BASIC CONCEPT OF CHEMISTRY - Lecture Notes

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 1 BASIC CONCEPTS OF CHEMISTRY Chemistry The branch of science which deals with the study of matter, its composition and changes in composition during various physical and chemical processes. Matter Anything which can occupy space and has mass is called matter Classification of matter according to the chemical nature I. Elements They are pure substances containing only a single type of atoms. On the basis of physical and chemical properties, elements are subdivided into metals, non metals & metalloids. II. Compounds They are formed by the combination of 2 or more elements in a fixed proportion by mass. The properties of compounds are entirely different from the corresponding elements. eg : Organic compounds such as, alcohols, hydrocarbons etc. Inorganic compounds such as copper sulphate, CO , NH etc. 2 3 III. Mixtures They are formed by the combination of 2 or more elements or compounds without any fixed mass ratio. The components retain their respective identity in a mixture 1) Homogenous mixtures Characteristics of homogenous mixture i) The components have uniform composition throughout the mixture ii) Components are in same phase in the mixture iii) The components are indistinguishable with naked eye or even with microscopes

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 2 All homogenous mixtures are called as solutions eg : 1. Rectified spirit (alcohol 95 % + water 5 %) 2. Gasoline (mixture of hydrocarbons) 3. Pure air 4. German silver [56 % - Cu + 24 % - Zn + 20 % - Ni] 5. 22 gold [91.6 % Au + 8.4 % (Cu + Cd + Cr)] ct 2) Heterogeneous mixture Characteristics of heterogeneous mixture i) Components have different compositions at different portion of the mixtures ii) The components generally are in different phases iii) Components are distinguishable with naked eye or with microscopes eg : 1. mixture of oil and water 2. milk, blood etc. (components are distinguishable with microscopes) 3. Iodised table salt (NaCl + KI + NaIO ) (components are also distinguishable with microscopes) 3 Laws of chemical combination 1. Law of conservation of mass (proposed by lavoisier) The total mass of reactants are equal to the total mass of products during any physical or chemical change. ex : 3 2 56g 44g 100g CaCO CaO CO    This law can also be stated as “matter can neither be created nor be destroyed, but can be transformed for one form to another.” Therefore, this law is also called as law of indestructibility. An exception for this law is nuclear reactions. Here, a par of the mass of reactants are converted into energy. 2. Law of constant or definite proportions (proposed by Joseph Proust) Same compounds always contain same elements combined in a fixed proportion by mass. ex : Pure water from any source contains H and O in 1.8 mass ratio. 2 2 Exceptions : 1. Compounds containing isotopes : In the case of compounds containing isotopes, 2 or more same compounds have same elements in different mass ratios. 12 14 2 2 CO CO 14:32 12:32

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 3 2. Isomers : In the case of isomers, 2 or more different compounds have same elements in same mass ratio. 3 2 2 6 2 2 3 3 CH CH OH C H O C: H :O 24:6:16 CH O CH          Qn: A sample of CaCO has the following percentage composition Ca- 40%, C - 12%, O - 48%. If law of 3 2 constant proportion is true, mass of carbon in another sample of CaCO having mass 12 gm is 3 1) 2.88 gm 2) 1.44 gm 3) 1.94 gm 4) 2.34 gm According to the law of constant proportions, the second sample of CaCO will also contain 12 % of 3 carbon. Therefore the required answer is 12% of 12 gm. Therefore mass of carbon = 12 12 1.44gm 100   Law of multiple proportions (proposed by Dalton) When 2 elements combine together to form more than one compound, then the ratio of mass of the element which combined with a fixed mass of the other, bear a simple whole number ratio. ex : 2 CO & CO 12:32 12:16 Here, we can fix the mass of carbon as 12 g, then the ratio of masses of oxygen that combine with 12g of C = 16 : 32 = 1 : 2 ex : 2 3 2 4 2 5 2 N O , N O & N O N O, NO, 28:80 28:16 14:16 28: 48 28:64 Here, we can fix the mass of nitrogen as 14 g, then the ratio of masses of oxygen that combine with 14g of N = 8 : 16 : 24 : 32 : 40 = 1 : 2 : 3 : 4 : 5 2 Law of reciprocal proportions (proposed by Ritcher) The ratio of the masses of the 2 elements, A and B which combined with a fixed mass of a third element C is either the same or a simple multiple of the ratios of masses of A & B which combined with each other. ex : HCl & NaH NaCl 23:1 23:35.5 1:35.5 Ratio masses of Na & H combined with 35.5 g of Cl = 23 : 1 2 2 Ratio of mass of Na & H in NaH = 23 : 1 2 ex : 2 2 2 SO & CO CS 12:32 6:32 32:32

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 4 Ratio of masses of C & O combined with 32 g of S = 6 : 32 2 ....(a) Ratio of mass of C & O in CO = 12 : 32 2 2 ....(b) a & b are multiplies Qn2: Which of the following compounds will obey the law of reciprocal proportions ? i) CO , SO & CS 2 2 2 ii) H O, CH & H O 2 4 2 2 iii) CH , H O & CO 4 2 2 iv) CO, SO & CO 2 2 v) H O, H S & H O 2 2 2 2 vi) H S, H O & SO 2 2 2 Ans : i, iii, vi Gay Lussac’s law of combining volumes When reactants in the gaseous state, reacts together to products in their gaseous state, volume reactants to products bear a simple whole no. ratio under identical conditions of temperature and pressure.   2(g) 2(g) 3(g) 2(g) (g) 2 g N 3H 2NH H Cl HCl vol. 2 : 1 : 3 2 1 : 1 : ratio        Therefore, it is clear that the volume ratios of gaseous reactants and products are equivalent to the mole ratios under identical conditions of temperature & pressure. Atoms and Molecules Atoms are the smallest particles of elements. Atoms may or may not have independent existence. ex : H, N, O, He, Ne,....... Molecules are the smallest particles of compounds or some types of elements, with independent existence. ex : CO , NH , H , N , O ,...... 2 3 2 2 2 Homoatomic molecules They contain only a single type of atoms. ex : H , N , O , ...... 2 2 2 Heteroatomic molecules They contain more than one type of atoms. ex : CO , NH ..... 2 3

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 5 Atomicity of a molecule The total number of atoms present in a molecule is called its atomicity. ex : Atomicity of O = 2 2 Atomicity of CO = 3 2 Atomicity of NH = 4 3 Atomicity of CaCO = 5 3 Atomic mass It is measured using mass spectrometer, and is expressed in terms of amu. 1 amu can be defined as the mass equivalent to th 1 12 the mass of a C-12 isotope. 1 amu =   1 mass of one C 12 isotope 12  = 24 24 10 1.66 10 g 10 g 6      27 1.66 10 kg    Atomic mass is a number, which expresses how many times the mass of an atom of an element is heavier than th 1 12 the mass of a C-12 isotope (1 amu)  atomic mass of an atom = massof thegiven atom massequivalent to1amu Therefore, it is clear that atomic mass is a unitless quantity ex : atomic mass of H - 1.008 Na - 23 Ca - 40 He - 4 Mg - 24 Mn - 55 C - 12 Al - 27 Fe - 56 N - 14 S - 32 Cu = 63.5 O - 16 Cl - 35.5 Br = 80 F - 19 K = 39 Ag = 108 Ne - 20 I = 126 Average atomic mass Chlorine has 2 isotopes Cl 35 and Cl 37 in 3 : 1 natural abundance ratio. Therefore, we take the average of these two values as the atomic mass of chlorine.

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 6 Therefore, average atomic mass of chlorine = 35 3 37 1 35.5 1 3      Consider n isotopes of an element having atomic masses M , M , M , ......M with percentage of 1 2 3 n abundance P , P , P ,.....P respectively. 1 2 3 n Therefore, average atomic mass = 1 1 2 2 3 3 n n 1 2 3 4 n M P M P M P ......... M P P P P P ......... P          i.e, average atomic mass = n i i i 1 M P 100   Consider 2 isotopes of an element having atomic masses M and M with percentage of abundance P 1 2 1 & P respectively. 2 Therefore, average atomic mass = 2 2 1 1 2 1 M P M P P P   But P + P = 100 1 2   2 1 P 100 P     average atomic mass =   1 1 2 1 M P M 100 P 100   Qn. Boron has 2 isotopes. B - 10 and B - 11 with an average atomic mass of 10.81. Calculate their percentage of abundance ratio. Average atomic mass =   1 1 2 1 M P M 100 P 100     1 1 10P 11 100 P 10.81 100    1 1 11P 1100 10P 1081    - 19 = - 1 P 1 P = 19 1 (100 - 19) = P 2 P = 81 2 P : P = 19 : 81 1 2

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 7 ALTERNATE METHOD 0.81 10.81 0.19 11 10 0.81 0.19 : 19 : 81 Molecular mass Molecular mass is a number, which expresses how many times the mass of a molecule of a compound is heavier than 1/12 the mass of a C-12 isotope. th Molecular mass of a molecule = Massof the given molecule Massequivalent to1amu ex : Molecular mass of CO = 44 2 Therefore, it is clear that the molecular mass of a molecules in the sum of the atomic masses of atoms of various elements present in a molecule. Mole Concept One mole is the collection of avogadro number (N ) of particles. A N = 6.023 × 10 A 23 1 mole books = 6.023 × 10 books 23 1 mole atoms = 6.023 × 10 atoms 23 1 mole molecules = 6.023 × 10 molecules 23 1 mole ions = 6.023 × 10 ions 23 Qn. Calculate the no. of years required to count avogadro number of rupees at the rate of 10 lakh rupees per second 10 Rs 6  1 sec 365 × 24 × 60 × 60 × 10 Rs 6  1 yr 6 1 yrs 1R s 365 24 60 60 10       23 23 6 6.022 10 6.022 10 Rs 365 24 60 60 10         yrs 10 2 10 yrs  

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 8 Gram atomic mass The atomic mass expressed in grams is called gram atomic mass. If the atomic mass of an atom is x, then, Mass of one atom of the element = x. amu = x × 1.66 × 10 g -24 24 10 x 10 g 6      Mass of 1 mole atoms = 24 23 10 x 10 6 10 g 6      = x × 1 g = x g One gram atom of the element = 1 mole atoms = x g ex : atomic mass of He is 4, then Mass of one He atom = 4 a.m.u Mass of one mole He atom = 4 g One gram atom of He = 1 mole He = 4 g He Gram molecular mass The molecular mass expressed in grams is called the gram molecular mass. If the molecular mass of a molecule is x, then Mass of one molecule of the compound = x a.m.u Mass of one mole molecule = x g One gram molecule of the compound = 1 mole molecules = xg Eg : Molecular mass of CO is 44, then 2 Mass of one CO molecule = 44 a.m.u 2 Mass of one mole CO = 44 g 2 One gram molecule of CO = 1 mole CO = 44 g CO 2 2 2 Molar mass It is the mass of one mole of particles Molar mass of atoms = gram atomic mass Molar mass of molecules = gram molecular mass The unit of molar mass is gram mole -1

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 9 Calculation of the no. of moles, no. of molecules and no. of atoms from the given mass of a substance. Let w be the mass of the substance in grams and M be its molar mass in g mole then, -1 1 w given mass in grams no. of moles M molar mass in g mole    A A w no. of molecules no. of moles N N M     A no. of molecules no. of moles N  A A w no. of atoms no. of moles N atomicity N atomicity M       Percentage composition of various elements present in a molecule The percentage of an element A in a compound is given by atomicmassof A no.of atoms of A % of A 100 molecular mass of the compound    Qn1. Calculate the no. of atoms of Ag present in 1404 a.m.u of Ag. Atomic mass of Ag = 108  Mass of 1 silver atom = 108 a.m.u 108 a.m.u Ag  1 Ag atom  1 a.m.u Ag  1 108 Ag atoms  1404 a.m.u Ag  1404 Ag 108 atoms = 13 Ag atoms Qn2. The mass of 1 H O molecule is 2 1) 30 × 10 g -23 2) 3 × 10 g -23 3) 3 × 10 amu -23 4) 18 × 10 g -23 Mass of 1 H O molecule = 18 a.m.u = 2 24 10 18 10 g 6    = 3 × 10 g -23

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 10 Qn3. From 320 mg of SO , 12 × 10 molecules are removed. The no. of moles of SO left are ? 2 20 2 1) 30 × 10 10 2) 18 × 10 20 3) 6 × 10 -3 4) 3 × 10 -3 Molar mass of SO = 64 g mole 2 -1 No. of SO molecules = 2 A w N M  = 3 23 1 1 320 10 g 6 10 moleculesmol 64g mole       = 30 × 10 molecules 20 No. of molecules removed = 12 × 10 20  No. of molecules remaining = 30 × 10 - 12 × 10 = 18 × 10 molecule 20 20 20 No. of moles of SO remaining = 2 20 3 23 1 A 18 10 molecules no. of molecules 3 10 mole 6 10 moleculesmol N        Qn4. Percentage of oxygen in the metal carbonate M CO in 45.28. The atomic mass of metal M is : 2 3 1) 46 2) 23 3) 18 4) 27 Let the atomic mass of metal M be A.  Molecular mass of M CO = 2 A + 60 2 3 Percentage of oxygen = 3 16 100 2A 60    48 100 45.28 2A 60    48 2A 60 100 45.28    100 2A 40  A 20  Ans : 23 Qn.5 The percentage of water in Na SO . x H O is 55.9. Calculate the value of x. 2 4 2 1) 6 2) 4 3) 7 4) 10 Molecular mass of Na SO . x H O = 142 + 18 x 2 4 2 % of H O in Na SO . x H O = 2 2 4 2 18x 100 55.9 142 18x   

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 11 18x 100 142 18x 55.9     33x 142 18x   15x 142  142 x 10 15   Avagadro’s Hypothesis Equal volumes of all gases contain equal number of moles or molecules under identical condition of temperature and pressure. Application 1. Relation between molecular mass and vapour density 2 Molecular mass 2 V.D Density of thegas Vapour Density Density of H gas    2. Molar volume of a gas at NTP or STP NTP  273 K and 1 atm pressure STP  273 K and 1 bar pressure Molar volume of a gas at NTP = 22.4 L Molar volume of a gas at STP = 22.7 L given volume of thegas at NTP or STP No.of moles of a gas molar volumeof thegasat NTPor STP  1m = 10 L = 10 dm = 10 cm = 10 mL 3 3 3 3 6 3 6 3. Loschmidt’s number : The number of gas molecules present in 1 mL of a gas at NTP is called Loschmidt’s number. 22400 mL of a gas at NTP  6.023 × 10 gas molecules 23  1 mL of a gas at NTP  23 6.023 10 22400  gas molecules = 2.687 × 10 gas molecules 19 Qn: V.D of a gas is equal to 4 times that of Cl . Calculate the mass of 0.25 moles of the gas. 2 V.D of Cl = 2 71 molecular mass 35.5 2 2  

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 12  V.D of the given gas = 4 × 35.5 = 142 Molecular mass of given gas = 2 × V.D = 2 × 142 = 284 Mass of 1 mol of the gas = 284 g  Mass of 0.25 mol of the gas = 0.25 × 284 = 71 g Empirical formula (EF) & Molecular formula (MF) E.F is the simplest formula which gives the simplest ratio of atoms of various elements present in a molecules. E.F. of Benzene is CH and that of glucose is CH O 2 M.F is the correct formula which gives the actual no. of atoms of various elements present in a molecule. MF of Benzene is C H and that of glucose is C H O . 6 6 6 12 6   MF n EF MFmass n EFmass 2xVD EFmass    Determination of EF from the percentage composition of various elements present in a molecule Step 1 : Determination of the relative no. of moles. To calculate its value, we devide the % of each element by its atomic mass. Step 2 : Determination of the simplest mole ratio. To calculate its value, we divide the relative no. of moles obtained in step 1 by the smallest value. Step 3 : Determination the simplest whole no. ratio. To calculate its value, we multiply the simplest mole ratios obtained in step 2 by a suitable no. The simplest whole number ratio obtained in step 3 gives the no. of atoms of various element present in the empirical formula Qn. Analysis of an organic compound shows that it contains 52.2 % of C, 13% H , and remaining % of 2 oxygen. Calculate the E.F of the compound Element At mass % Rel. no. of moles Simplest whole no. ratio Simplest whole no. ratio C 12 52.2 2 H 1 13 6 O 16 34.8 1 54.2 4.35 12  13 13 1  34.8 2.175 16  4.35 2 2.175  13 6 2.175  2.175 1 2.175  Note : The no. of atoms of various elements present in the E.F corresponds to the simplest ratio of relative no. of moles of the corresponding elements.

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 13 Qn. Analysis of an organic compound shows that it contains 52.2% of C, 13% of H and 34.8% of O . 2 2 Calculate the MF of the compound if its VD = 46 52.2 13 34.8 C : H :O : : 12 1 16  = 4.35 : 13 : 2.175 = 2 : 6 : 1  EF = C H O 2 6  EF mass = 24 + 6 + 16 = 46 2VD n EF mass  = 2 46 2 46   MF = 2 [EF] = 2 [C H O] = C H O 2 6 4 12 2 Balancing of chemical reactions I. Balancing of ionic reactions A y + + B x - y x x y A B Qn. 2 3 4 BaCl Na PO Products   BaCl 2 Ba 2+ + 2Cl - 3Na + + 3 4 PO  Na PO 3 4 Ba 2+ + 3 4 PO    3 4 2 Ba PO 2 3 Na Cl NaCl       2 3 4 3 4 2 3BaCl 2Na PO Ba PO 6NaCl    Balancing of combustion reaction of organic compounds

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 14 Balance the combustion reaction of following organic compounds i) Methane : 4 2 2 2 4 4 CH O 1CO H O 2 2    4 2 2 2 CH 2O CO 2H O    ii) Ethane : 2 6 2 2 2 7 6 C H O 2CO H O 2 2    2 2 2 6 2 7 3H O 2CO O C H 2    iii) Propane : 3 6 2 2 2 9 6 C H O 3CO H O 2 2    2 2 2 6 3 9 3H O 3CO O C H 2    iv) Methanol : 3 2 2 2 3 4 CH OH O 1CO H O 2 2    2 2 2 3 3 2H O CO O CH OH 2    v) Glucose : 6 12 6 2 2 2 C H O 6O 6CO 6H O    Stoichiometry of chemical reactions The co-efficients of reactants and products in a balance chemical equation is called the stoichiometric co-efficients of reactants and products in that chemical reaction. The simplest ratio of stoichiometric coefficients is called the stoichiometry of a chemical reaction ex : The stoichiometry of N , H and NH in the reaction 2 2 3 2 2 3 N 3H 2NH   is 1 : 3 : 2

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 15 Limiting reagent The reagent which react completely in a chemical reaction is called the limiting reagent and the reagent which react incompletely is called the excess reagent. The amount of product will be limited to a particular value by the limiting reagent. Qn. 3 moles of N and 6 moles of H are mixed. Calculate the maximum amount of NH produced. 2 2 3 3 2 2 2NH 3H N   1 mole N 2  3 mole H 2 3 mole N 2  9 mole H 2 But, given only 6 moles of H . H is the limiting reagent. 2 2 3 mole H 2  2 mole NH 3 6 mole H 2  4 mole NH 3 OR N 2 + 3 H 2 2NH 3 3 6 0 (3 - 2) (6 - 6) 4 1 0 No. of moles before reaction No. of moles after reaction Qn: 1 g of Mg react with 0.56 of O . Calculate the mass of excess reactant remaining unreacted. 2 Mg + 2 1 O 2 MgO 1 : 0.5 : 1 24 : 16 : 40 3 : 2 : 5 Mole ratio Mass ratio 2 3g Mg 2gO  1 g 2 2 2 Mg g O 0.66g O 3   But, given only 0.56 g of oxygen  oxygen is the limiting reagent & Mg is the excess reagent.

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 16 2 2g O 3g Mg 2 1g O 3/ 2g Mg 2 3 0.56g Mg 0.56g O 2   = 0.84 g Mg Mass of Mg remaining = 1 - 0.84 = 0.16 g Qn. 22.4 L of H & 11.2 L of Cl (each at NTP) are mixed. Calculate the no. of moles of HCl produced. 2 2 No. of moles of H = 2 22.4 1mol 22.4  No. of moles of Cl = 2 1 11.2 mol 2 22.4  H 2 + 2HCl 1 : 0.5 : 0 (1-0.5) : (0.5 - 0.5) : 1 0.5 : 0 No. of moles before reaction No. of moles after reaction Cl 2 Qn. Calculate the volume of O gas produced at NTP by heating 12.25 grams of KClO 2 3 Molecular mass of KClO = 39 + 35.5 + 48 = 122.5 3 No. of moles of KClO = 12.25 / 122.5 = 0.1 mole 3 1 mole KClO 3  2 3 mole O 2 0.1 mole KClO 3 2 3 0.1 mol O 2   = 2 3 0.1 22.4 of O at NTP 2   = 3.36 L Qn. 20 g of 50% pure CaCO reacts with excess dil. H SO . Calculate the volume of CO gas produced at 3 2 4 2 NTP. Mass of pure CaCO = 3 50 20 10g 100     3 2 4 4 2 2 excess CaCO H SO CaSO CO H O      no. of moles of CaCO = 3 10 0.1 mole 100 

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 17  0.1 mol of CaCO 3  0.1 mol of CO = 0.1 × 22.4 L of CO at NTP = 2.24 L 2 2 Equivalent mass 1. Equivalent mass of elements A) The mass of the element which combined with 1.008 g of H (1 eq. mass of H ) 2 2 2 Massof theelement Equivalent massof an element 1.008 Massof H combined   2 2 Equivalent massof an element Massof the element Massof H combined Equivalent massof H  B) Mass of the element which combined with 35.5 g of chlorine 2 Massof element Equivalent massof an element 35.5 Massof Cl combined   2 2 Equivalent massof an element Massof the element Massof Cl combined Equivalent mass of Cl  C) The mass of the element which combined with 80 g of Br (1 equivalent mass of Br ) 2 2 2 Massof the element Equivalent massof an element 80 Massof Br combined   2 2 Equivalent massof an element Massof the element Massof Br combined Equivalent massof Br  D) The mass of the element which combined with 8 g of O (1 eq. mass of O ) 2 2 2 Massof the element Equivalent massof an element 8 Massof O combined   2 2 Equivalent massof an element Massof the element Massof O combined Equivalent mass of O 

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 18 2. Equivalent mass of metals Eq. mass of metals = Atomic mass Valency 3. Equivalent mass of acids Eq. mass of acids = Molecular mass Basicity Basicity of acids The no. of moles of replaceable H ions from 1 mole of an acid is called its basicity. + ex : Basicity HCl = 1, H SO = 2 etc..... 2 4 4. Equivalent mass of bases Eq. mass of bases = Molecular mass Acidity Acidity of bases The no. of moles of replicable OH  ions from one mole of the base is called its acidity. . ex : Acidify of NaOH = 1, Ca(OH) = 2 , ....... 2 5. Equivalent of ions Eq. mass of ion = Formula mass Charge Eq. mass of Na = + 23 23 1  Eq. mass of Ca = 2+ 40 20 2  Eq. mass of Al = 3+ 27 9 3  Eq. mass of 2 3 60 30 CO 2    Eq. mass of 2 4 96 SO 48 2   

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 19 6. Equivalent mass of compounds Eq. mass of compounds = Molecular mass Totalchangeof anions or cations Note : Eq. mass of a compound = eq. mass of corresponding cation + eq. mass of corresponding anion ex : Eq. mass of Al (SO ) = eq. mass of Al + eq. mass of 2 4 3 3+ 2 4 SO  Experimental determination of atomic masses 1. Dulong and Petit’s law (applicable for solid elements with high metallic character) Atomic mass × specific heat  6.4 Atomic mass = 6.4 specific heat where the specific heat is in calories 2. Isomorphism method Isomorphous compounds According to Mitscherlisch’s law of isomorphism isomorphous compounds contains similar number of similar atoms united in a similar manner gives the similar crystalline structure. ex : 4 2 4 2 2 7H O 7H O and ZnSO Fe SO   KMnO and KClO 4 4 Consider a metal M with atomic mass A has a compound which on isomorphous with MgSO . 7H O, 4 2 then formula of the meta compound can be taken as MSO . 7H O 4 2 Molecular mass of MSO .7H O = A + 96 + 7 × 18 = A + 222 4 2 % of M = A 100 A 222   The % of M is determined experimentally. From this, we can calculate the value of A Qn: The carbonate of a metal is formed to be isomorphous with Magnesium carbonate & contains 6.091 % of carbon. The atomic mass of the metal will be 1) 192 2) 167 3) 137 4) 122 Let the metal = M Let At. mass of M = A The formula of the metal carbonate can be taken as MCO . 3 Molecular mass of MCO = A + 60 3

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 20 % of C = 12 100 A 60   = 1200 60 A  1200 6.91 A 60    1200 A 60 200 6.091    A 140  A 137   3. Volatile chloride method Consider a metal M with atomic mass A, equivalent mass E and valency n, then A E n  A En   The formula of the metal chloride can be taken as MCl . n Molecular mass of MCl = A + 35.5 n n 2VD En 35.5n    = n [E + 35.5] 2 VD n E 35.5    Then the correct atomic mass A = En. Qn. The VD of a metal chloride is 66.100 g of the metal oxide contains 53 g of the metal. Calculate the atomic mass of the metal. Let the metal be M, its its at mass-A, equivalent mass E, and valency n. 2 Mass of metal 8 E Mass of O combined     53 8 100 53    = 53 8 9 47  

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 21 2VD 2 66 132 n 3 9 35.5 44.5       A = En = 3 × 9 = 27 Experimental determination of equivalent masses 1. Hydride formation method Equivalent mass of the element = 2 Massof the compound 1.008 Mass of H combined  = 2 Mass of the element 11.2 Vol. of H combined in L at NTP  2 2 2.016 gH 22.4L NTP 1.008gH 11.2L      2. Chloride formation method Equivalent mass of an element = 2 Mass of theelement 35 Mass of Cl combined  2 Mass of theelement 11.2 Vol. of Cl combined in L at NTP   2 2 22.4L 71g Cl NTP 35.5gCl 11.2L      3. Oxide formation method Equivalent mass of an element = 2 Massof the element 8 Mass of O combined  = 2 Massof the element 5.6 Vol.of O combined in L at NTP  2 2 2 32g O 22.4L NTP 8g O 5.6L O      4. Metal displacement method More reactive metals can displace less reactive metals from their solutions. eg : 4 4 CuSO (aq) Zn(s) ZnSO (aq) Cu(s)    E+35.5

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 22 Eq.massof metal added Massof metal added Eq. mass of metaldisplaced Mass of metal displaced  5. Double displacement method 3 3 AB CD AD CB Eg : AgNO NaCl AgCl NaNO             Eq.mass of AB Mass of AB added Eq. mass of AD Mass of ADobtained  Qn. 0.84 g of a metal hydride contains 0.042 g of hydrogen. The equivalent mass of the metal is 1) 10 2) 20 3) 30 4) 40 Eq. mass of metal = 2 Massof the metal 1.008 Massof O combined  = 0.84 0.042 1.008 0.042    0.84 1 20 0.042   Qn. V.D of a metal chloride is 95 and specific heat of the metal is 0.13. Equivalent mass of the metal will be 1) 12 2) 17 3) 22 4) 9 Let the metal be M, its at mass be A, equivalent mass be E valency be ‘n’. The formula of metal chloride is MCl . n 6.4 6.4 A 50 Sp.heat 0.13    . Molecular mass of MCl = A + 35.5 n n 2VD 50 35.5n    2 × 95 = 50 + 35.5 n 35.5n 190 50   35.5 n = 140 140 n 35.5  4 n  50 A 12 E 4 n   

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 23 Qn. 3g of a metal oxide when converted to its chloride gives 5 g of the metal chloride. Calculate the equivalent mass of the metal. Eq.mass of metalchloride Massof metalchloride Eq.massof metal oxide Mass of metaloxide  metal chlorine metal oxygen E E 5 E E 3    5 35.5 E 3 8 E    3(E + 35.5) = 5(E + 8) 3E + 106.5 = 5E + 40 106.5 - 40 = 2E 2E = 66.5 E = 33.25 Qn. 0.4426 g of a metal chloride is dissolved in water and is made upto 1000 mL. 500 mL of this solution requires 0.51 grams of AgNO for the complete precipitation of Cl ions as AgCl. Calculate the valency 3 - of the metal, if its atomic mass is 112.3. 3 3 Mass of metal oxide Eq.mass of metal chloride Mass of AgNO Eq.massof AgNO  3 Ag NO 35.5 E E E    = 0.4426/ 2 0.51 0.2213 35.5 E 62 108 0.51 1 1    E 35.5 0.2213 170 0.51   0.2213 35.5 38 E 170 0.51           Atomic mass 112.3 Valency 3 38 Equivalent mass   

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 24 Solutions The homogeneous mixture of 2 or more non-reacting substances are called as solutions. Solute The component present in smaller amount Solvent The component present in larger amount Concentration of solutions Mass fraction Consider a binary solution of components A & B having masses M & M respectively, then A B A A B M Mass fraction of A M M   B A B M Mass fraction of B M M   Mass percentage Mass percentage = mass fraction × 100 For the above example, Mass % of A = A A B M 100 M M   Mass % of B = B B A M 100 M M   Mole fraction ( )  Consider a binary solution of components A & B, having number of moles n & n respectively. Then, A B mole fraction of A ( A  ) = A B A n n n mole fraction of B ( B  ) = B B A n n n A B 1   

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 25 Molarity (M) The number of moles of solute present per litre of the solution is called its molarity. no.of moles of solute Molarity(M) volume of solutionin L  Molarity law (Dilution law) M V = M V 1 1 2 2 Before dilution After dilution Molarity × volume of solution (in L) = Number of moles = w M Molarity × volume of solution (in mL) = Number of millimoles = w 100 M  w = given mass in g M = molar mass in g mole -1 Note : Number of moles or millimoles of reactants reacts according to the stoichiometry of the reaction and produce the number of moles or millimoles of products according to the corresponding stoichiometry. Qn. 20 ml of 0.1 M H SO is mixed with 30 ml of 0.2 M NaOH. Calculate the maximum amount of Na SO 2 4 2 4 produced (in g). Molecular mass of Na SO = 46 + 96 = 142 2 4 H SO 2 4 + 2NaOH Na SO 2 4 + 2H O 2 (0.1 × 20) (0.2 × 30) 0 0 2 6 (2 - 2) (6 - 4) 2 4 0 2 No. of millimoles before reaction Millimoles after reaction Millimoles of Na SO produced = 2 2 4 w 1000 2 M   

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 26 2 142 2M 0.284g w 1000 1000     Molality (m) The number of moles of solute present per kilogram of the solvent is called its molality. Molality (m) = Number of moles of solute Mass of solvent in kg Normality (N) The no. of equivalents of the solute present per litre of the solution. Normality (N) No of equivalents of the solute Volume of solution in L  No. of equivalents of the solute = Given mass w Equivalent mass E  Note: Both normality and molarity involves the volumes of solutions. Therefore both are temperature dependent. Relation between normality and molarity Molecular mass Normality Molarity Eqivalent mass   Molecular mass Basicity (for acids) Equivalent mass  = Acidity (for bases) Normality = Molarity × Basicity (for acids) Normality = Molarity × Acidity (for bases) Normality equation (Dilution law) N V 1 1 = N V 2 2 before dilution after dilution Normality × Volume of solution (in L) = no of equivalents Normality × Volume of solution (in mL) = no of milliequivalent = w 1000 E 

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 27 w = given mass in g q e g n i s s a m t n e l a v i u q e = E –1 Note: No. of equivalents or milliequivalents of reactants reacts in equal no and produce the same no. of equivalent of milli equivalents of products separately. Qn. 20 ml of 0.1 M H SO is mixed with 30 ml of 0.2 M NaOH. Calculate the max. amount of Na SO 2 4 2 4 produced (in grams) Normality of H SO = Molarity × Basicity = 0.1 × 2 = 0.2 N 2 4 Normality of NaOH = Molarity × Acidity = 0. 2 × 1= 0.2 N Equivalent mass of Na SO = 2 4 142 molecular mass 71 2 total charge of anions/cations   H SO 2 4 + NaOH Na SO 2 4 + H O 2 (0.2 × 20) (0.2 × 30) 0 0 4 6 (4 - 4) (6 - 4) 4 4 0 2 milliequivalents before reaction Milliequivalents after reaction Milliequivalents of Na SO produced = 4 2 4 w 1000 4 E    4 71 4 E 0.284g w 1000 100       Normality in a mixture of two or more non- reacting solutions Consider a mixture of 3 non- reacting solutions having normalities N , N , & N with volumes V , V & V 1 2 3 1 2 3 respectively. Then resulting normality N of the mixture having final total volume V is given by 4 4 N V + N V + N V = N V = N (V + V + V ) 1 1 2 2 3 3 4 4 4 1 2 3 Note : If a particular volume of water is added to the resulting mixture then, final total volume. V = V + V + V + V 4 1 2 3 water added Note : If N normal of an acid is mixed with N normal V ml of a base, then 1 2 2 i) If N V = N V 1 1 2 2  Resulting solution is neutral ii) If N V > N V 1 1 2 2  Resulting solution is acidic

BBrilliant STUDY CENTRE LT23-CHEMISTRY (ONLINE) -2021 28  Normality of resulting acidic solution = 1 1 2 2 1 2 N V N V V V   iii) If N V < N V 1 1 2 2  Resulting solutions is basic  Normality of resulting basic solution = 2 2 1 1 1 2 N V N V V V   Parts per million (ppm) The number of parts by mass of solute present per million parts by mass of the solution. 6 mass of solute ppm 10 mass of solution  