hydrogen, , atoms of H, , 1 mole of any particle, (atoms, molecules ion), , Relative mass of those, particles in grams, , number of, that particle, , 1 mole of, molecules, , Molecular mass in, grams, , number of, that particle, , Volume of gas, at STP, , Recapitulation, Divide by 22.4 L, , Multiply by 22.4 L, Multiply by, Avogadro’s number, Number of, moles (n), , Number of, molecules (N), , Divide by, Avogadro’s number, , Multiple by molar, mass (M), , Divide by molar mass (M), , Mass in, grams, , 17

Page 17 : CONCEPT MAP, Atoms Molecules, Ions, , Atoms:, Smallest unit of an, element, , Atomic mass, , Atomicity: The number of, atoms that a molecule of, an element contains., , Mole concept: A mole is, the amount of the, substance which, contains the same, number of particles as, there are, , atoms, , in 12 g of, isotope, , Relative Atomic, Mass: The mass, of one atom of, the element, relative to 1/12th, the mass of 1, atom of the, element C-12, , Hydrogen Scale:, The relative, atomic mass of an, element is the, number of times, an atom of the, element of the, element is, heavier than an, atom of hydrogen, , Gram Atomic, Mass:, The atomic mass of, an element, expressed in grams., , C-12 Scale:, The relative atomic, mass of an element, is the ratio of the, mass of an atom of, the element to, 1/12th the mass of, an atom C-12, , Molar volume: The, volume occupied by 1, mole of any gas at STP, = 22400 mL, , Relative, Molecular Mass:, , The mass of a, molecule of the, substance as, compared to, 1/12th the, mass of an, atom of C-12, , 18, , Gram, Molecular, Mass:, , Molecular mass, expressed in, grams, , Molecules: A, group of two, more atoms, , Ions: An atom, having +ve or, –ve charge, , Mass, , Formula Mass:, , The sum of, atomic, masses of all, atoms in a, formula unit, of the, compound, , Molecular, mass

Page 18 : Mole Concept for ionic Compounds, •, , Number of molecules, No. of particles ( N ), ( n) , Avogadro ' s number ( N0 ), N, , n, N0, Also Number of Moles, Mass of the subs tan ce (m), ( n) , Molar mass( M ), , , , n, , m, M, , From (1) and (2) n , , Law of Reciprocal Proportions, This law was given by Richter in 1794. The law states, that when definite mass of an element A combines, with two other elements B and C to form two, compounds and if B and C also combine to form a, compound, their combining masses are in same, proportion or bear a simple ratio to the masses of B, and C which combine with a constant mass of A., , …(1), , …(2), N, m, , N0 M, , - For example, hydrogen combines with sodium and, chlorine to form compounds NaH and HCl,, respectively., , m,  N0, M, Number of molecules, N, , , , , , In NaH ,, In HCl,, , Mass of the subs tan ce,  N0, Molar mass, , Sodium 23 parts, Hydrogen one part, Chlorine 35.5 parts Hydrogen one part, , - Sodium and chlorine also combine to form NaCl in, which 23 parts of sodium and 35.5 parts of chlorine, are present. These are the same parts which combine, with one part of hydrogen in NaH and HCl, respectively., , Law of Multiple Proportion, This law was put forward by Dalton in 1808. According, to this law if two elements combine to form more, than one compound, then the different masses of one, element which combine with a fixed mass of the other, element, bear a simple ratio to one another., Hydrogen and oxygen combine to form two, compounds H2 (water) and H2O2 (hydrogen peroxide), , - Hydrogen combines with sulphur and oxygen to, form compounds H 2 S and H 2 O respectively., In H 2 S , Hydrogen 2 parts, Sulphur 32 parts, In H 2 O , Hydrogen 2 parts Oxygen 16 parts, , In water,, Hydrogen 2 parts Oxygen 16 parts, In hydrogen peroxide, Hydrogen 2 parts Oxygen 32 parts, , - The masses of oxygen which combine with same, mass of hydrogen in these two compounds bear a, simple ratio 1:2., Nitrogen forms five stable oxides., N 2 O Nitrogen 28 parts Oxygen 16 parts, N 2 O2 Nitrogen 28 parts Oxygen 32 parts, N 2O3 Nitrogen 28 parts Oxygen 48 parts, N 2 O4 Nitrogen 28 parts Oxygen 64 parts, N 2O5 Nitrogen 28 parts Oxygen 80 parts, , Thus, according to this law, sulphur should combine, with oxygen in the ratio of 32 : 16 or a simple multiple, of it., Actually, both combine to form SO2 in the ratio of 32, : 32 or 1 : 1., - The law of reciprocal proportions is a special case of, a more general law, the law of equivalent masses,, which can be stated as under:, "In all chemical reactions, substances always react in, the ratio of their equivalent masses.", , - The masses of oxygen which combine with same, mass of nitrogen in the five compounds bear a ratio, 16 : 32 : 48 : 64 : 80 or 1 : 2 : 3 : 4 : 5., , Law of Gaseous Volumes, This law was enunciated by Gay-Lussac in 1808., According to this law, gases react with each other in, the simple ratio of their volumes and if the product is, 19

Page 19 : also in gaseous state, the volume of the product also, bears a simple ratio with the volumes of gaseous, reactants when all volumes are measured under, similar conditions of temperature and pressure., H 2  Cl2  2 HCl, ratio 1 : 1 : 2, 1 vol 1 vol 2 vol, 2 H 2  O2  2 H 2O, ratio 2 : 1 : 2, 2 vol 1 vol 2 vol, 2CO  O2  2CO2, ratio 2 : 1 : 2, 2 vol 1 vol 2 vol, N 2  3H 2  2 NH 3, ratio 1 : 3 : 2, 1 vol 3 vol 2 vol, , Structural formula, It represents the way in which atoms of various, elements present in the molecule are linked with each, other. For example, ammonia is represented as, , The formula indicates that three hydrogen atoms are, linked to one nitrogen atom by three single covalent, bonds., Determination of Empirical and Molecular Formulae, The following steps are followed to determine the, empirical formula of the compound:, - The percentage composition of the compound is, determined by quantitative analysis., - The percentage of each element is divided by its, atomic mass. It gives atomic ratio of the elements, present in the compound., - The atomic ratio of each element is divided by the, minimum value of atomic ratio so as to get the, simplest ratio of the atoms of elements present in the, compound., - If the simplest ratio is fractional, then values of, simplest ratio of each element is multiplied by a, smallest integer to get a simplest whole number for, each of the element., - To get the empirical formula, symbols of various, elements present are written side by side with their, respective whole number ratio as a subscript to the, lower right hand corner of the symbol., - The molecular formula of a substance may be, determined from the empirical formula if the, molecular mass of the substance is known. The, molecular formula is always a simple multiple of, empirical formula and the value of simple multiple is, obtained by dividing molecular mass with empirical, formula mass., Example: A compound of carbon, hydrogen and, nitrogen contains these elements in the ratio 9:1:3.5., Calculate the empirical formula. If its molecular mass, is 108, what is the molecular formula?, , Empirical formula, It represents the simplest relative whole number ratio, of atoms of each element present in the molecule of, the substance. For example, CH is the empirical, formula of benzene in which ratio of the atoms of, carbon and hydrogen is 1 : 1. It also indicates that the, ratio of carbon and hydrogen is 12 : 1 by mass., Molecular formula, Molecular formula of a compound is one which, expressed as the actual number of atoms of each, element present in one molecule. C6 H 6 is the, molecular formula of benzene indicating that six, carbon atoms and six hydrogen atoms are present in, the molecule of benzene., Thus, Molecular formula  n  Empirical formula, Molecular formula mass, where, n , Empirical formula mass, Molecular formula gives the following information’s:, - Various elements present in the molecule., - Number of atoms of various elements in the, molecule., - Mass ratio of the elements present in the molecule., e.g. The mass ratio of carbon and oxygen in CO2, molecule is 12 : 32 or 3 : 8., - Molecular mass of the substance., - The number written before the formula indicates the, number of molecules, e.g., 2CO2 means 2 molecules, of carbon dioxide., , 20

Page 20 : Soln.:, Element, , Element Atomic, ratio, mass, , Carbon, , 9, , 12, , Hydrogen, , 1, , 1, , Nitrogen, , 3.5, , 14, , Relative, number of, atoms, 9,  0.75, 12, 1, 1, 1, 3.5,  0.25, 14, , Simplest, ratio, , 0.75, 3, 0.25, 1, 4, 0.25, 0.25, 1, 0.25, , Empirical formula  C3 H 4 N, Empirical formula mass  (3 12)  (4 1)  14  54, Mol. mass 108, n, , 2, Emp. mass 54, Thus, molecular formula of the compound,  2  empirical formula,  2  C3 H 4 N  C6 H 8 N 2, , 21