Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University

Overview Stoichiometry Percent Yield

Stoichiometry Stoichiometry deals with the quantity of materials consumed and or produced in chemical reactions. Stoichiometry deals specifically with mass-mole relationships between reactants and products found in a chemical equation.

Stoichiometry Two Fundamental Principles Must always work from a balanced reaction. Axiom #3 is always foremost in out minds. When in doubt, convert to moles for they guide us everywhere we wish to go in chemistry.

Chapter 4, Unnumbered Figure 1, Page 130 5

Stoichiometry How to Solve Stoichiometry Problems? Use Handy-Dandy Five Step Method 1. Write out balanced chemical reaction. Place the given and unknown quantities (with units) above item in reaction. 2. Calculate molar masses for items needed to solve problem. Place these below items in reaction.

Stoichiometry Handy Dandy Five Step Method 3. Convert given quantity into moles (Axiom #3). 4. Using the balanced chemical equation (step 1), convert moles obtained in step 3 into moles of quantity you are after. 5. Convert moles of final quantity into the desired unit asked for in problem.

Stoichiometry: Concept Map

Stoichiometry 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O How many grams of water are produced from the combustion of 32.75 g of butane (C4H10)? The balanced chemical reaction for this process is shown below. 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

Stoichiometry 32.75 g ? g 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O Apply Handy Dandy Five Step Method Step 1: Equation is already balanced. Known and unknown values are placed above item in equation. 32.75 g ? g 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

Stoichiometry Apply Handy Dandy Five Step Method Step 2: Calculate Molar Masses for known and unknown. Values are placed below item in equation. C4H10 = (4 x 12.01 g/mol) + (10 x 1.01 g/mol) = 58.14 g/mol H2O = (2 x 1.01 g/mol) + (1 x 16.00 g/mol) = 18.02 g/mol

Stoichiometry 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O Apply Handy Dandy Five Step Method Step 2: Calculate Molar Masses for known and unknown. Values are placed below item in equation. 32.75 g ? g 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O 58.14 g/mol 18.02 g/mol

Stoichiometry Apply Handy Dandy Five Step Method Step 3: Convert given quantity into moles (Axiom #3) 32.75 g C4H10 X ( 1 mol C4H10 / 58.14 g C4H10 ) = 0.56329 mol C4H10

Stoichiometry Apply Handy Dandy Five Step Method Step 4: Using the balanced chemical equation (see Step 1), convert moles obtained in step 3 into moles of quantity you are after. Balanced reaction shows relationship between water and butane (2 C4H10 + 13 O2 → 8 CO2 + 10 H2O) 0.56329 mol C4H10 (10 moles H2O / 2 moles C4H10) = 2.8164 mol H2O

Stoichiometry Apply Handy Dandy Five Step Method Step 5: Convert moles of final quantity into the desired unit asked for in problem. 2.81647 mol H2O x (18.02 g H2O / 1 mol H2O) = 50.75 g H2O Check: Answer is reasonable with correct units.

Percent Yield Percent yield is a measure of how completely reactants are turned into products. Most reactions do not go to completion (i.e., reach 100% conversion) due to side reactions, transfer losses, etc.

% yield = (actual yield / theoretical yield) x 100% Percent Yield Percent yield is calculated using the following equation. % yield = (actual yield / theoretical yield) x 100% actual yield is amount obtained in the analysis. theoretical yield is amount obtained assuming 100 % complete reaction.

A Second Example for Practice According to the U.S. Dept. Energy if 3.4 x 1015 g of octane was burned in 2004, how many grams of carbon dioxide was released into the atmosphere that year? Use the Handy-Dandy Five Step Method to Solve this problem.

Figure: 04-03-03UN Title: Stoichiometry of fossil fuel combustion Caption: To calculate the mass of CO2 emitted upon the combustion of 3.4 x 1015 g of octane, use the setup shown.

Chapter 4, Unnumbered Figure 2, Page 130 20

Practice Problem Answer to Problem 1.0 x 1016 g CO2 This value is the maximum amount of carbon dioxide that would be produced assuming complete reaction (hence, this is the theoretical yield).