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A.P. Chemistry Chapter 3

A.P. Chemistry Chapter 3. 3.1 Balancing Chemical Equations. Law of Conservation of Mass A. Reactants B. Products Reactions A. Combustion comp. + O 2  B. Combination (synthesis) A + B  C. …Continue 3.1. C. Decomposition A  B + C D. Single Displacement

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A.P. Chemistry Chapter 3

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  1. A.P. ChemistryChapter 3

  2. 3.1 Balancing Chemical Equations • Law of Conservation of Mass A. Reactants B. Products • Reactions A. Combustion comp. + O2 B. Combination (synthesis) A + B  C

  3. …Continue 3.1 C. Decomposition A  B + C D. Single Displacement element + comp  element + comp E. Double Displacement AB + CD  AD + CB

  4. 3.3 • Atomic Mass Unit (amu) ~ 1 proton 1 amu = 1.66054 x 10-24 grams II. Average atomic mass = average of all isotopes Sample 3.4 75.53%35Cl x 34.969 amu = 26.41 24.47% 37Cl x 36.966 amu = 9.046 35.46 amu

  5. …Continue 3.3 • Formula Weight (ionic) Molecular Weight (covalent) • % Composition Sample 3.6 % of C12H22O11 %C= 12 (12.0)/342.0 x 100 = 42.1% %H= 22(1.01)/342.0 x 100= 6.4% %O= 11(16.0)/342.0 x 100= 51.5%

  6. 3.4 Moles 1 mole = 6.02 x 1023atoms molecules ions formula units Called Avogadro’s number

  7. …Continue 3.4 Sample 3.8 Calc. # of H atoms in 0.350 mol C6H12O6 0.350 mol x 6.02 x 10 23 x 12 atoms H 1 mol 1 molecule C6H12O6 = 2.53 x 1024 atoms of H

  8. …Continue 3.4 Sample 3.12 change 5.23g of C6H12O6 to molecules 5.23g x 1 mol x 6.02(1023) molecules 180.0g 1 mol =1.75 x 1022 molecules

  9. 3.5 Empirical Formulas Sample 3.12 C 40.92g x 1 mol C = 3.407mol/3.407 = 1 x 3 12.0g H 4.58g x 1 mol H = 4.54 mol/3.407 = 1.33 x 3 1.0g O 54.50g x 1 mol O = 3.407 mol/3.407 = 1 x 3 16.0g C3H4O3

  10. …Continue 3.5 II. Molecular Formula 40.0% C / 12.0 = 3.33/3.33 = 1 6.7% H / 1.01 = 6.63/3.33 = 2 53.3% O / 16.0 = 3.33/3.33 = 1 MW = 180.0 g/mol empirical formula = C1H2O1 molecular formula = C6H12O6

  11. …Continue 3.5 III. Combustion Analysis .255 g isopropyl alcohol 0.561g CO2x 1 mol CO2x 1 mol C x 12.0g C = 0.154g C 44.0g CO2 1 mol CO2 1 mol C 0.306g H2O x 1 mol H2O x 2 mol H x 1.01g H =0.0343g H 18.0g H2O 1 mol H2O 1 mol H 0.154g C / 12.0 = 0.0128 mol/.0042 = 3 .0343g H / 1.01 = .0340 mol/.0042 = 8 .067g O/16.0 = .0042 mol/.0042 = 1 C3H8O1 = C3H7OH

  12. 3.6 Mass-Mass Problems Sample 3.14 1.00g C6H12O6 ? g H2O C6H12O6 + 6O2 6CO2 + 6H2O 1.00g C6H12O6 x 1 mol x 6 mol H2O x 18.0g H2O 180.0g 1 mol 1 mol H2O = 0.600g water

  13. 3.7 Excess and Limiting Reagents 2H2 + O2 2H2O start 10 mol 7 mol 0 mol _____________________________________ end limiting reagent (reactant)- substance that is used up first in chemical reaction. It causes the reaction to cease. excess- substance left over. II. Percent yield = actual yield x 100 theoretical yield

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