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Page 1 STOICHIOMETRY ? Relative atomic mass (R.A.M) = atom carbon one of mass 12 1 element an of atom one of Mass ? = Total Number of nucleons ? Ymap Mole × 22.4 lt ? 22.4 lt Volume at STP ? N A × N A Number × mol. wt. × At. wt. ? ? At. wt. mol. wt. Mass Page 2 STOICHIOMETRY ? Relative atomic mass (R.A.M) = atom carbon one of mass 12 1 element an of atom one of Mass ? = Total Number of nucleons ? Ymap Mole × 22.4 lt ? 22.4 lt Volume at STP ? N A × N A Number × mol. wt. × At. wt. ? ? At. wt. mol. wt. Mass Density : Specific gravity = C 4 at water of density ce tan subs the of density ? For gases : Absolute density (mass/volume) = gas the of volume Molar gas the of mass Molar ? ? = RT PM Vapour density V.D.= 2 H gas d d = RT H RT gas 2 PM PM = 2 H gas M M = 2 M gas M gas = 2 V.D. Molemole analysis : Mass ??At. wt. / Mol. Wt. Mole Molemole relationship of equation Mole × 22.4 lt Volume at STP × mol. wt./At. wt. Mass Concentration terms : Molarity (M) : ? Molarity (M) = inml V ) solute of wt . Mol ( 1000 w ? ? Molality (m) : Molality = 1000 gram in solvent of mass solute of moles of number ? = 1000 w 1 / M 1 w 2 Mole fraction (x) : ? Mole fraction of solution (x 1 ) = N n n ? ? Mole fraction of solvent (x 2 ) = N n N ? x 1 + x 2 = 1 Page 3 STOICHIOMETRY ? Relative atomic mass (R.A.M) = atom carbon one of mass 12 1 element an of atom one of Mass ? = Total Number of nucleons ? Ymap Mole × 22.4 lt ? 22.4 lt Volume at STP ? N A × N A Number × mol. wt. × At. wt. ? ? At. wt. mol. wt. Mass Density : Specific gravity = C 4 at water of density ce tan subs the of density ? For gases : Absolute density (mass/volume) = gas the of volume Molar gas the of mass Molar ? ? = RT PM Vapour density V.D.= 2 H gas d d = RT H RT gas 2 PM PM = 2 H gas M M = 2 M gas M gas = 2 V.D. Molemole analysis : Mass ??At. wt. / Mol. Wt. Mole Molemole relationship of equation Mole × 22.4 lt Volume at STP × mol. wt./At. wt. Mass Concentration terms : Molarity (M) : ? Molarity (M) = inml V ) solute of wt . Mol ( 1000 w ? ? Molality (m) : Molality = 1000 gram in solvent of mass solute of moles of number ? = 1000 w 1 / M 1 w 2 Mole fraction (x) : ? Mole fraction of solution (x 1 ) = N n n ? ? Mole fraction of solvent (x 2 ) = N n N ? x 1 + x 2 = 1 % Calculation : (i) % w/w = 100 gm in solution of mass gm in solute of mass ? (ii) % w/v = mass of solute in gm 100 Volume of solution in ml ? (iii) % v/v = Volume of solute in ml 100 Volume of solution ? Derive the following conversion : 1. Mole fraction of solute into molarity of solution M = 2 2 1 1 2 x M M x 1000 x ? ? ? 2. Molarity into mole fraction x 2 = 2 1 MM 1000 1000 MM ? ? ? ? 3. Mole fraction into molality m = 1 1 2 M x 1000 x ? 4. Molality into mole fraction x 2 = 1 1 mM 1000 mM ? 5. Molality into molarity M = 2 mM 1000 1000 m ? ? ? 6. Molarity into Molality m = 2 MM 1000 1000 M ? ? ? M 1 and M 2 are molar masses of solvent and solute. ? is density of solution (gm/mL) M = Molarity (mole/lit.), m = Molality (mole/kg), x 1 = Mole fraction of solvent, x 2 = Mole fraction of solute Average/Mean atomic mass : A x = 100 x a ..... x a x a n n 2 2 1 1 ? ? ? Mean molar mass or molecular mass : M avg. = n 2 1 n n 2 2 1 1 n .... n n M n ...... M n M n ? ? ? ? or M avg. = ? ? ? ? ? ? n j 1 j j n j 1 j j j n M n Page 4 STOICHIOMETRY ? Relative atomic mass (R.A.M) = atom carbon one of mass 12 1 element an of atom one of Mass ? = Total Number of nucleons ? Ymap Mole × 22.4 lt ? 22.4 lt Volume at STP ? N A × N A Number × mol. wt. × At. wt. ? ? At. wt. mol. wt. Mass Density : Specific gravity = C 4 at water of density ce tan subs the of density ? For gases : Absolute density (mass/volume) = gas the of volume Molar gas the of mass Molar ? ? = RT PM Vapour density V.D.= 2 H gas d d = RT H RT gas 2 PM PM = 2 H gas M M = 2 M gas M gas = 2 V.D. Molemole analysis : Mass ??At. wt. / Mol. Wt. Mole Molemole relationship of equation Mole × 22.4 lt Volume at STP × mol. wt./At. wt. Mass Concentration terms : Molarity (M) : ? Molarity (M) = inml V ) solute of wt . Mol ( 1000 w ? ? Molality (m) : Molality = 1000 gram in solvent of mass solute of moles of number ? = 1000 w 1 / M 1 w 2 Mole fraction (x) : ? Mole fraction of solution (x 1 ) = N n n ? ? Mole fraction of solvent (x 2 ) = N n N ? x 1 + x 2 = 1 % Calculation : (i) % w/w = 100 gm in solution of mass gm in solute of mass ? (ii) % w/v = mass of solute in gm 100 Volume of solution in ml ? (iii) % v/v = Volume of solute in ml 100 Volume of solution ? Derive the following conversion : 1. Mole fraction of solute into molarity of solution M = 2 2 1 1 2 x M M x 1000 x ? ? ? 2. Molarity into mole fraction x 2 = 2 1 MM 1000 1000 MM ? ? ? ? 3. Mole fraction into molality m = 1 1 2 M x 1000 x ? 4. Molality into mole fraction x 2 = 1 1 mM 1000 mM ? 5. Molality into molarity M = 2 mM 1000 1000 m ? ? ? 6. Molarity into Molality m = 2 MM 1000 1000 M ? ? ? M 1 and M 2 are molar masses of solvent and solute. ? is density of solution (gm/mL) M = Molarity (mole/lit.), m = Molality (mole/kg), x 1 = Mole fraction of solvent, x 2 = Mole fraction of solute Average/Mean atomic mass : A x = 100 x a ..... x a x a n n 2 2 1 1 ? ? ? Mean molar mass or molecular mass : M avg. = n 2 1 n n 2 2 1 1 n .... n n M n ...... M n M n ? ? ? ? or M avg. = ? ? ? ? ? ? n j 1 j j n j 1 j j j n M n Calculation of individual oxidation number : Formula : Oxidation Number = number of electrons in the valence shell ? number of electrons left after bonding Concept of Equivalent weight/Mass : For elements, equivalent weight (E) = factor  Valency weight Atomic For acid/base, Acidity / Basicity M E? Where M = Molar mass For O.A/R.A, lost / gained e of moles of . no M E ? ? Equivalent weight (E) = v.f. weight moleculear or Atomic (v.f. = valency factor) Concept of number of equivalents : No. of equivalents of solute = M/n W E W wt. Eq. Wt ? ? No. of equivalents of solute = No. of moles of solute × v.f. Normality (N) : Normality (N) = litres) (in solution of Volume solute of s equivalent of Number Normality = Molarity × v.f. Calculation of valency Factor : nfactor of acid = basicity = no. of H + ion(s) furnished per molecule of the acid. nfactor of base = acidity = no. of OH ? ion(s) furnised by the base per molecule. At equivalence point : N 1 V 1 = N 2 V 2 n 1 M 1 V 1 = n 2 M 2 V 2 Page 5 STOICHIOMETRY ? Relative atomic mass (R.A.M) = atom carbon one of mass 12 1 element an of atom one of Mass ? = Total Number of nucleons ? Ymap Mole × 22.4 lt ? 22.4 lt Volume at STP ? N A × N A Number × mol. wt. × At. wt. ? ? At. wt. mol. wt. Mass Density : Specific gravity = C 4 at water of density ce tan subs the of density ? For gases : Absolute density (mass/volume) = gas the of volume Molar gas the of mass Molar ? ? = RT PM Vapour density V.D.= 2 H gas d d = RT H RT gas 2 PM PM = 2 H gas M M = 2 M gas M gas = 2 V.D. Molemole analysis : Mass ??At. wt. / Mol. Wt. Mole Molemole relationship of equation Mole × 22.4 lt Volume at STP × mol. wt./At. wt. Mass Concentration terms : Molarity (M) : ? Molarity (M) = inml V ) solute of wt . Mol ( 1000 w ? ? Molality (m) : Molality = 1000 gram in solvent of mass solute of moles of number ? = 1000 w 1 / M 1 w 2 Mole fraction (x) : ? Mole fraction of solution (x 1 ) = N n n ? ? Mole fraction of solvent (x 2 ) = N n N ? x 1 + x 2 = 1 % Calculation : (i) % w/w = 100 gm in solution of mass gm in solute of mass ? (ii) % w/v = mass of solute in gm 100 Volume of solution in ml ? (iii) % v/v = Volume of solute in ml 100 Volume of solution ? Derive the following conversion : 1. Mole fraction of solute into molarity of solution M = 2 2 1 1 2 x M M x 1000 x ? ? ? 2. Molarity into mole fraction x 2 = 2 1 MM 1000 1000 MM ? ? ? ? 3. Mole fraction into molality m = 1 1 2 M x 1000 x ? 4. Molality into mole fraction x 2 = 1 1 mM 1000 mM ? 5. Molality into molarity M = 2 mM 1000 1000 m ? ? ? 6. Molarity into Molality m = 2 MM 1000 1000 M ? ? ? M 1 and M 2 are molar masses of solvent and solute. ? is density of solution (gm/mL) M = Molarity (mole/lit.), m = Molality (mole/kg), x 1 = Mole fraction of solvent, x 2 = Mole fraction of solute Average/Mean atomic mass : A x = 100 x a ..... x a x a n n 2 2 1 1 ? ? ? Mean molar mass or molecular mass : M avg. = n 2 1 n n 2 2 1 1 n .... n n M n ...... M n M n ? ? ? ? or M avg. = ? ? ? ? ? ? n j 1 j j n j 1 j j j n M n Calculation of individual oxidation number : Formula : Oxidation Number = number of electrons in the valence shell ? number of electrons left after bonding Concept of Equivalent weight/Mass : For elements, equivalent weight (E) = factor  Valency weight Atomic For acid/base, Acidity / Basicity M E? Where M = Molar mass For O.A/R.A, lost / gained e of moles of . no M E ? ? Equivalent weight (E) = v.f. weight moleculear or Atomic (v.f. = valency factor) Concept of number of equivalents : No. of equivalents of solute = M/n W E W wt. Eq. Wt ? ? No. of equivalents of solute = No. of moles of solute × v.f. Normality (N) : Normality (N) = litres) (in solution of Volume solute of s equivalent of Number Normality = Molarity × v.f. Calculation of valency Factor : nfactor of acid = basicity = no. of H + ion(s) furnished per molecule of the acid. nfactor of base = acidity = no. of OH ? ion(s) furnised by the base per molecule. At equivalence point : N 1 V 1 = N 2 V 2 n 1 M 1 V 1 = n 2 M 2 V 2 Volume strength of H 2 O 2 : 20V H 2 O 2 means one litre of this sample of H 2 O 2 on decomposition gives 20 lt. of O 2 gas at S.T.P. Normality of H 2 O 2 (N) = 2 2 Volume,strengthof H O 5.6 Molarity of H 2 O 2 (M) = 2 . 11 O H of strength Volume 2 2 Measurement of Hardness : Hardness in ppm = 6 3 10 water of mass Total CaCO of mass ? Calculation of available chlorine from a sample of bleaching powder : % of Cl 2 = ) g ( W ) mL ( V x 55 . 3 ? ? where x = molarity of hypo solution and v = mL. of hypo solution used in titration.Read More

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