Basic Concepts from General Chemistry PDF

BASIC CONCEPTS FROM GENERAL CHEMISTRY Atomic Weight – refer to the relative weights of the atoms as compared with carbon. Carbon has atomic weight standard of 12. Gram atomic weight – refers to a quantity of the element in grams corresponding to the atomic weight. Gram molecular weight – GMW or MW...

BASIC CONCEPTS FROM GENERAL CHEMISTRY Atomic Weight – refer to the relative weights of the atoms as compared with carbon. Carbon has atomic weight standard of 12. Gram atomic weight – refers to a quantity of the element in grams corresponding to the atomic weight. Gram molecular weight – GMW or MW refers to the molecular weight in grams of any particular compound (referred also as mole). Molar solution – consists of 1-gram molecular weight dissolved in enough water to make 1 liter of solution. Molar concentrations are used for equilibrium calculations. Molal solution – consists of 1-gram molecular weight dissolved in 1 liter of water, resulting solutions having a volume slightly in excess of 1 liter. This is used when physical properties of solutions such as vapor pressure, freezing point, boiling point, are involved. MW Equivalent weight: EW = ---------- Z Where Z = absolute value of the ion charge = the number of H+ or OH- ions a species can react with or yield in an acid-base reaction = the absolute value of the change in valence occurring in an oxidation-reduction reaction Ex. 1. Calculating the Atomic Weight *Refer to Periodic Table Isotopes CL – 35: Atomic mass = 34.969, Abundance = 75.77% CL – 37: Atomic mass = 36.966, Abundance = 24.23% Solution: Atomic Weight of Cl = (34.969×0.7577) + (36.966×0.2423) = 26.504+8.957≈35.46 2. Determine equivalent weight of Ca2+ , CaCO3 a. EW = MW / Z = 40 g per mole/ 2 = 20 g per equivalent b. MW of CaCO3 = 40 x 1 + 12 x 1 + 16 x 3 = 100 g per mole Z=2 E = MW / Z = 100 g per mole / 2 = 50 g per equivalent 3. What concentration is 40 mg/l of Ca2+ when expressed in CaCO3? One equivalent of an ion or molecule is “chemically” equivalent to one equivalent of a different ion or molecule. Thus is concentrations are expressed in terms of equivalents per little (eq/l) they can be added, subtracted or converted easily. Eq/l Ca2+ = 40 mg/l x 1eq/20 g x 1 g/1000mg = 0.002 eq./l of Ca2+ - mg/L as CaCO3 = 0.002 eq/l x 50 g/eq x 1000mg/g = 100 mg/l as CaCO3 - Ionic bond – formed by the transfer of electrons from one atom to another Covalent bond – when electrons are not transferred by are shared between atoms e..gCl2, N2, O2 Valency or oxidation number – determined by the number of electrons that it can take on, give up or share with other atoms. If electrons are lost, the atom becomes positively charged ion, and if electrons are gained, the atom becomes negatively charged ion Reduction/Oxidation: - Metal or metal-like element loses electrons to gain or approach a stable condition with no electrons on its outer ring. - The nonmetal steals electrons from the metal to complete its outer ring to eight electrons, a stable configuration. Chemical Equations: Chemical reactions become equations only when they are balanced Mass must be conserved – the total number of each kind of atom must be the same on both sides of the equation The sum of the charge on one side of the equation must be equal that on the other Example: NaOH + HCl -----> NaCl + H2O (23 + 16 + 1) (1 + 35.5) (23 + 35.5) (2x1 + 16) 40 36.5 58.5 18 40 g of NaOH combined with 36.5 g HCl will yield 58.5 g NaCl and 18 g H2O Example 2: 2 HCl + Na2CO3 -----------> H2O + 2NaCl + CO2 73 106 18 117 44  GAS LAWS: 1. Boyle’s Law: The volume of a gas varies inversely with its pressure at a constant temperature 2. Charles Law: The volume of gas at constant pressure varies in direct proportion to the absolute temperature. 3. Generalized Gas Law: For a given quantity of gas, Boyle’s law and Charles’ law can be combined: PV=nRT Where n = to the number of moles of gas in a particular sample R = universal constant for all gases = depends on the units chosen for the measurement of P, V and T = 1 mole of gas at 1 atm pressure occupies a volume of 22. 414 liters at 273 K. = 0.082 liter atmosphere per mole per Kelvin (0.082 l-atm/mol-K) Example: What tank volume is required to hold 10,000 kg of methane gas (CH4) at 25oC and 2 atm pressure? Molecular weight of CH4 gas is 12 + 4x1 = 16 g Number of moles in 10,000 kg is 10,000,000 / 16 g = 625,000 mol General gas law: V = nRT/P = 625,000 (0.082) (273+ 25)/ 2 = 7.64 x 106 liters Volume of tank = 7.64 x 106 liters 4. Dalton’s law of Partial Pressure: In a mixture of gas, such as air, each gas exerts pressure independently of the others. The partial pressure of each gas is proportional to the amount (percent by volume) of the gas in the mixture, or in other words, it is equal to the pressure that gas would exert if it were the sole occupant of the volume available to the mixture. 5. Henry’s law: The weight of any gas that will dissolve in a given volume of a liquid, at constant temperature, is directly proportional to the pressure that the gas exerts above the liquid. C equi = KH x Pgas Where C equi = concentration of gas dissolved in the liquid at equilibrium Pgas = is the partial pressure of the gas above the liquid KH = is the Henry’s law constant for the gas at the given temperature. Example: KH for oxygen in water at 20oC is 43.8 mg/1-atm. Since air contains 21 percent by volume of oxygen, the partial pressure in air according to Dalton’s law would be 0.21 atm when the total air pressure is 1 atm. Therefore the equilibrium concentration of oxygen in water at 20oC and in the presence of 1 atm of air would be 43.8 x 0.21 = 9.2 mg/l Solution: Identify the Henry’s Law constant (KH) and the partial pressure (Pgas): KH for O = 20°C = 43.8 mg/L·atm Partial pressure of PO₂ (Pgas) = 0.21 atm. a. Apply Henry’s Law Formula: C equi = KH x Pgas C equi = 43.8×0.21 Cequi = 9.18 mg/L Application of Henry’s Law: 1. Aeration of water supply for the removal of gases 2. Aeration of domestic sewage for the application of oxygen 3. Removal of gases from industrial wastewater 6. Graham’s Law – the rates of diffusion of gases are inversely proportional to the square root of their density. Example: Using atomic weights of hydrogen, oxygen, chlorine, bromine: 1, 16, 36, 80 respectively. Oxygen diffuses about one-fourth, chlorine about 1/6, and bromine about 1/9 as fast as hydrogen. 7. Gay-Lussac’s Law of Combining Volume The volumes of all gases that react and that are produced during the course of reaction are related numerically to one another as a group of small, whole numbers. C + CO2 -----------> CO2 Solid 1 vol 1 vol CH4 + 2O2+ ----------> CO2 + 2 H2O 1 vol 2 vol 1 vol 2 vol (>100oC) 0 vol (< 100oC) SOLUTIONS Vapor Pressure – the presence of a non-volatile solute in a liquid always lowers the vapor pressure of the solution. Example: When sugar, sodium chloride, or similar substance is dissolved in water, the vapor pressure is decreased. Raoult’s Law: The extent of the physical blocking effect or depression of the vapor pressure is directly proportional to the concentration of the particles in solution. Raoult’s law is applicable only to dilute solutions. For solutes that do not ionize – the effect is proportional to the molal concentration For solutes that do ionize – the effect is proportional to the molal concentration times the number of ions formed per molecule of solute modified by the degree of ionization.