Question

An organic compound on analysis gave h=3.6 carbon 57.8 rest o2 vapour density of compound was 83 find molecular formula of the compund

Answer 1

mHey Dear,

◆ Answer -

Molecular formula = C8H6O4

◆ Explaination -

Consider 100 g of given compound.

No of moles of C are -

nC = 57.8/12

nC = 4.82 mol

No of moles of C are -

nH = 3.6/1

nH = 3.6 mol

No of moles of O are -

nC = 38.6/16

nC = 2.41 mol

Relative ratio of number of atoms is -

C : H : O = 4.8 : 3.6 : 2.4 = 4 : 3 : 2

Thus, empirical formula of the compound will be C4H3O2.

Empirical formula mass is thus -

E = 4×12 + 3×1 + 2×16

E = 83 g/mol

Molecular formula mass is calculated as -

M = 2 × vapour density

M = 2 × 83

M = 166 g/mol

Multiplication factor n is -

n = M / E

n = 166 / 83

n = 2

Therefore, molecular formula of the compound is (C4H3O2)2 = C8H6O4 .

Best luck dear...

Answer 2

Answer:

Multiplication factor n is -

n = M / E

n = 166 / 83

n = 2