“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 Module #4 Mass and Stoichiometry.
Download ReportTranscript “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 Module #4 Mass and Stoichiometry.
“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 Module #4 Mass and Stoichiometry Chemistry General FITCH Rules 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 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 1amu Invoke General Rule #5: 12 Chemists 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 Why a relative reference state of 126C? Represents a compromise between Physicists and Chemists Physicists used a reference Chemists used a reference state state of 11H to measure of 168O as an abundant element Relative velocities of gas everything (but gold) reacts with. phase atoms Measured mass changes Rock s airg RockOn,s fire Chemists must have Started out chunking Rocks in fire 12 6C required Least adjustments To put both Physics and chemistry On same scale 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 Expressing masses in terms of amu is more convenient Two Examples we Examined already of STABLE isotopes % Abundance 204 1.36 82 Pb 206 23.6 82 Pb 207 22.6 82 Pb 208 52.1 82 Pb 12 6 C 13 6 C % Abundance 98.89 1.11 Relative atomic mass (amu) 203.973 205.9745 206.9759 207.9766 Relative atomic masses (amu) 12.0000 13.00335 On the periodic table what is the reported atomic mass for lead? Atomic mass= 207.2 atomic massaverage % isotope amuisotope isotopes 100 On the periodic table what is the reported atomic mass for lead? Atomic mass= 207.2 Where did this number come from? atomicmassaverage % Abundance 1.36 23.6 22.6 52.1 99.98 % isotope amisotope isotopes 100 atomic mass (amu) 203.973 205.9745 206.9759 207.9766 Weighted Sum Average amu %x(amu) 301.88 4860.9982 4677.65534 10877.17618 20717.70976 207.1770976 207.2 atomicmassPb,average 20717709 . amu On the periodic table what is the reported amu for C? amu=12.01 Where did this number come from? atomic massaverage % Abundance 98.89 1.11 100 % isotope amuisotope isotopes 100 atomic masses (amu) 12.0000 13.00335 %(amu)/100 11.8668 0.144337185 12.01113719 atomicmassC,average 12.01amu How many particles are there? 12 1 1amu 6 C atom 12 23 12 6 . 022 x 10 12 g 6 C 12amu 166053873 24 6 C atoms . x10 g General Rule #4: Chemists Are Lazy: make this number unique N Avogadro mole 6.022x1023 atoms mole 1 6.022 x10 23 atoms 12 1 1amu 1mole 6 C atom 12 1mole 12 g 6 C 12amu 166053873 . x10 24 g 6.022 x1023 atoms ( x amu) g X x elementX 1mole atoms Mole Latin mass=moles of stuff Amedeo Avogadro Italian (Turin) 1756-1856 Conte di Quaregna e di Cerreto Galen, 170 Anders Celsius 1701-1744 1825-1898 Johann Balmer Marie the Jewess, 300 Amedeo Avogadro 1756-1856 James Maxwell 1831-1879 Jabir ibn Hawan, 721-815 John Dalton 1766-1844 Johannes Diderik Van der Waals 1837-1923 Galileo Galili 1564-1642 Evangelista Torricelli 1608-1647 Jacques Charles 1778-1850 Johannes Rydberg 1854-1919 TV Mad scientist B. P. Emile Clapeyron 1799-1864 J. J. Thomson 1856-1940 Abbe Jean Picard Daniel Fahrenheit 1620-1682 1686-1737 Germain Henri Hess Thomas Graham 1802-1850 1805-1869 Heinrich R. Hertz, 1857-1894 Max Planck 1858-1947 Blaise Pascal 1623-1662 Justus von Liebig (1803-1873 Robert Boyle, 1627-1691 James Joule (1818-1889) Wolfgang Pauli 1900-1958 Isaac Newton 1643-1727 Rudolph Clausius 1822-1888 Werner Karl Heisenberg 1901-1976 Charles Augustin Coulomb 1735-1806 William Thompson Lord Kelvin, 1824-1907 Linus Pauling 1901-1994 Fitch Rule G3: Science is Referential Example 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 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 aamu bamu ...... zamu i i i A B Z Example. Lead carbonate is principle component of white lead used in all white paints prior to WWII. Calculate the molar mass of lead carbonate Element #atoms amu Lead carbonate = PbCO3 Pb 1 207.2 principle component paint C 1 12.01 before WWII O 3 16.00 Molar mass= unknown Total amu 207.2 12.01 48.00 267.21 267.2 MM n amu aamu bamu ...... zamu i i i A B Z Example: Acetylsalicylic acid, C9H 8O4, is the active ingredient of aspirin. What is the mass in grams of 0.509 mol acetylsalicylic acid? Acetylsalicylic acid C9H 8O4 Active ingredient of aspirin 0.509 mol acetylsalicylic acid Mass? MM m MM n MM n m n amu aamu bamu ...... zamu i i A i MM 912.01 81008 . 416.00 g MM 18015 . mole g . . g 18015 0.509mole m 917 mole B Z Mass % atomic weight of element %element # atoms of that element 100% molar mass of compound 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 O 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 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 %Ca 1 100% 100.09 %Ca 4004 . 1 Ca 12.01 %C 1 100% 100.09 %C 119992% . 1200% . 16.00 % O 3 100% 100.09 %O 4795683885 . 4796% . 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. Are We Done? 47.96 12.00 40.04 100% Check answer Very Important! Converting between g, moles, and number of particles nN A particles Mass=m Molar grams Moles=n Avogadro’s Moles Mass m MM n particles Formula Units Number The Golden Bridge Stoichiometry or #of moles 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 m MM n MM Pb C O 207.2 12.01 3(16.00) 207.2 12.01 48.00 267.21 267.2 m n MM n amu aamu bamu ...... zamu i i i A B Z 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 14.8g n g 267.2 mole n 0.055389221mole n 00554 . mole Simplest Formula from Chemical Analysis (Empirical) 1. 2. Mass scale is based on atomic number of C 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) 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 m n 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 m n MM 6.64 gK n 01698 . molK 0170 . molK gK 39.10 1mole 8.84 gCr n 01700 . molCr 0170 . molCr gCr 52.00 mol 9.52 gO n 0.59500molO 0.595molO gO 16.00 mol 0.170molK 0.170molCr 0.595mol O Simplest Formula from Chemical Analysis (Empirical) 1. 2. Mass scale is based on atomic number of C 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 Divide by Smallest # 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 2.00molCr 2molCr 2.00molK 2molCr 7.00molO 2molCr Formula: Cations First Make this non-fractional, multiply x2 2K : 2Cr : 7 O K2Cr2O7 Context for the next problem: The Romans weren’t really Drunkards engaging in orgies Lead acetate is easily made by boiling wine in a lead pot = sapa, a sweetner used in many Roman recipes. 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. A recipe of Apicius which Uses ½ cup sapa Example : The compound that the Romans supposedly drank in excess 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 What is the simplest formula of ethyl alcohol? Romans Orgies Fall of empire 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 Example : The compound that the Romans supposedly drank in excess 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 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 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 5gtotal (C H O) 9.55gCO2 587 . gH2 O Example : The compound that the Romans supposedly drank in excess 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 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 gCO 1 moleCO 2 2 moleC moleCO2 nC 9.55gCO2 ( 12 . 01 2 ( 16 )) gCO moleCO 2 2 mole CO2 moleC nC 9.55gCO2 (44.01gCO2 moleCO2 nC 0.217mole Example : The compound that the Romans supposedly drank in excess 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 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 moleH2 O 2moleH nh gH2 O amu gH O 1 moleH O 2 2 2moleH mole H2 O n H 587 . gH2 O ( 16 . 00 2 ( 1008 . )) gH O moleH O 2 2 mole H2 O 2moleH n H 587 . gH2 O (18.016 gH2 O moleH2 O nH 0.6517mole nH 0.652mole Example : The compound that the Romans supposedly drank in excess 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 What is the simplest formula of ethyl alcohol? 5.00g ethyl alcohol How are we going to get it? gc gCO amugC amug CO 9.55 g CO2 gO in original sample? 12.01gC 5.87 g H2O 2.61g C g c 9.55gCO 44.01gCO simplest formula? gtotal gC g H gO 3 sig fig 2 2 2 2 amugH amug H2 O . gtotal 2.61gC 0.657 g H gO gH gH O elements C, H, O 500 3 sig fig Now know: gO 5.00 2.61 0.657 2(1008 . ) gC 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 0.657 g C Example : The compound that the Romans supposedly drank in excess 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 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 500 . gtotal 2.61gC 0.657 g H gO gO 5.00 2.61 0.657 moleO nO gO amu gO 1moleO nH 173 . gO (16.00gO nO 0108125 . mole nO 0108 . mole gO 173 . Example : The compound that the Romans supposedly drank in excess 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 What is the simplest formula of ethyl alcohol? Divide by smallest 5.00g ethyl alcohol 9.55 g CO2 5.87 g H2O simplest formula? elements C, H, O molO=0.108 molH=0.652 molC=0.217 0108 . molO 1 0108 . molO 0.652mol H 6.037 6 0108 . molO 0.217molC 2.009 2 0108 . molO C2H6O Mass Relations and Stoichiometry Chemical Formulas Write reactions Predict Change in amounts Relate to Mass Ratios and Stoichiometry Consider what happens when one part necessary for the Reaction (recipe) is limiting Atoms are Conserved: no alchemy Mass Relations In Reactions Writing and Balancing Chemical Reactions Lead Hand Side Reactants number atoms A as reactants = number atoms Bas reactants = number atoms C as reactants = . . . Number atoms Z as reactants = Right Hand Side Products number atoms A as products number atoms B as products number atoms C as products number atoms Z as products 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 galena PbS Burned in O2 To make Pb3 O4 and SO2 PbS O2 Pb3O4 SO2 PbS Pb3O4 1S 1S PbS O2 Pb3O4 SO2 1Pb 3Pb 3S Number in Front indicates Number of molecules 1S 3 PbS O2 Pb3O4 3Pb 3S Subscript indicating number SO2 Pb within a single molecule 3Pb 3S 3PbS O2 Pb3O4 3SO2 2O Count up total O 4O+(3x2)O=10 O 3PbS O2 Pb3O4 3SO2 From previous slide 4O+3x2O=10O 3PbS 5O2 Pb3O4 3SO2 4O+3x2O=10O 5x2O=10O Be sure to indicate the physical state of the reactants and products 3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( g ) 3 5 1 Sets up proportionalities 3 STOICHIOMETRY BUT How doe we make use Of those proportionalities? Mass Relations and Stoichiometry Chemical Formulas Write reactions Predict Change in amounts Relate to Mass Ratios and Stoichiometry Consider what happens when one part necessary for the Reaction (recipe) is limiting Mass Relations and Stoichiometry Chemical Formulas Write reactions Predict Change in amounts 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 Step wise 1. Easy conceptually = series of proportionalities 1. Longer time to compute 2. Easy to lose track of units 3. Leads to rounding errors 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!!! 3PbS( s) 5O2 ( g ) Pb3O4 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 ) 5 molO 2 ( g) gSO2 gSO2 0.804molSO2 molSO 2 Samu 2(Oamu) gSO2 0.804molSO2 molSO 2 m nMM 32.07 2(16.00) gSO2 gSO2 0804 . molSO2 molSO2 64.07 gSO2 gSO2 0.804molSO2 molSO 2 gSO2 5151228 . gSO2 515 . gSO2 3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( g ) Determine: a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts Alternative Strategy is to fold all calculations together x 134 . molO2 3molSO2 ( g ) g amus 2 g amuO2 molSO2 5molO2 ( g ) 3molSO2 ( g ) gSO 2 5 molO molSO 2 ( g) 2 3molSO2 ( g ) 64.07 gSO 2 . g 515 . g 5151228 5molO2 ( g ) molSO2 x 134 . molO2 x 134 . molO2 3PbS( s) 5O2 ( g ) Pb3O4 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 m nMM m n MM 10 3 g 1kgPb3 O4 kg molPb3 O4 formed gPb3 O4 molPb O 3 4 10 3 g 1kgPb3 O4 kg molPb3 O4 formed 3(207.2) 4(16.00) gPb3 O4 molPb3 O4 10 3 g 1kgPb3 O4 kg molPb3 O4 formed 685.6 gPb3 O4 molPb O 3 4 molPb3O4 formed 1458576429 . 1459 . molPb3O4 formed 1459 . 3PbS( s) 5O2 ( g ) Pb3O4 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 5molO2 x 1459 . molPb3O4 molPb3O4 x 1459 . 5molO2 7.295molO2 m n MM gO2 x 7.295molO2 molO 2 32.00gO2 x 7.295molO2 233.4 g molO2 3PbS( s) 5O2 ( g ) Pb3O4 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 x 1kg Pb3O4 kgPbPb xx 11kg PbO O4 3O 33 44 x 1kg Pb3O4 x 233.37 g 333 g mol 10 10 g molPbPb3O3O4 4 5molO2 g O2 kg gg kg PbPb3O3O4 4 mol Pb3O4 molO2 5molO2 mol Pb3O4 32.00 g O2 molO2 3PbS( s) 5O2 ( g ) Pb3O4 s 3SO2 ( 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 6.00gO2 32.00gO2 molO 2 01875 . molO2 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 01875 . molO2 5molO2 x 01125 x 26.9 g 3PbS( s) 5O2 ( g ) Pb3O4 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 molO molO22 33molPbS molPbS 239.27 gPbS x 6 . 00 gO xx 66.00 .00gO gO222 32 32 . 00 . 00 gO gO 5 5 molO molO molPbS 22 22 x 26.917 g x 26.9 g Chemical Formulas Write reactions Predict Change in amounts Relate to Mass Ratios and Stoichiometry Consider what happens when one part necessary for the Reaction (recipe) is limiting 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 is 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. Expelled from grammar school for detonating an explosive Limiting Reagents Made from chemicals obtained from his father’s business Justus von Liebig’s Law of the Limiting (1803-1873 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? 1pkgchips 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 192cookie 12eggs 2eggs 2 eggs Smallest yield = 72 cookies Limiting reagent = chips While I wasn’t looking, my son snitched 4 cookies and ran off to Panama. yield exp erimental % yield 100% yield theoretical % yield cookies 72 4 100% 94.44444 94% 72 Free at last Actual Yield=Theoretical Yield (most of the time) “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