When 4.8 g of methanol was reacted with 7.25 g of acetic acid, 7.4 g of ester was obtained.

When 4.8 g of methanol was reacted with 7.25 g of acetic acid, 7.4 g of ester was obtained. Calculate the mass fraction of the ether output.

Let’s write the reaction equation:
CH3OH + CH3COOH = CH3COOCH3 + H2O.
We have given the amount of two starting substances, 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 acetic acid is required for the reaction:
As you can see from the reaction, the amount of substance:
ν (CH3OH) = ν (CH3COOH).
Knowing that ν (in-va) = m (in-va) / M (in-va), we get:
m (CH3OH) / M (CH3OH) = m (CH3COOH) / M (CH3COOH).
Let’s define the molar masses:
M (CH3OH) = 12 + 1 * 3 + 16 + 1 = 32 g / mol.
M (CH3COOH) = 12 + 1 * 3 + 12 + 16 + 16 + 1 = 60 g / mol.
M (CH3COOHCH3) = 12 + 1 * 3 + 12 + 16 + 16 + 12 + 1 * 3 = 74 g / mol.
Let us express the mass of the acid:
m (CH3COOH) = m (CH3OH) * M (CH3COOH) / M (CH3OH).
Substituting the numerical values, we get:
m (CH3COOH) = m (CH3OH) * M (CH3COOH) / M (CH3OH) = 4.8 * 60/32 = 9 g.
We do not have enough acid to react with all the methanol, so we will calculate by acid.
As you can see from the reaction, the amount of substance:
ν (CH3COOCH3) = ν (CH3COOH).
m (CH3COOCH3) / M (CH3COOCH3) = m (CH3COOH) / M (CH3COOH).
Let’s express the mass of the ether:
m (CH3COOCH3) = m (CH3COOH) * M (CH3COOCH3) / M (CH3COOH).
Substituting the numerical values, we get:
m (CH3COOCH3) = m (CH3COOH) * M (CH3COOCH3) / M (CH3COOH) = 7.25 * 74/60 = 8.94 g.
Theoretically, 8.94 g of ether will be released during the reaction, we find the reaction yield:
w = (m pr / m theory) * 100% = (7.25 / 8.94) * 100% = 82.8%
Answer: the ether yield is 82.8%.