1. AP CHEMISTRYNOTES Ch 3 Stoichiometry

2. 3.1 Counting by Weighing Find the average mass = total mass of substance / number of substance

3. 3.2 Atomic Masses Find the average atomic mass = total % of amu of isotopes Average Atomic mass Ex: Average atomic mass of carbon 98.89%x12amu+1.11%x13amu=12.01amu (Given atomic mass)

5. 3.3 The mole Abbreviated mol Avogadro’s number –6.022 x 10 23 Conversion factor –Ex. Carbon 6.022 x 10 23 atoms)(12 amu/atom) = 12g 6.022 X 10 23 amu = 1g

6. Mole Molecule 1 mole= 6.022x10 23 particles Atoms 1 molecule= ____ atoms

8. 3.4 Molar Mass Calculating molar mass Molar mass and numbers of molecules

9. Mole Particle 1 mole= 6.022x10 23 particles Mass 1 mole= molar mass Atoms 1 molecule= ____ atoms

10. 3.5 Percent Composition of Compounds Mass percent 1 (from the formula) = Mass of an element in a compound mass of 1 mol of compound Mass percent of C, H, O in C 2 H 5 OH

11. 3.5 Percent Composition of Compounds Mass percent of C, H, O in C 2 H 5 OH –Mass of C in 1 mol C 2 H 5 OH x 100% = Mass of 1 mol C 2 H 5 OH –Mass of H in 1 mol C 2 H 5 OH x 100% = Mass of 1 mol C 2 H 5 OH –Mass of O in 1 mol C 2 H 5 OH x 100% = Mass of 1 mol C 2 H 5 OH 52.14% 13.13% 34.73% Try Ex 3.9 Add up 100.00%

12. 3.6 Determining the Formula of a Compound From an experiment to find mass of C, H, O

13. Unknown Compound (0.1156g) with C, H, N CO 2 : 0.1638g (all C converted to CO 2 ) H 2 O : 0.1676g (all H converted to H 2 O Mass of C in a compound = 0.1638g CO 2 x 12.01g C = 44.01 g CO 2 Find mass % of C 0.04470g C x 100% = 0.1156g compound 0.04470g C 38.67 %

14. Mass % of H CO 2 : 0.1638g (all C converted to CO 2 ) H 2 O : 0.1676g (all H converted to H 2 O Mass of H in a compound = 0.1676 g H 2 O x 2.016g H = 18.02 g H 2 O Find mass % of H 0.01875g H x 100% = 0.1156g compound 0.01875g H 16.22 %

15. Mass % of N Find mass % of N 100.00% - (% of C + % of H) 100.00% - (38.67% + 16.22%) = Assume there is a 100.00 g compound Mass of carbon Mass of hydrogen Mass of nitrogen 45.11 %N 38.67 g 16.22 g 45.11 g

16. Moles of C, H, N 38.67 g C x 1 mol C= 12.01 g C 16.22 g H x 1 mol H = 1.008 g H 45.11 g N x 1 mol N = 14.01 g N 3.220 mol C 16.09 mol H 3.219 mol N

17. Smallest whole number ratio of atoms divide by the smallest mol C: 3.220mol = 3.220mol H: 16.09 mol H = 3.220mol N: 3.219 mol N = 3.220mol 1.000 = 1 4.997 = 5 1.000 = 1

18. Determine the formula CH 5 N as an empirical formula C 2 H 10 N 2, C 3 H 15 N 3 might be a molecular formula Represented by (CH 5 N) n

19. Empirical Formula Determination Since mass percentage gives the number of grams of a particular element per 100 grams of compound, base the calculation on 100 grams of compound. Each percent will then represent the mass in grams of that element. Determine the number of moles of each element present in 100 grams of compound using the atomic masses of the elements present.

20. Empirical Formula Determination Divide each value of the number of moles by the smallest of the values. If each resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the elements in the empirical formula. If the numbers obtained in the previous step are not whole numbers, multiply each number by an integer so that the results are all whole numbers.

21. Molecular Formula Determination-1 Obtain the empirical formula. Compute the mass corresponding to the empirical formula. Calculate the ratioMolecular mass Empirical formula mass The integer from the previous step represents the number of empirical formula units in one molecule. When the empirical formula subscripts are multiplied by the integer, the molecular formula results. Molecular formula = (empirical formula) x molar mass empirical formula mass

22. Molecular Formula Determination-2 Using the mass percentages and the molar mass, determine the mass of each element present in one mole of compound. Determine the number of moles of each element present in one mole of compound. The integers from the previous step represent the subscripts in the molecular formula.

23. Pg 119 problems 70 72 74 76 78 80

24. 3.7 Chemical equation Chemical equation –The atoms have been reorganized –Bonds have been broken –New ones have been formed Reactants Products States –(s), (l), (g), (aq)

25. 3.8 Balancing Chemical Equations _ NH 3 (g)+_O 2 (g)  _NO(g)+_H 2 O(g) 4 5 46

26. 3.9 Stoichiometric Calculations: Amounts of reactants and Products C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) +4H 2 O (g)

27. Mole A Mole B Mass A Mass B Calculate from left to right 1 mol A Moler mass A Moler mass B 1 mol B Mole B Mole A

28. 3.10 Calculations Involving a Limiting Reactant Harber process CH 4 (g) + H 2 O (g)  3H 2 (g) + CO(g) N 2 (g) + 3H 2 (g)  2NH 3 (g)

29. 3.10 Calculations Involving a Limiting Reactant Calculate the mass of water to react with2.50 x 10 3 kg of methane. CH 4 (g) + H 2 O (g)  3H 2 (g) + CO(g) 1:1 ratio CH 4 :H 2 O

30. 3.10 Calculations Involving a Limiting Reactant 2.50 x 10 6 g CH 4 x 1mol CH 4 = 1.56 x 10 5 mol CH 4 16.04g CH 4 1.56 x10 5 mol CH 4 x 18.02g H 2 O = 2.81x10 6 gH 2 O 1 mol H 2 O

31. Which is limiting, CH 4 or H 2 O CH 4 is limiting

32. Assume there are 25.0kg N 2 and 5.0kg H 2. Which is limiting? N 2 (g) + 3H 2 (g)  2NH 3 (g) Which is limiting? 5.0kg H 2

33. How to determine which is limiting or excess? Find the number of moles of product from given number of grams of reactant to find which is limiting and which is excess.

34. Percent yield a.The ratio of the actual yield to the theoretical yield expressed as a percent. Percent yield = Actual yield x100% Theoretical yield

37. Exercise: pg 121 #99 BaO 2 (s) +2HCl(aq)  H 2 O 2 (aq) +BaCl 2 (aq) a. What mass of H 2 O 2 should result when 1.50g of BaO 2 is treated with 25.0 mL of HCl solution containing 0.0272 g of HCl per mL? b. What mass of which reagent is left unreacted?

38. Exercise: pg 121 #99 BaO 2 (s) +2HCl(aq)  H 2 O 2 (aq) +BaCl 2 (aq) a. What mass of H 2 O 2 should result when 1.50g of BaO 2 is treated with 25.0 mL of HCl solution containing 0.0272 g of HCl per mL? -Find which reactant is limiting. -Mass of BaO 2 to mass of H 2 O 2 -Mass of HCl to mass of H 2 O 2

39. BaO 2 (s) +2HCl(aq)  H 2 O 2 (aq) +BaCl 2 (aq) 1.50g of BaO 2 25.0 mL of HCl solution containing 0.0272 g of HCl per mL -Mass of BaO 2 to mass of H 2 O 2 1.50g BaO 2 x 1molBaO 2 x 1mol H 2 O 2 x 34.02gH 2 O 2 = 169.3g BaO 2 1molBaO 2 1mol H 2 o 2 -Mass of HCl to mass of H 2 O 2 0.0272g HCl x 25mL = 0.680g HCl 1 mL 0.680g HCl x 1mol HClx 1 mol H 2 O 2 x 34.02gH 2 O 2 = 36.46g HCl 2molHCl 1mol H 2 O 2 0.301g 0.317g limiting

40. BaO 2 (s) +2HCl(aq)  H 2 O 2 (aq) +BaCl 2 (aq) a. 1.50g BaO 2 is limiting b. What mass of which reagent is left unreacted? From the limiting reactant, find the mass of excess reactant (HCl) used, then subtract from given amount.

41. BaO 2 (s) +2HCl(aq)  H 2 O 2 (aq) +BaCl 2 (aq) From the limiting reactant, find the mass of excess reactant (HCl) used, then subtract from given amount. 1.50g BaO 2 x 1molBaO 2 x 2mol HCl x 36.46gHCl = 169.3g BaO 2 1molBaO 2 1mol HCl 0.680g – 0.646g=0.034g HCl left 0.646g HCl used

42. Pg 122 problems 102 106

43. Additional problem Writing balanced equations: 1.Calcium hydroxide solution reacts with phosphoric acid solution to produce liquid water and solid calcium phosphate. 2.The combustion reaction of ethanol (C 2 H 5 OH) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen gas.

44. Answers Writing balanced equations: 1.Calcium hydroxide solution reacts with hydrophosphoric acid solution to produce liquid water and solid calcium phosphate. 3Ca(OH) 2 (aq)+2H 3 PO 4 (aq)  6H 2 O(l)+1Ca 3 (PO 4 ) 2 (s) 2.The combustion reaction of ethanol (C 2 H 5 OH) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen gas. 1C 2 H 5 OH(l)+3O 2 (g)  2CO 2 (g)+3H 2 O(g)

45. Instruction Challenging group  Take your back packs and go to table 1 for easy ones, table 3 for medium ones, table 5 for hard ones  If you find the answer, take out the paper from the plastic cover, then label the question number, write your answer and your initial by pencil. Put the paper back to the cover

46. Instruction Reviewing group  take your back packs and go to the front desks to review

47. Additional problem 1. How many moles of calcium carbonate is required to produce 1.57 moles of carbon dioxide? ___CaCO 3 (s)  ___ CaO (s)+ ___ CO 2 (g)

48. Additional problem 2. How many grams of calcium carbonate is required to produce 1.57 moles of carbon dioxide? ___CaCO 3 (s)  ___ CaO (s)+ ___ CO 2 (g)

49. Additional problem 3. For the reaction shown below, 1.26 grams of methane, CH 4, were formed, how many grams of aluminum carbide was consumed? ___ Al 4 C 3 (s) + __ HCl(g)  __ CH 4 (g) +__ AlCl 3 (s)

50. Additional problem 4. For the reaction shown below, 1.26 grams of aluminum carbide and 3.50 grams HCl gas were reacted. How many CH 4 gas were formed? ___ Al 4 C 3 (s) + __ HCl(g)  __ CH 4 (g) +__ AlCl 3 (s)

51. Additional problem 5. For the reaction shown below, 1.26 grams of aluminum carbide and 3.50 grams HCl gas were reacted. How many grams of excess reactants were left? ___ Al 4 C 3 (s) + __ HCl(g)  __ CH 4 (g) +__ AlCl 3 (s)