1. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

2. Reaction Types

3. Combination Reactions Examples: N 2 (g) + 3 H 2 (g)  2 NH 3 (g) C 3 H 6 (g) + Br 2 (l)  C 3 H 6 Br 2 (l) 2 Mg (s) + O 2 (g)  2 MgO (s) Two or more substances react to form one product

4. 2 Mg (s) + O 2 (g)  2 MgO (s) Link to Video

5. Decomposition Reactions Examples: CaCO 3 (s)  CaO (s) + CO 2 (g) 2 KClO 3 (s)  2 KCl (s) + O 2 (g) 2 NaN 3 (s)  2 Na (s) + 3 N 2 (g) One substance breaks down into two or more substances Link to Video

6. Combustion Reactions Examples: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) Rapid reactions that produce a flame Most often involve hydrocarbons reacting with oxygen in the air

7. Balancing Reactions

8. Law of Conservation of Mass “We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” --Antoine Lavoisier, 1789

9. Dalton’s Postulates Atoms of an element are not changed into atoms of a different element by chemical reactions; atoms are neither created nor destroyed in chemical reactions. They just rearrange!

10. 3.7 A process in which one or more substances is changed into one or more new substances is a chemical reaction. A chemical equation uses chemical symbols to show what happens during a chemical reaction. reactantsproducts

11. How to “Read” Chemical Equations 2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO 3.7

12. Balancing Chemical Equations 1.Write the correct formula(s) for the reactants and the product(s). Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O 2.Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 3.7 2C 2 H 6 NOT C 4 H 12

13. Balancing Chemical Equations 3.Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O 3.7 start with C or H but not O 2 carbon on left 1 carbon on right multiply CO 2 by 2 C 2 H 6 + O 2 2CO 2 + H 2 O 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3 C 2 H 6 + O 2 2CO 2 + 3H 2 O

14. Balancing Chemical Equations 4.Balance those elements that appear in two or more reactants or products. 3.7 2 oxygen on left 4 oxygen (2x2) C 2 H 6 + O 2 2CO 2 + 3H 2 O + 3 oxygen (3x1) multiply O 2 by 7 2 = 7 oxygen on right C 2 H 6 + O 2 2CO 2 + 3H 2 O 7 2 remove fraction multiply both sides by 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O

15. Balancing Chemical Equations 5.Check to make sure that you have the same number of each type of atom on both sides of the equation. 3.7 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O ReactantsProducts 4 C 12 H 14 O 4 C 12 H 14 O 4 C (2 x 2)4 C12 H (2 x 6)12 H (6 x 2) 14 O (7 x 2)14 O (4 x 2 + 6)

16. Remember 2 Mg + O 2 2 MgO IS NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO 3.7 So How do we relate masses of reactants and products? That’s what we can observe!

17. Formula Weights

18. Formula Weight (FW) Sum of the average atomic weights for the atoms in a chemical formula So, the formula weight of calcium chloride, CaCl 2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu) 111.1 amu These are generally reported for ionic compounds

19. Molecular Weight (MW) Sum of the average atomic weights of the atoms in a molecule For the molecule ethane, C 2 H 6, the molecular weight would be C: 2(12.0 amu) + H: 6(1.0 amu) 30.0 amu

20. Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % element = (number of atoms)(atomic weight) (FW of the compound) x 100

21. Percent Composition So the percentage of carbon in ethane is… %C = (2)(12.0 amu) (30.0 amu) 24.0 amu 30.0 amu = x 100 = 80.0%

22. Moles

23. Using Moles Moles provide a bridge from the molecular scale to the real-world scale

24. Avogadro’s Number 6.02 x 10 23 1 mole of 12 C has a mass of 12 g

25. The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C 3.2 1 mol = N A = 6.0221367 x 10 23 Avogadro’s number (N A )

26. Molar mass is the mass of 1 mole of in grams eggs shoes marbles atoms 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams/mol) 3.2

27. Percent composition revisited n x molar mass of element molar mass of compound x 100% n is the subscript number … Assume one mole of substance and the math goes like this. C2H6OC2H6O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14%H = 6 x (1.008 g) 46.07 g x 100% = 13.13%O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% 3.5

28. Question What is the mass on one atom of carbon 12? What is the mass in grams of one amu?

29. Mole Relationships One mole of atoms, ions, or molecules contains Avogadro’s number of those particles One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound

30. Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? 3.2

31. Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 0.551 g K 1 mol K 39.10 g K x 3.2 = 0.0141 moles K

32. Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2 1S32.07 amu 2O+ 2 x 16.00 amu SO 2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2 3.3

33. Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C 3 H 8 O ? 3.3

34. Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C 3 H 8 O ? 1 mol C 3 H 8 O = (3 x 12.01) + (8 x 1.008) + 16.00 = 60.09 g C 3 H 8 O 1 mol H = 6.022 x 10 23 atoms H or 5.82 x 10 24 atoms H 3.3 1 mol C 3 H 8 O molecules = 8 mol H atoms 72.5 g C 3 H 8 O 1 mol C 3 H 8 O 60.09 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O x = 9.65 moles H

35. 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units Mass Changes in Chemical Reactions 3.8

36. Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x = 235 g H 2 O 3.8

37. Finding Empirical Formulas

38. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition

39. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

40. Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C:61.31 g x = 5.105 mol C H: 5.14 g x= 5.09 mol H N:10.21 g x= 0.7288 mol N O:23.33 g x = 1.456 mol O 1 mol 12.01 g 1 mol 14.01 g 1 mol 1.01 g 1 mol 16.00 g

41. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= 7.005  7 H:= 6.984  7 N:= 1.000 O:= 2.001  2 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 1.458 mol 0.7288 mol

42. Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2

43. Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this –C is determined from the mass of CO 2 produced –H is determined from the mass of H 2 O produced –O is determined by difference after the C and H have been determined

44. Elemental Analyses Compounds containing other elements are analyzed using methods analogous to those used for C, H and O

45. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products

46. Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

47. Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O

48. 3.6 Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O What is the percent composition of an organic substance called ethanol? Real Life Analysis

49. 3.6 g CO 2 mol CO 2 mol Cg C g H 2 O mol H 2 Omol Hg H g of O = g of sample – (g of C + g of H) Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O

50. Limiting Reactants

51. How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredients Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

52. How Many Cookies Can I Make? In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

53. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount

54. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount –In other words, it’s the reactant you’ll run out of first (in this case, the H 2 )

55. Limiting Reactants In the example below, the O 2 would be the excess reagent

56. Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. g Almol Almol Fe 2 O 3 neededg Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al neededg Al needed 124 g Al 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = 367 g Fe 2 O 3 Start with 124 g Alneed 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent 3.9

57. Use limiting reagent (Al) to calculate amount of product that can be formed. g Almol Almol Al 2 O 3 g Al 2 O 3 124 g Al 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 x = 234 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 3.9

58. Theoretical Yield The theoretical yield is the amount of product that can be made –In other words it’s the amount of product possible as calculated through the stoichiometry problem This is different from the actual yield, the amount one actually produces and measures

59. Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Theoretical Yield Percent Yield =x 100