Combustion analysis | Fundamentals of Chemical Reactions

Chemical equation:

C_xH_yO_z + O₂ → CO₂ + H₂O (unbalanced)
C_xH_yO_z + (X + $\frac{\mathrm{Y}}{4}$ - $\frac{\mathrm{Z}}{2}$) O₂ → X CO₂ + $\frac{\mathrm{Y}}{2}$ H₂O (balanced)

Determine the mass of carbon and hydrogen in the original sample:

mass of carbon in original sample = mass of carbon in the carbon dioxide formed
m_C = m_CO2 x $\frac{{\mathrm{M}}_{\mathrm{C}}}{{\mathrm{M}}_{\mathrm{CO}2}}$
m_C = 3.350 x $\frac{12.0}{44.0}$ = 9.14 x 10^-1 g

mass of hydrogen in original sample = mass of hydrogen in the water formed
m_H = m_H2O x $\frac{{\mathrm{M}}_{\mathrm{H}}}{{\mathrm{M}}_{\mathrm{H}2\mathrm{O}}}$
m_C = 0.823 x $\frac{2.02}{18.0}$ = 9.24 x 10^-2 g

Determine mass % of carbon and hydrogen in cpd:

mass % of C = $\frac{{\mathrm{m}}_{\mathrm{C}}}{{\mathrm{m}}_{\mathrm{compound}}}$ x 100 = $\frac{0.914}{1.250}$ x 100 = 73.1%
mass % of H = $\frac{{\mathrm{m}}_{\mathrm{H}}}{{\mathrm{m}}_{\mathrm{compound}}}$ x 100 = $\frac{0.0924}{1.250}$ x 100 = 7.39 %

Determine mass % of oxygen in cpd:

mass % of O = 100 - mass % of C - mass % of H = 19.5 %

Determine the empirical formula:

In a 100 g sample: 73.1 g of C, 7.39 g of H and 19.5 g of O
⇒ 6.09 mol of C, 7.32 mol of H and 1.219 mol of O
⇒ For 1 mole of O, we have 5 moles of C and 6 moles of H.

The empirical formula of the compound is C₅H₆O