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EMPIRICAL FORMULAS AND MOLECULAR FORMULAS

Curiosity,

Jeff Jordan, 2003

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WHAT IS A MOLECULAR FORMULA?

The formula of a compound that uses subscripts to show how many atoms of each element is in a molecule of the compound.

Ex: C6H12O6

Ex: CO2

Ex: N2H4

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WHAT IS AN EMPIRICAL FORMULA?

A molecular formula whose subscripts have been reduced to the smallest whole number ratio.

Ex: C6H12O6 becomes:

      • CH2O

Ex: CO2 becomes:

      • CO2

Ex: N2H4 becomes:

      • NH2

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WHY DO WE NEED TO KNOW EMPIRICAL FORMULAS

  • Empirical formulas are one of the experimental steps needed in determining the identity of unknown substances.
  • Once you identify the ratio of each element in a substance, you can use mass spectrometry to determine the molar mass of the substance.
  • Then using the empirical formula and the molar mass, you can determine the molecular formula of the unknown substance.

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HOW DO YOU DETERMINE THE EMPIRICAL FORMULA OF A COMPOUND?

  1. Convert all grams or percents to moles. (If you are given the percent composition of a compound, assume you have 100 g of the compound and then all the percentages are now the grams of each element (e.g., 18% carbon becomes 18 g of carbon)).
  2. Divide all of the moles by the smallest of the number of moles to determine the mole ratios.
  3. Use the mole ratios (as whole numbers*) as subscripts in the empirical formula.
  4. *Multiply all mole ratios by 2 if one of the mole ratios ends in 0.5 or 3 if one of the mole ratios ends in 0.33.

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EXAMPLE

 

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EXAMPLE

 

/ 0.0998 = 1

/ 0.0998 ≈ 2

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EXAMPLE

 

/ 0.0998 = 1

/ 0.0998 ≈ 2

Ca Br

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EXAMPLE

 

/ 0.0998 = 1

/ 0.0998 ≈ 2

Ca Br2

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EXAMPLE

 

/ 0.0998 = 1

/ 0.0998 ≈ 2

Ca Br2

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EXAMPLE

A compound contains 48.63% carbon, 8.18% hydrogen, and 43.19% oxygen. What is its empirical formula?

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EXAMPLE

A compound contains 48.63% carbon, 8.18% hydrogen, and 43.19% oxygen. What is its empirical formula?

​

Step 1:

 

 

 

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EXAMPLE

A compound contains 48.63% carbon, 8.18% hydrogen, and 43.19% oxygen. What is its empirical formula?

​

Step 2:

 

 

 

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EXAMPLE

A compound contains 48.63% carbon, 8.18% hydrogen, and 43.19% oxygen. What is its empirical formula?

​

Step 3:

 

 

 

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EXAMPLE

A compound contains 48.63% carbon, 8.18% hydrogen, and 43.19% oxygen. What is its empirical formula?

Step 4:

 

 

 

C3H6O2

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WHAT DO YOU NEED TO DETERMINE THE MOLECULAR FORMULA OF A COMPOUND?

  1. The empirical formula of the compound.
  2. The molar mass of the compound.

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EXAMPLE 1

The empirical formula for a compound is NO2. Its molar mass is 92.02 g/mol. What is its molecular formula?

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3 EASY, PAINFREE STEPS TO SOLVE

1. Determine the molar mass of the empirical formula.

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3 EASY, PAINFREE STEPS TO SOLVE

1. Determine the molar mass of the empirical formula.

NO2 = 46.01 g/mol

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3 EASY, PAINFREE STEPS TO SOLVE

2. Divide molar mass of molecular formula by molar mass of empirical formula.

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3 EASY, PAINFREE STEPS TO SOLVE

92.02 g/mol / 46.01 g/mol = 2

2. Divide molar mass of molecular formula by molar mass of empirical formula.

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3 EASY, PAINFREE STEPS TO SOLVE

3. Multiply each subscript in empirical formula by factor determined in step 2.

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3 EASY, PAINFREE STEPS TO SOLVE

N1x2O2x2

so molecular formula is N2O4

3. Multiply each subscript in empirical formula by factor determined in step 2.

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EXAMPLE 2

A compound is composed of 7.20 g of carbon, 1.20 g of hydrogen, and 9.60 g of oxygen. The molar mass of the compound is 180. g/mol. Find the molecular formula of the compound.

Answer: C6H12O6

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HOMEWORK:

See website for assignment over this material.