Question

complete combustion of 8.00g a hydrocarbon produced 26.0g of CO2 and 7.99g of H2O. what is the empirical formula for the hydrocarbon?

Answer 1

Consider the general balanced equation for combustion of hydrocarbon:

$C_xH_y+(x+\frac{y}{4})O_2\to xCO_2+\frac{y}{2}H_2O$

Molar mass of $CO_2=44\frac{g}{mol}$

So the number of moles of $CO_2$ formed$=x=\frac{26}{44}=0.59$$mol$

Molar mass of $H_2O=18\frac{g}{mol}$

So the number of moles of $H_2O$ formed$=\frac{y}{2}=\frac{7.99}{18}=0.44$$mol$

So $y=2\times0.44=0.88mol$

For $C: x=0.59\approx0.60$

For$H:y=0.88\approx0.90$

So there will be 3 $H$ atoms per 2 $C$ atoms

Hence the empirical formula of the given hydrocarbon is $C_2H_3$