As a result of dehydration of a monoatomic saturated alcohol weighing 370 g, water with a mass of 90 g

As a result of dehydration of a monoatomic saturated alcohol weighing 370 g, water with a mass of 90 g was released. Alcohol formula: a) CH3OH; b) C2H5OH; c) C3H7OH; d) C4H9OH.

Given:
m (alcohol) = 370 g
m (H2O) = 90 g

To find:
alcohol formula -?

Solution:
1) R-CH2-OH = (H2SO4, 180C) => R = CH2 + H2O;
2) n (H2O) = m (H2O) / M (H2O) = 90/18 = 5 mol;
3) n (alcohol) = n (H2O) = 5 mol;
4) Mr (alcohol) = M (alcohol) = m (alcohol) / n (alcohol) = 370/5 = 74;
5) The general formula of saturated monohydric alcohols is C (n) H (2n + 1) OH;
6) Mr (alcohol) = Ar (C) * n + Ar (H) * (2n + 1) + Ar (O) + Ar (H) = 12 * n + 1 * (2n + 1) + 16 + 1 = 14n + 18;
7) 14n + 18 = 74;
14n = 56;
n = 4;
C4H9OH – butanol.

Answer: Unknown alcohol – C4H9OH – butanol