Derive the molecular formula of alcohol, the reaction of 32 g of which with an excess of sodium

| September 3, 2021 | education

Derive the molecular formula of alcohol, the reaction of 32 g of which with an excess of sodium releases 11.2 hydrogen.

Unless otherwise stated, let us assume that the alcohol is monohydric.

The equation for the reaction of alcohol with sodium:

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

Amount of substance H2:

v (H2) = V (H2) / Vm = 11.2 / 22.4 = 0.5 (mol).

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

v (ROH) = v (H2) * 2 = 0.5 * 2 = 1 (mol).

Thus, the molar mass of alcohol ROH:

M (ROH) = m (ROH) / v (ROH) = 32/1 = 32 (g / mol).

Let’s find the molar mass of the unknown hydrocarbon radical:

M (R) = M (ROH) – M (OH) = 32 – 17 = 15 (g / mol), which corresponds to the methyl radical (-CH3), whence the alcohol formula: CH3OH (methanol).



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