# What ester of a monobasic organic acid was taken, if the complete hydrolysis of 14.8 g of this ester required

What ester of a monobasic organic acid was taken, if the complete hydrolysis of 14.8 g of this ester required 45.5 ml of a 20% solution of nastrium hydroxide with a density of this solution of 1.2 g / ml, if the alkali was taken with a 25% excess.

Given:
monobasic organic acid ester
m (ether) = 14.8 g
V solution (NaOH) = 45.5 ml
ω (NaOH) = 20%
ρ solution (NaOH) = 1.2 g / ml

To find:
ether formula -?

Decision:
1) Esters of monobasic organic acids have the general formula:
C (n) H (2n) O2
2) C (n) H (2n) O2 + NaOH => R-COONa + R’-OH;
3) m solution (NaOH) = ρ solution (NaOH) * V solution (NaOH) = 1.2 * 45.5 = 54.6 g;
4) m (NaOH) = ω (NaOH) * m solution (NaOH) / 100% = 20% * 54.6 / 100% = 10.9 g;
5) n (NaOH) = m (NaOH) / M (NaOH) = 10.9 / 40 = 0.27 mol;
6) Since the alkali was taken with a 25% excess, only 75% of the total amount of NaOH reacted with ether:
n react. (NaOH) = 75% * n (NaOH) / 100% = 75% * 0.27 / 100% = 0.2 mol;
7) n (ether) = n react. (NaOH) = 0.2 mol;
8) M (ether) = m (ether) / n (ether) = 14.8 / 0.2 = 74 g / mol;
9) M (C (n) H (2n) O2) = Mr (C (n) H (2n) O2) = Ar (C) * n + Ar (H) * 2n + Ar (O) * 2 = 12 * n + 1 * 2n + 16 * 2 = (14n + 32) g / mol;
10) 14n + 34 = 74;
14n = 42;
n = 3;
monobasic organic acid ester – C3H6O2 – methyl acetate or ethyl formate.

Answer: Unknown monobasic organic acid ester – C3H6O2 – methyl acetate or ethyl formate.

One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.