Empirical and Molecular Formulas

• GFM


  Gram Formula Mass
• Find the gram formula mass of C2H6O

  1. C = 12.01 x 2 = 24.02

  2. H = 1.01 x 6 = 6.06

  3. O = 16.00 x 1 = 16.01

  4. Total = 46.09 g/mol
• Molar Mass


  The mass of one mole of a substance
• Calculating Percent Composition


  1. Find the mass of each element

  2. Take the part divided by the whole

  3. The total should add up to 100%
• Empirical Formula


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


  The actual number of atoms of each element in a molecular compound.
• Empirical Formula is the simplest whole-number ratio of the atoms of each element in a compound.
• Molecular Formula gives the actual number of atoms of each element in a molecular compound.
• Why is it important to know the difference between molecular and empirical formulas?
• Steps to Determine Empirical Formula

  1. Find mole amounts.

  2. Divide each mole by the smallest mole.
• Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca.

  Find mole amounts.

  2.128 g Cl x 1 mol Cl / 35.45 g Cl = 0.0600 mol Cl

  1.203 g Ca x 1 mol Ca / 40.08 g Ca = 0.0300 mol Ca

  2. Divide each mole by the smallest mole.

  Cl = 0.0600 mol Cl / 0.0300 = 2.00 mol Cl

  Ca = 0.0300 mol Ca / 0.0300 = 1.00 mol Ca

  Ratio - 1 Ca: 2 Cl

  Empirical Formula = CaCl2
• 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/24.3 g) = 8.86 mole, N - 82.88 g * (1 mole/14.01 g) = 5.92 mole

  Divide by small: Mg - 8.86 mole/5.92 mole = 1.50, N - 5.92 mole/5.92 mole = 1.00

  Multiply 'til whole: Mg - 1.50 x 2 = 3.00, N - 1.00 x 2 = 2.00

  Empirical Formula = Mg3N2
• If the problem does not give you how many grams, assume 100 grams of the sample.
• Empirical formula gives the simplest whole-number ratio of the atoms of each element in a compound.
• Steps to calculate the empirical formula:

  Find the mole amounts of each element

  2. Divide each mole by the smallest mole

  3. Multiply the ratios to get whole numbers
• Steps to find the molecular formula:

  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 that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.

  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 that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.


  Empirical formula: A. Find mole amounts: 4.90 g N x 1 mol N/14.01 g = 0.350 mol N, 11.2 g O x 1 mol O/16.00 g = 0.700 mol O

  B. Divide each mole by the smallest mole: N = 0.350/0.350 = 1.00 mol N, O = 0.700/0.350 = 2.00 mol O

  Empirical Formula = NO2

  Empirical Formula Mass = 46.01 g/mol

  Molecular formula: Molar Mass = 92.0 g/mol, Emp. Formula Mass = 46.01 g/mol

  Molecular Formula = 2 x Emp. Formula = N2O4
• 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

  mole C - 255.64 g * (1 mole/12.01 g) = 21.29 mol

  mole H - 42.91 g * (1 mole/1.01 g) = 42.49 mol

  mole O - 229.85 g * (1 mole/16.00 g) = 14.37 mol

  C - 21.29/14.27 = 1.49, H - 42.49/14.27 = 2.98 (essentially 3), O - 14.27/14.27 = 1.00

  Empirical formula = C3H6O2

  Molecular formula = 3(C3H6O2) = C9H18O6