Question

A hydrocarbon contain 81.82% carbon and 18.18% hydrogen suggest the molecular formula if the v.d of hydrocarbon is 22

Answer 1

Let the empirical formula of the hydrocarbon be $\sf{C_cH_h.}$

Atomic mass of Carbon = 12 g.

Atomic mass of Hydrogen = 1 g.

The hydrocarbon contains 81.82% carbon so the mass percentage of carbon in this compound is 81.82%,

$\longrightarrow\sf{\dfrac{12c}{12c+h}=\dfrac{81.82}{100}\quad\quad\dots(1)}$

And that of hydrogen is 18.18%.

$\longrightarrow\sf{\dfrac{h}{12c+h}=\dfrac{18.18}{100}\quad\quad\dots(2)}$

Dividing (1) by (2), we get,

$\longrightarrow\sf{\dfrac{12c}{h}=\dfrac{81.82}{18.18}}$

$\longrightarrow\sf{\dfrac{12c}{h}=4.5}$

$\longrightarrow\sf{\dfrac{c}{h}=\dfrac{4.5}{12}}$

$\longrightarrow\sf{\dfrac{c}{h}=\dfrac{3}{8}}$

$\longrightarrow\sf{c:h=3:8}$

We know empirical formula of a compound is the formula with its elements in simple whole number ratio.

So the empirical formula of our hydrocarbon is $\sf{C_3H_8}$ since it's in simple whole number ratio.

Let the molecular formula of the compound be $\sf{(C_3H_8)_k}$ where $\sf{k=\dfrac{M}{E}.}$

Empirical formula is,

$\longrightarrow\sf{E=12\times3+8\times1}$

$\longrightarrow\sf{E=44\ g}$

Given that vapour density of this compound is 22.

$\longrightarrow\sf{V=22}$

So molecular mass is,

$\longrightarrow\sf{M=2V}$

$\longrightarrow\sf{M=44\ g}$

Thus,

$\longrightarrow\sf{k=\dfrac{M}{E}}$

$\longrightarrow\sf{k=1}$

∴ Molecular formula is $\sf{C_3H_8.}$

Answer 2

Answer: C3H8

Explanation:

Provided that, vapour density = 22

=> M. wt. = 22 × 2 = 44g

(Vapour density = M. wt. of molecule at STP/M. wt. of H2 (g) at STP)

Or simply V.D. = (M. wt. of molecule)/2

Assume we have 100 g sample of that compound.

Therefore, abundance of C = 81.82 g

Abundance of H = 18.18 g

Moles of C = 81.82/12 mol ≈ 7 mol

Moles of H = 18.18/1 mol ≈ 18 mol

Now, mole ratio: Moles of C : Moles of H = 7 : 18

Therefore, Empirical formula = C7H18

We know,

n = (mass of Ori F)/(mass of Emp F)

n = (44)/(84+18) = 0.431

Now, Original formula = (EF)_n = (C7H18)_0.431 = C3H8 (taken nearest values).