Question

A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 92.5 mg produced 251 mg of CO2 and 51.4 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

Answer 1

CxHyOz

w(C/CO2) = Ar(C)/Mr(CO2) = 12g/mol / 44g/mol = 0.2727.

m(C) = m(CO2) × w(C/CO2) = 251mg × 0.2727 = 68.4545mg.

w(H/H2O) = Ar(H)/Mr(H2O) = 1g/mol / 18g/mol = 0.1111.

m(H) = m(H2O) × w(H/H2O) = 51.4mg × 0.1111 = 5.7111mg.

m(O) = m(CxHyOz) – m(C) – m(H) = 92.5mg – 68.4545mg – 5.7111mg = 18.3344mg.

x:y:z = m(C)/M(C) : m(H)/M(H) : m(O)/M(O) = 68.4545/12 : 5.7111/1 : 18.3344/16 = 5.7045 : 5.7111 : 1.1459 = 4.9782 : 4.9839 : 1 ≈ 5:5:1.

The empirical formula is C5H5O. M(C5H5O) = 81g/mol.

The molecular formula is C10H10O2, because the molar mass of the compound was 162 g/mol.