2. Stoichiometry Unit V

3. I. The Mole a) Definition: An arbitrary unit used in chemistry to unite three measured properties of matter: –Mass –Volume –# of particles Similar to a dozen, except instead of 12, it’s Similar to a dozen, except instead of 12, it’s 6.02 X 10 23 (in scientific notation) This number is named in honor of Amedeo Avogadro (1776 – 1856)

4. a) Definition: t he mass of 1 mole (in grams) Equal to the numerical value of the average atomic mass (get from periodic table), or add the atoms together for a molecule 1 mole of C atoms= 12.0 g 1 mole of Mg atoms =24.3 g 1 mole of O 2 molecules =32.0 g II. Molar Mass

5. b) Molar Mass of Compounds The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) –Ex. Molar mass of CaCl 2 Element # of atomsAtomic Mass –Calcium = 1X40.08g =40.08g –Chlorine = 2X35.45g = 70.90g –Molar Mass of calcium chloride=110.98 g/mol

6. Practice Calculate the Molar Mass of calcium phosphate –Formula = –Gram Atomic Mass –Ca: 3 X 40.1 = –P: 2 X 31.0 = –O: 8 X 16.0 = –Molar Mass = Ca 3 (PO 4 ) 2 120.3 g 62.0 g 128.0 g 120.3g + 62.0g +128.0g310.3 g/mol

7. 1. Converting Moles to Mass Find the molar mass of the substance. Multiply the molar mass by the number of moles of the substance. Ex: Find the mass of 4.5 moles of Calcium Sulfate Chemical Name:Calcium Sulfate Chemical Formula:CaSO 4 Ca =1X40.08=40.08 S =1X32.06=32.06 O =4X15.9994=63.9976 Molar Mass =136.1 g/mol 136.1 g/molX4.5 mol=612.6 g

8. 2. Converting Mass to Moles Find the molar mass of the substance. Divide the given mass by the molar mass to determine the number of moles. Ex: Find the number of moles in 450 g of glucose (C 6 H 12 O 6 ). C =6X12.0111=72.0666 H =12X1.00794=12.09528 O =6X15.9994=95.9964 Molar Mass=180.16 g/mol Given Mass/Molecular Mass = 450 g ÷ 180.16 g/mol = 2.5 moles

9. III. Molar Volume a) Definition: the volume of 1 mole of a substance (in liters) One mole of any gas at STP occupies a volume of 22.4 liters. 1 mole of CO 2 (g) molecules= 22.4 liters 1 mole of N 2 (g) molecules =22.4 liters 1 mole of O 2 (g) molecules =22.4 liters

10. 1. Converting moles to volume Multiply the number of moles of the chemical (given) by 22.4. Practice: 2.0 moles of CO 2 (g) =____________liters 0.5 moles of O 2 (g) =____________liters 4.0 moles of CH 4 (g) = ________________ liters

11. 2. Converting volume to moles Divide the volume of the chemical (given) by 22.4. Practice: _____moles of N 2 (g) =33.6 liters _____moles of H 2 (g) = 5.6liters _____moles of CCl 4 (g) =56.0liters

12. IV. Molar Particles a)Definition: the # of particles (atoms/molecules/ions/electrons/etc.) in 1 mole. 1 mole of C atoms= 6.02 X 10 23 atoms 2 mole of Mg atoms =12.04 X 10 23 atoms 2 mole of O 2 molecules =1.204 X 10 24 molecules 1 mole of C 6 H 12 O 6 =6.02 X 10 23 molecules of glucose =6 moles of C atoms = 12 moles of H atoms =6 moles of O atoms =(6) · 6.02 X 10 23 atoms of carbon =36.12 X 10 23 atoms of carbon =(12) · 6.02 X 10 23 atoms of hydrogen =72.24 X 10 23 atoms of hydrogen =(6) · 6.02 X 10 23 atoms of oxygen = 36.12 X 10 23 atoms of oxygen

13. b) Conversions: Multiply the number of moles (given) by 6.02 x 10 23 Ex #1: 2.5 moles of CO 2 =__________________ molecules 1. Moles to Particles 6.02 X 10 23 molecules=x molecules 1 mole 2.5 moles X = 15.05 x 10 23 molecules of CO 2

14. 2. Particles to Moles Divide the number of particles (given) by 6.02 x 10 23. Ex: 1.806 X 10 24 molecules of CO 2 is equal to _____________________moles. X = 3 moles of CO 2 6.02 X 10 23 molecules=1.806 x 10 24 molecules 1 mole x moles

15. V. Conversions: a) Mass to Volume/Volume to Mass: One must first convert the given measurement to the mole. After converting to the mole, convert to the desire measurement.

16. Ex: At STP, 88 grams of CO 2 will occupy a volume of________liters. 1. Find the molar mass of CO 2. 2. Find the number of moles represented by the given mass. 3. Determine the number of liters represented by the # of moles.

17. 1. Find the molar mass of CO 2. CO 2 C = 1 X 12 = 12 O = 2 X 16 = 32 = 44 grams/mole 2. Find the number of moles represented by the given mass. # of moles = Given Mass Molecular Mass = 88 g 44 g = 2 moles

18. 3. Determine the number of liters represented by the # of moles. 22.4 liters=x liters 1 mole 2 moles X = 44.8 liters of CO 2

19. Ex: Find the mass of 56.0 liters of SO 2 (g) at STP. 1. Determine the molar mass of SO 2. 2. Convert from liters to moles. 3. Calculate the mass from the determined # of moles and the molar mass.

20. Ex: Find the mass of 56.0 liters of SO 2 (g) at STP. 1. Determine the molar mass of SO 2. 2. Convert from liters to moles. 3. Calculate the mass from the determined # of moles and the molar mass. SO 2 S = 1 X 32 = 32 O =2X16=32 64 grams 1 mole 22.4 liters 1 mole known 56.0 liters X moles = X = 2.5 moles 64 grams = x grams 1 mole 2.5 moles X = 160 grams

21. b) Converting Mass to Particles/Particles to Mass One must first convert the given measurement to the mole. After converting to the mole, convert to the desire measurement.

22. Ex #1: 88g of carbon dioxide = ________________molecules 1. Find the gram formula mass of CO 2 2. Find the number of moles represented by the determined mass 3. Determine the number of particles (molecules) based upon the calculated number of moles and the known ratio of moles to particles

23. Ex #1: 88g of carbon dioxide = ________________molecules CO 2 = 44 g/mol 1. Find the gram formula mass of CO 2 C = 1X12=12 O = 2X16=32 44g/mol

24. 2. Find the number of moles represented by the determined mass 88g of CO 2 =44g of CO 2 x moles 1 mole X = 2 moles of carbon dioxide 3. Determine the number of particles (molecules) based upon the calculated number of moles and the known ratio of moles to particles 6.02 X 10 23 =x molecules 1 mole 2 moles X = 12.04 X 10 23 molecules or 1.204 X 10 24 molecules

25. Ex #2: 9.03 X 10 23 molecules of SO 2 = ________________grams 1. Find the gram formula mass of SO 2 2. Find the number of moles represented by the given number of particles using the ratio 6.02 x 10 23 particles is to one mole. 3. Determine the number of grams be multiplying the number of moles (calculated in step 2) by the gram formula mass

26. Ex #2: 9.03 X 10 23 molecules of SO2 = ________________grams SO 2 = 64 g/mol 1. Find the gram formula mass of SO 2 S = 1X32=32 O = 2X16=32 64g/mol

27. 2. Find the number of moles represented by the given number of particles using the ratio 6.02 x 10 23 particles is to one mole 64 g of SO 2 = x g of SO 2 1 mole 1.5 moles X = 1.5 moles of sulfur dioxide 3. Determine the number of grams be multiplying the number of moles (calculated in step 2) by the gram formula mass X = 96 grams of SO 2 6.02 X 10 23 =9.03 x 10 23 molecules 1 mole x moles

28. Atoms or Molecules Moles Gram Atomic (Formula) Mass Divide by 6.02 X 10 23 Multiply by 6.02 X 10 23 Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table Volume Multiply By 22.4 Divide By 22.4

29. 1. How many atoms of Cu are present in 35.4 g of Cu?

30. 2. What is the mass (in grams) of 1.20 X 10 24 molecules of glucose (C 6 H 12 O 6 )?

31. VI. Chemical Equations A reaction indicates the concentration of reactants that are needed to get a product. The reactions must be balanced to establish ratios. Example: 2 Na + Cl 2  2 NaCl This reaction tells us that by mixing 2 moles of sodium atoms with 1 mole of chlorine molecules we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

32. Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H 2 + O 2  2 H 2 O –How many moles of reactants are needed? –What if we wanted 4 moles of water? –What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? –What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? 2 mol H 2 1 mol O 2 4 mol H 2 2 mol O 2 6 mol H 2, 6 mol H 2 O 25 mol O 2, 50 mol H 2 O

33. a) Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl 2  2 NaCl

34. 1. Mole-Mole Conversions How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl 2  2 NaCl

35. 2. Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2  2 NaCl

36. Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. 2 Al + 3 I 2  2 AlI 3

37. 3. Mass-Mole We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest –Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water – 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 0

38. Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide 4 Al + 3 O 2  2 Al 2 O 3

39. 4. Mass-Mass Conversions Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it. Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

40. Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N 2 + 3 H 2  2 NH 3 2.00g N 2 1 mol N 2 2 mol NH 3 17.06g NH 3 28.02g N 2 1 mol N 2 1 mol NH 3 = 2.4 g NH 3

41. Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

42. 5. Mass-Volume Conversions In calculating mass-volume questions, one must first convert the given mass to moles (calculate molar mass). Then, establish a ratio knowing the number of liters one mole of a gas occupies at STP and the number of calculated moles.

43. Ex: Mass-Volume Conversions 2 C 8 H 18 + 25 O 2 → 16 CO 2 + 18 H 2 O How many liters of CO 2 at STP are produced by the combustion of 342 grams of octane according the above equation? How many grams of O 2 at STP are required to completely combust 89.6 liters of octane according the above equation?

44. VIII. Percent Composition, Molecular and Empirical Formulas a) Percent Composition : the composition of a compound in terms of the percentage of each component present with respect to the whole. Ex #1: What is the percent composition by mass of the elements in sodium sulfate? Find the formula of sodium sulfate. Find the molar mass of sodium sulfate. Using the formula for % composition by mass (reference table T), calculate the % composition of each element in the compound.

45. Sodium Sulfate -Na 2 SO 4 Molar Mass = 142 g/mol % composition = part/whole X 100 Na 2 SO 4 Na=2X23=46 S=1X32=32 O=4X16=64 Na=46/142 X 100 =32.4% S=32/142 X 100 =22.5% O=64/142 X 100 =45.1% 100%

46. Ex #2: What is the percent composition by mass of the elements in Potassium permanganate? Ex #3: What is the percent composition by mass of the water in Copper (II) sulfate pentahydrate?

47. b) Types of Formulas 1.Molecular Formulas Represents the type and number of elements in a covalent compound.Represents the type and number of elements in a covalent compound. The subscripts do not need to be reduced in a molecular formula.The subscripts do not need to be reduced in a molecular formula. Ex: C 2 H 4 Ethene C 6 H 12 O 6 Glucose C 6 H 12 O 6 Glucose

48. 2. Empirical Formulas Represents the lowest whole number ratio of atoms in a compound. Represents the lowest whole number ratio of atoms in a compound. For ionic compounds, the formula is always empirical. This is due to the fact that one must reduce the subscripts in an ionic compound to the lowest whole number ratio. For ionic compounds, the formula is always empirical. This is due to the fact that one must reduce the subscripts in an ionic compound to the lowest whole number ratio. For covalent compounds (molecules), reducing its molecular formula to an empirical formula only shows the ratio of elements within the compound. For covalent compounds (molecules), reducing its molecular formula to an empirical formula only shows the ratio of elements within the compound. Ex: C 6 H 12 O 6 Molecular Formula C 1 H 2 O 1 Empirical Formula Ratio: C:H:O = 1:2:1

49. c) Determining empirical formulas from percent composition One can determine the chemical formula for a substance if given the percent composition of its elements. Ex: What is the empirical formula of a compound that consists of 58.80% barium, 13.75% sulfur, and 27.45% oxygen by mass? 1. Convert the percentages to grams. Ba58.80%58.80g S13.75%13.75g O27.45%27.45g

50. 2. Convert from Mass to Moles Ba137g=58.80g=0.43 mol 1 mol x mol S 32g=13.75g=0.43 mol 1 mol x mol O 16g=27.45g=1.71 mol 1 mol x mol

51. 3. Simplify the mole ratios to the lowest whole number Ba = 0.43/0.43=1 S = 0.43/0.43=1 O = 1.71/0.43=4 4. The simplified ratio of elements is equal to the subscripts of the atoms within the compounds. Ba 1 S 1 O 4 or BaSO 4

52. Practice: Given the following % compositions, determine the formulas for the following compounds. a)27.3 % carbon; 72.7 % oxygen b)30.43 % nitrogen; 69.57% oxygen

53. c) 37.6% carbon; 12.5% hydrogen; 49.9% oxygen d) 32.4 % sodium; 22.5% sulfur; 45.1% oxygen

54. c) Determining molecular formulas from percent composition KEY IDEA: The procedures for determining the molecular formula of a compound (given its elements % composition) is the same as finding its empirical formula with one exception. YOU MUST DETERMINE THE EMPIRICAL MASS OF THE COMPOUND. YOU MUST DIVIDE THE MOLECULAR MASS BY THE EMPIRICAL MASS TO DETERMINE THE ACTUAL NUMBER OF MOLES OF ATOMS WITHIN THE COMPOUND.

55. Ex #1 : In analyzing 30g of a compound, it was determined to consist of 80% carbon and 20% hydrogen by mass. Determine the compound’s empirical and molecular formulas. 1. Convert percentages to grams 80% C=80g C 20% H=20g H 2. Convert Mass to Moles 80g C/12g per mole=6.66 moles 20g H/1g per mole=20 moles 3. Simplify mole ratios C=6.66/6.66 = 1 H =20/6.66 = 3

56. 4. The mole ratios represent the subscripts for the empirical formula C1H3C1H3 5. Calculate the empirical mass C=1X12=12 H=3X 1= 3 Empirical Mass15g/mole 6. Divide the molecular mass (given) by the empirical mass to find the formula unit Molecular Mass=30= 2 Empirical Mass15

57. 7. Distribute the formula unit 2 C 1 H 3 =C 2 H 6 MOLECULAR FORMULA = C2H6C2H6

58. IX. Balancing Equations To abide by the Law of Conservation of Matter and Energy, the number of atoms within the reactants of a chemical equation must be equal to the number of atoms found within the products. a) Chemical Equation: - Indicates the concentration (moles) of reactants needed to form products with a specific concentration (# of moles). 1C 1 H 4 (g) + 2O 2 (g) → 1C 1 O 2 (g) + 2H 2 O 1 (g) THE NUMBERS IN RED ARE CALLED COEFFICIENTS. THEY INDICATE THE NUMBER OF MOLES OF THE CHEMICALS. COEFFICIENTS CAN BE USED TO ESTABLISH MOLE RATIOS. THE NUMBERS IN BLUE ARE CALLED SUBSCRIPTS. THEY INDICATE THE NUMBER OF ATOMS FOUND WITHIN THE COMPOUND. Reactants Products

59. b) Balancing Equations The combustion of propane (C 3 H 8 ) in the presence of molecular oxygen yields carbon dioxide and water vapor. 1. Write out the formulas for all chemicals involved in the reaction Propane C 3 H 8 Molecular Oxygen O 2 Carbon Dioxide CO 2 Water H 2 O 2.Arrange the reactants on the left side and the products on the right side of the yield sign __C 3 H 8 (g) + __O 2 (g) → __CO 2 (g) + __H 2 O (g) Reactants Products

60. 3. Create a checklist of the types and amounts of atoms present in the reactants and products __C 3 H 8 (g) + __O 2 (g) → __CO 2 (g) + __H 2 O (g) CHOCHO 382382 123123 4. Manipulate the coefficient to balance the # of atoms in the reactants and products. 3 X 3 X 7 4 X 8 X 10 5 10 X 1

61. X. Limiting Reactant Many times in chemistry the actual ratio of reactants is different from the theoretical one. This results in having an excess of a given reactant. (there is an overabundance of it). The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

62. Limiting Reactant To find the limiting reagent, one must calculate the number of moles of reactants and products. You then must calculate how much of a product can be produced from each of the reactants to determine which reactant is the limiting one. The lower amount of a product is the correct answer. The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

63. Limiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2  2 AlCl 3 Start with Al: Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.5 g AlCl 3 27.0 g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl 3 35.0g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 71.0 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3

64. LR Example Continued We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete.

65. Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

66. Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the previous problem?

67. Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

68. Limiting Reactant: Recap 1.You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2.Convert ALL of the reactants to the SAME product (pick any product you choose.) 3.The lowest answer is the correct answer. 4.The reactant that gave you the lowest answer is the LIMITING REACTANT. 5.The other reactant(s) are in EXCESS. 6.To find the amount of excess, subtract the amount used from the given amount. 7.If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!