How much acetic acid (by weight) is required to obtain 139.05 g of aminoacetic acid ethyl ester at 90% yield?

| August 4, 2021 | education

Aminoacetic acid ethyl ester formula: CH2 (NH2) – COOC2H5

Acetic acid formula – CH3COOH

Aminoacetic acid formula – CH2 (NH2) – COOH

η is the mass fraction of the practical yield of the product from the theoretically possible one.

Let’s write down given:

m (CH2 (NH2) – СООС2Н5) = 139.05 g

η (СН2 (NH2) – СООС2Н5) = 90% or 0.9

Find: m (CH3 COOH) -?

Solution

1) Find the mass of the theoretical yield of ethyl ester of aminoacetic acid:

η = m (practical output) * 100% / m (theoretical output),

m (theoretical output) = m (practical output) * 100% / η

m (theoretical output СН2 (NH2) – СООС2Н5) = 139.05 g * 100% / 90%

m (theoretical output СН2 (NH2) – СООС2Н5) = 154.5 g

2) Calculate the mass of aminoacetic acid, which is necessary to obtain 154.5 g of ether:

Let’s write the reaction equation:

CH2 (NH2) – COOH + C2H5OH → CH2 (NH2) – COOC2H5 + H2O

M (CH2 (NH2) – COOH) = 12 * 2 + 14 + 16 * 2 + 5 = 75 g / mol

М (СН2 (NH2) – СООС2Н5) = 12 * 4 + 16 * 2 + 14 + 9 = 103 g / mol

Let’s compose the ratio:

75 g (CH2 (NH2) – COOH) forms 103 g (CH2 (NH2) – COOC2H5)

X g (CH2 (NH2) – COOH) forms 154.5 g (CH2 (NH2) – COOC2H5)

X = 75g * 154.5g / 103g

X = 112.5 g (CH2 (NH2) – COOH)

3) Calculate the mass of acetic acid:

To obtain 1 mol (CH2 (NH2) – COOH), 1 mol (CH3 COOH) is needed, hence

M (CH3 COOH) = 12 * 2 + 16 * 2 + 4 = 60 g / mol

60 g (CH3 COOH) – 75 g (CH2 (NH2) – COOH)

Y g (CH3 COOH) – 112.5 g (CH2 (NH2) – COOH)

Y = 60g * 112.5g / 75g

Y = 90 g (CH3COOH)

Answer: 90 g (CH3COOH)



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