1. Stoichiometry Goals: 1.Perform stoichiometry calculations. 2.Understand the meaning of limiting reactant. 3.Calculate theoretical and percent yields of a chemical reaction. 4.Use stoichiometry to analyze a mixture of compounds or to determine the formula of a compound. 5.Define and use molarity in solution stoichiometry. 6.Perform calculations for pH and titration problems. Stoichiometry Stoichiometry: study of the quantitative relations between amounts of reactants and products.

2. What is STOICHIOMETRY? The study of the quantitative aspects of chemical reactions. It rests on the principle of the _________ ________________. 2 Al(s) + 3 Br 2 (liq) ------> Al 2 Br 6 (s) You must always begin with a balanced equation before carrying out a stoichiometry calculation.

3. Write a Chemical Equation Phosphine, PH 3 (g), combusts in oxygen gas to form gaseous water and solid tetraphosphorus decoxide. Always check (and REcheck) the balancing

4. Information from a Balanced Equation Equation: 2 H 2 (g) + O 2 (g) 2 H 2 O (l) Molecules: 2 molecules H 2 + 1 molecule O 2 2 molecules H 2 O Mass (amu): 4.0 amu H 2 + 32 amu O 2 36.0 amu H 2 O Amount (mol): Mass (g):

5. General Plan for Stoichiometry Calculations Students should become familiar with stoichiometry calculations.

6. Mole relationships in Chemical Equations Stoichiometric factor – relates the amounts of any two substances involved in a chemical reaction, on a mole basis. C 3 H 8 + O 2 CO 2 + H 2 O 5 mol of O 2 are required to burn ____ mol of C 3 H 8 ____ mol of H 2 O are produced for every 1 mol of C 3 H 8 burned ____ mol of CO 2 are produced for every 1 mol of C 3 H 8 burned propane 5 34

7. If 454 g of NH 4 NO 3 decomposes, how much N 2 O and H 2 O are formed? What is the theoretical yield of products? STEP 1 STEP 1 Write the balanced chemical equation. STEP 2 STEP 2 Convert mass of reactant to moles of reactant. (454 g) --> moles STEP 3 STEP 3 Convert moles reactant to moles product. A. Relate moles using coefficients, write a STOICHIOMETRIC FACTOR.

8. STEP 3 B) Multiply moles of reactant by the stoichiometric factor. STEP 4 Convert moles product to mass product = ___________________ YIELD

9. STEP 5 STEP 5 How much N 2 O is formed? Total mass of reactants = total mass of products

10. STEP 6 STEP 6 Calculate the percent yield. If you isolated only 131 g of N 2 O, what is the percent yield? % yield = Actual yield Theoretical yield * 100 Students should become familiar with % yield calculations.

11. What is a Limiting Reactant? In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply _______ the quantity of product that can be formed.

12. Which is the Limiting Reactant? Limiting reactant is Excess reactant is 2 NO (g) + O 2 (g) - - -> 2 NO 2 (g) eactants R eactantsProducts

13. How many grams of CO 2 are produced from 50 g of propane? (theoretical yield) How much O 2 is required to completely burn the 50 g of propane? How much H 2 O is formed? If the actual yield of CO 2 was 145 g. What is the % yield of the reaction? C 3 H 8 + O 2 CO 2 + H 2 O 50 g C 3 H 8 (1 mol/ 44 g) = 1.14 mol C 3 H 8 * (4 mol H 2 O / 1 mol C 3 H 8 )

14. If you begin with 99.5 g of C 3 H 8 and 211 g of O 2, which one is the limiting reactant? Assuming all the limiting reactant has reacted, how much CO 2 will be formed? C 3 H 8 + O 2 CO 2 + H 2 O

15. Limiting Reactant React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl 2 + H 2 123 (See CD Screen 4.8) Rxn 1: Balloon inflates fully, some Zn left * More than enough Zn to use up the 0.100 mol HCl Rxn 2: Balloon inflates fully, no Zn left * Right amount of each (HCl and Zn) Rxn 3: Balloon does not inflate fully, no Zn left. * Not enough Zn to use up 0.100 mol HCl

16. Zn + 2 HCl ---> ZnCl 2 + H 2 0.10 mol of HCl, need ? mol Zn. 0.10 mol HCl 1 mol Zn 2 mol HCl = 0.050 mol Zn Rxn 1Rxn 2Rxn 3 Rxn 1Rxn 2Rxn 3 mass Zn (g)7.003.271.31 mol Zn0.1070.0500.020 mol HCl0.1000.1000.100 mol HCl/mol Zn Lim Reactant 123

17. Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? Mass reactant Stoichiometric factor Moles reactant Moles product Mass product ___Al + ____Cl 2 ---> Al 2 Cl 6

18. Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? STEP 1 FIND THE LIMITING REAGENT. C STEP 1 FIND THE LIMITING REAGENT. Compare actual mole ratio of reactants to theoretical mole ratio. ___ Al + ___Cl 2 ---> Al 2 Cl 6 Students should become familiar with calculations using the concept of limiting reagent.

19. Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? Calculate moles of each reactant. Find the mole ratio of reactants:

20. Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? Limiting reactant = _____ BASE ALL CALCULATIONS on LR ____ moles ____ moles Al 2 Cl 6 mass ____ mass Al 2 Cl 6 __ Al + ___ Cl 2 ---> Al 2 Cl 6 Write conversion factor:

21. Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? STEP 2 STEP 2 CALCULATE THE MASS OF THE PRODUCT. Calculate moles of Al 2 Cl 6 expected based on LR. __ Al + __ Cl 2 ---> Al 2 Cl 6

22. Mix 5.40 g of Al with 8.10 g of Cl 2. How much of each reactant will remain when reaction is complete? __ Al + __ Cl 2 ---> Al 2 Cl 6

23. Chemical Analysis An impure sample of the mineral thenardite contains Na 2 SO 4. A mass of mineral sample weights 0.123 g. The Na 2 SO 4 in the sample is converted to insoluble BaSO 4 by adding BaCl 2. The recovered mass of BaSO 4 is 0.177 g. What is the mass percent of Na 2 SO 4 in the mineral?

24. Chemical Analysis

25. General Plan for Stoichiometry Calculations

26. Chemical Analysis Balanced equation: Grams to moles Moles to moles Moles to grams Mass %

27. Combustion Analysis of Hydrocarbons Active Figure 4.9

28. Procedure for Calculating Empirical Formula Grams of each element Use Molar mass Moles of each element Empirical Formula Calculate mole ratio The central part of the calculation is determining the number of moles of each element in the compound. Remember: in the mole ratio, divide by smaller number, then multiply ‘til whole.

29. What is the empirical formula of a hydrocarbon, C x H y, if 0.115 g burn and produce 0.379 g of CO 2 and 0.1035 g of H 2 O. C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O First, recognize that all C in CO 2 and all H in H 2 O is from C x H y. 1. Calculate amount (in moles) of C in CO 2 2. Calculate amount (in moles) of H in H 2 O 3. Find ratio of mol H/mol C to find values of x and y in C x H y.

30. What is the empirical formula of a hydrocarbon, C x H y, if 0.115 g burn and produce 0.379 g of CO 2 and 0.1035 g of H 2 O. 1. Calculate amount (moles) of C in CO 2 2. Calculate amount (moles) of H in H 2 O 3. Ratio of mol H/mol C to find values of x and y in C x H y.

31. Practice 4.67 Titanium (IV) oxide, TiO 2, is heated in hydrogen gas to give water and a new titanium oxide, Ti x O y. If 1.598 g of TiO 2 produces 1.438 g of Ti x O y, what is the formula of the new oxide? TiO 2 + H 2 H 2 O + Ti x O y 1.Calculate amount of Ti:

32. Practice… 2. Calculate amount of O: 3. Calculate the molar ratio of O to T

33. How are Reactions in Solution Quantified? In solution we need to define the - SOLVENT the component whose physical state is ____________ when solution forms SOLUTE the other solution component

34. What is Molarity? The amount of solute in a solution is given by its concentration. Molarity = Students should become familiar with calculations using MOLARITY.

35. Calculate molarity of a solution of 5.00 g of NiCl 2 6 H 2 O dissolved in enough water to make 250 mL of solution. STEP 1 STEP 1 Calculate the number of moles of solute STEP 2 STEP 2 Calculate the molarity of the solution

36. How many IONS are in the Solution?

37. What mass of oxalic acid, H 2 C 2 O 4, is required to make 250 mL of a 0.0500 M solution? STEP 1 Calculate moles of acid required. STEP 2 Calculate mass of acid required.

38. Preparing Solutions Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is __________ concentrated.

39. Preparing Solutions

40. You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution! M = moles/L Notice that the amount of NaOH (moles of NaOH) present did not change.

41. You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add? Amount of NaOH in original sol. = Amount of NaOH in final sol. must also = Volume of final solution = moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution moles of NaOH in FINAL solution

42. You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add ______ of water to the 50.0 mL of 3.0 M NaOH to make ________ of 0.50 M NaOH. A shortcut: Cinitial Vinitial = Cfinal Vfinal Principle of dilution: addition of solvent does not change the amount of solute in a solution but does change its concentration.

43. Solution Stoichiometry Zinc reacts with acids to produce H 2 gas.Zinc reacts with acids to produce H 2 gas. Have 10.0 g of ZnHave 10.0 g of Zn What volume of 2.50 M HCl is needed to convert the Zn completely? What volume of 2.50 M HCl is needed to convert the Zn completely?

44. General Plan for Solution Stoichiometry M = moles / volume Moles = M * volume

45. Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Stoichiometric factor

46. Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 1: Write the balanced equation Step 2: Calculate amount of Zn in moles Step 3: Use the stoichiometric factor

47. Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 4: Calculate volume of HCl req’d * Use molarity (M) as conversion factor

48. What is pH? It is a concentration scale. pH:pH: a way to express acidity – the concentration of H + in solution. Acidic solutionpH < 7 Neutral pH = 7 Basic solution pH > 7 Acidic solutionpH < 7 Neutral pH = 7 Basic solution pH > 7 Low pH: ______ [H + ] High pH: ____ [H + ]

49. The pH Scale pH = log (1/ [H + ]) = - log [H + ] In a neutral solution, [H + ] = [OH - ] = 1.00 x 10 -7 M at 25 o C pH = - log [H + ] = -log (1.00 x 10 -7 ) = - [0 + (-7)] = 7 See CD Screen 5.17 for a tutorial See book Appendix A.3 for more on logs

50. pH and [H+] If the [H + ] of soda is 1.6 x 10 -3 M, the pH is ____?

51. pH and [H+] If the pH of Coke is 3.12, what is the [H+]?

52. What is a Titration? Titration – procedure in which two reactants in solution react in the precise proportions shown by the chemical equation for the reaction. Buret – a calibrated instrument used in a titration. It is a graduated, long glass tube calibrated to deliver precise volumes of solution through a stopcock valve. Equivalence point – the point in a titration at which one reactant has been exactly consumed by addition of the other reactant.

53. Acid-Base Titration How do we measure the concentration of an acid in a solution? 1)A measured volume of a solution of an acid of unknown concentration is transferred to a flask. 2)A solution of a base of known concentration is added carefully from a buret until the reaction of the acid with the base is just complete. 3)Equivalence point of the titration – the point at which the acid is just neutralized. At that point, the number of moles of OH - added equals the number of moles of H + that were in the sample of acid. 4)The equivalence point is determined with an indicator dye – a substance that changes color as the reaction is completed (litmus, phenolphthalein).

54. What is a Titration? Phenolphthalein indicator

55. 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 2: Step 3: Step 1:

56. 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

57. Practice What volume of 0.155 M FeCl 3 contains 12.5 g FeCl 3 ?

58. Apples contain malic acid, H 2 C 4 H 4 O 5. 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? __H 2 C 4 H 4 O 5 (aq) + __NaOH(aq) ---> __Na 2 C 4 H 4 O 5 (aq) + __H 2 O(l) ?

59. 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 2: Step 3: Step 1:

60. 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 4: Step 5:

61. Remember Go over all the contents of your textbook. Practice with examples and with problems at the end of the chapter. Practice with OWL tutor. Work on your assignment for OWL Chapters 4 and 5. Practice with the quiz on CD of Chemistry Now.