Title Hand Book of Chemistry WITHOUT LOGO.pdf ✅

1 SOME USEFUL CONVERSION FACTORS 1 Å =10–10m, 1nm = 10–9 m 1 pm =10–12m 1 litre =10–3 m3 = 1 dm3 1 atm =760 mm or torr =101325 Pa or Nm–2 1 bar =105 Nm–2 = 105 Pa 1 calorie =4.184 J 1 electron volt(eV) = 1.6022 ×10–19 J (1 J =107 ergs) (1 cal >1 J > 1 erg > 1 eV) ATOMIC MASS OR MOLECULAR MASS Mass of one atom or molecule in a.m.u. C → 12 amu NH3 → 17 amu Actual Mass Mass of one atom or molecule in grams C → 12 ×1.6 × 10–24 g CH4 → 16 ×1.6 × 10–24 g Relative Atomic Mass or Relative Molecular Mass Mass of one atom or molecule w.r.t. 1/12th or 12C atom C → 12 CH4 → 16 It is unitless 1 Some Basic Concept of Chemistry Chapter
Hand Book (Chemistry) 2 GRAM ATOMIC MASS OR GRAM MOLECULAR MASS Mass of one mole of atom or molecule C → 12 g CO2 → 44 g It is also called molar mass DEFINITION OF MOLE One mole is a collection of that many entities as there are number of atoms exactly in 12 gm of C-12 isotope. The number of atoms present in exactly 12 gm of C-12 isotope is called Avogadro’s number [NA = 6.022 × 1023] 1u = 1amu = (1/12)th of mass of 1 atom of C12 = 1 A g N = 1.66 ×10–24 g For Elements ● 1 g atom = 1 mole of atoms = NA atoms ● g atomic mass (GAM) = mass of NA atoms in g ● Mole of atoms = Mass ( ) GAM or molar mass g For Molecule ● 1g molecule = 1 mole of molecule = NA molecule ● g molecular mass (GMM) = mass of NA molecule in g. ● Mole of molecule = Mass ( ) GMM or molar mass g 1 Mole of Substance ● Contains 6.022 × 1023 particles ● Weighs as much as molecular mass/ atomic mass/ionic mass in grams ● If it is a gas, one mole occupies a volume of 22.4 L at 1 atm & 273 K or 22.7 L at STP For Ionic Compounds ● 1 g formula unit = 1 mole of formula unit = NA formula unit. ● g formula mass (GFM) = mass of NA formula unit in g. ● Mole of formula unit = Mass ( ) GMM or molar mass g
3 Some Basic Concept of Chemistry VAPOUR DENSITY Ratio of density of vapour to the density of hydrogen at similar pressure and temperature. Molar mass Vapour density 2 = STOICHIOMETRY BASED CONCEPT aA + bB → cC + dD ● a,b,c,d, represents the ratios of moles, volumes [for gaseous] molecules in which the reactants react or products formed. ● a,b,c,d does not represent the ratio of masses. ● The stoichiometic amount of components may be related as Molesof A reacted Molesof Breacted Molesof Creacted Molesof Dreacted abcd = = = 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 their respective stoichiometric coefficient. The reactant having minimum ratio will be L.R. then find the moles of product 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 Actualmassof compound %purity 100 Totalmassof sample = × If impurity is unknown, it is always considered as inert (unreactive) material.
Hand Book (Chemistry) 4 EMPIRICAL AND MOLECULAR FORMULA ● Empirical formula: Formula depicting constituent atoms in their simplest ratio. ● Molecular formula: Formula depicting actual number of atoms in one molecule of the compound. ● The molecular formula is generally an integral multiple of the empirical formula. i.e. molecular formula = empirical formula × n molecular formula mass where n empiricalformula mass = ● For determination of atomic mass: Dulong’s & Petit’s Law: Atomic weight of metal × specific heat capacity (cal/gm-°C) ≈ 6.4. It should be remembered that this law is an empirical observation and this gives an approximate value of atomic weight. This law gives better result for heavier solid elements, at high temperature conditions. CONCENTRATION TERMS Concentration Type Mathematical Formula Concept Percentage by mass w Massof solute 100 % w Massof solution   ×   =   Mass of solute (in gm) present in 100 gm of solution. Volume percentage v Volumeof solute 100 % v Volumeof solution   ×   =   Volume of solute (in cm3 ) present in 100 cm3 of solution. Mass-volume percentage w Massof solute 100 % v Volumeof solution   ×   =   Mass of solute (in gm) present in 100 cm3 of solution. Parts per million 6 Massof solute 10 ppm Massof solution × = Parts by mass of solute per million parts by mass of the solution