Question

Given 8.50 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

Answer 1

Amount of ethyl butyrate formed = 11.3 g

Step-by-step explanation:

The reaction between butanoic acid (C3H7COOH) and excess ethanol (C2H5OH) is:

`C3H7COOH + C2H5OH \rightarrow C3H7COOC2H5 + H2O`

Since ethanol is the excess reagent, the formation of the product i.e. ethyl butyrate is influenced by the amount of butanoic acid present

Based on the reaction stoichiometry:

1 mole of butanoic acid produces 1 mole of ethyl butyrate

Mass of butanoic acid = 8.50 g

Molar mass of butanoic acid = 88 g/mol

`Moles\ of \ butanoic\  acid = (Mass)/(Molar\ mass) =(8.50g)/(88g/mol)=0.097 moles`

Moles of ethyl butyrate = 0.097

Molar mass of ethyl butyrate= 116 g/mol

`Mass\ of\ ethyl\  butyrate= 0.097*116 = 11.3 g`

Answer 2

Ethyl butyrate is also more commonly known as the ethyl butanoate with chemical formula of C₆H₁₂O₂. The equation of the butanoic acid (C₄H₈O₂) and ethanol (C₂H₆O) to form this substance can be written off as,
C₄H₈O₂ + C₂H₆O --> C₆H₁₂O₂ + H₂O
Note that the given equation is already balanced. Using dimensional analysis and conversion factors,
(8.5 g C₄H₈O₂)(1 mol/88.11 g)(1 mol C₆H₁₂O₂/1 mol C₄H₈O₂)(116.16 g / mol)
= 112.06 grams of ethyl butyrate

ANSWER: 112. 06 grams