MOLECULAR FORMULA AND EMPIRICAL FORMULA

Molecular Formula is a formula indicating the actual number of atoms of each element making up a molecule.  The molecular formula must accurately state the exact number of atoms of all of the elements in one molecule of the substance.

Empirical formula is the formula giving the simplest ratio between the atoms of the elements present in a compound.  You must find the ratio of atom to atom in a molecule, and then reduce it.

The formula for ionic compounds is a empirical formula. 

A molecular formula is a whole number multiple of an empirical formula.

Converting to empirical formula from molecular formula is more difficult than viceversa.

Converting Molecular Formula to Empirical Formula

1.      Determine the molecular formula. If the molecular formula is written structurally, then convert it to standard form (Example 2).

2.      Divide the entire equation by the largest whole number that all subscripts are divisible by,

3.      Write the new equation.

Example 1: Hydrogen Peroxide

Step 1: Molecular Formula is H2O2

Step 2: All subscripts are divisible by 2

Step 3: Empirical Formula is HO

Example 2: Glucose

Step 1: Molecular Formula is C6H12O6

Step 2: All subscripts are divisible by 6

Step 3: Empirical Formula is CH2O

 


Creating an Empirical Formula From Mass or Percent Composition

From Percent Composition:

Step 1: Create a chart with six columns and a number of rows equal to the number of elements in the compound.

Step 2: Write the elements in the first column.

Step 3: Write the percent composition of each element in the second column.

Step 4: Using the percent composition as the mass, divide each by the molecular mass of the respective element.

Step 5: Divide each of those numbers by the smallest of the numbers in that column to reduce the ratio.  If one or more numbers in the ratio is still distant from a whole number, try multiplying the entire ratio by a whole number. If the number is close, then round it to the nearest whole number.  Each one of the numbers is the subscript for the corresponding element.

Example 1: Determine the empirical formula for a compound containing 74.0% carbon, 8.65% hydrogen, and 17.3% nitrogen by mass.

C

74.0%

74.0 g / 12.0 g

6.16 / 1.24

4.96

5

H

8.65%

8.65 g / 1.01 g

8.56 / 1.24

6.90

7

N

17.3%

17.3 g / 14.0 g

1.24 / 1.24

1

1

 

Therefore the empirical formula is C5H7N

From Mass:

Follow the same procedure as fir Percent Composition but first divide the mass of each element by the mass of the total sample. The quotient is the percent composition.

Example 1: A 9.2 g sample of a compound is 2.8 g of nitrogen and 6.4 g of oxygen. Find the empirical formula of the compound.

N

2.8 g / 9.2 g

30%

30 g / 14 g

2.143 g / 2.143 g

1

1

O

6.4 g / 9.2 g

70%

70 g / 16 g

4.375 g / 2.143 g

2.04

2

 

Therefore, the empirical formula is NO2.

Creating a Molecular Formula From an Empirical Formula and Molecular Mass

Step 1: Determine the molecular mass in grams.

Step 2: Divide the molecular mass of the compound by the molecular mass (g) of the empirical formula

Step 3: Round the quotient to the closest integer.

Step 4: Multiply the rounded number by all the subscripts using the product as the new subscripts.

Example 1: A compound has an empirical formula of CH2 and a molecular mass of 42 g.  Determine its molecular mass.

Step 1: Carbon = 12 + Hydrogen = 2 (1.01)

            Molecular Mass (g) = 14.02

Step 2: 2.42 / 14.02 = 2.999

Step 3: 3

Step 4: CH2  *  3 = C3H6


Worksheet

1.      Write the empirical formula for the following compounds:

a)     P4O6

b)     C6H9

c)     CH2OHCH2OH

d)     BrCl2

e)     C6H8O6

f)       CuC2O4

g)     Hg2F2

2.      Write the empirical formula for each of the following compounds:

a)      A compound composed of 72% iron and 27.6% oxygen by mass.

 

 

 

 

 

 

 

 

 

 

 

 

 

b)     A compound composed of 9.93% carbon, 58.6% chlorine and 31.4% fluorine.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c)     A compound composed of 55.6 g carbon and 0.0933 g hydrogen.

 

 

 

 

 

 

 

 

 

 

 

 

 

3.      Write the molecular formula for the following compounds:

a)     A compound with a molecular mass of 70 g and an empirical formula of CH2.

b)     A compound with a molecular mass of 46.0 g and an empirical formula of NO2.