Question

a compound on analysis gave the following percentage composition by mass H-9.09, o-36.36, c-54.55. mole mass of compound is 88. Find its molecular Formula?​

Answer 1

Answer:

C8H16O4

Explanation:

carbon =its percentage 54.55 &its Atomic weight 12& relative no of atom= 54.55/12 = 4.53 & it's ratio 4.53/ 2.27=2

Hydrogen = its percentage 9.09 and it's Atomic weight 1 &its relative no of atom= 9.09/1=9.09 & its ratio= 9.09/2.27=4

oxygen= its percentage 36.36 &its Atomic weight 16 & its relative no of atom= 36.36/16 =2.27 & its ratio = 2.27/2.27=2

Empirical formula= C2H4O

Empirical formula weight= 44

molecular weight= 2×88=176, n= 176/44=4

So, molecular formula= 4 ×empirical formula

= 4(C2H4O)

= C8H16O4

Answer 2

Explanation:

a compound on analysis gave the following percentage composition by mass H-9.09, o-36.36, c-54.55. mole mass of compound is 88. Find its molecular Formula?