Empirical Formula Calculations

   Finding the empirical formula is somewhat the reverse of finding percentage   composition.  First you will be given the percentages of each element in a   particular compound.

   Assuming that you will always be given a 100 gram sample we will convert these percentages to grams.  The next step is to find the ratio of moles of each element and then calculate the simplest ratio of subscripts that maintains that ratio.

   It is much easier done than said!  Take notes as we work through this sample problem.

    Lets start with this data:   36.5% Na,   25.4% S,   and 38.1% O

 First, convert each percentage to grams:  36.5 g Na, 25.4 g S, 38.1 g O.

 Next, divide each by the grams/mole of that element:

     Na:    36.5 g
                    -----       =  1.58 mol Na
              23.0 g/mol

       S:    25.4 g
                   ------      =   0.791 mol S
              32.1 g/mol

       0:    38.1 g      =   2.38 mol O
                    ------
               16.0 g/mol

Now we can set up the ratio of moles of each element:

    Na      S       O
        1.58   0.791   2.38

 To convert these decimal numbers into whole numbers and maintain the same ratio between them, just divide each by the smallest of the subscripts.

   Na       S          O
        1.58    0.791    2.38
         ----     -----       ----
         0.791  0.791    0.791

 Our final formula would look like: Na₂SO₃   -->  sodium sulfite

 Following will be two problems for you to work.

Practice Problem #1

Given the following data, find the correct empirical formula:

      49.0% C,  2.70% H,  48.2% Cl

 The correct formula would be:  CH₃Cl₂

To solve:

C:  49.0g                     H:   2.70g                           Cl:   48.2g
       ----- = 4.08 mol            -----  =   2.67 mol                ----- =     1.36 mol
       12.0 g/mol                    1.01 g/mol                           35.5 g/mol

C          H         Cl
  4.08      2.57       1.36
    ----        ----         ----
   1.36      1.36        1.36       gives the final formula:  CH_3­Cl₂

                                                       Practice Problem #2

Given the following data find the empirical formula:

   N = 26.2%,     H = 7.50%,    Cl = 66.4%

 The correct formula would be: NH₄Cl    (ammonium chloride)

 The work:

 N:  26.2g                       H:  7.50g                            Cl:  66.4g
          ----- = 1.87 mol            -----  =   7.50 mol               -----  =         1.87 mol
      14.0 g/mol                     1.01 g/mol                       35.5 g/mol

N         H          Cl
   1.87     7.50       1.87
     ----      ----         ---      -->   NH₄Cl
    1.87      1.87      1.87             

Page Last Updated: Friday March 02, 2007           Webmaster: Larry Jones                 Pickens County School District