Question 7Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.

Question

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioCarbon80128012\dfrac{80}{12}1280​ = 6.666.666.66\dfrac{6.66}{6.66}6.666.66​ = 1Hydrogen201201\dfrac{20}{1}120​ = 20206.66\dfrac{20}{6.66}6.6620​ = 3Simplest ratio of whole numbers = C : H = 1 : 3Hence, empirical formula is CH3Empirical formula weight = 12 + 3(1) = 15V.D. = 15Molecular weight = 2 x V.D. = 2 x 15n=Molecular weightEmpirical formula weight=2×1515=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{2 \times 15}{15} = 2n=Empirical formula weightMolecular weight​=152×15​=2∴ Molecular formula = n[E.F.] = 2[CH3] = C2H6

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	80	12	$\dfrac{80}{12}$ = 6.66	$\dfrac{6.66}{6.66}$ = 1
Hydrogen	20	1	$\dfrac{20}{1}$ = 20	$\dfrac{20}{6.66}$ = 3

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 7

Question 7