1. Chemical Reactions

2.  A physical change alters the physical state of a substance without changing its composition ◦ Examples  Boiling  melting  Freezing  Vaporization  Condensation  Sublimation  Breaking a bottle  Crushing a can

3.  A chemical change (a chemical reaction) converts one substance into another ◦ Breaking bonds in the reactants (starting materials) ◦ Forming new bonds in the products

4. aA ( physical state ) + bB ( state )  cC ( state ) + dD ( state ) A, B = reactants C, D = products a, b, c, d = coefficients to indicate molar ratios of reactants and products

5. CH 4 and O 2 CO 2 and H 2 O CH 4 ( g ) + 2O 2 ( g )  CO 2 ( g ) + 2H 2 O ( g )

6. 2 molecules of C 4 H 10 13 molecules of O 2 10 molecules of C 4 H 10 8 molecules of CO 2 Balancing Chemical Equations: Unbalanced equation: C 4 H 10 + O 2  CO 2 + H 2 O Balanced equation: 2C 4 H 10 + 13O 2  8CO 2 + 10H 2 O

7. H 2 + O 2 → H 2 0  Reactants are on the left ◦ Things that are used ◦ H 2 + O 2  Product(s) are on the right ◦ Things that are made ◦ H 2 0  This equation is not yet balanced

8.  Do the number of atoms of each element on either side of the arrow balance? ◦ Compounds consist of more than one element ◦ Examples: NaCl, H 2 SO 4, H 2 0 ◦ Look at numbers of each atoms within the compounds

9.  Correct any imbalances with a coefficient ◦ Coefficient = the large number to the left of a substance in the equation ◦ Don’t change subscripts  This will change what the molecule is

10.  Example: to balance an equation you need two atoms of oxygen from water (H 2 O) ◦ If you change the subscript you change the compound ◦ H 2 O 2 does provide two atoms of oxygen but  H 2 O (water) ≠ H 2 O 2 (hydrogen peroxide) ◦ 2 H 2 O provides 2 atoms of oxygen and keeps the compound as water ◦ It also gives you 4 atoms of hydrogen that you should then make sure is balanced

11.  Relax and calmly go through what is on either side of the equation  There are many different ways to start balancing the equation  Once you start the remaining coefficients should fall into place  Where to start? ◦ Compounds (NaCl, H 2 SO 4, H 2 0)  See what elements they have in common ◦ Molecular elements (O 2, H 2 )  More than one atom present ◦ Monoatomic elements (Ca, Cl)  Only one atom present

12. H 2 + O 2 → H 2 0  List out what is on each side  Oxygen does not balance LeftRight 2 hydrogen 2 oxygen1 oxygen

13. H 2 + O 2 → 2H 2 0  Multiply right side by 2 to balance oxygen  Now hydrogen does not balance LeftRight 2 hydrogen4 hydrogen 2 oxygen

14. 2H 2 + O 2 → 2H 2 0  Multiply hydrogen by 2 to balance hydrogen  Equation is now balanced LeftRight 4 hydrogen 2 oxygen

15.  List out what is on each side  Carbon does not balance C 6 H 12 O 6 + O 2 → CO 2 + H 2 O LeftRight 6 carbon1 carbon 12 hydrogen2 hydrogen 8 oxygen3 oxygen

16.  Look at compounds first  Multiply CO 2 on right by 6 to balance C  Hydrogen does not balance C 6 H 12 O 6 + O 2 → 6CO 2 + H 2 O LeftRight 6 carbon 12 hydrogen2 hydrogen 8 oxygen13 oxygen

17.  Multiply water on right side by 6 to balance hydrogen  Oxygen does not balance C 6 H 12 O 6 + O 2 → 6CO 2 + 6H 2 O LeftRight 6 carbon 12 hydrogen 8 oxygen18 oxygen

18.  Multiply oxygen on left side by 6 to balance oxygen  Equation is now balanced C 6 H 12 O 6 + 6O 2 → 6CO 2 + 6H 2 O LeftRight 3 carbon6 carbon 8 hydrogen12 hydrogen 2 oxygen18 oxygen

19. Propane + oxygen → carbon dioxide + water C 3 H 8 + O 2 → CO 2 + H 2 O  List out what is on each side  Carbon does not balance LeftRight 3 carbon1 carbon 8 hydrogen2 hydrogen 2 oxygen3 oxygen

20. Propane + oxygen → carbon dioxide + water C 3 H 8 + O 2 → 3CO 2 + H 2 O  L ook at compounds first  Multiply CO 2 on right side by 3 to balance carbon  Hydrogen does not balance LeftRight 3 carbon 8 hydrogen2 hydrogen 2 oxygen7 oxygen

21. Propane + oxygen → carbon dioxide + water C 3 H 8 + O 2 → 3CO 2 + 4H 2 O  Multiply water on right side by 4 to balance hydrogen  Oxygen does not balance LeftRight 3 carbon 8 hydrogen 2 oxygen10 oxygen

22. Propane + oxygen → carbon dioxide + water C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O  Multiply oxygen on left side by 5 to balance oxygen  Equation is now balanced LeftRight 3 carbon 8 hydrogen 10 oxygen

23. Sodium azide → Sodium + nitrogen NaN 3 → Na + N 2  Identify what you have  Nitrogen does not balance  How do we balance nitrogen when left side has 3 and the right has 2? o find the lowest common multiple for both o In this case 6 o Multiply each side to make 6 atoms LeftRight 1 sodium 3 nitrogen2 nitrogen

24. Sodium azide → Sodium + nitrogen 2NaN 3 → Na + N 2  Multiply sodium azide on left side by 2 to get 6 nitrogen atoms  Still need to make 6 atoms of nitrogen on right side LeftRight 2 sodium1 sodium 6 nitrogen2 nitrogen

25. Sodium azide → Sodium + nitrogen 2NaN 3 → Na + 3N 2  Multiply nitrogen by 3 on left side to get 6 nitrogen atoms  Sodium is not balanced LeftRight 2 sodium1 sodium 6 nitrogen

26. Sodium azide → Sodium + nitrogen 2NaN 3 → 2Na + 3N 2  Multiply sodium on right side by 2 to balance sodium  Equation is now balanced LeftRight 2 sodium 6 nitrogen

27.  A mole is a quantity that contains 6.02 X 10 23 items (usu. atoms, molecules or ions) ◦ An amount of a substance whose weight, in grams is numerically equal to what its molecular weight was in amu ◦ Just like a dozen is a quantity that contains 12 items ◦ 1 mole of C atoms = 6.02 x 10 23 C atoms ◦ 1 mole of CO 2 molecules = 6.02 x 10 23 CO 2 molecules ◦ 1 mole of H 2 O molecules = 6.02 x 10 23 H 2 O molecules  The number 6.02 X 10 23 is Avogadro’s number 1 mol 6.02 x 10 23 atoms 1 mol 6.02 x 10 23 molecules

28.  How many molecules are in 2.5 moles of penicillin 2.5 moles penicillin X 6.02 x 10 23 molecules = 1 mole 1.5 X 10 24 molecules

29.  The formula weight is the sum of the atomic weights of all the atoms in a compound, reported in atomic mass units  The molar mass is the mass of one mole of any substance, reported in grams ◦ The value of the molar mass of a compound in grams equals the value of its formula weight in amu.

30.  The sum of the atomic weights of all the atoms in a compound, reported in atomic mass units ◦ This may be called molecular weight for covalent compounds  Example ◦ H 2 O  Contains 2 Hydrogen atoms and 1 Oxygen atom  Hydrogen weighs 1.01 amu (1.01 g H/mole)  Oxygen weighs 16.0 amu (16.0 g O/mole)  Formula/molecular weight = 2(1.01 amu) +16.0 amu = 18.0 amu  Looking only at one molecule  Molar mass = 2(1.01 g/mol) +16.0 g/mol = 18.0 g/mol  Looking at one mole of the substance

31.  Stoichiometry is the study of the quantitative relationships that exist between substances involved in a chemical reaction  Mole ratios within molecules: A x B y Mole ratio of A:B = x:y Example: H 2 O 2 ⇒ H:O = 2:2 = 1:1

32.  Mole ratios between molecules  A balanced equation tells us the number of moles of each reactant that combine and number of moles of each product formed aA + bB  cC + dD Mole ratio of A:B:C:D = a:b:c:d Example: 2H 2 + O 2 → 2H 2 0 H 2 :O 2 :H 2 0 = 2:1:2

33.  The coefficients in the balanced chemical equation can represent the ratio of molecules of the substances that are consumed or produced  The coefficients in the balanced chemical equation can represent the ratio of moles of the substances that are consumed or produced 2H 2 + O 2 → 2H 2 0

34.  There are 4 basic types of stoichiometry problems: ◦ Moles to moles ◦ Moles to grams ◦ Grams to moles ◦ Grams to grams  However, all stoichiometry problems are really very similar, and the same general approach can be used to solve any of them ◦ So really only one type of problem

35.  Grams to moles to moles to grams mol mol g g

36. N 2 +3H 2 → 2NH 3  How many moles of H 2 are required to produce 3.89 mol of NH 3 ? ◦ Equation says H 2 :NH 3 is 3:2 ◦ 3.89 mol NH 3 * 2 mol H 2 = 5.84 moles H 2 3 mol NH 3 mol mol g g

37. N 2 +3H 2 → 2NH 3  How many grams of NH 3 are produced from 3.44 mol of N 2 ? ◦ Use mole ratio between NH 3 and N 2 ◦ Then use molar mass to convert moles to g ◦ 3.44 mol N 2 * 2 mol NH 3 * 17 g NH 3 =117 g NH 3 1 mol N 2 1 mol NH 3 mol mol g g

38. N 2 +3H 2 → 2NH 3  How many moles H 2 react with 6.77 g of N 2 ?  Convert grams to moles using molar mass  Use molar ratio between N 2 and H 2 ◦ 6.77 g N 2 * 1 mol N 2 * 3 mol H 2 = 0.725 mol H 2 28.0 g N 2 1 mol N 2 mol mol g g

39. N 2 +3H 2 → 2NH 3  How many grams of NH 3 are produced from 8.23 g of H 2 ?  Use molar mass to convert g to moles  Use molar ratio to convert between moles  Use molar mass to convert moles to g 8.23 g H2 * 1 mol H 2 * 2 mol NH 3 * 17.0 g NH 3 = 46.2 g NH 3 2.02 g H 2 3 mol H 2 mol NH 3 mol mol g g

40.  Once you have converted things into moles you can use molar ratios in balanced equations to convert between different elements, compounds, and molecules  Problems may ask for g, molecules, atoms, or many other units mol mol unit given unit requested

41.  If I gave you a value in g how would you convert it to mol? ◦ Use formula weight (mass)  Calculate using the periodic table ◦ Example: 53.21 g C 6 H 12 O 6 Formula weight/molecular mass: 6*(12.01 g/mol) + 12*(1.01 g/mol) +6*(16.00 g/mol) = 180.18 g/mol ◦ Use value to calculate moles (may have to invert value) 53.21 g C 6 H 12 O 6 * 1 mol C 6 H 12 O 6 = 0.2953 mol C 6 H 12 O 6 180.18 g C 6 H 12 O 6

42.  If I give you a value in mol how would you compare it to mol of another substance? ◦ Use a balanced equation  Example 5 mol CO 2 to mol O 2 ◦ 6CO 2 + 6H 2 O  C 6 H 12 O 6 + 6O 2 ◦ Use the stoichiometric coefficients to convert between any of the substances in this equation 6CO 2 6CO 2 6CO 2 6H 2 O C 6 H 12 O 6 6O 2 ◦ In our case only interested in the relationship with O 2 5 mol CO 2 * 6CO 2 = 5 mol O 2 6O 2

43.  All questions will build on the examples on the previous two slides ◦ May ask you to go g  mol  mol  g ◦ May ask you to convert mol to number of atoms or number of molecules  Use Avagadro’s number 6.02 X 10 23 ◦ May ask you to use some other value in the future, but it will be something that you can relate to g or to mol

44.  INGREDIENTS: ◦ 3 cups all-purpose flour ◦ 1 teaspoon salt ◦ 1 cup shortening ◦ 1/2 cup cold water ◦ 2 cups pumpkin ◦ 2 eggs, beaten ◦ 3/4 cup packed brown sugar ◦ 2 teaspoon spices  If you want to make multiple pies - the amount of pie that you can bake depends on which of the ingredients you have the “least” of – what will you run out of first?

45. Ingredient Recipe In pantry# of recipes it can make flour3 cups14 cups4.67 x salt1 tsp 20 tsp20 x shortening1 cup7 cup7 x water½ cup ∞∞∞ pumpkin2 cups19 cups9.5 x eggs2189x sugar¾ cup3 cups4X spice2 tsp21 tsp10.5 x Limiting ingredient

46. N 2 +3H 2 → 2NH 3  You have 4 moles N 2 and 9 moles H 2  How many moles of NH 3 could be produced?  H 2 is the limiting reactant and limits how much NH 3 can be made 9 moles H 2 * 2 mol NH 3 = 6 mol NH 3 3 mol H 2 ActualRequiredRx N2N2 414 H2H2 933

47.  Compare the actual amount of each reactant to the amount required in the balanced equation to determine how many times the “reaction can be run”  Use the amount of the limiting reactant to calculate how much product can be produced

48.  How many g of S can be produced if we attempt to react 9.00g of Bi 2 S 3 with 16.00g of HNO 3 ?  9.00 g Bi 2 S 3 * 1 mol Bi 2 S 3 = 0.0175 mol Bi 2 S 3 514.3 g Bi 2 S 3 0.0175 mol Bi 2 S 3 * 1 Rx = 0.0175  allows Rx to run 0.0175 times 1 mol Bi 2 S 3  16.00 g HNO 3 * 1 mol HNO 3 = 0.254 mol HNO 3 63.0 g HNO 3 0.254 mol HNO 3 * 1 Rx = 0.3175  allows Rx to run 0.0318 times 8 mol HNO 3  Therefore Bi 2 S 3 is the limiting reactant

49.  Use Bi 2 S 3 - the limiting reactant to calculate S ◦ How many g of S can be produced if we attempt to react 9.00 g of Bi 2 S 3 with 16.00 g of HNO 3 ?  To see how much S can be produced this is gram to mole to mole to gram problem  9.00 g Bi 2 S 3 * 1 mol Bi 2 S 3 = 0.0175 mol Bi 2 S 3 514.3 g Bi 2 S 3  0.0175 mol Bi 2 S 3 * 3 mol S * 32.1 g S = 1.69 g S mol Bi 2 S 3 mol S

50.  If 3.00 mol H 2 reacts with 2.00 mol O 2 how many moles of H 2 O will form?  Solution 1: ◦ 3.00 mol H 2 * 1 Rx = 1.50 Rx 2 mol H 2 ◦ 2.00 mol O 2 * 1 Rx = 2.00 Rx 1 mol O 2 ◦ H 2 is limiting reactant so use it to calculate mol of H 2 O ◦ 3.00 mol H 2 * 2 mol H 2 O = 3.00 mol H 2 O 2 mol H 2  H 2 is the limiting reactant so 3.00 mol H 2 O will theoretically form

51.  If 3.00 mol H 2 reacts with 2.00 mol O 2 how many moles of H 2 O will form?  Solution 2: ◦ 3.00 mol H 2 * 2 mol H 2 O = 3.00 mol H 2 O 2 mol H 2 ◦ 2.00 mol O 2 * 2 mol H 2 O = 4.00 mol H 2 O 1 mol O 2  H 2 is the limiting reactant so 3.00 mol H 2 O will theoretically form

52.  Actual yield is determined experimentally, it is the mass of the product that is measured  Theoretical yield is the calculated mass of the products based on the initial mass or number of moles of the reactants

53.  If 3.00 mol H 2 reacts with 2.00 mol O 2 how many moles of H 2 O will form?  We just saw that H 2 is the limiting reactant so 3.00 mol H 2 O will theoretically form  If we run this in the lab and end up producing 1.75 mol H 2 O what is the percent yield? 1.75 mol H 2 O * 100% = 58.3 % 3.00 mol H 2 O

54.  Both processes occur together in a single reaction called an oxidation−reduction or redox reaction.  Thus, a redox reaction always has two components, one that is oxidized and one that is reduced  A redox reaction involves the transfer of electrons from one element to another.

55.  Oxidation is the loss of electrons from an atom. ◦ Reducing agents are oxidized  Reduction is the gain of electrons by an atom ◦ Oxidizing agents are reduced. LEO says GER

56. Zn + Cu 2+ Zn 2+ + Cu Zn loses 2 e – Zn loses 2 e − to form Zn 2+, so Zn is oxidized. Cu 2+ gains 2 e − to form Cu, so Cu 2+ is reduced. Cu 2+ gains 2 e −

57. Zn + Cu 2+ Zn 2+ + Cu Zn loses 2 e – Cu 2+ gains 2 e − Oxidation half reaction:ZnZn 2+ + 2 e − Each of these processes can be written as an individual half reaction: loss of e − Reduction half reaction:Cu 2+ + 2e − Cu gain of e −

58.  A compound that is oxidized while causing another compound to be reduced is called a reducing agent  Zn acts as a reducing agent because it causes Cu 2+ to gain electrons and become reduced 58 Zn + Cu 2+ Zn 2+ + Cu oxidizedreduced

59.  A compound that is reduced while causing another compound to be oxidized is called an oxidizing agent  Cu 2+ acts as an oxidizing agent because it causes Zn to lose electrons and become oxidized 59 Zn + Cu 2+ Zn 2+ + Cu oxidizedreduced

60. 60

61. 61 Iron Rusting 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) Fe 3+ O 2– neutral Feneutral O Fe loses e – and is oxidized. O gains e – and is reduced.

62. 62 Inside an Alkaline Battery Zn + 2 MnO 2 ZnO + Mn 2 O 3 neutral ZnMn 4+ Zn 2+ Mn 3+ Zn loses e − and is oxidized. Mn 4+ gains e − and is reduced.

63. 63 Zn + 2 MnO 2 ZnO + Mn 2 O 3

64. 64 Oxidation results in the:Reduction results in the: Gain of oxygen atoms Loss of hydrogen atoms Loss of oxygen atoms Gain of hydrogen atoms