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Chapter 9 – Mathematics of Chemical Equations (Stoichiometry) I. Fundamentals of Stoichiometry A. Why do we balance chemical reactions and what do coefficients mean? 1. Unbalanced chemical equation shows the correct reactant/s and correct product/s. 2. Balanced chemical equation demonstrates the Law of Conservation of Mass or Matter because the number of atoms/ions of each chemical substance is the same on both sides of the arrow. a. Coefficients show the number of particles of each chemical substance: molecules, formula units, atoms, or ions depending upon the composition of the substance. b. Coefficents also represent the number of moles since moles “count” particles. 3. Substances react in specific ratios or proportions based upon the number of particles or moles, not grams. B. Mole – Mole problems 1. general formula: mole A (starting) x mole B (ratio) mole A 2. examples: C. Mass – Mass problem 1. general formula: mass x given 1 mol x molar mass mole ratio unknown given x molar mass unknown quantity of moles of moles of quantity of given given unknown unknown (g) (mole ratio) (g) 2. examples: D. Mass – Volume of a gas problems 1. general formula using molar volume: mass x given 1 mol x molar mass mole ratio x molar volume unknown unknown given quanity of moles of moles of volume of given given unknown unknown (mole ratio) 2. examples: 3. general formula using gas densities quantity of given (g) moles of given moles of quantity of volume unknown unknown (gas density) (mole ratio) (g) E. Volume – Volume problems 1. Equal volumes of gases under the same temperature and pressure contain equal numbers of particles and 2. therefore moles (Avogadro’s Hypothesis) 3. general formula: vol (mol) A x vol B (mol) vol A (mol) 3. examples: II. Limiting Reactants A. The quantity of the product/s is determined by the quantity of the reactant that is in short supply (limiting). Any “extra” reactant is termed in “excess.” B. To determine the limiting reactant, perform two massmass calculations. Whichever reactant produces the smaller quantity of the same product is limiting. C. Examples: III. Percent Yield A. Compares the actual amount of product produced in an experiment to the theoretical amount calculated using stoichiometry B. Formula: % yield = experimental yield x 100 theoretical yield 1. Usually the experimental amount produced is less than the theoretical amount as calculated through stoichiometry 2. example: 3. the higher the percent yield, the more successful or efficient the procedure C. Actual yield (experimental yield) versus theoretical yield 1. actual or experimental yield is the amount of product produced from conducting the actual reaction, and is always less than the calculated amount 2. theoretical yield is calculated from performing a stoichiometric calculation 3. Actual yield can be predicted from knowing the percent yield and theoretical yield a. Examples (pg 318): see sample problem G IV. Stoichiometric calculations involving particles A. To calculate particles you must use Avogadro’s number 1. one mole of particles (ions, molecules, atoms, or formula units contains) 6.02 x 1023 particles. B. Example calculations 1. 2.