Ch3 z53 stoich

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Ch3 z53 stoich

  1. 1. Section 3.1 Atomic Mass Atoms are so small, it is difficult to discuss how much they weigh in grams. Use atomic mass units. An atomic mass unit (amu) is one twelfth the mass of a carbon-12 atom. This gives us a basis for comparison. The decimal numbers on the table are atomic masses in amu.
  2. 2. Figure 3.1 Schematic Diagram of a Mass Spectrometer
  3. 3. Figure 3.2 Neon Gas
  4. 4. They are not whole numbers Because they are based on averages of atoms and of isotopes. can figure out the average atomic mass from the mass of the isotopes and their relative abundance. add up the weighted percents as decimals times the masses of the isotopes. %(m 1 ) + %(m 2 ) + . . . = Avg. mass
  5. 5. Examples of average mass There are two isotopes of carbon 12 C with a mass of 12.00000 amu (98.892%), 13 C with a mass of 13.00335 amu (1.108%). What is the average mass? . . . (12.00000)(.98892) + (13.00335)(.01108) = 12.01amu
  6. 6. Examples of average atomic mass pp There are two isotopes of nitrogen , one with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is the percent abundance of each? . . . Hint : the abundances have to had up to 100% (or 1.00 in decimal form). So, set one of them equal to “x” and the other equal to 1-x . . . Where would you get the value of the average atomic mass? . . . From the periodic table = 14.007 Now, set up the problem . . .
  7. 7. Examples pp Two isotopes of N, one of 14.0031 amu & one of 15.0001 amu. What’s % abundance? . . . 14.0031(x) + 15.0001(1-x) = ? You get . . . 99.61% of 14.0031 and 0.39% of 15.0001 . . .
  8. 8. Finding An Isotopic Mass pp A sample of boron consists of 10 B (mass 10.0 amu) and 11 B (mass 11.0 amu). If the average atomic mass of B is 10.8 amu, what is the % abundance of each boron isotope? . . .
  9. 9. Assign X and Y values: pp X = % 10 B Y = % 11 B Determine Y in terms of X X + Y = 100 Y = 100 - X Solve for X: X (10.0) + (100 - X )(11.0) = 10.8 100 100 Multiply through by 100 10.0 X + 1100 - 11.0X = 1080
  10. 10. Collect X terms pp 10.0 X - 11.0 X = 1080 - 1100 - 1.0 X = -20 X = -20 = 20 % 10 B - 1.0 Y = 100 - X % 11 B = 100 - 20% = 80% 11 B
  11. 11. Learning Check Copper has two isotopes 63 Cu (62.9 amu) and 65 Cu (64.9 amu). What is the % abundance of each isotope? (Hint: Check periodic table for atomic mass) 1) 30% 2) 70% 3) 100%
  12. 12. Solution AT8 2) 70% . . . Solution 62.9X + 64.90 (100-x) = 63.50 100 100 62.9x + 64.90(100-x) = 6350 -2.0 X = -140 X = 70%
  13. 13. 3.2 The Mole The mole is a number. A very large number, but still, just a number. 6.022 x 10 23 of anything is a mole A large dozen. The number of atoms in exactly 12 grams of carbon-12.
  14. 14. The Mole Makes the numbers on the periodic table the mass of the average atom.
  15. 15. More Stoichiometry
  16. 16. 3.3 Molar mass Mass of 1 mole of a substance. Often called molecular weight. To determine the molar mass of an element , look on the periodic table. To determine the molar mass of a compound , add up the molar masses of the elements that make it up.
  17. 17. Find the molar mass of CH 4 Mg 3 P 2 Ca(NO 3 ) 3 Al 2 (Cr 2 O 7 ) 3 CaSO 4 · 2H 2 O 16.043 134.862 226.095 701.926 172.12
  18. 18. Calculating atoms, moles and mass A silicon chip has a mass of 5.68 mg. How many atoms of Si are in the chip? (Use spider) . . . (5.68 mg)(1 g/1000 mg)(1 mol /28.086 g)(6.02 x 10 23 atoms /1mol) = 1.22 x 10 20 atoms of Silicon.
  19. 19. Calculating atoms, moles and mass Cobalt is added to steel to improve corrosion resistance. Calculate the moles and the mass of Co in a sample of Co containing 5.00 x 10 20 atoms of Co. . . Answers? . . . 8.30 x 10 -4 moles Co; 4.89 x 10 -2 g Co
  20. 20. 3.4 Percentage Composition Like all percents Part x 100% whole Strategy . . . Find the mass of each component, Divide by the total mass. Multiply by 100%
  21. 21. Example pp Calculate the percent composition of a compound that is composed only of 29.0 g of Ag and 4.30 g of S. The answer is . . . 87.1% Ag from [29.0 ÷ (29.0 + 4.30)] x 100%. What is the % of S? . . . 12.9% (must add up to 100%).
  22. 22. Getting % Composition from the formula If we know the formula, then we assume we have 1 mole. From that we can then figure out the pieces and the whole.
  23. 23. Example pp Calculate the percent composition of C 2 H 4 ? Assume one mole. How many moles of carbon in that one mole? How many grams of carbon is that? How many moles of hydrogen? How many grams of hydrogen is that? What is total grams of C 2 H 4 ? What is % comp. of carbon? Of hydrogen? . . . 85.60% carbon, 14.40% hydrogen.
  24. 24. Example Calculate the percent of aluminum in Aluminum sulfate. . . Figure out the formula, which is . . . Al 2 (SO 4 ) 3 (M = 342 g/ mol) Then do the procedure to get an answer of . . . 15.77% -- Did you get half of this? If so, you forgot there were two aluminums in the formula.
  25. 25. 3.4 Percent Composition Review pp Percent of each element a compound is composed of. Find the mass of each element, divide by the total mass, multiply by 100. Easiest if use one mole of the compound . Find the percent composition of CH 4 . . . 12/16 x 100 = 75% of carbon, so H is 100% - 75% = 25%
  26. 26. 3.4 Percent Composition Review Try on your own Find the percent composition of Al 2 (Cr 2 O 7 ) 3 Al 7.69%; Cr 44.45%; O 47.87% % Composition of CaSO 4 · 2H 2 O Ca 23.28%; S 18.62%; H 2.34%; O 55.76%
  27. 27. The Empirical Formula The lowest whole number ratio of elements in a compound. The molecular formula is the actual ratio of elements in a compound. The two can be the same (but may not be). CH 2 is an empirical formula C 2 H 4 is a molecular formula C 3 H 6 is a molecular formula H 2 O is both .
  28. 28. Calculating Empirical Just find the lowest whole number ratio C 6 H 12 O 6 is ? CH 3 N is? It is not just the ratio of atoms, it is also the ratio of moles of atoms. In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen. In one molecule of CO 2 there is 1 atom of C and 2 atoms of O.
  29. 29. Calculating Empirical pp We can get the ratio from percent composition. 1st, assume you have 100 g. 2nd, the percentages become grams. 3rd, turn grams to moles (spider). 4th, find lowest whole number ratio by dividing each “subscript” by the smallest. 5th, clear fractions by multiplying with the correct factor.
  30. 30. Example pp Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so 38.67 g C x 1mol C = 3.220 mole C 12.01 g C 16.22 g H x 1mol H = 16.09 mole H 1.01 g H 45.11 g N x 1mol N = 3.219 mole N 14.01 g N
  31. 31. Example pp The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol N C 1 H 5 N 1 CH 5 N
  32. 32. Short cut way pp A compound is 43.64 % P and 56.36 % O. What is the empirical formula? P (43.64 g/30.97 g/mol) O (56.36g/16.00 g/mol) P 1.41 O 3.52 P 1.41/1.41 O 3.52/1.41 P 1 O 2.50 Multiply by 2 to “clear the fraction.” We get P 2 O 5
  33. 33. Clearing the Fractions pp If after dividing each ratio by the smallest you get a fraction and . . . The last two decimals are approximately .33, .67, .25, .50, .75 or .20, then multiply all the ratios as follows . . .
  34. 34. Clearing the Fractions pp .33 = 1/3 multiply by 3 .67 = 2/3 3/2 .25 = 1/4 4 .50 = 1/2 5 .75 = 3/4 4/3 .20 = 1/5 5
  35. 35. Clearing the Fractions pp Example, get to X 4 Y 1.67 Z 2 Multiply each ratio by what? . . . 3/2 (the reciprocal of 2/3, which is .67) You get what? . . . X 6 Y 2.5 Z 3 so now you have to multiply by what? . . . Multiply by 2 to get what? . . . X 12 Y 5 Z 6
  36. 36. Working backwards From percent composition, you can determine the empirical formula. Empirical Formula the lowest ratio of atoms in a molecule. Based on mole ratios. A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O. What is its empirical formula? Ans. . . C 13 H 14 N 2 O 4
  37. 37. CO 2 is absorbed H 2 O is absorbed Sample is burned completely to form CO 2 and H 2 O Pure O 2 in
  38. 38. Figure 3.5 Schematic Diagram of the Combustion Device Used to Analyze Substances for Carbon and Hydrogen
  39. 39. Empirical Example A 0.2000 gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O 2 . 0.2998 g of CO 2 and 0.0819 g of H 2 O are produced. What is the empirical formula? Answer . . . C 3 H 4 O 3
  40. 40. 3.5 Determinine the Formula of a Compound: Empirical To Molecular Formulas Empirical is lowest ratio. Molecular is actual molecule. Need Molar mass also to solve this. Ratio of molar to empirical mass will tell you the molecular formula . Must be a whole number because...
  41. 41. Example A compound is made of only sulfur and oxygen. It is 69.6% S by mass. Its molar mass is 192 g/mol. What is its formula? Answer . . . S 4 O 4
  42. 42. 3.6 Chemical Equations Are sentences. Describe what happens in a chemical reaction. Reactants  Products Equations should be balanced. Have the same number of each kind of atoms on both sides because . . .
  43. 43. 3.7 Balancing equations pp CH 4 + O 2  CO 2 + H 2 O Reactants Products C 1 1 O 2 3 H 4 2
  44. 44. Balancing equations pp CH 4 + O 2  CO 2 + 2 H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4
  45. 45. Balancing equations pp CH 4 + O 2  CO 2 + 2H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4 4
  46. 46. Balancing equations pp CH 4 + 2 O 2  CO 2 + 2H 2 O Reactants Products C 1 1 O 2 3 H 4 2 4 4 4
  47. 47. Another Example H 2 + H 2 O O 2  Make a table to keep track of where you are at
  48. 48. Example pp H 2 + H 2 O O 2  Need twice as much O in the product R P H O 2 2 2 1
  49. 49. Example pp H 2 + H 2 O O 2  Changes the O R P H O 2 2 2 1 2
  50. 50. Example pp H 2 + H 2 O O 2  Also changes the H R P H O 2 2 2 1 2 2
  51. 51. Example pp H 2 + H 2 O O 2  Need twice as much H in the reactant R P H O 2 2 2 1 2 2 4
  52. 52. Example pp H 2 + H 2 O O 2  Recount R P H O 2 2 2 1 2 2 4 2
  53. 53. Example pp H 2 + H 2 O O 2  The equation is balanced, has the same number of each kind of atom on both sides R P H O 2 2 2 1 2 2 4 2 4
  54. 54. Example H 2 + H 2 O O 2  This is the answer R P H O 2 2 2 1 2 2 4 2 4 Not this
  55. 55. Abbreviations ( s )  ( g )  ( aq ) heat  catalyst
  56. 56. Practice Ca(OH) 2 + H 3 PO 4  H 2 O + Ca 3 (PO 4 ) 2 Hint: Start with most complicated molecule first. Answer . . . 3Ca(OH) 2 +2 H 3 PO 4  6H 2 O + Ca 3 (PO 4 ) 2 Cr + S 8  Cr 2 S 3 . . . 16Cr + 3S 8  8Cr 2 S 3 KClO 3 ( s )  KCl + O 2 ( g ) 2KClO 3 ( s )  KCl + 3O 2 ( g )
  57. 57. Practice Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. Fe 2 S 3 + 6HCl  2FeCl 3 + 3H 2 S Fe 2 O 3 ( s ) + Al( s )  Fe( s ) + Al 2 O 3 ( s ) Fe 2 O 3 ( s ) + 2Al( s )  2Fe( s ) + Al 2 O 3 ( s )
  58. 58. Meaning A balanced equation can be used to describe a reaction in molecules and atoms. Not grams. Chemical reactions happen by molecules at a time . . . or dozens of molecules at a time . . . or moles of molecules (but not by grams).
  59. 59. 3.8 Stoichiometry Given an amount of either starting material or product, can determine the other quantities. use conversion factors from molar mass (g - mole) balanced equation (mole - mole) keep track (use “spider equation”).
  60. 60. For example... pp If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form? . . . Fe + CuSO 4  Fe 2 (SO 4 ) 3 + Cu (unbalanced ) 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 10.1 g Fe 55.85 g Fe 1 mol Fe = 0.181 mol Fe
  61. 61. 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu pp 0.181 mol Fe 2 mol Fe 3 mol Cu = 0.272 mol Cu 0.272 mol Cu 1 mol Cu 63.55 g Cu = 17.3 g Cu
  62. 62. Could have done it via “spider” pp 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu = 17.3 g Cu
  63. 63. pp
  64. 64. Examples pp One way of producing O 2 ( g ) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O 2 (g) are produced? Ans. 0.314 moles O2 How many grams of potassium chloride? 15.51g KCl How many grams of oxygen? 9.98g O 2
  65. 65. Examples pp A piece of aluminum foil 5.11 in x 3.23 in x 0.0381 in is dissolved in excess HCl (aq). How many grams of H 2 ( g ) are produced? . . . Al: find volume, convert to mL, find mass using density 2.702 g/ml to get . . . 27.84 g. Now write the equation . . . 2Al + 6HCl  2AlCl 3 + 3H 2 Since the g of Al just happens to be the same as its molar mass and the mole ratio would be 3/2, so get 3/2 moles H 2  3 g H 2
  66. 66. Examples pp How many grams of Al are needed to produce 15 grams of iron from the following reaction? Fe 2 O 3 ( s ) + Al( s )  Fe( s ) + Al 2 O 3 ( s ) Ans . . . 7.2 g Al
  67. 67. Examples if time pp K 2 PtCl 4 ( aq ) + NH 3 ( aq )  Pt(NH 3 ) 2 Cl 2 ( s )+ KCl( aq ) what mass of Pt(NH 3 ) 2 Cl 2 can be produced from 65 g of K 2 PtCl 4 ? How much KCl will be produced? How much from 65 grams of NH 3 ?
  68. 68. 3.9 Limiting Reagent pp Reactant that determines the amount of product formed. See p. 116 # 3 (assigned as HW) The one you run out of first. http://www.chemcollective.org/tutorials.php Makes the least product. Book shows you a ratio method. It works. So does mine
  69. 69. Limiting reagent pp To determine the limiting reagent requires that you do two stoichiometry problems. Figure out how much product each reactant makes. The one that makes the least is the limiting reagent.
  70. 70. If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? (steps to follow) pp 2Cu + S  Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.07g S 1 mol S 1 mol S 1 mol Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is Limiting Reagent
  71. 71. Your turn If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many L of gas will be produced at STP? Grams of solid? Excess remaining? . . . (Write balanced equation, then do 2 stoichs to find LR) . . . Mg + 2HCl  MgCl 2 + H 2 LR is . . . (10.1 g Mg )(1 mol/24.3 g)(1/1)(22.4 L/mol) = 9.3 L H 2 (2.87 L HCl )(1 mol/22.4 L)(1/2)(22.4 L/mol) = 1.43 L H 2 HCl is the LR , so use HCl for all calculations.
  72. 72. Your turn If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? Grams of solid? Excess remaining? Mg + 2HCl  MgCl 2 + H 2 . ( HCl is the LR ) How many liters of gas produced? . . . (2.87 L)(1 mol/22.4 l)(1/2)(22.4 L/mol) = 1.43 Liters of H 2 is produced. How many grams of solid produced? . . . (2.87 L)(1 mol/22.4L)(1/2)(95.2 g/mol) = 6.10 g MgCl 2 is produced .
  73. 73. Your turn If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? Grams of solid? XS remaining? . . . Mg + 2HCl  MgCl 2 + H 2 . Mg is xs, so calc how much reacted with HCl. (2.87 L HCl)(1 mol/22.4L)(1mol B/2 mol A)(24.3 g) = 1.57 g Mg consumed . So, xs = . . .? 10.1 g - 1.57 g = 8.53 g Mg left over (but SF) 8.5 g Mg remaining
  74. 74. Example - Class does Ammonia is produced by the following reaction : N 2 + H 2  NH 3 What mass of NH 3 can be produced from a mixture of 100. g N 2 and 500. g H 2 ? Ans. . . . 121g NH 3 (Did you balance the reaction?) How much unreacted material remains? 21.4 g H 2 reacted so 478 g H 2 remain.
  75. 75. Excess Reagent The reactant you don’t run out of. The amount of stuff you make is the yield. The theoretical yield is the amount you would make if everything went perfect. The actual yield is what you make in the lab.
  76. 76. Percent Yield % yield = Actual x 100% Theoretical % yield = what you got x 100% what you could have got
  77. 77. Example pp 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu What is the actual yield? (read question) What is the theoretical yield? (do stoich) . . . (3.92g A)(1mol/27g A)(3/2)(63.5 g B/mol) = 13.84 g Cu What is the percent yield? (use formula) AY = 6.78 g; TY = 13.84 g (Al is LR) % yield = (6.78/13.84)(100) = 49.0%.
  78. 78. Another Example 2ZnS + 3O 2  2ZnO + 2SO 2 If the typical yield is 86.78%, how much SO 2 should actually be expected if 4897 g of ZnS are used with an excess of O 2 ? First, calculate theoretical yield (stoich) . . . ( 4897)(1/97.46)(2/2)(64.07/1) = 3219 g SO 2 Solve for actual yield Since %Y = AY/TY, AY = %Y x TY So, 3219 g SO 2 x 86.78% = . . . 2794 g
  79. 79. Examples Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the 3 types of yield & their values? Theoretical = 59.4 g Actual = 50.3 g Percent = 84.7%
  80. 80. Examples Years of experience have proved that the percent yield for the following reaction is 74.3% Hg + Br 2  HgBr 2 If 10.0 g of Hg and 9.00 g of Br 2 are reacted, how much actual HgBr 2 will be produced? Ans. . . . 10g Hg = .05 mole; 9g Br2 = .0562 moles; so Hg = LR TY = 18.97 g; actual = .743 x TY = 14.10 g If the reaction did go to completion, how much excess reagent would be left? Xs left = .0062 mol Br2 x 160 g/mol Br2 = .992 g
  81. 81. Example (lab 2) Commercial brass is an alloy of Cu and Zn. It reacts with HCl by the following reaction Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) Cu does not react. When 0.5065 g of brass is reacted with excess HCl, 0.0985 g of ZnCl 2 are eventually isolated. What is the composition of the brass? . . . Find g of Zn using stoich, then do % comp for Zn, then 100% - %Zn to get % for Cu Ans. 9.32% Zn (from 0.047g Zn); 90.68% Cu You will do something similar to this (using Ag and Cu) with Lab #1.

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