Stoichiometry (Chapter 9) & Molarity (Part of Chapter 14)Test Review Topics (Test is Thursday, January27th)

  • Determining molar mass properly (round each elements’ mass to 1 number after the decimal before multiplying by the number of moles of that element in the compound
  • Mole-mole conversions
  • Mixed mass and mole conversions
  • Mass-mass conversions (using molar mass)
  • Stoichiometry conversions using
  • Avogadro’s # (1 mol = 6.02 x 1023 atoms, ions, molecules or formula units)
  • Molar volume (1 mol = 22.4 L at STP)
  • Use stoichiometry to determine
  • Limiting reactant – amount that you start with that makes a smaller amount of the product
  • Excess reactant – amount that you start with that makes a larger amount of the product
  • Amount of product that can be produced – you must change both reactants to grams of the same product
  • Amount of excess leftover after reaction stops – you must find the difference in the amounts of the two products produced and change it back to grams of the excess reactant. The amount leftover must be less than the initial amount.
  • Excess reactant used up – original amt. – excess leftover
  • Amount more of limiting reactant needed – change excess leftover to grams of the limiting reactant
  • Determine theoretical yield for a reaction to use in a percent yield calculation – change the amount in the reactants to the amount of the product that is the actual yield.
  • Determine percent yield of a reaction given the amount of one reactant (use to determine the theoretical yield) and the amount of one product (actual yield, do not change!)
  • % Yield = (Actual Yield/Theoretical Yield) x 100
  • Actual Yield – what you actually get when you do an experiment
  • Theoretical Yield – what you should get if the experiment you performed had no errors and went to completion.
  • Reasons you might not get 100% yield in an experiment.
  • The closer your percent yield is to 100% the better, an amount over 100% is not better, it just means that you got more than you were supposed to end up getting from the reaction.
  • Molarity (a.k.a. concentration) – dilute vs. concentrated
  • Molarity (M) = moles of solute / liters of solution
  • Be able to solve for molarity, moles, grams, liters or milliliters
  • 1 L = 1000 mL
  • Solution Stiochiometry – using molarity to determine a volume, mass, molarity or number of moles of another substance using a balanced chemical equation
  • Making solutions of specific concentration from stock solutions or concentrated solutions (dilutions)
  • M1V1 = M2V2
  • Keep sets of conditions together, you will be given 3 of the 4 parts of the problem and be asked to solve for either a second volume or second molarity