1. STOICHIOMETRY AND CHEMICAL ANALYSIS Let the fun begin…

2. SCIENCE 10 REVIEW: When counting elements, don’t forget to look at both the subscript and the coefficient. When counting elements, don’t forget to look at both the subscript and the coefficient. For example: For example: PO = has 2 phosphorus atoms and 5 oxygen atoms P 2 O 5 = has 2 phosphorus atoms and 5 oxygen atoms = has 4 phosphorus atoms and 10 oxygen atoms 2P 2 O 5 = has 4 phosphorus atoms and 10 oxygen atoms Because there are 2 molecules (indicated by the coefficient) and 2 atoms in each molecule (indicated by the subscript) – So you multiply!!

3. REMEMBER: Never change a subscript to balance an equation! Never change a subscript to balance an equation! O 2(g) + H 2(g)  H 2 O (l) Is unbalanced – but you can’t change it the following way! O 2(g) + H 2(g)  H 2 O 2(l) X Make sure the coefficients are the lowest whole-number ratio : Make sure the coefficients are the lowest whole-number ratio : 2O 2(g) + 4H 2(g)  4H 2 O (l) This is a balanced formula but these are not the lowest numbers you could use: O 2(g) + 2H 2(g)  2H 2 O (l) O 2(g) + 2H 2(g)  2H 2 O (l)

4. 1) Write the chemical formulas for the reactants and products including the states Cu (s) + AgNO 3(aq)  Ag (s) + Cu(NO 3 ) 2(aq) Cu (s) + AgNO 3(aq)  Ag (s) + Cu(NO 3 ) 2(aq) 2) Balance the element (atom or ion) present in the greatest number by multiplying by the lowest coefficient possible (NO 3 ) 2(aq) = 2 present (lowest coefficient possible to balance = 2) (NO 3 ) 2(aq) = 2 present (lowest coefficient possible to balance = 2) Cu (s) + 2AgNO 3(aq)  Ag (s) + Cu(NO 3 ) 2(aq) Cu (s) + 2AgNO 3(aq)  Ag (s) + Cu(NO 3 ) 2(aq) 3) Repeat step 2 for the rest of the elements Now we have 2 Ag, so balance the other side Now we have 2 Ag, so balance the other side Cu (s) + 2AgNO 3(aq)  2Ag (s) + Cu(NO 3 ) 2(aq) Cu (s) + 2AgNO 3(aq)  2Ag (s) + Cu(NO 3 ) 2(aq) 4) Count elements on each side of the final equation to ensure they balance: 1 Cu (s) = 1 Cu (s) ; 2Ag = 2Ag (s) ; 2 NO 3 = (NO 3 ) 2(aq) 1 Cu (s) = 1 Cu (s) ; 2Ag = 2Ag (s) ; 2 NO 3 = (NO 3 ) 2(aq) BALANCING CHEMICAL EQUATIONS

5. CONSERVATION OF MASS After a chemical reaction the total mass of the products is equal to the total mass of the reactants.

6. REACTION TYPES Five Reactions Types Formation Decomposition Single-Replacement Double-Replacement Complete Combustion Prediction of products of a reaction

7. CLASSIFYING CHEMICAL REACTIONS: COMPOSITION (FORMATION) 2Mg (s) + O 2(g)  2MgO (s) element + element  compound Predict and balance the following: Li(s) + Cl2(g)  Na(s) + F2(g)  Ba(s) + N2(g) 

8. CLASSIFYING CHEMICAL REACTIONS: DECOMPOSITION 2H 2 O (l)  O 2(g) + 2H 2(g) compound  element + element Predict and balance the following: NaCl (s)  Sr 3 P 2(s)  Cs 2 O (s) 

9. CLASSIFYING CHEMICAL REACTIONS: COMBUSTION CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g) fuel + oxygen  carbon dioxide + water Predict and balance the following: C 3 H 8(g) + O 2(g)  C 2 H 6(g) + O 2(g)  C 4 H 10(g) + O 2(g) 

10. CLASSIFYING CHEMICAL REACTIONS: SINGLE REPLACEMENT 2AgNO 3(aq) + Cu (s)  Cu(NO 3 ) 2(aq) + 2Ag (s) compound + element  compound + element 2NaI (aq) + Cl 2 (g)  I 2 (s) + 2NaCl (aq) Predict and balance the following: NaBr (aq) + O 2(g)  CuCl 2(aq) + Al (s)  Li 2 CO 3(aq) + K (s) 

11. CLASSIFYING CHEMICAL REACTIONS: DOUBLE REPLACEMENT 2AgNO 3(aq) + CuCl 2(aq)  Cu(NO 3 ) 2(aq) + 2AgCl (aq) compound + compound  compound + compound CaCI 2(aq) + Na 2 CO 3 (aq)  2NaCl (aq) + CaCO 3(s) Predict and balance the following: NaBr (aq) + MgO (aq)  CuCl 2(aq) + AlF 3(aq)  Li 2 CO 3(aq) + K 2 O (aq) 

12. USING THE SOLUBILITY TABLE:

13. UNIT PREPARATION Calculating Mole Amounts: Of a Solid n = m / M Of a Gas n = PV / RT Of a Solution n = c V

14. CHEMICAL REACTION EQUATIONS

15. REACTION ASSUMPTIONS Reactions are spontaneous – reactions will occur when all the reactants are mixed together Reactions are spontaneous – reactions will occur when all the reactants are mixed together Reactions are fast – the reaction must occur within a reasonable time (see pg. 280) Reactions are fast – the reaction must occur within a reasonable time (see pg. 280) Reactions are quantitative – one that is more than 99% complete; in other words, at least one reactant is completely used up Reactions are quantitative – one that is more than 99% complete; in other words, at least one reactant is completely used up Reactions are stoichiometric – means that there is a simple whole-number ratio of chemical amounts of reactants and products (the coefficients for a balanced equation do not change) Reactions are stoichiometric – means that there is a simple whole-number ratio of chemical amounts of reactants and products (the coefficients for a balanced equation do not change)

16. THREE TYPES OF EQUATIONS 1. Complete Balanced All reactants and products w/o dissociation 2. Ionic Shows highly soluble ionic compounds as ions 3. Net Ionic Cancel spectators from Ionic equation

17. NET IONIC EQUATIONS A chemical reaction equation that includes only reacting entities (molecules, atoms and/or ions) and omits any that do not change Writing Net Ionic Equations: 1) Write a complete balanced chemical equation 2) Dissociate all high-solubility ionic compounds, and ionize all strong acids to show the complete ionic equation 3) Cancel identical entities that appear on both the reactant and product sides 4) Write the net ionic equation, reducing coefficients if neccessary

18. PRACTICE Write the net ionic equation for the reaction of aqueous barium chloride and aqueous sodium sulfate. (Refer to the solubility table) 1) BaCl 2(aq) + Na 2 SO 4(aq)  BaSO 4(s) + 2NaCl (aq) 2) Ba 2+ (aq) + 2Cl - (aq) +2Na + (aq) + SO 4 2- (aq)  BaSO 4(s) + 2Na + (aq) + 2Cl - (aq) (Complete ionic equation) 3) Ba 2+ (aq) + 2Cl - (aq) +2Na + (aq) + SO 4 2- (aq)  BaSO 4(s) + 2Na + (aq) + 2Cl - (aq) 4) Ba 2+ (aq)) + SO 4 2- (aq)  BaSO 4(s) (Net ionic equation) Ions that are present but do not take part in (change during) a reaction are called spectator ions (like spectators at a sports game: they are present but do not take part in the game) When cancelling spectator ions, they must be identical in every way: chemical amount, form (atom, ion, molecule) and state of matter

19. PRACTICE Write the net ionic equation for the reaction of zinc metal and aqueous copper (II) sulfate (Refer to the solubility table) 1) Zn (s) + CuSO 4(aq)  Cu (s) + ZnSO 4(aq) 2) Zn (s) + Cu 2+ (aq) + SO 4 2- (aq)  Cu (s) + Zn 2+ (aq) + SO 4 2- (aq) (Complete ionic equation) 3) Zn (s) + Cu 2+ (aq) + SO 4 2- (aq)  Cu (s) + Zn 2+ (aq) + SO 4 2- (aq) 4) Zn (s) + Cu 2+ (aq)  Cu (s) + Zn 2+ (aq) (Net ionic equation)

20. PRACTICE Write the net ionic equation for the reaction of hydrochloric acid and barium hydroxide 1) 2HCl (aq) + Ba(OH) 2(aq)  BaCl 2(aq) + 2HOH (l) aka: 2H2O (l) 2) 2H + (aq) + 2Cl - (aq) + Ba 2+ (aq) + 2OH - (aq)  Ba 2+ (aq) + 2Cl - (aq) + 2H 2 O (l) (Complete ionic equation) 3) 2H + (aq) + 2Cl - (aq) + Ba 2+ (aq) + 2OH - (aq)  Ba 2+ (aq) + 2Cl - (aq) + 2H 2 O (l) 4) H + (aq)) + OH - (aq)  H 2 O (l) (Net ionic equation) – coefficients reduced to 1 For entities involving strong acids, H + (aq) is used as a matter of convenience over H 3 O + (aq)

21. STOICHIOMETRY & INDUSTRY Companies use stoichiometry to make the most money from the least amount of product Example: Natural gas

22. QUALITATIVE VS QUANTITATIVE ANALYSIS Qualitative Determining if a substance is present Quantitative How much of a substance is present

23. QUALITATIVE ANALYSIS Ion color in solution Some ions (anions or cations) have specific colors when in solution

24. QUALITATIVE ANALYSIS Flame Tests Metal ions may produce colors when thermally excited (e- returning to ground state)

25. QUALITATIVE ANALYSIS Precipitation Reactions Using replacement reactions to produce a precipitate can identify certain ions in solution

26. STOICHIOMETRY Measurement and relationship of quantities in chemical equations

27. STOICHIOMETRY A method of problem-solving using known quantities of one entity in a chemical reaction to find out unknown quantities of another entity in the same reaction Always requires a balanced chemical equation Series of steps for measurement calculations Always involves a mole ratio to “switch substances” Writing chemical reactions, net ionic equations and dissociation equations is essential Converting grams to moles (and vice versa) is also essential

28. MASS RATIOS In theory mass ratios could be used in stoichiometric calculations BUT it would be very messy (Example: not whole number ratios)

29. MOLE RATIOS  Balanced chemical equations provide a means of predicting moles of one substance given moles of another

31. TYPES OF STOICHIOMETRIC CALCULATIONS 1.Gravimetric – calculations involving mass calculations & predictions 2.Solution – using concentrations & volumes to calculate & predict 3. Gas – using volume, temperature and pressures are used to predict

32. GRAVIMETRIC STOICHIOMETRY Gravimetric Stoichiometry – method used to calculate the masses of reactants or products in a chemical reaction Gravimetric = mass measurement Using gravimetric stoichiometry, we can apply our knowledge of balanced chemical reactions and mass to mol conversions to: Predict the mass of product we will get from a reaction Estimate the amount of reactant we need for a reaction to produce a certain mass of product

33. GRAVIMETRIC STOICHIOMETRY STEPS: Step 1: Write a balanced chemical reaction equation List the measured mass (given), the unknown quantity (required) and conversion factors (M = molar mass) Step 2: Convert the mass of the measured substance to moles Step 3: Calculate the moles of the unknown substance using the mole ratio required given Step 4: Convert from moles of the required substance to its mass.

34. GRAVIMETRIC STOICHIOMETRY STEP 1Write a balanced chemical equation representing the reaction. Under the equation write the mass that is given and indicate what is required. Write the molar masses for both substances. STEP 2Convert the given mass to moles.

35. GRAVIMETRIC STOICHIOMETRY STEP 3Using the mole ratio from the balanced equation calculate the amount in moles of the required substance. STEP 4Convert the amount in moles of the given substance to mass.

36. HOW MANY GRAMS OF OXYGEN ARE REQUIRED TO COMPLETELY BURN 10.0 G OF PROPANE (C 3 H 8(G) )? The balanced chemical equation is: C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g) The number of moles of propane reacting is: (MASS to MOLES) 10.0 g x 1 mol = 0.227 mol 44.11 g The number of moles of oxygen produced is: (MOLE RATIO) 0.227 mol x 5 mol = 1.13 mol 1 mol The mass of oxygen produced is: (MOLES TO MASS) 1.1335 mol x 32.00 g = 36.3 g 1 mol

37. PRACTICE 20.0 g of butane (C 4 H 10(g) ) are completely burned in a lighter. How many grams of CO 2(g) are produced?

38. PRACTICE #2 (TEAM UNIT ANALYSIS) 2 C 4 H 10(g) + 13 O 2(g)  8 CO 2(g) + 10 H 2 O (g) m = 20.0g m = ? M = 58.14 g/mol M = 44.01 g/mol 2) Mass to moles, mole ratio, moles to mass 20.0 g x 1 mol x. 8 mol CO 2 x 44.01 g = 60.6 g 58.14 g 2 mol C 4 H 10 1 mol

39. PRACTICE What mass of iron (III) oxide is required to produce 100.0 g of iron? Fe 2 O 3(s) + 3 CO (g)  2 Fe (s) + 3 CO 2(g)

40. GRAVIMETRIC STOICHIOMETRY ExampleWhen iron and a silver nitrate solution are mixed they react in a single replacement reaction to produce solid silver. If 100.0 g of iron reacts completely with excess silver nitrate solution what mass of silver is produced.

41. Remember to keep the unrounded values in your calculator for further calculation until the final answer is reported.

42. STEP 1 Fe (s) + 3 AgNO 3(aq) -> Fe(NO 3 ) 3(aq) + 3 Ag (s) 100.0 g? g 55.85 g/mol 107.87 g/mol

43. STEP 2 & STEP 3 n = M 55.85 g/mol x 100g = 1.7905 mol Fe Use mole ratio -> mole Ag 3 mol Ag x 1.7905 mol Fe = 5.3715 mol Ag 1 mol Fe

44. STEP 4 m = n M = 5.3715 mol Ag x 107.87 g/mol = 579.4 g If 100.0 g of iron reacts completely in excess silver nitrate solution 579.4 g of silver can be produced.

45. GRAVIMETRIC PRACTICE Sample Problems p 276 – 278 Practice Problems # 8 -15 p 278 Read “Waste Water Treatment” p 279 Ticket to the TEST

46. If you ever need some Chem help feel free to give me a call 602 - 1023

47. PERCENT YIELD FOR REACTIONS We can use stoichiometry to test experimental designs, technological skills, purity of chemicals, etc. We evaluate these by calculating a percent yield. This is the ratio of the actual (experimental) quantity of product obtained and the theoretical (predicted) quantity of product obtained from a stoichiometry calculation Percent yield = actual yield x 100 predicted yield Some forms of experimental uncertainties: All measurements (limitations of equipment) Purity of chemical used (80-99.9% purity) Washing a precipitate (some mass is lost through filter paper) Estimation of reaction completion (qualitative judgements i.e. color)

48. PERCENT YIELD EXAMPLE Example: In a chemical analysis, 3.00 g of silver nitrate in solution was reacted with excess sodium chromate to produced 2.81 g of filtered, dried precipitate. Predicted value : Percent yield=

49. TESTING THE STOICHIOMETRIC METHOD Stoichiometry is used to predict the mass of precipitate actually produced in a reaction. Filtration is used to separate the mass of precipitate actually produced in a reaction

50. TESTING THE STOICHIOMETRIC METHOD What mass of lead is produced by the reaction of 2.13 g of zinc with an excess of lead(II) nitrate solution? Design : A known mass of zinc is place in a beaker with an excess of lead(II) nitrate solution. The lead is produced in the reaction is separated by filtration and dried. The mass of the lead is determined Prediction:

51. TESTING THE STOICHIOMETRIC METHOD Prediction: 6.75 g of lead will be produced (Stoichiometric calculation) Analysis : Evaluation : Could you also calculate the % yield ? Percent yield =

52. SOLUTION STOICHIOMETRY

53. The majority of work in research and industry involves solutions. Recall that solutions are easy to handle and are usually easier to control in reactions. The major difference compared to gravimetric and gas stoichiometry is that we use molar concentration (mol/L) as a conversion factor rather than molar mass or molar volume Solution stoichiometry – the procedure for calculating the molar concentration or volume of solution products or reactants

54. 2. SOLUTION STOICHIOMETRY Stoichiometric calculations for solutions involve molar concentration and the volume of a solution to determine an amount in moles (rather than M and m)

55. SOLUTION STOICHIOMETRY STEP 1 Write a balanced chemical equation including states. Under the equation list both the given and required measurements and conversion factors. STEP 2 Convert the given information to an amount in moles.

56. SOLUTION STOICHIOMETRY STEP 3 Use the mole ratio to calculate the amount of moles of the required substance. STEP 4 Convert the amount in moles to the requested quantity in the question.

57. SOLUTION STOICHIOMETRY Example If 10.0 mL of 0.020 mol/L HCl react exactly with 12.0 mL of Ba(OH) 2 solution, find the molar concentration of the basic solution. What was the pOH of the basic solution? pH?

58. EXAMPLE Ammonia and phosphoric acid solutions are used to produce ammonium hydrogen phosphate fertilizer. What volume of 14.8mol/L NH 3(aq) is needed to react with 1.00kL of 12.9mol/L of H 3 PO 4(aq) ?

59. EXAMPLE In an experiment, a 10.00 mL sample of sulfuric acid solution reacts completely with 15.9 mL of 0.150 mol/L potassium hydroxide. Calculate the amount concentration of the sulfuric acid.

60. SOLUTION STOICHIOMETRY 2HCl (aq) + Ba(OH) 2(aq) -> 2HOH (l) + BaCl 2(aq) 10.0 mLC = ? 0.020 mol/L12.0 mL n HCl = Cv = 0.020 mol x 10.0 mL x 1 L L 1000 mL n HCl = 2.00 x 10 -4 mol

61. SOLUTION STOICHIOMETRY Mole ratio 1 mole Ba(OH) 2 = ? mole Ba(OH) 2 2 mole HCl 2.00 x 10 -4 mol HCl n Ba(OH)2 = 2.00 x 10 -4 mol HCl x 1/2 n Ba(OH)2 = 1.00 x 10 -4 mol Ba(OH) 2

62. SOLUTION STOICHIOMETRY Molar concentration of Ba(OH) 2 solution => C = n/v [ Ba(OH) 2 ] = 1.00 x 10 -4 mol x 1000 mL 12.0 mL1 L [ Ba(OH) 2 ] = 0.0083 mol/L pOH = 2.08pH = 11.92

63. GAS STOICHIOMETRY

64. Many chemical reactions involve gases as a reactant or a product Gas Stoichiometry – the procedure for calculating the volume of gases as products or reactants Gases also have a molar volume (L/mol) rather than concentration. This is the conversion factor used to convert (liters of gas) to (moles of gas) The Ideal Gas Law (PV = nRT) may also be required to: A) find the number of moles of reactant B) Find the V, P, or T of the product

65. GAS STOICHIOMETRY Using gas laws and molar volumes stoichiometric calculations of reactions involving gases is possible Note: Law of Combining Volumes

66. GAS STOICHIOMETRY The mole ratio from a balanced chemical equation for a gas reaction can be used as a volume ratio.

67. IDEAL GAS LAW & STOICHIOMETRY If temperature and pressure are not constant stoichiometric calculations must utilize the Ideal Gas Law

68. EXAMPLE #1 If 300g of propane burns in a gas barbecue, what volume of oxygen measured at SATP is required for the reaction? Remember: 24.8L/mol for SATP24.8L/mol for SATP **Remember – molar volume is the conversion factor for gases just like molar mass is the conversion factor in gravimetric stoichiometry

69. EXAMPLE Hydrogen gas is produced when sodium metal is added to water. What mass of sodium is necessary to produce 20.0L of hydrogen at STP? Remember: 22.4L/mol for STP22.4L/mol for STP **Remember – molar volume is the conversion factor for gases just like molar mass is the conversion factor in gravimetric stoichiometry

70. EXAMPLE What volume of ammonia at 450kPa and 80 o C can be obtained from the complete reaction of 7.5kg of hydrogen with nitrogen? If the conditions are not STP or SATP, the molar volume cannot be used! You must use the ideal gas law to find the gas values using moles determined from stoichiometry

71. GAS STOICHIOMETRY SUMMARY 1.Write a balanced chemical equation and list the measurements, unknown quantity symbol, and conversion factors for the measured and required substances. 2.Convert the measured quantity to a chemical amount using the appropriate conversion factor 3.Calculate the chemical amount of the required substance using the mole ratio from the balanced chemical equation. 4.Convert the calculated chemical amount to the final quantity requested using the appropriate conversion factor.

72. GRAVIMETRIC, GAS AND SOLUTION STOICHIOMETRY SUMMARY 1.Write a balanced chemical equation and list the quantities and conversion factors for the given substance and the one to be calculated. 2.Convert the given measurement to its chemical amount using the appropriate conversion factor 3.Calculate the amount of the other substance using the mole ratio from the balanced chemical equation. 4.Convert the calculated chemical amount to the final quantity requested using the appropriate conversion factor. Remember to use the Ideal Gas Law for all gases not at STP or SATP

73. IDEAL GAS LAW & STOICHIOMETRY Recall: PV = nRT Sample Problem p 286 Practice Problems # 24 – 27 p 287 [See pg 104 for pressure conversions] Ticket to the TEST