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Solution


Q.

Quantitative analysis of an organic compound (X) shows following % composition.

C : 14.5 %

Cl : 64.46 %

H : 1.8 %

(Empirical formula mass of the compound (X) is _____ ×10-1)

(Given molar mass in g mol-1 of C : 12, H : 1, O : 16, Cl : 35.5)                             [2025]

Ans.

(1655)

Element Mass percentage Moles of atoms in 100 g compound = mass percentage / molar mass Simple ratio of atoms
C 14.5 14.5 / 12 = 1.2 2
Cl 64.46 64.46 / 35.5 = 1.8 3
H 1.8 1.8 / 1 = 1.8 3
O 100 – (14.5 + 64.46 + 1.8) = 19.24 19.24 / 16 = 1.2 2

 

Empirical formula =C2H3Cl3O2

Empirical formula mass=2×12+3×1+3×35.5+2×16= 165.5=1655×10-1