Question

An organic compound contains carbon 40%hydrogen6.67%and remainingoxygen .Derive the molecular formula of the compound if its molecular mass is180.

Answer 1

Step 1 : calculation of Empirical Formula

Element	Percentage of Element	Atomic Mass of Element	Moles Of Atoms	Mole Ratio
C	40	12	$\frac{40}{12}=3.33$	$\frac{3.33}{3.33}=1$
H	6.67	1	$\frac{6.67}{1}=6.67$	$\frac{6.67}{3.33}=2$
O	53.33	16	$\frac{53.33}{16}=3.33$	$\frac{3.33}{3.33}=1$

Thus the simple ratio of C:H:O is 1:2:1

Therefore the empirical formula of the compound is CH₂O.

Step 2 : Calculation of the Molecular Formula

Empirical Formula Mass = 12+1$\times$2+ 16
= 30 amu
Molecular mass = 180 amu

Molecular Formula = (Empirical Formula )_n

$\mathrm{n}=\frac{\mathrm{Molecular}\mathrm{mass}}{\mathrm{Empirical}\mathrm{Formula}\mathrm{Mass}}$
= $$\begin{gathered} \frac{180}{30} \\ =6 \end{gathered}$$

Thus Molecular formula is (CH₂O)₆ = C₆H₁₂O₆