1. CHAPTER 3 Chemical E quations & Reaction Stoichiometry

2. 2 Objectives  Understand how to write chemical equations  Perform calculations based on chemical equations  Calculate percent yields from chemical reactions  Understand the concept of sequential reactions

3. 3 Equations Consider a simple equation: The question is: what happens to 2 if we multiply it by 3? The answer: it is converted to 6 In chemistry, we try to answer similar problems using chemical equations

4. 4 Chemical Equations We ask: what happens to methane when it reacts with oxygen (burns)? We know the answer from the chemical experiment The only thing we have to check is the Law of Conservation of Matter CH 4 + O 2  ? CH 4 + O 2  CO 2 + H 2 O

5. 5 Chemical Equations The Law of Conservation of Matter: in any physical or chemical change the total mass of matter remains constant which means the number of atoms of each element involved remains unchanged CH 4 + O 2  CO 2 + H 2 O unbalanced equation

6. 6 Chemical Equations CH 4 + O 2  CO 2 + H 2 O  We need to: (1) balance the equation (2) make sure that we have the same number of atoms for each element on the left and on the right side CH 4 + 2O 2  CO 2 + 2H 2 O reactants products

7. 7 Chemical Equations Symbolic representation of a chemical reaction that shows:  reactants on the left side  products on the right side  relative amounts of each using stoichiometric coefficients CH 4 + 2O 2  CO 2 + 2H 2 O

8. 8 Balancing Chemical Equations  Fe + O 2  Fe 3 O 4  Al + H 2 SO 4  Al 2 (SO 4 ) 3 + H 2  C 6 H 12 O 6 + O 2  CO 2 + H 2 O

9. 9 Quantitative Aspects Let’s go back to If we multiply both sides of the equation by some number, they still will be equal We can treat chemical equations in the same way

10. 10 Quantitative Aspects What happens in the reaction between methane and oxygen numerically? CH 4 + 2O 2  CO 2 + 2H 2 O 1 molecule2 molecules1 molecule2 molecules We can multiply both left and right sides by the same number – they will remain equal: CH 4 + 2O 2  CO 2 + 2H 2 O 1 molecule2 molecules1 molecule2 moleculesx 8 8 molecules16 molecules8 molecules16 molecules

11. 11 Example 1 How many O 2 molecules are required to react with 81 atoms of Fe? Fe + O 2  Fe 3 O 4

12. 12 Quantitative Aspects Let’s multiply the equation by Avogadro’s number: CH 4 + 2O 2  CO 2 + 2H 2 O 1 molecule2 molecules1 molecule2 molecules 6.022x10 23 molecules 2x(6.022x10 23 molecules) 6.022x10 23 molecules 2x(6.022x10 23 molecules) ==== 1 mole2 moles1 mole2 moles In the same way as we talk about chemical equations in terms of molecules, we can consider them in terms of moles

13. 13 Example 2 How many moles of H 2 is produced when 70 moles of aluminum react with excess sulfuric acid? Al + H 2 SO 4  Al 2 (SO 4 ) 3 + H 2

14. 14 Quantitative Aspects Since we know the molar mass of each substance we can also establish the mass relationships: CH 4 + 2O 2  CO 2 + 2H 2 O 1 mole2 moles1 mole2 moles x16.0 g/mol32.0 g/mol44.0 g/mol18.0 g/mol 16.0 g64.0 g44.0 g36.0 g=++ The total mass of products should be the same as the total mass of reactants

15. 15 Example 3 What mass of CO is required to react with 146 g of iron(III) oxide? Fe 2 O 3 + CO  Fe + CO 2

16. 16 What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron(III) oxide with excess carbon monoxide? Example 4 Fe 2 O 3 + 3CO  2Fe + 3CO 2 We should be concerned only with iron(III) oxide and carbon dioxide since carbon monoxide is in excess. Therefore, we calculate the number of moles of CO 2 which can be produced from 0.540 mole of Fe 2 O 3 : 1 mole of Fe 2 O 3 produces 3 moles of CO 2, hence 0.540 mole of Fe 2 O 3 will produce (0.540 x 3) mole = 1.62 mole of CO 2

17. 17 Example 4 (continued) Now we can calculate the mass of CO 2 produced: mass(CO 2 ) = #moles(CO 2 ) x molar mass (CO 2 ) mass(CO 2 ) = 1.62 mole x 44.0 g/mol = 71.28 g

18. 18 What mass of iron(III) oxide reacted with excess carbon monoxide if carbon dioxide produced by the reaction had a mass of 8.65 grams? Fe 2 O 3 + 3CO  2Fe + 3CO 2 Example 5 We should be concerned only with iron(III) oxide and carbon dioxide. First, we calculate their molar masses: M r (Fe 2 O 3 ) = 159.69 g/mol M r (CO 2 ) = 44.01 g/mol

19. 19 Example 5 From the equation we see that 1 mole of Fe 2 O 3 affords 3 moles of CO 2. We can convert these quantities to grams: m (Fe 2 O 3 ) = 1 mole x 159.69 g/mole = 159.69 g m (CO 2 ) = 3 mole x 44.01 g/mole = 132.03 g Now we can write the following relationship: 132.03 g CO 2 is obtained from 159.69 g Fe 2 O 3 8.65 g CO 2 is obtained from ? g Fe 2 O 3 Therefore,

20. 20 Limiting Reactant Concept Example: A box contains 87 bolts, 110 washers, and 99 nuts. How many sets, each consisting of one bolt, two washers, and one nut, can you construct from the contents of one box? 87 bolts  110 washers  99 nuts 

21. 21 Example 6 What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

22. 22 Example 7 Calculate the mass of carbon tetrachloride which can be produced by the reaction of 10.0 g of carbon with 100.0 g of chlorine. Determine the mass of excess reagent left unreacted. 1)Balance the equation and calculate molar masses of all substances (since all of them are included in the formulation of the problem): M r :12.01 g/mol 70.91 g/mol 153.82 g/mol C + Cl 2  CCl 4 C + 2Cl 2  CCl 4

23. 23 Example 7 (continued) 2)Calculate the amount of reactants in moles by dividing the mass of each by its molar mass: M r :12.01 g/mol 70.91 g/mol C + 2Cl 2  CCl 4 m : 10.0 g 100.0 g 0.833 mol 1.41 mol It follows from the equation that 1 mole of C reacts with 2 moles of Cl 2, therefore 0.833 mole of C should react with 0.833 x 2 = 1.67 mole of Cl 2. The amount of Cl 2 participating in the reaction is smaller (1.41 mol) which means that Cl 2 is the limiting reactant

24. 24 3)Now, knowing the amount of the limiting reactant, we can easily calculate the amount of product. According to the equation, from 2 moles of Cl 2 we can obtain 1 mole of CCl 4. Hence, from 1.41 mol of Cl 2 we can obtain 1.41/2 = 0.705 mol of CCl 4. C + 2Cl 2  CCl 4 m (CCl 4 ) = #moles(CCl 4 ) x M r (CCl 4 ) = = 0.705 mol x 153.82 g/mol = 108 g Example 7 (continued)

25. 25 4)According to the equation, the number of moles of carbon consumed in the reaction equals to the number of moles of CCl 4 formed: C + 2Cl 2  CCl 4 #moles (C) = 0.705 mol m (C) = #moles(C) x M r (C) = 0.705 mol x 12.01 g/mol = 8.47 g This means that the amount of carbon that didn’t react is m (C-unreacted) = 10.0 g – 8.47 g = 1.53 g Example 7 (continued)

26. 26 Percent Yields from Reactions Theoretical yield is calculated by assuming that the reaction goes to completion It is the maximum yield possible for the given reaction BUT in many reactions the reactants are not completely converted to products a particular set of reactants may undergo two or more reactions simultaneously sometimes it is difficult to separate the desired product from other products in the reaction mixture

27. 27 Percent Yields from Reactions Actual yield is the amount of a specified pure product actually obtained from a given reaction Percent yield: There are many reactions that do not give the 100% yield When calculating the percent yield, always make sure that the actual and theoretical yields are expressed in the same units

28. 28 Example 8 A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What is the percent yield? 1)We start by calculating the molar masses of C 2 H 5 OH and CH 3 COOC 2 H 5 : M r (C 2 H 5 OH) = 46.08 g/mol M r (CH 3 COOC 2 H 5 ) = 88.12 g/mol 2)The amount of C 2 H 5 OH in moles is C 2 H 5 OH + CH 3 COOH  CH 3 COOC 2 H 5 + H 2 O

29. 29 Example 8 (continued) 3)According to the equation, from 1 mole of C 2 H 5 OH we can obtain 1 mole of CH 3 COOC 2 H 5, therefore from 0.217 mol of C 2 H 5 OH we can obtain 0.217 mol of CH 3 COOC 2 H 5, or theoretical yield = 0.217 mol x 88.12 g/mol = 19.1 g The actual yield, however, is 14.8 g and C 2 H 5 OH + CH 3 COOH  CH 3 COOC 2 H 5 + H 2 O

30. 30 Example 9 10.6 g of Fe reacts with 25 g of Br 2 to form 18 g of FeBr 3. What is the percent yield? 1)Calculate the number of moles of each reactant to determine which reactant is the limiting one: 2Fe + 3Br 2  2FeBr 3 According to the equation, we need 3 moles of Br 2 per each 2 mole of Fe. The calculation shows that we have more moles of Fe than of Br 2. Obviously, Br 2 is the limiting reagent.

31. 31 Example 9 (continued) 2)In this case we don’t even need to perform any further calculations in order to see that Br 2 is the limiting reactant. Here’s why: According to the equation, we need 3 moles of Br 2 per each 2 mole of Fe. The calculation performed in step (1) shows that we have more moles of Fe than of Br 2. Obviously, Br 2 is the limiting reactant. 2Fe + 3Br 2  2FeBr 3 3)From the equation we see that each 3 moles of Br 2 result in 2 moles of FeBr 3. We can convert these quantities to grams (multiplying by the molar mass): m (Br 2 ) = 3 mole x 159.81 g/mole = 479.43 g m (FeBr 3 ) = 2 mole x 295.56 g/mole = 591.12 g

32. 32 Example 9 (continued) 4)Now we can write the following relationship: 479.43 g Br 2 gives 591.12 g FeBr 3 25 g Br 2 gives ? g FeBr 3, Therefore, this is theoretical yield

33. 33 Sequential Reactions A set of reactions required to convert starting materials into the desired product The amount of the desired product from each reaction is taken as the starting material for the next reaction. magnesium perchlorate sodium hydroxide

34. 34 Example 10 13 g of decane, C 10 H 22, is burned in a C-H combustion train. The CO 2 gas formed reacts with sodium hydroxide, NaOH, and is converted into sodium carbonate, Na 2 CO 3. What mass of Na 2 CO 3 will be formed if all reactions proceed to completion? 2C 10 H 22 + 31O 2  20CO 2 + 22H 2 O 2NaOH + CO 2  Na 2 CO 3 + H 2 O 1)The balanced sequential equations are: 2)We calculate the amount of C 10 H 22 in moles:

35. 35 Example 10 (continued) 2C 10 H 22 + 31O 2  20CO 2 + 22H 2 O 2NaOH + CO 2  Na 2 CO 3 + H 2 O 3)Now, looking at the first equation, we can see that when 2 moles of C 10 H 22 are burned, 20 moles of CO 2 are formed, that is to find the number of moles of CO 2 formed we need to multiply the given number of moles of C 10 H 22 by 10: #moles(CO 2 ) = 0.091 mol x 10 = 0.91 mol 4)From the second equation it follows that every mole of CO 2 produces 1 mole of Na 2 CO 3, that is the amount of Na 2 CO 3 obtained from 0.91 mol of CO 2 will also be equal to 0.91 mol.

36. 36 Example 10 (continued) 5)Finally, knowing the number of moles of Na 2 CO 3 we can calculate its mass: m(Na 2 CO 3 ) = #moles (Na 2 CO 3 ) x M r (Na 2 CO 3 ) m(Na 2 CO 3 ) = 0.91 mol x 105.99 g/mol = 96 g

37. 37 What mass of (NH 4 ) 3 PO 4 can be produced in the result of the following reactions if we start with 10 moles of N 2 and excess hydrogen? Example 11 N 2 + H 2  NH 3 (yield = 44%) NH 3 + H 3 PO 4  (NH 4 ) 3 PO 4 (yield = 95%)

38. 38 Reading Assignment Read Chapter 3 Learn Key Terms (p. 112) Take a look at Lecture 5 notes (will be posted on the web not later than Monday morning) If you have time, read Chapter 4 Homework #1 due by 9/13