Question

A combustion device was used to analyze a 0.6349 g compound containing only carbon, hydrogen, and oxygen. The experiment produced 1.603 g carbon dioxide and 0.2810 g of water. Determine the empirical formula of the compound

Answer 1

0.281g H₂O ($\frac{2g}{18g}$) = 0.0312gH

and 1.603gCO₂ ($\frac{12g}{44g}$) = 0.437gC

0.6349gTotal - 0.437gC - 0.0312gH = 0.167gO

0.437gC($\frac{1mol}{12g}$) = 0.0364molC $\implies$ $\frac{0.0364molC}{0.0104}$ = 3.5 (2) = 7C

0.0312gH($\frac{1mol}{1g}$) = 0.0312molH $\implies$ $\frac{0.0312molH}{0.0104}$ = 3 (2) = 6H

0.167gO($\frac{1mol}{16g}$) = 0.0104molO $\implies$ $\frac{0.0104molO}{0.0104}$ = 1(2) = 2O

the empirical formula of the compound is C₇H₆O₂