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

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.