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CHEMISTRY 161 Chapter 4 The Mole

CHEMISTRY 161 Chapter 4 The Mole. Macroscopic versus Microscopic Worlds. 2 H 2 + O 2 2 H 2 O. 1 liter water contains about 3.3 X 10 25 molecules. CHEMICAL MASS SCALE. standard / calibration. atomic mass unit (amu, u). one atom of carbon-12 12 u (exactly).

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CHEMISTRY 161 Chapter 4 The Mole

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  1. CHEMISTRY 161 Chapter 4 The Mole

  2. Macroscopic versus Microscopic Worlds 2 H2 + O2 2 H2O 1 liter water contains about 3.3 X 1025 molecules

  3. CHEMICAL MASS SCALE standard / calibration atomic mass unit (amu, u) one atom of carbon-12 12 u (exactly) we have to correlate u with kg

  4. H2O PSE O: 15.999 u H: 1.008 u H: 1.008 u H2O: 18.015 u formula mass: weight of one molecule

  5. CaO O: 15.999 u Ca: 40.08 u CaO: 56.08 u formula mass: weight of one molecule correlation between u and kg

  6. MOLE

  7. one mole of a compound contains the same number of molecules/atoms as the number of atoms in exactly 12 g of 12C Avogadro’s number Na 6.023 x 1023

  8. 6.023 x 1023 1 mole of H2O molecules 6.023 x 1023 atoms 1 mole of 12C atoms 6.023 x 1023 1 mole of Na molecules 6.023 x 1023 1 mole of NaCl Avogadro’s number links micro and macroscopic world

  9. 6.023 x 1023 1 mole of H2O molecules 1 molecule of H2O – 2 H atoms and 1 O atom 6.023 x 1023 molecules of H2O 6.023 x 1023 atoms of O 2 x6.023 x 1023 atoms of H 1 mole of H2O – 2 mole H atoms and 1 mole O atoms

  10. H2O Na O: 15.999 u H: 1.008 u H: 1.008 u H2O: 18.015 u 15.999 g/mol Na 1.008 g/mol Na 1.008 g/mol 1 mole of H2O – 18.015 g

  11. CaO O: 15.999 u Ca: 40.08 u CaO: 56.08 u 1 mole of CaO – 56.08 g

  12. 1 mole of H2O – 18.015 g 6.023 x 1023 molecules of H2O – 18.015 g one molecule of H2O – 2.99 x 10-23 g molecular weight of one mole of H2O 18.015 g mol-1

  13. 2 H2 + O2 2 H2O 2 molecules 1 molecule 2 molecules 2 moles 1 mole 2 moles 4.03176g 31.9988g 36.03g x g y g 70.0g STOICHIOMETRY

  14. 2 H2 + O2 2 H2O 2 molecules 1 molecule 2 molecules 2 moles 1 mole 2 moles 4.03176g 31.9988g 36.03g x g y g 70.0g STOICHIOMETRY

  15. Example I: How many grams of iron are in a 15.0 g sample of iron(III) oxide? 1. molecular formula Fe2O3 2. weight of one molecule 159.7 u 3. weight of one mole Fe2O3 159.7 g 4. 1 molecule Fe2O3 contains 2 atoms of Fe 5. 1 mole Fe2O3 contains 2 moles of Fe 159.7 g 111.69 g 15.0 g x g x = 10.5 g

  16. Example II: How many grams of oxygen are in a 10.0 g sample of nickel (II) nitrate? 1. molecular formula 2. weight of one molecule 3. weight of one mole Ni(NO3)2 4. 1 molecule Ni(NO3)2 contains 6 atoms of O 5. 1 mole Ni(NO3)2 contains 6 moles of O x = 5.25 g

  17. Example III: How many atoms are in 10 kg of sodium? 1 mole sodium = 22.98977 g 6.023 x 1023 atoms = 22.98977 g x atoms = 10,000 g x = 2.6 x 1026 atoms

  18. Example IV How heavy are 1 million gold atoms? 1 mole gold = 196.96654 g 6.023 x 1023 atoms = 196.96654 g 1,000,000 atoms = x g x = 3.2 x 10-16 g = 0.32 fg

  19. Mass Percentage Composition P4O10 = X 100 % = X 100 %

  20. Example V A sample was analyzed and contains 0.1417 g of nitrogen and 0.4045 g of oxygen. Calculate the percentage composition. = X 100 % 1. mass of whole sample 2. percentages of elements

  21. MOLECULAR FORMULA H2O EMPIRICAL FORMULA H2O2 HO H2O H2O P2O5 P4O10 P2*2O2*5

  22. Example A sample contains 0.522 g of nitrogen and 1.490 g of oxygen. Calculate its empirical formula. N2O5

  23. COMBUSTION CxHy CO2 H2O

  24. COMBUSTION C3H8 + O2 + H2O CO2 0.013068 g How many grams of oxygen are consumed? • balance equation • convert to moles

  25. The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO2and 4.512 g of H2O. Calculate the empirical formula of the compound. How many grams of Al2O3 are produced when 41.5 g Al react? 2Al(s) + Fe2O3(s) Al2O3(s) + 2 Fe(s)

  26. LIMITING REACTANT C2H4 + H2O C2H5OH

  27. limiting reactant excess reactant

  28. How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to 4 NH3 + 5 O2 4 NO + 6 H2O 2 C2H2 + 5 O2 4 CO2 + 2 H2O

  29. THE YIELD 2 C2H2 + 5 O2 4 CO2 + 2 H2O theoretical yield of CO2: 10 g actual yield of CO2: 8 g X 100 % =

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