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.
Answers
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 =)