Empirical and Molecular Formulas

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Empirical and Molecular Formulas
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Empirical Formula

• Empirical Formula
• A formula that gives the simplest whole-number
  ratio of the atoms of each element in a compound.

Molecular Formula Empirical Formula
H2O2 HO
C6H12O6 CH2O
CH3O
CH3O
CH2O
CH3OOCH C2H4O2
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• Determine the empirical formula for a compound
  containing 2.128 g Cl and 1.203 g Ca.
• Steps
• 1. Find mole amounts.
• 2. Divide each mole by the smallest mole.

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• 1. Find mole amounts.
• 2.128 g Cl x 1 mol Cl 0.0600 mol Cl
• 35.45 g Cl
• 1.203 g Ca x 1 mol Ca 0.0300 mol Ca
• 40.08 g Ca

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• 2. Divide each mole by the smallest mole.
• Cl 0.0600 mol Cl 2.00 mol Cl
• 0.0300
• Ca 0.0300 mol Ca 1.00 mol Ca
• 0.0300
• Ratio 1 Ca 2 Cl
• Empirical Formula CaCl2

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A compound weighing 298.12 g consists of 72.2
magnesium and 27.8 nitrogen by mass. What is the
empirical formula?
Hint Percent to mass Mass to mole Divide by
small Multiply til whole
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A compound weighing 298.12 g consists of 72.2
magnesium and 27.8 nitrogen by mass. What is the
empirical formula?
Percent to mass Mg (72.2/100)298.12 g
215.24 g
N (27.8/100)298.12 g 82.88 g
Mass to mole Mg 215.24 g ( 1 mole ) 8.86
mole
24.3 g
N 82.88 g ( 1 mole ) 5.92 mole
14.01 g
Divide by small Mg - 8.86 mole/5.92 mole 1.50
N - 5.92 mole/5.92 mole 1.00 mole
Multiply til whole Mg 1.50 x 2 3.00
Mg3N2
N 1.00 x 2 2.00
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Molecular Formula

• The molecular formula gives the actual number
  of atoms of each element in a molecular compound.
• Steps
• 1. Find the empirical formula.
• 2. Calculate the Empirical Formula Mass.
• 3. Divide the molar mass by the EFM.
• 4. Multiply empirical formula by factor.

Find the molecular formula for a compound whose
molar mass is 124.06 and empirical formula is
CH2O3.
2. EFM 62.03 g 3. 124.06/62.03
2 4. 2(CH2O3) C2H4O6
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• Find the molecular formula for a compound that
  contains 4.90 g N and 11.2 g O. The molar mass of
  the compound is 92.0 g/mol.
• Steps
• 1. Find the empirical formula.
• 2. Calculate the Empirical Formula Mass.
• 3. Divide the molar mass by the EFM.
• 4. Multiply empirical formula by factor.

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• Empirical formula.
• A. Find mole amounts.
• 4.90 g N x 1 mol N 0.350 mol N
• 14.01 g N
• 11.2 g O x 1 mol O 0.700 mol O
• 16.00 g O

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• B. Divide each mole by the smallest mole.
• N 0.350 1.00 mol N
• 0.350
• O 0.700 2.00 mol O
• 0.350
• Empirical Formula NO2
• Empirical Formula Mass 46.01 g/mol

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• Molecular formula
• Molar Mass 92.0 g/mol 2.00
• Emp. Formula Mass 46.01 g/mol
• Molecular Formula 2 x Emp. Formula N2O4

13
A 528.39 g compound containing only carbon,
hydrogen, and oxygen is found to be 48.38 carbon
and 8.12 hydrogen by mass. The molar mass of
this compound is known to be 222.25 g/mol. What
is its molecular formula?
14
A 528.39 g compound containing only carbon,
hydrogen, and oxygen is found to be 48.38 carbon
and 8.12 hydrogen by mass. The molar mass of
this compound is known to be 222.25 g/mol. What
is its molecular formula?
g C (48.38/100)528.39 g 255.64 g g H
(8.12/100)528.39 g 42.91 g g O
(43.5/100)528.39 g 229.85 g
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A 528.39 g compound containing only carbon,
hydrogen, and oxygen is found to be 48.38 carbon
and 8.12 hydrogen by mass. The molar mass of
this compound is known to be 222.25 g/mol. What
is its molecular formula?
From last slide 21.29 mol C, 42.49 mol H, 14.27
mol O
C 21.29/14.27 1.49 H 42.49/14.27 2.98
(esentially 3) O 14.27/14.27 1.00
C 1.49 x 2 3 H 3 x 2 6 O 1 x 2 2
C3H6O2
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A 528.39 g compound containing only carbon,
hydrogen, and oxygen is found to be 48.38 carbon
and 8.12 hydrogen by mass. The molar mass of
this compound is known to be 222.25 g/mol. What
is its molecular formula?
From last slide Empirical formula C3H6O2
EFM 74.09 Molar mass 222.24 3 EFM
74.09 3(C3H6O2) C9H18O6