1. IV Stoichiometry

2. Stoichiometry The relationship (mole ratio) between elements in a compound and between elements and compounds in a reaction e.g. H 2 O means: 1 molecule H 2 O contains 2 atom H and 1 atom O and 1 mole H 2 O contains 2 moles H atoms and 1 mole O atoms

3. Chemical Compounds Combination of elements –A compound is made up of specific elements in a specific ratio - Law of Constant Composition Chemical Formula –Written representation of a chemical compound. So, a specific compound has a specific formula

4. Terms Formula Unit –Involves the lowest subscript which describes the ratio. –e.g. H 2 O; CH; CH 2 ; CH 4 –May or may not actually exist Formula Weight –The mass of the formula unit Empirical Formula –Written representation of the formula unit

5. Terms continued Molecule –Integral multiple of the formula unit (integer may be 1) that actually exists –e.g. C 2 H 2 ; C 2 H 4 ; C 2 H 6 Molecular Weight (Molar Mass) –Mass of the molecule Molecular Formula –Written representation of the molecule

6. Mass to mole conversions Stoichiometry is in mole ratios Most measurements are made in grams So, you need to be able to get from grams to moles and moles to grams The atomic weight listed on the periodic table is listed without units. Why?

7. Units depend on what you want. If you are looking on the atomic scale, atoms or molecules, units are amu. If you are working on the macroscopic scale, moles of material, units are in grams.

8. Convert 34.0 grams NH 3 to moles. 1)Determine M of NH 3. M = AW N + 3(AW H) = 14.0 + 3(1.01) = 17.0 2) Determine the moles of NH 3

9. How many molecules in 32.0 g of oxygen gas. A)2.0 B)1.0 C)0.5 D)6.02 x 10 23 E)1.20 x 10 24

10. How many molecules in 32.0 g of oxygen gas. Oxygen occurs as O 2 gas. AW of O = 16.00, so O 2 = 32.00

11. % Composition (% weight/ weight; %w/w)

12. 1. Calculate the % composition form the Molecular Formula (or the Empirical Formula) What is the % composition by weight of C 2 H 4 O 2 ? C = 12.01; H = 1.01; O = 16.00 Step 1: Find the molecular weight of the compound. MW = 2(AW C) + 4(AW H) + 2(AW O) = 2(12.01) + 4(1.01) + 2(16.00) = 60.06 g/ mol

13. Step 2: Find the % of each element.

14. Find the Chemical Formula from the % composition Which Chemical formula will you get? –Empirical Formula, to get the molecular formula you would need more information than just the % composition

15. A compound containing only carbon, hydrogen, and oxygen was found to contain 62.02 % C and 10.42% H. What is the formula of the compound? C = 12.01H = 1.01O = 16.00 1)Find the amount of O %O = 100 - %C - %H = 100 - 62.02 - 10.42 = 27.57 % O

16. 2) Determine the moles of each element (assume a 100 g sample)

17. 3) Divide by the smallest number to get whole number ratio Mol C = 5.163 mol Mol H = 10.3168 mol Mol O = 1.723 mol / 1.723 = 2.996 = 3 / 1.723 = 5.9877 = 6 / 1.723 = 1 So, the compound has the ratio 3C:6H:1 O And the Empirical formula: C 3 H 6 O

18. A further analysis of the compound found that the molar mass was 115.99 g/mol. What is the molecular formula of the compound? Molecular formula = (empirical formula)(# of formula units) You can find the # of formula units from MW/ FW

19. Formula weight = weight of empirical formula: C 3 H 6 O FW = 3(12.01) + 6(1.01) + 16.00 = 58.048 # formula units = So the molecular formula contains 2 formula units. Multiply the subscripts by 2 2(C 3 H 6 O) = C 6 H 12 O 2

20. A compound is 84.80% uranium and the balance oxygen. What is the Empirical Formula of the compound? U = 238.0O = 15.9994 A)U 2 O 5 B)U 3 O 8 C)UO 3 D)UO 2 E)U 3 O

21. A compound is 84.80% uranium and the balance oxygen. What is the Empirical Formula of the compound? U = 238.0O = 15.9994 1.Determine the amount of O 100 - 84.80 = 15.20 2.Determine moles and mole ratio / 0.3563 = 1 / 0.3563 = 2.666 0.666 = 2/3 so O is 2 2/3 = 8/3 U 1 O 8/3 Clear denominator, multiply by 3 U3O8U3O8

22. 3) Determine an unknown element (X) from the % composition A compound XO 2 is 78.8% X. What element is X? O = 16.0 A)Ni B)Co C)P D)Sn

23. 3) Determine and unknown element (X) from the % composition A compound XO 2 is 78.8% X. What element is X? O = 16.0 % O = 100 - 78.8 = 21.2 Since Compound is 78.8% X, 0.6625 mol X = 78.8 g X Go to Periodic Table 118.9 = Sn

24. Reactions Chemical Reaction is represented by a Chemical Equation General Form: aA + bB cC + dD A/B are ? Reactants C/D are ? Products a,b,c,d are ?Stoichiometric coefficients

25. Law of Conservation of Mass says? –No mass lost or gained Total mass of reactants = total mass of products Elements are re - arranged not changed This allows us to “Balance” Equations The number and kinds of atoms in the reactants have to show up as the same number and kind of atoms in the product

26. Ca + H 2 OCa(OH) 2 + H 2 1 Ca + H 2 O1 Ca(OH) 2 + H 2 1Ca + 2H 2 O1 Ca(OH) 2 + H 2 1 Ca + 2 H 2 O1Ca(OH) 2 + 1 H 2 Ca + 2 H 2 OCa(OH) 2 + H 2

27. C 3 H 7 OH + O 2 CO 2 + H 2 O C 3 H 7 OH + O 2 3 CO 2 + H 2 O C 3 H 7 OH + O 2 3 CO 2 + 4 H 2 O C 3 H 7 OH + 9/2 O 2 3 CO 2 + 4 H 2 O 2 C 3 H 7 OH + 9 O 2 6 CO 2 + 8 H 2 O

28. SiF 4 + H 2 O HF + SiO 2 SiF 4 + H 2 O 4 HF + SiO 2 SiF 4 + 2 H 2 O 4 HF + SiO 2

29. When the reaction: C 2 H 8 N 2 + N 2 O 4 N 2 + H 2 O + CO 2 Is balanced using the smallest whole numbers, what is the coefficient of N 2 ? A)1 B)2 C)3 D)4 E)5

30. When the reaction: C 2 H 8 N 2 + N 2 O 4 N 2 + H 2 O + CO 2 Is balanced using the smallest whole numbers, what is the coefficient of N 2 ? C 2 H 8 N 2 + N 2 O 4 N 2 + H 2 O + CO 2 2 4 2 3

31. Hydrates A compound (solid) that contains intact water molecules as part of the compound. The water can be removed by heating to leave an anhydrous residue (solid). Water can then be re - added to the anhydrate to yield the original hydrate

32. CaSO 4 2H 2 O Calcium sulfate dihydrate Each mole of compound contains: 1 mole calcium sulfate and 2 mole water or 1 mole Ca 1 mole S 6 mol O 4 mol H

33. CaSO 4 2H 2 O(s) CaSO 4 (s) + 2 H 2 O(g) 172 136 + 2(18) CaSO 4 (s) CaSO 4 2H 2 O(s) heat Water vapor

34. 17.0 g of CaSO 4 2H 2 O(s) was heated to remove the water. What mass of residue (CaSO 4 ) is left? A)13.4 B)13.44 C)3.56 D)15.0 E)15.03

35. 17.0 g of CaSO 4 2H 2 O(s) was heated to remove the water. What mass of residue (CaSO 4 ) is left? What happened to the difference (17.0 - 13.4) = 3.6 grams of material? Lost as water vapor

36. 2.15.00 grams of the hydrate Na 2 SO 4 XH 2 O(s) was heated to remove the water. After heating, 7.95 grams of material remained. What is the formula of the hydrate: (find the value of X) A)Na 2 SO 4 H 2 O B)Na 2 SO 4 2H 2 O C)Na 2 SO 4 3H 2 O D)Na 2 SO 4 5H 2 O E)Na 2 SO 4 7H 2 O

37. Na 2 SO 4 XH 2 O(s) Na 2 SO 4 (s) + XH 2 O 15.007.9515.00-7.95 7.05 Find moles of both products and compare / 0.0559 = 1 / 0.0559 = 6.99 = 7 So, X = 7, formula = Na 2 SO 4 7H 2 O

38. Limiting Reactant Limiting Reactant is that element or compound that determines the amount of product that you get. It is the reactant that is used up

39. You are the owner of a bike shop. A shipment came in with 183 frames, 150 seats, 252 pedals, 131 brake assemblies. How many bikes can you sell? (enter the number)

40. Each bike needs, 1 frame, 1 seat, 2 pedals, 1 brake. Given: 183 frames 150 seats 252 pedals 131 brake assemblies Pedals will allow you to make only 126 bikes so that is the limiting reactant

41. NH 3 + O 2 N 2 + H 2 O Balance NH 3 + O 2 N 2 + H 2 O 2 3 3/2 4 6 3 2

42. 4 6 3 2 NH 3 + O 2 N 2 + H 2 O Remember: Stoichiometry is mole ratios How many moles of N 2 can be formed from 4 mol NH 3 and 4 mol O 2 ? 1. Determine the LR. Mol O 2 given > mol O 2 required. So, NH 3 is LR 2. Determine amount of product

43. 4 6 3 2 NH 3 + O 2 N 2 + H 2 O How many moles of N 2 can be formed from 6 mol NH 3 and 4 mol O 2 ? 1. Determine the LR. Mol O 2 given < mol O 2 required. So, O 2 is LR 2. Determine amount of product

44. 4 6 3 2 NH 3 + O 2 N 2 + H 2 O How many moles of N 2 can be formed from 5 mol NH 3 and 4 mol O 2 ? A)5 B)4 C)3.75 D)2.67 E)2.5

45. 4 6 3 2 NH 3 + O 2 N 2 + H 2 O How many moles of N 2 can be formed from 5 mol NH 3 and 4 mol O 2 ? 1. Determine the LR. Mol O 2 given > mol O 2 required. So, NH 3 is LR 2. Determine amount of product

46. If 10.0 grams each of NH 3 and O 2 are reacted, how many grams of water and N 2 are formed? 1.Find moles of each reactant. 2.Determine the Limiting Reactant Determine mole of O 2 needed Compare to what was given: 0.442 mol required > 0.312 mol given So, O 2 = LR 4 6 3 2 NH 3 + O 2 N 2 + H 2 O

47. 3. Determine the amount of product based on the LR total mass of products = 17.0 g What happened to conservation of mass? 3.0 g un-reacted NH 3

48. 10.0 grams Cr and 10.0 grams S are reacted to give chromic sulfide. How many grams of chromic sulfide are formed? Cr = 52.0S = 32.0 A)20.9 g B)20.0 g C)19.2 g D)3.12 g E)0.80 g

49. 10.0 grams Cr and 10.0 grams S are reacted to give chromic sulfide. How many grams of chromic sulfide are formed? Cr = 52.0S = 32.0 1.Write and balance the equation: 2Cr + 3S --> Cr 2 S 3 2. Determine moles of each

50. 3. Determine LR 4. Determine the amount of product based on the LR

51. Determination of an unknown element A metal oxide has the formula XO 3 and reacts with H 2 to form free metal X and H 2 O. If 15.99 grams XO 3 yields 6.00 grams H 2 O, what element is X? A) Nd B) Ti C) S D) Mo E) H

52. Determination of an unknown element A metal oxide has the formula XO 3 and reacts with H 2 to form free metal X and H 2 O. If 15.99 grams XO 3 yields 6.00 grams H 2 O, what element is X? XO 3 + H 2 --> X + H 2 O 3 3 Find moles XO 3 from stoichiometry and moles of H 2 O Determine MW XO 3 AW X = 143.9 - 3(16.0) = 95.9 = MO

53. Theoretical Yield –Maximum amount of product that can obtained if all the reactant is converted to product Actual Yield –Actual amount of product obtained % Yield

54. One hundred grams of potassium chlorate was heated. What is the final state of affairs? K = 39.1Cl = 35.45O = 16.00

55. One hundred grams of potassium chlorate was heated. What is the final state of affairs? K = 39.1Cl = 35.45O = 16.00 1.Determine the formula for potassium chlorate A. KClOB. KClO 2 C. KClO 3 D. KClO 4

56. One hundred grams of potassium chlorate was heated. What is the final state of affairs? K = 39.1Cl = 35.45O = 16.00 1.Determine the formula for potassium chlorate A. KClOB. KClO 2 C. KClO 3 D. KClO 4 2.Write and Balance the reaction 3. Determine Theoretical Yields of Products KClO 3 -->KCl + O2O2 2 2 3

57. KClO 3 -->KCl + O2O2 2 2 3 100. gramsConvert to moles-mole ratio-convert to g Or: Conservation of mass: 100.0 g - 39.2 g = 60.8 g KCl Part B. If the reaction only produced 50.37 g KCl, 1)What is the %Yield and 2) how many grams of O 2 2)were produced.

58. Consecutive Reactions –Sequence of reactions (steps) that are required to reach desired products

59. FeS 2 + O 2 --> Fe 2 O 3 + SO 2 SO 2 + O 2 --> SO 3 SO 3 + H 2 O --> H 2 SO 4 FeS 2 + O 2 --> Fe 2 O 3 + SO 2 SO 2 + O 2 --> SO 3 SO 3 + H 2 O --> H 2 SO 4 4 112 8 22 4 2 8 88 4( 4 8( 888 4 FeS 2 + 15 O 2 + 8 H 2 O --> 2 Fe 2 O 3 + 8 H 2 SO 4 Now you can have any kind of problem

60. 4 FeS 2 + 15 O 2 + 8 H 2 O --> 2 Fe 2 O 3 + 8 H 2 SO 4 How many moles of sulfuric acid can be made from 5 mol FeS 2 and 17 mol O 2 ? Have LR problem. 1) Determine the LR Mol O 2 given < mol O 2 required so O 2 LR 2) Determine the amount of product

61. Combustion Analysis Combustion analysis –Reaction of a compound that contains carbon, hydrogen and sometimes oxygen burned in air (O 2 ) to produce CO 2 and H 2 O Can be used to determine empirical formula and % composition

62. A 0.3000 gram sample containing only carbon, hydrogen and oxygen was burned in air to produce 0.440 grams CO 2 and 0.180 grams H 2 O. Determine the empirical formula of the sample. C x H y O z + O 2 --> CO 2 + H 2 O x ? Determine moles of C and H from moles of CO 2 and H 2 O Determine moles of O from g of O in sample g O = g sample - g C - g H g O = 0.3000 -(0.01 mol x 12.0 g/mol) - (0.02 mol x 1.01g/mol) g O = 0.16 g mol O = 0.16g/ 16.0 g/mol = 0.01 mol

63. Mol C = 0.01 Mol H = 0.02 Mol O = 0.01 /0.01 = 1 /0.01 = 2 /0.01 = 1 Empirical formula = CH 2 O

64. A) impossible B) Very hard C) able to be done D) un understandable E) hard but workable How will the exam on Thursday be?