Question 6(ii)Calculate:(a) A compound contains 69.5% oxygen, 30.5% nitrogen and its molecular weight is 92. Then what will be the molecular formula of the compound?(b) If the empirical formula of a compound is CHO and its vapour density is 29, find the molecular formula of the compound.

Question

Question 6(ii)Calculate:(a)	A compound contains 69.5% oxygen, 30.5% nitrogen and its molecular weight is 92. Then what will be the molecular formula of the compound?(b) If the empirical formula of a compound is CHO and its vapour density is 29, find the molecular formula of the compound.

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioO69.51669.516\dfrac{69.5}{16}1669.5​ = 4.344.342.17\dfrac{4.34}{2.17}2.174.34​ = 2N30.51430.514\dfrac{30.5}{14}1430.5​ = 2.172.172.17\dfrac{2.17}{2.17}2.172.17​ = 1Simplest ratio of whole numbers = O : N = 2 : 1Hence, empirical formula is NO2Molecular weight = 92Empirical formula weight = 14 + 2(16) = 14 + 32 = 46n=Molecular weightEmpirical formula weight=9246=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{92}{46} = 2n=Empirical formula weightMolecular weight​=4692​=2Molecular formula = n[E.F.] = 2[NO2] = N2O4Hence, molecular formula is N2O4.(b) Empirical formula of the compound = CHOEmpirical formula mass = 12 + 1 + 16 = 29Molecular formula mass = 2 x vapour density= 2 x 29 = 58 gMolecular formula = (Empirical formula)n∴ Molecular formula = (CHO)2 = C2H2O2

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
O	69.5	16	$\dfrac{69.5}{16}$ = 4.34	$\dfrac{4.34}{2.17}$ = 2
N	30.5	14	$\dfrac{30.5}{14}$ = 2.17	$\dfrac{2.17}{2.17}$ = 1

Solved Sample Paper 2 — Question 14

Question 6(ii)