JEE Main 2020 — Some Basic Concepts Of Chemistry Question with Solution

Let the organic compound X is = C_xH_yO_z

Here moles of C = x, moles of H = y, moles of O = z

Given, $$\frac{W_C}{W_H}=\frac{4}{1}$$

$$\therefore$$ $$\frac{x}{y}=\frac{\frac{W_C}{12}}{\frac{W_H}{1}}$$ = $$\frac{4}{1}\times \frac{1}{12}$$ = $$\frac{1}{3}$$

Also Given, $$\frac{W_C}{W_O}=\frac{3}{4}$$

$$\therefore$$ $$\frac{x}{z}=\frac{\frac{W_C}{12}}{\frac{W_O}{16}}$$ = $$\frac{3}{4}\times \frac{16}{12}$$ = 1

$$\Rightarrow$$ x = z

$$\therefore$$ Empirical formula = C_xH_3xO_x = CH₃O

As compound is saturated acyclic organic
compound, so molecular formula = C₂H₆O₂

C₂H₆O₂ + $$\frac{5}{2}$$O₂ $$\to$$ 2CO₂ + 3H₂O

For 1 mole of C₂H₆O₂ number of moles of O₂ required = $$\frac{5}{2}$$

$$\therefore$$ For 2 mole of C₂H₆O₂ number of moles
of O₂ required = $$\frac{5}{2}$$ $$\times$$ 2 = 5