Question 31A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride (M = 56).

Question

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioMetal M34.55634.556\dfrac{34.5}{56}5634.5​ = 0.6160.6160.616\dfrac{0.616}{0.616}0.6160.616​ = 1chlorine65.535.565.535.5\dfrac{65.5}{35.5}35.565.5​ = 1.841.840.616\dfrac{1.84}{0.616}0.6161.84​ = 2.98 = 3Simplest ratio of whole numbers = M : Cl = 1 : 3Hence, empirical formula is MCl3Empirical formula weight = 56 + 3(35.5) = 162.5 gMolecular weight = 2 x V.D. = 2 x 162.5 = 325n=Molecular weightEmpirical formula weight=325162.5=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{ 325}{162.5} = 2n=Empirical formula weightMolecular weight​=162.5325​=2Molecular formula = n[E.F.] = 2[MCl3] = M2Cl6∴ Molecular formula = M2Cl6

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Metal M	34.5	56	$\dfrac{34.5}{56}$ = 0.616	$\dfrac{0.616}{0.616}$ = 1
chlorine	65.5	35.5	$\dfrac{65.5}{35.5}$ = 1.84	$\dfrac{1.84}{0.616}$ = 2.98 = 3

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 32

Question 31