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Stoichiometry, Percentage Composition, Empirical Formulas

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Stoichiometry, Percentage Composition, Empirical Formulas. E. A. Mottel (with ... The element silver (Ag) has two naturally occurring isotopes, 107 and 109. Its ... – PowerPoint PPT presentation

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Title: Stoichiometry, Percentage Composition, Empirical Formulas


1
Stoichiometry,Percentage Composition, Empirical
Formulas
  • E. A. Mottel (with modifications by JEF)
  • Integrated First-Year Curriculum in Science,
    Engineering and Mathematics

2
Topics
  • Isotopes - Percentage Composition
  • Molecules - Percentage Composition
  • Molecules - Empirical Formulas

3
Isotope Composition
  • Isotope1mass Isotope1percentage
  • Isotope2mass Isotope2percentage
  • Isotope3mass Isotope3percentage
  • Atomic Mass

4
The element silver (Ag) has two naturally
occurring isotopes, 107 and 109. Its average
atomic mass is 107.868. Calculate the mass of
the heavier isotope
  • (mass 109Ag)( 109Ag)
  • (mass 107Ag)( 107Ag) avg. atomic mass

5
Solve Equation
  • sln solve(
  • m109Ag(1 - 0.5182) 106.9050.5182
  • 107.868,
  • m109Ag)
  • sln 108.9037547

6
Chemical Analysis and Empirical Formulas
  • Percentage composition by mass of a compound
  • Determination of the empirical formula from
    percentage by mass information
  • Empirical versus molecular formulas

7
Percentage Composition
  • Find the percentage composition by mass of
    phosphorus in Na3PO4.
  • (mass P)
  • (mass P) (mass P)/(molar mass)
  • mass P 130.97
  • molar mass 322.99 130.97 416
  • (mass P) 0.1889

8
Percentage Composition
  • Silver Nitrate AgNO3
  • Silver Mass 1(AW Ag)
  • Nitrogen Mass 1(AW N)
  • Oxygen Mass 3(AW O)
  • Molar Mass 1(AW Ag) 1(AW N) 3(AW O)
  • ( Ag) 1(AW Ag)/(Molar Mass)
  • ( N) 1(AW N)/(Molar Mass)
  • ( O) 3(AW O)/(Molar Mass)

9
Empirical Formula
  • Empirical formula problems are the reverse of
    percentage composition problems.
  • A compound that contains only carbon, hydrogen
    and oxygen is 48.38 C and 8.12 H. What is the
    empirical formula of this compound?

10
Empirical Formula CnCHnHOnO
  • eqn1 mmass nC12.01nO16.00nH1.008
  • eqn2 nC12.01/mmass 0.4838
  • eqn3 nO16.00/mmass (1 - 0.4838 - 0.0812)
  • eqn4 nH1.008/mmass 0.0812
  • sln solve(eqn1, eqn2, eqn3, eqn4, mmass,
    nH, nC, nO)
  • sln nO .6749108619 nC, mass 24.82430757
    nC, nH 1.999735887 nC, nC nC
  • (CH2O2/3)nC -----gt (C3H6O2) nU

11
Empirical vs. Molecular Formulas
  • Empirical Formula (C3H6O2) nU
  • Possible Molecular Formulas
  • C3H6O2 nU 1
  • C6H12O4 nU 2
  • C9H18O6 nU 3
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