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Matter and Measurement _ Atoms_ Molecules_ and Ions_Stoichiometry

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Matter and Measurement _ Atoms_ Molecules_ and Ions_Stoichiometry Powered By Docstoc
					   Matter and Measurement / Atoms, Molecules, and
Ions/Stoichiometry: Calculations with Chemical Formulas
                    and Equations

           H Advanced Chemistry
                  Unit 1
   Objective #6 Isotopes / Calculating
Average Atomic Mass / Mass Spectrometer

*average atomic masses can be determined
  from the masses of the various isotopes
  that naturally occur for an element and
  their percent of abundance in nature
*examples:
*operation of mass spectrometer:
*instrument that provides the atomic mass
  and percent of abundance of the isotopes
  that make up a sample of an element
Operation of Mass Spectrometer
*gaseous sample is introduced into instrument and the
   particles are bombarded with high energy electrons
*positive ions are produced and these are passed
   through a magnetic field
*the amount of deflection is measured; the more
   massive the particle the less the deflection
*a graph of the intensity of the detector signal vs.
   the particle mass is called a mass spectrum
*it is from the data provided from this graph that
   one obtains the atomic mass and percent abundance
   necessary to calculate the average atomic mass for
   an element
Example of Mass Spectrograph for Hg
 Most Common Isotopes of Mercury Based on Mass
                 Spectrograph

Isotope              Percent Abundance
*mercury-196                    .15 %
*mercury-198                   9.97 %
*mercury-199                  16.87 %
*mercury-200                  23.1 %
*mercury-201                  13.18 %
*mercury-202                  29.86 %
*mercury-204                   6.87 %
Objective #12 Description of Combustion
               Analysis
    Objective #12 Combustion Analysis

*Example I: An unknown compound is
  composed of carbon, nitrogen, and
  hydrogen. When .1156 grams of this
  compound is reacted with oxygen and
  burned, the following components are
  produced: .1638 g of carbon dioxide
  and .1676 g of water. Calculate the
  percent composition and empirical
  formula for this unknown compound.
Step I All of the carbon that was in the original
   sample is now contained in the carbon dioxide.
   Convert from grams of carbon dioxide to grams
   of carbon.
.1638 g CO2 X 12.0 g C/44.0 g CO2 = .04467 g C

Step II Divide the resulting mass of carbon by the
  original mass of the compound and multiply by
  100% to obtain the percentage of carbon in the
  sample
% C = .04467 g/.1156 g X 100 = 38.64% C
Step III All of the hydrogen that was in
  the original sample is now contained in
  the water. Convert from grams of
  water to grams of hydrogen.
.1676 g H2O X 2.0 g H / 18.0 g H2O =
  .01862 g
Step IV Divide the resulting mass of
  hydrogen by the original mass of the
  compound and multiply by 100% to obtain
  the percentage of the hydrogen in the
  sample
%H = .01862 g/.1156 g X 100 = 16.11% H
Step V The mass and percentage of the
  final component can be determined by
  mass and percentage difference.
.1156 g - .04467 g - .01862 g = .05231 g N
100% - 38.64% - 16.11% = 45.25% N
Step VI Use the percentages or masses
  of components to compute the empirical
  formula of the compound.
.04467 g C X 1 mole C/12.0 g C = .003723
  moles C
.01862 g H X 1 mole H/1.0 g H = .01862
  moles H
.05231 g N X 1 mole N/14.0 g N = .003736
  moles N
.003723 moles C/.003723 = 1
.01862 moles H/.003723 = 5
.003736 moles N/.003723 = 1
CH5N
Example II Caproic acid, which is
 responsible for the foul odor of dirty
 socks, is composed of C, H, and O atoms.
 Combustion analysis of a .225 g sample
 of this compound produces .512 g of
 carbon dioxide and .209 g of water.
 What is the emprical formula of caproic
 acid? Determine the molecular formula
 if the molecular mass is 116 g.
.512 g CO2 X 12.0 g C/44.0 g CO2 = .140 g C
.209 g H2O X 2.0 g H/18.0 g H2O = .0232 g H
By difference .0618 g O
.140 g C X 1 mole C/12.0 g C = .0117 g C
.0232 g H X 1 mole H/1.0 g H = .0232 mole H
.0618 g O X 1 mole O/16.0 g O
 =.00386mole O
.0117 mole C/.00386 = 1 mole C
.0232 mole H/.00386 = 6 mole H
.00386 mole O / .00386 = 1 mole O
C3H6O empirical formula
116 g/58 g = 2
C6H12O2 molecular formula
     Objectives #13-14 Stochiometry

1. Calculate the mass of water produced
   from the decomposition of 5.00 g of
   ammonium dichromate.
5.00 g X 1mole A.D. / 252.0 g A.D. X 4
   mole H2O / 1 mole A.D. X 18.0 g H2O =
   14.4 g
2. Calculate the number of water
   molecules produced from 10. g of A.D.
10. g A.D. X 1 mole A.D. / 252.0 g A.D. X
 4 mole H2O / 1 mole A.D.X
6.02 X 1023 molecules / 1 mole H2O =
9.56 X 1022 molecules
3. Calculate the amount of A.D. required
   to produce 5500 J of energy if the
   change in energy of the reaction is
   2750 J.
5500 J X 1 mole A.D. / 2750 J X
252.0 g A.D. / 1 mole A.D. = 5.0 X 102 g
4. Calculate the number of chromium
   atoms produced from the
   decomposition of 10.0 g of A.D.
10.0 g A.D. X 1 mole A.D. / 252.0 g A.D. X
   1 mole Cr2O3 / 1 mole A.D. X
6.02 X 1023 f. units Cr2O3 / 1 mole Cr2O3
   X 2 Cr atoms / 1 f. unit Cr2O3 =
4.78 X 1022 atoms
Example II: The reaction below produced 4.50 g
  of aluminum sulfate in the laboratory. Calculate
  the theoretical yield and the percent yield
  given 5.00 g of sulfuric acid reacted with 5.00
  g of aluminum hydroxide.
5.00 g H2SO4 X 1 mole H2SO4 / 98.1 g X 1 mole
  Al2(SO4)3 / 3 mole H2SO4 X 342.3 g Al2(SO4)3 /
  1 mole = 5.82 g
5.00 g Al(OH)3 X 1 mole Al(OH)3 / 78.0 g X
1 mole Al2(SO4)3 / 2 mole Al(OH)3 X 342.0 g
  Al2(SO4)3 = 11.0 g
%Yield = 4.50 g / 5.82 g X 100 = 77%
Use the above information to determine
  the amount of the excess reactant
  leftover.
5.00 g H2SO4 X 1 mole H2SO4 / 98.1 g X 2
  mole Al(OH)3 / 3 mole H2SO4 X 78.0 g
  Al(OH)3 = 2.65 g (amount used)
5.00 g – 2.65 g = 2.35 g leftover
   Objectives #15-16 Molarity, Molality,
    Solution Stoichiometry and Titrations

Example I: Determine the molarity of a
 solution containing 2.37 moles of
 potassium nitrate in enough water to
 give 650 ml of solution.
M = 2.37 moles / .650 L = 3.65 M
Example II: Determine the molarity of a
  solution containing 25.0 g of sodium
  hydroxide dissolved in enough water to
  give 2.50 L of solution.
M= 25.0 g NaOH / 40.0 g /
2.50 L = .250 M
Example III: How many grams of sucrose
  are present in 125 ml of a 1.07 M
  sucrose solution?
Moles = .125 ml X 1.07 M = .134 moles
.134 moles sucrose X 342.0 g / 1 mole
= 45.8 g
*Molality
Example I: Calculate the molality of a solution
   containing 5.0 g of sodium chloride dissolved
   in 250 g of water
5.0 g NaCl X 1mole NaCl / 58.5 g / .250 kg H2O
= .34 m
Example II: Calculate the molality of a
  solution containing 5.0 moles of NaCl
  dissolved in 250 g of water.
5.0 moles NaCl / .250 kg H2O = 20. m
*Dilution
*moles before dilution = moles after dilution
*formula: M1V1 = M2V2
Example I: What is the molarity of a
  solution prepared by diluting 12.0 ml of
  .405 M NaCl to a final volume of
 80.0 ml?
12.O ml X .405 M / 80.0 ml = .0608 M
Example II: What volume of .25 M HCl
  solution must be diluted to prepare
1.00 L of .040 M HCl?
.040 M X 1.00 L / .25 M = .16 L
*Precipitation Reactions and Solution
  Stoichiometry
*key concept: to find moles of substance in
  solution multiply moles by volume
Example I: How many grams of precipitate can be
  produced from the reaction of 1.00 L of .375 M
  silver nitrate solution with an excess of sodium
  chloride solution?
.375 M X 1.00 L silver nitrate = .375 moles
.375 moles AgNO3 X 1 mole AgCl / 1 mole AgNO3
  X 143.4 g AgCl / 1 mole AgCl = 53.8 g
Example II: How many grams of precipitate
  can be produced from the reaction of 2.50
  L of a .200 M calcium chloride solution
  with an excess of phosphoric acid?
.200 M calcium chloride X 2.50 L = .500
  moles CaCl2
.500 moles CaCl2 X 1 mole Ca3(PO4)2 / 3
  moles CaCl2 X 310.3 g Ca3(PO4)2 X 1 mole =
  51.7 g
                      .
Example III: What volume in ml of a .300 M
   silver nitrate solution is needed to react
   with 40.0 ml of a .200 M potassium
   phosphate?
.200 M potassium phosphate X .0400 L =
   .00800 moles
.00800 moles K3PO4 X 3 mole AgNO3 / 1
   mole K3PO4 = .0240 moles AgNO3
moles / M = L
 .0240 moles AgNO3 / .300 M = 80.0 ml
*Titration Problems:
Example I: What is the molarity of a 37.5
  ml sample of a sulfuric acid solution that
  will completely react with 23.7 ml of a .100
  M sodium hdyroxide solution?
.100 M NaOH X .0237 L = .00237 moles
  NaOH
.00237 moles NaOH X 1 mole H2SO4 / 2 mole
  NaOH = .00119 moles H2SO4
M = .00119 moles H2SO4 / .0375 L = .0317 M
Example II: What volume in liters of a 1.00
  M hydrogen sulfide solution is needed to
  react completely with .500 L of a 4.00 M
  nitric acid solution?
4.00 moles HNO3 X .500 L = 2.00 moles
  HNO3
2.00 moles HNO3 X 3 moles H2S / 2 moles
  HNO3 = 3.00 moles H2S
L = moles / M
3.00 moles H2S / 1.00 L = 3.00 L

				
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posted:11/23/2012
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