What is the molecular formula of a compound that is 51.02 % carbon, 13.80 % hydrogen, and 35.18 % oxygen? The molecular mass is 88.0 g/mol.

Molecular formula = C₄H₁₂O₂

Explanation:

Given data:

Percentage of hydrogen = 13.80%

Percentage of carbon = 51.02%

Percentage of oxygen = 35.18%

Molecular formula = ?

Solution:

Number of gram atoms of H = 13.80 / 1.01 = 13.6

Number of gram atoms of O = 35.18 / 16 = 2.2

Number of gram atoms of C = 51.02 / 12 = 4.25

Atomic ratio:

            C                      :      H             :         O

           4.25/2.2           :     13.6/2.2     :       2.2/2.2

            2                      :        6          :        1

C : H : O = 2 : 6 : 1

Empirical formula is C₂H₆O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = C₂H₆O = 46 g/mol

n = 88 / 46

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 (C₂H₆O)

Molecular formula = C₄H₁₂O₂

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