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Stoichiometry, Percentage Composition, Empirical Formulas

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

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Stoichiometry, Percentage Composition, Empirical Formulas

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  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) + 106.905*0.5182 = 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 = 1*30.97 • molar mass = 3*22.99 + 1*30.97 + 4*16 • (%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 = nC*12.01+nO*16.00+nH*1.008; • eqn2 := nC*12.01/mmass = 0.4838; • eqn3 := nO*16.00/mmass = (1 - 0.4838 - 0.0812); • eqn4 := nH*1.008/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 -----> (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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