Four grams of hydrocarbon $C_x{H}_y$ on complete combustion gave $12$ grams of $C{O}_2$. What is the empirical formula of the hydrocarbon?

$C_x{H}_y$ + $(x+y/4) O_2$ $\to$ $xC{O}_2$ + $y/2H_{2}O$

1 mole of $C_x{H}_y$ gives x moles of $C{O}_2$

given, $\frac{4}{12x+y}$moles of $C_x{H}_y$ gives $\frac{12}{44}$moles of 

$4x/ 12x + y = 12/ 44$

or, $8x = 3y$

$x/ y = 3/ 8$

Hence, the empirical formula is $\mathrm{C}{}_3{\mathrm{H}}_8$.

Thus, the correct option is D.