Given the mass of CO2 =15.02 g
Moles of [tex]CO_{2}[/tex]=[tex]15.02 gCO_{2} *\frac{1 molCO_{2} }{44.01 gCO_{2} } =0.341 mol CO_{2}[/tex]
Mass of H2O = 2.458 g
Moles of [tex]H_{2}O[/tex]=[tex]2.458 g H_{2}O*\frac{1molH_{2}O }{18.02g H_{2}O } =0.136molH_{2}O[/tex]
Moles of C = [tex]0.341 mol CO_{2}*\frac{1 molC}{1 molCO_{2} } =0.341molC[/tex]
Moles of H = [tex]0.136 mol H_{2}O * \frac{2 mol H}{1 mol H_{2}O } =0.272 mol H[/tex]
Mass of C in the sample = [tex]0.341 mol C*\frac{12.01g C}{1 mol C} = 4.095 g C[/tex]
Mass of H = [tex]0.272 mol H *\frac{1.01 g H}{1 molH}[/tex]=0.275 gH
Mass of O in the sample = 5.467 g - (4.095 g +0.275 g) = 1.097 g O
Moles of O = [tex]1.097 g O * \frac{1 molO}{16 g O} =0.0686 mol O[/tex]
Simplest mole ratios of the elements in the compound:
[tex]Cx_{\frac{0.341mol}{0.0686mol}} H_{\frac{0.272mol}{0.0686mol} }O_{\frac{0.0686mol}{0.0686mol} }[/tex]
Therefore the empirical formula of the compound is [tex]C_{5}H_{4}O[/tex]
