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The Mole

The Mole. Review Write the formula. Sodium sulfide Calcium oxide Aluminum oxide. Na 2 S. CaO. Al 2 O 3. Write the formula. Potassium phosphate Calcium hydroxide Chromium(III) sulfate. K 3 PO 4. Ca (OH) 2. Cr 2 (SO 4 ) 3. Name the following. BaS NaCl K 2 SO 4. Barium Sulfide.

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The Mole

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  1. The Mole

  2. ReviewWrite the formula Sodium sulfide Calcium oxide Aluminum oxide Na2S CaO Al2O3

  3. Write the formula Potassium phosphate Calcium hydroxide Chromium(III) sulfate K3PO4 Ca(OH)2 Cr2(SO4)3

  4. Name the following BaS NaCl K2SO4 Barium Sulfide Sodium Chloride Potassium Sulfate

  5. Name the following FeS Cr2O3 Al2 (SO3)3 Iron (II) Sulfide Chromium (III) Oxide Aluminum Sulfite

  6. Multiplying Fractions = 1.25 3 a2 5 c = = × × 3 a 4 b =

  7. The Conversion Factor 12 inches = 1 foot Convert: 1297 ft = _____ inches The Factor Label Method (using the bridge model) 15564 in

  8. Example 1 345 meters = _____ cm = 34,500 cm Example 2 45.7 inches = _____ feet = 3.81 ft

  9. Mole, Atoms, Formula Units, Molecules Conversions • The student will be able to interconvert between moles and atoms, formula units, and molecules.

  10. Moles A mole is a number like a dozen. 1 dozen = 12 things 1 gross = 144 things

  11. Moles 1 mole = 6.02 × 1023 particles That is 602,000,000,000,000,000,000,000. Why is it so big?

  12. Moles Because in chemistry we’re dealing with atoms and molecules, which are very small.

  13. Moles How much space would 1 mole of marbles take up?

  14. Moles It would cover the earth… 50 miles deep!

  15. Moles Representative Particles Atom Ion Molecule Formula Unit 1 mole = 6.02 × 1023atoms (single elements) 1 mole = 6.02 × 1023ions (ion elements w/ a charge) 1 mole = 6.02 × 1023molecules (covalent compounds) 1 mole = 6.02 × 1023formula units (ionic compounds)

  16. Mole Map 1 Moles 1 step (1 conversion factor) 1 step (1 conversion factor) Molar Mass 6.02 × 1023 Mass (g) Particles 2 steps (2 conversion factors) Atoms, Ions, Molecules, or Formula Units

  17. Moles to ParticlesParticle to MolesConversions(use bridge model to setup problems)(1 step problems)

  18. Example 1 How many moles are contained in 5.43 × 1024 atoms of lead? 5.43 × 1024 atoms Pb 1 molPb = 1 6.02 × 1023 atoms Pb 5.43 × 1024molPb = 9.02 molPb 6.02 × 1023

  19. Example 2 How many formula units are contained in 0.00233 moles of KCl? formula units KCl 6.02 × 1023 0.00233 molKCl = 1 1 molKCl 1.40 × 1021 formula units KCl 1.40 × 1021 formula units KCl = 1

  20. Example 3 How many moles are contained in 2.20 x 1024 molecules of CO2? 2.20 × 1024molecules CO2 molCO2 1 = 6.02 × 1023 1 molecules CO2 2.20 × 1024molCO2 3.65 molCO2 = 6.02 × 1023

  21. Practice Book Page 324 #5-6 a & b

  22. Mole to MassMass to Mole Conversions(1 step problem)

  23. Mass-Mole Conversion for Atoms(use the Periodic Table to find mass) • 1 mole = atomic mass (in grams) • 1 mole Ne = 20.2 g Ne • 1 mole Sr = 87.6 g Sr • 1 mole U = 238.0 g U

  24. Example 1 Convert: 5.0 mol Li = ______ g Li 5.0 mol Li 6.9 g Li = 1 1 mol Li 35 g Li = 35 g Li 1

  25. Example 2 Convert: 15.0 g Be = ______ mol Be 1 15.0 g Be mol Be = 1 9.0 g Be 15.0 mol Be = 1.67 mol Be 9.0

  26. Example 3 How many grams of vanadium are contained in 0.0129 mol vanadium? g V 0.0129 mol V 50.9 = 1 1 mol V 0.657 g V = 0.657 g V 1

  27. Practice Book Page 328 #15-16 a & b

  28. Mass to ParticlesParticles to MassElements(2 step problems)

  29. Example 1 What is the mass of 3.1 × 1025 molecules of N? 3.1 × 1025 molecules N 1 molN g N 14.0 = 6.02 × 1023 1 molN 1 molecules N 4.3 × 1026molN = 720 g N 6.02 × 1023

  30. Example 2 How many atoms of N are in 321 grams of N? atoms N 321 grams N 1 molN 6.02 × 1023 = 14.0 1 molN 1 grams N 1.93 × 1026 atoms N = 1.38 × 1025 atoms N 14.0

  31. Practice Book Page 331 # 19-20 a-c & 21 a

  32. Molar Mass • Definition: The mass (in grams) of one mole of any pure substance.

  33. Example 1 Determine the molar mass of glucose: C6H12O6. = = = C H O 6 12 6 × × × 12.0 1.01 16.0 72.0 12.1 96.0 molar mass = 180.1 g/mol

  34. Example 2 Determine the molar mass of calcium phosphate. PO43- Ca2+ Ca3(PO4) 2 3 2 = = = Ca P O 3 2 8 × × × 40.1 31.0 16.0 120.3 62.0 128 molar mass = 310 g/mol

  35. Practice Book Page 335 # 35-36 a-c

  36. Mole to Mass Mass to MoleCompounds Practice Book Page 336 #37-39 Page 337 # 40 a-c

  37. Mass to ParticlesParticles to Mass Compounds Conversions(2 step problems) • Remember Avogadro’s number: • 1 mole = 6.02 × 1023 particles • Particles are atoms, ions, molecules and formula units

  38. Remember • Atoms are used for single elements. (Cu, H) • Ions are used for charged atoms (Cu2+, H+) • Molecules are used for covalent compounds. (CO2, H2O) • Formula units are used for ionic compounds. (NaCl, K2SO4)

  39. Mass to Particles Particles to Mass Conversions(2 step problems) Equalities 1 mol = 6.02 × 1023 particles = (molar mass) g 1 mol C = 6.02 × 1023 atoms C = 12.0 g C 1 mol CO2=6.02 × 1023molecules CO2=44.0 g CO2

  40. Mole Map 1 Moles 1 step 1 step Molar Mass 6.02 × 1023 Mass (g) Particles 2 steps Atoms, Ions, Molecules, or Formula Units

  41. Example 1 How many molecules are in 25 grams of dinitrogenpentoxide (N2O5)? molecules N2O5 25 g N2O5 1 mol N2O5 6.02 × 1023 = 108.0 1 mol N2O5 1 g N2O5 1.5 × 1025 molecules N2O5 = 1.4 x 1023 molecules N2O5 108.0

  42. Example 2 What is the mass of 3.1 × 1025 molecules of dinitrogenpentoxide? 3.1 × 1025 molecules N2O5 1 mol N2O5 g N2O5 108.0 = 6.02 × 1023 1 mol N2O5 1 molecules N2O5 3.3 × 1027mol N2O5 = 5600 g N2O5 6.02 × 1023

  43. Practice Book Page 360 #139,141,142

  44. Percent Composition, Molecular, and Empirical Formulas • The student will be able to: • Determine the percent composition of a compound. • Determine the empirical formula from molecular formula. • Determine the empirical formula from % data. • Determine the molecular formula from the empirical formula and molar mass.

  45. Percent Composition • Definition: The percent by mass of each element in a compound.

  46. Percent Composition % = × 100 # atoms × atomic mass molar mass

  47. Example 1 What is the percent composition of K2SO4? %K = × 100 = 44.9%K %S = × 100 = 18.4%S %O = × 100 = 36.7%O 2 × 39.1 1 × 32.1 4 × 16.0 174.3 174.3 174.3

  48. Example 2 What is the percent composition of aluminum carbonate? Al2(CO3) 3 %Al = × 100 = 23.1%Al %C = × 100 = 15.4%C %O = × 100 = 61.5%O 2 × 27.0 3 × 12.0 9 × 16.0 234.0 234.0 234.0

  49. Practice Book Page 344 #54-57

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