1. Foundations of College Chemistry, 14 th Ed. Morris Hein and Susan Arena Accurate measurement and dosage calculations are critical in dispensing medicine to patients all over the world. Calculations from Chemical Equations Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

2. 9.1 Introduction to Stoichiometry 9.2 Mole-Mole Calculations 9.3 Mole-Mass Calculations 9.4 Mass-Mass Calculations 9.5 Limiting Reactant and Yield Calculations Chapter Outline © 2014 John Wiley & Sons, Inc. All rights reserved.

3. Molar Mass (MM): sum of the atomic mass of the atoms in an element, compound, or formula unit. Mole: Avogadro’s number (6.022 x 10 23 ) of units (atoms, molecules, ions etc.) grams of a substance moles of the substance Useful Conversion Factors Molar mass = number of units of substance 6.022 x 10 23 units of substance MM allows conversion between g and mol of a substance. Moles of a substance = Mole/Molar Mass Review © 2014 John Wiley & Sons, Inc. All rights reserved.

4. Equations must always be balanced before calculation of any mass, moles, or volume of a reactant or product! Stoichiometry: area of chemistry that deals with quantitative relationships between products and reactants in chemical equations. a A + b B c C + d D Example Using X.X g of A, how much C will be formed? Solving stoichiometry problems always requires the use of: 1. A balanced chemical equation (coefficients must be known!) 2. Conversion factors in units of moles (i.e. mole ratios) Introduction to Stoichiometry © 2014 John Wiley & Sons, Inc. All rights reserved.

5. Mole ratio: ratio (conversion factor) between any two species in a chemical reaction. 2 H 2 (g) + O 2 (g) 2 H 2 O (l) Example The coefficients of a balanced chemical equation are used to generate mole ratios. 6 possible mole ratios exist: 2 mol H 2 1 mol O 2 2 mol H 2 2 mol H 2 O 1 mol O 2 2 mol H 2 2 mol H 2 O 2 mol H 2 1 mol O 2 2 mol H 2 O 1 mol O 2 Mole Ratios © 2014 John Wiley & Sons, Inc. All rights reserved.

6. The mole ratio can be used as a conversion factor to convert between moles of one substance and another. Example If 4.0 mol of oxygen are present, how many moles of H 2 O could be formed? = 8.0 mol H 2 O 4.0 mol O 2 × 2 mol H 2 O 1 mol O 2 2 H 2 (g) + O 2 (g) 2 H 2 O (l) Mole Ratios © 2014 John Wiley & Sons, Inc. All rights reserved.

7. Given the following balanced chemical equation, write the mole ratio need to calculate: a. The moles of H 2 O produced from 3 moles of CO 2 b. The moles of H 2 needed to produce 3 moles of H 2 O. CO 2 (g) + 4 H 2 (g) CH 4 (g) + 2 H 2 O (l) = 6.0 mol H 2 O 3.0 mol CO 2 × 2 mol H 2 O 1 mol CO 2 a. Mole ratio Desired quantity in the numerator of the mole ratio: known quantity in the denominator Mole Ratios Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

8. 3.0 mol H 2 O × 4 mol H 2 2 mol H 2 O Mole ratio = 6.0 mol H 2 Given the following balanced chemical equation, write the mole ratio need to calculate: a. The moles of H 2 O produced from 3 moles of CO 2 b. The moles of H 2 needed to produce 3 moles of H 2 O. CO 2 (g) + 4 H 2 (g) CH 4 (g) + 2 H 2 O (l) b. Desired quantity in the numerator of the mole ratio: known quantity in the denominator Mole Ratios Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

9. Given the following balanced chemical equation, what is the mole ratio needed to calculate the following: the moles of KCl produced when 4.5 moles of O 2 are formed? 2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g) 2 mol KCl 3 mol O 2 = 3.0 mol KCl 4.5 mol O 2 × 2 mol KCl 3 mol O 2 Mole ratio 3 mol KCl 2 mol O 2 3 mol O 2 2 mol KCl 2 mol O 2 3 mol KCl Calculate Mole Ratios Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

10. 1. Make sure the equation is balanced! 2. If needed, convert the quantity of known substance to moles. 3. Convert the moles of known substance to desired substance using a mole ratio. 1 mol substance molar mass substance Moles = (grams) x Mole ratio = moles of desired substance moles of known substance Moles desired substance = Moles of known substance x Mole ratio From Step 2 Problem Solving for Stoichiometry Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

11. 4. Convert moles of desired substance to the desired units from the problem. Molar mass (in g) 1 mole grams = (moles) x If answer is in moles, you are finished. If answer is in grams, multiply by the compound’s molar mass. 6.022 x 10 23 molecules 1 mole Atoms/molecules = (moles) x If answer is in atoms/molecules, multiply by Avogadro’s number. Problem Solving for Stoichiometry Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

12. Flow Chart for Stoichiometry Problems Grams of Known Atoms/Molecules of Known Moles of Known Moles of Desired Step 2 Step 3 Using the Mole Ratio Atoms/Molecules of Desired Grams of Desired Step 2 Step 4 Problem Solving for Stoichiometry Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

13. Known substance is given in moles; desired substance is requested in moles. How many moles of CO 2 will be produced by reaction of 2.0 mol of glucose, given the following balanced equation? Calculate = 12 mol CO 2 2.0 mol C 6 H 12 O 6 × Solution Map mol C 6 H 12 O 6 mol CO 2 The mole ratio needed relates mol C 6 H 12 O 6 to mol CO 2. 6 mol CO 2 1 mol C 6 H 12 O 6 C 6 H 12 O 6 + 6 O 2 6 CO 2 + 6 H 2 O 6 mol CO 2 1 mol C 6 H 12 O 6 Mole-Mole Calculations © 2014 John Wiley & Sons, Inc. All rights reserved.

14. a. 8.3 x 10 -27 molecules b. 3.3 x 10 -26 molecules c. 3.0 x 10 21 molecules d. 1.2 x 10 22 molecules How many H 2 O molecules are produced when 0.010 mol O 2 react, given the following balanced equation? Calculate = 1.2 x 10 22 molecules 0.010 mol O 2 × Solution Map mol O 2 mol H 2 O molecules H 2 O Mole ratio: 2 H 2 + O 2 2 H 2 O 2 mol H 2 O 1 mol O 2 2 mol H 2 O 1 mol O 2 × 6.022 x 10 23 molecules H 2 O 1 mole H 2 O Mole-Mole Calculations Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

15. a. 0.38 moles b. 0.67 moles c. 1.0 moles d. 0.25 moles How many moles of Al are produced when 0.5 mol of O 2 react, given the following balanced equation? Calculate = 0.67 mol Al 0.5 mol O 2 × Solution Map mol O 2 mol Al Mole ratio: 4 Al + 3 O 2 2 Al 2 O 3 4 mol Al 3 mol O 2 4 mol Al 3 mol O 2 Mole-Mole Calculations Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

16. What mass of H 2 can be produced when 6.0 mol of Al reacts with HCl? = 18 g H 2 Solution Map mol Al mol H 2 g H 2 The mole ratio and molar mass of H 2 are needed: 3 mol H 2 2 mol Al 2 Al + 6 HCl 2 AlCl 3 + 3 H 2 6.0 mol Al 3 mol H 2 2 mol Al × 2.016 g H 2 1 mol H 2 2.016 g H 2 1 mol H 2 Calculate × Mole-Mass Calculations © 2014 John Wiley & Sons, Inc. All rights reserved.

17. How many moles of water are produced when 325 g of octane (C 8 H 18 ) are burned? = 25.6 mol H 2 O Solution Map g C 8 H 18 mol C 8 H 18 moles H 2 O The mole ratio and molar mass of C 8 H 18 are needed: 18 mol H 2 O 2 mol C 8 H 18 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O 325. g C 8 H 18 × 114.2 g C 8 H 18 Calculate 1 mol C 8 H 18 114.2 g C 8 H 18 1 mol C 8 H 18 18 mol H 2 O 2 mol C 8 H 18 × Mole-Mass Calculations Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

18. 2 mol AgNO 3 1 mol Ag 2 S 1 mol AgNO 3 How many grams of AgNO 3 are needed to produce 0.25 mol of Ag 2 S? = 85.0 g AgNO 3 Solution Map mol Ag 2 S mol AgNO 3 g AgNO 3 The mole ratio and molar mass of AgNO 3 are needed: 2 mol AgNO 3 1 mol Ag 2 S 2 AgNO 3 + H 2 S Ag 2 S + 2 HNO 3 0.25 mol Ag 2 S × 169.9 g AgNO 3 Calculate 1 mol AgNO 3 a. 42.5 g b. 57.1 g c. 2.19 x 10 -3 g d. 85.0 g × 169.9 g AgNO 3 Mole-Mass Calculations Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

19. How many grams of HNO 3 are required to produce 8.75 g of N 2 O from the following reaction? = 125. g HNO 3 Solution Map g N 2 O mol N 2 O mol HNO 3 g HNO 3 The mole ratio and molar masses of N 2 O and HNO 3 are needed: 10 mol HNO 3 1 mol N 2 O 4 Zn + 10 HNO 3 4 Zn(NO 3 ) 2 + N 2 O + 5 H 2 O 8.75 g N 2 O × 44.02 g N 2 O Calculate 1 mol N 2 O 1 mol HNO 3 63.02 g HNO 3 44.02 g N 2 O 1 mol N 2 O10 mol HNO 3 1 mol N 2 O × 1 mol HNO 3 63.02 g HNO 3 × Mass-Mass Calculations © 2014 John Wiley & Sons, Inc. All rights reserved.

20. a. 204 g b. 249 g c. 22.6 g d. 27.6 g How many grams of CrCl 3 are required to produce 75.0 g of AgCl using the following reaction? = 27.6 g CrCl 3 Solution Map g AgCl mol AgCl mol CrCl 3 g CrCl 3 Mole ratio/molar masses needed: 1 mol CrCl 3 3 mol AgCl CrCl 3 + 3 AgNO 3 Cr(NO 3 ) 3 + 3 AgCl 75.0 g AgCl × 143.3 g AgCl Calculate 1 mol AgCl 1 mol CrCl 3 158.4 g CrCl 3 × 143.3 g AgCl 1 mol AgCl 1 mol CrCl 3 3 mol AgCl 1 mol CrCl 3 158.4 g CrCl 3 × Mass-Mass Calculations Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

21. In many chemical reactions, one reactant is used in excess. The maximum amount of product formed depends on the amount of reactant not in excess (the limiting reactant). A Nonchemical Analogy To put together a bicycle, you need several parts. The number of bicycles is limited by the part you have the least of. Limiting Reactants © 2014 John Wiley & Sons, Inc. All rights reserved.

22. Chemical Example If you started with 5 molecules of H 2 and 3 molecules of Cl 2, how much HCl could you make? Because you need 1 molecule of H 2 for each molecule of Cl 2, the Cl 2 limits the reaction. With 3 molecules of Cl 2,, you can make a total of 6 molecules of HCl (because of the reaction coefficients). When the coefficients of the balanced equation are more complex, a general method should be used. H 2 + Cl 2 2 HCl 2 molecules of H 2 remain unused (are in excess). Limiting Reactants © 2014 John Wiley & Sons, Inc. All rights reserved.

23. Problem Solving Strategy for Limiting Reactant Problems 1.Calculate the amount of product formed from each reactant present. 2.The reactant that gives the least amount of product is limiting; the other reactant is in excess. 3.The amount of product is determined by the calculation from Step 1 with the limiting reactant. 4.If the amount of excess reactant is desired, determine the amount of excess reactant needed to consume the limiting reactant and subtract from the initial quantity present. Limiting Reactants © 2014 John Wiley & Sons, Inc. All rights reserved.

24. H 2 + Cl 2 2 HCl How many moles of HCl can be produced from 4.0 mol of H 2 and 3.5 mol of Cl 2 ? What is the limiting reactant? Step 1 Calculate the moles of HCl formed from each reactant using the appropriate mole ratios. 2 mol HCl 1 mol H 2 2 mol HCl 1 mol Cl 2 4.0 mol H 2 3.5 mol Cl 2 = 8.0 moles HCl = 7.0 moles HCl Step 2 Less HCl is formed with Cl 2, so it is the limiting reactant. The maximum amount of product is 7.0 moles. × × Limiting Reactant Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

25. 3 Fe + 4 H 2 O Fe 3 O 4 + 4 H 2 How many moles of Fe 3 O 4 can be produced from 16.8 g Fe and 10.0 g H 2 O? What is the limiting reactant? Step 1 Calculate the moles of Fe 3 O 4 formed from each reactant using the appropriate molar masses and mole ratios. 55.45 g Fe 16.8 g Fe 10.0 g H 2 O = 0.1001 mol Fe 3 O 4 Step 2 Less Fe 3 O 4 is formed with Fe, so it is the limiting reactant. The maximum amount of product is 0.1001 moles. 1 mol Fe 3 O 4 3 mol Fe 1 mol Fe 18.02 g H 2 O = 0.139 mol Fe 3 O 4 1 mol Fe 3 O 4 4 mol H 2 O 1 mol H 2 O × × × × Limiting Reactant Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

26. 184.1 g MgBr 2 50.0 g MgBr 2 2 mol AgBr 1 mol MgBr 2 187.4 g AgBr 1 mol AgBr = 102 g AgBr How many grams of AgBr can be produced from 50.0 g MgBr 2 and 100.0 g AgNO 3 ? How much excess reactant remains? Step 1 Calculate the mass of AgBr formed from each reactant using the appropriate molar masses and mole ratios. MgBr 2 + 2 AgNO 3 2 AgBr + Mg(NO 3 ) 2 × ×× Limiting Reactant Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

27. How many grams of AgBr can be produced from 50.0 g MgBr 2 and 100.0 g AgNO 3 ? How much excess reactant remains? Step 1 Calculate the mass of AgBr formed from each reactant using the appropriate molar masses and mole ratios. Step 2 Less AgBr is formed with MgBr 2, so it is the limiting reactant. The maximum amount of product is 102 g. 100.0 g AgNO 3 169.9 g AgNO 3 2 mol AgBr 2 mol AgNO 3 1 mol AgNO 3 187.4 g AgBr 1 mol AgBr = 110.3 g AgBr × × × MgBr 2 + 2 AgNO 3 2 AgBr + Mg(NO 3 ) 2 Limiting Reactant Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

28. From the problem on the previous slide, how much excess reactant remains? Step 3 Calculate how much AgNO 3 reacts with the limiting reactant, assuming all MgBr 2 reacts. MgBr 2 + 2 AgNO 3 2 AgBr + Mg(NO 3 ) 2 184.1 g MgBr 2 50.0 g MgBr 2 2 mol AgNO 3 1 mol MgBr 2 169.9 g AgNO 3 1 mol AgNO 3 = 92.3 g AgNO 3 Unreacted AgNO 3 = initial – amount reacted = 100.0 g – 92.3 g = 7.7 g AgNO 3 unused ××× Limiting Reactant Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

29. Ba(NO 3 ) 2 + Na 2 SO 4 BaSO 4 + 2 NaNO 3 How many grams of BaSO 4 can be produced from 200.0 g of Ba(NO 3 ) 2 and 100.0 g of Na 2 SO 4 ? Step 1 Calculate the mass of BaSO 4 formed from each reactant using the appropriate molar masses and mole ratios. 200.0 g Ba(NO 3 ) 2 261.4 g Ba(NO 3 ) 2 1 mol BaSO 4 1 mol Ba(NO 3 ) 2 233.4 g BaSO 4 1 mol BaSO 4 = 178.6 g BaSO 4 × ×× Limiting Reactant Problems © 2014 John Wiley & Sons, Inc. All rights reserved.

30. Ba(NO 3 ) 2 + Na 2 SO 4 BaSO 4 + 2 NaNO 3 How many grams of BaSO 4 can be produced from 200.0 g of Ba(NO 3 ) 2 and 100.0 g of Na 2 SO 4 ? Step 1 Calculate the mass of BaSO 4 formed from each reactant using the appropriate molar masses and mole ratios. Step 2 Less BaSO 4 is formed with Na 2 SO 4, so it is the limiting reactant. The maximum amount of product is the smaller amount, 164.4 g. 100.0 g Na 2 SO 4 142.0 g Na 2 SO 4 1 mol BaSO 4 1 mol Na 2 SO 4 233.4 g BaSO 4 1 mol BaSO 4 = 164.4 g BaSO 4 × ×× Limiting Reactant Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

31. The amount of products formed calculated by stoichiometry are the maximum yields possible (100%). Yields are often lower due to side reactions, loss of product while isolating/transferring the material, etc. Theoretical yield: maximum possible yield for a reaction, calculated based on the balanced chemical equation. Actual yield: actual yield obtained from the reaction. Percent yield: ratio of the actual and theoretical yield Actual yield Theoretical yield % yield = 100 × Reaction Yield © 2014 John Wiley & Sons, Inc. All rights reserved.

32. Calculate the percent yield of AgBr if 375.0 g of the compound are prepared from 200.0 g of MgBr 2. MgBr 2 + 2 AgNO 3 Mg(NO 3 ) 2 + 2 AgBr 200.0 g MgBr 2 184.1 g MgBr 2 2 mol AgBr 1 mol MgBr 2 187.8 g AgBr 1 mol AgBr With theoretical yield, we can calculate % yield Actual yield Theoretical yield % yield = 100 = 375.0 g 408.0 g 100 = 408.0 g AgBr To calculate the % yield, calculate the theoretical yield. Solution Map g MgBr 2 mol MgBr 2 mol AgBr g AgBr × × = 91.91 % ××× Calculate % Yield © 2014 John Wiley & Sons, Inc. All rights reserved.

33. Calculate the percent yield of Al 2 O 3 if 125.0 g of Al give 100.0 g of Al 2 O 3. 2 Al + 3 CrO Al 2 O 3 + 3 Cr Solution Map g Al mol Al mol Al 2 O 3 g Al 2 O 3 125.0 g Al 26.98 g Al 1 mol Al 2 O 3 2 mol Al 1 mol Al102.0 g Al 2 O 3 1 mol Al 2 O 3 To calculate the % yield, calculate the theoretical yield. With theoretical yield, we can calculate % yield Actual yield Theoretical yield % yield = 100 = 100.0 g 236.3 g 100 = 42.32 % = 236.3 g Al 2 O 3 × × ×× × % Yield Practice © 2014 John Wiley & Sons, Inc. All rights reserved.

34. Define stoichiometry and describe the strategy required to solve problems based on chemical equations. 9.1 Introduction to Stoichiometry Solve problems where the reactants and products are both in moles. 9.2 Mole-Mole Calculations Solve problems where known mass is given and the answer is to be determined in moles or the moles of known are given and mass is determined. 9.3 Mole-Mass Calculations Learning Objectives © 2014 John Wiley & Sons, Inc. All rights reserved.

35. Solve problems where mass is given and the desired unit to be determined is mass. 9.4 Mass-Mass Calculations Solve problems involving limiting reactants and yield. 9.5 Limiting Reactant and Yield Calculations Learning Objectives © 2014 John Wiley & Sons, Inc. All rights reserved.