Single Choice

An organic compound contains $$49.3\%$$ carbon $$6.84\%$$ hydrogen and its vapour density is $$73$$. Molecular formula of the compound is

A$$C_3H_5O_2$$
B$$C_6H_{10}O_4$$
Correct Answer
CC3H10O2
D$$C_4H_{10}O_2$$

Solution

$$n_C : n_H : n_O = \dfrac{49.3}{12} : \dfrac{6.84}{1} : \dfrac{43.86}{16}= 1.5 : 2.5 : 1 = 3 : 5 : 2$$

Empirical formula $$=C_3H_{5}O_2$$

Empirical formula mass $$=12\times 3+5+16\times 3=73$$

V. D. $$=73$$

Molecular mass $$=2\times V.D.=2\times 73=146$$

$$2\times$$ Empirical formula mass $$=$$ Molecular mass

Molecular formula $$=2\times$$ Empirical formula $$=C_6H_{10}O_4$$

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