1.

In an organic compound, 40% carbon, 6.66% hydrogen and rest oxygen is present. Its vapour destiny is 30. Calculate the empirical formula and molecular formula.

Answer»

Percentage of oxygen = 100 - (40 + 6.66) = 100 - 46.6 = 53.4%

Element%Atomic massRelative ratio of atomsSimplest ratio
C401240/12 = 3.33.3/3.3 = 1
H6.661.0086.61/1.008 = 6.56.5/3.3 = 2
O53.41653.4/16 = 3.33.3/3.3 = 1

Hence, empirical formula = CH2O
Now, empirical formula mass
= 12 + (2 × 1) + 16 = 12 + 2 + 16 = 30″
Molecular mass = 2 x Vapour density = 2 × 30 = 60
n = Molecular mass / Empirical formula mass
= \(\frac{60}{30}\) = 2
Thus, Molecular formula = (Empirical formula) × n = (CH2O) × 2 = C2H4O2


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