{:("Column-I","Column-II"),("A) "NH_(3),"P) EF = MF"),("B) "N_(2)H_(4),"Q) MF = "(EF)_(2)),("C) "N_(3)H,"R) Maximum percentage of nitrogen by mass "),("D) "C_(2)N_(2),"S) Least percentage of nitrogen by mass"):}

{:("Column-I","Column-II"),("A) "NH_(3),"P) EF = MF"),("B) "N_(2)H_(4)

1.	{:("Column-I","Column-II"),("A) "NH_(3),"P) EF = MF"),("B) "N_(2)H_(4),"Q) MF = "(EF)_(2)),("C) "N_(3)H,"R) Maximum percentage of nitrogen by mass "),("D) "C_(2)N_(2),"S) Least percentage of nitrogen by mass"):}
Answer» Solution :`N_(2)H_(4)`, EMPIRICAL formula = `NH_(2)` `N_(3)H` (hydrazoic acid) % of nitrogen = `(42)/(43)XX100` `C_(2)N_(2)` (cyanogen) % of nitrogen = `(28)/(52)xx100`