Empirical Formulas

• opposite of percent composition
• use % to find formula for compound

EMPIRICAL FORMULA:  simplest formula;  #'s of subscripts are reduced to lowest terms

MOLECULAR FORMULA:  subscripts are multiples of empirical formula subscripts

MOLECULAR FORMULA                            EMPIRICAL FORMULA
C6H6
C6H12O6
C12H16O4N8
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TO SOLVE EMPIRICAL FORMULA PROBLEMS:
A sample of a compound is found to contain 36.0 % calcium and 64.0 % chlorine.  Calculate the empirical formula.

Step 1:  Rewrite % as grams.
36.0 g Ca                64.0 g Cl

Step 2:  Find moles of each element.
Ca:  36.0 g Ca | 1 mole Ca = 0.898 moles Ca            Cl64.0 g Cl | 1 mole Cl = 1.80 moles Cl
| 40.1 g Ca                                                          | 35.5 g Cl

Step 3:  Find mole ratio.  (Divide by smallest number of moles.)
Ca:  0.898 moles = 1        Cl1.80 moles = 2
0.898                            0.898
* These whole numbers are subscripts in formula.*

Step 4:  Write the formula.
Ca1Cl2  ====>   CaCl2

Example 2:  A sample of a compound contains 66.0 % calcium and 34.0 % phosphorus.  What is the empirical formula?
Ca:  66.0 g Ca | 1 mole Ca  = 1.65 moles Ca        P:  34.0 g P | 1 mole P = 1.10 moles P
| 40.1 g Ca                                          | 31.0 g P

Ca:  1.65 = 1.5                P:  1.10 = 1
1.10                             1.10
Q: So, what happens now?  I can't write Ca1.5P1.  And 1.5 is not close enough to round to 2.

A: The easiest way to get 1.5 to a whole # is to multiply by 2.  Remember to multiply both #'s by 2 to get your answer.
Ca:  1.5 x 2 = 3                P:  1 x 2 = 2                So, formula is Ca3P2

PRACTICE - A compound contains 43.4 % sodium, 11.3 % carbon, and 45.3 % oxygen.  What is the empirical formula for this compound?