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____________. Chapter 3. Objectives. Atomic Mass. The Mole Concept. Molar Mass. Percentage Composition. Empirical & Molecular Formulas. Chemical Reactions. Balancing Chemical Equations. Mass - Mass Calculations. Limiting Reactant. Percent Yield. Chapter 3. Key Terms.

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  1. ____________ Chapter 3 Objectives • Atomic Mass • The Mole Concept • Molar Mass • Percentage Composition • Empirical & Molecular Formulas • Chemical Reactions • Balancing Chemical Equations • Mass - Mass Calculations • Limiting Reactant • Percent Yield

  2. Chapter 3 Key Terms Atomic mass is the mass of an atom in atomic mass units. The atomic mass unit is exactly 1/12 of the mass of the carbon-12 isotope, which has been assigned the mass exactly 12.00000 amu. The average atomic mass of all naturally occurring isotopes of a given element is called the atomic weight of the element. This number is unique for each element.

  3. Use the Periodic Table to identify the atomic weight of each element and then round off to the nearest tenth. Example #1 28.0855 28.1 silicon = sodium 22.98977 23.0 = = chromium 51.996 52.0 magnesium 24.305 = 24.3

  4. The Mole Concept A moleof an element is that quantity of the element, which consists of its atomic weight expressed in grams. 1 mol Ag = 107.868 grams 1 mol sulfur = 32.066 grams

  5. A mole of an element contains the Avogadro Number of atoms. This number is equal to 6.022 x 1023 atoms. 1 mol Ag = 107.9 g = 6.022 x 1023 atoms 1 mole = gram atomic weight 1 mole = 6.022 x 1023 atoms 1 mol Sn= 118.7 g = 6.022 x 1023 atoms 1 mol K = 39.1 g = 6.022 x 1023 atoms

  6. Example #2 A sample of sulfur weighs 1.28 grams. How many moles of sulfur is this? 1.28 g S = moles S 1 mol 1.28 g S x 32.1 g Answer = 0.0399 mol S

  7. Example #3 83.5 grams C = atoms 1 mol 6.022 X 1023 atoms 83.5 g C x x 12.0 g 1 mol 4.19 x 1024 C atoms

  8. ____ ___ The formula weight of a compound is the sum of the atomic masses of all atoms represented in the formula of the compound. The formula weight of a compound expressed in grams is called the molar mass of the compound. 1 mol = g-FW = molar mass 1 mol = 6.022 X 1023 molecules

  9. Example #4 Calculate the molar mass of caffeine, C8H10N4O2. MM C8H10N4O2 = 8(12.0) + 10(1.0) + 4(14.0) + 2(16.0) = Answer: 194.0 g/mole

  10. Example #5 How many moles of Na2CO3 are represented by 1.00 x 103 grams of the compound? 1.00 x 103 g Na2CO3 = mol Na2CO3 1 mol 1.00 x 103 g x = 9.43 mol Na2CO3 106.0 g MM Na2CO3 = 2(23.0) + 12.0 + 3(16.0) = 106.0 g/mol

  11. _________ __________ Percentage composition refers to the amount of each element in a compound expressed as a percent. ____ __ _______ = _______ _ ____ __ ________

  12. Example #6 Calculate the percentage composition of ammonia, NH3. MM NH3 = 14.0 + 3(1.0) = 17.0 14.0 %N = X 100 = 82.4% 17.0 3.0 %H = X 100 = 17.6% 17.0 or % H = 100% - 82.4% = 17.6%

  13. Example #7 What is the percentage of calcium in calcium acetate? 40.1 %Ca = X 100 = 25.4% 158.1 MM Ca(C2H3O2)2 = 40.1 + 4(12.0) + 6(1.0) + 4(16.0) = 158.1 g/mol

  14. Example #8 How many grams of silver can be obtained from a 17.85 g sample of silver nitrate, AgNO3? = g Ag 17.85 g AgNO3 107.9 g Ag 17.85 g AgNO3 x = 11.34 g Ag 169.9 g AgNO3 MM AgNO3 = 107.9 + 14.0 + 3(16.0) = 169.9 g/mol

  15. Hydrates A hydrate is a compound that contains water as part of its chemical composition. CuSO4 5H2O Cupric sulfate, pentahydrate Na2CO3 10H2O Sodium carbonate, decahydrate The water in the hydrate increases its molar mass.

  16. Example #9 Calculate the percentage of water in nickel(II) chloride, hexahydrate. 108.0 %H2O = X 100 = 237.7 45.44% H2O 108.0 H2O MM NiCl2 6H2O = 58.7 + 2(35.5) + 6( 18.0) = 237.7 g/mol

  17. Empirical & Molecular Formulas The empirical formula of a compound indicates the simplest whole number mole ratio of atoms in a molecule. The molecular formula of a compound shows the actual number of atoms in a molecule. For benzene the molecular formula is C6H6. Its empirical formula would be CH, showing the one to one ratio between the atoms of carbon and hydrogen.

  18. Relationship Between Molecular Formulas and Empirical Formulas Molecular formula Butene = C4H8 Empirical formula Butene = CH2 So that C4H8 = (CH2)4 Structural formula In general then MF = (EF)n MM where n = EFW The molecular formula subscripts are always a whole number multiple of the empirical formula subscripts.

  19. Rules for Calculating the Empirical Formula of a Compound • 1. Calculate the number of moles of each element. (Conversion Problems) • 2. Calculate the whole number mole ratio for each element by dividing by the smallest number of moles in part 1. • 3. Use the whole number mole ratios as the subscripts in the empirical formula.

  20. Example #10 Calculate the empirical formula for the compound that consists of 77.73% iron and 22.27% oxygen. 1 mol 77.73 g Fe x = 1.392 mol =1 g 55.85 1.392 mol 1 mol 22.27 g O x = 1.392 mol = 1 g 16.00 1.392 mol EF = FeO

  21. Example #11 Calculate the empirical formula of the compound which consists of 10.91 g phosphorus and 14.09 g oxygen. 1 mol = 0.3523 mol 10.91 g P x =1 x 2 = 2 30.97 g 0.3525 mol 1 mol = 0.8806 mol 14.09 g O x = 2.50 x 2 = 5 16.00 g 0.3525 mol EF = P2O5

  22. Example #12 If the molar mass of the compound in example # 11 is 283.88 g/mol, what is it’s molecular formula? From example #11: EF = P2O5 EFW P2O5 = 2(30.97) + 5(16.00) = 141.94 MM 283.88 Since MF = (EF)n and n= = = 2 EFW 141.94 The MF = (P2O5) = P4O10 2

  23. Chemical Reactions A chemical reaction is a change in matter in which new substances with new properties are formed. A beginning material in a chemical reaction is called a reactant. Any material formed as a result of the chemical change is called a product of the reaction. yield reactants products form produce change to

  24. Chemical Equations A word equation represents the reactants and the products in a chemical reaction with their names. aluminum(s) + oxygen(g) aluminum oxide(s) A chemical equation represents the reactants and products in a chemical reaction with their symbols and formulas. 4Al(s) + 3O2(g) 2Al2O3(s) (s) = solid (g) = gas (l) = liquid (aq) = aqueous

  25. Writing Chemical Equations • Determine the reactants, the products, and the physical states involved. • Use the correct symbols or formulas of the reactants and products to write the unbalanced chemical equation. • Balance the equation by applying the law of conservation of matter to determine the correct coefficients.

  26. The Law of Conservation of Matter The law of conservation of matter states that matter cannot be created nor destroyed in any chemical reaction. Implication: The number of atoms of reactants must be equal to the number of atoms of products.

  27. Balancing Equations Balancing an equation is the process of using multiplication to make the number of atoms of products equal to the number of atoms of reactants. A coefficient is a number placed in front of some chemical species to represent that number of chemical units. iron(s) + oxygen(g) iron(III) oxide(s) Fe(s) + O2(g) Fe2O3(s) 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

  28. The Diatomic Elements The following elements exist in the gaseous state as diatomic (two atom) molecules and not as single atom substances. F fluorine(g) = O 2 oxygen(g) = 2 Cl chlorine(g) = 2 H hydrogen(g) = 2 Br bromine(g) = 2 nitrogen(g) = N 2 I iodine(g) = 2

  29. Examples: Balance each equation. #13. Hydrogen gas burns in oxygen gas to produce water vapor. 2 H2(g) + O2(g) 2 H2O(g) #14. Pure iridium metal reacts with oxygen gas to produce solid iridium(III) oxide. 4 Ir(s) + 3 O2(g) 2 Ir2O3(s) #15. Butane gas, (C4H10), burns in air to produce carbon dioxide and water vapor. 2( C4H10(g) + 13 O2(g) 4 CO2(g) + 5 H2O(g) ) 2 2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)

  30. Reaction Stoichiometry The stoichiometryof a chemical reaction deals with calculations involving the amounts of reactants and products in the reaction. Such calculations are often referred to as mass-mass problems if only masses are involved. These problems can also be referred to as mass-volume, volume-mass, or volume-volume problems depending on the data provided by the problem.

  31. Mole Relationship in a Chemical Reaction The coefficients of a balanced chemical equation indicate the mole ratio between any reactant and/or product. 4Fe(s) + 3O2(g) 2Fe2O3(s) Implication 4 moles iron react with 3 moles of oxygen to produce exactly 2 moles of ferric oxide.

  32. Refer to the equation described below to answer the following oral questions. 4Fe(s) + 3O2(g) 2Fe2O3(s) 4 mol 3 mol 2 mol 8 mol 6 mol 4 mol 2 mol 1.5 mol 1 mol 12 mol 9 mol 6 mol Too much oxygen 4 mol 6 mol 2 mol 3 mol in excess Too much oxygen 6 mol 7 mol 3 mol 2.5 mol in excess Only 4.5 mol oxygen used

  33. Process for Solving Mass-Mass Problems 1. Write a balanced chemical equation. 2.Convert the given data to moles using the molar mass of the given substance. 3. Use the coefficients of the balanced equation to convert moles given to molesasked for. 4. Use the molar mass of the substanceasked for to convert to grams. grams given mol mol asked grams molar mass coefficients molar mass

  34. Example #16 Co(s) + 2 HCl(aq) CoCl2(aq) + H2(g) Cobalt metal reacts with hydrochloric acid according to the following unbalanced equation. If you begin with 2.56 g of cobalt metal and excess hydrochloric acid, what mass of cobalt(II) chloride can be obtained? = g CoCl2 5.65 g CoCl2 2.56 g Co 1 mol 1 mol CoCl2 129.9 g 2.56 g Co x x x = 58.9 g 1 mol Co 1 mol MM CoCl2 = 58.9 + 2( 35.5) = 129.9 g/mol

  35. Example #17a C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) Propane, C3H8, can be used as a fuel in your home, car, or barbeque grill because it is easily liquefied and transported. If 454 g of propane is burned, [a] what mass of oxygen is required and how much [b] carbon dioxide and [c] water are formed? = g O2 1,650 g O2 454 g C3H8 1 mol 5 mol O2 32.0 g 454 g C3H8 x x x = 44.0 g 1 mol C3H8 1 mol MM C3H8 = 3( 12.0) + 8( 1.0) = 44.0 g/mol MM O2 = 2(16.0) = 32.0 g/mol

  36. Example #17b Student Presentation 1,360 g CO2

  37. Example #17c Student Presentation 743 g H2O

  38. _______ _______ ___ __________ ____ • The reactant determining the amount of productformed in a chemical reaction is called the limitingreactant. • The other reactant(s) is/are said to be in excess. • The amount of product that should be formed in a chemical reaction is called theoretical yield. • Theoretical yield also corresponds to the maximumamount of product possible.

  39. Example #18 A tool set consists of 4 wrenches, 3 screwdrivers, and 2 pliers. The manufacturer has in stock 1,000 pliers, 2,000 screwdrivers, and 1,500 wrenches. A.) Can an order for 500 tool sets be filled out? No! 375 B.) How many tool sets can be made? C.) How many of which tools will remain after the maximum number of tool sets are made? 875 screwdrivers 0 wrenches 250 pliers

  40. 1,000 pliers / 2 pliers per set = 500 sets 2000 screwdrivers / 3 screwdrivers per set = 666 sets 1500 wrenches / 4 wrenches per set = 375 sets Theoretical yield Once 375 sets have been put together, all the wrenches will be used up so that 0 wrenches will remain. Limiting reactant More sets cannot be made even though some pliers and screwdrivers remain. Reactants in excess Making the 375 tool sets with 3 screwdrivers per set will require only 1,125 screwdrivers and this means that (2,000 - 1,125) = 875 screwdrivers will remain. Reactants in excess that remain Making the 375 tool sets with 2 pliers per set, uses up 750 pliers. This means that (1,000 - 750) = 250 pliers will remain.

  41. Example #19 Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperature. The other products are solid copper and water vapor. If a sample containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed? 2 NH3(g) + CuO(s) 3 N2(g) + Cu(s) + H2O(g) 3 3 L. R. g N2 produced 1 mol 1 mol N2 18.1 g NH3 x x = 0.5324 mol N2 17.0 g 2 mol NH3 1 mol 1 mol N2 28.0 g 90.4 g CuO x L.R. = CuO x = 0.3790 mol N2 x = 79.5 g 3 mol CuO 1 mol 10.6 g N2 MM NH3 = 14.0 + 3( 1.0) = 17.0 g/mol Choose smaller value to determine the L. R. MM CuO = 63.5 + 16.0 = 79.5 g/mol MM N2 = 2(14.0) = 28.0 g/mol

  42. ______ ____ The maximum amount of product possible, theoreticalyield, is seldom obtained in a chemical reaction due to side reactions and other complications. The actual yield of product is the amount of product that is obtained in a chemical reaction and it is always less than the theoretical yield. The actual yield of product is often expressed as the percent yield. _____ ____ _ ____

  43. Example #20 Methanol, CH3OH, is used as a fuel in race cars and is a potential replacement for gasoline. It can be prepared by combining gaseous carbon monoxide with hydrogen gas. A student reacts 68.5 g CO(g) with 8.60 g H2(g). Calculate the T.Y. of methanol and if 35.7 g CH3OH are actually produced, what is the %yield of methanol? Choose smaller number as the T. Y. 2 H2(g) + CO(g) CH3OH(l) = = 8.60 g H2 g CH3OH 68.5 g CO g CH3OH 1 mol 1 mol CH3OH 32.0 g = 8.60 g H2 x x x 68.8 g CH3OH 2.0 g 2 mol H2 1 mol 1 mol 1 mol CH3OH 32.0 g = 68.5 g CO x x x 78.3 g CH3OH 28.0 g 1 mol CO 1 mol T.Y. = 68.8 g CH3OH MM H2 = 2(1.0) = 2.0 g/mol MM CH3OH = 12.0 + 4(1.0) + 16.0 = 32.0 g/mol MM CO = 12.0 + 16.0 = 28.0 g/mol

  44. Example #21 _____ ____ ____ _ 35.7 g %Yield = X 100 68.8 g %Yield = 51.9%

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