Empirical and Molecular Formulas

1
Empirical and Molecular Formulas

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Determining Empirical Formulas

• Definition the lowest whole number ratio of
  elements in a compound.
• Steps to determine empirical formulas
• 1) find moles of each element find the
  moles of H20 if compound is hydrated.
• 2) divide by smallest number of moles to
  get a whole number.
• 3) multiply by a common whole number, if
  necessary.

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Examples

• Calculate EF of a compound with 25.9 N and 74.1
  O.
• assume mass
  mass? mols
• 25.9g N x 1 mol N 1.85 mols N
• 14.007g N
• 74.1g O x 1 mol O 4.63 mols O
• 16 g O
• 2)
  N 1.85 1
• 
  1.85
• 
  O 4.63 2.5
• 
  1.85
• 3) Multiply by common whole number
• N 1 x 2 2
• O 2.5 x 2 5 N2O5
• 
  

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Just remember this!!

• Percents to grams
• Grams to moles
• Divide by smallest
• Multiply til whole

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examples continued

• Given 36.5g Na, 25.4g S, and 38.1g O
• 36.5g Na x 1 mol Na 1.59 mols Na
• 22.989g Na
• 25.4g S x 1 mol S .79 mols S
• 32g S
• 38.1g O x 1 mol O 2.4 mols O
• 16g O
• Na 1.59 2 S .79 1
  O 2.4 3
• .79 .79
  .79
• 
  
  Na2SO3

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• An oxide of aluminum is formed by the reaction of
  4.151g of aluminum with 3.692g of oxygen. Find
  EF.
• 4.151g Al x 1 mol Al 0.1539 mol Al atoms
• 26.98 g Al
• 3.692g O x 1 mol O 0.2308 mol O atoms
• 16 g O
• 0.1539 mol Al 1.000 mol Al atoms
• 0.1539
• 0.2308 mol O 1.500 mol O atoms
• 0.1539
• 1.500 O x 2 3.000 3 O atoms
• 1.000 Al x 2 2.000 2 Al atoms
• Al2O3

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Determining Molecular Formulas

• Molecular formulas show the number of atoms in a
  compound. In order to determine the molecular,
  you must have the empirical formula first.
• The molecular mass of molecule will always be
  given.
• Multiple molecular mass x
• empirical formula (whole
  number)
• The molecular formula is always an integer
  multiple of the empirical formula.

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Examples

• Determine the molecular formula of a compound
  whose EF is CH2O and molecular mass is 120 g/mol
• 120 g/mol 4
• 30 g/mol
• -Distribute that 4 throughout the empirical
  formula
• C4H8O4

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continued

• A compound is 64.9 carbon, 13.5 hydrogen, and
  21.6 oxygen. Its molc mass is 74 g/mol. What is
  its MF?
• 64.9g C x 1 mol C 5.40 mols C 4 C
• 12.01 g C
• 13.5g H x 1 mol H 13.37 mols H 10 H
• 1.01 g H 1.35
• 21.6g O x 1 mol O 1.35 mols O 1 O
• 16.0 g O 1.35
• 74 g/mol 1
• 74 g/mol
  C4H10O

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• Ex) a compound is 54.5 carbon, 9.1 hydrogen,
  and 36.4 oxygen. Its molc mass is 88 g/mol.
  What is its molecular formula?
• 54.5 C x 1 mol C 4.54 mols C
• 12.01 g C
• 9.1 g H x 1 mol H 9 mols H
• 1.01 g H
• 36.4 g O x 1 mol O 2.28 mols O
• 16 g O
• 4.54 mols C 2 C
• 2.28 mols
• 9 mols H 4H
  
• 2.28 mols
• 2.28 mols O 1O
• 2.28 mols
  
• 
  C4H8O2

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• Ex) A compound has an empirical formula of ClCH2
  and a molecular weight of 98.96 g/mol. What is
  its molecular formula?
• Mass Cl mass C 2(mass H)
• Mass of empirical unit 35.45 12.00 2(2.008)
  49.47 g/mol
• 98.96 g/mol 2.000
• 49.47 g/mol
• Cl2C2H4