Question

an organic compound was found to contain 60% Carbon 13.4 % Hydrogen and remaining is Oxygen. Find the empirical and molecular formula of the compound. the vapour density of the compound is 30.

Answer 1

Step-by-Step Explanation:

Percentage composition of;

Carbon: 60%

Hydrogen: 13.4%

Oxygen: 100-(C+H)= 100-(60+13.4) =100-73.4= 26.6%

Calculating Empirical Formula:

Number of gram atoms of element= percentage composition of element

                                                            atomic mass of element

Carbon= 60/12= 5 gram atoms

Hydrogen= 13.4/1.008= 13.29≈13 gram atoms

Oxygen= 26.6/16= 1.66≈2 gram atoms

Atomic ratio = gram atoms of element

                         smallest number of gram atoms

Carbon: 5/2= 2.5≈ 3

Hydrogen: 13/2= 6.5≈7

Oxygen: 2/2=1

Ratio: C: H :O

          3: 7 : 1

Empirical Formula= C₃H₇O

Molecular Formula= n(Empirical formula)

where n= molecular mass/empirical formula mass

Empirical formula mass=  C₃H₇O= (12x3)+(1x7)+(16x1)= 36+7+16= 59

Now to find Molecular mass; we know that vapor density of a compound is its molecular mass divided by the molar mass of H₂. So,

Molecular mass= 30x2.016= 60.48≈60.5≈61

n= 61/59 = 1.033

Molecular Formula= 1.03(C₃H₇O)

Molecular Formula= C₃H₇O

HOPE IT HELPS   =)