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.

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