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Chapter 12   Stoichiometry
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Chapter 12 Stoichiometry

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  • 1. STOICHIOMETRY Or How I Learned to Love Counting Atoms
  • 2. STOICHIOMETRY Nearly everything you use is manufactured from chemicals - soaps, shampoos and conditioners, 8-tracks, and clothes. Obviously, for the manufacturer to make a profit, the cost of making any of these items cannot exceed the money paid for them. Therefore, the chemical processes used in manufacturing must be carried out economically. That is where balanced chemical equations help.
  • 3. Equations are the recipes that tell chemists what amounts of reactants to mix and what amounts of products to expect. You can determine the quantities of reactants and products in a reaction from the balanced equation. When you know the quantity of one substance in a reaction, you can calculate the quantity of any other substance consumed or created in the reaction. (Quantity usually means the amount of a substance expressed in grams or moles. But quantity could just as well be in liters, tons, or molecules.) The calculation of quantities in chemical reactions is a subject of chemistry called stoichiometry .
  • 4. Calculations using balanced equations are called stoichiometric calculations . For chemists, stoichiometry is a form of bookkeeping.
  • 5. INTERPRETING CHEMICAL EQUATIONS As you may recall, ammonia is widely used as a fertilizer. Ammonia is produced industrially by the reaction of nitrogen with hydrogen. What kinds of information can be derived from this equation? p.356 N 2 ( g ) + 3H 2 ( g ) 2NH 3 ( g )
  • 6. Do you see that mass and atoms are conserved in this chemical reaction? Mass and atoms are conserved in every chemical reaction. The mass of the reactants equals the mass of the products. The number of atoms of each reactant equals the number of atoms for that reactant in the product(s). Unlike mass and atoms, however, molecules, formula units, moles and volumes of gases will not necessarily be conserved - although they may be. Only mass and atoms are conserved in every chemical reaction.
  • 7.
    • Quiz
    • Write a balanced equation for the combination reaction of the formation of water from its gases.
    • What is one interpretation of the balanced equation based on what you read last night?
    • Si, Se Puede!
  • 8. Interpreting Chemical Reactions Again, the formation of ammonia from hydrogen and nitrogen is 3H 2 + N 2 2 NH 3
  • 9. Mole-Mole Calculations
    • The most important interpretation of this equation is that 1 mol of nitrogen reacts with 3 mol of hydrogen to form 2 mol of ammonia.
    • With this interpretation, you can relate moles of reactants to moles of product.
    • The coefficients from the balanced equation are used to write conversion factors called mole ratios.
  • 10. The mole ratios are used to calculate the number of moles of product from a given number of moles of reactant or to calculate the number of moles of reactant from a given number of moles of product. Three of the mole ratios for this equation are 1 mol N 2 2 mol NH 3 3 mol H 2 3 mol H 2 1 mol N 2 2 mol NH 3
  • 11. In the mole ratio below, W is the unknown quantity. The value of a and b are the coefficients from the balanced equation. Thus a general solution for a mole-mole problem is given by Given Mole ratio Calculated x mol G x = mol W b mol W xb a mol G a From balanced equation
  • 12. Using the ammonia reaction, answer the following question. How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? 0.60 mol N 2 x = 1.2 mol NH 3 Given Mole Ratio 2 mol NH 3 1 mol N 2
  • 13. MASS-MOLES CALCULATIONS Balances don’t tell you numbers in moles but in grams. As such, there are two related stoichiometry calculations: Moles - Mass & Mass - Moles
  • 14. In a mole-mass problem you are asked to calculate the mass (usually in grams) of a substance that will react with or be produced from a given number of moles of a second substance. (If, in an example, you are told something in is excess, just ignore that substance and solve the problem with the needed substances.) moles A moles B mass B moles A x mole ratio of x molar mass of B B A
  • 15. Example: Plants use carbon dioxide and water to form glucose (C 6 H 12 O 6 ) and oxygen. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide? Answer: 1. Write the balanced equation 6CO 2 (g) + 6H 2 O(l) -> C 6 H 12 O 6 (s) + 6O 2 (g)
  • 16. 2. Determine what you need to find/know. Unknown: mass of C 6 H 12 O 6 produced Given: amount of H 2 O = 3.00 mol = grams C 6 H 12 O 6 3. Determine conversion factors moles H 2 O x x moles C 6 H 12 O 6 moles H 2 O grams C 6 H 12 O 6 1 mole C 6 H 12 O 6
  • 17. 4. Solve 3.00 moles H 2 O x x = 90.0 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 6 moles H 2 O 180 g C 6 H 12 O 6 1 mole C 6 H 12 O 6
  • 18. In a mass-mole problem you are asked to calculate the moles of a substance that will react with or be produced from a given number of grams of a second substance. mass A moles A moles B 1 mole A molar mass A mass A x x mole ratio B A
  • 19. Worksheet #21 Mol  mass mol A  mol B  mass B 75.0 mol C 7 H 6 O 3 = 13500 g C 9 H 8 O 4 = 13.5 kg C 9 H 8 O 4 x 1 mol C 9 H 8 O 4 1 mol C 7 H 6 O 3 180 g C 9 H 8 O 4 1 mol C 9 H 8 O 4 x x 1 kg 1000 g
  • 20. Mass-Mass Calculations
    • No laboratory balance can measure substance directly in moles
    • Instead, the amount of a substance is usually determined by measuring its mass in grams
    • From the mass of a reactant or product, the mass of any other reactant or product in a given chemical equation can be calculated
    • The mole interpretation of a balanced equation is the basis for this conversion
  • 21. If the given sample is measured in grams, the mass can be converted to moles by using the molar mass Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown If it is the mass of the unknown that needs to be determined, the number of moles of the unknown can be multiplied by the molar mass. As in mole-mole calculations, the unknown can be either a reactant or a product
  • 22. Mass-mass problems can be solved in basically the same way as mole-mole problems. 1. The mass G is changed to moles of G (mass G mol G) by using the molar mass of G. Mass G X = mol G 2. The moles of G are changed to moles of W (mol G mol W ) by using the mole ratio from the balanced equation. Mol G X = mol W 1 mol G molar mass G b mol W a mol G
  • 23. 3. The moles of W are changed to grams of W (mol W mass W )
  • 24. mass A x x x mole ratio molar mass of A molar mass of B given 1 mole A grams A moles B moles A grams B 1 mole B The route for solving mass-mass problems is: mass A moles A moles B mass B
  • 25. Example: Calculate the number of grams of NH 3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. 2. Write what you know: Unknown: g NH 3 ; g H 2 -> g NH 3 Given: 5.40 g H 2 Solution: 1. Write the balanced equation N 2 + 3H 2 2NH 3
  • 26. 4. Solve 5.40 g H 2 x x x given = 30.6 g NH 3 changes given to moles mole ratio change moles of wanted to grams 3. Determine conversion factors g H 2 mol H 2 mol NH 3 g NH 3 1 mol H 2 2.0 g H 2 2 mol NH 3 3 mol H 2 17.0 g NH 3 1 mol NH 3
  • 27. P.361, #13 5.00 g CaC 2 x x x = 2.03 g C 2 H 2 1 mole CaC 2 64.1 g CaC 2 1 mole C 2 H 2 1 mole CaC 2 26.0 g C 2 H 2 1 mole C 2 H 2
  • 28. Worksheet #15 a. 384 g O 2 = 1104 g NO 2 = 1.10x10 3 g NO 2 given Molar mass A Mole ratio Molar mass B x 1 mol O 2 32g O 2 x 2 mol NO 2 1 mol O 2 x 46g NO 2 1 mol NO 2
  • 29. OTHER STOICHIOMETRIC CALCULATIONS As you already know, a balanced equation indicates the relative number of moles of reactants and products. From this foundation, stoichiometric calculations can be expanded to include any unit of measurement that is related to the mole. The given quantity can be expressed in number of representative particles, units of mass, or volumes of gases at STP.
  • 30. The following equation summarizes these steps for a typical stoichiometric problem aG bW (given quantity) (wanted quantity)
  • 31.  
  • 32. Using the ammonia reaction equation, determine the number of liters of ammonia that can be produced from 5 grams of nitrogen at STP. 5g N 2 x x x = N 2 + 3H 2 2NH 3 1 mole N 2 28g N 2 2 mole NH 3 1 mole N 2 22.4 L NH 3 1 mole NH 3
  • 33. If the law of conservation of mass is true, how is it possible to make 30.6 g NH 3 from only 5.40 g H 2 ? Looking back at the equation for the reaction, you will see that hydrogen is not the only reactant. Another reactant, nitrogen, is also involved. If you were to calculate the number of grams of nitrogen needed to produce 30.6 g NH 3 and then compare the total masses of reactants and products, you would have an answer to this question
  • 34. LIMITING REAGENT As you know, a balanced equation is a chemist’s recipe - a recipe that can be interpreted on a microscopic scale (interacting particles) or on a macroscopic scale (interacting moles). The coefficients used to write the balanced equation give both the ratio of representative particles and the mole ratio. Recall the equation for the preparation of ammonia. N 2 (g) + 3H 2 (g) 2NH 3 (g)
  • 35. When one molecule of N 2 reacts with three molecules of H 2 , two molecules of NH 3 are produced. What would happen if two molecules of N 2 reacted with three molecules of H 2 ? Would more than two molecules of NH 3 be formed? Before the reaction takes place, nitrogen and hydrogen are present in a 2:3 molecule ratio. According to the balanced equation, one molecule of N 2 reacts with three molecules of H 2 to produce two molecules of NH 3 . At this point, all the hydrogen has been used up, and the reaction stops. One molecule of unused nitrogen is left over, in addition to the two molecules of ammonia that have been produced.
  • 36. In this reaction, only the hydrogen is completely used up. It is called the limiting reagent (reactant). As the name implies, the limiting reagent (reactant) limits or determines the amount of product that can be formed in a reaction. The reaction occurs only until the limiting reagent (reactant) is used up. By contrast, the reactant that is not completely used up in a reaction is called the excess reagent (reactant). In this example, nitrogen is the excess reagent (reactant) because some nitrogen will remain unreacted.
  • 37. Example: Sodium chloride can be prepared by the reaction of sodium metal with chlorine gas. 2Na(s) + Cl 2 (g) 2NaCl(s) Suppose that 6.70 mol of Na reacts with 3.20 mol Cl 2 . a. what is the limiting reagent (reactant)? b. how many moles of NaCl are produced?
  • 38. Solution: List the knowns and unknowns for a . known: moles of sodium = 6.70 mol Na moles of chlorine = 3.20 mol Cl 2 2 mol Na = 1 mol Cl 2 (from balanced equation) unknown: limiting reagent (reactant) Given Mole Required amount ratio amount 2. Solve for the unknown: 6.70 mol Na x = 3.35 mol Cl 2 1 mol Cl 2 2 mol Na
  • 39. This calculation indicates that 3.35 mol Cl 2 is needed to react with 6.70 mol Na. Because only 3.20 mol Cl 2 is available, however, chlorine must be the limiting reagent (reactant). Sodium, then, must be in excess. Now, then, to find the amount of product (NaCl) made. To find the amount of product, you first have to identify the limiting reactant. Once that is done, you use that reactant to determine the amount of product made, because once the limiting reagent runs out, no more product is made.
  • 40. 1. List the knowns and unknowns for b. known: amount of limiting reagent: 3.20 mol Cl 2 . 1 mol Cl 2 = 2 mol NaCl (from balanced equation) unknown: amount (moles) of NaCl produced. 2. Solve 3.20 mol Cl 2 = 6.40 mol NaCl x 2 mol NaCl 1 mol Cl 2
  • 41. P.381, #66 4FeS 2 + 15O 2 --> 2Fe 2 O 3 + 8SO 3 20.0 g 16.0 g x 20.0 g FeS 2 = .17 mol FeS 2 16.0 g O 2 x 1 mol O 2 32 g O 2 = .5 mol O 2 Next, determine the limiting reagent once you know the mole amounts of each reagent. 1 mol FeS 2 119.8 g FeS 2
  • 42. PERCENT YIELD When an equation is used to calculate the amount of product that will form during a reaction, a value representing the theoretical yield is obtained. The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants. In contrast, the amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield . The actual yield is often less than the theoretical yield.
  • 43. The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent. The percent yield measures the efficiency of the reaction. A percent yield should not normally be larger than 100%. Many factors can cause percent yields to be less than 100%. Percent yield = x 100 % actual yield theoretical yield
  • 44. Example: Calcium carbonate is decomposed by heating, as shown in the following equation. CaCO 3 (s) CaO(s) + CO 2 (g) a. what is the theoretical yield of CaO if 24.8 g of CaCO 3 is heated? b. What is the percent yield if 13.1 g CaO is produced?
  • 45. Solution: 1. List the knowns and unknowns in a. known: mass of CaCO 3 = 24.8 g 1 mol CaCO 3 = 1 mol CaO (from balanced equation) 1 mol CaCO 3 = 100.1 g (molar mass) 1 mol CaO = 56.1 g (molar mass) unknown: theoretical yield of CaO = ? g CaO
  • 46. 2. Solve for the unknown. 24.8 g CaCO 3 x x x given amount molar mass mole ratio molar mass = 13.9 g CaO Again, this is the theoretical yield, the amount you would make if the reaction were 100% accurate. 1 mol CaCO 3 100.1 g CaCO 3 1 mol CaO 1 mol CaCO 3 56.1 g CaO 1 mol CaO
  • 47. 3. Determine % yield for b. actual yield = 13.1 g CaO theoretical yield = 13.9 g CaO Percent yield = x 100% actual yield theoretical yield Percent yield = x 100% = 94.2% 13.1 g CaO 13.9 g CaO