Chapter 5 How Chemists Measure Atoms and Molecules

CHAPTER 5 HOW CHEMISTS MEASURE ATOMS AND MOLECULES Chapter 5 Keywords Glossary mass spectrometry mole mole ratio atomic weight Avogadro constant theoretical yield atomic mass unit molar mass actual yield Dalton percent limiting reactant isotopes mass percent weighted average stoichiometry Supplementary Chapter 5 Check for Understanding questions Chapter 5 Exercises (You may use a periodic table as needed.) Answers 1. If a single molecule of a substance weighs 7.64 x 10-23 g, what is its molar mass? 2. Which of the following contains the smallest number of molecules? A. B. C. D. E. 1.0 g H2O 1.0 g NH3 1.0 g H2 1.0 g O2 all of the above are the same 3. Ethanol (C2H5OH) has a density of 0.789 g/mL. How many ethanol molecules are in a teaspoon (~ 5 mL) of ethanol? 4. 1023 atoms of zinc weigh nearly the same as: A. B. C. D. a mole of zinc (Zn). a half of a mole of carbon (C). two moles of xenon (Xe). a mole of boron (B). CHAPTER 5 HOW CHEMISTS MEASURE ATOMS AND MOLECULES 5. In your own words describe what the atomic weight of an element means. 6. What is the molar mass of C3H5(NO3)3? 7. How many moles of ammonia are in 61 g of NH3? 8. Create a solution map for calculating how many hydrogen atoms are present in a 75-g sample of CH3CH2OH and then make this calculation. 9. What mass of N2O contains 0.50 mol of nitrogen atoms? 10. A 1979 penny is 95.05% copper by mass. If the penny weighs 3.098 g, how many copper atoms does the penny contain? 11. What is the mass percent of oxygen in K2SO4? 12. A substance was analyzed and found to contain only sulfur and fluorine. If there was 0.1570 g fluorine present in a 0.2011-g sample of this substance, what is the mass percent composition of this material? 13. How many moles of HNO3 does a 60.0-g sample of a nitric acid solution that is 70.4% HNO3 (by mass) contain? 14. How many grams of Fe2O3 contain 15.0 g of iron? CHAPTER 5 15. HOW CHEMISTS MEASURE ATOMS AND MOLECULES The balanced equation 2Cu(s) + O2(g) ÷ 2CuO(s) tells us that 1 mol of Cu A. B. C. D. E. reacts with 1 mol O2. produces 1 mol CuO. reacts with 32 g of O2. produces 2 mol CuO. does both A and C. 16. Create a solution map for calculating the mass (in grams) of ammonia that can be produced from 5 kg of hydrogen gas and an excess of nitrogen according to the reaction N2(g) + 3H2(g) ÷ 2NH3(g). 17. According to the reaction 4Li(s) + O2(g) ÷ 2Li2O(s), how many grams of Li2O can be produced from each gram of lithium that reacts? 18. The unbalanced equation for the reaction of Na2O2 with water is: Na2O2 + H2O ÷ NaOH + O2 What mass of oxygen gas will be produced if 3.25 g of sodium peroxide reacts with a large excess of water? 19. What mass of O2 is needed to react completely with 4.20 g CH4 in a combustion reaction? 20. How many grams of oxygen gas can be formed from the decomposition of 0.437 mol HgO? 21. For the reaction 2CH4(g) + 3O2(g) + 2NH3(g) ÷ 2HCN(g) + 6H2O(g), if you start with 5.0 g of each reactant, which substance is the limiting reactant? CHAPTER 5 HOW CHEMISTS MEASURE ATOMS AND MOLECULES 22. How many moles of NH3 can be produced from the reaction of 4.00 mol N2 and 10.0 mol H2? 23. For the reaction CaCO3(s) + 2HCl(aq) ÷ CaCl2(aq) + CO2(g) + H2O(l), how many grams of CO2 can be produced if 68.1 g CaCO3 (molar mass =100.09 g/mol) is mixed with 51.6 g HCl (molar mass =36.46 g/mol)? 24. How does the actual yield of a chemical reaction compare to the theoretical yield of the reaction? 25. The unbalanced equation below describes the formation of C2H2. CaC2(s) + H2O(l) ÷ Ca(OH)2(s) + C2H2(g) What is the maximum number of grams of C2H2 that can be produced if 8.00 g of CaC2 and 8.00 g of H2O are allowed to react? 26. When 2.34 grams of aluminum reacts with 1.50 grams of hydrochloric acid, which reactant is in excess? How much of this reactant is left over? APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS Chapter 5 Check for Understanding 5.1 1. The atomic weight of sodium is 22.99. What is the mass, in grams, of a billion sodium atoms? Answer: 3.817 x 10-14 g Solution What we know: number of Na atoms; atomic weight of Na Desired answer: g Na The number of sodium atoms can be converted into mass using the atomic weight of sodium. Recall that the numerical value of the atomic weight of sodium refers to the mass of a single sodium atom in atomic mass units. First, convert this value to an equivalent mass in grams. 22.99 u Na 1.6605 x10-24 g Na 3.817 x10-23 g Na x = 1Na atom 1 u Na Na atom Now use this result as the conversion factor between atoms of sodium and mass of sodium. 1x109 atoms Na x 3.817 x10-23 g Na = 3.817 x10-14 g Na atom Na Check for Understanding 5.2 1. Write the solution map for the conversion of grams of sodium to atoms of sodium. Indicate the numerical ratio that is the conversion factor in each step. Answer: The solution map is: g Na ö mol Na ö atoms Na The conversion factor for the first step is 1mol Na . 22.99 g Na The conversion factor for the second step is 6.022 x1023 atoms Na . 1mol Na APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS Solution Each step of the solution map requires a conversion factor. For g Na ö mol Na, we need the molar mass of sodium expressed in the form 1mol Na which will cancel g Na and introduce mol Na into the numerator. The 22.99 g Na numerical value for the molar mass is obtained from the periodic table. For mol Na ö atoms Na the conversion factor needed is the Avogadro constant 6.022 x1023 atoms Na expressed in the form so that mol Na will cancel. 1mol Na Check for Understanding 5.3 1. What is the mass, in grams, of one cobalt atom? Answer: 9.786 x 10-23 g Solution What we know: number of Co atoms Desired answer: g Co/atom The solution map for this calculation is: atom Co ö mol Co ö g Co The conversion factor needed in the first step is the Avogadro constant expressed in the 1mol Co form . 6.022 x1023 atoms Co The conversion factor needed in the second step is the molar mass of cobalt. The numerical value for the molar mass is obtained from the atomic weight of cobalt (58.93) 58.93g Co and is expressed in the form . 1mol Co Putting these together yields: 1 atom Co x 1 mol Co 58.93g Co x = 9.786 x10-23 g Co 23 6.022 x10 atoms Co 1 mol Co A. APPENDIX B 2. SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS How many atoms of aluminum are present in a piece of aluminum foil that weighs 6.27 g? Answer: 1.40 x 1023 atoms Al Solution What we know: g Al Desired answer: number of Al atoms The solution map for this calculation is: g Al ö mol Al ö atoms Al The conversion factor needed in the first step is the molar mass of aluminum. The numerical value for the molar mass is obtained from the periodic table and is expressed in 1mol Al the form . 26.98g Al The conversion factor needed in the second step is the Avogadro constant in the form 6.022 x10 23 atoms Al . 1mol Al Putting these together yields: 1 mol Al 6.022 x1023 atoms Al x 6.27 g Al x = 1.40 x1023 atoms Al 26.98 g Al 1 mol Al Check for Understanding 5.4 1. Calculate the molar mass of sodium phosphate. Answer: 163.94 g/mol Solution What we know: sodium phosphate; atomic weights of elements Desired answer: g sodium phosphate/mol APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS The chemical formula for sodium phosphate is Na3PO4. The formula indicates that in 1 mole of sodium phosphate there are 3 moles of sodium, 1 mole of phosphorus and 4 moles of oxygen. From the periodic table we see that 1 mole of sodium atoms weighs 22.99 g, 1 mole of phosphorus atoms weighs 30.97 g and 1 mole of oxygen atoms weighs 16.00 g. Thus, one mole of Na3PO4 will weigh: 3 Na 3 mol Na x 22.99 g = 68.97 g 1 mol Na 30.97 g = 1 mol P 1P 1 mol P x 4O 4 mol O x 30.97 g 16.00 g = 1 mol O 64.00 g 163.94 g The molar mass of sodium phosphate is 163.94 g/mol. Check for Understanding 5.5 1. Calculate the number of moles in a sample of sodium bicarbonate that weighs 6.22 g. Answer: 0.0740 mol Solution What we know: g sodium bicarbonate; atomic weights of elements Desired answer: mol sodium bicarbonate The solution map for this calculation is: g sodium bicarbonate ö mol sodium bicarbonate APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed is the molar mass of sodium bicarbonate. The chemical formula for sodium bicarbonate is NaHCO3. From the periodic table we see that 1 mole of sodium atoms weighs 22.99 g, 1 mole of hydrogen atoms weighs 1.008 g, 1 mole of carbon atoms weighs 12.01 g and 1 mole of oxygen atoms weighs 16.00 g. Thus, one mole of NaHCO3 will weigh: 1 Na 1 mol Na x 22.99 g = 22.99 g 1 mol Na 1H 1 mol H x 1.008g = 1 mol H 1C 1 mol C x 12.01g = 1 mol C 12.01g 3O 3 mol O x 16.00 g = 1 mol O 48.00 g 1.008g 84.01 g The molar mass of sodium bicarbonate is 84.01 g/mol. Using this result yields: 6.22 g NaHCO3 x 2. 1mol NaHCO3 = 0.0740 mol NaHCO3 84.01 g NaHCO3 How many molecules are present in 1.05 g of iodine (I2)? Answer: 2.49 x 1021 molecules Solution What we know: g I2 Desired answer: number of I2 molecules The solution map for this calculation is g I2 ö mol I2 ö molecules I2 APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS The conversion factor needed in the first step is the molar mass of I2. From the periodic table we see that 1 mole of iodine atoms weighs 126.9 g. Thus, one mole of I2 will weigh: 126.9 g 2 mol I x = 253.8g 2I 1 mol I This is needed in the form 1mol I 2 . 253.8g I 2 The conversion factor needed in the second step is the Avogadro constant in the form 6.022 x1023 molecules I 2 . 1mol I 2 Putting these together yields: 1.05 g I 2 x 3. 1 mol I 2 6.022 x1023 molecules I 2 x = 2.49 x1021molecules I 2 253.8 g I 2 1 mol I 2 What is the mass in grams of 25 molecules of CCl2F2? Answer: 5.0 x 10-21 g Solution What we know: molecules CCl2F2 Desired answer: g CCl2F2 The solution map for this calculation is: molecules CCl2F2 ö mol CCl2F2 ö g CCl2F2 The conversion factor needed in the first step is the Avogadro constant in the form 1mol CCl2 F2 . 6.022 x1023 molecules CCl2 F2 The conversion factor needed in the second step is the molar mass of CCl2F2. From the periodic table we can get the molar masses of carbon, fluorine and chlorine and add them as follows. 12.01 g C + 2(35.45 g) Cl + 2(19.00 g) F = 120.91 g. Thus the molar mass 120.91g CCl2 F2 of CCl2F2 is 120.91 g/mol. This is used in the form to convert units 1mol CCl2 F2 properly. APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS Putting these together yields: 25 molecules CCl 2 F2 x 1 mol CCl2 F2 120.91g CCl2 F2 x = 5.0 x10-21 g CCl2 F2 23 6.022 x10 molecules CCl2 F2 1 mol CCl2 F2 Check for Understanding 5.6 1. What is the numerical ratio between the number of hydrogen atoms and oxygen atoms in (NH4)3PO4? Answer: 12 H atoms 4 O atoms Solution In each formula unit of (NH4)3PO4 there are 12 H atoms and 4 O atoms. 2. What is the numerical ratio between the number of moles of sulfur and moles of Al2(SO4)3? Answer: 3molS 1mol Al2 (SO 4 )3 Solution There are 3 S atoms in each formula unit of Al2(SO4)3. This ratio can be converted to a ratio of moles by multiplying each quantity by the Avogadro constant. Check for Understanding 5.7 1. Calculate the number of moles of barium nitrate that contain 0.81 mol of oxygen atoms. Answer: 0.14 mol Solution What we know: mol oxygen atoms Desired answer: mol barium nitrate APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS The solution map for this calculation is: mol oxygen atoms ö mol barium nitrate The chemical formula for barium nitrate is Ba(NO3)2. The formula indicates that in 1 mole of barium nitrate there are 6 moles of oxygen atoms. This relationship can be applied as follows. 0.81 mol O x 2. 1mol Ba(NO3 ) 2 = 0.14 mol Ba(NO3 ) 2 6 mol O How many carbon atoms are present in 16.0 g C3H8? Answer: 6.56 x 1023 atoms Solution What we know: g C3H8 Desired answer: number of carbon atoms The solution map for this calculation is: g C3H8 ö mol C3H8 ö mol C ö atoms C The conversion factor needed in the first step is the molar mass of C3H8. From the periodic table we can get the molar masses of carbon and hydrogen and add them as follows. 3(12.01 g) C + 8(1.008 g) H = 44.09 g. Thus the molar mass of C3H8 is 40.09 1mol C3 H8 g/mol. This is needed in the form . 44.09 g C3 H8 The formula of C3H8 indicates that in 1 mole of C3H8 there are 3 moles of carbon atoms. This relationship can be applied as the conversion factor in the second step in the form 3mol C . 1mol C3 H8 The conversion factor needed in the last step is the Avogadro constant in the form 6.022 x10 23 atoms C . 1mol C Putting these together yields: 16.0 g C3 H8 x 1 mol C3 H8 3 mol C 6.022 x1023 atoms C x x = 6.56 x1023 atoms C 44.09 g C3 H8 1 mol C3 H8 1 mol C A. APPENDIX B 3. SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS How many grams of silver are present in 25.0 g AgBr? Answer: 14.4 g Solution What we know: g AgBr Desired answer: g Ag The solution map for this calculation is: g AgBr ö mol AgBr ö mol Ag ö g Ag The conversion factor needed in the first step is the molar mass of AgBr. From the periodic table we can get the molar masses of silver and bromine and add them as follows. 107.9 g Ag + 79.90 g Br = 187.8 g. Thus the molar mass of AgBr is 187.8 1mol AgBr g/mol. This is needed in the form . 187.8g AgBr The formula of AgBr indicates that in 1 mole of AgBr there is 1 mole of silver. This relationship can be applied as the conversion factor in the second step in the form 1mol Ag . 1mol AgBr The conversion factor needed in the last step is the molar mass of Ag expressed in the 107.9 g Ag form . 1mol Ag Putting these together yields: 25.0 g AgBr x 1 mol AgBr 1 mol Ag 107.9 g Ag x x = 14.4 g Ag 187.8 g AgBr 1 mol AgBr 1 mol Ag Check for Understanding 5.8 1. What is the mass percent of carbon in C6H12O6? Answer: 40.00 % Solution What we know: C6H12O6; atomic weights of elements Desired answer: mass % for C in C6H12O6 APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS Since the mass percent values apply to any sample of C6H12O6, it is convenient to consider one mole of this compound. From the periodic table we can get the molar masses of carbon, hydrogen and oxygen and add them as follows. 6(12.01 g) C + 12(1.008 g) H + 6(16.00 g) O = 180.16 g. Thus the molar mass of C6H12O6 is 180.16 g/mol, so 180.16 g represents the total mass of material. The formula indicates that 1 mole of C6H12O6 contains 6 moles of carbon. Therefore the mass of carbon (=component mass) present is: 6 mol C x 12.01g C = 72.06 g C 1 mol C These values are used to obtain the mass percent for carbon. mass % C = 72.06 g C x 100 = 40.00 % 180.16 g C6 H12 O6 Check for Understanding 5.9 1. Based on the reaction between iron(III) oxide and carbon, how many moles of carbon are needed to produce 1.0 kmol of iron? Fe2O3(s) + 3C(s) ÷ 2Fe(s) + 3CO(g) Answer: 1500 mol Solution What we know: kmol Fe; balanced equation relating Fe and C Desired answer: mol C The solution map for this problem is: kmol Fe ö mol Fe ö mol C The conversion factor needed in the first step is that between kmol and mol in the form 103 mol Fe . 1kmol Fe The conversion factor needed in the second step is the mole ratio for these two 3mol C substances from the balanced equation in the form . 2 mol Fe APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS Putting these together yields: 103 mol Fe 3mol C x 1.0 kmol Fe x = 1500 mol C 1 kmol Fe 2 mol Fe 2. What mass of NaHCO3 will react with 0.00257 mol HCl? NaHCO3(s) + HCl(aq) ÷ NaCl(aq) + H2CO3(aq) Answer: 0.216 g Solution What we know: mol HCl; balanced equation relating HCl and NaHCO3 Desired answer: g NaHCO3 The solution map for this problem is: mol HCl ö mol NaHCO3 ö g NaHCO3 The conversion factor needed in the first step is the mole ratio for these two substances 1mol NaHCO3 from the balanced equation in the form . 1mol HCl The conversion factor needed in the second step is the molar mass of NaHCO3 expressed 84.01g NaHCO3 in the form . 1mol NaHCO3 Putting these together yields: 0.00257 mol HCl x 3. 1 mol NaHCO3 84.01g NaHCO3 x = 0.216 g NaHCO3 1 mol HCl 1 mol NaHCO3 How many grams of sodium metal are need to produce 125 g hydrogen gas by the reaction below? 2Na(s) + 2H2O(l) ÷ 2NaOH(aq) + H2(g) Answer: 2.85 x 103 g APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS Solution What we know: g H2; balanced equation relating H2 and Na Desired answer: g Na The solution map for this problem is: g H2 ö mol H2 ö mol Na ö g Na The conversion factor needed in the first step is the molar mass of H2 expressed in the 1mol H 2 form . 2.016 g H 2 The conversion factor needed in the second step is the mole ratio for these two 2 mol Na substances from the balanced equation in the form . 1mol H 2 The conversion factor needed in the last step is the molar mass of Na expressed in the 22.99 g Na form . 1mol Na Putting these together yields: 125 g H 2 x 1 mol H 2 2 mol Na 22.99 g Na x x = 2.85 x103 g Na 2.016 g H 2 1 mol H 2 1 mol Na Check for Understanding 5.10 1. What is the maximum number of moles of Al2(SO4)3 that can be produced from the reaction of 95 moles Al(OH)3 and 120 moles H2SO4? First write a balanced chemical equation for the reaction? Answer: 40. mol Solution The balanced equation for the reaction is: 2Al(OH)3(s) + 3H2SO4(aq) ÷ Al2(SO4)3(aq) + 6H2O(l) APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS What we know: mol Al(OH)3; mol H2SO4; balanced equation relating Al(OH)3, H2SO4 and Al2(SO4)3 Desired answer: maximum moles Al2(SO4)3 produced First determine the limiting reactant by calculating how many moles of Al2(SO4)3 can form from each starting amount of reactant. The solution maps for these calculations are: mol Al(OH)3 ö mol Al2(SO4)3 mol H2SO4 ö mol Al2(SO4)3 For the first calculation the conversion factor needed is the Al2(SO4)3/Al(OH)3 mole ratio from the balanced equation. Applying this yields: 95 mol Al(OH)3 x 1mol Al2 (SO 4 )3 = 48 mol Al2 (SO 4 )3 2 mol Al(OH)3 For the second calculation the conversion factor needed is the Al2(SO4)3/H2SO4 mole ratio from the balanced equation. Applying this yields: 120 mol H 2SO 4 x 1mol Al2 (SO 4 )3 = 40.mol Al2 (SO 4 )3 3 mol H 2SO 4 Since the starting amount of H2SO4 produces the smaller amount of Al2(SO4)3, sulfuric acid is the limiting reactant and the maximum moles of Al2(SO4)3 is 40. moles. 2. What is the theoretical yield in grams of silicon for the reaction between 2.5 kg SiCl4 and 1.6 kg Mg? SiCl4(s) + 2Mg(s) ÷ Si(s) + 2MgCl2(s) Answer: 4.1 x 102 g Solution What we know: kg SiCl4; kg Mg; balanced equation relating SiCl4, Mg and Si Desired answer: maximum g Si produced APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS First determine the limiting reactant by calculating how many grams of silicon can form from each starting amount of reactant. The solution maps for these calculations are: kg SiCl4 ö g SiCl4 ö mol SiCl4 ö mol Si ö g Si kg Mg ö g Mg ö mol Mg ö mol Si ö g Si For the first calculation the conversion factors needed are that between kg and g, the molar mass of SiCl4, the Si/SiCl4 mole ratio and finally the molar mass of Si. Putting these together yields: 103 g SiCl4 1 molSiCl 4 1 molSi 28.09 g Si x x x 2.5 kg SiCl4 x = 4.1x102 g Si 1 kg SiCl4 169.89 g SiCl4 1 molSiCl4 1 molSi For the second calculation the conversion factors needed are that between kg and g, the molar mass of Mg, the Si/Mg mole ratio and finally the molar mass of Si. Putting these together yields: 2.5 kg Mg x 103 g Mg 1 mol Mg 1 molSi 28.09 g Si x x x = 9.2 x102 g Si 1 kg Mg 24.31 g Mg 2 mol Mg 1 molSi Since the starting amount of SiCl4 produces the smaller amount of silicon, SiCl4 is the limiting reactant and the theoretical yield is 4.1 x 102 g Si. 3. When a mixture of 58 g CO and 9.2 g H2 reacts, which reactant is in excess? How many grams of the reactant in excess remain? CO(g) + 2H2(g) ÷ CH3OH(l) Answers: H2 0.7 g Solution What we know: g CO; g H2; balanced equation relating CO, H2 and CH3OH Desired answer: g excess reactant remaining APPENDIX B SOLUTIONS TO CHECK FOR UNDERSTANDING PROBLEMS To find which reactant is in excess, determine the limiting reactant by calculating how many moles of CH3OH can form from each starting amount of reactant. The solution maps for these calculations are: g CO ö mol CO ö mol CH3OH g H2 ö mol H2 ö mol CH3OH For the first calculation the conversion factors needed are the molar mass of CO and the CH3OH/CO mole ratio. Putting these together yields: 58 g CO x 1mol CH 3 OH 1 mol CO x = 2.1mol CH 3 OH 28.01 g CO 1 mol CO For the second calculation the conversion factors needed are the molar mass of H2 and the CH3OH/H2 mole ratio. Putting these together yields: 9.2 g H 2 x 1mol CH 3 OH 1 mol H 2 x = 2.3mol CH 3 OH 2.016 g H 2 2 mol H 2 Since the starting amount of CO produces the smaller amount of CH3OH, CO is the limiting reactant and H2 is the reactant in excess. Now use the theoretical yield of CH3OH to calculate how much H2 was used and subtract this from the starting quantity. The solution maps for these calculations are: mol CH3OH ö mol H2 ö g H2 used g H2 remaining = g H2 initially - g H2 used For the first calculation the conversion factors needed are the H2/CH3OH mole ratio and the molar mass of H2. Putting these together yields: 2.1 mol CH 3 OH x 2 mol H 2 2.016 g H 2 x = 8.5g H 2 used 1 mol CH 3 OH 1 mol H 2 Therefore, the mass of H2 that remains equals 9.2 g - 8.5 g = 0.7 g. APPENDIX C ANSWERS TO SELECTED END-OF-CHAPTER EXERCISES Chapter 5 1. 46.0 g/mol 2. B 3. 5 x 1022 molecules 4. D 6. 227.10 g/mol 7. 3.6 mol 8. g CH3CH2OH ö mol CH3CH2OH ö mol H 5.9 x 1024 atoms 9. 11 g 10. 2.790 x 1022 atoms 11. 36.72% 12. 21.93% S 78.07% F 13. 0.670 mol 14. 21.4 g 15. B ö atoms H A. APPENDIX C ANSWERS TO SELECTED END-OF-CHAPTER EXERCISES 16. kg H2 ö g H2 ö mol H2 ö mol NH3 17. 2.152 g 18. 0.667 g 19. 16.8 g 20. 6.83 g 21. O2 22. 6.67 mol 23. 29.9 g 24. The actual yield is always less than the theoretical yield. 25. 5.78 g 26. Al is in excess 1.97 g remain ö g NH3