Some Basic Concepts of Chemistry Short Notes PDF

CHAPTER 1 Some Basic Concepts of Chemistry Some Useful Conversion Factors The number of atoms present in exactly 12 gm of C-12 isotope is...

CHAPTER 1 Some Basic Concepts of Chemistry Some Useful Conversion Factors The number of atoms present in exactly 12 gm of C-12 isotope is called Avogadro’s number [NA = 6.022 × 1023] 1 Å = 10–10m, 1nm = 10–9 m 1g 1 pm = 10–12m 1u = 1amu = (1/12)th of mass of 1 atom of C12 = NA 1 litre = 10–3 m3 = 1 dm3 –24 = 1.66 ×10 g 1 atm = 760 mm or torr For Elements = 101325 Pa or Nm–2 ™ 1 g atom = 1 mole of atoms = NA atoms 1 bar = 105 Nm–2 = 105 Pa ™ g atomic mass (GAM) = mass of NA atoms in g 1 calorie = 4.184 J Mass ( g ) 1 electron volt(eV) = 1.6022 ×10–19 J ™ Mole of atoms = GAM or Molar mass (1 J = 107 ergs) (1 cal > 1 J > 1 erg > 1 eV) For Molecule Atomic Mass or Molecular Mass ™ 1g molecule = 1 mole of molecule = NA molecule Mass of one atom or molecule in a.m.u. ™ g molecular mass (GMM) = mass of NA molecule in g. C → 12 amu NH3→ 17 amu Mass ( g ) ™ Mole of molecule = GMM or Molar mass Actual Mass Mass of one atom or molecule in grams: 1 Mole of Substance C → 12 ×1.6 × 10–24 g ™ Contains 6.022 × 1023 particles. CH4 → 16 ×1.6 × 10–24 g ™ Weighs as much as molecular mass/ atomic mass/ionic mass Relative Atomic Mass or Relative Molecular Mass in grams. Mass of one atom or molecule w.r.t. 1/12th or 12C atom: C → 12 ™ If it is a gas, one mole occupies a volume of 22.4 L at 1 atm CH4 → 16 & at 273 K or 22.7 L at STP. It is unitless. For Ionic Compounds Gram Atomic Mass or ™ 1 g formula unit = 1 mole of formula unit = NA formula unit. Gram Molecular Mass ™ g formula mass (GFM) = mass of NA formula unit in g. Mass of one mole of atom or molecule: C → 12 g Mass ( g ) ™ Mole of formula unit = GMM or Molar mass CO2 → 44 g It is also called as molar mass. Vapour density Ratio of density of vapour to the density of hydrogen at similar Definition of Mole pressure and temperature. One mole is a collection of that many entities as there are number of atoms exactly in 12 gm of C-12 isotope.  Molar mass their respective stoichiometric coefficient. The reactant having Vapour density = minimum ratio will be L.R. then find the moles of product 2 formed or excess reagent left by comparing it with L.R. through stoichiometric concept. Percentage Purity The percentage of a specified compound or element in an impure sample may be given as: Actual mass of compound =%purity × 100 Total mass of sample If impurity is unknown, it is always considered as inert (unreactive) material. Stoichiometry Based Concept Empirical and Molecular Formula aA + bB → cC + dD ™ Empirical formula: Formula depicting constituent atoms in their simplest ratio. ™ a,b,c,d, represents the ratios of moles, volumes [for gaseous] ™ Molecular formula: Formula depicting actual number of molecules in which the reactants react or products formed. atoms in one molecule of the compound. ™ a,b,c,d does not represent the ratio of masses. ™ The molecular formula is generally an integral multiple of the empirical formula. ™ The stoichiometic amount of components may be related as: i.e. molecular formula = empirical formula × n Moles of A reacted Moles of Breacted Moles of C reacted Moles of D reacted = = = a b c d molecular formula mass where n = empirical formula mass Concept of Limiting Reagent If data of more than one reactant is given then first convert all the data into moles then divide the moles of reactants with Concentration Terms Concentration Type Mathematical Formula Concept  w  Mass of solute × 100 Percentage by mass %  = Mass of solute (in gm) present in 100 gm of solution.  w Mass of solution  v  Volumeof solute × 100 Volume percentage %  = Volume of solute (in cm3) present in 100 cm3 of solution.  v Volumeof solution Mass-volume  w  Mass of solute × 100 %  = Mass of solute (in gm) present in 100 cm3 of solution. percentage  v  Volumeof solution Mass of solute × 106 Parts per million ppm = Parts by mass of solute per million parts by mass of the solution. Mass of solution Moleof A XA = Moleof A + Moleof B + Moleof C +... Mole fraction Ratio of number of moles of one component to the total number of moles. Moleof B XB = Moleof A + Moleof B + Moleof C +... Moleof solute Molarity M= Moles of solute in one liter of solution. Volumeof solution (in L) Moleof solute Molality m= Moles of solute in one kg of solvent. Mass of solvent (Kg) 2 J EE (XI) Module-1 P W Mixing of Solutions It is based on law of conservation of moles. (i) Two solutions having same solute: Total moles M V + M 2 V2    Final molarity = = 1 1 Total volume V1 + V2 M1V1 (ii) Dilution Effect: Final molarity, M 2 = V1 + V2 P W Some Basic Concepts of Chemistry 3