Chemistry

An organic compound containing C, H and O has 49.3% carbon, 6.84% hydrogen, 43.86% oxygen and its vapour density is 73. What will be the molecular formula of the compound?

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Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
C	49.3	12	$\dfrac{49.3}{12}$ = 4.10	$\dfrac{4.10}{2.74}$ = 1.5 x 2 = 3
H	6.84	1	$\dfrac{6.84}{1}$ = 6.84	$\dfrac{6.84}{2.74}$ = 2.5 x 2 = 5
O	43.86	16	$\dfrac{43.86}{16}$ = 2.74	$\dfrac{2.74}{2.74}$ = 1 x 2 = 2

Simplest ratio of whole numbers = C : H : O = 3 : 5 : 2

Hence, empirical formula is C₃H₅O₂

Empirical formula weight = 3(12)+ 5(1) + 2(16) = 36 + 5 + 32 = 73

Vapour density (V.D.) = 73

Molecular weight = 2 x V.D. = 2 x 73 = 146

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{146}{73} = 2$

∴ Molecular formula = n[E.F.] = 2[C₃H₅O₂] = C₆H₁₀O₄

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