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Unit 9 Stoichiometry

Unit 9 Stoichiometry. Chemistry I Mr. Patel SWHS. Topic Outline. MUST have a scientific calculator (not graphing)!!! Stoichiometry (12.1) Mole to Mole Stoichiometry (12.2) Mass to Mole Stoichiometry (12.2) Mass to Mass Stoichiometry (12.2) Gas Volume Stoichiometry (12.2)

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Unit 9 Stoichiometry

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  1. Unit 9Stoichiometry Chemistry I Mr. Patel SWHS

  2. Topic Outline • MUST have a scientific calculator (not graphing)!!! • Stoichiometry (12.1) • Mole to Mole Stoichiometry (12.2) • Mass to Mole Stoichiometry (12.2) • Mass to Mass Stoichiometry (12.2) • Gas Volume Stoichiometry (12.2) • Percent Yield (12.3) • Solution Concentration (16.2)

  3. Part I:stoichiometry

  4. Consider a recipe • 2 eggs (E) + 5 cups of flour (F) + 2 cups oil (O) + 4 cups water (W) produce 1 cake (E2F5O2W4 ): 2 E + 5 F + 2 O + 4 W  1 E2F5O2W4 • A recipe can be considered a chemical equation • How many eggs are needed to make 5 cakes? • 10 eggs • How many cups of water are needed to make 5 cakes? • 20 cups of water

  5. Balanced Equations • We can obtain that same type of information from a balanced chemical reaction • Thus it is important that a chemical equation is always balanced before doing any stoichiometry. • Just as you learned previously, balance equations by checking one element at a time starting with the leftmost element • Goal: The atoms before and after the arrow are the same!

  6. (This reaction is utilized in the Haber Process.) Ex: Balance the following equation. ___N2 + ___ H2 ___ NH3 ___N2 + ___ H2 ___ NH3 3 2 There are two N on the left and one on the right…so we put a 2 in front of NH3. There are two H on the left and six (2x3) on the right…so we put a 3 in front of H2.

  7. Ex: Balance the following equation. ___ CaCl2 + ___ Pb(NO3)3 ___ Ca(NO3)2 + ___ PbCl3 3 2 3 2 ___ CaCl2 + ___ Pb(NO3)3 ___ Ca(NO3)2 + ___ PbCl3 There is one Ca on the left and one on the right…so we do not need to add a coefficient. There are twoCl on the left and three on the right…so we put a three in front of CaCl2 and two in front of the PbCl3 (2 and 3 both go into 6). There is onePb on the left and two on the right…so we put a two in front of Pb(NO3)3. There are six NO3 on the left and two on the right…so we put a three in front of Ca(NO3)2.

  8. Stoichiometry • Stoichiometry – calculation of quantities in a chemical reaction • Greek “measure elements” • Quantities = reactants and/or products • Predict how much reactant needed and product produced • Allows for efficient use of resources (save money) • Aids in safety precautions (volume expansion)

  9. Stoichiometry • Atoms and mass are always the same before and after a reaction (conservation of mass) 1 N2 + 3 H2  2 NH3 +  2 atom 8 atoms 6 atoms Atoms: +  1 molec 2 molec 3 molec Molecules: +  1x(6.02x1023) 2x(6.02x1023) 3x(6.02x1023) Molecules: +  Mole Ratio: 1 mole 2 mole 3 mole

  10. Stoichiometry • Stoichiometry allows for the conversion within a balanced chemical equation using the coefficients as the mole ratios! • Conversion Factor 5: Chemical Equation (mole to mole) 1 N2 + 3 H2 2 NH3 1 mol N2 2 mol NH3 3 mol H2 Mole Ratios: 3 mol H2 1 mol N2 2 mol NH3

  11. Chemical Equation 1 mole MOLES PARTICLES 1 mole LITERS AvogadroNumber 22.4 L 1 mole Molar Mass Periodic Table GRAMS

  12. Stoichiometry • Three types of stoichiometry problems: • Mole to Mole (1 step) • Mass to Mole (2 steps) • Mass to Mass (3 steps) • These are all conversion problems (use factor label method). All the same rules still apply in setting up and solving factor label problems!

  13. Part II:Mole to Molestoichiometry

  14. Ex: How many mol of NH3 are produced from 0.600 mol N2? ___N2 + ___H2 ___ NH3 3 1 2 This is Mole ratiofrom chemical equation 1 mol N2 = 2 mol NH3 2 mol NH3 0.600 mol N2 = 1.20 mol NH3 1 mol N2 Math: (0.600) x (2) / (1) = 1.20

  15. Ex: How many mol of Al are produced from 3.70 mol Al2O3? ___Al + ___ O2 ___ Al2O3 3 4 2 This is Mole ratiofrom chemical equation 2 mol Al2O3 = 4 mol Al 4 mol Al 3.70 mol Al2O3 = 7.40 mol Al 2 mol Al2O3 Math: (3.70) x (4) / (2) = 7.40

  16. Ex: How many mol of Al are required to consume 6.00 mol O2? ___Al + ___ O2 ___ Al2O3 3 4 2 This is Mole ratiofrom chemical equation 2 mol Al2O3 = 4 mol Al 4 mol Al 6.00 mol O2 = 8.00 mol Al 3 mol O2 Math: (6.00) x (4) / (3) = 8.00

  17. Ex: How many mol of LiC2H3O2 are required to consume 21.5 molMg3(PO4)2? ___LiC2H3O2+ ___ Mg3(PO4)2 ___ Li3PO4 + ___Mg(C2H3O2)2 6 1 2 3 This is Mole ratiofrom chemical equation 1 molMg3(PO4)2 = 6 molLiC2H3O2 6 molLiC2H3O2 21.5 molMg3(PO4)2 = 129 molLiC2H3O2 1 mol Mg3(PO4)2 Math: (21.5) x (6) / (1) = 8.00

  18. Part III:Mass to Molestoichiometry

  19. Ex: How many moles of NH3 are produced from 126.0 g H2? ___N2 + ___H2 ___ NH3 3 1 2 This is Molar Massfrom periodic table 1 mol H2 = 2.02 g H2 This is Mole ratiofrom chemical equation 3 mol H2= 2 molNH3 1 mol H2 2 mol NH3 126.0 g H2 = 41.58 mol NH3 mol H2 3 g H2 2.02 Math: (126.0) x (1) / (2.02) x (2) / (3) = 41.58

  20. Ex: How many grams of MgCl2 are required to produce 8.645 molAgCl? ___ MgCl2+ ___ Ag3(PO3)  ___ Mg3(PO3)2 + ___ AgCl 2 6 3 1 This is Mole ratiofrom chemical equation 6 molAgCl= 3 molMgCl2 This is Molar Massfrom periodic table 1 mol MgCl2 = 95.20 g MgCl2 3 mol MgCl2 95.20 g MgCl2 8.645 molAgCl = 411.5 g MgCl2 mol MgCl2 1 molAgCl 6 Math: (8.645) x (3) / (6) x (95.20) / (1) = 411.5

  21. Ex: How many grams of Ag3(PO3) are required to consume 1.29 mol MgCl2? ___ MgCl2 + ___ Ag3(PO3)  ___ Mg3(PO3)2 + ___ AgCl 2 6 3 1 This is Mole ratiofrom chemical equation 3 molMgCl2= 2 molAg3(PO3) This is Molar Massfrom periodic table 1 molAg3(PO3) = 402.58 g Ag3(PO3) 2 mol Ag3(PO3) 186.87 g Ag3(PO3) 1.29 mol MgCl2 = 346 g Ag3(PO3) mol Ag3(PO3) 1 mol MgCl2 3 Math: (1.29) x (2) / (3) x (186.87) / (1) = 346

  22. Ex: How many moles of Fe2O3 are produced from 1.15 g FeCl3? ___ K2O + ___FeCl3 ___ KCl + ___ Fe2O3 3 2 6 1 This is Molar Massfrom periodic table 1 mol FeCl3 = 162.20 g FeCl3 This is Mole ratiofrom chemical equation 2 molFeCl3= 1 molFe2O3 1 mol FeCl3 1 mol Fe2O3 1.15 g FeCl3 = 3.55 x 10-3mol Fe2O3 mol FeCl3 2 162.20 g FeCl3 Math: (1.15) x (1) / (162.20) x 1) / (2) = 0.00355 = 3.55 x 10-3

  23. Part IV:Mass to Massstoichiometry

  24. Ex: How many grams of CaCl2 are required to produce 6.90g AgCl? ___ AgNO3 + ___ CaCl2 ___ Ca(NO3)2 + ___ AgCl 2 1 1 2 This is Molar Massfrom periodic table 1 molAgCl = 143.35 g AgCl This is Mole Ratiofrom chemical equation 2 molAgCl = 1 mol CaCl2 This is Molar Massfrom periodic table 1 mol CaCl2 = 110.98 g CaCl2 mol CaCl2 1 molAgCl 1 g CaCl2 110.98 6.90 g AgCl 2.67gCaCl2 = 1 mol CaCl2 g AgCl molAgCl 2 143.35 Math: (6.90) x (1) / (143.35) x (1) / (2) x (110.98) / (1) = 2.67

  25. Ex: How many grams of AgNO3 are required to consume 68.10g CaCl2? ___ AgNO3 + ___ CaCl2 ___ Ca(NO3)2 + ___ AgCl 2 1 1 2 This is Molar Massfrom periodic table 1 mol CaCl2 = 110.98 g CaCl2 This is Mole Ratiofrom chemical equation 1 mol CaCl2 = 2 mol AgNO3 This is Molar Massfrom periodic table 1 mol AgNO3 = 169.91 g AgNO3 mol AgNO3 1 mol CaCl2 2 g AgNO3 169.91 68.10 g CaCl2 208.5gAgNO3 = 1 mol AgNO3 g CaCl2 mol CaCl2 1 110.98 Math: (68.10) x (1) / (110.98) x (2) / (1) x (169.91) / (1) = 208.5

  26. Ex: How many grams of O2 are required to produce 55.6g NO? ___ NH3 + ___ O2 ___ NO + ___ H2O 4 4 5 6 This is Molar Massfrom periodic table 1 mol NO = 30.01 g NO This is Mole Ratiofrom chemical equation 4 mol NO = 5 mol O2 This is Molar Massfrom periodic table 1 mol O2 = 32.00 g O2 mol O2 1 mol NO 5 g O2 32.00 55.6 g NO = 74.1gO2 1 mol O2 g NO mol NO 4 30.01 Math: (55.6) x (1) / (30.01) x (5) / (4) x (32.00) / (1) = 74.1

  27. Part V:Gas volumestoichiometry

  28. Mole to Volume Conversion • Gases are often measured in volume rather than grams • A conversion is available between mole and volume only at specific conditions • Only for gases (ideal) • Standard Temperature and Pressure (STP) • 0oC and 1 atm • Conversion Factor 4: 1 mole = 22.4 L

  29. Ex: Convert 12.5 mol Ar to liters of Ar at STP. Use the factor-label method. 22.4 L Ar 12.5 mol Ar = 280 L Ar 1 mol Ar Math: (12.5) x (22.4) / (1) = 280

  30. Ex: How many liters of NH3 are produced from 16.5 mol O2 at STP? ___ N2 + ___ O2 ___ NH3 1 3 2 This is Mole Ratiofrom chemical equation 3 mol O2 = 2 mol NH3 This is Molar Volumeat STP 1 mol NH3 = 22.4 L NH3 mol NH3 2 L NH3 22.4 16.5 mol O2 = 246 L NH3 1 mol NH3 mol O2 3 Math: (16.5) x (2) / (3) x (22.4) / (1) = 554

  31. Ex: How many liters of CO will be liberated/produced from 172.0g Fe2O3 at STP? ___ Fe2O3 + ___ C  ___ Fe + ___ CO 1 2 3 3 This is Molar Massfrom periodic table 1 mol Fe2O3 = 159.70 g Fe2O3 This is Mole Ratiofrom chemical equation 1 mol Fe2O3 = 3 mol CO This is Molar Volumeat STP 1 mol CO = 22.4 L CO mol CO 1 mol Fe2O3 3 L CO 22.4 172.0 g Fe2O3 = 72.38 L CO 1 mol CO g Fe2O3 mol Fe2O3 1 159.70 Math: (172.0) x (1) / (159.70) x (3) / (1) x (22.4) / (1) = 72.38

  32. Part VI:Percent Yield

  33. Percent Yield • Measures the efficiency of a reaction • Similar to a grade on an assignment • How well you scored based upon best score • Tell how “well” the reaction proceeded • Can determine the applicability of the reaction process: • High Yield = Less Money Wasted = More Money Earned

  34. Percent Yield • Theoretical Yield • maximum amount of product that could have been formed (calculated) • Actual Yield • amount of product actually obtained when doing the reaction % Yield Actual Yield = X 100% Theoretical Yield Note: % Yield is never 100% - usually less due to error, heat loss, etc.

  35. Ex: How many moles of Fe are produced from 95.1g CO. ___ Fe2O3 + ___ CO  ___ Fe + ___ CO2 1 2 3 3 This is Molar Massfrom periodic table 1 mol CO = 28.01 g CO This is Mole Ratiofrom chemical equation 3 mol CO = 2 mol Fe mol Fe 1 mol CO 2 95.1 g CO = 2.26 mol Fe g CO mol CO 3 28.01 Math: (95.1) x (1) / (28.01) x (2) / (3) = 1.59

  36. Ex: How many grams of CO2 are produced from 84.8g Fe2O3. ___ Fe2O3 + ___ CO  ___ Fe + ___ CO2 1 2 3 3 This is Molar Massfrom periodic table 1 mol Fe2O3 = 159.70 g Fe2O3 This is Mole Ratiofrom chemical equation 1 mol Fe2O3 = 3 mol CO This is Molar Massfrom periodic table 1 mol CO2 = 44.01 g CO2 mol CO2 1 mol Fe2O3 3 44.01 g CO2 84.8 g Fe2O3 = 70.1 g CO2 1 mol CO2 g Fe2O3 mol Fe2O3 1 159.70 Math: (84.8) x (1) / (159.70) x (3) / (1) x (44.01) / (1) = 1.59

  37. Ex: Calculate the percent yield if 27.5L CO2 are actually obtained from 84.8g Fe2O3. ___ Fe2O3 + ___ CO  ___ Fe + ___ CO2 1 2 3 3 This is Molar Massfrom periodic table 1 mol Fe2O3 = 159.70 g Fe2O3 This is Mole Ratiofrom chemical equation 1 mol Fe2O3 = 3 mol CO This is Molar Volumeat STP 1 mol CO2 = 22.4 L CO2 mol CO2 1 mol Fe2O3 3 L CO2 22.4 84.8 g Fe2O3 = 35.7 L CO2 1 mol CO2 g Fe2O3 mol Fe2O3 1 159.70 27.5 L = = 77.0 % % Yield X 100% 35.7 L

  38. Ex: Calculate the percent yield if 5.124g Cu3N are actually obtained from 12.85g CuCl. ___ CuCl + ___ Sr3N2 ___ Cu3N + ___ SrCl2 6 2 1 3 This is Molar Massfrom periodic table 1 molCuCl = 99.00 g CuCl This is Mole Ratiofrom chemical equation 6 molCuCl = 2 mol Cu3N This is Molar Volumeat STP 1 mol Cu3N = 204.66 g Cu3N mol Cu3N 1 molCuCl 2 g Cu3N 204.66 12.85 g CuCl 8.855 gCu3N = 99.00 1 mol Cu3N g CuCl molCuCl 6 5.124g = = 57.87 % % Yield X 100% 8.855g

  39. Ex: The theoretical yield for a reaction is 25.0g. When performed, only 21.0g of product was obtained. Calculate the percent yield of the reaction. % Yield = Actual Yield X 100% Theoretical Yield 21.0g = = 84.0 % X 100% 25.0g

  40. Part VII:Solutionconcentration

  41. Concentration • Measure of the amount of solute in a given volume of solvent • Solute = lesser quantity particle • Solvent = greater quantity particle • Dilute = solution with lesser concentration • Concentrated = solution with greater concentration • These are all qualitative terms

  42. Molarity • In chemistry, there are many quantitative measurements to describe concentration • Molarity (M) • Molality (m) • Parts per million (ppm) • Parts per billion (ppb)

  43. Molarity • Molarity (M or [X]) is determined by the following equation: • Consider: molarity x liters = moles  grams • Written as “3.00 M” and read as “3.00 Molar” molarity moles solute = liters solution

  44. Ex: What is the concentration of a 3.2 L solution containing 9.3 mol CaCl2? 9.3 mol CaCl2 mol M = = = 2.9 M CaCl2 L 3.2 L

  45. Ex: What is the molarity when 0.90g NaCl are dissolved to a volume of 0.100 L? 0.015 mol NaCl mol M = = = 0.15 M NaCl L 0.100 L 1 mol NaCl 0.90 g NaCl = 0.015 mol NaCl 58.50 g NaCl

  46. Ex: How many grams of Al(C2H3O2)3 will I need to dissolve to prepare a 1850mL solution with a concentration that is 0.75M? 1 L 1850 mL Step 1: Convert mL to LWe can only use L = 1.85 L 1000 mL mol mol = M x L M =  Step 2: Determine #of moles L = (0.75M)(1.85L) = 1.40 molAl(C2H3O2)3 g Al(C2H3O2)3 204.11 1.40 molAl(C2H3O2)3 286 g Al(C2H3O2)3 = Step 3: Convert molesto grams molAl(C2H3O2)3 1

  47. Try the following. • What is the molarity of a 0.725 L solution containing 0.673g K2C2O4? • 5.59 x 10-3 M K2C2O4or 0.00559 M K2C2O4

  48. Dilutions • In a lab, chemicals are stored as stock solutions (concentrated) • We need to dilute these stock solutions with solvent in order to obtain the amount and concentration we desire • Note: moles do not change • Dilution Equation: • M = Molarity • V = Volume M1V1 = M2V2

  49. Ex: How many mL of 2.00 M MgSO4 must be diluted to prepare 100.0 mL of 0.400 M MgSO4? M1V1 = M2V2 (2.00M)(V1) = (0.400M)(100.0mL) V1 = 20.0 mL Preparation of solution: Need to mix 20.0 mL of the concentrated MgSO4 with about 80 mL of H2O to get the desired 0.400M MgSO4.

  50. Ex: How many mL of 12.0 M HCl(aq)must be diluted to prepare 250.0 mL of 3.80 M HCl(aq)? Explain how to prepare this solution. M1V1 = M2V2 (12.0M)(V1) = (3.80M)(250.0mL) V1 = 79.2 mL Preparation of solution: Mix 79.2 mL of concentrated HCl(aq) with about (250-79.2)= 170.8 mL H2O to get the desired 250.0 mL of 3.80M HCl(aq). Note: Always add acid to water…NEVER add water to acid – very exothermic.

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