Derive the formula of a monohydric alcohol, if, by reacting 4.14 g of it with sodium metal, 1.008 liters

Derive the formula of a monohydric alcohol, if, by reacting 4.14 g of it with sodium metal, 1.008 liters of hydrogen were obtained.

The equation for the reaction of a monohydric alcohol with metallic sodium in general form:

2ROH + 2Na = 2RONa + H2, where R is an unknown hydrocarbon radical.

Amount of substance H2:

v (H2) = V (H2) / Vm = 1.008 / 22.4 = 0.045 (mol).

According to the reaction equation, from 2 mol of ROH, 1 mol of H2 is formed, therefore

v (ROH) = v (H2) * 2 = 0.045 * 2 = 0.09 (mol).

Molecular weight of an unknown monohydric alcohol:

M (ROH) = m (ROH) / v (ROH) = 4.14 / 0.09 = 46 (amu).

Then:

M (R) = M (ROH) – M (OH) = 46 – 17 = 29 (amu), which corresponds to the ethyl radical (-C2H5).

Thus, the formula for alcohol is: C2H5OH (ethanol).