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Chapter 9 – Chem 160

Chapter 9 – Chem 160. Chemical Calculations: The Mole Concept and Chemical Formulas. The Law of Definite Proportions. Compounds are pure substances and They are a chemical combination They can be broken down They have a definite, constant elemental composition.

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Chapter 9 – Chem 160

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  1. Chapter 9 – Chem 160 Chemical Calculations: The Mole Concept and Chemical Formulas

  2. The Law of Definite Proportions • Compounds are pure substances and • They are a chemical combination • They can be broken down • They have a definite, constant elemental composition

  3. Problem 9.5 – which two of three experiments produced the same compound?

  4. Problem 9.5 cont’d

  5. Calculation of Formula Masses • Definition • Calculation

  6. Example 9.3 a) C7H6O2

  7. Significant Figures andAtomic Mass • Uncertainty in data vs. uncertainty in formula mass • We’ll use atomic masses rounded to hundredths • Consider formula mass calculations as a pure addition

  8. Percent composition • Definition • Calculation - % composition of Au(NO3)3

  9. The Mole • Avogadro’s number • # of molecules in 1.20 moles of CO • 1.20 moles CO x6.022x1023 CO molecules 1 mole CO • Cancel “moles CO” • Answer 7.2264 x 1023 – what units? • Significant figures?

  10. Mass of a Mole • Molar mass of an element is … • Molar mass of a compound is …

  11. Problem 9.37 (d) • Mass of 1.357 moles of Na3PO4 • 3 x Na = 3 x 22.99 = 68.97 g • P = 30.97 g • 4 x O = 4 x 16.00 = 64.00g • 1 mole 68.97+30.97+64.00=163.94 g • 1.357 mole x 163.94 g = 222.466 g • Note significant figures

  12. Significant Figures and Avogadro’s Number • The mole is the amount of substance … • Avogadro’s number should never be the limiting factor in s.f. considerations

  13. A.M.U. and Gram Units • 6.022 x 1023 amu = 1.000 g • Proof • 6.022x1023 atoms N x 1 mole N x 14.01 amu 1 mole N 14.01 g N 1 atom N • 6.022 x 1023 amu/g

  14. A.M.U. and Gram Units (cont’d) • What is the mass, in grams, of a molecule whose mass on the amu scale is 104.00 amu? (Example 9.8) • 104.00 amu x 1.000 g 6.022 x 1023 amu • 1.7270009 x 10-22 g • 1.727 x 10-22 g accounting for s.f.

  15. Counting Particles by Weighing • Atomic ratio to mass ratio • Table 9.2 • Cl (35.45) / Na (22.99) • Extension to molecules

  16. Counting Particles by Weighing – Problem 9.58 b) • Grams of Cu that will contain twice as many atoms as 20.00 g of Zn • 20.00 g Zn x 6.022 x 1023 atoms x 63.55 g Cu 65.41 g Zn 6.02 x 1023 atoms • = 19.43128 g Cu contains ….. • as many atoms as 20.00 g Zn • Answer is 19.43 x 2 = 38.86 g Cu

  17. Mole and Chemical Formulas • Microscopic level interpretation • Macro level

  18. Mole and Chemical Formulas – Problem 9.60 • 6 mole to mole conversion factors from the formula K2SO4 • 2 moles of K / 1 mole of K2SO4 • 1 mole of S / 1 mole of K2SO4 • 4 moles of O / 1 mole of K2SO4 • 2 moles of K / 1 mole of S • 2 moles of K / 4 moles of O • 1 mole of S / 4 moles of O

  19. Mole andChemical Calculations

  20. Mole & Chemical Calculations Problem 9.70 d) • Mass of 989 molecules of H2O • {(2x1.01)+16.00} gx 989 molecules 6.022 x 1023 molecules • = 2.95945 x 10-20 g

  21. Purity of Samples • Definition • Problem 9.86 a) calculate the mass in grams of Cu2S present in a 25.4 g sample of 88.7% pure Cu2S • 25.4 g of sample Cu2S x 88.7 g Cu2S 100 g of sample Cu2S • = 22.5298 g

  22. Empirical andMolecular Formulas • Empirical formula – smallest whole number ratio of atoms • Molecular formula – actual number of atoms in a formula unit

  23. Empirical andMolecular Formulas – cont’d • Problem 9.96 a) write empirical formula for P4H10 • P2H5 • 9.96 d) C5H12? • No change

  24. Determination ofEmpirical Formulas • Elemental composition data • Empirical formula + molecular mass

  25. Empirical andMolecular Formulas – cont’d • Problem 9.98 b) determine the empirical formula if 40.27% K, 26.78% Cr and 32.96% O

  26. Empirical andMolecular Formulas – cont’d • Empirical formula for 9.98 b) is K2CrO4

  27. Determination ofMolecular Formulas • Molecular mass information needed • molecular formula = (empirical formula)x; where x is a whole number • x= molecular formula (experimental) empirical formula(calculat’d from atomic masses) • For example (CH2)5 = C5H10

  28. Determination ofMolecular Formulas – 9.116 b) • P2O3, empirical formula; molecular mass is 220 amu • P: 2 x 30.97 = 61.94 amu • O: 3 x 16.00 = 48.00 amu • Empirical formula = 109.94 amu • Molecular/Empirical = 2.00 • Molecular formula is P4O6

  29. Determination ofMolecular Formulas – 9.120 b) • Citric acid, molecular mass 192 amu; 37.50% C, 4.21% H and 58.29% O

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