Answer to Question #90688 in General Chemistry for Stacey

Question

A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 81.9 mg produced 222 mg of CO2 and 45.5 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

Answer 1

Calculating first the amount of carbon in CO₂ that is equal to amount of carbon in an unknown compound:

$n(C)=n(CO_2)=\frac{m(CO_2)}{M(CO_2)}=\frac{0.222g}{44\frac{g}{mol}}=5.05\times10^{-3}mol$

The amount of H is two times the amount of water produced:

$n(H)=2n(H_2O)=2\frac{m(H_2O)}{M(H_2O)}=2\frac{0.0455g}{18\frac{g}{mol}}=5.06\times 10^{-3}mol$

The amount of O is a remnant:

$n(O)=\frac{m(O)}{M(O)}=\frac{m(compound)-m(C)-m(H)}{M(O)}=\frac{m(compound)-n(C)M(C)-n(H)M(H)}{M(O)}=\frac{0.0819g-5.05\times 10^{-3}mol \cdot 12\frac{g}{mol}-5.06\times 10^{-3}mol \cdot 1\frac{g}{mol}}{16\frac{g}{mol}}=1.02\times 10^{-3}mol$

n(C):n(H):n(O)=5.05x10^-3:5.06x10^-3:1.02x10^-3=5:5:1

Thus empirical formula is C₅H₅O

Molar mass of C₅H₅O is 81g/mol that is two times less than given 162g/mol. Thus molecular formula of compound is C₁₀H₁₀O₂.