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Chem 208 Exam #1
  • Solution
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Problem #1 (10 pt)
  • Ethylene glycol has a density of 1.11 g/mL at 25 °C.


  • What is the mass, in kg, of 12.0 L of ethylene glycol?


  • 12.0 L * 1000 mL * 1.11 g * 1 kg  = 13.32 kg
  •                1 L          1 mL     1000 g
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Problem #1 (10 pt)
  • Ethylene glycol has a density of 1.11 g/mL at 25 °C.


  • What is the volume, in L, occupied by 33.4 kg of ethylene glycol?


  • 33.4 kg * 1000 g * 1 mL * 1 L  = 30.09 L
  •                1 kg     1.11 g  1000 mL
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Problem #2 (20 pt)
  • Bananas cost $0.45 per pound.  A single banana weighs 15 grams.  Ignoring sales tax, how many bananas can I buy for $6.50


  • 6.50 $ * 1 lb * 453.6 g * 1 banana  = 436.8 bananas
  •            0.45 $  1 lb          15 g
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Problem #3 (20 pt)
  • The reaction given below proceeds with a 72.6% yield.  You initially have 8.00 g of AgNO3 and 4.00 g of MgO.  What is the mass of Ag2O that you recover?


  • 2 AgNO3 (aq) + MgO (s) → Ag2O (s) + Mg(NO3)2 (aq)


  • 8.00 g AgNO3 * 1 mol AgNO3 * 1 mol Ag2O  = 0.0235 mol Ag2O
  •                        169.9 g AgNO3  2 mol AgNO3
  • 4.00 g MgO  * 1 mol MgO * 1 mol Ag2O  = 0.0993 mol Ag2O
  •                     40.3 g MgO  1 mol MgO


  • AgNO3 is the limiting reagent!
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Problem #3 (cont’d)
  • 0.0235 mol Ag2O * 231.8 g Ag2O = 5.447 g Ag2O
  •                             1 mol Ag2O


  • 5.447 g Ag2O theor * 72.6 g actual = 3.96 g Ag2O
  •                                100 g theor
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Problem #4 (20 pt)
  • Joe completely combusts 3 g of a chemical compound containing only carbon and hydrogen.  After combustion, he isolates 8.30 g of CO2 and 7.11 g of H2O.  The molar mass of the compound is determined to be 112 g/mol.  What is the molecular (not empirical) formula of Joe’s compound?



  • 8.30 g CO2 * 1 mol CO2 *   1 mol C  = 0.188 mol C
  •                     44.01 g CO2  1 mol CO2
  • 7.11 g H2O  * 1 mol H2O * 2 mol H  = 0.789 mol H
  •                     18.01 g H2O 1 mol H2O


  • C0.188H0.789
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Problem #4 (cont’d)
  • C0.188H0.789


  • C0.188H0.789
  •    0.188   0.188


  • C1H4.16


  • CH4  molar mass = 12.01 + 4*1.008 = 16


  • 112 g/mol  = 7
  • 16 g/mol


  • C7H28
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Problem #6 (20 pt)
  • Aluminum (III) sulfate reacts with sodium nitrate to give the double replacement products sodium sulfate and aluminum (III) nitrate.  If I have 11.00 g aluminum (III) sulfate , how many grams of sodium nitrate would I need for a complete reaction to occur? If, after complete reaction, I recover 1.50 g of aluminum (III) nitrate, what was the % yield of the reaction?


  • Al2(SO4)3  + 6  NaNO3 → 3 Na2SO4 + 2 Al(NO3)3


  • 11.00 g Al2(SO4)3 * 1 mol Al2(SO4)3 * 6 mol NaNO3 * 85.0 g NaNO3 = 16.4 g NaNO3
  •                        342.2 g Al2(SO4)3  1 mol Al2(SO4)3    1 mol NaNO3
  • 11.00 g Al2(SO4)3 * 1 mol Al2(SO4)3 * 2 mol Al(NO3)3 * 213 g Al(NO3)3 = 13.6 g Al(NO3)3
  •                        342.2 g Al2(SO4)3  1 mol Al2(SO4)3    1 mol  AL(NO3)3
  • 1.50 g actual * 100 = 10.9% yield
  • 13.6 g theoretical