Question

Calculate the empirical and molecular formula of an organic compound whose percentage composition is Carbon =70.54%, Hydrogen=5.94%, 23.52%.

The molecular weight of the compound is 136?

Answer 1

• The empirical formula represents the simplest whole-number ratio of the atoms of each different elements comprising the compound.
• The molecular formula denotes the actual number of atoms of different elements present.
• Molecular formula = Empirical formula × n

Where n=$\frac{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{a}\mathrm{compound}}{\mathrm{Empirical}\mathrm{formula}\mathrm{mass}}$

• The given hydrocarbon contains, percentage composition of Carbon =70.54%, Hydrogen=5.94%, 23.52%.
• We can calculate the empirical formula as:

Elements	Percentage	Atomic Mass	Atomic ratio	Simple ratio
Carbon	70.54	12	$\frac{70.54}{12}$= 5.87	$\frac{5.87}{1.47}$=3.99≈4
Hydrogen	5.94	1	$\frac{5.94}{1}$= 5.94	$\frac{5.94}{1.47}$= 4
Oxygen	23.52	16	$\frac{23.52}{16}$= 1.47	$\frac{1.47}{1.47}$=1

Therefore the empirical formula of the given hydrocarbon =C₄H₄O.

• The empirical formula mass= (4×12) + (4×1) + 16 = 48+4+16 =68g.
• The given molecular weight of the compound =136g.
• Molecular formula = Empirical formula × n

Therefore n= $\frac{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{compound}}{\mathrm{Empirical}\mathrm{formula}\mathrm{mass}}$

=$\frac{136}{68}$=2

• Molecular formula of the compound =(C₄H₄O)× 2 = C₈H₈O₂
• Hence, the Empirical formula is C₄H₄O, and Molecular formula is C₈H₈O₂