Question

8. A compound contain 54.55% carbon, 9.09% hydrogen, 36.36% oxygen. The empirical formula of this compound is- 1 point C3H5O C4H8O2 C2H4O2 C2H4O

Answer 1

Let amount of organic = 100 gm

amount of carbon = 54.55 g

amount of hydrogen = 9.09 g

Amount of oxygen = 36.26 g

number of mole of carbon =

12

34.55

=4.54583

number of mole of hydrogen =

1.018

9.09

=9

number of mole of oxygen =

16

36.26

=2.26

4.54833:9:2.26

2:4:1

organic compounds Empirical formula

=C

2

H

4CHEMISTRY

A compound has the following percentage composition by mass:

Carbon - 54.55%, Hydrogen - 9.09% and Oxygen - 36.26%. Its vapour density is 44. Find the Empirical and Molecular formula of the compound. (H = 1; C = 12; O = 16).

O

Molecular weight = 24+4+16=44

(for empirical)

Molecular weight = 2× vapour density

=2×44=88 g

To molecular formula be = C

4

H

8

O