# On alkaline hydrolysis of 6 g of a certain ester, 6.8 g of the sodium salt of the saturated monobasic acid and 3.2 g of alcohol

On alkaline hydrolysis of 6 g of a certain ester, 6.8 g of the sodium salt of the saturated monobasic acid and 3.2 g of alcohol were obtained. Establish the molecular formula of the ester. Solution: CnH2n + 1COOCmH2m + 1 + NaOH → CnH2n + 1COONa + CmH2m + 1OH
According to the law of conservation of mass:
6 + m (NaOH) = 6.8 + 3.2
⇒ m (NaOH) = 10 – 6 = 4 g
n (NaOH) = 4/40 = 0.1 mol
⇒ n (ether) = n (alcohol) = n (salt) = n (NaOH) = 0.1 mol
CnH2n + 1COONa: 0.1 = 6.8 / (14n + 1 + 12 + 32 + 23)
⇔ 6.8 = 1.4n + 6.8
⇔ 1,4n = 0
⇒ HCOONa – sodium formate
CmH2m + 1OH: 0.1 = 3.2 / (14m + 18)
⇔ 1.4m = 1.4
⇔ m = 1 ⇒ CH3OH – methyl alcohol.
This means that the methyl ester of formic acid entered the reaction.
Answer: The molecular formula of the ester is HCOOCH3. Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.