Question 5(b)A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula.[C=12, H=1]

Question

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioCarbon82.761282.7612\dfrac{82.76}{12}1282.76​ = 6.896.896.89\dfrac{6.89}{6.89}6.896.89​ = 1Hydrogen17.24117.241\dfrac{17.24}{1}117.24​ = 17.2417.246.89\dfrac{17.24}{6.89}6.8917.24​ = 2.5Simplest ratio of whole numbers = C : H = 1 : 2.5 = 2 : 5Hence, empirical formula is C2H5Empirical formula weight = 2(12) + 5(1) = 29V.D. = 29Molecular weight = 2 x V.D. = 2 x 29n=Molecular weightEmpirical formula weight=2×2929=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{2 \times 29}{29} =2n=Empirical formula weightMolecular weight​=292×29​=2∴ Molecular formula = n[E.F.] = 2[C2H5] = C4H10

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	82.76	12	$\dfrac{82.76}{12}$ = 6.89	$\dfrac{6.89}{6.89}$ = 1
Hydrogen	17.24	1	$\dfrac{17.24}{1}$ = 17.24	$\dfrac{17.24}{6.89}$ = 2.5

Solved 2016 Question Paper ICSE Class 10 Chemistry — Question 13

Question 5(b)