Stoichiometry

1
Stoichiometry

• AP review
• December 11, 2003
• NSHS

2
The study of quantities in chemical reactions

• Formula stoichiometry .. composition,
  empirical and molecular formulas ..
• Mole ratios and molar mass
• Mole, mass stoichiometry .. Mole ratios from
  balanced equation
• Gas stoichiometry .. pVnRT
• Solution stoichiometry nMV, M1V1 M2V2

3
Mole Map(the state of chemistry)
Particleberg
x
Avogadros Hwy 6.02 x 1023
Molar mass express (periodic table)
Ideal road pVnRT
/
Mole City
Masstown
Volumeville
x
/
X by L
Molarity trail (moles/L)
/ by L
Camp Concentration
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Mole mass stoichiometry

• Balanced equation and mole map are the keys to
  success
• Steps for the Jones/McCormick method
• 1. Write balanced equation
• 2. Identify the given(s) and convert to
  moles (mole map).
• 3. Identify the unknown
• 4. Determine limiting reactant if needed
  (mole ratios)
• 5. Use mol ratio determined by coefficients in
  balanced equation to solve for moles of unknown.
• 6. Convert moles of unknown to the units
  required (mole map)

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Formula Stoichiometry

• composition
• Empirical formula
• Molecular formula

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59. When a 1.00-gram sample of limestone was
dissolved in acid, 0.22 gram of CO2 was
generated. If the rock contained no carbonate
other than CaCO3, what was the percent of CaCO3
by mass in the limestone? (A) 17(B) 51(C)
64(D) 50(E) 100
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• CaCO3 . 1 g / 100 g/mol .01 mol CaCO3
• .01 mol CaCO3 x 1mol CO2 .01
  mol CO2
• 1mol CaCO3
• CaCO3 2HCl ? CaCl2 CO2 H2O
  
• .01 mol CO2 x 44 g/mol .44 g CO2
• .22 g CO2 x 100 50.
• .44g CO2
• . Choice d (50)

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2) An unknown compound contains only the three
elements C,H, and O. A pure sample of the
compound is analyzed and found to be 65.60
percent C and 9.44 percent H by mass. (a)
Determine the empirical formula of the
compound. (b) When 1.570 grams of the compound
is vaporized at 300 C and 1.00
atmosphere, the gas occupies a volume of
577 milliliters. What is the molar mass of the
compound based on this result? (c) What is
the molecular formula of the compound?
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• 65.6 g C/ 12.01 g/mol C 5.462 mol C
• 9.44 g H / 1.01 g/mol H 9.35 mol H
• 24.96 g O/ 16.0 g/mol O 1.56 mol O
• 100g-(65.6g 9.44g) 24.96 g O
• _________________________________________
• 1.56 mol/1.56 mol 1 . X 2 2
• 5.462 mol/1.56 mol 3.5 x 2 7
• 9.35 mol/1.56 mol 6 x 2 12
• Multiply all by 2 and the empirical formula is
  C7H12O2

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• Molar mass is found from the gas density
• A useful formula for molar mass (gfm) and gas
  density is gfm DRT/p
• D density in g/L R .0821 Latm/Kmol T Kelvin
  temperature
• P pressure in atmospheres.
• gfm (1.57g/.577L).0821 Latm/ K mol (573K)
  128 g/mol
• 1atm.
• gfm/efm multiplier to change empirical to
  molecular formula
• 128/128 1
• The molecular formula is C7H12O2

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Mole mass stoichiometry Remember this ??

• Balanced equation and mole map are the keys to
  success
• Steps for the Jones/McCormick method
• 1. Write balanced equation
• 2. Identify the given(s) and convert to
  moles (mole map).
• 3. Identify the unknown
• 4. Determine limiting reactant if needed
  (mole ratios)
• 5. Use mol ratio determined by coefficients in
  
• balanced equation to solve for moles of
  unknown.
• 6. Convert moles of unknown to the units
  required (mole
• map)

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• 2) A mixture of H2(g), and O2(g) is present in
  0.500-liter rigid container at 25 C. The number
  of moles of H2 and the number of moles of O2 are
  equal. The total pressure is 1,120 millimeters of
  mercury. The mixture is sparked, and H2 and O2
  react until one reactant is completely consumed.
• (a) Identify the reactant remaining and calculate
  the number of moles of the reactant remaining.
• (b) Calculate the number of moles of water
  produced.

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2H2 (g) O2(g)
2 H2O (l)
1120 mm Hg / 2 560. mm Hg pressure of each
gas 560.mm Hg .737 atm 760 mm Hg/atm.
a. Use pVnRT to convert to moles n .737
atm(.500L) .0301 mol H2 and
O2 .0821 Latm/kmol x 298K With 2
givens, a limiting reactant must be calculated.
2 mol hydrogen with 1 mol oxygen are the rules of
the game, so the hydrogen will be consumed
first .0301 mol H2 x 1 mol H2O/ 1 mol H2
.0301 mol H2O formed .0301 mol H2 x 1 mol O2/ 2
mol H2 .0151 mol O2 used in reaction .0301 mol
O2 - .0151 mol O2 .0151 mol O2 remains after
reaction
14

• A 25.00 g mixture of potassium chlorate and
  potassium chloride is strongly heated until the
  all the potassium chlorate has decomposed into
  oxygen and solid potassium chloride. The mass of
  the product is 21.45g.
• Write the reaction for the complete decomposition
  of potassium chlorate.
• Calculate the mass of potassium chloride in the
  original mixture.

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• (a). 2KClO3(s) ? 2 KCl(s) 3O2(g)
• --------------------------------------------------
  -------------------
• (b). 25.00g 21.45 g 3.55 g O2
• 3.55 g O2 .111 mol O2
• 32.0 g/mol
• .111 mol O2 x 2 mol KClO3 .0740 mol KClO3
  (in the orig. mix.)
• 3 mol O2
• ..0740 mol KClO3 x 123 g/mol 9.10 g KClO3
  (in the orig. mix.)
• 25.00g 9.10g 15.9 g KCl (in the orig. mix.)
• 15.9g KCl x 100 63.6 KCl
• 25.00

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Gas Stoichiometry

• Balanced equation and pVnRT
• As you have noted, gas stoichiometry is
  frequently part of AP stoichiometry so we have
  already used the ideal gas law in example
  problems.

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Solution Stoichiometry

• Molarity (mol/L) x volume (L) moles is the key

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Look at Free Response problem 7 on your handout
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(a). 1 mol O2 x 2 mol MnO2 2 mol MnO2
equation 1 1 mol
O2 2 mol MnO2 x 1 mol I2
2 mol I2 equation 2
1 mol MnO2
2 mol I2 x 2 mol S2O32- 4 mol S2O32-
equation 3
1 mol I2
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(b). .00486 L x .0112 mol/L 5.44 x 10-5
mol S2O32- REMEMBER MOLARITY X LITERS
MOLES 5.44 X 10-4 mol S2O32- x 1 mol O2
1.36 x 10-5 mol O2

4 mol S2O32- The 14 mol ratio was
determined in part 1
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C) If some iodine is lost, then fewer moles of
thiosulfate will be required for the titration.
When the moles thiosulfate are reduced, so will
the moles oxygen detected will be reduced.
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D). Starch solution would be a good indicator.
The solution will be dark blue due to the
Starch-I3- complex until the iodine is converted
to I- when the blue color will disappear. (You
will probably see the starch indicator in the
kinetics experiment later in the course.)