1. Ch. 3 Stoichiometry: Calculations with Chemical Formulas

2. Law of Conservation of Mass Atoms are neither created nor destroyed during any chemical reaction. Atoms are simply rearranged.

3. Stoichiometry The quantitative nature of chemical formulas and chemical reactions

4. Reactants The chemical formulas on the left of the arrow that represent the starting substances 2H 2 + O 2  2H 2 O Reactants

5. Products The substances that are produced in the reaction and appear to the right of the arrow 2H 2 + O 2  2H 2 O Products

6. Because atoms are neither created nor destroyed in any reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

7. Balancing Chemical Equations CH 4 + O 2  CO 2 + H 2 O C=1 C=1 H=4 H=2 O=2 O=3

8. Balancing Chemical Equations CH 4 + O 2  CO 2 + 2H 2 O C=1 C=1 H=4 H=2 X 2 =4 O=2 O=3

9. Balancing Chemical Equations CH 4 + O 2  CO 2 + 2H 2 O C=1 C=1 H=4 H=2 x 2 = 4 O=2 O= 2 + 2x1 = 4

10. Balancing Chemical Equations CH 4 + 2O 2  CO 2 + 2H 2 O C=1 C=1 H=4 H=2 x 2 = 4 O=2 x 2 = 4 O= 2 + 2x1 = 4

11. Combustion Reactions Rapid reactions that produce a flame. Most combustion reactions involve O 2 as a reactant Form CO 2 and H 2 O as products

12. Combustion Reactions C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (l) C= 3 C=1 X 3 = 3 H=8 H=2 X 4 = 8 O= 2 X 5 = 10 O=(2 X3)+(1X4)=10

13. Combination Reactions (synthesis) 2 or more substances react to form 1 product.

14. Combination Reactions (synthesis) 2Mg(s) + O 2 (g)  2MgO(s) Mg=1 x 2=2 Mg= 1 x 2=2 O= 2 O=1 x 2 = 2

15. Decomposition Reaction 1 substance undergoes a reaction to produce 2 or more substances

16. Decomposition Reaction CaCO 3 (s)  CaO (s) + CO 2 (g) Ca=1 Ca=1 C=1 C=1 O=3 O=1+2=3

17. 3 Methods of Measuring Counting Mass Volume

18. Example 1 If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of.050 bushel of apples?

19. Example 1 Count: 1 dozen apples = 12 apples Mass: 1 dozen apples = 2.0 kg apples Volume: 1 dozen apples = 0.20 bushels apples Conversion Factors: 1 dozen 2.0 k.g 1 dozen 12 apples 1 dozen 0.20 bushels

20. Example 1 0.50 bushel x 1 dozen x 2.0 kg = 0.20 bushel 1 dozen = 5.0 kg

21. Avogadro’s Number Named after the Italian scientist Amedo Avogadro di Quaregna 6.02 x 10 23

22. Mole (mol) 1 mol = 6.02 x 10 23 representative particles Representative particles: atoms, molecules ions, or formula units (ionic compound)

23. Mole (mol) Moles = representative x 1 mol particles 6.02 x 10 23

24. Example 2 (atoms  mol) How many moles is 2.80 x 10 24 atoms of silicon?

25. Example 2 2.80 x 10 24 atoms Si x 1 mol Si 6.02 x 10 23 atoms Si = 4.65 mol Si

26. Example 3 (mol  molecule) How many molecules of water is 0.360 moles?

27. Example 3 0.360 mol H 2 O x 6.02 x 10 23 molecules H 2 O 1 mol H 2 O =2.17 molecules H 2 O

28. The Mass of a Mole of an Element The atomic mass of an element expressed in grams = 1 mol of that element = molar mass Molar mass S Molar mass C Molar mass Hg Molar mass Fe

29. 6.02 x 10 23 atoms S 6.02 x 10 23 atoms C 6.02 x 10 23 atoms Hg 6.02 x 10 23 atoms Fe

30. Example 4 (mol  gram) If you have 4.5 mols of sodium, how much does it weigh?

31. Example 4. 45 mol Na x 23 g Na = 10.35 g Na = 1.0 x 10 2 g Na 1 mol Na

32. Example 5 (grams  atoms) If you have 34.3 g of Iron, how many atoms are present?

33. Example 5 34.3 g Fe x 1 mol Fe x 6.02 x 10 23 atoms 55.8 g Fe 1 mol Fe =3.70 x 10 23 atoms Fe

34. The Mass of a Mole of a Compound To find the mass of a mole of a compound you must know the formula of the compound H 2 O  H= 1 g x 2 O= 16 g 18 g = 1 mole = 6.02 x 10 23 molecules

35. Example 6 (gram  mol) What is the mass of 1 mole of sodium hydrogen carbonate?

36. Example 6 Sodium Hydrogen Carbonate = NaHCO 3 Na=23 g H=1 g C=12 g O=16 g x3 84 g NaHCO 3 = 1 mol NaHCO 3

37. Mole-Volume Relationship Unlike liquids and solids the volumes of moles of gases at the same temperature and pressure will be identical

38. Avogadro’s Hypothesis States that equal volumes of gases at the same temperature and pressure contain the same number of particles Even though the particles of different gases are not the same size, since the gas particles are spread out so far the size difference is negligible

39. Standard Temperature and Pressure (STP) Volume of a gas changes depending on temperature and pressure STP= 0 o C (273 K) 101.3 kPa (1 atm)

40. Standard Temperature and Pressure (STP) At STP, 1 mol = 6.02 X 10 23 particles = 22.4 L of ANY gas= molar volume

41. Conversion Factors AT STP 1 mol gas 22.4 L gas 22.4 L gas 1 mol gas

42. Example 7 At STP, what volume does 1.25 mol He occupy?

43. Example 7 1.25 mol He x 22.4 L He = 28.0 L He 1 mol He

44. Example 8 If a tank contains 100. L of O 2 gas, how many moles are present?

45. Example 8 100. L O 2 X 1 mol O 2 = 4.46 mol O 2 22.4 L O 2

46. Calculating Molar Mass from Density The density of a gas at STP is measured in g/L This value can be sued to determine the molar mass of gas present

47. Example 9 A gaseous compound of sulfur and oxygen has a density of 3.58 g/L at STP. Calculate the molar mass.

48. Example 9 1 mol gas x 22.4 L gas X 3.58 g gas = 1 mol gas 1 L gas Molar Mass= 80.2 g

49. Percent Composition The relative amounts of the elements in a compound These percentages must equal 100

50. Percent Composition %element = mass of element x 100 mass of compound

51. Example 10 Find the mass percentage of each element present in Al 2 (CO 3 ) 3

52. Example 10 Al 2 (CO 3 ) 3 Al= 27 g x 2 = 54 g / 234 g x 100=23% C= 12 g x 3 = 36 g/ 234 g x 100= 15% O = 16 g x 9 = 144 g / 234 g x 100=62% 234 g Al 2 (CO 3 ) 3

53. Empirical Formula The simplest whole number ratio of atoms in a compound The formula obtained from percentage composition Ex CH, CH 4, H 2 O, C 3 H 8 NOT C 2 H 4, or C 6 H 12 O 6 these could be simplified

54. Example 11 Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical fromula.

55. Example 11 Assume that you have 100 grams of the compound therefore: Hg = 73.9 %  73.9 g Cl= 26.1%  26.1 g

56. Example 11 Step 2: Change grams of your compound to moles Hg = 73.9 g x 1 mol =0.368 mol Hg 200.6g Cl= 26.1 g x 1 mol = 0.735 mol Cl 35.5 g

57. Example 11 Step 3: Find the lowest number of moles present Hg = 73.9 g x 1 mol =0.368 mol Hg 200.6g Cl= 26.1 g x 1 mol = 0.735 mol Cl 35.5 g 0.368 < 0.735

58. Example 11 Step 4: Divide by the lowest number of moles to obtain whole numbers Hg = 73.9 g x 1 mol = 0.368 mol = 1 200.6g 0.368 mol Cl= 26.1 g x 1 mol = 0.735 mol= 1.99=2 35.5 g 0.368 mol

59. Example 11 Step 5: Put the whole numbers into the empirical formula Hg = 73.9 g x 1 mol = 0.368 mol = 1 200.6g 0.368 mol Cl= 26.1 g x 1 mol = 0.735 mol= 1.99=2 35.5 g 0.368 mol HgCl 2

60. Molecular Formulas The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula We can obtain the molecular formula from the empirical formula IF we know the molecular weight of the compound.

61. Example 12 The empirical formula of ascorbic acid is C 3 H 4 O 3. The molecular weight of ascorbic acid is 176 amu. Determine the molecular formula.

62. Example 12 Step 1: First determine the molecular weight of the empirical formula C 3 H 4 O 3 C= 12 amu x 3 H= 1 amu x 4 O= 16 amu x 3 88 amu

63. Example 12 Step 2: Divide the molecular weight of the molecular formula by the molecular weight of the empirical formula C 3 H 4 O 3 176 amu = 2 C= 12 amu x 3 88 amu H= 1 amu x 4 O= 16 amu x 3 88 amu

64. Example 12 Step 3: Multiply the empirical formula by the number calculated in step 2 176 amu = 2 88 amu (C 3 H 4 O 3 ) x 2 = C 6 H 8 O 6

65. Quantitative Information from a Balanced Equation 2 H 2 (g) + O 2 (g)  2 H 2 O (l) 2 molecules 1 molecule2 molecules Or since we can’t count out 2 molecules 2 mol 1 mol 2 moles The coefficients in a chemical reaction can be interpreted as either the relative number of molecules (formula units) involved in the reaction OR the relative number of moles

66. Example 13 (mol  mol) 2C 4 H 10 (l) + 13 O 2 (g)  8CO 2 (g) + 10H 2 O(l) How many moles of O 2 do you need to react with 5 moles of C 4 H 10 ?

67. Example 13 2C 4 H 10 (l) + 13 O 2 (g)  8CO 2 (g) + 10H 2 O(l) How many moles of O 2 do you need to react with 5 moles of C 4 H 10 ? 5 mol C 4 H 10 x 13 mol O 2 = 32.5 mol O 2 2 mol C 4 H 10

68. Example 14 (g  g) 2C 4 H 10 (l) + 13 O 2 (g)  8CO 2 (g) + 10H 2 O(l) How many grams of O 2 do you need to react with 50.0 g of C 4 H 10 ?

69. Example 14 (g  g) 2C 4 H 10 (l) + 13 O 2 (g)  8CO 2 (g) + 10H 2 O(l) How many grams of O 2 do you need to react with 50.0 g of C 4 H 10 ? 50.0 g C 4 H 10 x 1 mol C 4 H 10 x 13 mol O 2 x 32 g O 2 =179 g O 2 58 g C 4 H 10 2 mol C 4 H 10 1 mol O 2

70. Limiting Reactants (Reagents) The reactant that is completely consumed It determines or limits the amount of product that forms The other reactant(s) are called excess reagents

71. Example 14 (limiting reactants) 2C 4 H 10 (l) + 13 O 2 (g)  8CO 2 (g) + 10H 2 O(l) If you have 25.0 g O 2 and 25.0 g of C 4 H 10, what is the limiting reactant?

72. Example 14 (limiting reactants) 2C 4 H 10 (l) + 13 O 2 (g)  8CO 2 (g) + 10H 2 O(l) 25.0 g C 4 H 10 x 1 mol C 4 H 10 x 8mol CO 2 = 1.72 mol CO 2 58 g C 4 H 10 2 mol C 4 H 10 25.0 g C 4 H 10 x 1 mol C 4 H 10 x 8 mol CO 2 = 0.481 mol CO 2 32 g O 2 13 mol O 2 0.481 mol < 1.72 mol C 4 H 10 is the limiting reactant

73. Theoretical Yield The quantity of the product that is calculated to form when all of the limiting reactant reacts.

74. Example 15 (Theoretical Yield) 2C 4 H 10 (l) + 13 O 2 (g)  8CO 2 (g) + 10H 2 O(l) 25.0 g C 4 H 10 x 1 mol C 4 H 10 x 8mol CO 2 = 1.72 mol CO 2 58 g C 4 H 10 2 mol C 4 H 10 25.0 g C 4 H 10 x 1 mol C 4 H 10 x 8 mol CO 2 = 0.481 mol CO 2 32 g O 2 13 mol O 2 0.481 mol < 1.72 mol C 4 H 10 is the limiting reactant Calculate the theoretical yeild

75. Example 15 (Theoretical Yield) 25.0 g C 4 H 10 x 1 mol C 4 H 10 x 8 mol CO 2 = 0.481 mol CO 2 32 g O 2 13 mol O 2 0.481 mol CO 2 X 44 g CO 2 = 21.2 g CO 2 1 mol CO 2 If all of the limiting reactant (25.0 g C 4 H 10 ) reacts than 21.2 g of CO 2 will form.

76. Percent Yield Percent Yield = Actual Yield X 100 Theoretical yield

77. Example 15 (% Yield) A student calculates that they should theoretically make 105 g of iron in an experiment. When they perform the experiment only 87.9 g of iron were produced. What is the percent yield?

78. Example 15 (% Yield) A student calculates that they should theoretically make 105 g of iron in an experiment. When they perform the experiment only 87.9 g of iron were produced. What is the percent yield? 87.9 g / 105 g x 100 = 83.7% (actual) (theoretical)