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Chemistry

A volatile chloride of metal M contains 34.5% of the metal. If the density of metal chloride relative to hydrogen is 162.5. Calculate the molecular formula of metal chloride.

[M = 56, Cl = 35.5].

Stoichiometry

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Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
M34.55634.556\dfrac{34.5}{56} = 0.6160.6160.616\dfrac{0.616}{0.616} = 1
Cl100 - 34.5 = 65.535.565.535.5\dfrac{65.5}{35.5} = 1.841.840.616\dfrac{1.84}{0.616} = 2.98 = 3

Simplest ratio of whole numbers between M : Cl = 1 : 3

Hence, empirical formula is MCl3

Empirical formula weight = 56 + 3(35.5) = 56 + 106.5 = 162.5

Vapour density (V.D.) = 162.5

Molecular weight = 2 x V.D. = 2 x 162.5 = 325

n=Molecular weightEmpirical formula weight=325162.5=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{325}{162.5} = 2

Molecular formula = n[E.F.] = 2[MCl3] = M2Cl6

Hence, Molecular formula = M2Cl6

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