A 50.00 g sample of a compound containing only carbon, hydrogen, and oxygen was partially analyzed. The sample contained 24.66 g carbon, and 3.43g of hydrogen. The molecular weight of the compound was determined to be 146.0 amu. Determine emperical the molecular formula of the compound

1. Empirical formula => C₂H₃O

2. Molecular formula => C₆H₉O₃

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 50 g

Mass of Carbon = 24.66 g

Mass of Hydrogen = 3.43 g

Molecular weight of compound = 146.0 amu

Empirical formula =?

Molecular formula =?

Next, we shall determine the mass of oxygen in the compound. This can be obtained as follow:

Mass of compound = 50 g

Mass of C = 24.66 g

Mass of H = 3.43 g

Mass of O =?

Mass of O = mass of compound – ( mass of C + mass of H)

= 50 – (24.66 + 3.43)

= 50 – 28.09

= 21.91 g

1. Determination of the empirical formula.

Mass of C = 24.66 g

Mass of H = 3.43 g

Mass of O = 21.91 g

Divide by their molar mass

C = 24.66 / 12 = 2.055

H = 3.43 / 1 = 3.43

O = 21.91 / 16 = 1.369

Divide by the smallest

C = 2.055 / 1.369 = 2

H = 3.43 / 1.369 = 3

O = 1.369 / 1.369 = 1

Therefore, the empirical formula of the compound is C₂H₃O

2. Determination of the molecular formula.

Molecular weight of compound = 146.0 amu

Empirical formula => C₂H₃O

Molecular formula =?

Molecular formula = [C₂H₃O]ₙ = molecular weight

Thus,

[C₂H₃O]ₙ = 146

[(12×2) + (3×1) + 16]n = 146

[24 + 3 + 16]n = 146

43n = 146

Divide both side by 43

n = 146 / 43

n = 3

Molecular formula = [C₂H₃O]ₙ

Molecular formula = [C₂H₃O]₃

Molecular formula = C₆H₉O₃

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