Empirical

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Empirical Molecular Formulas

• Law of Constant Composition a compound contains
  elements in a certain fixed proportions (ratios),
  regardless of how the compound is prepared or
  where it is found in nature.
• If you have one molecule of methane gas, you will
  always have 1 carbon atoms and 4 hydrogen atoms.

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Empirical Formula

• Empirical Formula is the formula that gives the
  lowest ratio of atoms in a compound. It does not
  necessarily tell you the exact number of each
  type of atom.
• Example 1 The percent composition of a compound
  is 69.9 iron and 30.1 oxygen. What is the
  empirical formula of a compound?

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Example 1 The percent composition of a compound
is 69.9 iron and 30.1 oxygen. What is the
empirical formula of a compound?

• Step 1 List the given values
• Fe69.9 and O 30.1
• Step 2 Calculate the mass (m) of each element in
  a 100g sample.
• mFe 69.9 x 100g 69.9g
• 100
• mO 30.1 x 100g 30.1g
• 100

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• Step 3 Convert Mass (m) into moles (n)
• nFe m/M 69.9g/55.86g/mol 1.25 mol Fe
• nO m/M 30.1g/16.00g/mol 1.88 mol O
• Step 4 State the Amount Ratio
• nFe nO
• 1.25mol 1.88 mol
• Step 5 Calculate lowest whole number ratio
• 1.25mol 1.88 mol
• 1.25mol 1.25 mol
• 1 1.5
• 2 3

When you dont get a whole number, multiply
entire ratio by 2, 3, 4 etc. until you get a
whole number
Empirical Formula is Fe2O3
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Example 2 The percent composition of a compound
is 21.6 sodium, 33.3 chlorine, and 45.1
oxygen. What is the empirical formula of the
compound?
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• Step 1 List the given values
• Cl33.3, Na 21.6 and O 45.1
• Step 2 Calculate the mass (m) of each element in
  a 100g sample.
• mCl 33.3 x 100g 33.3g Cl
• 100
• mNa 21.6 x 100g 21.6g Na
• 100
• mO 45.1 x 100g 45.1g O
• 100

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• Step 3 Convert Mass (m) into moles (n)
• nCl m/M 33.3g/35.5g/mol 0.94 mol Cl
• nNa m/M 21.6g/23.0g/mol 0.94 mol Na
• nO m/M 45.1g/16.00g/mol 2.82 mol O
• Step 4 State the Amount Ratio
• nFe nNa nO
• 0.94mol 0.94mol 2.82 mol
• Step 5 Calculate lowest whole number ratio
• 0.94mol 0.94mol 2.82 mol
• 0.94mol 0.94mol 0.94 mol
• 1 1 3

Empirical Formula is NaClO3
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Molecular Formula

• Molecular Formula of a compound tells you exact
  number of atoms in one molecule of a compound.
  This formula may be equal to the empirical
  formula or may be a multiple of this formula.
• To determine, you need
• The empirical formula
• The molar mass of the compound

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• Molecular Formula
• - shows the actual number of atoms
• Example C6H12O6
• Empirical Formula
• - shows the ratio between atoms
• Example CH2O

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The empirical formula of a compound is CH3O and
its molar mass is 93.12g/mol. What is the
molecular formula?

• Step 1 List given values
• Empirical FormulaCH3O
• Mcompound 93.12 g/mol
• Step 2 Determine the molar mass for the
  empirical formula, CH3O.
• MEmpirical 12.01g/mol 3(1.01g/mol)
  16.00g/mol
• 31.04 g/mol

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• Step 3. Divide the molar mass by the empirical
  formula molar mass.
• 3
• Step 4. Calculate Molecular Formula by
  multiplying this number by the empirical formula.
• Molecular formula x (empirical formula)
• 3 x CH3O
• Therefore, the molecular formula is C3H9O3

93.12 g/mol 31.04 g/mol
Molecular formula molar mass Empirical formula
molar mass
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Example 2 The percent composition of a compound
is determined by a combustion and analyzer is a
40.03 carbon, 6.67 hydrogen, 53.30 oxygen.
The molar mass is 180.18g/mol. What is the
molecular formula

• Step 1 List given values
• C 40.03, O53.30, H6.67
• Mcompound 180.18 g/mol
• Step 2 Calculate the mass of each element in a
  100g sample
• mC40.03g mO53.30g mH6.67g

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• Step 3 Convert Mass (m) into moles (n)
• nC m/M 40.03g/12.01g/mol 3.33 mol C
• nH m/M 6.67g/1.01g/mol 6.60 mol H
• nO m/M 53.30g/16.00g/mol 3.33 mol O
• Step 4 State the Amount Ratio
• nC nH nO
• 3.33mol 6.60mol 3.33 mol
• Step 5 Calculate lowest whole number ratio
• 3.33mol 6.60mol 3.33 mol
• 3.33mol 3.33mol 3.33 mol
• 1 2 1

Empirical Formula is CH2O
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• Step 6 Determine the molar mass for the
  empirical formula
• MEmpirical 12.01g/mol 2(1.01g/mol)
  16.00g/mol
• 30.03 g/mol
• Step 7. Divide the molar mass by the empirical
  formula molar mass.
• 6
• Step 8. Calculate Molecular Formula by
  multiplying this number by the empirical formula.
• Molecular formula x (empirical formula)
• 6 x (CH2O)
• Therefore, the molecular formula is C6H12O6

180.18 g/mol 30.03 g/mol
Molar mass Empirical formula molar
mass
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Example 3 The percent composition of a compound
is determined by a combustion and analyzer is a
32.0 carbon, 6.70 hydrogen, 42.6 oxygen
18.7 nitrogen. The molar mass is 75.08g/mol.
What is the molecular formula?

• Calculate the mass of each element in a 100g
  sample
• mC32.0g mO42.6g mH6.70g mN18.7g
• Convert Mass (m) into moles (n)
• nC m/M 32.0g/12.01g/mol 2.66 mol C
• nH m/M 6.70g/1.01g/mol 6.65 mol H
• nO m/M 42.6g/16.00g/mol 2.66 mol O
• nN m/M 18.7g/14.01g/mol 1.33 mol N

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• State the Amount Ratio
• nC nH nO nN
• 2.66mol 6.65mol 2.6 mol 1.33mol
• Step 5 Calculate lowest whole number ratio
• 2.66mol 6.65mol 2.6 mol 1.33mol
• 1.33mol 1.33mol 1.33 mol 1.33mol
• 2 5 2 1

Empirical Formula is C2H5O2N
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• Determine the molar mass for the empirical
  formula
• MEmpirical 75.08g
• Divide the molar mass by the empirical formula
  molar mass.
• 1
• Calculate Molecular Formula by multiplying this
  number by the empirical formula.
• Molecular formula x (empirical formula)
• 1 x (C2H5O2N)
• Therefore, the molecular formula is C2H5O2N

75.08 g/mol 75.08 g/mol
Molar mass Empirical formula molar
mass