Section 6.5: Mass and Number of Entities Tutorial 1 Practice, page 280 1. Given: mPb = 8.16 kg Required: number of atoms of lead, N Pb Solution: Step 1. Look up the molar mass of lead. M Pb = 207.20 g/mol Step 2. Calculate the amount by multiplying the mass of lead by a conversion factor derived from the molar mass. mPb = 8.16 kg = 8.16 ! 103 g " 1 mol % nPb = (8.16 ! 103 g) $ ' # 207.2 g & nPb = 39.382 mol (2 extra digits carried) Step 3. Calculate the number of atoms using an appropriate conversion factor. " 6.02 ! 1023 atoms % N Pb = (39.382 mol ) $ ' # & 1 mol N Pb = 2.37 ! 1025 atoms Statement: A typical car battery contains 2.37 × 1025 atoms of lead. 2. Given: mC = 2.00 ! 10"1 g Required: number of atoms of carbon, N C Solution: Step 1. Look up the molar mass of carbon. M C = 12.01 g /mol Step 2. Calculate the amount by multiplying the mass of carbon by a conversion factor derived from the molar mass. # 1 mol & nC = (2.00 ! 10"1 g) % ( $ 12.01 g ' nC = 0.016 653 mol (2 extra digits carried) Step 3. Calculate the number of atoms using an appropriate conversion factor. " 6.02 ! 1023 atoms % N C = (0.016 653 mol ) $ ' # & 1 mol N C = 1.00 ! 1022 atoms Statement: A one-carat diamond contains 1.00 × 1022 atoms of carbon. Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-1 3. Given: mC H = 7.3 g 2 4 Required: number of molecules of ethylene, N C H 2 4 Solution: Step 1. Calculate the molar mass of ethylene. ! ! g $ g $ M C H = 2 # 12.01 + 4 # 1.01 & 2 4 mol % mol &% " " g 2 4 mol Step 2. Calculate the amount by multiplying the mass of ethylene by a conversion factor derived from the molar mass. ! 1 mol $ nC H = (7.3 g) # & 2 4 " 28.06 g % M C H = 28.06 nC H = 0.2602 mol [2 extra digits carried] 2 4 Step 3. Calculate the number of molecules using an appropriate conversion factor. " 6.02 ! 1023 molecules % N C H = (0.2602 mol ) $ ' 2 4 # & 1 mol N C H = 1.6 ! 1023 molecules 2 4 Statement: A 7.3 g sample of ethylene contains 1.6 × 1023 molecules. 4. Given: mCaCl = 1.5 kg 2 Required: number of formula units of calcium chloride, N CaCl 2 Solution: Step 1. Calculate the molar mass of calcium chloride. ! ! g $ g $ M CaCl = # 40.08 + 2 # 35.45 & 2 mol % mol &% " " g 2 mol Step 2. Calculate the amount by multiplying the mass of calcium chloride by a conversion factor derived from the molar mass. mCaCl = 1.5 kg = 1.5 ! 103 g M CaCl = 110.98 2 " 1 mol % nCaCl = (1.5 ! 103 g) $ ' 2 # 110.98 g & nCaCl = 13.52 mol [2 extra digits carried] 2 Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-2 Step 3. Calculate the number of formula units using an appropriate conversion factor. " 6.02 ! 1023 formula units % N CaCl = (13.52 mol ) $ ' 2 # & 1 mol N CaCl = 8.1 ! 1024 formula units 2 Statement: A 1.5 kg pail of calcium chloride contains 8.1 × 1024 formula units of calcium chloride. Tutorial 2 Practice, page 282 1. Given: mH O = 15.0 g 2 2 Required: number of atoms of hydrogen, N H ; number of atoms of oxygen, N O Solution: Step 1. Calculate the molar mass of hydrogen peroxide. ! ! g $ g $ M H O = 2 # 1.01 + 2 16.00 #" 2 2 mol &% mol &% " g mol Step 2. Calculate the amount by multiplying the mass of hydrogen peroxide by a conversion factor derived from the molar mass. ! 1 mol $ nH O = (15.0 g) # & 2 2 " 34.02 g % M H O = 34.02 2 2 nH O = 0.440 92 mol [2 extra digits carried] 2 2 Step 3. Calculate the number of molecules of hydrogen peroxide using an appropriate conversion factor. " molecules % N H O = 0.440 92 mol $ 6.02 ! 1023 ' 2 2 # 1 mol & ( ) N H O = 2.6543 ! 1023 molecules [2 extra digits carried] 2 2 Step 4. Determine the number of each type of atom present. " 2 atoms % N H = (2.6543 ! 1023 molecules ) $ ' # 1 molecule & N H = 5.31 ! 1023 atoms " 2 atoms % N O = (2.6543 ! 1023 molecule s) $ ' # 1 molecule & N O = 5.31 ! 1023 atoms Statement: A 15.0 g sample of hydrogen peroxide contains 5.31 × 1023 atoms of hydrogen and 5.31 × 1023 atoms of oxygen. Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-3 2. Given: mCa 3 (PO 4 )2 = 2.5 g Required: number of ions calcium, N Ca 2+ Solution: Step 1. Calculate the molar mass of calcium phosphate. M Ca (PO ) = 3M Ca + 2( M P + 4 M O ) 3 4 2 ! ! g $ g g $ = 3# 40.08 + 2 # 30.97 + 4 ' 16.00 & mol % mol mol &% " " g mol Step 2. Calculate the amount by multiplying the mass of calcium phosphate by a conversion factor derived from the molar mass. ! 1 mol $ nCa (PO ) = (2.5 g) # & 3 4 2 " 310.18 g % M Ca nCa 3 (PO 4 )2 3 (PO 4 )2 = 310.18 = 0.008 060 mol [2 extra digits carried] Step 3. Calculate the number of formula units of calcium phosphate using an appropriate conversion factor. " formula units % N Ca (PO ) = (0.008 060 mol ) $ 6.02 ! 1023 ' 3 4 2 # & 1 mol N Ca 3 (PO 4 )2 = 4.8521 ! 1021 formula units [2 extra digits carried] Step 4. Determine the number of calcium ions present. " % 3 ions N Ca 2+ = (4.8521 ! 1021 formula units ) $ ' # 1 formula unit & N Ca 2+ = 1.5 ! 1022 ions Statement: A 2.5 g sample of calcium phosphate contains 1.5 × 1022 calcium ions. 3. Given: N CO = 8.4 ! 1024 molecules 2 Required: mass of carbon dioxide, mCO 2 Solution: Step 1. Calculate the amount of carbon dioxide using an appropriate conversion factor. " % 1 mol nCO = (8.4 ! 1024 molecules ) $ ' 2 # 6.02 ! 1023 molecules & nCO = 13.95 mol [2 extra digits carried] 2 Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-4 Step 2. Calculate the molar mass of carbon dioxide. M CO = M C + 2 M O 2 ! ! g $ g $ = # 12.01 + 2 # 16.00 & mol % mol &% " " g 2 mol Step 3. Use the amount of carbon dioxide and its molar mass to calculate the mass of carbon dioxide. ! 44.01 g $ mCO = (13.95 mol ) # & 2 " 1 mol % M CO = 44.01 mCO = 6.1 ' 102 g 2 Statement: The mass of 8.4 × 1024 molecules of carbon dioxide is 6.1 × 102 g. Mini Investigation: Determining Numbers in Everyday Entities, page 282 Answers are determined from experimental data. Sample answer: Mass of chalk required to write your name: 0.08 g The mass of water that you can hold in your hand: 125.45 g The mass of water drops that will fit on a penny without overflowing: 2.50 g A. (i) Given: mCaCO = 0.08 g 3 Required: amount of calcium carbonate, nCaCO 3 Solution: Step 1. Calculate the molar mass of calcium carbonate, M CaCO . 3 ! ! g $ ! g $ g $ M CaCO = # 40.08 + 12.01 + 3 16.00 #" 3 mol &% #" mol &% mol &% " g mol Step 2. Use the mass of calcium carbonate and its molar mass to calculate the amount of calcium carbonate. ! 1 mol $ nCaCO = (0.08 g) # & 3 " 100.09 g % M CaCO = 100.09 3 nCaCO = 7.99 ' 10(4 mol [2 extra digits carried] 3 Statement: The amount of calcium carbonate in a 0.08 g piece of chalk is 8 × 10−4 mol. Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-5 (ii) Given: mH O = 125.45 g 2 Required: amount of water, nH O 2 Solution: Step 1. Calculate the molar mass of water, M H O . 2 ! g $ ! g $ M H O = 2 # 1.01 + # 16.00 & 2 mol % " mol &% " g 2 mol Step 2. Use the mass of water and its molar mass to calculate the amount of water. ! 1 mol $ nH O = (125.45 g) # & 2 " 18.02 g % M H O = 18.02 nH O = 6.96171 mol [2 extra digits carried] 2 Statement: The amount of water in 125.45 g of water is 6.962 mol. (iii) Given: mH O = 2.50 g 2 Required: amount of water, nH O 2 Solution: Use the mass of water and its molar mass to calculate the amount of water. ! 1 mol $ nH O = (2.50 g) # & 2 " 18.02 g % nH O = 0.13873 mol [2 extra digits carried] 2 Statement: The amount of water in 2.50 g of water is 0.139 mol. B. (i) Given: nCaCO = 7.99 ! 10"4 mol 3 Required: number of formula units of calcium carbonate, N CaCO 3 Solution: Calculate the number of formula units of calcium carbonate using an appropriate conversion factor. " formula units % N CaCO = (7.99 ! 10-4 mol ) $ 6.02 ! 1023 ' 3 # & 1 mol N CaCO = 5 ! 1020 formula units 3 Statement: A 0.08 g piece of chalk contains 5 × 1020 formula units of calcium carbonate. Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-6 (ii) Given: nH O = 6.96171 mol 2 Required: number of molecules of water, N H O 2 Solution: Calculate the number of molecules of water using an appropriate conversion factor. " molecules % N H O = (6.96171 mol ) $ 6.02 ! 1023 ' 2 # 1 mol & N H O = 4.191 ! 1024 molecules 2 Statement: 125.45 g of water contains 4.191 × 1024 molecules of water. (iii) Given: nH O = 0.138 73 mol 2 Required: number of molecules of water, N H O 2 Solution: Calculate the number of molecules of water using an appropriate conversion factor. " molecules % N H O = (0.138 73 mol ) $ 6.02 ! 1023 ' 2 # 1 mol & N H O = 8.35 ! 1022 molecules 2 Statement: 2.50 g of water contains 8.35 × 1022 molecules of water. Section 6.5 Questions, page 283 1. (a) Given: nNH = 2.0 mol 3 Required: number of molecules of ammonia, N NH 3 Solution: Calculate the number of molecules of ammonia using an appropriate conversion factor. " molecules % N NH = (2.0 mol ) $ 6.02 ! 1023 ' 3 # 1 mol & N NH = 1.2 ! 1024 molecules 3 Statement: A 2.0 mol sample of ammonia contains 1.2 × 1024 molecules. (b) Given: mH O = 1.6 mg 2 Required: number of molecules of water, N H O 2 Solution: Step 1. Calculate the molar mass of water, M H O . 2 ! g $ ! g $ M H O = 2 # 1.01 + # 16.00 & 2 mol % " mol &% " M H O = 18.02 2 g mol Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-7 Step 2. Use the mass of water and its molar mass to calculate the amount of water. mH O = 1.6 mg = 1.6 ! 10-3 g 2 # 1 mol & nH O = (1.6 ! 10"3 g) % ( 2 $ 18.02 g ' nH O = 8.879 ! 10"5 mol [2 extra digits carried] 2 Step 3. Calculate the number of molecules of water using an appropriate conversion factor. " molecules % N H O = (8.879 ! 10-5 mol ) $ 6.02 ! 1023 ' 2 # 1 mol & N H O = 5.3 ! 1019 molecules 2 Statement: A 1.6 mg sample of water contains 5.3 × 1019 molecules. 2. (a) Given: N Cu = 3.01 ! 1023 atoms Required: amount of copper, nCu Solution: Calculate the amount of copper using an appropriate conversion factor. " % 1 mol nCu = (3.01 ! 1023 atoms ) $ ' # 6.02 ! 1023 atoms & nCu = 0.500 mol Statement: The amount of copper present is 0.500 mol. (b) Given: N HCl = 6.02 ! 1025 molecules Required: amount of hydrochloric acid, nHCl Solution: Calculate the amount of hydrochloric acid using an appropriate conversion factor. " % 1 mol nHCl = (6.02 ! 1025 molecules ) $ ' 23 # 6.02 ! 10 molecules & nHCl = 1.00 ! 102 mol Statement: The amount of hydrochloric acid present is 1.00 × 102 mol. (c) Given: mCaCO = 42.0 g 3 Required: amount of calcium carbonate, nCaCO 3 Solution: Step 1. Calculate the molar mass of calcium carbonate, M CaCO . 3 ! ! g $ ! g $ g $ M CaCO = # 40.08 + # 12.01 + 3# 16.00 & & 3 mol % " mol % mol &% " " M CaCO = 100.09 3 g mol Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-8 Step 2. Use the mass of calcium carbonate and its molar mass to calculate the amount of calcium carbonate. ! 1 mol $ nCaCO = (42.0 g) # & 3 " 100.09 g % nCaCO = 0.420 mol 3 Statement: The amount of calcium carbonate in a 42.0 g sample is 0.420 mol. 3. (a) Given: N CO = 4.2 ! 1024 molecules 2 Required: number of atoms of carbon, N C ; number of atoms of oxygen, N O Solution: Multiply the number of molecules by the number of each type of atom per molecule. " 1 atom % N C = (4.2 ! 1024 molecules ) $ ' # 1 molecule & N C = 4.2 ! 1024 atoms " 2 atoms % N O = (4.2 ! 1024 molecules ) $ ' # 1 molecule & N O = 8.4 ! 1024 atoms Statement: A sample of 4.2 × 1024 carbon dioxide molecules contains 4.2 × 1024 atoms of carbon and 8.4 × 1024 atoms of oxygen. (b) Given: N CO = 4.2 ! 1024 molecules 2 Required: mass of carbon dioxide, mCO 2 Solution: Step 1. Calculate the amount of carbon dioxide using an appropriate conversion factor. " % 1 mol nCO = (4.2 ! 1024 molecules ) $ ' 2 # 6.02 ! 1023 molecules & nCO = 6.977 mol [2 extra digits carried] 2 Step 2. Calculate the molar mass of carbon dioxide. ! ! g $ g $ M CO = # 12.01 + 2 16.00 #" 2 mol &% mol &% " M CO = 44.01 2 g mol Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-9 Step 3. Use the amount of carbon dioxide and its molar mass to calculate the mass of carbon dioxide. ! 44.01 g $ mCO = (6.977 mol ) # & 2 " 1 mol % mCO = 310 g 2 Statement: The mass of 4.2 × 1024 molecules of carbon dioxide is 310 g. 4. Given: mNaOCl = 120.0 g Required: number of formula units of sodium hypochlorite, N NaOCl Solution: Step 1. Calculate the molar mass of sodium hypochlorite, M NaOCl . ! g $ ! g $ ! g $ M NaOCl = # 22.99 + # 16.00 + # 35.45 & & mol % " mol % " mol &% " g mol Step 2. Use the mass of sodium hypochlorite and its molar mass to calculate the amount of sodium hypochlorite. ! 1 mol $ nNaClO = (120.0 g) # & " 74.44 g % M NaOCl = 74.44 nNaClO = 1.612 04 mol [2 extra digits carried] Step 3. Calculate the number of formula units of sodium hypochlorite using an appropriate conversion factor. " formula units % N NaOCl = (1.612 04 mol ) $ 6.02 ! 1023 ' # & 1 mol N NaOCl = 9.70 ! 1023 formula units Statement: A 120.0 g sample of sodium hypochlorite contains 9.70 × 1023 formula units. 5. (a) Given: mC H O = 250.0 mg 9 8 4 Required: amount of ASA in one tablet, nC H O 9 8 4 Solution: Step 1. Calculate the molar mass of ASA, M C H O 9 8 4 ! ! ! g $ g $ g $ M C H O = 9 # 12.01 + 8 # 1.01 + 4 # 16.00 & & 9 8 4 mol % mol % mol &% " " " M C H O = 180.17 9 8 4 g mol Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-10 Step 2. Use the mass of ASA and its molar mass to calculate the amount of ASA. mC H O = 250.0 mg = 2.500 ! 10"1 g 9 8 4 # 1 mol & nC H O = (2.500 ! 10"1 g) % ( 9 8 4 $ 180.17 g ' nC H O = 1.387 58 ! 10"3 mol [2 extra digits carried] 9 8 4 Statement: A 250.0 mg sample of ASA contains 1.388 × 10−3 mol. (b) Given: mC H O = 250.0 mg ; nC H O = 1.387 58 mol 9 8 4 9 8 4 Required: number of hydrogen atoms, N H Solution: Step 1. Calculate the number of molecules of ASA using an appropriate conversion factor. " molecules % N C H O = (1.387 58 ! 10-3 mol ) $ 6.02 ! 1023 ' 9 8 4 # 1 mol & N C H O = 8.3532 ! 1020 molecules [2 extra digits carried] 9 8 4 Step 2. Determine the number of hydrogen atoms present. " 8 atoms % N H = (8.3532 ! 1020 molecules ) $ ' # 1 molecule & N H = 6.68 ! 1021 atoms Statement: There are 6.68 × 1021 hydrogen atoms in 250.0 mg of ASA. 6. (a) Given: mHe = 0.327 g ; VHe = 2.00 L Required: number of atoms of helium, N He Solution: Step 1. Use the mass of helium and its molar mass to calculate the amount of helium. g M He = 4.00 mol ! 1 mol $ nHe = (0.327 g) # & " 4.00 g % nHe = 0.081 750 mol [2 extra digits carried] Step 2. Calculate the number of helium atoms present. " atoms % N He = (0.081 750 mol ) $ 6.02 ! 1023 ' # 1 mol & N He = 4.9214 ! 1022 atoms [2 extra digits carried] Statement: A 0.327 g sample of helium contains 4.92 × 1022 atoms. Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-11 (b) Given: VHe = 2.00 L ; Vblimp = 5.74 ! 106 L ; N He = 4.9214 ! 1022 atoms Required: number of helium atoms in blimp, N HeBlimp Solution: Use the volume of the blimp and the number of helium atoms per 2.00 L to calculate the number of helium atoms in the blimp. " 4.9214 ! 1022 atoms % N HeBlimp = (5.74 ! 106 L) $ ' 2.00 L # & N HeBlimp = 1.41 ! 1029 atoms [2 extra digits carried] Statement: There are 1.41 × 1029 atoms of helium in the blimp. 7. (a) Given: mAs O = 14 mg 2 3 Required: number of formula units of arsenic(III) oxide, N As O 2 Solution: Step 1. Calculate the molar mass of As2O3, M As O 2 3 3 ! ! g $ g $ M As O = 2 # 74.92 + 3 16.00 #" 2 3 mol &% mol &% " g mol Step 2. Use the mass of As2O3 and its molar mass to calculate the amount of As2O3. mAs O = 14 mg = 1.4 ! 10"2 g M As O = 197.84 2 2 3 3 # 1 mol & nAs O = (1.4 ! 10"2 g) % ( 2 3 $ 197.84 g ' nAs O = 7.076 ! 10"5 mol (2 extra digits carried) 2 3 Step 3. Calculate the number of formula units of As2O3 using an appropriate conversion factor. " formula units % N As O = (7.076 ! 10-5 mol ) $ 6.02 ! 1023 ' 2 3 # & 1 mol N As O = 4.260 ! 1019 formula units [2 extra digits carried] 2 3 Statement: There are 4.3 × 1019 formula units of As2O3 in a 14 mg sample of the compound. Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-12 (b) Given: N As O = 4.260 ! 1019 formula units 2 3 Required: number of oxide ions, N O2Solution: Calculate the number of oxide ions present using an appropriate conversion factor. " % 3 ions N O2- = (4.260 ! 1019 formula units ) $ # 1 formula units '& N O2- = 1.3 ! 1020 ions Statement: There are 1.3 × 1020 oxide ions present in a 14 mg sample of As2O3. 8. (a) Given: mass of toothpaste containing Na2FPO3 = 120.0 g; percentage of Na2FPO3 in toothpaste = 0.76 % Required: amount of sodium monofluorophosphate, nNa FPO 2 3 Solution: Step 1. Calculate the mass of Na2FPO3 in the toothpaste sample. 0.76 mNa FPO = ! 120.0 g 2 3 100 = 0.9120 g [2 extra digits carried] Step 2. Calculate the molar mass of Na2FPO3, M Na FPO 2 M Na 2 FPO3 3 ! ! g $ ! g $ ! g $ g $ = 2 # 22.99 + # 19.00 + # 30.97 + 3# 16.00 & & & mol % " mol % " mol % mol &% " " g mol Step 3. Use the mass of Na2FPO3 and its molar mass to calculate the amount of Na2FPO3. ! 1 mol $ nNa FPO = (0.9120 g) # & 2 3 " 143.95 g % M Na nNa 2 FPO3 2 FPO3 = 143.95 = 6.336 ' 10(3 mol [2 extra digits carried] Statement: A 120.0 g tube of toothpaste contains 6.3 × 10−3 mol of sodium monofluorophosphate. (b) Given: nNa FPO = 6.336 ! 10"3 mol 2 3 Required: number of sodium ions, N Na + , number of monofluorophosphate ions, N FPO 2! 3 Solution: Step 1. Calculate the number of formula units of Na2FPO3 using an appropriate conversion factor. " formula units % N Na FPO = (6.336 ! 10 –3 mol ) $ 6.02 ! 1023 ' 2 3 # & 1 mol N Na 2 FPO3 = 3.814 ! 1021 formula units [2 extra digits carried] Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-13 Step 2. Determine the number of sodium ions and monofluorophosphate ions present. " % 2 ions N Na + = (3.814 ! 1021 formula units ) $ ' # 1 formula units & N Na + = 7.6 ! 1021 ions # & 1 ion N FPO 2! = (3.814 " 1021 formula units ) % ( 3 $ 1 formula units ' N FPO 2! = 3.8 " 1021 ions 3 Statement: A 120.0 g tube of toothpaste contains 7.6 × 1021 sodium ions and 3.8 × 1021 monoflurophosphate ions. 9. (a) Given: mAg = 509.0 g ; mAu = 6.0 g Required: number of silver atoms, N Ag ; number of gold atoms, N Au Solution: Step 1. Look up the molar mass of silver, M Ag , and the molar mass of gold, M Au . g mol g M Au = 196.97 mol Step 2. Use the mass and the molar mass to calculate the amount. ! 1 mol $ nAg = (509.0 g) # & " 107.87 g % M Ag = 107.87 nAg = 4.718 64 mol [2 extra digits carried] ! 1 mol $ nAu = (6.0 g) # & " 196.97 g % nAu = 3.046 ' 10(2 mol [2 extra digits carried] Step 3. Calculate the number of each type of atoms present. " atoms % N Ag = (4.718 64 mol ) $ 6.02 ! 1023 ' 1 mol & # N Ag = 2.84 ! 1024 atoms " atoms % N Au = (3.046 ! 10 –2 mol ) $ 6.02 ! 1023 ' # 1 mol & N Au = 1.8 ! 1022 atoms Statement: The gold medal contains 2.84 × 1024 atoms of silver and 1.8 × 1022 atoms of gold. Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-14 (b) Answers may vary. Answers are based on $44.53 per gram of gold and $0.943 per gram of silver. $44.53 $ 0.943 Given: cost of Au = ; cost of Ag = ; mmedal = 515.0 g ; mAg = 509.0 g ; mAu = 6.0 g 1g 1g Required: cost of pure gold medal, CAuMedal ; cost of actual medal, Cmedal Solution: Cost of gold medal: ! $44.53 $ CAuMedal = (515 g) # & " 1g % CAuMedal = $22 900 Cost of actual medal: ! $44.53 $ ! $0.943 $ Cmedal = (6.0 g) # + (509.0 6.0 g) # & & " 1g % " 1 6.0 g % Cmedal = $750 Statement: A medal made of pure gold would cost $22 930 while the actual medal would cost $750, based on $44.53 per gram of gold and $0.943 per gram of silver. Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6.5-15