Question

A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 29.5 mg produced 80.1 mg of CO2 and 16.4 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

Answer 1

Answer:

The empirical formula is $C_5H_5O$.

Explanation:

Given that, the combustion of 29.5 mg produced 80.2 mg CO₂ and 16.4 mg of H₂O.

Mass of C  $=80.1\times 10^{-3} g CO_2\times \frac{\textrm{1 mol C}O_2}{44 g}\times \frac{12 g CO_2}{\textrm{1 mol C}O_2}$

                   $=0.022$ g

Mass of H $=16.4\times 10^{-3} g HO_2\times \frac{\textrm{1 mol }H_2O}{18 g}\times \frac{2 g H}{\textrm{1 mol }H_2O}$

                 $=1.8 \times 10^{-3}$ g

Mass of O = $29.5\times 10^{-3}g-(0.022 g+1.8 \times 10^{-3} g)$

                $=5.68 \times 10^{-3}$ g

Number of mole of C $=\frac{Mass}{\textrm{ Atomic mass}}$

                                  $=\frac{0.022g}{12 g/mol}$

                                  $=1.83\times 10^{-3}$ mol

Number of mole of H $=\frac{Mass}{\textrm{ Atomic mass}}$

                                    $=\frac{1.8\times 10^{-3}g}{1g/mol}$

                                   $=1.8 \times 10^{-3}$ mol

Number of mole O $=\frac{Mass}{\textrm{ Atomic mass}}$

                                $=\frac{5.68 \times 10^{-3}g}{16 g/mol}$

                                $=0.355\times 10^{-3}$ mol

Dividing by lowest integer

$C:\frac{1.83\times 10^{-3}}{0.355\times 10^{-3}}=5$

$H:\frac{1.8\times 10^{-3}}{0.355\times 10^{-3}}=5$

The empirical formula is $C_5H_5O$.