12.1 – What is Stoichiometry? Objectives: Describe stoichiometry calculations. Write mole ratios from balanced equations. Complete mole-mole stoichiometry problems.

12.1 – Mole-Mass Relationships Stoichiometry – study of quantitative relationships that exist in chemical formulas and/or chemical reactions. Law of Conservation of Mass – Matter can neither be created nor destroyed in a chemical reaction The mass of the reactants must equal the mass of the products

12.1 – Using Mole Ratios Mole Ratio - The relationship between moles of different substances in a balanced chemical equation. 2H2O  2H2 + 1O2 In this chemical reaction, the mole ratio of H2O to H2 is 2:2 (or simplifying 1:1). For each mole of H2O reacted, 1 mole H2 gas will be produced. The mole ratio can be expressed as a conversion factor The mole ratio of H2O to O2 is 2:1. For each mole of H2O reacted, ½ mole of O2 will be produced. 2 mol H2O or 2 mol H2 2 mol H2O or 1 mol O2

12.1 – Mole Ratios Examples: Given the unbalanced equation, Sb(s) + Cl2(g)  SbCl3(s) what is the mole ratio of Cl2 to SbCl3? 2Sb(s) + 3Cl2(g)  2SbCl3(s) 3:2 Pb(C2H3O2)2 (aq) + KI (aq)  PbI2(s) + KC2H3O2 (aq) what is the mole ratio of Pb(C2H3O2)2 to KI? 1Pb(C2H3O2)2 (aq) + 2KI (aq)  1PbI2(s) + 2KC2H3O2 (aq) 1:2

12.2 – What is Stoichiometry? Objectives: Use mole ratios to complete stoichiometric calculations.

12.2 – Solving Stoichiometry Problems The 4 Steps Balance the equation. Convert the amount of known substance into moles if necessary. Convert the number of moles of known substance into moles of unknown substance using the mole ratio. Convert the number of moles of the unknown substance into the desired units if necessary. Amount given x 1 mol moles of unknown molar mass 6.02 x 1023 p. 22.4 L = Amount of unknown   moles of known

12.2 – Mole to Mole Problems The 4 Steps Balance the equation. Convert the amount of known substance into moles if necessary. Convert the number of moles of known substance into moles of unknown substance using the mole ratio. Convert the number of moles of the unknown substance into the desired units if necessary. moles known x moles of unknown =   moles of known

12.2 – Mole to Mole Problems Stoichiometry Conversions Mass (g) Mass molar mass mass x 1 mol Mole Ratio Step 1 mol moles x molar mass molar 22.4 L volume x 1 mol 1 mol moles x 22.4 L Volume (L) Volume (L) moles x moles unknown moles known Moles of Known Substance Moles of Unknown Substance 6.02 x 10 23 part. part. x 1 mol moles x 6.02 x 1023 part. This ratio is derived from the coefficients of the balanced equation. 1 mol Particles (molecules, ions, atoms, formula units, etc.) Particles (molecules, ions, atoms, formula units, etc.)

12.2 – Mole to Mass / Mass to Mole The 4 Steps Balance the equation. Convert the amount of known substance into moles if necessary. Convert the number of moles of known substance into moles of unknown substance using the mole ratio. Convert the number of moles of the unknown substance into the desired units if necessary. Mole to Mass Mass to Mole moles given x moles of unknown molar mass = Amount of unknown   moles of known 1 mol mass given x 1 mol moles of unknown = Amount of unknown   molar mass moles of known

12.2 – Mass to Particle / Particle to Mass The 4 Steps Balance the equation. Convert the amount of known substance into moles if necessary. Convert the number of moles of known substance into moles of unknown substance using the mole ratio. Convert the number of moles of the unknown substance into the desired units if necessary. Mass to Particles Particles to Mass mass given x 1 mol moles of unknown 6.02 x 1023 p. = particles of unknown   molar mass moles of known particles given x 1 mol moles of unknown molar mass = mass of unknown   6.02 x 1023 p. moles of known

12.2 – Mass to Mass The 4 Steps Balance the equation. Convert the amount of known substance into moles if necessary. Convert the number of moles of known substance into moles of unknown substance using the mole ratio. Convert the number of moles of the unknown substance into the desired units if necessary. Mass to mass mass given x 1 mol moles of unknown molar mass = mass of unknown   moles of known

12.3 – Limiting Reactants Objectives: Describe how a limiting reactant affects a chemical reaction. Determine the limiting reactant in a chemical reactions. Calculate the mass of product for a reaction using limiting reactants. Calculate the excess reactant remaining after a chemical reaction.

12.3 – Limiting Reactants Why Do Reactions Stop? Reactions usually stop because one reactant has run out. Stoichiometric Proportion – The reactants are in exactly the right proportion so that one does not limit the other(s). Limiting Reactant – The reactant that will be used up first in a chemical reaction. The quantities of products in a chemical formed in a chemical reaction are determined by the amount of the limiting reactant. All other reactants are considered to be “in excess” meaning some will remain when the reaction has stopped

12.3 – Limiting Reactants Determining the Limiting Reactant To find the limiting reactant, carry out the stoichiometric calculations with each reactant. The calculation that provides the least product is based on the limiting reactant Example: H2 reacts with O2 to form H2O. If 50. grams of H2 reacts with 50. grams of O2, what mass of water will be produced? 2H2 + O2 → 2H2O 50. g 50. g ? g 50. g H2 x 1 mol H2 2 mol H2O 18.015 g H2O = 450 g H2O   2.016 g H2 2 mol H2 1 mol H2O 50. g O2 x 1 mol O2 2 mol H2O 18.015 g H2O = 56 g H2O   31.998 g O2 1 mol H2O

12.4 – Percent Yield Objectives: Differentiate between theoretical yield and actual yield. Calculate the percent yield for a reaction given the theoretical yield.

12.4 – Percent Yield Yields Theoretical yield – the maximum amount of a product that can be produced based on calculation (also called calculated yield or mathematical yield) Actual yield – the amount of a product that is actually produced through experiment (also called laboratory yield, experimental yield, or real yield). Percent yield – the % of the expected yield that was actually produced. percent yield = actual yield x 100 theoretical yield