Question

A compound on analysis
was found to contain C=
54.24% , H= 9.05% and
O=36.71% Calculate Empirical
formulae and Molecular
formulae? Molecular mass =
88u
​

Answer 1

Answer:

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Answer 2

Given :-

• C = 54.24 %
• H = 9.05 %
• O = 36.71 %

➢ Relative no of atoms

• Divide the % composition of the elements with their atomic masses

C = 54.24 / 12 = 4.52

H = 9.05 / 1 = 9.05

O = 36.71 / 16 = 2.29

➢ Simple ratio

• Now divide all the values obtained above with the least possible value (i.e. 2.29 in this case )

➜ C = 4.52 / 2.29 = 2

➜ H = 9.05 / 2.29 = 4

➜ O = 2.29 / 2.29 = 1

Empirical formula = C2 H4 O

➜ Empirical mass = 2 x C + 4 x H + O

➜ Empirical mass =  24 + 4 + 16

➜ Empirical mass = 44 u

Molecular mass = 88u

We know that ,

➜ Molecular mass = n x empirical mass

➜ n = 88 / 44

➜ n = 2

Similarly ,

➜ Molecular formula = n x Empirical formula

➜ Molecular formula = 2 ( C2 H4 O )

Molecular formula =  C4 H8 O2