Chemistry

An organic compound X has the following composition: O = 71.19% H= 2.22 %
[At. mass C = 12 H = 1 O = 16 ]
(a) Find its empirical formula.
(b) If in the gaseous state, its vapour density is 45. Find its molecular formula.

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(a) Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Hydrogen	2.22	1	$\dfrac{2.22}{1}$ = 2.22	$\dfrac{2.22 }{2.21}$ = 1
Oxygen	71.19	16	$\dfrac{71.19}{16}$ = 4.44	$\dfrac{4.44}{2.21 }$ = 2
Carbon	26.59	12	$\dfrac{26.59}{12}$ = 2.21	$\dfrac{ 2.21 }{2.21}$ = 1

Simplest ratio of whole numbers = H : O : C = 1 : 2 : 1

Hence, empirical formula is CHO₂

(b) Empirical formula weight = 12 + 1 + (2 x 16) = 13 + 32 = 45

V.D. = 45

Molecular weight = 2 x V.D. = 2 x 45 = 90

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{90}{45} = 2$

So, molecular formula = 2[E.F.] = 2(CHO₂) = C₂H₂O₄

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