Given 7.75 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield? express your answer in grams to three significant figures.

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Chemistry

6 February, 04:19

Given 7.75 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield? express your answer in grams to three significant figures.

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Siena Mcpherson · Answer 1

Answer is: 9.69 grams of ethyl butyrate would be synthesized.

Chemical reaction: C₃H₇COOH + C₂H₅OH → C₃H₇COOC₂H₅ + H₂O.

m (C₃H₇COOH) = 7.35 g; mass of butanoic acid.

n (C₃H₇COOH) = m (C₃H₇COOH) : M (C₃H₇COOH).

n (C₃H₇COOH) = 7.35 g : 88.11 g/mol.

n (C₃H₇COOH) = 0.0834 mol; amount of butanoic acid.

From chemical reaction: n (C₃H₇COOH) : n (C₃H₇COOC₂H₅) = 1 : 1.

n (C₃H₇COOC₂H₅) = 0.0834 mol; amount of ethyl butyrate.

m (C₃H₇COOC₂H₅) = n (C₃H₇COOC₂H₅) · M (C₃H₇COOC₂H₅).

M (C₃H₇COOC₂H₅) = 6·Ar (C) + 2·Ar (O) + 12·Ar (H) · g/mol.

M (C₃H₇COOC₂H₅) = 6·12.01 + 2·15.99 + 12·1.01 · g/mol.

M (C₃H₇COOC₂H₅) = 116.16 g/mol; molar mass of ethyl butyrate.

m (C₃H₇COOC₂H₅) = 0.0834 mol · 116.16 g/mol.

m (C₃H₇COOC₂H₅) = 9.69 g.