Calculate the mass of the ester, which is formed by the interaction of acetic acid with a mass of 9 g with pentan-1-ol

Calculate the mass of the ester, which is formed by the interaction of acetic acid with a mass of 9 g with pentan-1-ol with a mass of 7.04 g

Let’s write the reaction equation:
CH3COOH + C5H11OH = CH3COOC5H11 + H2O.
We have given the amount of two starting materials, one of them may be in excess and will not fully react, we will determine that in excess, for this, we will find how much acid is required for the reaction:
As you can see from the reaction, the amount of substance:
ν (CH3COOH) = ν (C5H11OH).
Knowing that ν (in-va) = m (in-va) / M (in-va), we get:
m (CH3COOH) / M (CH3COOH) = m (C5H11OH) / M (C5H11OH).
Let’s define the molar masses:
M (CH3COOH) = 12 + 1 * 3 + 12 + 16 + 16 + 1 = 60 g / mol.
M (C5H11OH) = 12 * 5 + 1 * 11 + 16 + 1 = 88 g / mol.
Let us express the mass of alkali:
m (CH3COOH) = m (C5H11OH) * M (CH3COOH) / M (C5H11OH).
Substituting the numerical values, we get:
m (CH3COOH) = m (C5H11OH) * M (CH3COOH) / M (C5H11OH) = 7.04 * 60/88 = 4.8 g.
We need 4.8 g of acid, and we have 9 g, which means that there is not enough alcohol, to neutralize the acid, we will carry out the calculations for alcohol.
As you can see from the reaction, the amount of substance:
ν (C5H11OH) = ν (CH3COOC5H11).
m (C5H11OH) / M (C5H11OH) = m (CH3COOC5H11) / M (CH3COOC5H11).
Determine the molar mass:
M (СН3СООС5Н11) = 12 + 1 * 3 + 12 + 16 + 16 + 12 * 5 + 1 * 11 = 130 g / mol.
Let us express the mass of the ester:
m (CH3COOC5H11) = m (C5H11OH) * M (CH3COOC5H11) / M (C5H11OH).
Substituting the numerical values, we get:
m (CH3COOC5H11) = m (C5H11OH) * M (CH3COOC5H11) / M (C5H11OH) = 7.04 * 130/88 = 10.4 g.
Answer: 10.4 g of ester is formed.