When 12 g of methanol reacted with 10 g of acetic acid, an ester weighing 10.5 g

When 12 g of methanol reacted with 10 g of acetic acid, an ester weighing 10.5 g was formed. Determine the ether yield (you should get 83.3%)

Let’s implement the solution:

In accordance with the condition of the problem, we write down the equation of the process:
CH3OH + CH3COOH = CH3COO – CH3 + H2O – esterification, methyl acetate was formed.

Calculations:
M (CH3OH) = 32 g / mol;

M (CH3COOH) = 60 g / mol;

M (ether) = 74 g / mol;

Y (CH3OH) = m / M = 12/32 = 0.375 mol (substance in excess); Y (CH3COOH) = m / M = 10/60 = 0.16 mol (deficient substance).

We will make calculations for the substance in deficiency.

We find the ether exit:
m (ether) = Y * M = 0.16 * 74 = 11.8 g (theoretical weight);

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

W = 10.5 / 11.8 * 100 = 88.98%

Answer: the yield of methyl acetate was 88.98%