By heating methanol with a mass of 2.4 g and acetic acid with a mass of 3.6 g,

By heating methanol with a mass of 2.4 g and acetic acid with a mass of 3.6 g, methylethanate with a mass of 3.7 g was obtained. Determine the yield of ether.

To solve, we write down the equation of the process:

CH3OH + CH3COOH = CH3COO – CH3 + H2O – esterification, methylethanate was formed;
Calculations:
M (CH3OH) = 32 g / mol;

M (CH3COOH) = 60 G / mol;

Y (CH3OH) = m / M = 2.4 / 32 = 0.075 mol (substance in excess);

Y (CH3COOH) = m / M = 3.6 / 60 = 0.06 mol (deficient substance);

Y (ether) = 0.06 mol since the amount of substances is 1 mol.

Calculations are made for the substance in deficiency.

M (methylethanate) = 74 g / mol.

Find the mass and the ether output:
m (methylethanate) = Y * M = 0.06 * 74 = 4.44 g;

W = m (practical) / m (theoretical) * 100;

W = 3.7 / 4.44 * 100 = 83.3%

Answer: the yield of methylethanate is 83.3%