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General FITCH Rules “A” students work (without solutions manual) ~ 10 problems/night. Office Hours W – F 2-3 pm Particle Neutron, n Proton, p Electron, e Symbol 1 0 1 1 0 −1 n H e Mass g 1.6749285x10-24 1.6726231x10-24 1.093898x10-28 Amu 1.00867 1.00728 0.00055 These numbers are inconvenient mass of 1 126 C atom 1 amu = Invoke General Rule #5: 12 Chemist’s Are Lazy Create some unit (not g) that is more useful (atomic mass unit) Invoke General Rule #3 19926 x10 − 26 g . There must be some reference state 1amu = 12 1amu = 166053873 . x10 − 24 g Chemistry Alanah Fitch Flanner Hall 402 508-3119 [email protected] G1: Suzuki is Success G2. Slow me down G3. Scientific Knowledge is Referential G4. Watch out for Red Herrings G5. Chemists are Lazy C1. It’s all about charge C2. Everybody wants to “be like Mike” (grp.18) C3. Size Matters C4. Still Waters Run Deep C5. Alpha Dogs eat first Why a relative reference state of C12? Represents a compromise between Physicists and Chemists Physicists used a reference Chemists used a reference state state of H1 to measure of O16 as an abundant element Relative velocities of gas everything reacts with. phase atoms Measured precipitate mass changes C12 required Least adjustments To put both Physics and chemistry On same scale 1 Properties and Measurements Property Size Volume Weight Unit m cm3 gram Temperature oC, 1.66053873x10-24g amu K Reference State size of earth m mass of 1 cm3 water at specified Temp (and Pressure) boiling, freezing of water (specified Pressure mass of 1C12 atom/12 Two Examples we Examined already of STABLE isotopes Relative % Abundance atomic mass (amu) 204 1.36 203.973 Pb 82 206 25.4 205.9745 Pb 82 207 21.1 206.9759 82 Pb 208 52.1 207.9766 82 Pb Expressing masses in terms of amu is more convenient On the periodic table what is the reported atomic mass for lead? Atomic mass= 207.2 Where did this number come from? ⎛ % isotope ⎞ ⎟ am atomic massaverage = ∑ ⎜ 100 ⎠ isotope isotopes ⎝ 148% . ( 203.973) atomic mass Pb ,average = 100 23.6% ( 205.9745) + 100 22.6% ( 206.9759) + 100 atomic mass Pb ,average = 207.17709amu 52.3% ( 207.9766) + 100 12 6 % Abundance 98.89 1.11 C 13 6 C Relative atomic masses (amu) 12.0000 13.00335 On the periodic table what is the reported amu for C? amu=12.01 Where did this number come from? atomic massaverage = atomic massC ,average = ⎛ % isotope ⎞ ⎜ ⎟ amuisotope 100 ⎠ isotopes ⎝ ∑ 11% . 98.9% (13.00006) + (1199671 ) . 100 100 atomic massC ,average = 12.0077amu atomic massC ,average = 12.01amu 2 We saw that Example ⎛ 1 126 C atom ⎞ ⎛ ⎞ 1amu 23 12 ⎟⎜ − 24 ⎟ = 6.022 x10 6 C atoms 12amu ⎠ ⎝ 166053873 . x10 g ⎠ (12 g C )⎜⎝ 12 6 General Rule #4: Chemists Are Lazy: make this number unique Element H C O Pb Amu 1.008 12.01 16.00 207.2 Mass (g) 1.008 24.02 48.00 207.2 Moles (n) 1 2 3 1 # atoms 6.022x1023 12.044x1023 18.066x1023 6.022x1023 N Avogadro = mole = 6.022 x10 23 atoms 1= ⎛ mole 6.022 x10 23 atoms 12 6 ⎞⎛ atom 1amu ⎟⎜ (12 g C )⎜⎝ 1 12Camu . x10 ⎠ ⎝ 166053873 12 6 (( x = amu) g X − 24 x elementX ⎞⎛ 1mole ⎞ ⎟⎜ ⎟ = 1mole g ⎠ ⎝ 6.022 x10 23 atoms ⎠ ) = 1mole atoms Properties and Measurements Property Size Volume Weight Unit m cm3 gram Temperature oC, 1.66053873x10-24g 6.022x1023 amu mole K Reference State size of earth m mass of 1 cm3 water at specified Temp (and Pressure) boiling, freezing of water (specified Pressure mass of 1C12 atom/12 atomic mass of an element in grams Molar Mass in grams is numerically equal to the sum of the masses (in amu) of the atoms in the formula MM = ∑ n (amu ) = a(amu ) + b(amu )+ ...... z(amu ) i i A B Z i 3 Example: Acetylsalicylic acid, C9H 8O4, is the active ingredient of aspirin. What is the mass in grams of 0.509 mol acetylsalicylic acid? Example. Lead carbonate is principle component of white lead used in all white paints prior to WWII. Calculate the molar mass of lead carbonate Lead carbonate = PbCO3 principle component paint before WWII Molar mass= unknown Element #atoms amu Pb 1 207.2 C 1 12.01 3O 3 16.00 Total amu 207.2 12.01 48.00 267.21 Acetylsalicylic acid C9H 8O4 Active ingredient of aspirin 0.509 mol acetylsalicylic acid Mass? MM = 267.2 MM = ∑ n (amu ) = a(amu ) + b(amu )+ ...... z(amu i i A B i Mass % ⎡ atomic weight of element ⎤ % element = (# atoms of that element )⎢ ⎥ (100% ) ⎣ molar mass of compound ⎦ Z ( MM ) n = m ∑ n (amu ) = a(amu ) + b(amu )+ ...... z(amu ) i i MM = m n g ⎞ ⎛ . . g ⎜ 18015 ⎟ 0.509mole = m = 917 ⎝ mole ⎠ i A B Z MM = 9(12.01) + 8(1008 . ) + 4(16.00) g MM = 18015 . mole Calcium carbonate, commonly called is used in many commercial products to relieve an upset stomach. It has the formula of CaCO3. Because lead ore bodies form by substitution of lead onto old coral reefs (calcite or calcium carbonate) some antacid materials have been tested for their lead composition. What are the mass percents of Ca, C, and O in calcium carbonate? CaCO3 commercial product upset stomach mass percent Ca mass percent C mass percent O old coral reefs lead in antacids Ca C 3O MM = 40.08 12.01 3(16.00) 40.08 12.01 48.00 100.09 ∑ n (amu ) i i i ⎡ atomic weight of element ⎤ % element = (# atoms of that element )⎢ ⎥ (100% ) ⎣ molar mass of compound ⎦ 4 Calcium carbonate, commonly called is used in many commercial products to relieve an upset stomach. It has the formula of CaCO3. What are the mass percents of Ca, C, and O in calcium carbonate? Because lead ore bodies form by substitution of lead onto old coral reefs (calcite or calcium carbonate) some antacid materials have been tested for their lead composition. Ca C 3O 40.08 12.01 3(16.00) 40.08 12.01 48.00 100.09 ⎡ 40.08 ⎤ 100% % Ca = (1) ⎢ ⎣ 100.09 ⎥⎦ % Ca = 40.04 Calcium carbonate, commonly called is used in many commercial products to relieve an upset stomach. It has the formula of CaCO3. What are the mass percents of Ca, C, and O in calcium carbonate? Because lead ore bodies form by substitution of lead onto old coral reefs (calcite or calcium carbonate) some antacid materials have been tested for their lead composition. 1 Ca ⎡ 12.01 ⎤ 100% % C = 1⎢ ⎣ 100.09 ⎥⎦ % C = 119992% . = 12.00% ⎡ 16.00 ⎤ % O = 3⎢ 100% ⎣ 100.09 ⎥⎦ % O = 47.95683885 = 47.96% Are We Done? 47.96 12.00 40.04 100% Converting between g, moles, and number of particles Example. Lead carbonate is principle component of white lead used in all white paints prior to WWII. Determine the number of moles of lead carbonate in a sample of 14.8 g lead carbonate nN A = particles Mass=m Molar grams Avogadro’s Moles Mass MM = Moles=n m n Very Important! Check answer particles Formula Units Number The Golden Bridge Stoichiometry or #of moles Lead carbonate = PbCO3 principle component paint before WWII 14.8 g PbCO3 number of moles, n = unknown m MNM = n MM = Pb C 3O 207.2 12.01 3(16.00) 207.2 12.01 48.00 267.21 267.2 m n= MM ∑ n (amu ) = a(amu ) + b(amu )+ ...... z(amu ) i i A B Z i 5 Example. Lead carbonate is principle component of white lead used in all white paints prior to WWII. Determine the number of moles of lead carbonate in a sample of 14.8 g lead carbonate Lead carbonate = PbCO3 principle component paint before WWII 14.8 g PbCO3 number of moles, n = unknown Pb C 3O 207.2 12.01 3(16.00) 207.2 12.01 48.00 267.21 267.2 n= 14.8 g g 267.2 mole Simplest Formula from Chemical Analysis (Empirical) 1. Mass scale is based on atomic number of C 2. Mass scale is therefore proportional to number of atoms or moles 3 Convert mass to moles = whole units of atoms 4. If moles are fractional this implies that atoms are fractional NOT ALLOWED by chemistry (remember we chemists do not break up atoms - that is reserved for physicists) 5. So multiply until we get whole number ratios n = 0.055389221mole n = 0.0554mole Example A 25.00-g sample of an orange compound contains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g of oxygen. Find the simplest formula (Empirical) Total =25.00g K= 6.64g Cr=8.84 g O=9.52 g unknown - simplest formula orange compound CHECK THAT THE COMPOUND IS PURE!! 6.64 8.84 9.52 25.00 n( MM )) = m n= m MM This means that we have accounted for the total weight of The compound and we are confident in assigning weight Ratios. Example : A 25.00-g sample of an orange compound contains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g of oxygen. Find the simplest formula (Empirical) Sig fig n= m MM 6.64 gK . . molK molK = 0170 = n = 01698 gK 39.10 (1)mole 8.84 gCr = n = 01700 . . molCr molCr = 0170 gCr 52.00 mol 9.52 gO = n = 0.59500molO = 0.595molO gO 16.00 mol 0.170molK 0.170molCr 0.595mol O 6 Simplest Formula from Chemical Analysis (Empirical) 1. Mass scale is based on atomic number of C 2. Mass scale is therefore proportional to number of atoms or moles 3 Convert mass to moles = whole units of atoms 4. If moles are fractional this implies that atoms are fractional NOT ALLOWED by chemistry (remember we chemists do not break up atoms - that is reserved for physicists) 5. So multiply until we get whole number ratios Example : A 25.00-g sample of an orange compound contains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g of oxygen. Find the simplest formula (Empirical) 0.170molK 0.170molCr 0.595mol O 0170 . molCr 100 . molCr = 0170 . molCr 1molCr 0170 . molK 100 . molK = 0170 . molCr 1molCr 0.595molO 350 . molO = 0170 . molCr 1molCr 7 O : 2K : 2Cr Formula: Cations First 2.00molCr 2molCr 2.00molK 2molCr 7.00molO 2molCr 2K : 2Cr : 7 O K2Cr2O7 Make this non-fractional, multiply x2 Example : The compound that gives fermented grape juice, malt liquor, and vodka their intoxicating properties is ethyl alcohol, which contains the elements carbon, hydrogen, and oxygen. When a sample of ethyl alcohol is burned in air, it is found that Example : The compound that gives fermented grape juice, malt liquor, and vodka their intoxicating properties is ethyl alcohol, which contains the elements carbon, hydrogen, and oxygen. When a sample of ethyl alcohol is burned in air, it is found that 5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O 5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O What is the simplest formula of ethyl alcohol? What is the simplest formula of ethyl alcohol? 5.00g ethyl alcohol 9.55 g CO2 5.87 g H2O simplest formula? Fermented grape juice malt liquor vodka intoxicating properties ethyl alcohol elements C, H, O 5.00g ethyl alcohol 9.55 g CO2 5.87 g H2O simplest formula? burned in air elements C, H, O Burned in air All the C and H in the sample is converted In air (contains O2) to CO2 and H2O so g C in CO2 represents g C in original ethyl alcohol And g H in H2O represents g H original O2 5g total ( C + H + O ) ⎯⎯ → 9.55gCO2 + 587 . gH 2 O 7 Example : The compound that gives fermented grape juice, malt liquor, and vodka their intoxicating properties is ethyl alcohol, which contains the elements carbon, hydrogen, and oxygen. When a sample of ethyl alcohol is burned in air, it is found that 5.00 g ethyl alcohol goes to 9.55 g COg plus 5.87g H2O Example : The compound that gives fermented grape juice, malt liquor, and vodka their intoxicating properties is ethyl alcohol, which contains the elements carbon, hydrogen, and oxygen. When a sample of ethyl alcohol is burned in air, it is found that 5.00 g ethyl alcohol goes to 9.55 g COg plus 5.87g H2O What is the simplest formula of ethyl alcohol? What is the simplest formula of ethyl alcohol? 5.00g ethyl alcohol 9.55 g CO2 5.87 g H2O simplest formula? elements C, H, O m ⎛ 1 ⎞ n= = m⎜ ⎟ ⎝ MM ⎠ MM ⎛ moleCO2 ⎞ ⎛ 1moleC ⎞ ⎟ ⎟⎜ nC = ( gCO2 )⎜ ⎝ amu gCO2 ⎠ ⎝ 1moleCO2 ⎠ 5.00g ethyl alcohol 9.55 g CO2 5.87 g H2O simplest formula? ⎞ ⎛ moleC ⎞ ⎛ mole CO2 ⎟ ⎟⎜ nC = (9.55gCO2 )⎜ ⎝ (12.01 + 2(16)) gCO2 ⎠ ⎝ moleCO2 ⎠ elements C, H, O m ⎛ 1 ⎞ n= = m⎜ ⎟ ⎝ MM ⎠ MM ⎛ mole CO2 ⎞ ⎛ moleC ⎞ ⎟ ⎟⎜ nC = (9.55gCO2 )⎜ ⎝ (44.01gCO2 ⎠ ⎝ moleCO2 ⎠ nC = 0.217mole ⎛ moleH 2 O ⎞ ⎛ 2moleH ⎞ ⎟ ⎟⎜ nh = ( gH 2 O)⎜ ⎝ amu gH 2 O ⎠ ⎝ 1moleH 2 O ⎠ ⎞ ⎛ 2moleH ⎞ ⎛ mole H 2 O ⎟ ⎟⎜ n H = (587 . gH 2 O)⎜ . )) gH2 O ⎠ ⎝ moleH2 O ⎠ ⎝ (16.00 + 2(1008 ⎛ mole H 2 O ⎞ ⎛ 2moleH ⎞ ⎟ ⎟⎜ n H = (587 . gH 2 O)⎜ ⎝ (18.016 gCO2 ⎠ ⎝ moleH 2 O ⎠ nC = 0.6517mole nC = 0.652mole Example : The compound that gives fermented grape juice, malt liquor, and vodka their intoxicating properties is ethyl alcohol, which contains the elements carbon, hydrogen, and oxygen. When a sample of ethyl alcohol is burned in air, it is found that 5.00 g ethyl alcohol goes to 9.55 g COg plus 5.87g H2O Example : The compound that gives fermented grape juice, malt liquor, and vodka their intoxicating properties is ethyl alcohol, which contains the elements carbon, hydrogen, and oxygen. When a sample of ethyl alcohol is burned in air, it is found that 5.00 g ethyl alcohol goes to 9.55 g COg plus 5.87g H2O What is the simplest formula of ethyl alcohol? What is the simplest formula of ethyl alcohol? ⎛ amugC ⎞ ⎜⎜ ⎟⎟ ⎝ amug CO2 ⎠ 5.00g ethyl alcohol How are we going to get it? gc = ( gCO ) 9.55 g CO2 gO in original sample? ⎛ 12.01gC ⎞ 5.87 g H2O ⎟⎟ g c = (9.55g CO )⎜⎜ ⎝ 44.01g CO ⎠ simplest formula? g total = g C + g H + g O 3 sig fig 2 2 = 2.61g C 2 ⎛ amugH ⎞ ⎟⎟ g H = ( g H O )⎜⎜ elements C, H, O 5.00 g total = 2.61g C + 0.657 g H + g O ⎝ amug H O ⎠ 3 sig fig Now know: gO = 5.00 − 2.61 − 0.657 ⎛ 2(1008 . ) gC ⎞ ⎟⎟ = 0.657 g C g c = (587 . g CO )⎜⎜ mol C 0.217 gO = 1733 . ⎝ 18.016 g CO ⎠ mol H 0.652 gO = 173 . Need to know mol O. 2 2 2 2 5.00g ethyl alcohol 9.55 g CO2 5.87 g H2O simplest formula? elements C, H, O m ⎛ 1 ⎞ n= = m⎜ ⎟ ⎝ MM ⎠ MM 5.00 g total = 2.61g C + 0.657 g H + g O gO = 5.00 − 2.61 − 0.657 ⎛ moleO ⎞ ⎟ nO = ( gO)⎜ ⎝ amu gO ⎠ gO = 173 . ⎛ 1mole O ⎞ n H = (173 . gO)⎜ ⎟ ⎝ (16.00 gO ⎠ nO = 0108125 . mole nO = 0108 . mole 8 Example : The compound that gives fermented grape juice, malt liquor, and vodka their intoxicating properties is ethyl alcohol, which contains the elements carbon, hydrogen, and oxygen. When a sample of ethyl alcohol is burned in air, it is found that 5.00 g ethyl alcohol goes to 9.55 g COg plus 5.87g H2O What is the simplest formula of ethyl alcohol? 5.00g ethyl alcohol 9.55 g CO2 5.87 g H2O simplest formula? molO=0.108 molH=0.652 molC=0.217 elements C, H, O 0108 . molO =1 0108 . molO 0.652mol H = 6.037 = 6 0108 . molO 0.217molC = 2.009 = 2 . molO 0108 C2H6O Mass Relations and Stoichiometry Sugar of lead (lead acetate) was used from 0 A.D. to 1750s A.D. to “sweeten” ethyl alcohol (wine) and to prevent wine from going bad One (unproven) theory is that the Roman leaders were poisoned by excessive lead acetate consumption leading to the fall of Rome. Mass Relations In Reactions Writing and Balancing Chemical Reactions Chemical Formulas Lead Hand Side Right Hand Side Write reactions Predict Change in amounts Relate to Mass Ratios and Stoichiometry Reactants number atoms A as reactants = number atoms Bas reactants = number atoms C as reactants = . . . Number atoms Z as reactants = Products number atoms A as products number atoms B as products number atoms C as products number atoms Z as products Consider what happens when one part necessary for the Reaction (recipe) is limiting 9 Steps to Balance Reaction Equations 1. 2 3. 4. 5. 6. Write a “skeleton” equation with molecular formulas of reactants on left, products on right Indicate the physical state of the reactants and products a. (g) for a gas b. (l) for a liquid c. (s) for a solid d. (aq) for an ion or molecule in water (aqueous) solution Chose an element that appears in only one molecular formula on each side of the equation Balance the equation for mass of that element a. placing coefficients in front of the molecular formula NOT by changing subscripts in the molecular formula Continue for the other elements The best answer is the one which is simplest whole-number coefficients Example: Lead ore, Lead sulfide (with the common name “galena”), was “calcined” by early metallurgists to form a lead oxide used to purify silver from other metals. Calcined means to burn in the presence of oxygen. At low temperatures the yellow lead oxide PbO, litharge, is formed. At higher temperatures the a red lead oxide, minium, is formed, Pb3O4. The sulfur is converted in the reaction to the gas SO2. Minium was the predominate source of red paints and pigments for illuminated bibles from which we derive the word miniature. Write a balanced chemical equation for the reaction of lead ore, galena with oxygen gas (O2) to form minium and SO2 gas. minium PbS Burned in O2 To make Pb3 O4 and SO2 PbS + O2 → Pb3 O4 + SO2 1S PbS 1S PbS + O2 → Pb3 O4 + SO2 1Pb 3Pb 3S 1S 3( PbS ) + O2 → Pb3 O4 + Pb3O4 3Pb 3S Subscript indicating number SO2 Pb within a single molecule 3Pb 3S 3 PbS + O2 → Pb3 O4 + 3SO2 2O 4O+(3x2)O=10 O 10 3 PbS + O2 → Pb3 O4 + 3SO2 From previous slide Mass Relations and Stoichiometry 4O+3x2O=10O 3 PbS + 5O2 → Pb3 O4 + 3SO2 Chemical Formulas 4O+3x2O=10O 5x2O=10O Write reactions Be sure to indicate the physical state of the reactants and products 3 PbS ( s ) + 5O2 ( g ) → Pb3 O4( s ) + 3SO2 ( g ) 3 5 1 3 Sets up proportionalities Predict Change in amounts STOICHIOMETRY BUT How doe we make use Of those proportionalities? Relate to Mass Ratios and Stoichiometry Consider what happens when one part necessary for the Reaction (recipe) is limiting 2 MAIN Strategies to solving these problems Mass Relations and Stoichiometry Chemical Formulas Write reactions Step wise 1. Easy conceptually = series of proportionalities Predict Change in amounts Relate to Mass Ratios and Stoichiometry Consider what happens when one part necessary for the Reaction (recipe) is limiting 1. Longer time to compute 2. Easy to lose track of units Will model this one To begin with String wise 1. Not easy conceptually when you are starting out 1. Shorter time to compute 2. Easy to keep track of units BUT will eventually Only use this method!!! 11 3 PbS ( s ) + 5O2 ( g ) → Pb3 O4( s ) + 3SO2 ( g ) 3 PbS ( s ) + 5O2 ( g ) → Pb3 O4( s ) + 3SO2 ( g ) Determine: a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts 1. 2. Find moles of SO2 by Stoichiometry Find grams of SO2 from those moles 3molSO2 ( g ) x = 134 . molO2 5molO2 ( g ) STOICHIOMETRY ⎡ 3molSO2 ( g ) ⎤ ⎥ 134 x= ⎢ . molO2 = 0.804molSO2 ( g ) ⎢⎣ 5molO2 ( g ) ⎥⎦ ( ) ⎡ gSO2 ⎤ gSO2 = 0.804molSO2 ⎢ ⎥ ⎣ molSO2 ⎦ ⎡ Samu + 2(Oamu) ⎤ gSO2 = 0.804molSO2 ⎢ ⎥ molSO2 ⎣ ⎦ Alternative Strategy is to fold all calculations together m = nMM ⎡ (32.07 + 2(16.00)) gSO2 gSO2 = 0.804molSO2 ⎢ molSO2 ⎢⎣ ⎡ 64.07 gSO2 ⎤ gSO2 = 0.804molSO2 ⎢ ⎥ ⎣ molSO2 ⎦ gSO2 = 5151228 . gSO2 = 515 . gSO2 3 PbS ( s ) + 5O2 ( g ) → Pb3 O4( s ) + 3SO2 ( g ) Determine: a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts b) The mass in grams of O2 required to form 1.000 kg of Pb3O4 Stepwise calculations 1. Find moles of Pb3O4 formed 2. Use Stoichiometry to find moles of O2 3. Find grams of O2 from those moles m = nMM n= m MM ⎡ 10 3 g ⎤ 1kgPb3 O4 ⎢ ⎥ ⎣ kg ⎦ molPb3 O4 formed = ⎡ gPb3 O4 ⎤ ⎢ ⎥ ⎣ molPb3 O4 ⎦ ⎡ 10 3 g ⎤ 1kgPb3O4 ⎢ ⎥ ⎣ kg ⎦ molPb3 O4 formed = ⎡ 3(207.2) + 4(16.00) gPb3 O4 ⎤ ⎢ ⎥ molPb3O4 ⎣ ⎦ Determine: a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts ( ⎡ 3molSO2 ( g ) ⎤ ⎡ amus + 2 amuO2 ⎥⎢ x = 134 . molO2 ⎢ molSO2 ⎢⎣ 5molO2 ( g ) ⎥⎦ ⎢ ⎣ ( ) ) ⎤⎥ ⎥ ⎦ ⎡ 3molSO2 ( g ) ⎤ ⎡ gSO ⎤ 2 ⎥⎢ x = 134 . molO2 ⎢ ⎥ ⎢⎣ 5molO2 ( g ) ⎥⎦ ⎣ molSO2 ⎦ ( ) ⎡ 3molSO2 ( g ) ⎤ ⎡ 64.07 gSO ⎤ 2 ⎥⎢ x = 134 . molO2 ⎢ . g = 515 . g ⎥ = 5151228 ⎢⎣ 5molO2 ( g ) ⎥⎦ ⎣ molSO2 ⎦ ( ) 3 PbS ( s ) + 5O2 ( g ) → Pb3 O4( s ) + 3SO2 ( g ) Determine: a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts b) The mass in grams of O2 required to form 1kg of Pb3O4 1. Find moles of Pb3O4 formed 2. Use Stoichiometry to find moles of O2 3. Find grams of O2 from those moles ⎡ 10 3 g ⎤ 1kgPb3 O4 ⎢ ⎥ ⎣ kg ⎦ molPb3 O4 formed = ⎡ 685.6 gPb3 O4 ⎤ ⎢ ⎥ ⎣ molPb3 O4 ⎦ molPb3 O4 formed = 1458576429 . = 1459 . molPb3 O4 formed = 1459 . 5molO2 x = 1459 . molPb3 O4 molPb3 O4 x = (1459 . )(5molO2 ) = 7.295molO2 m = n( MM ) ⎡ gO2 ⎤ x = 7.295molO2 ⎢ ⎥ ⎣ molO2 ⎦ ⎡ 32.00 gO2 ⎤ x = 7.295molO2 ⎢ ⎥ = 233.4 g ⎣ molO2 ⎦ 12 3 PbS ( s ) + 5O2 ( g ) → Pb3 O4( s ) + 3SO2 ( g ) 3 PbS ( s ) + 5O2 ( g ) → Pb3 O4( s ) + 3SO2 ( g ) Determine: a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts b) The mass in grams of O2 required to form 1kg of Pb3O4 Alternative string strategy ((( ))) ( ⎡ 10 3 g ⎤ ⎡ mol Pb3O4 ⎢ kg ⎥ ⎢ 685.6 g ⎣ ⎦ ⎢⎣ Pb3O4 molPbPb3O3O44 ⎤⎤⎡ 5molO2 ⎤ ⎡ g O2 ⎤ 10 gg⎤⎤⎡⎡mol x = 1kg Pb3O4 ⎡⎡10 kgPbPb xx == 11kg ⎥ ⎢ ⎢ O ⎢ ⎢ ⎥ 3 4 Pb3O O 3 44 kg ⎦⎦⎢⎢ ggPbPbOO ⎥⎥⎥⎥⎢⎢ mol Pb O ⎥⎥ ⎢⎢ molO ⎥⎥ ⎣⎣ kg 3 3 4 4 ⎦⎦⎣ 3 4 ⎦⎣ 2 ⎦ ⎣⎣ x = 1kg Pb3O4 ) 333 ⎤ ⎡ 5molO2 ⎥⎢ ⎥⎦ ⎢⎣ mol Pb3O4 ⎤ ⎡ 32.00 g O2 ⎥⎢ ⎥⎦ ⎢⎣ molO2 ⎤ ⎥ ⎥⎦ x = 233.37 g Stepwise method Determine: a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts b) The mass in grams of O2 required to form 1kg of Pb3O4 c) The mass in grams of PbS required to react with 6.00 g of O2 1. Get moles of O2 2. Use Stoichiometry to get moles of PbS 3. Get grams of PbS m m = nMM n= MM x= 6.00 gO2 ⎡ 32.00 gO2 ⎤ ⎢ ⎥ ⎣ molO2 ⎦ molO2 . = 01875 x = ( 01125 . molPbS ) M PbS ⎡ 207.2amu Pb + 32.07amuS ⎤ x = 01125 . molPbS ⎢ ⎥ molPbS ⎣ ⎦ ⎡ 239.27 gPbS ⎤ x = 01125 . molPbS ⎢ = 26.917875g ⎣ molPbS ⎥⎦ x 3molPbS = . mol molO2 01875 . 5molO2 x = 01125 x = 26.9 g 3 PbS ( s ) + 5O2 ( g ) → Pb3 O4( s ) + 3SO2 ( g ) Stringwise Determine: a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts b) The mass in grams of O2 required to form 1kg of Pb3O4 c) The mass in grams of PbS required to react with 6.00 g of O2 x = (6.00 gO2 )⎡ molO2 ⎤ ⎡ 3molPbS ⎤ ⎡ 239.27 gPbS ⎤ x = (6.00 gO2 )⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ 32.00 gO2 ⎦ ⎣ 5molO2 ⎦ ⎣ molPbS ⎦ Chemical Formulas Write reactions x = 26.917 g x = 26.9 g Predict Change in amounts Relate to Mass Ratios and Stoichiometry Consider what happens when one part necessary for the Reaction (recipe) is limiting 13 Rules for Limiting Reagent 1. Calculate the amount of product that would be formed if the first reactant were completely consumed 2. Repeat for the second reactant 3. Choose the smaller of the two amounts. This it the theoretical yield of the product. 4. The reactant that produces the smaller amount of the product is the limiting reagent. 5. The other reagent is in excess, only part of it is consumed. Limiting Reagents Liebig’s Law of the Limiting (1810-1820 Germany) I have 1 dozen eggs, 2 packages of chocolate chips, and An entire carton of flour, an entire carton of sugar, a new bottle Of vanilla, and a new box of baking powder. The chocolate chip recipe calls for 2 eggs, 3 cups flour, 1 cup Sugar, 1 package chocolate chips, 1 tbsp vanilla, and 2 Tbsp baking powder. The recipe results in 36 Chocolate chip cookies. a. What is the limiting reagent? 1 pkgchips + 2eggs + 3cflour + 1csugar + 1tbspvanilla + 2tbsppowder ⎯heat ⎯ ⎯→ 36cookies Chips 1pkg Eggs 2 Flour 3c Sugar 1c vanilla 1tbsp Powder 2tbsp Yield 36 cookies 2pkg 12 eggs >3cups >1cup >1tbsp >2tbsp Yield?? ? 36cookies ⎡ 36cookies ⎤ = ? = 2 pkg ⎢ ⎥ = 72cookies 2 pkg 1 pkg ⎣ 1 pkg ⎦ ? 36cookies ⎡ 36cookies ⎤ = ? = 12eggs⎢ ⎥ = 192cookies 12eggs 2eggs ⎣ 2eggs ⎦ Smallest yield = 72 cookies Limiting reagent = chips While I wasn’t looking, my son snitched 4 cookies. ⎡ yield exp erimental ⎤ % yield = ⎢ ⎥ 100% ⎣ yield theoretical ⎦ ⎡ 72 − 4 ⎤ % yield cookies = ⎢ 100% = 94.44444 = 94% ⎣ 72 ⎥⎦ “A” students work (without solutions manual) ~ 10 problems/night. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours W-F 2-3 pm Actual Yield=Theoretical Yield (most of the time) 14