QJEEM13Ch8889: JEE Main 2013 Chemistry - Class 11 JEEM13Ch8889 Solution: A gaseous hydrocarbon gives upon combustion 0.72 g of water and …

Required reaction,

C_$$x$$H_$$y$$ + $$\left(x+\frac{y}{4}\right)$$O₂ $$\to$$$$x$$CO₂ + $$\frac{y}{2}$$H₂O

0.72 gm of H₂O = $$\frac{0.72}{18}$$ mole of H₂O = 0.04 mole of H₂O

In one H₂O molecule 2 hydrogen atoms present.

So in 0.04 mole of H₂O molecules 2$$\times$$0.04 = 0.08 moles of H atoms present.

0.72 gm of CO₂ = $$\frac{3.08}{44}$$ mole of CO₂ = 0.07 mole of CO₂

And in one CO₂ molecule 1 C atom present.

So in 0.07 mole of CO₂ molecules 0.07$$\times$$1 = 0.07 moles of C atoms present

$$\therefore$$ C : H = 0.07 : 0.08 = 7 : 8

$$\therefore$$ Empirical formula of hydrocarbon = C₇H₈