Amount of Substance 0.0 / 5 ? ChemistryequationsASAQA Created by: Laura J Thomas - Team GRCreated on: 10-04-14 19:55 Relative atomic and molecular mass Relative atomic mass - average mass of one atom of an element/one twelfth mass of an atomic of carbon-12 Relative molecular mass - average mass of one molecule/one twelfth mass of an atomic of carbon-12 Avogadro constant the number of atoms in 12g of carbon-12 6.022 x10^23 Mole - the amount of substance that contains 6.022 x10^23 particles number of moles = mass in grams/relative molecular mass 1 of 6 Ideal gas equation Boyle's Law - PV = constant Charles' Law - P/T = constant Gay-Lussac's Law - PV/T = constant for fixed mass of gas Ideal gas equation - PV = nRT P = pressure - Pa V = volume - m^3 n = number of moles R = constant - 8.31 J/K/mol T = temperature - K 2 of 6 Empirical and molecular formulae An empirical formula is one which represents the simplest ratio of the atoms of each element present in a compound Empirical find masses of each of the elements present work out number of moles of atoms of each element convert number of moles of each element into a whole number ratio by dividing each number by the smallest number Molecular divide the relative molecular mass by the relative mass of the empirical formula 3 of 6 Moles in solutions Concentrations tells us how much solute is present in a known volume of solution measured in mol/dm^3 number of moles/volume in dm^3 Moles (concentration*volume)/1000 when concentration is given in mol/dm^3 and volume in cm^3 4 of 6 Ionic equations In some reactions, the equation can be simplified by considering the ions present. Example HCl + NaOH ---> NaCl + H2O Ions present: H+ Cl- Na+ OH- Na+ Cl- H+ + Cl- + Na+ + OH- ---> Na+ + Cl- + H20 H+ + OH- ---> H2O 5 of 6 Atom economies and percentage yields Atom economy = (100*mass of desired product)/total mass of reactants Percentage yield = (100*number of moles of specified product)/theoretical maximum number of moles of product 6 of 6
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