3 - Stoichiometry Lecture

3 - Stoichiometry Lecture - Stoichiometry Chapter 3 Atomic...

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Unformatted text preview: Stoichiometry Chapter 3 Atomic Mass Unit Golf ball = 46.0 g set mass = 1 Baseball = 149 g 149/46.0 = 3.24 Volleyball = 270 g 270/46.0 = 5.87 Atomic Mass Unit (amu) A convenient unit for describing the mass of an atom The mass of a carbon12 atom = 12 amu 1 amu = 1.660 x 1024 g 1 amu approximately = mass of a proton 1 amu approximately = mass of a neutron Isotopes Oxygen has three natural isotopes O16 99.757% O17 0.038% O18 0.205% Breathe in all three isotopes in the above ratios Atomic Mass The weighted average mass of an element's atoms To calculate: Atomic Mass = Sum [(Abundance) x (Isotopic Mass)] Atomic Mass The weighted average mass of an element's atoms To calculate: Atomic Mass = Sum [(Abundance) x (Isotopic Mass)] Example: From O-16: (0.99757)(15.99949) = 15.9606 amu O-17: (0.00038)(16.99913) = 0.00646 amu O-18: (0.00205)(17.99916) = 0.03690 amu 16.0040 amu Atomic Mass Naturally occurring copper is a mixture of 69.17% 63Cu with a mass of 62.93 amu and 30.83% 65Cu with a mass of 64.93 amu. What is the atomic mass of copper? Atomic Mass Naturally occurring copper is a mixture of 69.17% 63Cu with a mass of 62.93 amu and 30.83% 65Cu with a mass of 64.93 amu. What is the atomic mass of copper? Atomic mass = Atomic Mass Naturally occurring copper is a mixture of 69.17% 63Cu with a mass of 62.93 amu and 30.83% 65Cu with a mass of 64.93 amu. What is the atomic mass of copper? Atomic mass = (0.6917)(62.93) = 43.53 Atomic Mass Naturally occurring copper is a mixture of 69.17% 63Cu with a mass of 62.93 amu and 30.83% 65Cu with a mass of 64.93 amu. What is the atomic mass of copper? Atomic mass = (0.6917)(62.93) = 43.53 (0.3083)(64.93) = 20.02 Atomic Mass Naturally occurring copper is a mixture of 69.17% 63Cu with a mass of 62.93 amu and 30.83% 65Cu with a mass of 64.93 amu. What is the atomic mass of copper? Atomic mass = (0.6917)(62.93) = 43.53 (0.3083)(64.93) = 20.02 63.55 Mole Mole = 6.02214 x 1023 atoms Avogadro's Number mol = # of C12 atoms in exactly 12 grams of pure C12 1 mole of an element = that elements atomic mass Units Dozen = 12 items Gross = 144 items Mole of C = 12 g C12 Mole of O = 16 g O Mole of ? = AW g of ? Mole How large is that number??? Mole How large is that number??? Travel along a mole of hydrogen atoms (6.022 10 23 -10 m 10 13 atoms) = 6.022 10 m atom Mole Travel along a mole of hydrogen atoms -10 m 10 23 13 (6.022 10 atoms) = 6.022 10 m atom 1km mi (6.022 10 m) = 3.74 x1010 mi 1000 m 1.609km 13 3.74 billion miles!!! Avogadro's Number Amount of water in world's oceans (liters) Age of Earth in seconds Population of earth 602,200,000,000,000,000,000,000 Average college tuition Distance from earth to sun (cm) Avogadro's Number in Calculations Calculate the mass of 500 atoms of iron (Fe). Avogadro's Number in Calculations Calculate the mass of 500 atoms of iron (Fe). Atomic mass of Fe = 55.85 g/mol Avogadro's Number in Calculations Calculate the mass of 500 atoms of iron (Fe). Atomic mass of Fe = 55.85 g/mol 500 atoms x 1 mol____ = 8.3029 x 1022 mol 6.022 x 1023atoms Avogadro's Number in Calculations Calculate the mass of 500 atoms of iron (Fe). Atomic mass of Fe = 55.85 g/mol 500 atoms x 1 mol____ = 8.3029 x 1022 mol 6.022 x 1023atoms 8.3029 x 1022 mol x 55.85 g_ = 4.63716 x 1020 g 1 mol Avogadro's Number in Calculations Calculate the mass of 500 atoms of iron (Fe). Atomic mass of Fe = 55.85 g/mol 8.3029 x 1022 mol x 55.85 g_ = 4.63716 x 1020 g 1 mol Rounds to 4.64 x 1020 g Molar Mass Mass in grams of one mole of compound Number of grams/ mole Relates mass to moles Molar mass of CaCO3 = ? Molar Mass CaCO3 40.078g Ca = mol Molar Mass CaCO3 40.078g Ca = mol 12.01107g C= mol Molar Mass CaCO3 40.078g Ca = mol 12.01107g C= mol O= 15.9994g mol Molar Mass CaCO3 40.078g Ca = mol 12.01107g C= mol O= 15.9994g mol MM CaCO3 40.078 gCa 12.01107 gC 15.9994 gO = 1 + + 1 3 mol mol mol Molar Mass CaCO3 40.078g Ca = mol 12.01107g C= mol O= 15.9994g mol MM CaCO3 40.078 gCa 12.01107 gC 15.9994 gO = 1 + + 1 3 mol mol mol = 100.08727 g CaCO3 / mol = 100.09 g CaCO3 / mol Molar Mass Example #2 A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg. How many silicon (Si) atoms are present in this chip? Molar Mass Example #2 A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg. Steps to solving: Molar Mass Example #2 A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg. Steps to solving: Convert mg to g g to mol mol to atoms Molar Mass Example #2 A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg. Steps to solving: Convert mg to g 1 g Si 5.68mg Si = 5.68 x 10-3 g Si 1000 mg Si Molar Mass Example #2 A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg. Steps to solving: g to mol 1 mol Si 5.68 x 10 g Si = 2.02 x 10-4 mol Si 28.0855 g Si -3 Molar Mass Example #2 A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg. Steps to solving: mol to atoms 6.022 x 1023 atoms 2.02 x 10-4 mol Si = 1.22 x 1020 atoms 1 mol Si Molar Mass Example #3 Cobalt (Co) is a metal that is added to steel to improve its resistance to corrosion. Calculate both the number of moles in a sample of cobalt containing 5.00 x 1020 atoms and the mass of the sample. Molar Mass Example #3 Calculate both the number of moles in a sample of cobalt containing 5.00 x 1020 atoms and the mass of the sample. Steps to solving: Molar Mass Example #3 Calculate both the number of moles in a sample of cobalt containing 5.00 x 1020 atoms and the mass of the sample. Steps to solving: Convert atoms to mol Molar Mass Example #3 Calculate both the number of moles in a sample of cobalt containing 5.00 x 1020 atoms and the mass of the sample. Steps to solving: Convert atoms to mol mol to g Molar Mass Example #3 Calculate both the number of moles in a sample of cobalt containing 5.00 x 1020 atoms and the mass of the sample. Steps to solving: Convert atoms to mol 20 5.00 x 10 1 mol Co atoms Co = 8.30 x 10-4 mol Co 6.022 x 1023 atoms Co Molar Mass Example #3 Calculate both the number of moles in a sample of cobalt containing 5.00 x 1020 atoms and the mass of the sample. Steps to solving: mol to g 58.9332 g Co 8.30 x 10 mol Co = 4.89 x 10-2 g Co 1 mol Co -4 Percent by Mass Multiply element's molar mass by its subscript Divide by the compound's molar mass Multiply by 100 (subscript #)(element's molar mass) % of element = x 100 compounds molar mass Percent by Mass Example #1 Bicarbonate of soda (sodium hydrogen carbonate) is used in many commercial preparations. Its formula is NaHCO3. Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate. Percent by Mass Example #1 NaHCO3. Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate. Steps to solving: Look up molar mass of each element mol to mass Calc molar mass of compound Calc mass percentages of elements Percent by Mass Example #1 NaHCO3. Steps to solving: Look up molar mass of each element Na 22.99 g/mol Percent by Mass Example #1 NaHCO3. Steps to solving: Look up molar mass of each element Na 22.99 g/mol H 1.008 g/mol Percent by Mass Example #1 NaHCO3. Steps to solving: Look up molar mass of each element Na 22.99 g/mol H 1.008 g/mol C 12.01 g/mol Percent by Mass Example #1 NaHCO3. Steps to solving: Look up molar mass of each element Na 22.99 g/mol H 1.008 g/mol C 12.01 g/mol O 16.00 g/mol Percent by Mass Example #1 NaHCO3. Steps to solving: mol to mass Na 22.99 g/mol 22.99 g Percent by Mass Example #1 NaHCO3. Steps to solving: mol to mass Na 22.99 g/mol H 1.008 g/mol 22.99 g 1.008 g Percent by Mass Example #1 NaHCO3. Steps to solving: mol to mass Na 22.99 g/mol H 1.008 g/mol C 12.01 g/mol 22.99 g 1.008 g 12.01 g Percent by Mass Example #1 NaHCO3. Steps to solving: mol to mass Na 22.99 g/mol H 1.008 g/mol C 12.01 g/mol O 3(16.00) g/mol 22.99 g 1.008 g 12.01 g 48.00 g Percent by Mass Example #1 NaHCO3. Steps to solving: Calc molar mass of compound 22.99 g + 1.008 g + 12.01g + 48.00 g = 84.01g Percent by Mass Example #1 NaHCO3. Steps to solving: Calc mass percentages of elements Na 22.99 g/84.01 g x 100 = 27.36% Percent by Mass Example #1 NaHCO3. Steps to solving: Calc mass percentages of elements Na 22.99 g/84.01 g x 100 = 27.36% H 1.008 g/84.01 g x 100 = 1.20% Percent by Mass Example #1 NaHCO3. Steps to solving: Calc mass percentages of elements Na 22.99 g/84.01 g x 100 = 27.36% H 1.008 g/84.01 g x 100 = 1.20% C 12.01 g/84.01 g x 100 = 14.30% Percent by Mass Example #1 NaHCO3. Steps to solving: Calc mass percentages of elements Na 22.99 g/84.01 g x 100 = 27.36% H 1.008 g/84.01 g x 100 = 1.20% C 12.01 g/84.01 g x 100 = 14.30% O 48.00 g/84.01 g x 100 = 57.14% Empirical Formula Formulas represent the symbolic language of chemistry Empirical Formula: the SIMPLEST WHOLE NUMBER ratio of the various atoms in a compound. Molecular Formula: specifies the EXACT FORMULA of the molecule Empirical Formula H2F10 Molecular Formula Empirical Formula Benzene C 6 H6 (molecular formula) CH (empirical formula) Empirical Formula Twohanded person Threehanded person H2F10 H3F15 Empirical Formula Benzene Acetylene C 6 H6 C2H2 CH Empirical Formula Calc. Determine the empirical and molecular formula for a compound that gives the following analysis (in mass percents) 71.65% Cl 24.27% C 4.07% H The molecular weight (molar mass) is known to be 98.96 Empirical Formula Calc. 71.65% Cl 24.27% C 4.07% H The molecular weight is known to be 98.96 Steps to solving: Percent to mass Mass to mol Divide by small Multiply by whole Empirical Formula Calc. 71.65% Cl 24.27% C 4.07% H The molecular weight is known to be 98.96 Steps to solving: Percent to mass Assume 100 g sample 71.65 g Cl 24.27 g C 4.07 g H Empirical Formula Calc. 71.65% Cl 24.27% C 4.07% H The molecular weight is known to be 98.96 Steps to solving: Mass to mol mol 71.65 g Cl = 2.021 mol Cl 35.4527 g Cl mol 24.27 g C = 2.021 mol C 12.011 g C mol 4.07 g H = 4.04 mol H 1.008 g H Empirical Formula Calc. 71.65% Cl 24.27% C 4.07% H The molecular weight is known to be 98.96 Steps to solving: Divide by small 1 2.021 mol Cl = 1 2.021 mol 1 2.021 mol C = 1 2.021 mol 1 4.04 mol H = 2 2.021 mol Empirical Formula Calc. 71.65% Cl 24.27% C 4.07% H The molecular weight is known to be 98.96 Steps to solving: Multiply by whole Cl = 1 C = 1 CH2Cl H = 2 Empirical Formula Calc Empirical Formula CH Cl 2 Molecular Formula Compare weight of empirical formula to molecular formula Empirical formula weight 1( 35.453) + 1( 12.011) + 2 ( 1.008 ) = 49.48 g / mol Empirical Formula Calc Empirical Formula CH Cl 49.48 g/mol 2 Molecular Formula molecular weight 98.96 = =2 empirical formula weight 49.48 C2H4Cl2 Empirical Formula Calc Empirical Formula CH Cl 49.48 g/mol 2 Molecular Formula C2H4Cl2 Empirical Formula Calc. #2 A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. The compound has a molecular weight (molar mass) of 283.88. What are the compound's empirical and molecular formula? Empirical Formula Calc. #2 A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. Step 1: Percent to mass Empirical Formula Calc. #2 A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. Step 1: Percent to mass 43.64 g P 56.36 g O Empirical Formula Calc. #2 A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. Step 2: Mass to mol mol 43.64 g P = 1.409 mol P 30.97 g P mol 56.36 g O = 3.523 mol O 15.999 g O Empirical Formula Calc. #2 A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. Step 3: Divide by small 1 1.409 mol P = 1P 1.409 1 3.523 mol O = 2.5 O 1.409 Empirical Formula Calc. #2 A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. Step 4: Multiply till whole 1P x2=2P 2.5 O x 2 = 5 O P2O5 Empirical Formula Calc. #2 P 2 O5 The compound has a molecular weight of 283.88. What are the compound's empirical and molecular formula? Empirical Formula Calc. #2 P 2 O5 Calculate empirical formula weight 2(30.974 g/mol P) + 5(15.999 g/mol O) = 141.94 g/mol Empirical Formula Calc. #2 P 2 O5 Calculate empirical formula weight Compare the empirical formula weight to molecular weight (molar mass) molecular weight 283.88 = =2 empirical formula weight 141.94 Empirical Formula Calc. #2 P 2 O5 Calculate empirical formula weight Compare the empirical formula weight to molecular weight (P2O5) x 2 = P4O10 Chemical Reactions Chemical Reactions 2 1 1/4 1 Chemical Reactions Sodium bicarbonate sodium carbonate + water + carbon dioxide Heat Chemical Reactions Sodium bicarbonate sodium carbonate + water + carbon dioxide Heat 2 NaHCO3 uuuuu Na 2 CO3 +H 2 O+CO 2 Heat r Chemical Reactions Sodium bicarbonate sodium carbonate + water + carbon dioxide Heat 2 NaHCO3 uuuuu Na 2 CO3 +H 2 O+CO 2 Heat r Reactants Products Chemical Reactions Sodium bicarbonate sodium carbonate + water + carbon dioxide Heat 2NaHCO Reactants 3 Heat uuuuu Na 2 CO3 +H 2 O+CO 2 r Products Chemical Reactions Sodium bicarbonate sodium carbonate + water + carbon dioxide Coefficient Heat 2NaHCO Reactants 3 Heat uuuuu Na 2 CO3 +H 2 O+CO 2 r Products Chemical Reactions 2NaHCO 3 two units of sodium bicarbonate 2 Na atoms 2 H atoms 2 C atoms 6 O atoms Chemical Reactions Sodium bicarbonate sodium carbonate + water + carbon dioxide Heat 2NaHCO Reactants 3 Heat uuuuu Na 2 CO3 +H 2 O+CO 2 r Products Chemical Reactions Solids Liquids Gases Aqueous solution Catalyst (s ) (l ) (g ) (aq ) over arrow Chemical Reactions Sodium bicarbonate sodium carbonate + water + carbon dioxide Heat 2 NaHCO3 ( s) uuuuu Na 2 CO3 ( s ) + H 2 O(l ) + CO2 ( g ) Heat r Balancing Chemical Reaction Reaction used in extracting lead metal from its ores 2 PbS ( s ) + 3O2 ( g ) 2 PbO ( s ) + 2 SO2 ( g ) Balancing Chemical Reaction Reaction used in extracting lead metal from its ores 2 PbS ( s ) + 3O2 ( g ) 2 PbO ( s ) + 2 SO2 ( g ) Right Side 2 Pb 2 S 2 O + (2 x 2)O = 6 O Left side 2 Pb 2 S (3 x 2) O = 6 O Balancing Chemical Reaction Step 1: Write an unbalanced equation, using the correct formulas for all reactants and products. Step 2: Add appropriate coefficients to balance the numbers of atoms of each element. Balancing Chemical Reaction Step 1: Write an unbalanced equation, using the correct formulas for all reactants and products. Step 2: Add appropriate coefficients to balance the numbers of atoms of each element. H 2 SO4 + NaOH Na 2 SO4 + H 2 O Balancing Chemical Reaction Step 1: Write an unbalanced equation, using the correct formulas for all reactants and products. Step 2: Add appropriate coefficients to balance the numbers of atoms of each element. H 2 SO4 + 2 NaOH Add this coefficient Na 2 SO4 + H 2 O Balance these 2 Na Balancing Chemical Reaction Step 1: Write an unbalanced equation, using the correct formulas for all reactants and products. Step 2: Add appropriate coefficients to balance the numbers of atoms of each element. H 2 SO4 + NaOH One sulfate here ... Na 2 SO4 + H 2 O ... and one sulfate here Balancing Chemical Reaction Step 1: Write an unbalanced equation, using the correct formulas for all reactants and products. Step 2: Add appropriate coefficients to balance the numbers of atoms of each element. H 2 SO4 + 2 NaOH 4 H and 2 O here Na 2 SO4 + 2 H 2 O 4 H and 2 O here Balancing Chemical Reaction Step 3: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are the same. H 2 SO4 + 2 NaOH Na 2 SO4 + 2 H 2 O Balancing Chemical Reaction Step 4: Make sure the coefficients are reduced to their lowest wholenumber values. 2 H 2 SO4 + 4 NaOH 2 Na 2 SO4 + 4 H 2 O H 2 SO4 + 2 NaOH Na 2 SO4 + 2 H 2 O Stoichiometry Greek Stoicheion Metron element or part measure Stoichiometry Study of the quantitative aspects of chemical formulas and reactions Relate moles and grams for reactants or products Reaction Stoichiometry Coefficients = how many moles of each substance are needed for the reaction Moles combine in the same ratios that molecules do. 3H 2 + 1N 2 This number of moles of hydrogen ... 2 NH 3 to yield this number of moles of ammonia ... reacts with this number of moles of nitrogen ... Example + Ethylene + Hydrogen chloride C2H4(g) + HCl(g) Ethyl chloride C2H5Cl(l) anesthetic We have 42.0 g HCl How much ethylene would we need for this reaction? How many grams of ethyl chloride would we produce? Example + Ethylene + Hydrogen chloride C2H4(g) + HCl(g) 42.0 g Ethyl chloride C2H5Cl(l) anesthetic We cannot convert grams of HCl directly into grams of C2H4 or C2H5Cl. However, we can convert grams of HCl into moles, and use the coefficients of the balanced equation to obtain moles of the other species Example + Ethylene + Hydrogen chloride C2H4(g) + HCl(g) 42.0 g Ethyl chloride C2H5Cl(l) anesthetic Calculate molar mass for products and reactants: C2 H 4 = 2(12.01107) + 4(1.008) = 28.05414 HCl = 1.008 + 35.4527 = 36.4607 C2 H 5Cl = 2(12.01107) + 5(1.008) + 35.4527 = 64.51484 Example + Ethylene + C2H4(g) + 28.05414 g/mol Hydrogen chloride HCl(g) 42.0 g 36.4607 g/mol Ethyl chloride C2H5Cl(l) 64.51484 g/mol Convert g HCl to mol HCl: 1 mol HCl 42.0 g HCl = 1.151925 mol HCl 36.4607 g HCl Example + Ethylene + C2H4(g) + 28.05414 g/mol Hydrogen chloride HCl(g) 42.0 g 36.4607 g/mol Ethyl chloride C2H5Cl(l) 64.51484 g/mol Convert mol HCl to mol C2H4: Coefficient of what we are interested in 1 mol C2 H 4 1.151925 mol HCl 1 mol HCl Coefficient of what we are cancelling out = 1.151925 mol C2 H 4 Example + Ethylene + C2H4(g) + 28.05414 g/mol Hydrogen chloride HCl(g) 42.0 g 36.4607 g/mol Ethyl chloride C2H5Cl(l) 64.51484 g/mol Convert mol C2H4 to g C2H4: 28.05414g C2 H 4 1.151925 mol C 2 H 4 1 mol C2 H 4 = 32.31627 32.3 g C2 H 4 Example + Ethylene + C2H4(g) + 28.05414 g/mol Hydrogen chloride HCl(g) 42.0 g 36.4607 g/mol Ethyl chloride C2H5Cl(l) 64.51484 g/mol Do in steps or put calculation all together: 1 mol HCl 1 mol C2 H 4 42.0 g HCl 36.4607 g HCl 1 mol HCl 28.05414g C 2 H 4 1 mol C H 2 4 = 32.3 g C 2 H 4 Example + Ethylene + C2H4(g) + 28.05414 g/mol Hydrogen chloride HCl(g) 42.0 g 36.4607 g/mol Ethyl chloride C2H5Cl(l) 64.51484 g/mol How many grams of ethyl chloride will we produce? 1 mol HCl 1 mol C2 H 5Cl 64.51484g C2 H 5Cl 42.0 g HCl = 74.31627 36.4607 g HCl 1 mol HCl 1 mol C 2 H 5Cl 74.3 g C2 H 5Cl Stoichiometry Flowchart Examples C3 H 8 ( g ) + 5O2 ( g ) 3CO2 ( g ) + 4 H 2 O( g ) How many moles of CO2 can we make from 2.0 moles of C3H8? Examples C3 H 8 ( g ) + 5O2 ( g ) 3CO2 ( g ) + 4 H 2 O( g ) How many moles of CO2 can we make from 2.0 moles of C3H8? mol CO 2 3 (2.0 mol C3 H 8 ) = 6 mol CO 2 1 mol C3 H 8 Examples C3 H 8 ( g ) + 5O2 ( g ) 3CO2 ( g ) + 4 H 2 O( g ) How many moles of H2O can we make from 2.0 moles of C3H8? Examples C3 H 8 ( g ) + 5O2 ( g ) 3CO2 ( g ) + 4 H 2 O( g ) How many moles of H2O can we make from 2.0 moles of C3H8? mol H 2 O 4 (2.0 mol C3 H 8 ) = 8 mol H 2 O 1 mol C3 H 8 Examples C3 H 8 ( g ) + 5O2 ( g ) 3CO2 ( g ) + 4 H 2 O( g ) How many moles of CO2 can be produced from 3.5 moles of O2? Examples C3 H 8 ( g ) + 5O2 ( g ) 3CO2 ( g ) + 4 H 2 O( g ) How many moles of CO2 can be produced from 3.5 moles of O2? mol CO 2 3 (3.5 mol O 2 ) 5 mol O 2 = 2.1 mol CO 2 Examples C3 H 8 ( g ) + 5O2 ( g ) 3CO2 ( g ) + 4 H 2 O( g ) How many grams of CO2 are produced from 50.0 g of C3H8? 150 g CO 2 1 mol C3 H8 3 mol CO 2 44.00987 g CO 2 (50.0 g C3 H8 ) = 149.70291 44.09721 g C3 H8 1 mol C3 H8 1 mol CO 2 Examples Aqueous sodium hypochlorite (NaOCl), best known as household bleach, is prepared by reaction of sodium hydroxide with chlorine: 2 NaOH (aq ) + Cl2 ( g ) NaOCl (aq) + NaCl (aq) + H 2 O How many grams of NaOH are needed to react with 25.0 g of Cl2? Examples How many grams of NaOH are needed to react with 25.0 g of Cl2? For the balanced equation: 2 NaOH + Cl2 NaOCl + H2O GIVEN FIND Grams of Cl2 Moles of Cl2 Moles of NaOH Grams of NaOH Use molar mass of Cl2 as a conversion factor Use coefficients in the balanced equation to find mole ratios Use molar mass of NaOH as a conversion factor Examples Thermite reaction 2 Al ( s ) + Fe2 O3 ( s ) Al2 O3 ( s ) + 2 Fe(l ) Examples Thermite reaction 2 Al ( s ) + Fe2 O3 ( s ) Al2 O3 ( s ) + 2 Fe(l ) If 86.0 g of Fe is produced, what is the minimum mass in grams of Fe2O3 that must be used? Examples Thermite reaction 2 Al ( s ) + Fe2 O3 ( s ) Al2 O3 ( s ) + 2 Fe(l ) If 86.0 g of Fe is produced, what is the minimum mass in grams of Fe2O3 that must be used? 123g Fe 2 O3 mol Fe 1 mol Fe 2 O3 159.6882 g Fe 2 O3 (86.0 g Fe) = 122.958 55.845 g Fe 2 mol Fe mol Fe 2 O3 Limiting Reactants Limiting Reactants N2(g) + 3H2(g) 2NH3(g) [Haber process] 1.00 mol 5.00 mol What is the maximum amount of NH3 that can be produced? How much H2 will be left over? Limiting Reactants N2(g) + 3H2(g) 2NH3(g) [Haber process] 1.00 mol 5.00 mol N2 reacts: mol NH 3 17.0307 g NH 3 2 1.00 mol N 2 = 34.1g NH 3 1 mol N 2 mol NH 3 H2 reacts: mol NH 3 17.0307 g NH 3 2 5.00 mol H 2 = 56.8g NH3 3 mol H 2 mol NH 3 Limiting Reactants N2(g) + 3H2(g) 2NH3(g) [Haber process] How much H2 will be left over? 3 mol H 2 2.016 g H 2 2.00 mol NH 3 = 6.048 g H 2 2 mol NH 3 mol H 2 2.0158 g H 2 5.00 mol H 2 = 10.079 g H 2 mol H 2 Limiting Reactants N2(g) + 3H2(g) 2NH3(g) [Haber process] How much H2 will be left over? 10.079 6.048 = 4.0310 4.03 g H2 Chemical Reaction Yields Loss of product Glassware contamination Reactant impurities Incomplete reactions ??? Side reactions Chemical Reaction Yields Theoretical yield Maximum amount that can be produced Amount actually obtained Actual yield Percent yield actual yield % yield = x 100 theoretical yield Percent Yield Example 2 1 1/4 1 Supplies: 16 graham crackers 8 marshmallows 2 chocolate bars Percent Yield Example 2 1 1/4 1 Supplies: 16 graham crackers 8 marshmallows 2 chocolate bars 16/2 = 8 8/1 = 8 2/(1/4) = 8 Percent Yield Example The combustion of acetylene gas (C2H2) produces carbon dioxide and water: 2C2 H 2 ( g ) + 5O2 ( g ) 4CO2 ( g ) + 2 H 2 O( g ) Percent Yield Example The combustion of acetylene gas (C2H2) produces carbon dioxide and water: 2C2 H 2 ( g ) + 5O2 ( g ) 4CO2 ( g ) + 2 H 2 O( g ) When 26.0 g of acetylene is burned in sufficient oxygen for complete reaction, the theoretical yield of CO2 is 88.0 g. Calculate the percent yield if the actual yield is only 72.4 g CO2. Percent Yield Example 2C2 H 2 ( g ) + 5O2 ( g ) 26.0 g Actual 4CO2 ( g ) + 2 H 2 O( g ) 88.0 g 72.4 g Theoretical 72.4 g CO 2 % yield = 88.0 g CO 2 x 100 = 82.3% Percent Yield Example B2O3(l) + 3Mg(s) 2B(s) + 3MgO(s) What is the percent yield if 675 g of boron is obtained from the reaction of 2350 g of boric oxide? Percent Yield Example B2O3(l) + 3Mg(s) 2B(s) + 3MgO(s) What is the percent yield if 675 g of boron is obtained from the reaction of 2350 g of boric oxide? mol B2 O3 2 mol B 10.811 g B 2350 g B2 O3 = 729 g B 69.6 g B2 O3 1 mol B2 O3 mol B Percent Yield Example B2O3(l) + 3Mg(s) 2B(s) + 3MgO(s) mol B2 O3 2 mol B 10.811 g B 2350 g B2 O3 = 729 g B 69.6 g B2 O3 1 mol B2 O3 mol B g B 675 % yield of B = x 100 = 92.6% 729 g B Summary Atomic mass unit Atomic mass Mole Molar Mass Percent by Mass Empirical / Molecular Formula Chemical Reactions Balancing Stoichiometry Limiting Reactant % yield ...
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This note was uploaded on 11/22/2010 for the course CHM 121 taught by Professor Scheru during the Spring '09 term at Oakton.

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