Saturated monohydric alcohol weighing 15 g reacted with metallic sodium, forming 2.8 l of H2

Saturated monohydric alcohol weighing 15 g reacted with metallic sodium, forming 2.8 l of H2 Determine the spiral formula

Let us compose the equation for the interaction of the limiting monohydric alcohol with metallic sodium:
2CnH2n + 1OH + 2Na = 2CnH2n + 1ONa + H2;

Let’s find the amount of substance in hydrogen:
2.8 / 22.4 = 0.125 mol;

There is a factor 2 times greater before alcohol than before hydrogen. This means that the amount of substance in alcohol will be equal to 0.25 mol. Let’s find the molecular weight of the alcohol:
15 / 0.25 = 60 g / mol;

The relative atomic mass of carbon is 12, hydrogen is 1, oxygen is 16. To find the formula, compose the equation:
12n + 2n + 1 + 16 + 1 = 60;

14n = 60-18;

14n = 42;

n = 42/14;

n = 3;

Answer: the formula for alcohol is C3H7OH (propanol).