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The Mole Chapter 11 Chemistry RiverDell High School Ms. C. Militano

The Mole Chapter 11 Chemistry RiverDell High School Ms. C. Militano. What is a mole in chemistry? What conversion factors are associated with the mole? Types of conversions involving mole equalities. I. What is a Mole?. SI base unit that measures amount of a substance

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The Mole Chapter 11 Chemistry RiverDell High School Ms. C. Militano

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  1. The Mole Chapter 11ChemistryRiverDell High SchoolMs. C. Militano What is a mole in chemistry? What conversion factors are associated with the mole? Types of conversions involving mole equalities

  2. I. What is a Mole? • SI base unit that measures amount of a substance B. 1 mol = Avogadro Number of particles (particles can be atoms, molecules or ions) 6.02 x 1023 is Avogadro's Number • Molar Mass – mass of one mole of atoms of an element 1. ex. C = 12.0amu N = 14.0 amu D. Mole Equalities - 1 mole = molar mass - 1 mole = 6.02 x 1023 particles

  3. II. Mole Conversions [mass-mole-atoms]A. Type of Problem Equality 1. MOLES  MASS 2. MASS  MOLES 1 mole= molar mass (g) 3. MOLES  ATOMS 4.ATOMS  MOLES 1 mole = 6.02 X 1023 atoms 5.MASS  ATOMS 1 mole = 6.02 x 1023 atoms 6.ATOMS MASS 1 mole = molar mass (g)

  4. Making Conversions – use quantities in the Atom and Mass box as conversion factors 6.02x1023 molar mass ATOMS MOLE MASS(g) 1 mole 1 mole

  5. B. Using Factor Label to Make Mole Conversions 1 mole  molar mass 6.02 x 1023 1 mole PARTICLES<----> MOLES<----> MASS 6.02 x 10231 mole 1 mole molar mass

  6. C. Solving Mole Problems EXAMPLES • 1.00 mole of He = 4.00 g. • 2.00 mole of He = _____g 2.00 mol He X 4.00g He = 1 1 mole He 8.00 g He

  7. EXAMPLES • 1.00 mole He = 6.02 X 1023 atoms • 2.00 mole He = ________atoms He 2.00 mole He x 6.02 x 1023 atoms He = 12.04 x 1023 1 1 mole He 1.20 x 1024 atoms He • 16.00g He = _____ moles He 16.0 g He x 1 mole He = 1 4.00 g He 4.00g He

  8. EXAMPLES • 3.01 X 1023 atoms He = _____ moles 3.01 x 1023 atoms He x 1 mole He = 1 6.02 x 1023 atoms He .500 mol He • 8.00g He =______atoms He 8.00 g He x 1 mole He x 6.02 x 1023 atoms He = 1 4.00g He 1 mole He 12.04 x 1023 atoms He =1.20 x 1024 atoms He

  9. Sample Problems – More Practice • Moles to mass. Find the mass of 3.50 moles of carbon. • Mass to Moles How many moles of carbon are contained in 60.0 g of carbon? • Moles to Atoms How many atoms of carbon are found in 4.00 moles of carbon?

  10. Sample Problems More Practice • Atoms to Moles How many moles of carbon are represented by 1.806 x 1024 atoms of carbon? • Mass to Atoms How many carbon atoms are found in 36.0g of carbon? • Atoms to Mass What is the mass of 1.204 x 1024 atoms of carbon?

  11. Answers to Sample Problems 1. 42.0g C 2. 5.00 mol C 3. 24.1 x 1023 = 2.41 x 1024 atoms C 4. .300 x 10-1 = 3.00 mol C 5. 18.06 x 1023 = 1.81 x 1024 atoms C 6. 24.0 g C

  12. More Sample Problems 2.00 moles of Cu = atoms of Cu 60.0 grams of C = moles of C 3.00 x 1023 atoms He = moles of He 2.50 moles Al = grams of Al 28.0 grams N = atoms of N 1.80 x 1023 atoms Mg = grams of Mg

  13. Answers to More Sample Problems • 12.04 x 1023 = 1.20 x 1024 atoms Cu • 5.00 mol C • .498 mol He or 4.98 x 10-1 mol He • 67.5 g Al • 12.04 x 1023 = 1.20 x 1024 atoms N • 7.27 g Mg

  14. Compounds and Diatomic Molecules • Convert 4.00 mol NaOH to grams. a. Na = 23.0 O = 16.0 H = 1.00 formula mass = 40.0 b. 4.00mol NaOH x 40.0g NaOH = 160.g NaOH 1 1 mol NaOH

  15. 2. How many molecules are found in 2.00 mol of H2SO4? (sulfuric acid) 2.00 mol H2SO4 x 6.02 x 1023 molecules H2SO4 1 1 mol H2SO4 12.04 x 1023 = 1.20 x 1024 molecules H2SO2

  16. How many moles are found in 64.0 g of oxygen gas (O2)? 64.0g O2 x 1mol O2 = 2.00 mol O2 1 32.0g O2 4. How many formula units are found in 117.0g of sodium chloride (NaCl)? 117.0g NaCl x 1 mol NaCl x 6.02x1023 units NaCl 1 58.5 g NaCl 1 mol NaCl 12.04 x 1023 = 1.20 x 1024 formula units NaCl

  17. III. Percent Composition • Procedure 1. Determine total mass for each element 2. Determine the molar mass (formula mass) 3. Divide mass of each element by the molar mass (formula mass) 4. Multiply by 100%

  18. B. Problem Solving • Determine the percent composition of each element in carbon dioxide (CO2) C = 12.0 2O = 2 x 16.0 = 32.0 Sum = 44.0 (formula mass) Carbon 12.0/44.0 = .273 = 27.3% Oxygen 32.0/44.0 = .727 72.7%

  19. B. Problem Solving • Determine the percent composition of each element in calcium hydroxide Ca(OH)2 Ca = 40.1 O = 16.0 x 2 = 32.0 H = 1.00 x 2 = 2.00 sum is 74.1(molar mass) % Ca = 40.1/74.1 = 54.1% % O = 32.0/74.1 = 43.2% % H = 2.00/74.1 = 2.70%

  20. B. Problem Solving • Determine the precent composition of each element in TNT (trinotrotoluene) C7H5(NO2)3 7C = 7(12.0) = 84.0 5H = 5(1.00) = 5.00 3N = 3(14.0) = 42.0 6O = 6(16.0) = 96.0 sum =227(molar mass) % C = 84.0/227 =37.0% % H = 5.00/227 = 2.20% % N = 42.0/227 = 18.5% % O = 96.0/227 = 42.3%

  21. IV. Hydrates • Definitions 1. hydrate – compound that has a specific number of water molecules in its crystal (solid state) 2. water of hydration – water molecules that are part of the crystal (solid state) 3. anhydride – compound without water of hydration

  22. B. Naming Hydrates 1. CaCl2 2H2Ocalcium chloride dihydrate 2. NaC2H3O23H2Osodium acetate trihydrate 3. CuSO4 5H2Ocopper sulfate pentahydrate 4. MgSO4 7H2Omagnesium sulfate heptahydrate 5. Na2CO3 10H2Osodium carbonate decahydrate

  23. C. Heating a Hydrate ∆ Hydrate anhydride+water ∆ Na2CO3 10H2O Na2CO3+10 H2O ∆ MgSO4 7H2O MgSO4+7 H2O ∆ Na2CO3 10H2O  Na2CO3+ 10H2O

  24. D. Problem Solving - % Water in Hydrates • Procedure a. determine formula mass of the compound b. divide mass of only water molecules by the formula mass of the compound c. multiply answer by 100%

  25. Examples – Find % water in the hydrate 2a. CaCl2 2 H2O Ca = 40.1 Cl = 35.5 (2) = 71.0 H = 1.00(4) = 4.00 O = 16.0(2) = 32.0 Total mass is 147.1 Mass of water 2(18.0) = 36.0 % H2O mass water = 36 = .2447 = mass of compound 147.1 24.5%

  26. 2b. Find the % water in the hydrate MgSO4 7H2O Mg = 24.3 S = 32.1 O(4) = 64.0 H2O(7) = 18(7) = 126 sum = 246.4 % water = 126/246.4 = 51.1%

  27. V. Empirical Formulas A. lowest whole number ratio of subscripts B. How to Determine the Empirical Formula 1. if given % composition write % as a number of grams without the % sign 2. divide # grams by molar mass to get the number of moles 3. Divide # moles for each element by smallest 4. Round to nearest whole number when possible 5. Multiply by 2,3,or 4 to get whole numbers- if necessary 6. Use resulting numbers as subscripts in the formula

  28. C. Sample Problems 1. Find the empirical formula of a compound containing 19.55 gof potassium (39.08) and 2.00gof oxygen (16.00). Determine # moles of each element # moles Oxygen –2.00/ 16.00 = .125 # moles Potassium -19.55/ 39.08 = .500 Divide # moles of each by the smallest .125/.125 = 1 .500/.125 = 4.00 Formula is K4O

  29. Find the empirical formula of a compound containing 5.41g Fe (55.85), 4.64g Ti (47.88), and 4.65g O (16.00). # moles Iron(Fe) - 5.41/55.85 = .0968 # moles Titanium(Ti) - 4.64/47.88 = .0969 # moles Oxygen(O) - 4.65/16.00 = .291 Fe = .0968/.0968 = 1.00 Ti = .0969/.0968 = 1.00 O = .291/.0968 = 3.01 Formula is FeTiO3

  30. 3. Find the empirical formula of methyl acetate which contains 48.64% C, 8.16% H, 43.20% O. # moles C = 48.64/12.00 = 4.053 # moles H = 8.16/1.00 = 8.16 # moles O = 43.20/16.00 = 2.700 Oxygen – 2.700/2.700 = 1.000 x 2 = 2.000 Hydrogen – 8.16/2.700 = 3.02 x 2 = 6.04 Carbon - 4.053/2.700 = 1.500 x 2 = 3.000 Formula is C3H6O2

  31. VI. Determining Molecular Formulas • Procedure 1. Determine empirical formula 2. Calculate empirical formula mass 3. Divide actual formula mass to get “X” empirical formula mass 4. Multiply subscripts in the empirical formula by “X”

  32. B. Determine Molecular Formula 1.Determine the molecular formula of succinic acid. (Molar mass is 118.1g) C =40.68% H = 5.05%, O =54.24% Carbon = 40.68/12.0 = 3.39 Hydrogen = 5.08/1.00 = 5.08 # of moles Oxygen = 54.24/16.0 = 3.39 C = 3.39/3.39 = 1.0 O = 3.39/3.39 = 1 mole ratio H = 5.08 / 3.39 = 1.49 Multiply by 2 to get the smallest whole number ratio

  33. Empirical formula = C2H3O2 Calculate empirical formula mass Empirical formula mass = 2(12.00) + 3(1.00) + 2(16.00) =59.00g Divide formula mass by empirical formula mass 118.1/59.0 = 2.00 = “X” Multiply each subscript in the empirical formula by X The molecular formula is C4H6O4

  34. 2. Determine the molecular formula for styrene. C = 92.25%, H = 7.75%. (Molar mass – 104.00g) 92.25/ 12.0 = 7.69 mol C 7.75/1.00 = 7.75 mol H 7.69/7.69 = 1.00 7.75/7.69 = 1.01 Empirical formula is CH Empirical formula mass is 12.0 + 1.00 = 13.0 104.00/13.0 = 8 ( “X”) Formula is C8H8

  35. 3. Determine the molecular formula for ibuprofen. C=75.7%, H=8.80%, O=15.5%. Molar Mass-206.00g Determine the # of moles 75.7/12.0 = 6.31 mol C 8.80/ 1.00 = 8.80 mol H 15.5/16.0 = .969 mol O Divide by the smallest number of moles .969/.969 = 1.00 6.31/.969 = 6.51 8.80/.969 = 9.08 Multiply by 2 to get whole number ratio Empirical Formula is C13H18O Empirical Formula Mass 13(12.0)+18(1.00)+2(16.0) = 206.0 Divide formula mass by empirical formula mass to get X 206/206 = 1.00 (“X”) Molecular Formula is C13H18O2

  36. Given the molecular formula, determine the empirical formula for the following. a. C6H6 (benzene) b. C2H6 (ethane) c. C10H8 (naphthalene) d. C8H10N4O2 (caffeine) e. C14H18N2O5 (aspartame) Answers a) CH b) CH3 c) C5H4 d) C4H5N2O e) C14H18N2O5

  37. 5. Determine the molecular formula if the empirical formula is CH and the molar mass is78.00g. Empirical formula mass is 12.0 + 1.00 = 13.O 78.00/13.0 = 6.00 (“X”) The molecular formula is C6H6 • Determine the molecular formula for butane if the empirical formula is C2H5 and the molar mass is 58.00g. 2(12.0) + 5(1.00) = 29.0 (empirical formula mass) 58.0/29.0 = 2.00 (“X”) The molecular formula is C4H10

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