The Mole. Empirical & Molecular Formulas. Chapter 3

Transcription

1 The Mole Empirical & Molecular Formulas Chapter 3 I am always looking for good wrong answers for future use. So if the answer you really wanted wasn t among the choices, please tell me (or at the very least, write down the slide # and submit your suggestion.)!1

2 How many green squares should be placed on the right side of the bottom scale in order to make it balanced when released?!2

3 How many green squares should be placed on the right side of the bottom scale in order to make it balanced when released? A. 3 B. 4 C. 5 D. 6 E. 7!3

4 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 2. N3O6 3. NO2 4. NO 5. NO3!4

5 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 2. N3O6 3. NO2 4. NO 5. NO3 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.!5

6 The following diagram represents the collection of carbon dioxide and water formed by the combustion of a hydrocarbon. What was the empirical formula of the original hydrocarbon? 1. C4H16 2. C2H4 3. C2H8 4. CH4 5. CH2!6

7 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 2. C2H4 3. C2H8 4. CH4 5. CH2 6. 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 formula must have been CH4.!7

8 Mass Percent Problems Staple your Data Table to the back of your LAD A1 sheets Hand in over by the window Clickers Today!8

9 Analysis of a tellurium oxide compound indicated % tellurium. The molar mass is approximately 600 g/mole. Determine the molecular formula. Yes, Calculators 1. TeO 2. Te2O3 3. Te2O 4. TeO2 5. TeO3 6. Te3O 7. Te4O 8. TeO4 9. Te4O6!9

10 Analysis of a tellurium oxide compound indicated % tellurium. The molar mass approximately 600 g/mole. Determine the molecular formula. The Molecular Formula is Te4O = = 1 2 = = = = 3 Te2O3 MM = thus Te4O6.!10

11 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, % F. Yes, Calculators The Formula is CvHwOxPyFz.!11

12 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, % F. The Empirical Formula is C6H14O3PF = = 6 C = = 14 H = = 3 O = = 1 P = = 1 F!12

13 How many grams of zinc nitrate(189 g/ mol) contain 48 grams of oxygen atoms? No Calculator g g g g g 6. none of the above!13

14 How many grams of zinc nitrate(189 g/ mol) contain 48 grams of oxygen atoms? g ( ) 2 48gO 1mol 16g 1Zn NO 3 6O's 189g 1mol = Look for easy math. Zn(NO3)2 3 O s = 48g, since 2 NO3 s per zinc nitrate, you only need 0.5 mol of zinc nitrate = half of molar mass g g g No Calculator g!14

15 In which compound below is the mass ratio of copper to sulfur closest to 2:1? No Calculators 1. CuS2 2. CuS 3. Cu2S 4. Cu2S3 5. Cu3S2!15

16 In which compound below is the mass ratio of copper to sulfur closest to 2:1? 1. CuS2 ~1:1 2. CuS ~2:1 3. Cu2S ~4:1 Expect easy math. MM Cu=63.6 and S=32. For calculation purposes assume 64 and Cu2S3 ~4:3 5. Cu3S2 ~6:4 No Calculators!16

17 A certain compound contains only one sodium atom and is 5% sodium by mass. What is the molar mass of the compound? No Calculators g/mol g/mol g/mol g/mol 5. none of the above!17

18 A certain compound contains only one sodium atom and is 5% sodium by mass. What is the molar mass of the compound? No Calculators g/mol make the math easy. molar mass of Na is 23, so 23 is 10% of 230, and thus 5% of g/mol g/mol g/mol 5. none of the above!18

19 Combustion Analysis Determining Empirical Formula by Combustion!19

20 Combustion analysis is used to determine the amount of carbon, hydrogen, and oxygen in a combustible compound. 1. Measure the mass of compound to be combusted. 2. Measure the mass of water produced. 3. Measure the mass of carbon dioxide produced 4. oxygen will make up any remaining mass in original compound. 5. CxHyOz + O2 H2O + CO2!20

21 A combustion device was used to determine the empirical formula of an organic compound. A g sample was burned and produced g of carbon dioxide and g of water and no other oxides. Determine the empirical formula for the compound. watch out for compounds that may contain oxygen. Yes, Calculators change to moles of each element determine mass of C, mass of H check to see if there there is any missing mass that would be oxygen!21

22 A combustion device was used to determine the empirical formula of an organic compound. A g sample was burned and produced g of carbon dioxide and g of water and no other oxides. Determine the empirical formula for the compound gCO 2 1mol 44g = molC 12g 1mol = 0.437gC 0.281gH 2 O 1mol 18g = molH 2O 2H 's 1H 2 O = molH 1g 1mol = gH gTotal 0.437gC gH = 0.167gO 0.167gO 1mol 16g = molO = 1 Oxy 2 = 2 Oxy molC molH = = 7 C = 3 H 2 = 6 H C7H6O2!22

23 A combustion device was used to determine the empirical formula of an organic compound. A g sample was burned and produced g of carbon dioxide and g of water and no other oxides. Determine the empirical formula for the compound. Alternatively you could solve for the missing mass first 0.281gH 2 O 2g 18g = gH 1.603gCO 2 12g 44g = 0.437gC gTotal 0.437gC gH = 0.167gO 0.437gC 1mol 12g = molC gH 1mol 1g 0.167gO 1mol 16g = molO molC = molH molH molO = = 7 C = 3 H 2 = 6 H = 1 Oxy 2 = 2 Oxy C7H6O2!23

24 Embrace the millimole Convert molar mass of CO2 into mg per mmole The Formula is CwHxOyNz.!24

25 A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis. Combustion of a mg sample yielded mg of carbon dioxide and 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.!25

26 A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis. Combustion of a mg sample yielded mg of carbon dioxide and mg of water. Analysis for nitrogen showed that the compound contained 4.62 % N by mass. Yes, Calculators Calculate the empirical formula. Calculate millimoles and mass of C from CO2 and millimoles and mass of H from H2O mass and millimoles of N from % from mass of the above 3, determine mass of the oxygen, then convert to millimoles and calculate the empirical formula!26

27 A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis. Combustion of a mg sample yielded mg of carbon dioxide and mg of water. Analysis for nitrogen showed that the compound contained 4.62 % N by mass mgCO 2 Calculate the empirical formula. 1mol 44g 19.96mgH 2 O 1mol 18g 1C 1CO 2 = 1.80mmolC 2H 1H 2 O = 2.22mmolH 4.62% 32mg = 1.47mg 1mol 14g = 0.106mmolN 2.22mmolH 1g 1mol C17H21O4N = 2.22mg H and 1.80mmolC 12g 1mol = 21.6mg C 32mg 21.6mgC 2.22mgH 1.47mgN = 6.71mg O 6.71mgO 1mol 16g = 0.42mmol O 1.80mmolC 0.106mmolN = 17 C 2.22mmolH 0.106mmolN = 21 H 0.106mmolN 0.106mmolN = 1 N 0.42mmolO 0.106mmolN = 4 O!27

28 A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis. Combustion of a mg sample yielded mg of carbon dioxide and mg of water. Analysis for nitrogen showed that the compound contained 4.62 % N by mass. Calculate the empirical formula. C17H21O4N Cocaine!28

29 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!29

30 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. 16.2mgH O 2 2g 18g = 1.8mgH 39.6mgCO mgTotal 10.8mgC 1.8mgH = 7.2mgO 10.8mgC 1mol 12g = 0.9mmolC 1.8mgH 1mol 1g = 1.8mmolH 7.2mgO 1mol 16g = 0.45mmolO 12g 44g thus C2H4O1 MM=44 thus C4H8O2 Now let s draw a possible structural formula of this organic acid. Hint, put all the C s in a row, and put oxygens on the same C. 0.9mmolC = 2C mmolH = 4H mmolO 0.45 = 1O = 10.8mgC!30

31 A possible structural formula of an organic acid with the formula C4H8O2 All organic acids have a -COOH group. Thus the formula below would be an option. Name? CH3CH2CH2 O = -C-O-H!31

32 A possible structural formula of an organic acid with the formula C4H8O2 All organic acids are carbonroot-oic acid This 4-carbon acid would be butanoic acid O CH3CH2CH2 = -C-O-H!32

33 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, g of Dianabol is burned, and g CO2 and g H2O are formed. In the second experiment, the molecular mass of Dianabol is found to be g/mole. What is the molecular formula for Dianabol Again, type in a numerical answer in order CwHx(and)Oy(if necessary).!33

34 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, g of Dianabol is burned, and g CO2 and g H2O are formed. In the second experiment, the molecular mass of Dianabol is found to be g/mole. What is the molecular formula for Dianabol C20H28O2!34

35 One of the additives in unleaded gasoline that replaced tetraethyl lead in leaded gasoline is called MTBE. When g MTBE is burned completely, g CO2 and g H2O form. In a separate experiment the molecular mass of MTBE is found to be What is the molecular formula for MTBE? Again, type in a numerical answer in order CwHx(and)Oy(if necessary).!35

36 One of the additives in unleaded gasoline that replaced tetraethyl lead in leaded gasoline is called MTBE. When g MTBE is burned completely, g CO2 and g H2O form. In a separate experiment the molecular mass of MTBE is found to be What is the molecular formula for MTBE? C5H12O!36

37 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? No Calculators 1. CH4 2. C2H6 3. C3H4 4. C3H6 5. C2H2!37

38 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. CH4 No Calculators 2. C2H6 3. C3H4 130gCO 2 1mol 44g! 3molCO = 3molC 2 4. C3H6 54gH 2 O 1mol 18g! 3molH O = 6molH 2 5. C2H2!38

39 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 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. Determine moles of C Determine moles of H on P A.4 empirical ratio then molar mass Calculate for molecular formula!39

40 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 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 mgCO 2 1mol 44g 1C 1CO 2 = 3.56mmolC 3.56mmolC 3.56 = 1 3 = 3C 42.8mgH 2 O 1mol 18g 2H 1H 2 O = 4.76mmolH 4.76mmolH 3.56 = = 4H thus C3H4 C 3 H 4 = 40 g mol 120g 40mol = 3 thus C9H12!40

41 Stoichiometry Limiting Reactants Percent Yield Chapter 4 Sections 4.3 & 4.4 I am always looking for good wrong answers for future use. So if the answer you really wanted wasn t among the choices, please tell me (or at the very least, write down the slide # and submit your suggestion.)!41

42 What weight will the 4th scale display? A. 25 B. 26 C. 27 D. 28 E. 29 F. 30!42

43 What weight will the 4th scale display? A. 25 B. 26 C. 27 D. 28 E. 29 F. 30!43

44 The following diagram represents a high-temperature reaction between and H2O. Based on this diagram, CH4 write a balanced chemical equation to represent this reaction.!44

45 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. 2CH4 + 2H2O 2CO + 6H2 It is more appropriate to write chemical equations in the lowest whole number ratio. CH4 + H2O CO + 3H2!45

46 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? reactants Break out the scrap paper to sketch a response or write right on the desk!46

47 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. N2 + 3H2 2NH3 8N s, 4N2 require 24 H s, 12H2 for a complete reaction. Only 9H2 are present, thus H2 limits. 9H2 require 3N2, one N2 in excess, and 6NH3 are produced.!47

48 Nitrogen monoxide and oxygen gas 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. On scrap paper or the whiteboards, or on the desk with chalk to sketch a response.!48

49 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. 2NO + O2 2NO2 8NO require 4O2 for a complete reaction. 5O2 are present, thus O2 is in excess and NO limits. 8NO require 4O2, one O2 in excess, and 8NO2 are produced.!49

50 Working Problems that Involve Carbonates!50

51 Metal carbonates will nearly always react into something AND carbon dioxide. In the AP curriculum, we will see carbonates decomposing upon heating or reacting with acid. Write an equation to represent the decomposition buy strong heating of copper(ii) carbonate into copper(ii) oxide and carbon dioxide!51

52 Write an equation to represent the decomposition of copper(ii) carbonate into copper(ii) oxide and carbon dioxide. CuCO3 CuO + CO2 In the AP curriculum, we will usually see metal carbonates decomposing upon heating into the metal oxide and carbon dioxide. (Alkali carbonates do NOT decompose upon heating.)!52

53 Metal carbonates will nearly always react into something AND carbon dioxide. In the AP curriculum, we will see carbonates decomposing upon heating OR reacting with acid. Write an equation to represent the reaction of potassium carbonate with hydrochloric acid to produce potassium chloride, water, and carbon dioxide.!53

54 Write an equation to represent the reaction of potassium carbonate with hydrochloric acid to produce potassium chloride, water, and carbon dioxide. + 2HCl 2KCl + H2O + K2CO3 CO2 Let s take a closer look at what s really happening when metal carbonates react with acid.!54

55 On first inspection, this reaction should look like a double replacement reaction: K2CO3 + 2HCl 2KCl + H2CO3 however, carbonic acid is a phantom a substance that decomposes: H2CO3 H2O + CO2!55

56 The Phantoms Molecules that decompose into gases when they form as a product. As we just saw, H2CO3 decomposes into H2O and CO2 NH4OH decomposes into H2O and NH3 H2SO3 decomposes into H2O and SO2 H + with S 2 produces a gas, H2S!56

57 Write an equation to represent the reaction between nitric acid and solid calcium carbonate. Let s write the overall rxn, and let s then turn that into a net ionic rxn!57

58 Write an equation to represent the reaction between nitric acid and calcium carbonate. CaCO3 + 2HNO3 Ca(NO3)2 + H2O + CO2 CaCO3 + 2H + Ca 2+ + H2O + CO2!58

59 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. 25% 2. 30% 3. 50% 4. 75% First, we should think about what happens when hydrochloric acid is poured on calcium chloride? 5. 90%!59

60 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? Resist the temptation of just writing everything you see into an equation. CaCO3 + CaCl2 + 2HCl think about what s happening between the CaCl2 and the HCl nothing.so what about the acid and carbonate?!60

61 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? What is happening to the carbonate? CaCO3 + 2HCl CaCl2 + H2O + CO2 Look at the stoichiometry between CO2 and CaCO3 1:1 back to the question!61

62 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. 25% 2. 30% 3. 50% 4. 75% 5. 90%!62

63 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. 25% 2. 30% 3. 50% 4. 75% 5. 90% CaCO3 + 2HCl H2O + CO2 + CaCl2 0.66g 1mol 44g = 0.015molCO 2 thus 0.015molCaCO molCaCO 3 100g 1mol = 1.5gCaCO 3 1.5gCaCO = 75% 2gMixture Could you do this without a calculator?!63

64 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.!64

65 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. Type in the atomic number of the element.!65

66 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. Hint #1 Since the element is in group 6A, we know that the formula must be Al2X3!66

67 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. 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. 54/MM = 18.56/100% MM = 291!67

68 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. And finally... Knowing the molar mass of Al, solve for the MM of X = 237 X3 = 237 so MM of X = 79 Thus it must be Se!68

69 Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of g of XCl2 with excess chlorine forms g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. Type in the atomic number of the element.!69

70 Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of g of XCl2 with excess chlorine forms g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. 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!70

71 be 2.55 g Cl2 in the XCl2!71 Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of g of XCl2 with excess chlorine forms g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. Hint #2 10 g of XCl2 + Cl2 --> g of XCl4 Thus there must be 2.55 g Cl2 in the XCl4 and thus there must also

72 Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of g of XCl2 with excess chlorine forms g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. 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. 2.55g of Cl2 * 1 mole Cl2/71 g * 1 mole X/1mole Cl2 = moles X in XCl2,!72

73 Special Challenge Problem #2 An element X forms a compound with two chlorine attached (XCl2) and with 4 chlorines attached (XCl4). Treatment of g of XCl2 with excess chlorine forms g of XCl4. Calculate the atomic mass of X and identify which element it is likely to be. and finally... Since 2.55g of the 10g of XCl2 is Cl, 7.45g of the 10g must be X. Calculate the molar mass of X 7.45 g / moles = ~207g/mole. Thus X might be lead.!73

74 Special Challenge Problem #3 A g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce 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?) Type in the % of the CuO compound.!74

75 Special Challenge Problem #3 A g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce 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: The mass of each compound in the mixture will equal the total mass. Second equation: The moles of copper in each compound will equal the moles of copper in the total mixture.!75

76 Special Challenge Problem #3 A g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce 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: You could name the mass of Cu2O as x and the mass of CuO as y. The two substances make up the total mass of mixture. xg + yg = 1.500g!76

77 Special Challenge Problem #3 A g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce 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?) 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 g Cu * 1 mole/63.55 g = total moles Cu x y = !77

78 Special Challenge Problem #3 A g sample of a mixture containing only CuO and Cu2O was treated with hydrogen to produce 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 x y = x + y = Subtract the other x y equation x + y = (-x -y = ) equals: x = thus X = 0.6 g Cu2O 0.6 g Cu2O / g = 40% Cu2O in mixture Thus 60% CuO in mixture!78