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Stoichiometry Molarity and more Some questions require whiteboards. If you want to draw, you can come after school and draw on the big board. Bonus points may be available for suggesting good wrong answers for future use. So if you were thinking that the answer you really wanted wasn’t among the choices, put it on a piece of paper to submit. Write down the slide # and submit your suggestion. 1 The following diagram represents the collection of elements formed by the decomposition of a compound. The blue spheres represent nitrogen atoms and the red spheres represent oxygen atoms, what was the empirical formula of the original compound? 1. N6O12 • N3O6 • NO2 2 The following diagram represents the collection of elements formed by the decomposition of a compound. The blue spheres represent nitrogen atoms and the red spheres represent oxygen atoms, what was the empirical formula of the original compound? 1. N6O12 • N3O6 • NO2 • We can not know the molecular formula from the information given. • We only know the ratio of N’s to O’s in the original compound. 3 The following diagram represents the collection of carbon dioxide and water formed by the decomposition of a hydrocarbon. What was the empirical formula of the original hydrocarbon? 1. C4H16 • C2H8 • CH4 4 The following diagram represents the collection of carbon dioxide and water formed by the decomposition of a hydrocarbon. What was the empirical formula of the original hydrocarbon? • C4H16 • C 2H 8 • CH4 • While the diagram indicates 4 carbons, and you might think there could have been 1 C4H16, 2 C2H8, or 4 CH4. • However, the maximum number of H’s that can attach to C’s is CnH2n+2. Thus to achieve the 1:4 C:H ratio, both the empirical and molecular 5 The following diagram represents a high-temperature reaction between CH4 and H2O. Based on this diagram, write a balanced chemical equation to represent this reaction. 6 The following diagram represents a high-temperature reaction between CH4 and H2O. Based on this diagram, write a balanced chemical equation to represent this reaction. 1. CH4 + H2O → CO2 + 3H2 2. While the diagram actually represents • 2CH4 + 2H2O → 2CO2 + 6H2 3. It is more appropriate to write chemical equations is the lowest whole number ratio. 7 Nitrogen and hydrogen react to form ammonia (NH3). Consider the model of the mixture shown below. Draw a representation of the product mixture, assuming the reaction goes to completion. Which color sphere best represents nitrogen and which color for hydrogen? Break out the scrap paper to sketch a response. 8 Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3). Consider the model of the mixture shown below. Blue spheres = N and white spheres = H. Draw a representation of the product mixture, assuming the reaction goes to completion. 1. N2 + 3H2 → 2NH3 2. 8N’s, 4N2 require 24 H’s, 12H2 for a complete 3. 4. reaction. Only 9H2 are present, thus H2 limits. 9H2 require 3N2, one N2 in excess, and 6NH3 are produced. 9 Nitrogen monoxide and oxygen react to form nitrogen dioxide. Consider the model of the mixture shown below. Blue spheres = N and red spheres = O. Draw a representation of the product mixture, assuming the reaction goes to completion. Break out the whiteboards to sketch a response. 10 Nitrogen monoxide and oxygen react to form nitrogen dioxide. Consider the model of the mixture shown below. Blue spheres = N and white spheres = O. Draw a representation of the product mixture, assuming the reaction goes to completion. 1. 2NO + O2 → 2NO2 2. 8NO require 4O2 for a complete reaction. 3. 5O2 are present, thus O2 is in excess and NO 4. limits. 8NO require 4O2, one O2 in excess, and 8NO2 are produced. 11 The molarity an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 ml of solution would be No calculator 1. 2.1 M 2. 3. 4. 5. 6. 0.20 M 0.05 M 50 M 0.0084 M 8.4 M 12 The molarity an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 ml of solution moles 1. 2.1 be M would Molarity, M = 2. 0.20 M 1mol 4.2g 0.1mol • 42g 3. 4. 5. 6. 0.05 M 50 M 0.0084 M 8.4 M V (L) 0.1mol 0.20M 0.5L 13 The molarity of the sodium ions in an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 ml of solution. No calculator 1. 2. 3. 4. 5. 2.1 M 4.2 M 0.20 M 0.10 M 0.05 M 14 The molarity of the sodium ions an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 ml of solution. 1. 2.1 M 1mol 0.1mol 4.2g 0.1mol 0.20M 2. 4.2 M 42g 0.5L 3. 0.20 M of the solution • the molarity of each ion is present in a 1:1 ratio with the “molecule” and thus has the same molarity, 0.20 M for each ion, Na+ and F− + − • NaF → Na + F 4. 0.10 M 5. 0.05 M 15 In a 0.010 M solution of K2SO4 the total concentration of all the ions is 1. 2. 3. 4. 5. No calculator 0.010 M 0.020 M 0.030 M 0.070 M impossible to determine. 16 In a 0.010 M solution of K2SO4 the total concentration of the ions is 1. 0.010 M 2. 0.020 M 3. 0.030 M • When dissolves, 3 ions are produced. + 2K 2− • K2SO4 → • resulting in 3 ions per particle dissolved ✓ 0.020 M solution of K+ ions and 0.010 M solution of SO42− ions + SO4 4. 0.070 M 5. impossible to determine. 17 When making a 1.0 M aqueous solution of NaCl. Select all that apply. 1. It is best to dissolve 58.44 g of NaCl in a 2000 ml beaker and add 1000 ml of water with a graduated cylinder. 2. It is best to dissolve 29.22 g of NaCl in a 2000 ml beaker and use a graduated cylinder to add 1000 ml of water. 3. It is best to put 58.44 g of NaCl in a volumetric flask and then add 1000 ml of water. 4. It is best to dissolve 58.44 g of NaCl in some water in a volumetric flask and then fill the volumetric flask up to the 1 L mark. 18 When making a 1.0 M aqueous solution of NaCl. Select all that apply. 1. It is best to dissolve 58.44 g of NaCl in a 2000 ml beaker and add 1000 ml of water with a graduated cylinder. 2. It is best to dissolve 58.44 g of NaCl in a 2000 ml beaker and use a graduated cylinder to add 1000 ml of water. 3. It is best to put 58.44 g of NaCl in a volumetric flask and then add 1000 ml of water. 4. It is best to dissolve 58.44 g of NaCl in some water in a volumetric flask and then fill the volumetric flask up to the 1 L mark. • This of course will allow for the room that the dissolved salt will take up as part of the total volume moles of the solution. Molarity, M = This is volume of the solution, not volume of water added. V (L) 19 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M solution of Na2CO3. 1. Write out balanced “overall” equation on your Mega-white boards. 2. Be sure and indicate the precipitate. 20 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 1. Write out balanced “overall” equation on paper. 2. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 3. Net ionic equation? 21 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 1. Mega-white boards 2. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 2− + 3+ 2Al → Al2(CO3)3(ppt) 3. 3CO3 4. sodium and nitrate ions are spectator ions 22 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the mass of the dried precipitate? 1. 7.8 g 2. 70.2 g 3. 0.19 g 4. 0.28 g 5. 0.56 g 6. 187 g MM (g/mol) 3Na2CO3 = 106 Al(NO3)3 = 213 Al2(CO3)3 = 234 NaNO3 = 85 23 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the mass of the dried precipitate? 1. #3 0.19 g 2. 2.4 mmol each, and because of the 3:2 ratio between the reactants, the 3Na2CO3 limits 1Al2 (CO3 )3 234g 2.4mmol 0.8mmol 187mg 3Na2CO3 1mol 3. which equals 0.19 g Al2(CO3)3 MM (g/mol) 3Na2CO3 = 106 Al(NO3)3 = 213 Al2(CO3)3 = 234 NaNO3 = 85 24 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the molar concentration of each ion in solution after the reaction? 1. Mega-white boards MM (g/mol) 3Na2CO3 = 106 Al(NO3)3 = 213 Al2(CO3)3 = 234 NaNO3 = 85 25 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3. 3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the molar concentration of each ion in solution after the reaction? Remember - the volume of the solution has doubled which affects the concentrations 2. Treat the spectator ion concentrations as completely unreacted in twice the volume. 3. Assume the limiting ion in the precipitate is effectively gone from solution. 4. Calculate the mmoles left over (this means a subtraction) of the excess ion that forms the precipitate and calculate its molarity in the total 26 volume. 1. 24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of 0.10 M Na2CO3.3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3 What is the molar concentration of each ion in solution after the reaction? 2Na 4.8mmol 2.4mmolNa2CO3 0.10MNa Still in Solution 1Na2CO3 48ml 3NO3 7.2mmolNO3 2.4mmolAl(NO3 )3 0.15M NO3 Still in Solution Al(NO3 )3 48ml 1. Assume the CO32− ≃ 0 1Al 3 2.4mmol1Al(NO3 )3 2.4mmolAl 3inSolution 1Al(NO3 )3 1Al(NO3 )3 2.4mmolNa2CO3 0.8mmolAl(NO3 )3 needed 3Na2CO3 3 1.6mmolAl Left 3 3 2.4mmolAl 0.8mmolAl 0.033MAl 3 Still in Solution 48ml 27 − Calculate the Molarity of NO3 ions in 40 ml of a 0.20 M solution of Mg(NO3)2 after reacting with 60 ml of a 0.10 M solution of Al(NO3)3 No Calculator 1. Type in a numerical answer. 28 − ions Calculate the Molarity of NO3 in 40 ml of a 0.20 M solution of Mg(NO3)2 after reacting with 60 ml of a 0.10 M solution of Al(NO3)3 3 2NO 40ml 16mmol NO3 1Mg(NO3 )2 No Calculator 3 3NO 60ml 18mmol NO3 1Al(NO3 )3 3 34NO total 3.4M NO3 100ml Total 29 What is the mass of copper(II) sulfate (molecular weight = 159.6 g/mol) in 40. mL of 2.0 M copper(II) sulfate? 1. 2. 3. 4. 5. 6. 3.2 g 5.5 g 8.8 g 13 g 16 g 32 g No Calculator 30 What is the mass of copper(II) sulfate (molecular weight = 159.6 g/mol) in 40. mL of 2.0 M copper(II) sulfate? 1. 2. 3. 4. 3.2 g No Calculator 5.5 g Learn to estimate using easy math 8.8 g 0.04L 0.08mol NO ; 0.1mol 3 13 g 5. 0.1 mol would mean 16g, half that or 0.05 mol would be ~8g, thus the 0.08 mol must be in between≃13g 5. 16 g 6. 32 g 31 How many grams of zinc nitrate(189 g/mol) contain 48 grams of oxygen atoms? 1. 2. 3. 4. 5. No Calculator 95 g 125 g 145 g 165 g none of the above 32 How many grams of zinc nitrate(189 g/mol) contain 48 grams of oxygen atoms? No Calculator 1. 95 g 2. Look for easy math. Zn(NO3)2 3. 3 O’s = 48g, since 2 NO3‘s per zinc nitrate, you only need 0.5 mol of zinc nitrate = half of molar mass 2. 125 g 3. 145 g 4. 165 g 5. 189 g 33 A 2.00 gram mixture of calcium carbonate and calcium chloride are treated with an excess of hydrochloric acid and 0.66 grams of carbon dioxide (44 g/mol) are produced. What is the percent of CaCO3 (100 g/mol) by mass in the original mixture? No Calculator 1. 2. 3. 4. 5. 25% Write a net ionic equation to 30% represent the hydrochloric acid 50% + calcium carbonate reaction 75% 90% 34 A 2.00 gram mixture of calcium carbonate and calcium chloride are treated with an excess of hydrochloric acid and 0.66 grams of carbon dioxide (44 g/mol) are produced. What is the percent of CaCO3 (100 g/mol) by mass in the original mixture? 1. 2. 3. 4. 25% 30% 50% 75% No Calculator CaCO3 + + H ➙ H2O + CO2 + 2+ Ca 0.66 0.33 3 60ml 0.015 1.5, move the decimal... • 2 44 22 • thus 0.015 mol of CaCO3 to start = 1.5 g and 1.5 is 75% of 2 g total 5. 90% 35 In which compound below is the mass ratio of copper to sulfur closest to 2:1? 1. 2. 3. 4. 5. CuS2 CuS Cu2S Cu2S3 Cu3S2 No Calculators 36 In which compound below is the mass ratio of copper to sulfur closest to 2:1? 1. 2. 3. 4. 5. CuS2 ~1:1 CuS ~2:1 Cu2S ~4:1 Cu2S3 ~4:3 Cu3S2 ~6:4 Expect easy math. MM Cu=63.6 and S=32. For calculation purposes assume 60 and 30 No Calculators 37 A certain compound contains only one sodium atom and is 5% sodium by mass. What is the molar mass of the compound? No Calculators 1. 2. 3. 4. 5. 460 g/mol 230 g/mol 110 g/mol 55 g/mol none of the above 38 A certain compound contains only one sodium atom and is 5% sodium by mass. What is the molar mass of the compound? No Calculators 1. 460 g/mol • 2. 3. 4. 5. make the math easy. molar mass of Na is 23, so 23 is 10% of 230, and thus 5% of 460. 230 g/mol 110 g/mol 55 g/mol none of the above 39 Analysis of a tellurium oxide compound indicated 84.22 % tellurium. The molar mass is between 580 and 610 g/mole. Determine the molecular formula. Yes, Calculators 1. TeO 2. Te2O3 3. Te2O 40 Analysis of a tellurium oxide compound indicated 84.22 % tellurium. The molar mass is between 580 and 610 g/mole. Determine the molecular formula. 1. 2. 3. 46 The Molecular Formula is Te4O6. 84.22 0.66 Te2O3 MM 0.66 = 303.2 1thus 2 2 Te4O6. 127.6 0.66 15.78 0.986 16 0.986 1.5 2 3 0.66 41 Determine the empirical formula of a nerve gas that gave the following analysis: 39.10% C, 7.67% H, 26.11% O, 16.82% P, Yes, Calculators 10.32 % F. 1. 2. 3. The Formula is CvHwOxPyFz. Type your answer in as a number vwxyz (no spaces, no letters, just one number) i.e. for C3H12O4PF2, you would type in 312412 42 Determine the empirical formula of a nerve gas that gave the following analysis: 39.10% C, 7.67% H, 26.11% O, 16.82% P, 10.32 % F. 1. 614311 39.1 3.26 3.26 6C 12 0.543 2. The Empirical 7.67 7.67 7.67 14 H Formula is 1 0.543 C6H14O3PF 26.11 1.63 1.63 3O 16 0.543 16.82 0.543 0.543 1 P 31 0.543 10.32 0.543 0.543 1 F 19 0.543 43 Combustion analysis is used to determine the amount of carbon, hydrogen, and oxygen in a combustible compound. 1. 2. 3. 4. 5. Measure the mass of compound to be combusted. Measure the mass of water produced. Measure the mass of carbon dioxide produced oxygen will make up any remaining mass in original compound. CxHyOz + O2 ➙ H2O + CO2 44 Cumene is a organic compound contains only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of 47.6 mg of cumene produces 156.8 mg of carbon dioxide and 42.8 mg of water. The molar mass of cumene is between 115 and 125 g/mole. Determine the molecular formula. Yes, Calculators 1. Determine moles of C Determine moles of H 3. empirical ratio ➙ then molar mass 4. Calculate for molecular formula 2. 45 Cumene is a organic compound contains only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of 47.6 mg of cumene produces 156.8 mg of carbon dioxide and 42.8 mg of water. The molar mass of cumene is between 115 and 125 g/mole. Determine the molecular formula. 1mol 1C 3.56mmolC 156.8mgCO2 3.56mmolC 1 3 3C 44g 1CO2 3.56 1mol 2H 4.76mmolH 42.8mgH 2O 4.76mmolH 1.33 3 4H 18g 1H 2O 3.56 1. thus C3H4 2. thus C9H12 g C3 H 4 40 mol 120g 3 40mol 46 A combustion device was used to determine the empirical formula of an organic compound. A 0.6349 g sample was burned and produced 1.603 g of carbon dioxide and 0.2810 g of water and no other oxides. Determine the empirical formula for the compound. 1. 2. 3. 4. 5. Compounds that may contain oxygen are a bit trickier. Yes, Calculators determine mass of C, mass of H check to see if there there is any “missing mass” that would be oxygen back to moles of each element determine empirical formula, molar mass, then molecular formula 47 A combustion device was used to determine the empirical formula of an organic compound. A 0.6349 g sample was burned and produced 1.603 g of carbon dioxide and 0.2810 g of water and no other oxides. Determine the empirical formula for the compound. 2g 0.281gH 2O 0.0312gH 18g 12g 1.603gCO2 0.437gC 44g 0.6349gTotal 0.437gC 0.0312gH 0.167gO 1mol 0.0364molC 0.437gC 0.0364molC 3.5 2 = 7 C 12g 0.0104 1mol 0.0312molH 0.0312gH 0.0312molH 3H 2=6H 1g 0.0104 1mol 0.167gO 0.0104molO 16g 1. thus C7H6O2 0.0104molO 1 Oxy 2 = 2 Oxy 0.0104 48 A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis. Combustion of a 32.00 mg sample yielded 78.99 mg of carbon dioxide and 19.96 mg of water. Analysis for nitrogen showed that the compound contained 4.62 % N by mass. Yes, Calculators Calculate the empirical formula. • • • The Formula is CwHxOyNz. Type your answer in as a number wxyz (no spaces, no letters, just one number) i.e. for C3H12O4N2, you would type in 31242 49 A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis. Combustion of a 32.00 mg sample yielded 78.99 mg of carbon dioxide and 19.96 mg of water. Analysis for nitrogen showed that the compound contained 4.62 % N by mass. Calculate the empirical formula. 1mol 1C 78.99mgCO2 1.80mmolC 44g 1CO2 1mol 2H 19.96mgH 2O 2.22mmolH 18g 1H 2O • • 172141 C17H21O4N 1mol 4.62% 32mg 1.47mg 0.106mmolN 14g 1g 12g 2.22mmolH 2.22mg H and 1.80mmolC 21.6mg C 1mol 1mol 32mg 21.6mgC 2.22mgH 1.47mgN 6.71mg O 1.80mmolC 17 C 0.106mmolN 2.22mmolH 21 H 0.106mmolN 0.106mmolN 1 N 0.106mmolN 0.42mmolO 4O 0.106mmolN 1mol 6.71mgO 0.42mmol O 16g 50 50. grams of an unknown hydrocarbon are burned in an excess of oxygen to form 130 grams of carbon dioxide and 54 grams of water. What might this hydrocarbon be? 1. 2. 3. 4. 5. CH4 C2H6 C3H4 C3H6 C2H2 No Calculators 51 50. grams of an unknown hydrocarbon are burned in an excess of oxygen to form 130 grams of carbon dioxide and 54 grams of water. What might this hydrocarbon be? 1. 2. 3. 4. 5. CH4 C2H6 C3H4 C3H6 C2H2 No Calculators 1mol 130gCO2 ; 3molCO2 3molC 44g 1mol 54gH 2O ; 3molH 2O 6molH 18g 52 The combustion analysis of 19.8mg of an organic acid produced 39.6 mg of carbon dioxide and 16.2 mg of water. The molar mass is ~88 g/mole. Determine the molecular formula. Yes, Calculators 53 The combustion analysis of 19.8 mg of an organic acid produced 39.6 mg of carbon dioxide and 16.2 mg of water. The molar mass is ~88 g/mole. Determine the molecular formula. 2g 12g 16.2mgH 2O 1.8mgH 39.6mgCO2 10.8mgC 18g 44g 19.8mgTotal 10.8mgC 1.8mgH 7.2mgO 1mol 0.9mmolC 10.8mgC 0.9mmolC 2C 12g 0.45 1mol 1.8mgH 1.8mmolH 1g 1mol 7.2mgO 0.45mmolO 16g 1. 2. 1.8mmolH 4H 0.45 0.45mmolO 1O 0.45 thus C2H4O1 MM=44 thus C4H8O2 Now let’s draw a possible structural formula of this 54 organic acid. = A possible structural formula of an organic acid with the formula C4H8O2 1. All organic acids have a -COOH group. 2. Thus the formula below would be an option. O 3. Name? CH3CH2CH2 -C-O-H 55 A possible structural formula of an organic acid with the formula C4H8O2 All organic acids are • 2. 3. carbonroot-oic acid This 4-carbon acid would be butanoic acid O CH3CH2CH2 -C-O-H = 1. 56 Dianabol is one of the anabolic steroids that has been used by some athletes to increase the size and strength of their muscles. It is similar to the male hormone testosterone. Some studies indicate that the desired effects of the drug are minimal, and the side effects, which include sterility, behavior changes, increased risk of liver cancer and heart disease, keep most people from using it. The molecular formula of Dianabol, which consists of carbon, hydrogen, and oxygen, can be determined using the data from two different experiments. In the first experiment, 14.765 g of Dianabol is burned, and 43.257 g CO2 and 12.395 g H2O are formed. In the second experiment, the molecular mass of Dianabol is found to be 300.44 g/mole. What is the molecular formula for Dianabol 1. Again, type in a numerical answer in order CwHx(and)Oy(if necessary). 57 The molecular formula of Dianabol, which consists of carbon, hydrogen, and oxygen, can be determined using the data from two different experiments. In the first experiment, 14.765 g of Dianabol is burned, and 43.257 g CO2 and 12.395 g H2O are formed. In the second experiment, the molecular mass of Dianabol is found to be 300.44 g/mole. What is the molecular formula for Dianabol 1. C20H28O2 58 One of the additives in unleaded gasoline that replaced tetraethyl lead in leaded gasoline is called MTBE. When 15.078 g MTBE is burned completely, 37.640 g CO2 and 18.489 g H2O form. In a separate experiment the molecular mass of MTBE is found to be 88.150. What is the molecular formula for MTBE? 1. Again, type in a numerical answer in order CwHx(and)Oy(if necessary). 59 One of the additives in unleaded gasoline that replaced tetraethyl lead in leaded gasoline is called MTBE. When 15.078 g MTBE is burned completely, 37.640 g CO2 and 18.489 g H2O form. In a separate experiment the molecular mass of MTBE is found to be 88.150. What is the molecular formula for MTBE? 1. C5H12O 60 For those of you who have found this last material too easy.... Try these next three challenge problems to stimulate your brain and keep you sharp. These would NOT likely show up on an AP exam. 61 Special Challenge Problem #1 When aluminum is heated with an element from Group 6A of the periodic table an ionic compound is formed. When an experiment is performed with an unknownı element of group 6A, the product is 18.56% aluminum by mass. Determine the identity of the reacting element and the formula of the compound. 1. Type in the atomic number of the element. 62 Special Challenge Problem When aluminum is heated with an element from Group 6A of the periodic table an ionic compound is formed. When an experiment is performed with an unknownı element of group 6A, the product is 18.56% aluminum by mass. Determine the identity of the reacting element and the formula of the compound. 1. 2. Hint #1 Since the element is in group 6A, we know that the formula must be Al2X3 63 Special Challenge Problem #1 When aluminum is heated with an element from Group 6A of the periodic table an ionic compound is formed. When an experiment is performed with an unknownı element of group 6A, the product is 18.56% aluminum by mass. Determine the identity of the reacting element and the formula of the compound. 1. Hint #2 Set up a ratio to determine the total molar mass of the compound compared to the total mass of aluminum, using the molar masses. 3. 54/MM = 18.56/100% 4. MM = 291 2. 64 Special Challenge Problem #1 When aluminum is heated with an element from Group 6A of the periodic table an ionic compound is formed. When an experiment is performed with an unknownı element of group 6A, the product is 18.56% aluminum by mass. Determine the identity of the reacting element and the formula of the compound. 1. 2. 3. 4. 5. And finally... Knowing the molar mass of Al, solve for the MM of X 291 - 54 = 237 X3 = 237 so MM of X = 79 Thus it must be Se 65 Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of 10.00 g of XCl2 with excess chlorine forms 12.55 g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. 1. Type in the atomic number of the element. 66 Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of 10.00 g of XCl2 with excess chlorine forms 12.55 g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. 1. 2. 3. 4. 5. Hint #1 XCl2 + Cl2 --> XCl4 Determine the mass of Cl2 in XCl4 Then the moles of Cl2 which is the same as the moles of X Which can lead you to the molar mass of X 67 Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of 10.00 g of XCl2 with excess chlorine forms 12.55 g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. Hint #2 2. 10 g of XCl2 + Cl2 --> 12.55 g of XCl4 3. Thus there must be 2.55 g Cl2 in the XCl4 and thus there must also be 2.55 g Cl2 in the XCl2 1. 68 Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of 10.00 g of XCl2 with excess chlorine forms 12.55 g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. 1. Hint #3 Determine the moles of Cl2 in XCl2, and since there is a 1:1 ratio of X:Cl2, we would know the moles of X. 3. 2.55g of Cl2 * 1 mole Cl2/71 g * 1 mole X/1mole Cl2 = 0.0359 moles X in XCl2, 2. 69 Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of 10.00 g of XCl2 with excess chlorine forms 12.55 g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. and finally... 2. Since 2.55g of the 10g of XCl2 is Cl, 7.45g of the 10g must be X. Calculate the molar mass of X 3. 7.45 g /0.0359 moles = ~207g/mole. 4. Thus X might be lead. 1. 70 Special Challenge Problem #3 A 1.500 g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce 1.252 g of pure copper metal. Calculate the percent composition of the mixture. (i.e. What percent of the mixture is each of the two compounds?) 1. Type in the % of the CuO compound. 71 Special Challenge Problem #3 A 1.500 g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce 1.252 g of pure copper metal. Calculate the percent composition of the mixture. (i.e. What percent of the mixture is each of the two compounds?) Hint #1 Solving this problem will require two variables, thus two equations. First equation: 1. 2. 3. • The mass of each compound in the mixture will equal the total mass. Second equation: 4. • The moles of copper in each compound will equal the moles of copper in the total mixture. 72 Special Challenge Problem #3 A 1.500 g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce 1.252 g of pure copper metal. Calculate the percent composition of the mixture. (i.e. What percent of the mixture is each of the two compounds?) First equation - total mass: 2. You could name the mass of Cu2O as x and the mass of CuO as y. The two substances make up the 1.500 total mass of mixture. 3. xg + yg = 1.500g 1. 73 Special Challenge Problem #3 A 1.500 g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce 1.252 g of pure copper metal. Calculate the percent composition of the mixture. (i.e. What percent of the mixture is each of the two compounds?) 1. 2. 3. 4. 5. 6. Second equation - total moles: We know that the moles of copper in the two compounds will add up to the total moles of copper. Xg * 1 mole Cu2O/143.1 g * 2 Cu/1 Cu2O = moles of copper in Cu2O Yg * 1 mole CuO/79.55 g * 1 Cu/1 CuO = moles of copper in CuO 1.252 g Cu * 1 mole/63.55 g = total moles Cu 0.01398 x + 0.01257 y = 0.0197 74 Special Challenge Problem #3 A 1.500 g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce 1.252 g of pure copper metal. Calculate the percent composition of the mixture. (i.e. What percent of the mixture is each of the two compounds?) and finally... Simply the last equation 1. 2. • • Subtract the other x y equation. 3. • • 4. 5. 0.01398 x + 0.01257 y = 0.0197 1.112 x + y = 1.567 1.112 x + y = 1.567 (-x -y = -1.500) equals: 0.112 x = 0.067 thus X = 0.6 g Cu2O 0.6 g Cu2O / 1.500 g = 40% Cu2O in mixture Thus 60% CuO in mixture 75 Oxidation Numbers Keeping Track of Electrons in Redox Reactions 76 Oxidation • Oxidation Numbers aka Oxidation States • A concept devised to keep track of electrons in a redox reaction. • Increase in oxidation number = oxidation ✓ LEO (Lose Electrons = Oxidation) • Decrease in oxidation number = reduction ✓ says GER (Gain Electrons = Reduction) • The sum of oxidation numbers in neutral compounds must = 0 • The sum of oxidation numbers in a polyatomic ion = the charge of the ion 77 Determining Oxidation # • The charge on a monotomic ion is the Ox # • Atoms in elemental form are zero • • H in H2, atoms in a lump of iron, Fe, P atoms in P4 Nonmetals usually have negative Ox # - but they can be positive. • • • oxygen is −2 in both ionic and molecular compounds except in peroxides in which the oxidation number is -1. hydrogen is always +1 when bonded to nonmetals and always −1 when bonded to metals. fluorine is −1 in all compounds. The other halogens 78 The many oxidation states of manganese, Mn 1.Mn , Mn = 0 2.MnCl2 , Mn = +2 3.MnF3 , Mn = +3 4.MnO2 , Mn = +4 5.K3MnO4 , Mn = +5 6.K2MnO4 , Mn = +6 7.KMnO4 , Mn = +7 8. Methylcyclopentadienyl manganese tricarbonyl MnC5H4CH3(CO)3 , Mn = +1 What?? 79 Determine the oxidation number of each element in the following substances 1. CaCl2 2. PbO2 3. Cl2 4. S8 5. SO4 280 Determine the oxidation number of each element in the following substances +2 -1 +4 -2 1. CaCl2 +4 -2 2. PbO2 0 +1 -2 +1 -1 +3 -1 +5 -2 3. Cl2 0 4. S8 +6 -2 5. SO4 2− 81 Determine the oxidation number of phosphorus in Mg2P2O7 1. Submit a numeric value. If it is negative, put on the - sign. If positive, just leave it. 82 Determine the oxidation number of phosphorus in Mg2P2O7 +2 +5 -2 1. Mg2P2O7 83 Determine the oxidation number of iron in K4Fe(CN)6 1. Submit a numeric value. If it is negative, put on the - sign. If positive, just leave it. 84 Determine the oxidation number of iron in K4Fe(CN)6 +1 +2 -1 1. K4Fe(CN)6 (CN= -1) 85 Determine the oxidation number of chromium in Na2Cr2O7 86 Determine the oxidation number of chromium in Na2Cr2O7 +1 +6 -2 1. Na2Cr2O7 87 Nitrogen is the master of multiple oxidation states. Determine the oxidation number of nitrogen in each of the following compounds. 1. 2. 3. 4. 5. NO N2O NO2 N2H4 NH 88 Nitrogen is the master of multiple oxidation states. Determine the oxidation number of nitrogen in each of the following compounds. +2 -2 +1 -3 1. NO -2 +1 0 2. N2O +4 -2 +3 -2 -2 +5 -2 3. NO2 +1 4. N2H4 -3 +1 5. NH3 89