Derive the formula of dihydric alcohol, the interaction of 9.3 g of which with calcium yields 3.36 liters of gas.

The equation for the reaction of dihydric alcohol with calcium in general form:

R (OH) 2 + Ca = R (O) 2Ca + H2, where R is an unknown hydrocarbon radical.

Amount of substance H2:

v (H2) = V (H2) / Vm = 3.36 / 22.4 = 0.15 (mol).

According to the reaction equation, 1 mole of R (OH) 2 forms 1 mole of H2, therefore:

v (R (OH) 2) = v (H2) = 0.15 (mol).

Hence the molecular weight of the unknown dihydric alcohol:

M (R (OH) 2) = m (R (OH) 2) / M (R (OH) 2) = 9.3 / 0.15 = 62 (amu)

and the molecular weight of the hydrocarbon radical:

M (R) = M (R (OH) 2) – M ((OH) 2) = 62 – 17 * 2 = 28 (amu),

which corresponds to two methylene groups (-CH2CH2-).

Thus, the formula for a dihydric alcohol is: HO-CH2CH2-OH (ethylene glycol).