3. A compound consists of 91.63 grams of carbon, 7.69 grams of hydrogen and 40.81 grams of
oxygen. Its molar mass is 220g/mol. Determine the empirical and molecular formulas.

Molecular formula = C₁₂H₁₂O₄

Empirical formula is C₃H₃O.

Explanation:

Given data:

Mass of C = 91.63 g

Mass of H = 7.69 g

Mass pf O = 40.81 g

Molar mass of compound = 220 g/mol

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 7.69 / 1.01 = 7.61

Number of gram atoms of O = 40.81 / 16 = 2.55

Number of gram atoms of C = 91.63 / 12 = 7.64

Atomic ratio:

            C                      :      H                :         O

           7.64/2.55          :    7.61 /2.55    :       2.55/2.55

               3                     :          3               :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass  = 3×12+ 3×1.01 +16 = 55.03

n = 220 / 55.03

n = 4

Molecular formula = 4 (empirical formula)

Molecular formula = 4 (C₃H₃O)

Molecular formula = C₁₂H₁₂O₄

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