Percent Composition - Formulas

 This program demonstrates how to find percentage composition as well as empirical and molecular formulas from that data.  Please read each section carefully.  You might wish to take notes. Remember that all atomic masses are rounded to 3 significant figures.

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 1] - Percentage Composition

2] - Empirical Formulas

3] - Molecular Formulas

   

                    Percentage Composition Calculations

   To calculate the correct percentage composition of each element in some compound you must correctly follow several steps.  It is not very difficult, once you get the hang of it.  Please take notes on the steps to follow.

Several problems follow this one for you to work.

    Lets look at a sample problem:  magnesium chloride   -->  MgCl₂                                                      

step 1:  calculate formula mass

         Mg:  1 x 24.3  =  24.3
             Cl:  2 x 35.4  =  70.8
                                           ----
           formula mass =  95.1

 step 2:  divide each component mass by the formula mass and multiply by 100

         Mg:   24.3/95.1  x 100 =  25.6%

         Cl:   70.8/95.1  x 100 =  74.4%

 step 3:  make certain the percentages add up to 100 (+/- 0.1)

         25.6 + 74.4 = 100

    This problem shows all the steps to follow.  Be careful to use three   significant figures at all times.  Remember to first find the correct   formula, then find formula mass, then calculate the percentage composition.

    The next two problems allow you some practice.  Work these out on paper   and then check your answer.

 Practice Problem #1

Find the percentage composition for each element in:  iron (III) silicate

    iron (III) silicate    -->   Fe₂(SiO₃)₃

    Fe:  32.9%       Si: 24.8%     O: 42.4%

 Remember that with significant figures your answer can be one above or below   the final digit.  Example:  iron could have been reported as 32.8%, 32.9%, or   even 33.0%.

 Work:   Fe:  2 x 55.8 = 112               112/340. x 100 = 32.9%
                Si:  3 x 28.1 =  84.3             84.3/340. x 100 = 24.8%
                 O:  9 x 16.0 = 144               144/340. x 100 = 42.4%
                                            ------                                          -----
                                             340.                                      100.1  (ok)

 Practice Problem #2

Find the percentage composition for each element in:  tin (IV) arsenate .  To see the correct answers and the method to solve the problem, please continue.

    tin (IV) arsenate    -->   Sn₃(AsO₄)₄

    Sn:   39.1%     As:  32.8%     O:  28.0%

 Work:  Sn:  3 x 119  =  357            357/913  x 100  =  39.1%
            As:  4 x 74.9 =  300.           300./913 x 100  =  32.8%
                O: 16 x 16.0 =  256            256/913  x 100  =  28.0%
                                              -----                                          -----
                                             913                                         99.9  (ok)