1. Unit 9 Stoichiometry 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 (E 2 F 5 O 2 W 4 ): 2 E + 5 F + 2 O + 4 W  1 E 2 F 5 O 2 W 4 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. Ex: Balance the following equation. ___N 2 + ___ H 2  ___ NH 3 (This reaction is utilized in the Haber Process.) ___N 2 + ___ H 2  ___ NH 3 There are two N on the left and one on the right…so we put a 2 in front of NH 3. There are two H on the left and six (2x3) on the right…so we put a 3 in front of H 2. 23

7. Ex: Balance the following equation. ___ CaCl 2 + ___ Pb(NO 3 ) 3  ___ Ca(NO 3 ) 2 + ___ PbCl 3 There is one Ca on the left and one on the right…so we do not need to add a coefficient. There are two Cl on the left and three on the right…so we put a three in front of CaCl 2 and two in front of the PbCl 3 (2 and 3 both go into 6). 23 ___ CaCl 2 + ___ Pb(NO 3 ) 3  ___ Ca(NO 3 ) 2 + ___ PbCl 3 There is one Pb on the left and two on the right…so we put a two in front of Pb(NO 3 ) 3. 2 There are six NO 3 on the left and two on the right…so we put a three in front of Ca(NO 3 ) 2. 3

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 N 2 +3 H 2  2 NH 3 2 atom + 6 atoms  8 atoms 1 molec + 3 molec  2 molec 1x(6.02x10 23 ) + 3x(6.02x10 23 )  2x(6.02x10 23 ) 1 mole + 3 mole  2 mole Atoms: Molecules: Mole Ratio:

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 N 2 + 3 H 2  2 NH 3 3 mol H 2 1 mol N 2 2 mol NH 3 3 mol H 2 1 mol N 2 2 mol NH 3 Mole Ratios:

11. LITERS GRAMS MOLESPARTICLES Avogadro Number 1 mole Molar Mass Periodic Table 1 mole 22.4 L Chemical Equation

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 MOLE STOICHIOMETRY

14. Ex: How many mol of NH 3 are produced from 0.600 mol N 2 ? ___N 2 + ___ H 2  ___ NH 3 mol N 2 2 0.600 mol N 2 = 1.20 mol NH 3 Math: (0.600) x (2) / (1) = 1.20 This is Mole ratio from chemical equation 1 mol N 2 = 2 mol NH 3 mol NH 3 1 3 1 2

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

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

17. Ex: How many mol of LiC 2 H 3 O 2 are required to consume 21.5 mol Mg 3 (PO 4 ) 2 ? ___LiC 2 H 3 O 2 + ___ Mg 3 (PO 4 ) 2  ___ Li 3 PO 4 + ___Mg(C 2 H 3 O 2 ) 2 mol Mg 3 (PO 4 ) 2 6 21.5 mol Mg 3 (PO 4 ) 2 = 129 mol LiC 2 H 3 O 2 Math: (21.5) x (6) / (1) = 8.00 This is Mole ratio from chemical equation 1 mol Mg 3 (PO 4 ) 2 = 6 mol LiC 2 H 3 O 2 mol LiC 2 H 3 O 2 1 6231

18. PART III: MASS TO MOLE STOICHIOMETRY

19. g H 2 1 126.0 g H 2 = 41.58 mol NH 3 Math: (126.0) x (1) / (2.02) x (2) / (3) = 41.58 This is Molar Mass from periodic table 1 mol H 2 = 2.02 g H 2 mol H 2 2.02 mol H 2 2 mol NH 3 3 This is Mole ratio from chemical equation 3 mol H 2 = 2 mol NH 3 Ex: How many moles of NH 3 are produced from 126.0 g H 2 ? ___N 2 + ___ H 2  ___ NH 3 3 1 2

20. mol AgCl 3 8.645 mol AgCl = 411.5 g MgCl 2 Math: (8.645) x (3) / (6) x (95.20) / (1) = 411.5 This is Molar Mass from periodic table 1 mol MgCl 2 = 95.20 g MgCl 2 mol MgCl 2 6 95.20 g MgCl 2 1 This is Mole ratio from chemical equation 6 mol AgCl = 3 mol MgCl 2 Ex: How many grams of MgCl 2 are required to produce 8.645 mol AgCl? ___ MgCl 2 + ___ Ag 3 (PO 3 )  ___ Mg 3 (PO 3 ) 2 + ___ AgCl 3 1 26

21. mol MgCl 2 2 1.29 mol MgCl 2 = 346 g Ag 3 (PO 3 ) Math: (1.29) x (2) / (3) x (186.87) / (1) = 346 This is Molar Mass from periodic table 1 mol Ag 3 (PO 3 ) = 402.58 g Ag 3 (PO 3 ) mol Ag 3 (PO 3 ) 3 186.87 g Ag 3 (PO 3 ) 1 This is Mole ratio from chemical equation 3 mol MgCl 2 = 2 mol Ag 3 (PO 3 ) Ex: How many grams of Ag 3 (PO 3 ) are required to consume 1.29 mol MgCl 2 ? ___ MgCl 2 + ___ Ag 3 (PO 3 )  ___ Mg 3 (PO 3 ) 2 + ___ AgCl 3 1 26

22. g FeCl 3 1 1.15 g FeCl 3 = 3.55 x 10 -3 mol Fe 2 O 3 Math: (1.15) x (1) / (162.20) x 1) / (2) = 0.00355 = 3.55 x 10 -3 This is Molar Mass from periodic table 1 mol FeCl 3 = 162.20 g FeCl 3 mol FeCl 3 162.20 mol FeCl 3 1 mol Fe 2 O 3 2 This is Mole ratio from chemical equation 2 mol FeCl 3 = 1 mol Fe 2 O 3 Ex: How many moles of Fe 2 O 3 are produced from 1.15 g FeCl 3 ? ___ K 2 O + ___ FeCl 3  ___ KCl + ___ Fe 2 O 3 3 62 1

23. PART IV: MASS TO MASS STOICHIOMETRY

24. g AgCl 1 6.90 g AgCl = 2.67g CaCl 2 Math: (6.90) x (1) / (143.35) x (1) / (2) x (110.98) / (1) = 2.67 This is Mole Ratio from chemical equation 2 mol AgCl = 1 mol CaCl 2 mol AgCl 143.35 mol AgCl 1 mol CaCl 2 2 This is Molar Mass from periodic table 1 mol CaCl 2 = 110.98 g CaCl 2 g CaCl 2 110.98 1mol CaCl 2 This is Molar Mass from periodic table 1 mol AgCl = 143.35 g AgCl Ex: How many grams of CaCl 2 are required to produce 6.90g AgCl? ___ AgNO 3 + ___ CaCl 2  ___ Ca(NO 3 ) 2 + ___ AgCl 2 1 12

25. g CaCl 2 1 68.10 g CaCl 2 = 208.5g AgNO 3 Math: (68.10) x (1) / (110.98) x (2) / (1) x (169.91) / (1) = 208.5 This is Mole Ratio from chemical equation 1 mol CaCl 2 = 2 mol AgNO 3 mol CaCl 2 110.98 mol CaCl 2 2 mol AgNO 3 1 This is Molar Mass from periodic table 1 mol AgNO 3 = 169.91 g AgNO 3 g AgNO 3 169.91 1mol AgNO 3 This is Molar Mass from periodic table 1 mol CaCl 2 = 110.98 g CaCl 2 Ex: How many grams of AgNO 3 are required to consume 68.10g CaCl 2 ? ___ AgNO 3 + ___ CaCl 2  ___ Ca(NO 3 ) 2 + ___ AgCl 2 1 12

26. g NO 1 55.6 g NO = 74.1g O 2 Math: (55.6) x (1) / (30.01) x (5) / (4) x (32.00) / (1) = 74.1 This is Mole Ratio from chemical equation 4 mol NO = 5 mol O 2 mol NO 30.01 mol NO 5 mol O 2 4 This is Molar Mass from periodic table 1 mol O 2 = 32.00 g O 2 g O 2 32.00 1mol O 2 This is Molar Mass from periodic table 1 mol NO = 30.01 g NO Ex: How many grams of O 2 are required to produce 55.6g NO? ___ NH 3 + ___ O 2  ___ NO + ___ H 2 O 4 4 56

27. PART V: GAS VOLUME STOICHIOMETRY

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) – 0 o C 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. mol Ar L Ar 12.5 mol Ar = 280 L Ar Math: (12.5) x (22.4) / (1) = 280 22.4 1

30. 16.5 mol O 2 = 246 L NH 3 Math: (16.5) x (2) / (3) x (22.4) / (1) = 554 This is Mole Ratio from chemical equation 3 mol O 2 = 2 mol NH 3 mol O 2 2 mol NH 3 3 This is Molar Volume at STP 1 mol NH 3 = 22.4 L NH 3 L NH 3 22.4 1mol NH 3 Ex: How many liters of NH 3 are produced from 16.5 mol O 2 at STP? ___ N 2 + ___ O 2  ___ NH 3 1 32

31. g Fe 2 O 3 1 172.0 g Fe 2 O 3 = 72.38 L CO Math: (172.0) x (1) / (159.70) x (3) / (1) x (22.4) / (1) = 72.38 This is Mole Ratio from chemical equation 1 mol Fe 2 O 3 = 3 mol CO mol Fe 2 O 3 159.70 mol Fe 2 O 3 3 mol CO 1 This is Molar Volume at STP 1 mol CO = 22.4 L CO L CO 22.4 1mol CO This is Molar Mass from periodic table 1 mol Fe 2 O 3 = 159.70 g Fe 2 O 3 Ex: How many liters of CO will be liberated/produced from 172.0g Fe 2 O 3 at STP? ___ Fe 2 O 3 + ___ C  ___ Fe + ___ CO 1 2 33

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 Theoretical Yield X 100% Note: % Yield is never 100% - usually less due to error, heat loss, etc.

35. g CO 1 95.1 g CO = 2.26 mol Fe This is Mole Ratio from chemical equation 3 mol CO = 2 mol Fe mol CO 28.01 mol CO 2 mol Fe 3 This is Molar Mass from periodic table 1 mol CO = 28.01 g CO Ex: How many moles of Fe are produced from 95.1g CO. ___ Fe 2 O 3 + ___ CO  ___ Fe + ___ CO 2 1 2 33 Math: (95.1) x (1) / (28.01) x (2) / (3) = 1.59

36. g Fe 2 O 3 1 84.8 g Fe 2 O 3 = 70.1 g CO 2 This is Mole Ratio from chemical equation 1 mol Fe 2 O 3 = 3 mol CO mol Fe 2 O 3 159.70 mol Fe 2 O 3 3 mol CO 2 1 This is Molar Mass from periodic table 1 mol CO 2 = 44.01 g CO 2 g CO 2 44.01 1mol CO 2 This is Molar Mass from periodic table 1 mol Fe 2 O 3 = 159.70 g Fe 2 O 3 Ex: How many grams of CO 2 are produced from 84.8g Fe 2 O 3. ___ Fe 2 O 3 + ___ CO  ___ Fe + ___ CO 2 1 2 33 Math: (84.8) x (1) / (159.70) x (3) / (1) x (44.01) / (1) = 1.59

37. g Fe 2 O 3 1 84.8 g Fe 2 O 3 = 35.7 L CO 2 This is Mole Ratio from chemical equation 1 mol Fe 2 O 3 = 3 mol CO mol Fe 2 O 3 159.70 mol Fe 2 O 3 3 mol CO 2 1 This is Molar Volume at STP 1 mol CO 2 = 22.4 L CO 2 L CO 2 22.4 1mol CO 2 This is Molar Mass from periodic table 1 mol Fe 2 O 3 = 159.70 g Fe 2 O 3 Ex: Calculate the percent yield if 27.5L CO 2 are actually obtained from 84.8g Fe 2 O 3. ___ Fe 2 O 3 + ___ CO  ___ Fe + ___ CO 2 1 2 33 = 27.5 L 35.7 L 77.0 %= X 100% % Yield

38. g CuCl 1 12.85 g CuCl = 8.855 g Cu 3 N This is Mole Ratio from chemical equation 6 mol CuCl = 2 mol Cu 3 N mol CuCl 99.00 mol CuCl 2 mol Cu 3 N 6 This is Molar Volume at STP 1 mol Cu 3 N = 204.66 g Cu 3 N g Cu 3 N 204.66 1mol Cu 3 N This is Molar Mass from periodic table 1 mol CuCl = 99.00 g CuCl Ex: Calculate the percent yield if 5.124g Cu 3 N are actually obtained from 12.85g CuCl. ___ CuCl + ___ Sr 3 N 2  ___ Cu 3 N + ___ SrCl 2 6 2 13 = 5.124g 8.855g 57.87 %= X 100% % Yield

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 Theoretical Yield = 21.0g 25.0g 84.0 %= X 100%

40. PART VII: SOLUTION CONCENTRATION

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 CaCl 2 ? M= mol L = 9.3 mol CaCl 2 3.2 L 2.9 M CaCl 2 =

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

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

47. Try the following. 1)What is the molarity of a 0.725 L solution containing 0.673g K 2 C 2 O 4 ? 1)5.59 x 10 -3 M K 2 C 2 O 4 or 0.00559 M K 2 C 2 O 4

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 M 1 V 1 = M 2 V 2

49. Ex: How many mL of 2.00 M MgSO 4 must be diluted to prepare 100.0 mL of 0.400 M MgSO 4 ? M 1 V 1 = M 2 V 2 (2.00M)(V 1 ) = (0.400M)(100.0mL) V 1 = 20.0 mL Preparation of solution: Need to mix 20.0 mL of the concentrated MgSO 4 with about 80 mL of H 2 O to get the desired 0.400M MgSO 4.

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. M 1 V 1 = M 2 V 2 (12.0M)(V 1 ) = (3.80M)(250.0mL) V 1 = 79.2 mL Preparation of solution: Mix 79.2 mL of concentrated HCl (aq) with about (250-79.2)= 170.8 mL H 2 O 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.