0.004 g of hydrocarbon on combustion liberates 0.011g of $C{O}_2$ at STP.  The empirical formula of the hydrocarbon is _______

$\%C=\frac{12}{44}\times \frac{{\text{wt.\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}CO}}_{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}liberated}}{\text{wt.\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}organic\hspace{0.17em}\hspace{0.17em}compound\hspace{0.17em}\hspace{0.17em}taken}}\times 100$

$\%C=\frac{12}{44}\times \frac{0.011}{0.004}\times 100 = 75\%$

$\% H=100-\% C=100–75=25$

$C:H =\frac{75}{12}:\frac{25}{1}=\frac{75}{12}\times \frac{1}{25}=1:4$

$\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}EF}={\text{CH}}_4$