1. 4-1 CHEM 100, FALL 2011, LA TECH Instructor: Dr. Upali Siriwardane e-mail: upali@chem.latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m Tu,Th,F 9:00 - 10:00 a.m. Test Dates : March 25, April 26, and May 18; Comprehensive Fina Exam: 9:30-10:45 am, CTH 328. Chemistry 100(02) Fall 2011 October 3, 2011 (Test 1): Chapter 1 & 2 October 26, 2011 (Test 3): Chapter 3 & 4 November 16, 2011 (Chapter 5 & 6) November 17, 2011 (Make-up test) comprehensive: Chapters 1-6 9:30-10:45:15 AM, CTH 328

2. 4-2 CHEM 100, FALL 2011, LA TECH Chapter 4. Quantities of Reactants and Products Chemical Equations Balanced chemical equation Patterns of Chemical Reactions Combination, decomposition and displacement Stoichiometric coefficients Stoichiometric Conversion Factors Moles of Reactants Mole ratios of reactants and products Converting moles of Reactants to Products Calculate grams of products from grams of reactants Stiochiometric Reactions Reactions in One reactant in Limited Supply Limiting reactant Evaluating Success of Synthesis Theoretical Yield Actual Yield Percent Yield Stoichiometric principles to find the empirical formula

3. 4-3 CHEM 100, FALL 2011, LA TECH Types of Reactions Synthesis reactions or combination reactions Decomposition reactions Displacement reactions Single Double(Exchange reactions) Combustion Reactions Formation Reactions

4. 4-4 CHEM 100, FALL 2011, LA TECH Types of Chemical Reactions

5. 4-5 CHEM 100, FALL 2011, LA TECH Reaction of H 2 and I 2

6. 4-6 CHEM 100, FALL 2011, LA TECH Synthesis or Combination Reactions Formation of a compound from simpler compounds or elements. Decomposition reactions Compound breaks up to from simpler compounds or elements. Displacement reactions Single displacement: In a compound an element is replaced by another element. Exchange reactions Double displacement: In a compound group or ion is replaced by another group or ion in another compound..

7. 4-7 CHEM 100, FALL 2011, LA TECH Formation Reactions Formation of a compound from elements at standard state. Combustion Reactions Compound reacts with oxygen to produce oxides: water and carbon dioxide for organic compounds Acid/Base (Neutralization)Reactions An acid and a base react to form water and salt ( most ionic compounds are salts) ( most ionic compounds are salts) Precipitation Reactions when two aqueous salt solutions are mixed an insoluble salt is formed when two aqueous salt solutions are mixed.

8. 4-8 CHEM 100, FALL 2011, LA TECH Combination Reaction

9. 4-9 CHEM 100, FALL 2011, LA TECH Decomposition Reactions

10. 4-10 CHEM 100, FALL 2011, LA TECH Dynamite

11. 4-11 CHEM 100, FALL 2011, LA TECH Electrolysis Decomposition caused by an electric current Anode electrode where oxidation occursCathode electrode where reduction occurs

12. 4-12 CHEM 100, FALL 2011, LA TECH Electrolysis

13. 4-13 CHEM 100, FALL 2011, LA TECH Displacement Reactions

14. 4-14 CHEM 100, FALL 2011, LA TECH Exchange Reactions They are double displacement or exchange reactions of ionic compounds where an insoluble salt is formed (Precipitation Reactions) when two aqueous salt solutions are mixed. Ba(NO 3 ) 2 (aq) +Na 2 SO 4 (aq)= BaSO 4 (s) + 2 NaNO 3 (aq)

15. 4-15 CHEM 100, FALL 2011, LA TECH Chemical Equations P 4 O 10 P 4 O 10 (s) + 6H 2 O 6H 2 O (l) = 4 H 3 PO 4 (l) reactants reactants enter into a reaction. products products are formed by the reaction. Parantheses Parantheses represent physical state Stoichiometric Coefficients Coefficients are numbers in front of chemical formula formula gives the amounts (moles) (moles) of each substance used and each substance produced. Equation Must be balanced!

16. 4-16 CHEM 100, FALL 2011, LA TECH Chemical Reactions Could be described in words Chemical equation: Reactants?Products? Reaction conditions? =, --->, or ? Stoichiometric coefficients? Number in front of substances representing moles, atoms, molecules

17. 4-17 CHEM 100, FALL 2011, LA TECH Balanced Chemical Equation Representation of a chemical reaction which uses coefficients (prefix numbers known as stoichiometric coefficients) to represent the relative amounts of reactants and products

18. 4-18 CHEM 100, FALL 2011, LA TECH Writing and Balancing Chemical Equations Write a word equation. Convert word equation into formula equation. Balance the formula equation by the use of prefixes (coefficients) to balance the number of each type of atom on the reactant and product sides of the equation.

19. 4-19 CHEM 100, FALL 2011, LA TECH Balancing Chemical Equations 1. Check for Diatomic Molecules - H 2 - N 2 - O 2 - F 2 - Cl 2 - Br 2 - I 2 If these elements appear By Themselves in an equation, they Must be written with a subscript of 2 2. Balance Metals 3. Balance Nonmetals 4. Balance Oxygen 5. Balance Hydrogen 6.Recount All Atoms 7. If EVERY coefficient will reduce, rewrite in the simplest whole-number ratio.

20. 4-20 CHEM 100, FALL 2011, LA TECH ExampleHydrogen gas reacts with oxygen gas to produce water. Step 1. hydrogen + oxygen  water Step 2. H2 + O2  H2O Step 3. 2 H2 + O2  2 H2O

21. 4-21 CHEM 100, FALL 2011, LA TECH Example Iron(III) oxide reacts with carbon monoxide to produce the iron oxide (Fe 3 O 4 ) and carbon dioxide. iron(III) oxide + carbon monoxide  Fe 3 O 4 + carbon dioxide Fe 2 O 3 + CO  Fe 3 O 4 + CO 2 3 Fe 2 O 3 + CO  2 Fe 3 O 4 + CO 2 3 Fe 2 O 3 + CO  2 Fe 3 O 4 + CO 2

22. 4-22 CHEM 100, FALL 2011, LA TECH Stoichiometry stoi·chi·om·e·try noun Calculations of the quantitative relationships between reactants and products in a chemical reaction.

23. 4-23 CHEM 100, FALL 2011, LA TECH Stoichiometric Relationships

24. 4-24 CHEM 100, FALL 2011, LA TECH Stoichiometric Converstion Factors The large numbers in front of chemical formulas. Coefficients represent the number of molecules of the substance in the reaction. They provide the convestion factor to conver moles of reactants to products or vice versa. 4 NH 3 (g) + 5 O 2 (g) ------> 4 NO(g) + 6 H 2 O(g) 4 mol NH 3 = 5 mol O 2 4 mol NH 3 = 5 mol O 2 5 mol O 2 = 6 mol H 2 O 5 mol O 2 = 6 mol H 2 O 4 mol NH 3 = 4 mol NO; 1mole NH 3 = 1 mole NO 4 mol NH 3 = 4 mol NO; 1mole NH 3 = 1 mole NO 1 mol NH 3 = 1 mol NO 1 mol NH 3 = 1 mol NO

25. 4-25 CHEM 100, FALL 2011, LA TECH ExamplesCalculate the following using the chemical equation given below: 4 NH 3 (g) NH 3 (g) + 5 O 2 (g) O 2 (g) ----> 4 NO(g) + 6 H 2 O(g) a) moles of NO(g) from 2 moles of NH 3 (g) and excess O 2 (g). b) moles of H 2 O(g) H 2 O(g) from 3 moles of O 2 (g) O 2 (g) and excess NH 3 (g).

26. 4-26 CHEM 100, FALL 2011, LA TECH The Mole and Chemical Reactions: The Macro-Nano Connection 2 H 2 + O 2  2 H 2 O 2 H 2 molecules 1 O 2 molecule 2 H 2 O molecules 2 H 2 molecules 1 O 2 molecule 2 H 2 O molecules 2 moles H 2 molecules 1 mole O 2 molecules 2 moles H 2 O molecules 2 kmoles H 2 molecules 1 kmole O 2 molecules 2 kmoles H 2 O molecules 2 mmoles H 2 molecules 1 mmole O 2 molecules 2 mmoles H 2 O molecules 4 g H 2 32 g O 2 36 g H 2 O 4 g H 2 32 g O 2 36 g H 2 O

27. 4-27 CHEM 100, FALL 2011, LA TECH Examples How many moles of H 2 O will be produced by 0.80 mole of O 2 with excess H 2 according to the equation? 2H 2 (g) + O 2 (g) = 2 H 2 O(l) 2H 2 (g) + O 2 (g) = 2 H 2 O(l)

28. 4-28 CHEM 100, FALL 2011, LA TECH Stoichiometric Reactions Reactions where mole ratio of the products and reactants are the same as mole ratio from the stoichiometric coefficients in the balanced chemical equation. All reactants will be completely converted into products.

29. 4-29 CHEM 100, FALL 2011, LA TECH Limiting Reactant

30. 4-30 CHEM 100, FALL 2011, LA TECH Limiting Reactant

31. 4-31 CHEM 100, FALL 2011, LA TECH Analogy in Recipe : Making Cheese Sandwiches You were given 20 slices bread, 5 slices of cheese, 4 slices of ham If you want to make sandwiches containing two slices bread and one slice of cheese and one slice of ham, How many sandwiches you could make? What is the limiting ingredient?

32. 4-32 CHEM 100, FALL 2011, LA TECH What is the limiting reagent? Limiting reagent is the reactant, which is used up first. To find the limiting reactant you have to compare the amounts of reactants in moles.

33. 4-33 CHEM 100, FALL 2011, LA TECH Ways to find Limiting Reactant There two ways to find the limiting ingredient: 1) Comparing mole ratio of the chemical equation to mole ratio calculated from the grams of reactants 2) Calculate moles of product and whichever reactant produces lowest moles of product is the limiting reactant

34. 4-34 CHEM 100, FALL 2011, LA TECH Examples A 300.0 g sample of phosphorus ( M.W. 123.88 g/mol) burns in 500.0 g of oxygen (M.W. 32.00 g/mol) according to following equation: P 4 (s) + 5O 2 (g) = P 4 O 10 (s) P 4 (s) + 5O 2 (g) = P 4 O 10 (s) What is the limiting reagent? What is the limiting reagent?

35. 4-35 CHEM 100, FALL 2011, LA TECH First Method moles of P 4 and O 2 moles of P 4 and O 2 300 500 300 500 ---------- = 2.42 mol P 4 ; ----------= 15.63 mol O 2 ---------- = 2.42 mol P 4 ; ----------= 15.63 mol O 2 123.88 32.00 123.88 32.00 P 4 and O 2 P 4 and O 2 theoretical mole ratio 1 : 5 actually mole ratio 1 : 6.46 P 4 is present in lower amount than required, it is used up first. Therefore, P 4 is the limiting reagent. Second Method 2.42 mol P41 mol P4O10 = 2.42 mole P4O10 1 mol P4 15.63 mole O21 mol P4O10 = 3.13 mole P4O10 5 mole O2

36. 4-36 CHEM 100, FALL 2011, LA TECH EXAMPLE How much H 2 O, in moles results from burning an excess of H 2 in 3.3 moles of O 2 ? 2 H2 + O2  2 H2O (3.3 mol O2) (2 mol H2O) #mol H2O = = 6.6 mol H2O (1 mol O2) Mole ratio from balanced chemical equation

37. 4-37 CHEM 100, FALL 2011, LA TECH Theoretical yield Theoretical yield yield is the amount (grams) of products formed according to chemical equation. Use the limiting reagent to calculate the moles of the product and then convert moles to grams.

38. 4-38 CHEM 100, FALL 2011, LA TECH Actual Yield Actual yield yield is the grams of the product obtained by an experiment. Actual yield should be less than the theoretical yield if the experiment was carried out meticulously. meticulously. If the products are contaminated with impurities or the formula of product was wrong the actual yield could be higher.

39. 4-39 CHEM 100, FALL 2011, LA TECH % Yield actual yield % yield = ------------------------ x 100 theoretical yield If the products are contaminated with impurities or the formula of product was wrong the % yield could be higher than 100%.

40. 4-40 CHEM 100, FALL 2011, LA TECH QuestionSulfur trioxide, SO 3 SO 3, is made from the oxidation of SO 2 SO 2 and the reaction is represented by the equation 2SO 2 2SO 2 + O 2 O 2 -----> 2SO 3 A 16.0-g sample of SO 2 SO 2 gives 18.0 g of SO 3. SO 3. The percent yield of SO 3 is

41. 4-41 CHEM 100, FALL 2011, LA TECH ExamplesA 300.0 g sample of phosphorus ( M.W. 123.88 g/mol) burns in 500.0 g of oxygen (M.W. 32.00 g/mol) according to following equation: P 4 (s) P 4 (s) + 5O 2 (g) 5O 2 (g) = P 4 O 10 (s) a) What is the limiting reagent? b) How many moles of P 4 O 10 P 4 O 10 are produced theoretically? c) If 612 g of is actually produced in this reaction, calculate the percent yield.

42. 4-42 CHEM 100, FALL 2011, LA TECH Steps in Stoichiometric Calculations Check whether chemical equation is balanced get the moles from grams of materials find the limiting reactant calculate moles of products from the limiting reactant convert moles of the products to grams find the actual yield of the reaction calculate % yield of the reaction

43. 4-43 CHEM 100, FALL 2011, LA TECH Question How much hydrogen gas is produced when 1 kg of sodium reacts with water?

44. 4-44 CHEM 100, FALL 2011, LA TECH Question 2Al(s) + 6HCl(aq)--> 2AlCl 3 (aq) +3H 2 (g) According to the equation above, how many grams of aluminum are needed to react with 0.582 mol of hydrochloric acid?

45. 4-45 CHEM 100, FALL 2011, LA TECH 5.23 g Al

46. 4-46 CHEM 100, FALL 2011, LA TECH EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. weld Photo by Mike Condren

47. 4-47 CHEM 100, FALL 2011, LA TECH EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132 #/yard) (1 yard/36 in) (454 g/#) = 1.67  10 3 g/in = 1.67  10 3 g/in The mass of iron in a weld adding 10% mass: #g = (1.67  10 3 g) (0.10) = 167 g

48. 4-48 CHEM 100, FALL 2011, LA TECH EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g = (1.67  10 3 g) (0.10) = 167 g Balanced chemical equation: Fe 2 O 3 + 2 Al  2 Fe + Al 2 O 3 What mass of Fe 2 O 3 is required for the thermite process? (1 mol Fe) (1 mol Fe 2 O 3 ) (159.7 g Fe 2 O 3 ) (1 mol Fe) (1 mol Fe 2 O 3 ) (159.7 g Fe 2 O 3 ) #g Fe 2 O 3 = (167 g Fe)  (55.85 g Fe) (2 mol Fe) (1 mol Fe 2 O 3 ) (55.85 g Fe) (2 mol Fe) (1 mol Fe 2 O 3 ) = 238 g Fe 2 O 3

49. 4-49 CHEM 100, FALL 2011, LA TECH EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe Balanced chemical equation: Fe 2 O 3 + 2 Al  2 Fe + Al 2 O 3 What mass of Fe 2 O 3 is required for the thermite process? #g Fe 2 O 3 = 238 g Fe 2 O 3 What mass of Al is required for the thermite process? (1 mol Fe) (2 mol Al) (26.9815 g Al) (1 mol Fe) (2 mol Al) (26.9815 g Al) #g Al = (167 g Fe)  (55.85 g Fe) (2 mol Fe) (1 mol Al) (55.85 g Fe) (2 mol Fe) (1 mol Al) = 80.6 g Al

50. 4-50 CHEM 100, FALL 2011, LA TECH EXAMPLE What is the number of moles of CaSO 4 (S) that can be produced by allowing 1.0 mol SO 2, 2.0 mol CaCO 3, and 3.0 mol O 2 to react? 2SO 2(g) + 2CaCO 3(s) + O 2(g)  2CaSO 4(S) + 2CO 2(g) balanced equation relates: 2SO 2(g)  2CaCO 3(s)  O 2(g) have only: 1SO 2(g)  2CaCO 3(s)  3O 2(g) not enough SO 2 to use all of the CaCO 3 or the O 2 not enough CaCO 3 to use of the O 2 SO 2 is the limiting reactant

51. 4-51 CHEM 100, FALL 2011, LA TECH EXAMPLE What is the number of moles of CaSO 4 (S) that can be produced by allowing 1.0 mol SO 2, 2.0 mol CaCO 3, and 3.0 mol O 2 to react? 2SO 2(g) + 2CaCO 3(s) + O 2(g)  2CaSO 4(S) + 2CO 2(g) have only: 1SO 2(g)  2CaCO 3(s)  3O 2(g) SO 2 is the limiting reactant if use all of SO2 #CaCO3 = (1 mol SO2)(2 mol CaSO4/2 mol SO2) = 1 mol CaSO4

52. 4-52 CHEM 100, FALL 2011, LA TECH EXAMPLE A rocket fuel, hydrazine, is produced by a reaction of Cl 2 that is reacted with excess NaOH and NH 3. What (a) theoretical yield can be produced from 1.00 kg of Cl 2 ? 2NaOH + Cl 2 + 2NH 3  N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00 kg Cl 2 ) #kg N 2 H 4 = (1 kmol Cl 2 ) (70.9 kg Cl 2 ) molar mass (1 kmol N 2 H 4 ) (1 kmol Cl 2 ) balanced equation (32.0 g N 2 H 4 ) (1 mol N 2 H 4 ) molar mass = 0.451 kg N 2 H 4

53. 4-53 CHEM 100, FALL 2011, LA TECH EXAMPLE (b) What is the actual yield if 0.299 kg of 98.0% N 2 H 4 is produced for every 1.00 kg of Cl 2 ? (98.0 kg N 2 H 4 ) (100 kg product) purity factor (0.299 kg product) # kg N 2 H 4 = = 0.293 kg N 2 H 4 2NaOH + Cl 2 + 2NH 3  N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = 0.451 kg N 2 H 4 (b) actual yield

54. 4-54 CHEM 100, FALL 2011, LA TECH EXAMPLE (c) What is the percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3  N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = 0.451 kg N 2 H 4 (b) actual yield# kg N 2 H 4 = 0.293 kg N 2 H 4 (c) percent yield 0.293 kg 0.293 kg % yield =  100 = 65.0 % yield 0.451kg 0.451kg

55. 4-55 CHEM 100, FALL 2011, LA TECH Combustion Analysis

56. 4-56 CHEM 100, FALL 2011, LA TECH Example Benzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO 2 and 2.85-mg of H 2 O formed. What is the empirical formula of benzoic acid? 4.48-mg C %C =  100 = 68.9% C 6.49-mg sample (16.4-mg of CO2 )(12.01-mg C) #mg C = = 4.48-mg C (44.01-mg CO2 )

57. 4-57 CHEM 100, FALL 2011, LA TECH Example Benzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO 2 and 2.85-mg of H 2 O formed. What is the empirical formula of benzoic acid? 0.319-mg H %C =  100 = 4.92% H 6.49-mg sample (2.85-mg of H2O )(2.02-mg H) #mg H = = 0.319-mg H (18.02-mg H2O)

58. 4-58 CHEM 100, FALL 2011, LA TECH Example Benzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO 2 and 2.85-mg of H 2 O formed. What is the empirical formula of benzoic acid? 68.9% C 4.92% H % O = (100 - (68.9% C + 4.92% H) = 26.2% O

59. 4-59 CHEM 100, FALL 2011, LA TECH Example Benzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO 2 and 2.85-mg of H 2 O formed. What is the empirical formula of benzoic acid? % C 68.9 H 4.92 O 26.2 (%/gaw) 68.9/12.0 = 5.75 4.92/1.01 = 4.87 26.2/16.0 = 1.64 Divide by Smallest 5.75/1.64 = 3.51 4.87/1.64 = 2.97 1.64/1.64 = 1.00 Multiply by Integer 3.51  2 = 7 2.97  2 = 6 1.00  2 = 2 Relative # Atoms C7H6O2C7H6O2