Question 5(i)An organic compound, whose vapour density is 45, has the following percentage compositionH = 2.22%, O = 71.19% and remaining carbon.Calculate,(a) it's empirical formula, and(b) it's molecular formula

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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

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(CHO₂) = C₂H₂O₄

Solved Sample Paper 3 — Question 9

Question 5(i)