1. Mass Relations in Chemistry: Stoichiometry

2. What is a Mole? In chemistry you will do calculations using a measurement called a mole. The mole, the SI unit that measures the amount of substances, is a unit just like the dozen. The mole can be related to: »the number of particles (ions, atoms, etc.) »the mass (in grams) »volume of an element (STP conditions)

3. The Number of Particles in a Mole Because atoms, molecules and ions are very small, the number of individual particles in an object is very large. Instead of counting these particles individually, you can count them using a term that represents a specific number of particles. 602 214 199 000 000 000 000 000. Avogadro’s number A mole is the Avogadro’s number of particles…. rather like a dozen is twelve, the mole is just that, an Avogadro’s number of particles Representive particlesRepresentive particles - species present in the substance (usually atoms, molecules, or formula units)

4. Examples: How many moles of a magnesium is 1.25 X 10 23 atoms of magnesium? How many moles is 2.80 X 10 24 atoms of silicon? When given # of particles, divide by 6.02 x 10 23

5. Mass of a Mole of a Compound Gram atomic mass - the atomic mass of an element expressed in grams (use the Periodic Table) The gram atomic mass of any element contains 1 mole of atoms of that element. 1 atom Carbon = 12 amu…atomic mass units 1 mole Carbon = 12 grams (6.02 x 10 23 atoms) Examples: The formula of hydrogen is H 2. What is the gram atomic mass? Note: The subscript = the # of atoms Pg. 179

6. Finding Formula Mass Element # of Atoms x Atomic Mass = Total Mass »Subscript use periodic table Examples: Find gram formula mass for: »Aluminum Chloride »Lithium Sulfide »Calcium Hydroxide

7. Relating Moles and Grams Moles = grams given Formula Mass »Use Periodic table to determine the formula mass of substance

8. The Molar Mass of a Substance Molar Mass- the mass in grams of a mole of any substance Mol = gr Fm Examples: 1.How many grams are in 9.45 mol of dinitrogen trioxide (N 2 O 3 )? 2.Find the number of moles are in 92.2 g of iron (III) oxide (Fe 2 O 3 ) 3.Find the number of grams for 2.5 moles of Calcium Bromide. 1. 718.2 gr. 2..578 moles 3. 500gr pg. 182

9. The Volume of a Mole of Gas The volume of gas changes when temperature or pressure changes. Because of the variation gas is measured in Standard Temperature and Pressure (STP) Standard temperature = 0 o C (273 0 Kelvin)(32 0 F) Standard pressure is 101.3 kPa (KiloPascals) -Molar Volume of gas= 22.4 L and is measured at STP -1 mole of any gas at STP will occupy 22.4 liters of volume Calculate the number of liters if you have16 grams of Sulfur Dioxide@STP

10. % Composition with Formulas Use the atomic masses on Periodic Table Find the % composition, by mass, for the formulas of : H 2 ONa 2 SO 4

11. CP CC Calculating the Percent Composition of a Compound Percent Composition- percent by mass of each element of a compound The percent composition of a compound has as many percent values as there are different elements in a compound grams of element A grams of compound Example: An 8.20-g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound? X 100% mass of element A=

12. Using Percent as a Conversion Factor You can use percent composition to calculate the number of grams of an element contained in a specific amount of a compound. Example: calculate the mass of a carbon in 82.0 g of propane (C 3 H 8 )

13. Calculating Empirical Formulas Empirical Formula- gives the lowest whole- number ratio of atoms (of the element) in a compound. Empirical formula for H 2 O 2 is HO Empirical formula for CO 2 is CO 2 Examples-Find the Empirical Formula if given: --6 gr of Magnesium and 4 gr of Oxygen --79.8 gr of Carbon and 20.2 gr of Hydrogen --67.6 gr.Mercury (Hg), 10.8 gr. Sulfur, 21.6 gr. Oxygen

14. Molecular Formulas Molecular Compounds- gives the actual ratio of atoms in a compound. This can be calculated once you have found the Empirical formula. Example: CO 2 Same molecular & Empirical C 6 H 12 O 6 = molecular CH 2 O = empirical Example: Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH 4 N

15. Calculating Molecular Formulas Steps to Solve: Find the Formula Mass of the Empirical Formula Divide total grams given Formula Mass of Empirical Multiply this answer by the Empirical Formula, this will equal the Molecular Formula

16. STOICHIOMETRY USING THE REACTION EQUATION LIKE A RECIPE

17. STOICHIOMETRY USING THE REACTION EQUATION LIKE A RECIPE

18. Nearly everything we use is manufactured from chemicals. –Soaps, shampoos, conditioners, cd’s, cosmetics, medications, and clothes. For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them. USING EQUATIONS

19. Chemical processes carried out in industry must be economical, this is where balanced equations help. Equations are a chemist’s recipe. –Eqs tell chemists what amounts of reactants to mix and what amounts of products to expect.

20. 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 meaning the amount of a substance in grams, liters, molecules, or moles. USING EQUATIONS

21. The calculation of quantities in chemical reactions is called stoichiometry. When you bake cookies you probably use a recipe. A cookie recipe tells you the amounts of ingredients to mix together to make a certain number of cookies. USING EQUATIONS

22. If you need a larger number of cookies than the yield of the recipe, the amounts of ingredients can be doubled or tripled. In a way, a cookie recipe provides the same kind of information that a balanced chemical equ. provides –Ingredients are the reactants –Cookies are the products.

23. Imagine you are in charge of manufacturing for Rugged Rider Bicycle Company. The business plan for Rugged Rider requires production of 128 custom-made bikes each day. One of your responsibilities is to be sure that there are enough parts available at the start of each day.

24. Assume that the major components of the bike are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P). The finished bike has a “formula” of FSW 2 HP 2. The balanced equation for the production of 1 bike is. F +S+2W+H+2P  FSW 2 HP 2

26. Now in a 5 day workweek, Rugged Riders is scheduled to make 640 bikes. How many wheels should be in the plant on Monday morning to make these bikes? What do we know? –Number of bikes = 640 bikes –1 FSW 2 HP 2 =2W (balanced eqn) What is unknown? –# of wheels = ? wheels

27. The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation. 640 FSW 2 HP 2 1 FSW 2 HP 2 2 W = 1280 wheels

28. We can derive the same kind of information from a normal chemical reaction equation. For instance the synthesis reaction of ammonia: N 2 (g) + 3H 2 (g)  2NH 3 (g) What kinds of information can we glean from this equation? Well for starters…

29. PARTICLES SUMMARY 1 molecule of N 2 reacts with 3 molecules of H 2 to produce 2 molecules of ammonia. N 2 and H 2 will always react to form ammonia in this 1:3:2 ratio of molecules. So if you started with 10 molecules of N 2 it would take 30 molecules of H 2 and would produce 20 molecules of NH 3

30. It isn’t possible to count such small numbers of molecules and allow them to react. You could react Avogadro’s number of N 2 molecules and make them react with 3 times Avogadro’s number of H 2 molecules forming 2 times Avogadro’s number of NH 3 molecules. PARTICLES SUMMARY

31. N2N2 +3H 2  2NH 3 2 atoms N+6 atoms H  2 atoms N & 6 atoms H 1 molecule N 2 +3 molecule H 2  2 molecule NH 3 10molecules N 2 +30molecules H 2  20 molecules NH 3 6.02X10 23 molecules N 2 + 6.02X10 23 molecules H 2  6.02X10 23 molecules NH 3 1X3X2X

32. MOLES SUMMARY We have recently learned that Avogadro’s number of particles is the same as a mole of a substance. On the basis of the particle interpretation we just discussed, the equation also tells you the number of moles of reactants and products.

33. 1 mole of N 2 molecules reacts with 3 moles of H 2 molecules to make 2 moles of NH 3 molecules. The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical reaction. MOLES SUMMARY

34. This is the most important information that a reaction equation provides. Using this information, you can calculate the amounts of reactants and products. MOLES SUMMARY

35. A balanced chemical equation must also obey the law of conservation of mass. –Mass can be neither created nor destroyed in ordinary chemical or physical processes. Remember that mass is related to the number of atoms in a compound through the mole. MASS SUMMARY

36. The mass of 1 mol of N 2 molecules is 28 g; the mass of 3 mols of H 2 molecules is 6 g for a total mass of reactnts of 34 g. The mass of 2 moles of NH 3 molecules is 2 * 17g or 34 g. As you can see the reactants mass is equal to the mass of the products. MASS SUMMARY

37. VOLUME SUMMARY Remember that 1 mole of any gas at STP occupies 22.4 L of space. It follows that 22.4 L of N 2 reacts with 67.2 L of H 2 to form 44.8 L of Ammonia gas.

38. 1 mol N 2 +3 mol N 2  2 mol NH 3 28 g N 2 +3 (2 g H 2 )  2 (17 g NH 3 ) 34 g reactants  34 g products +  22.4 L N 2 67.2 L H 2  44.8 L NH 3 22.4 L N2N2 +3H 2  2NH 3

39. USING EQUATIONS You can see how much information is stored in a simple balanced reaction eqn We can combine this information with our knowledge of mole conversions to perform important common stoichiometric calculations.

40. MOLE – MOLE CALCULATIONS A balanced rxn eqn is essential for all calculations involving amounts of reactants and products. If you know the number of moles of 1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn eqn.

41. Let’s go back to our synthesis of ammonia rxn. N 2 (g) + 3H 2 (g)  2NH 3 (g) The MOST important interpretation of this rxn is that 1 mole of N 2 reacts with 3 moles of H 2 to produce 2 moles of NH 3. MOLE – MOLE CALCULATIONS

42. These connections of the coefficients allows us to set up conversion factors called mole ratios. The mole ratios are used to calculate the connections in moles of compounds in our reaction equation. We can start calculating… MOLE – MOLE CALCULATIONS

43. Sample Mole – Mole problem: How many moles of ammonia are produced when.60 moles of N 2 are reacted with H 2 ? Given:.60 moles of N 2 Uknown: ____ moles of NH 3 N 2 (g) + 3H 2 (g)  2NH 3 (g) MOLE – MOLE CALCULATIONS

44. Sample Mole – Mole problem: How many moles of ammonia are produced when.60 moles of N 2 are reacted with H 2 ? Given:.60 moles of N 2 Uknown: ____ moles of NH 3 N 2 (g) + 3H 2 (g)  2NH 3 (g) MOLE – MOLE CALCULAT’NS

45. Sample Mole – Mole problem: How many moles of ammonia are produced when.60 moles of N 2 are reacted with H 2 ? Given:.60 moles of N 2 Uknown: ____ moles of NH 3 N 2 (g) + 3H 2 (g)  2NH 3 (g) MOLE – MOLE CALCULAT’NS

46. According to the reaction equation, for every 1 mole of N 2 reacted we form 2 mols of NH 3. To determine the number of moles of NH 3, the given quantity of N 2 is multiplied by the mole ratio from the rxn eqn in such a way that the units of “mol N 2 ” cancel MOLE – MOLE CALCULAT’NS

47. Solve for the unknown:.6 mol N 2 1 mol N 2 2 mol NH 3 = 1.2 mol NH 3 N 2 (g) + 3H 2 (g)  2NH 3 (g) MOLE – MOLE CALCULAT’NS

48. MOLE – MOLE EXAMPLE 2 This equation shows the formation of aluminum oxide. 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s) How many moles of aluminum are needed to form 3.7 mol Al 2 O 3 ? Given: 3.7 moles of Al 2 O 3 Uknown: ____ moles of Al

49. MOLE – MOLE EXAMPLE 2 This equation shows the formation of aluminum oxide. 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s) How many moles of aluminum are needed to form 3.7 mol Al 2 O 3 ? Given: 3.7 moles of Al 2 O 3 Uknown: ____ moles of Al

50. Solve for the unknown: 3.7 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al = 7.4 mol Al Coefficients in the balanced equation 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s) MOLE – MOLE CALCULAT’NS

51. MASS – MASS CALCULAT’NS No lab balance measures moles directly, instead the mass of a substance is usually measured 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.

52. The mole – mole connection is still vital to do these calcs. 1.If the given sample is measured in grams, the mass can be converted to moles by using the molar mass. 2.Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown. MASS – MASS CALCULAT’NS

53. 3.If it is mass of the unknown that needs to be determined, the number of moles of the unknown can be multiplied by the molar mass of the desired compound. As in mole-mole calcs, the unknow can be either a reactant or a product. MASS – MASS CALCULAT’NS

54. Again back to our synthesis of ammonia rxn. N 2 (g) + 3H 2 (g)  2NH 3 (g) Calculate the number of grams of NH 3 produced by the reaction of 5.4 g of H 2 with an excess of N 2. N 2 (g) + 3H 2 (g)  2NH 3 (g) MASS – MASS CALCULAT’NS

55. What do we know? –Mass of H 2 = 5.4 g H 2 –3 mol H 2 = 2 mol NH 3 (from balanced equation - AKA mole ratio) –Molar mass of H 2 = 2.0 g H 2 –Molar mass of NH 3 =17.0g NH 3 What are we asked for? Mass of ammonia produced MASS – MASS CALCULAT’NS

56. Step 1: convert mass of given to moles of given using MM of G 5.4 g H 2 2.0 g H 2 1 mol H 2 = 2.7 mol H 2 Coefficients in the balanced equation Molar mass of H 2 MASS – MASS CALCULAT’NS

57. Step 2: convert mols of G to mols of U using the mole ratio 2.7 mol H 2 3 mol H 2 2 mol NH 3 = 1.8 mol NH 3 Coefficients in the balanced equation Coefficients from the balanced equation N 2 (g) + 3H 2 (g)  2NH 3 (g) MASS – MASS CALCULAT’NS

58. Step 3: convert moles of desired compound to mass using MM of U 1.8 mol NH 3 1 mol NH 3 17.0 g NH 3 = 31.0 g NH 3 Coefficients in the balanced equation Molar mass of NH 3 MASS – MASS CALCULAT’NS

59. Mass to mass calculations always follow those same three steps. It uses the mole math that we have had lots of practice with (mass to moles and moles to mass) The only difference is the new middle step where we use our newly acquired mole ratio MASS – MASS CALCULAT’NS

60. Let’s do another one: Acetylene gas (C 2 H 2 ) is produced by adding water to calcium carbide (CaC 2 ). CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2 How many grams of acetylene are produced by adding water to 5.00 g CaC 2 ? CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2 MASS – MASS CALCULAT’ NS

61. What do we know? –Mass of CaC 2 = 5.0 g CaC 2 –1 mol CaC 2 = 1 mol C 2 H 2 (from balanced equation) –MM of CaC 2 = 64.0 g CaC 2 –MM of C 2 H 2 = 26.0g C 2 H 2 What are we asked for? –Mass of C 2 H 2 produced MASS – MASS CALCULAT’NS

62. Step 1: convert mass of given to moles of given using MM of G 5.0 g CaC 2 64.0 g CaC 2 1 mol CaC 2 =.078mol CaC 2 MASS – MASS CALCULAT’NS

63. Step 2: convert mols of G to mols of U using the mole ratio.078mol CaC 2 1 mol CaC 2 1 mol C 2 H 2 =.078mol C 2 H 2 CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2 MASS – MASS CALCULAT’NS

64. Step 3: convert moles of desired compound to mass using MM of U.078 mol C 2 H 2 1 mol C 2 H 2 26.0 g C 2 H 2 = 2.03 g C 2 H 2 MASS – MASS CALCULAT’NS

65. MORE MOLE CALCULAT’NS A balanced reaction equation indicates the relative number of moles of reactants and products. We can expand our stoichiometric calculations to include any unit of measurement that is related to the mole.

66. The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP. The problems can include mass- volume, volume-volume, and particle-mass calculations. MORE MOLE CALCULAT’NS

67. In any of these problems, the given quantity is first converted to moles. Then the mole ratio from the balanced eqn is used to convert from the moles of given to the number of moles of the unknown MORE MOLE CALCULAT’NS

68. Then the moles of the unknown are converted to the units that the problem requests. The next slide summarizes these steps for all typical stoichiometric problems MORE MOLE CALCULAT’NS

70. MORE MOLE EXAMPLES How many molecules of O 2 are produced when a sample of 29.2 g of H 2 O is decomposed by electrolysis according to this balanced equation: 2H 2 O  2H 2 + O 2

71. What do we know? –Mass of H 2 O = 29.2 g H 2 O –2 mol H 2 O = 1 mol O 2 (from balanced equation) –MM of H 2 O = 18.0 g H 2 O –1 mol O 2 = 6.02X10 23 molecules of O 2 What are we asked for? – molecules of O 2 MORE MOLE EXAMPLES

72. Step 1: convert mass of given to moles of given using MM of G 29.2 g H 2 O 18.0 g H 2 O 1 mol H 2 O = 1.62 mol H 2 O

73. Step 2: convert mols of G to mols of U using the mole ratio 1.62 mol H 2 O 2 mol H 2 O 1 mol O 2 =.811 mol O 2 2H 2 O  2H 2 + O 2

74. Step 3: convert moles of unknown compound to desired units.811 mol O 2 1 mol O 2 6.02X10 23 = 4.88X10 23 molecules O 2

75. The last step in the production of nitric acid is the reaction of NO 2 with H 2 O. 3NO 2 +H 2 O  2HNO 3 +NO How many liters of NO 2 must react with water to produce 5.00X10 22 molecules of NO? MORE MOLE EXAMPLES

76. What do we know? –Molecs NO=5.0X10 22 molecs NO –1 mol NO = 3 mol NO 2 (from balanced equation) –1 mol NO=6.02X10 23 molecs NO –1 mol NO 2 = 22.4 L NO 2 What are we asked for? – Liters of NO 2 MORE MOLE EXAMPLES

77. Step 1: convert molecules of given to moles of given using Avogadro’s number. 5.0X10 22 molecs NO 6.02X10 23 molecs 1 mol NO =.083 mol NO

78. Step 2: convert mols of G to mols of U using the mole ratio.083 mol NO 1 mol NO 3 mol NO 2 =.249 mol NO 2 3NO 2 +H 2 O  2HNO 3 +NO

79. Step 3: convert moles of unknown compound to desired units.249 mol NO 2 1 mol NO 2 22.4 L NO 2 = 5.58 L of NO 2

80. Assuming STP, how many liters of oxygen are needed to produce 19.8 L SO 3 according to this balanced equation? 2SO 2 +O 2  2SO 3 MORE MOLE EXAMPLES

81. What do we know? –Volume of SO 3 = 19.8 L –2 mol SO 3 = 1 mol O 2 (from balanced equation) –1 mol SO 3 = 22.4 L SO 3 –1 mol O 2 = 22.4 L O 2 What are we asked for? – volume of oxygen MORE MOLE EXAMPLES

82. Step 1: convert volume of G to moles of G 19.8 L SO 3 22.4 L SO 3 1 mol SO 3 =.884 mol SO 3

83. Step 2: convert mols of G to mols of U using the mole ratio.884 mol SO 3 2 mol SO 3 1 mol O 2 =.442 mol O 2 2SO 2 +O 2  2SO 3

84. Step 3: convert moles of unknown compound to desired units.442 mol O 2 1 mol O 2 22.4 L O 2 = 9.9 L of O 2 MORE MOLE EXAMPLES