2. Stoichiometry  is one where the substance retains its identity.   Examples of physical reactions: ◦ melting / freezing ◦ boiling / condensing ◦ subliming ◦ subdivision ◦ dissolving (solvation) 

3. Stoichiometry  the atoms of one or more substances are rearranged to produce new substances with new properties.  the atoms remain the same, but their arrangement changes:   H 2 O  H 2 + O 2   water is broken down into two substances, oxygen gas and hydrogen gas.  The two new substances bear no resemblance to the water of which they are made.

4. Stoichiometry a) a colour change b) gas production c) an energy change d) precipitate formation

5. Stoichiometry  refers to balancing chemical equations and the associated mathematics.  is possible because of the law of conservation of mass.

6. Stoichiometry Concise representations of chemical reactions

7. Stoichiometry CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

8. Stoichiometry Reactants appear on the left side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

9. Stoichiometry Products appear on the right side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

10. Stoichiometry The states of the reactants and products are written in parentheses to the right of each compound. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

11. Stoichiometry Coefficients are inserted to balance the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

12. Stoichiometry  Subscripts tell the number of atoms of each element in a molecule

13. Stoichiometry  Coefficients tell the number of molecules

14. Stoichiometry  Matter cannot be lost in any chemical reaction. ◦ all the atoms must balance in the chemical equation.  same number and type on each side of the equation ◦ can only change the stoichiometric coefficients in front of chemical formulas. ◦ Subscripts in a formula are never changed when balancing an equation.

15. Stoichiometry  Balance the following equation: Na (s) + H 2 O (l)  H 2 (g) + NaOH (aq)  count the atoms of each kind on both sides of the arrow: ◦ The Na and O atoms are balanced (one Na and one O on each side) ◦ there are two H atoms on the left and three H atoms on the right.

16. Stoichiometry  Balance the following equation: Na (s) + 2 H 2 O (l)  H 2 (g) + NaOH (aq)  the 2 increases the number of H atoms among the reactants.  O is now unbalanced.  have 2 H 2 O on the left; can balance H by putting a coefficient 2 in front of NaOH on the right:

17. Stoichiometry  Balance the following equation: Na (s) + 2 H 2 O (l)  H 2 (g) + 2 NaOH (aq)  H is balanced  O is balanced  Na is now unbalanced, with one on the left but two on the right. ◦ put a coefficient 2 in front of the reactant:

18. Stoichiometry  Balance the following equation: 2 Na (s) + 2 H 2 O (l)  H 2 (g) + 2 NaOH (aq)  Check our work: ◦ 2 Na on each side ◦ 4 H on each side ◦ 2 O on each side  Balanced !!

19. Stoichiometry  CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g)

20. Stoichiometry  Pb(NO 3 ) 2 (aq) + 2 NaI (aq)  PbI 2 (aq) + 2 NaNO 3 (aq)

21. Stoichiometry  Remember: ◦ when changing the coefficient of a compound, the amounts of all the atoms are changed. ◦ do a recount of all the atoms every time you change a coefficient. ◦ do a final check at the end. ◦ fractions are not allowed. Multiply to reach whole numbers.

22. Stoichiometry  (s) - solid  (l) - liquid  (g) - gas  (aq) -aqueous (meaning that the compound is dissolved in water.)  (ppt) -precipitate (meaning that the reaction produces a solid which falls out of solution.)   Reaction conditions occasionally appear above or below the reaction arrow ◦ e.g., "Δ" is often used to indicate the addition of heat).

23. Stoichiometry  Complete 3.11 to 3.14

25. Stoichiometry  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

26. Stoichiometry

27.  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

28. Stoichiometry  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

29. Stoichiometry  Complete Problems 3.15 to 3.20

31. Stoichiometry  Sum of the atomic weights for the atoms in a chemical formula  So, the formula weight of calcium chloride, CaCl 2, would be Ca: 1(40.08 amu) + Cl: 2(35.45 amu) 110.98 amu  These are generally reported for ionic compounds

32. Stoichiometry  Sum of the atomic weights of the atoms in a molecule  For the molecule ethane, C 2 H 6, the molecular weight would be C: 2(12.01 g/mol) + H: 6(1.01 g/mol) 30.08 g/mol

33. Stoichiometry  formula and molecular weights are often called molar mass.  The formula weight (in amu) is numerically equal to the molar mass (in g/mol).  Always report molar mass in g/mol.  Always round the atomic weights to 2 decimal places.

34. Stoichiometry  Complete questions 3.21 and 3.22

35. Stoichiometry Percentage composition is obtained by dividing the mass contributed by each element (number of atoms times AW) by the formula weight of the compound and multiplying by 100. % element = (number of atoms)(atomic weight) (molar mass of the compound) x 100

36. Stoichiometry So the percentage of carbon in ethane (C 2 H 6 ) is… %C = (2)(12.01 g/mol) (2(12.01) + 6(1.01))g/mol 24.02 g/mol 30.08 g/mol = x 100 = 79.85%

37. Stoichiometry the formula weight is 98.09 amu (do calculation) % H = 2(1.01 amu) x 100 = 2.06 % H 98.09 amu % S = 32.07 amu x 100 = 32.69 % S 98.09 amu % O = 4(16.00 amu) x 100 = 65.25 % O 98.09 amu the sum of the percent compositions will always equal 100 %

38. Stoichiometry  Complete questions 3.24 and 3.26  Calculate the % composition of each element in each compound.

40. Stoichiometry  6.022 x 10 23  1 mole of 12 C has a mass of 12 g

41. Stoichiometry  By definition, these are the mass of 1 mol of a substance (i.e., g/mol) ◦ is the atomic weight on the periodic table ◦ The formula weight (in amu’s) will be the same number as the molar mass (in g/mol)

42. Stoichiometry Moles provide a bridge from the molecular scale to the real-world scale

43. Stoichiometry  Using the mass and molar mass to calculate the number of MOLES of a compound or element. ◦ mMassgramsg ◦ namount of mattermolesmol ◦ MMolar massgrams per moleg/mol number of = mass molesmolar mass n= m M

44. Stoichiometry  If you have 20.2 grams of Potash, how many moles do you have?  Molar Mass KCl (calculated):74.55 g/mol  Mass KCl:20.2 g 20.2 g = 0.271 moles KCl 74.55 g/mol  if the units work out, you are correct.

45. Stoichiometry  How many moles are present in 0.750 g of ammonium phosphate?  Ammonium phosphate:(NH 4 ) 3 PO 4  Molar mass: (3(14.01) + 12(1.01) + 30.98 + 4(16.00)) g/mol = 149.13 g/mol  Number of moles = 0.750 g = 5.03 x 10 -3 mol 149.13 g/mol (NH 4 ) 3 PO 4

46. Stoichiometry  What is the mass of 12.31 mol of NaOH ? ◦ Step 1. Calculate molar mass  1 x 22.00 g/mol  1 x 16.00 g/mol  1 x 1.01 g/mol  40.00 g/mol ◦ Step 2. Calculate mass  (12.31 mol)(40.00 g/mol) = 492.4 g NaOH

47. Stoichiometry  What is the mass of 2.1 x 10 -4 mol of diarsenic trioxide?  Diarsenic trioxide: As 2 O 3  Molar mass: (2(74.92) + 3(16.00)) g/mol = 197.84 g/mol  Mass = (2.1 x 10 -4 mol)(197.84 g/mol) = 0.042 g As 2 O 3

48. Stoichiometry How many moles of caffeine (C 8 H 10 N 4 O 2 ) is required to reach the 12.0 g lethal dose in a 140 pound female?  Step 1. Molar Mass  8 x 12.01 g/mol  10 x 1.01 g/mol  4 x 14.01 g/mol  2 x 16.00 g/mol  194.22 g/mol  Step 2. Calculate Moles  12.00 g = 0.06179 mol caffeine  194.22 g/mol

49. Stoichiometry  We can also find the number of atoms of a substance using Avogadro’s Number:  6.022 x 10 23 atoms/mol  Example:  Calculate the number of atoms in 0.500 moles of silver? 0.500mol Ag x 6.022x10 23 atoms = 3.01 x 10 23 atoms mol of Ag

50. Stoichiometry  For molecules!  Avogadro’s number represents the number of particles, whether atoms or molecules.  Eg. Calculate the number of molecules in 4.99mol of methane. 4.99 mol x 6.022 x 10 23 molecules = 3.00 x 10 24 molecules mol of CH 4

51. Stoichiometry  How many molecules are present in 0.045 mol of H 2 O?  Number of atoms = (0.045 mol)(6.022 x 10 23 particles/mol) = 2.7 x 10 22 molecules H 2 O

52. Stoichiometry  How many moles are present in 8.75 x 10 26 molecules of Vitamin C?  Moles = 8.75 x 10 26 molecules = 1450 mol 6.022 x 10 23 particles/molVitamin C

53. Stoichiometry  What is the mass of 2.1 x 10 22 molecules of N 2 O?  Step 1. Convert molecules to moles: ◦ Moles = 2.1 x 10 22 molecules = 0.0349 mol N 2 O 6.022 x 10 23 particles/mol  Step 2. Convert moles to mass: ◦ Mass = (0.0349 mol)((2(14.01) + 16.00) g/mol) = 1.54 g N 2 O

54. Stoichiometry  How many molecules are present in 4.65 t of CO 2 ?  Step 1: Convert mass to moles ◦ Moles = (4.65 t)(1000 kg/t)(1000 g/kg) = 1.06 x 10 5 (12.01 + 2(16.00)) g/mol mol CO 2  Step 2: Convert moles to molecules ◦ Molecules = (1.06 x 10 5 mol)(6.022 x 10 23 particles/mol) = 6.36 x 10 28 molecules CO 2

55. Stoichiometry  Complete questions 3.32 to 3.42, even (leave 3.32 for last)

57. Stoichiometry One can calculate the empirical formula from the percent composition

58. Stoichiometry 1.Assume we start with 100 g of sample. 2.The mass percent then translates as the number of grams of each element in 100 g of sample. 3.From these masses, calculate the number of moles (use atomic weight from Periodic Table) 4.The lowest number of moles becomes the divisor for the others. (gives a mole ratio greater than 1) 5.Adjust mole ratios so all numbers are whole (1, 2, etc) 6.The lowest whole-number ratio of moles is the empirical formula.

59. Stoichiometry 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. Assumptions: 1.These are % by mass; in 100g of the compound, the percentages represent masses in grams. 2.Mass divided by molar mass gives moles of each atom.

60. Stoichiometry Assuming 100.00 g of para-aminobenzoic acid, each percentage converts into a mass: 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

61. Stoichiometry 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

62. Stoichiometry These are the subscripts for the empirical formula: C 7 H 7 NO 2

63. Stoichiometry  Complete questions 3.43 to 3.46

64. Stoichiometry  Once we know the composition of a compound the next step is to determine the molar mass.  From these two pieces of information the actual molecular formula can be determined: ◦ take the ratio of the actual molar mass divided by the molar mass determined by the empirical formula ◦ multiply the atoms in the empirical formula by the results of the ratio.

65. Stoichiometry  Complete questions 3.47 to 3.50

66. Stoichiometry  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

67. Stoichiometry Compounds containing other elements are analyzed using methods analogous to those used for C, H and O

68. Stoichiometry  Complete questions 3.51 & 3.52

69. Stoichiometry The coefficients in the balanced equation give the ratio of moles of reactants and products

70. Stoichiometry 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)

71. Stoichiometry C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O

72. Stoichiometry  Work on questions 3.57 to 3.66  Copy examples done in class.

74. Stoichiometry  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)

75. Stoichiometry  In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

76. Stoichiometry The limiting reactant is the reactant present in the smallest stoichiometric amount

77. Stoichiometry  The limiting reactant is the reactant you’ll run out of first (in this case, the H 2 )

78. Stoichiometry In the example below, the O 2 would be the excess reagent

79. Stoichiometry  Complete questions 3.69 to 3.76

80. Stoichiometry  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

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

82. Stoichiometry  Complete questions 3.77 to 3.80