Ethylene glycol, commonly used in automobile antifreeze, contains only carbon, hydrogen and oxygen. combustion analysis of a 23.46mg sample yields 20.42 mg of water and 33.27 mg of carbon dioxide. what is the empirical formula of ethylene glycol? if other experiments determine that it's molar mass is 62.0g/mol, what is it's molecular formula?

Molecular formula of ethylene glycol is C₂H₆O₂.

Explanation:

Given data:

mass of ethylene glycol = 23.4 mg

mass of water = 20.42 mg

mass of carbon dioxide = 33.27 mg

molar mass of ethylene glycol = 62.0 g / mol

Molecular formula of ethylene glycol = ?

Solution:

percentage of carbon = (33.27 mg/ 23.46 mg) × (12 /44) × 100

                                      =  (1.4182× 0.27) × 100 = 38.29

percentage of hydrogen =  (20.42 mg/ 23.46 mg) × (2 /18) × 100

                                          = (0.8704× 0.11) × 100 = 9.67

percentage of oxygen = 100 - (38.29 + 9.67 )

                                      = 100 - 47.96 = 52.04

Now we will determine the number of grams atoms of carbon, hydrogen and oxygen.

No. of gram atoms of carbon = 38.29 /12 = 3.19

No. of gram atoms of hydrogen = 9.67 / 1 = 9.67

No. of gram atoms of oxygen = 52.04 / 16 = 3.25

Atomic ratio:

     C :H :O               3.19/ 3.19    :   9.67 / 3.19   :  3.25 /3.19

     C :H :O                 1 : 3 : 1

Molecular formula:

    Molecular formula = n × (empirical formula)

    n = molar mass of compound / empirical formula mass

    empirical formula mass= 1 × 12 + 3 × 1  + 1 × 16

     empirical formula mass= 12+3 +16 = 31

                  n = 62 g /mol / 31 = 2

         Molecular formula = n × (empirical formula)

         Molecular formula = 2 × (CH₃O)

          Molecular formula = C₂H₆O₂

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