Question

Help plzzz... FAST
An organic compound on analysis gave following % composition C=
57.8% and H= 3.6% and rest is O. The molecular mass of the compound
is 166 g/mol. Find its empirical and molecular formula.

Answer 1

Answer:

We have,

An organic compound C=57.8%

H=3.6%

Thus the oxygen is 38.6% Since the total %is100

So, 

Molecular weight i=2×vapordensity

=2×83=166

So, Mass of C=\dfrac{57.8}{100}\times166$$

=95.948g

No pof carbon atoms =1295.948=7.99∼8atoms Since 

12=mass of the carbon

Mass of H=1003.6×166=5.976∼6 atoms

Mass of O=10038.6×166=64.076g

No of atoms =1664.076=4atoms

Thus. The formulaa will be C8H6O4

May this help you .....

Answer 2

Answer: empirical formula = C4H3O2

Molecular formula = C8H6O4

Explanation:

C= 57.8/12= 4.81

H=3.6/1= 3.6

O=38.6/16= 2.41

Dividing these with lowest number to get simple whole number ratio

C= 4.81/2.4= 2 ×2=4

H=3.6/2.14=1.5×2=3

O= 2.41/2.41=1×2=2

Empirical formula = C4H3O2

Given compound mass= 166

n = compound mass ÷ empirical formula mass

n = 166÷ 83= 2

Molecular formula = empirical formula × n

Molecular formula = C4H3O2× 2= C8H6O4