• # question_answer On complete combustion 1.4 g hydrocarbon gave 1.8 g water. Empirical formula of the hydrocarbon is A) $CH$ B) $C{{H}_{2}}$ C) $C{{H}_{3}}$ D) $C{{H}_{4}}$

$\because$ 1.8gm water obtained from 1.4gm hydrocarbon $\therefore$18gm water obtained from - $\frac{1.4}{1.8}\times 18=14$ gm.     Empirical formula Mass = 14 $\therefore$ Empirical formula = $C{{H}_{2}}$.