# Calculate the mass of the ether that is formed by the interaction of absolute ethyl alcohol with a mass of 18.4

**Calculate the mass of the ether that is formed by the interaction of absolute ethyl alcohol with a mass of 18.4 with glacial acetic acid with a mass of 18, take the ether yield equal to 70 of the theoretically possible.**

1.Let’s find the amount of acetic acid substance by the formula:

n = m: M.

M (CH3COOH) = 60 g / mol.

n = 18 g: 60 g / mol = 0.3 mol.

2.Let’s find the amount of the substance of ethyl alcohol.

M (C2H5OH) = 46 g / mol.

n = 18.4 g: 46 g / mol = 0.4 mol.

Acetic acid is deficient. We solve the problem by lack (by acid).

3. Let’s compose the reaction equation, find the quantitative ratios of substances.

С2Н5ОН + CH3 – COОH → CH3 – COО – С2Н5 + H2O.

According to the reaction equation, there is 1 mol of ethyl acetate per 1 mol of acid. Substances are in quantitative ratios 1: 1.

The amount of substance will be the same.

n (CH3COOH) = n (CH3 – CO O – C2H5) = 0.3 mol.

4.Let’s find the mass of ethyl acetate by the formula:

m = n × M,

M (CH3 – COO – C2H5) = 88 g / mol.

m = 0.3 mol × 88 g / mol = 26.4 g.

5. 26.4 g were calculated (theoretical yield).

According to the condition of the problem, the ether received 70%.

Let’s find a practical way out of the product.

26.4 g – 100%

x – 70%,

x = (26.4 × 70%): 100 g = 18.48 g.

Answer: 18.48 g.