Question

0.30 gm of an organic compound containing C, H & O on combustion gave 0.44 gm CO2 & 0.18 gm H2O. Determine the empirical formulae of compound.

Answer 1

Explanation:

Moles of C = moles of CO2 = 0.44/44 == 0.01 moles

mole of Hydrogen = 2 × moles of H2O = 2× 0.18/18 = 0.02 mole

Weight of oxygen = ( 0.30 -- Wt. of carbon and hydrogen) = (0.3 - 0.01×12--0.02×1) == 0.16 gm

moles of oxygen = 0.16/6 = 0.01 Mole

Empirical Formula = CH2O

Empirical Formula mass = 30 gm

n = 60/30 == 2

Formula = ( CH2O) n == C2H4O2

I hope it is correct