Percentage of nitrogen `= (1.4 xx N_(1) xx V_(1))/(W)`
`W =` Mass of organic compound `= 0.48g`
`N_(1) = (1)/(2) = 0.5, V_(1) = 19.2 mL`
Percentage of `N = (1.4 xx 0.5 xx 19.2)/(0.48) = 28`
Percentage of oxygen `= 100 - (48 +8 +28) = 16`
Calculation of empirical formula.
`{:("Element","Percentage","At.mass","Relative number of atoms","Simplest ratio of atoms"),("Carbon",48.0,12,(48)/(12)=4,(4)/(1)=4),("Hydrogen",8.0,1,(8)/(1)=8,(8)/(1)=8),("Nitrogen",28.0,14,(28)/(14)=2,(2)/(1)=2),("Oxygen",16.0,16,(16)/(16)=1,(1)/(1)=1):}`
Empirical formula `= C_(4)H_(8)N_(2)O`