Question

A hydrcarbon contains 82.8%of carbon.find its molecular formula if its vapour density is 29.

Answer 1

the rest is (100-82.8)% =17.2% H
with respect to number of atoms,
 C:H = (82.8/12):(17.2/1)
        = 6.9 : 17.2
        = 1:2.5(APPROX)
        = 2:5(multiplying both by 2 to get whole number ratio)
hence the empirical formula is C2H5
i.e. (12*2+1*5)*n=2*29 (M=2D)
i.e. n=2
the molecular formula is C4H10 (butane or its isomers)

Answer 2

The composition or molecular formula of the compound is $C_{4}H_{10}$.(Butane)

Step-by-step process

Given:

The carbon percentage of the hydrocarbon = 82.8%

Vapour density- 29

To find= molecular formula of the compound

Solution:

Composition of a hydrocarbon=  Hydrogen+ carbon

We know that,

The molecular mass of a compound= 2× vapour density

So, molar mass= 2×29

= 58gram

The mass of carbon = 82.8×58÷100

=82.8×0.58

=48 gram

Thus, the mass of hydrogen= total molar mass- the mass of carbon

= 58-48

= 10 gram

Since, the mas of one carbon atom is 12 gram,

number of carbon atoms= 48÷12= 4C

Now, the number of Hydrogen atoms= 10÷1= 10 H

Result:

Thus, the composition or molecular formula of the compound is $C_{4}H_{10}$(Butane).

(#SPJ2)