Question

A hydrocarbon with molecular mass 78 a.m.u. contains 7.7% by mass of hydrogen.Calculate the molecular formula of this compound.

Answer 1

Given : molecular mass of hydrocarbon = 78 amu
percentage of hydrogen = 7.7%
thus percentage of carbon = 100 - 7.7 %
=92.3%

Step 1 : calculation of Empirical Formula

Element	Percentage of Element	Atomic Mass of Element	Moles Of Atoms	Mole Ratio
C	92.3	12	$\frac{92.3}{12}=7.69$	$\frac{7.69}{7.69}=1$
H	7.7	1	$\frac{7.7}{1}=7.7$	$\frac{7.7}{7.69}=1$

Thus the simple ratio of C:H is 1:1

Therefore the empirical formula of the compound is CH.

Step 2 : Calculation of the Molecular Formula

Empirical Formula Mass = 12+1
= 13 amu
Given Molecular mass = 78 amu

Molecular Formula = (Empirical Formula )_n

$\mathrm{n}=\frac{\mathrm{Molecular}\mathrm{mass}}{\mathrm{Empirical}\mathrm{Formula}\mathrm{Mass}}$
= $$\begin{gathered} \frac{78}{13} \\ =6 \end{gathered}$$

Thus Molecular formula is (CH)₆ = C₆H₆