Section 1: Defining Stoichiometry The amount of each reactant present at the start of a chemical reaction determines how much product can form. K What I Know W What I Want to Find Out L What I Learned

Essential Questions Which relationships can be derived from a balanced chemical equation? How are mole ratios written from a balanced chemical equation? Copyright © McGraw-Hill Education Defining Stoichiometry

Vocabulary Review New reactant stoichiometry mole ratio Defining Stoichiometry Copyright © McGraw-Hill Education

Particle and Mole Relationships Chemical reactions stop when one of the reactants is used up. Stoichiometry is the study of quantitative relationships between the amounts of reactants used and amounts of products formed by a chemical reaction. Stoichiometry is based on the law of conservation of mass. The mass of reactants equals the mass of the products. Copyright © McGraw-Hill Education Defining Stoichiometry

Interpreting Chemical Equations KNOWN UNKNOWN C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Equation interpreted in terms of molecules = ? Equation interpreted in terms of moles = ? Equation interpreted in terms of mass = ? Use with Example Problem 1. Problem The combustion of propane (C3H8) provides energy for heating homes, cooking food, and soldering metal parts. Interpret the equation for the combustion of propane in terms of representative particles, moles, and mass. Show that the law of conservation of mass is observed. SOLVE FOR THE UNKNOWN The coefficients in the chemical equation indicate the number of molecules. 1 molecule C3H8 + 5 molecules O2 → 3 molecules CO2 + 4 molecules H2O The coefficients in the chemical equation also indicate the number of moles. 1 mol C3H8 + 5 mol O2 → 3 mol CO2 + 4 mol H2O Response ANALYZE THE PROBLEM The coefficients in the balanced chemical equation shown below represent both moles and representative particles, in this case molecules. Therefore, the equation can be interpreted in terms of molecules and moles. The law of conservation of mass will be verified if the masses of the reactants and products are equal. Defining Stoichiometry Copyright © McGraw-Hill Education

204.1 g reactants = 204.1 g products Interpreting Chemical Equations SOLVE FOR THE UNKNOWN To verify that mass is conserved, first convert moles of reactant and product to mass by multiplying by a conversion factor—the molar mass—that relates grams to moles. moles of reactant or product × grams reactant or product 1 mol reactant or product = grams of reactant or product Calculate the mass of the reactant C3H8. 1 mol C 3 H 8 × 44.09 g C 3 H 8 1 mol C 3 H 8 = 44.09 C 3 H 8 Calculate the mass of the reactant O2. 5 mol O 2 × 32.00 g O 2 1 mol O 2 = 160.0 g O 2 Calculate the mass of the reactant CO2. 3 mol C O 2 × 44.01 g CO 2 1 mol CO 2 = 132.0 g CO 2 Calculate the mass of the reactant H2O. 4 mol H 2 O × 18.02 g H 2 O 1 mol H 2 O = 72.08 g H 2 O Add the masses of the reactants. 44.09 g C3H8 + 160.0 g O2 = 204.1 g reactants Add the masses of the products. 132.0 g CO2 + 72.08 g H2O = 204.1 g products The law of conservation of mass is observed. 204.1 g reactants = 204.1 g products EVALUATE THE ANSWER The sums of the reactants and the products are correctly stated to the first decimal place because each mass is accurate to the first decimal place. The mass of reactants equals the mass of products, as predicted by the law of conservation of mass. Defining Stoichiometry Copyright © McGraw-Hill Education

Particle and Mole Relationships A mole ratio is a ratio between the numbers of moles of any two substances in a balanced equation. The number of mole ratios that can be written for any equation is (n)(n – 1) where n is the number of species in the chemical reaction. Copyright © McGraw-Hill Education Defining Stoichiometry

Practice Problems Each table will act as a group of 4 Grab a worksheet Each group member is to do one of the following problems: #1 (b and c) on page 371 #3 (b and c) on page 372

Using Stoichiometry All stoichiometric calculations begin with a balanced chemical equation. 4Fe(s) + 3O2(g) → 2Fe2O3(s) Steps to solve mole-to-mole, mole-to-mass, and mass-to-mass stoichiometric problems: Write a balanced chemical equation for the reaction. Determine where to start your calculations by noting the unit of the given substance. If mass (in grams) of the given substance is the starting unit, you must convert to moles. If amount (in moles) of the given substance is the starting unit, convert moles of the given substance to moles of the unknown. Copyright © McGraw-Hill Education Stoichiometric Calculations

Using Stoichiometry The end point of the calculation depends on the desired unit of the unknown substance. If the answer must be in moles, stop you are finished. If the answer must be in grams, convert moles of unknown to grams of unknown using the molar mass as the conversion factor. Example: The carbon dioxide exhaled by astronauts can be removed from a spacecraft by reacting with lithium hydroxide as follows: CO2(g) + LiOH(s) → Li2CO3(s) + H2O An average person exhales about 20 moles of CO2 per day. How many moles of LiOH would be required to maintain 2 astronauts in a spacecraft for three days? Copyright © McGraw-Hill Education Stoichiometric Calculations

Using Stoichiometry Balance equation: CO2 + 2LiOH → Li2CO3 + H2O 2. Determine moles of given substance: 20 moles per person, 2 people = 40 moles x 3 days = 120 moles of CO2 Convert moles of given substance to moles of unknown: 120 mol CO2 x 2 mol/1 mol = 240 moles of LiOH Copyright © McGraw-Hill Education Stoichiometric Calculations

Mole-to-Mole Stoichiometry KNOWN UNKNOWN moles C3H8 = 10.0 mol C3H8 moles CO2 = ? mol CO2 Use with Example Problem 2. Problem One disadvantage of burning propane (C3H8) is that carbon dioxide (CO2) is one of the products. The released carbon dioxide increases the concentration of CO2 in the atmosphere. How many moles of CO2 are produced when 10.0 mol of C3H8 are burned in excess oxygen in a gas grill? SOLVE FOR THE UNKNOWN Write the balanced chemical equation for the combustion of C3H8. Use the correct mole ratio to convert moles of known (C3H8) to moles of unknown (CO2). 10.0 mol ? mol C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Mole ratio: 3 mol CO 2 1 mol C 3 H 8 10.0 mol C 3 H 8 × 3 mol CO 2 1 mol C 3 H 8 = 30.0 mol CO 2 Burning 10.0 moles of C3H8 produces 30.0 moles CO2. Response ANALYZE THE PROBLEM You are given moles of the reactant, C3H8 and must find the moles of the product, CO2. First write the balanced chemical equation, then convert from moles of C3H8 to moles of CO2. The correct mole ratio has moles of unknown substance in the numerator and moles of known substance in the denominator. Stoichiometric Calculations Copyright © McGraw-Hill Education

Mole-to-Mass Stoichiometry Use with Example Problem 3. Problem Determine the mass of sodium chloride (NaCl), commonly called table salt, produced when 1.25 mol of chlorine gas (Cl2) reacts vigorously with excess sodium. KNOWN UNKNOWN moles of chlorine = 1.25 mol Cl2 mass of sodium chloride = ? g NaCl SOLVE FOR THE UNKNOWN Write the balanced chemical equation, and identify the known and the unknown values. 1.25 mol ? g 2Na(s) + Cl2(g) → 2NaCl(s) Mole ratio: 2 mol NaCl 1 mol Cl 2 Response ANALYZE THE PROBLEM You are given the moles of the reactant, Cl2, and must determine the mass of the product, NaCl. You must convert from moles of Cl2 to moles of NaCl using the mole ratio from the equation. Then, you need to convert moles of NaCl to grams of NaCl using the molar mass as the conversion factor. Stoichiometric Calculations Copyright © McGraw-Hill Education

Mole-to-Mass Stoichiometry EVALUATE THE ANSWER Because the given number of moles has three significant figures, the mass of NaCl also has three. To quickly assess whether the calculated mass value for NaCl is correct, perform the calculations in reverse: divide the mass of NaCl by the molar mass of NaCl, and then divide the result by 2. You will obtain the given number of moles of Cl2. SOLVE FOR THE UNKNOWN Multiply moles of Cl2 by the mole ratio to get moles of NaCl. 1.25 mol Cl 2 × 2 mol NaCl 1 mol Cl 2 = 2.50 mol NaCl 2.50 mol NaCl × 58.44 g NaCl 1 mol NaCl = 146 g NaCl Stoichiometric Calculations Copyright © McGraw-Hill Education

Mass-to-Mass Stoichiometry KNOWN UNKNOWN mass of ammonium nitrate = 25.0 g NH4NO3 mass of water = ? g H2O Use with Example Problem 4. Problem Ammonium nitrate (NH4NO3), an important fertilizer, produces dinitrogen monoxide (N2O) gas and H2O when it decomposes. Determine the mass of H2O produced from the decomposition of 25.0 g of solid NH4NO3. SOLVE FOR THE UNKNOWN Write the balanced chemical equation, and identify the known and unknown values. 25.0 g ? g NH4NO3(s) → N2O(g) + 2H2O(g) Multiply grams of NH4NO3 by the inverse of molar mass to get moles of NH4NO3. 25.0 g NH 4 NO 3 × 1 mol NH 4 NO 3 80.04 g NH 4 NO 3 = 0.312 mol NH 4 NO 3 Response ANALYZE THE PROBLEM You are given a description of the chemical reaction and the mass of the reactant. You need to write the balanced chemical equation and convert the known mass of the reactant to moles of the reactant. Then, use a mole ratio to relate moles of the reactant to moles of the product. Finally, use the molar mass to convert from moles of the product to the mass of the product. Stoichiometric Calculations Copyright © McGraw-Hill Education

Mole ratio: 2 mol H 2 O 1 mol NH 4 NO 3 Mass-to-Mass Stoichiometry EVALUATE THE ANSWER The number of significant figures in the answer, three, is determined by the given grams of NH4NO3. To verify that the mass of H2O is correct, perform the calculations in reverse. SOLVE FOR THE UNKNOWN Mole ratio: 2 mol H 2 O 1 mol NH 4 NO 3 Multiply moles of NH4NO3 by the mole ratio to get moles of H2O. 0.312 mol NH 4 NO 3 × 2 mol H 2 O 1 mol NH 4 NO 3 = 0.624 mol H 2 O Multiply moles of H2O by the molar mass to get grams of H2O. 0.624 mol H 2 O × 18.02 g H 2 O 1 mol H 2 O = 11.2 g H2O Stoichiometric Calculations Copyright © McGraw-Hill Education

Section 3: Limiting Reactants A chemical reaction stops when one of the reactants is used up. K What I Know W What I Want to Find Out L What I Learned

Why do reactions stop? Reactions proceed until one of the reactants is used up and one is left in excess. The limiting reactant limits the extent of the reaction and, thereby, determines the amount of product formed. The excess reactants are all the leftover unused reactants. Copyright © McGraw-Hill Education Limiting Reactants

Why do reactions stop? Determining the limiting reactant is important because the amount of the product formed depends on this reactant. Copyright © McGraw-Hill Education Limiting Reactants

Calculating the Product when a Reactant is Limiting Example: S8(l) + 4Cl2(g) → 4S2Cl2(l) If 200.0g of sulfur reacts with 100.0g of chlorine, what mass of disulfur dichloride is produced? Determine moles of reactants Copyright © McGraw-Hill Education Limiting Reactants

Calculating the Product when a Reactant is Limiting 2. Determine whether the two reactants are in the correct mole ratio, as given in the balanced chemical equation. Only 1.808 mol of chlorine is available for every 1 mol sulfur, instead of the 4mol of chlorine required by the balanced chemical equation, thus chlorine is the limiting reactant. Copyright © McGraw-Hill Education Limiting Reactants

Calculating the Product when a Reactant is Limiting 3. Calculate the amount of product formed. Copyright © McGraw-Hill Education Limiting Reactants

Calculating the Product when a Reactant is Limiting Now that you have determined the limiting reactant and the amount of product formed, what about the excess reactant, sulfur? How much of it reacted? You need to make a mole-to-mass calculation to determine the mass of sulfur needed to react completely with 1.410 mol of chlorine. Copyright © McGraw-Hill Education Limiting Reactants

Calculating the Product when a Reactant is Limiting 2. Next, obtain the mass of sulfur needed: 3. Knowing that 200.0g of sulfur is available and only 90.42g is needed, you can calculate the amount of sulfur left unreacted when the reaction ends. Using an excess reactant can speed up the reaction. Using an excess reactant can drive a reaction to completion. Copyright © McGraw-Hill Education Limiting Reactants

Determining the Limiting Reactant Use with Example Problem 5. Problem The reaction between solid white phosphorus (P4) and oxygen produces solid tetraphosphorus decoxide (P4O10). This compound is often called diphosphorus pentoxide because its empirical formula is P2O5. a. Determine the mass of P4O10 formed if 25.0 g of P4 and 50.0 g of oxygen are combined. b. How much of the excess reactant remains after the reaction stops? Response ANALYZE THE PROBLEM You are given the masses of both reactants, so you must identify the limiting reactant and use it to find the mass of the product. From moles of the limiting reactant, the moles of the excess reactant used in the reaction can be determined. The number of moles of the excess reactant that reacted can be converted to mass and subtracted from the given mass to find the amount in excess. KNOWN UNKNOWN mass of phosphorus = 25.0 g P4 mass of tetraphosphorus decoxide = ? g P4O10 mass of oxygen = 50.0 g O2 mass of excess reactant = ? g excess reactant Copyright © McGraw-Hill Education Limiting Reactants

Determining the Limiting Reactant SOLVE FOR THE UNKNOWN Determine the limiting reactant. Write the balanced chemical equation, and identify the known and the unknown. 25.0 g 50.0 g ? g P4(s) + 5O2(g) → P4O10(s) Determine the number of moles of the reactants by multiplying each mass by the conversion factor that relates moles and mass—the inverse of molar mass. Calculate the moles of P4. 25.0 g P4 × 1 mol P4 123.9 g P4 = 0.202 mol P4 Calculate the moles of O2. 50.0 g O2 × 1 mol O2 32.00 g O2 = 1.56 mol O2 Calculate the actual ratio of available moles of O2 and available moles of P4. Calculate the ratio of moles of O2 to moles of P4. 1.56 mol O2 0.202 mol P4 = 7.72 mol O2 1 mol P4 Limiting Reactants Copyright © McGraw-Hill Education

Determining the Limiting Reactant SOLVE FOR THE UNKNOWN Determine the mole ratio of the two reactants from the balanced chemical equation. Mole ratio: 5 mol O2 mol P4 Because 7.72 mol of O2 is available but only 5 mol is needed to react with 1 mol of P4, O2 is in excess and P4 is the limiting reactant. Use the moles of P4 to determine the moles of P4O10 that will be produced. Multiply the number of moles of P4 by the mole ratio of P4O10 (the unknown) to P4 (the known). Calculate the moles of product (P4O10) formed. 0.202 mol P4 × 1 mol P4O10 1 mol P4 = 0.202 mol P4O10 To calculate the mass of P4O10, multiply moles of P4O10 by the conversion factor that relates mass and moles—molar mass. 0.202 mol P4O10× 283.9 g P4O10 1 mol P4O10 = 57.3 g P4O10 Because O2 is in excess, only part of the available O2 is consumed. Use the limiting reactant, P4, to determine the moles and mass of O2 used. Limiting Reactants Copyright © McGraw-Hill Education

Determining the Limiting Reactant SOLVE FOR THE UNKNOWN Multiply the moles of limiting reactant by the mole ratio to determine moles of excess reactant needed. 0.202 mol P4 × 5 mol O2 1 mol P4 = 1.01 mol O2 Convert moles of O2 consumed to mass of O2 consumed. Multiply the moles of O2 by the molar mass. 1.01 mol O2 × 32.00 g O2 1 mol O2 = 32.3 g O2 Subtract the mass of O2 used from the mass available. 50.0 g O2 available - 32.3 g O2 consumed = 17.7 g O2 in excess EVALUATE THE ANSWER All values have a minimum of three significant figures, so the mass of P4O10 is correctly stated with three digits. The mass of excess O2 (17.7 g) is found by subtracting two numbers that are accurate to the first decimal place. Therefore, the mass of excess O2 correctly shows one decimal place. The sum of the O2 that was consumed (32.3 g) and the given mass of P4 (25.0 g) is 57.3 g, the calculated mass of the product P4O10. Limiting Reactants Copyright © McGraw-Hill Education

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How Much Product? Laboratory reactions do not always produce the calculated amount of products. Reactants stick to containers. Competing reactions form other products. Copyright © McGraw-Hill Education Percent Yield

How Much Product? The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. The actual yield is the amount of product actually produced when the chemical reaction is carried out in an experiment. The percent yield of a product is the ratio of the actual yield expressed as a percent. Copyright © McGraw-Hill Education Percent Yield

Percent Yield Problem Response KNOWN UNKNOWN mass of silver nitrate = 0.500 g AgNO3 theoretical yield = ? g Ag2CrO4 actual yield = 0.455 g Ag2CrO4 percent yield = ? % Ag2CrO4 Use with Example Problem 6. Problem Solid silver chromate (Ag2CrO4) forms when excess potassium chromate (K2CrO4) is added to a solution containing 0.500 g of silver nitrate (AgNO3). Determine the theoretical yield of Ag2CrO4. Calculate the percent yield if the reaction yields 0.455 g of Ag2CrO4. SOLVE FOR THE UNKNOWN Write the balanced chemical equation, and identify the known and the unknown. 0.500 g ? g 2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq) Use molar mass to convert grams of AgNO3 to moles of AgNO3. 0.500 g AgNO3 × 1 mol AgNO3 169.9 g AgNO3 = 2.94×10−3 mol AgNO3 Response ANALYZE THE PROBLEM You know the mass of a reactant and the actual yield of the product. Write the balanced chemical equation, and calculate theoretical yield by converting grams of AgNO3 to moles of AgNO3, moles of AgNO3 to moles of Ag2CrO4, and moles of Ag2CrO4 to grams of Ag2CrO4. Calculate the percent yield from the actual yield and the theoretical yield. Percent Yield Copyright © McGraw-Hill Education

Percent Yield EVALUATE THE ANSWER SOLVE FOR THE UNKNOWN The quantity with the fewest significant figures has three, so the percent is correctly stated with three digits. The molar mass of Ag2CrO4 is about twice the molar mass of AgNO3, and the ratio of moles of AgNO3 to moles of Ag2CrO4 in the equation is 2:1. Therefore, 0.500 g of AgNO3 should produce about the same mass of Ag2CrO4. The actual yield of Ag2CrO4 is close to 0.500 g, so a percent yield of 93.2% is reasonable. SOLVE FOR THE UNKNOWN Use the mole ratio to convert moles of AgNO3 to moles of Ag2CrO4. 2.94×10−3 mol AgNO3 × 1 mol Ag2CrO4 2 mol AgNO3 = 1.47×10−3 mol Ag2CrO4 Calculate the theoretical yield. 1.47×10−3 mol Ag2CrO4 × 331.7 g Ag2CrO4 1 mol Ag2CrO4 = 0.488 g Ag2CrO4 Calculate the percent yield. 0.455 g Ag2CrO4 0.488 g Ag2CrO4 X 100= 93.2% Ag2CrO4 Percent Yield Copyright © McGraw-Hill Education

Percent Yield in the Marketplace Percent yield is important in the cost effectiveness of many industrial manufacturing processes. Copyright © McGraw-Hill Education Percent Yield

Quiz 1. What is the maximum amount of product that can be produced from a given amount of reactant? percent yield theoretical yield actual yield mole ratio

1. What is the maximum amount of product that can be produced from a given amount of reactant? percent yield theoretical yield actual yield mole ratio

2. The reaction of 5.0 grams of fluorine with excess chlorine produced 5.6 grams of ClF3. What percent yield of ClF3 was obtained?      Cl2 + 3F2 → 2ClF3 0.58 0.69 0.76 0.86

2. The reaction of 5.0 grams of fluorine with excess chlorine produced 5.6 grams of ClF3. What percent yield of ClF3 was obtained?      Cl2 + 3F2 → 2ClF3 0.58 0.69 0.76 0.86

3. In the bunsen burner reaction of methane and oxygen, what is the percent yield if 64.0 g of methane yields 95.1 grams of carbon dioxide?      CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) 0.46 0.54 1.49 0.5

3. In the bunsen burner reaction of methane and oxygen, what is the percent yield if 64.0 g of methane yields 95.1 grams of carbon dioxide?      CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) 0.46 0.54 1.49 0.5

4. Percent yield can never be greater than ________. theoretical yield 100% both a and b 50%

4. Percent yield can never be greater than ________. theoretical yield 100% both a and b 50%

5. The statement that percent yield can never be greater than theoretical yield is another example of the ________. ideal gas law law of conservation of mass law of conservation of momentum law of conservation of energy

5. The statement that percent yield can never be greater than theoretical yield is another example of the ________. ideal gas law law of conservation of mass law of conservation of momentum law of conservation of energy