Question 4(i)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?

Question

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioC49.31249.312\dfrac{49.3}{12}1249.3​ = 4.104.102.74\dfrac{4.10}{2.74}2.744.10​ = 1.5 x 2 = 3H6.8416.841\dfrac{6.84}{1}16.84​ = 6.846.842.74\dfrac{6.84}{2.74}2.746.84​ = 2.5 x 2 = 5O43.861643.8616\dfrac{43.86}{16}1643.86​ = 2.742.742.74\dfrac{2.74}{2.74}2.742.74​ = 1 x 2 = 2Simplest ratio of whole numbers = C : H : O = 3 : 5 : 2Hence, empirical formula is C3H5O2Empirical formula weight = 3(12)+ 5(1) + 2(16) = 36 + 5 + 32 = 73Vapour density (V.D.) = 73Molecular weight = 2 x V.D. = 2 x 73 = 146n=Molecular weightEmpirical formula weight=14673=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{146}{73} = 2n=Empirical formula weightMolecular weight​=73146​=2∴ Molecular formula = n[E.F.] = 2[C3H5O2] = C6H10O4

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

Solved Sample Paper 4 — Question 5

Question 4(i)