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

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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

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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

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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)

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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

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B. Using Factor Label to Make Mole Conversions

• 1 mole ? molar mass
• 6.02 x 1023 1
  mole
• PARTICLES lt----gt MOLES lt----gt MASS
• 6.02 x 1023 ? 1 mole
• 1 mole molar
  mass

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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

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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
• 
  
  

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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

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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?

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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?

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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

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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

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Answers to More Sample Problems

1. 12.04 x 1023 1.20 x 1024 atoms Cu
2. 5.00 mol C
3. .498 mol He or 4.98 x 10-1 mol He
4. 67.5 g Al
5. 12.04 x 1023 1.20 x 1024 atoms N
6. 7.27 g Mg

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• 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

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• 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

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• 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
  

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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

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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

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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

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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

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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

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B. Naming Hydrates

• 1. CaCl2 ? 2H2O calcium chloride
  dihydrate
• 2. NaC2H3O2 ? 3H2O sodium acetate trihydrate
• 3. CuSO4 ? 5H2O copper sulfate
  pentahydrate
• 4. MgSO4 ? 7H2O magnesium sulfate heptahydrate
• 5. Na2CO3 ? 10H2O sodium carbonate decahydrate

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C. Heating a Hydrate

• ?
• Hydrate ? anhydride water
• ?
• Na2CO3 ? 10H2O ? Na2CO3 10 H2O
• ?
• MgSO4 ? 7H2O ? MgSO4 7 H2O
• ?
• Na2CO3 ? 10H2O ? Na2CO3 10H2O

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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

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• 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
• 
  

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• 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

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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

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C. Sample Problems

• 1. Find the empirical formula of a compound
• containing 19.55 g of potassium (39.08) and
• 2.00g of 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

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• 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

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• 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

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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

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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

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• 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

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• 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

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• 3. Determine the molecular formula for
  ibuprofen.
• C75.7, H8.80, O15.5. 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

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• 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

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• 5. Determine the molecular formula if the
  empirical
• formula is CH and the molar mass is 78.00g.
• 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