Question

A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 78.1 mg produced 212 mg of CO2 and 43.4 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

Answer 1

1. The empirical formula of the compound is C₅H₅O

2. The molecular formula of the compound is C₁₀H₁₀O₂

We'll begin by calculating the mass of C, H and O in the compound. This can be obtained as follow:

For C:

Mass of CO₂ = 212 mg

Molar mass of CO₂ = 44 g/mol

Molar mass of C = 12 g/mol

Mass of C =?

Mass of C = 12/44 × 212

Mass of C = 57.82 mg

For H:

Mass of H₂O = 43.4 mg

Molar mass of H₂O  = 18 g/mol

Molar mass of H₂ = 1 × 2 = 2 g/mol

Mass of H =?

Mass of H = 2/18 × 43.4

Mass of H = 4.82 mg

For O:

Mass of C = 57.82 mg

Mass of H = 4.82 mg

Mass of compound = 78.1 mg

Mass of O =?

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

Mass of O = 78.1 – (57.82 + 4.82)

Mass of O = 15.46 mg

1. Determination of the empirical formula of the compound.

Mass of C = 57.82 mg

Mass of H = 4.82 mg

Mass of O = 15.46 mg

Empirical formula =?

Divide by their molar mass

C = 57.82 / 12 = 4.818

H = 4.82 / 1 = 4.82

O = 15.46 / 16 = 0.966

Divide by the smallest

C = 4.818 / 0.966 = 5

H = 4.82 / 0.966 = 5

O = 0.966 / 0.966 = 1

Therefore, the empirical formula of the compound is C₅H₅O

2. Determination of the molecular formula of the compound.

Molar mass of compound = 162 g/mol.

Empirical formula = C₅H₅O

Molecular formula =?

Molecular formula = empirical formula × n = molar mass

[C₅H₅O]n = 162

[(12×5) + (5×1) + 16]n = 162

[60 + 5 + 16]n = 162

81n = 162

Divide both side by 81

n = 162 / 81

n = 2

Molecular formula = [C₅H₅O]n

Molecular formula = [C₅H₅O]₂

Molecular formula = C₁₀H₁₀O₂

Thus, the molecular formula of the compound is C₁₀H₁₀O₂

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