Question

A compound has the following percentage composition by mass:
Carbon - 54.55%, Hydrogen - 9.09% and Oxygen - 36.26%. Its vapour density is 44. Find the Empirical and Molecular formula of the compound. (H = 1; C = 12; O = 16).

Answer 1

Let amount of organic = $100gm$

amount of carbon = $54.55g$

amount of hydrogen = $9.09g$

Amount of oxygen = $36.26g$

number of mole of carbon = $\frac{34.55}{12}$

$=4.54583$

number of mole of hydrogen = $\frac{9.09}{1.018}$

$=9$

number of mole of oxygen = $\frac{36.26}{16}$

$=2.26$

$4.54833:9:2.26$

$2:4:1$

organic compounds Empirical formula

$=C_2{H}_{4}O$

Molecular weight = $24+4+16=44$

(for empirical)

Molecular weight = $2\times$ vapour density

$=2\times 44=88g$

To molecular formula be = $C_4{H}_8{O}_2$