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Q.

From a mixture which makes up crude oil, a particular hydrocarbon ingredient (that is one containing hydrogen and carbon atoms only) has been isolated. 10 g of this liquid is burnt in excess of oxygen and the products are 4.4 g of carbon dioxide and 1.8 g of water.

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Detailed Solution

Empirical formula: CH₂.

Work:

From 4.4 g CO₂ → $n_C=4.4/44=0.10$ $n$ $C$ $$ $=$ $4.4/44$ $=$ $0.10$ mol C.
From 1.8 g H₂O → $n_H=2\times(1.8/18)=0.20$ $n$ $H$ $$ $=$ $2$ $\times$ $($ $1.8/18$ $)$ $=$ $0.20$ mol H.
Ratio C:H = 0.10:0.20 = 1:2 ⇒ empirical formula CH₂ (i.e., $\mathrm{C_nH_{2n}}$ $C$ $n$ $$ $H$ $2n$ $$ , an alkene or cycloalkane).

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