Question

Given 7.15g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

Answer #1

Given the mass of butanoic acid(CH3CH2CH2COOH) = 7.15 g

Molecular mass of butanoic acid(CH3CH2CH2COOH) = 88.1 g/mol

Hence moles of butanoic acid(CH3CH2CH2COOH) = mass/molecular mass = 7.15 g / 88.1 g/mol = 0.08116 mol

When butanoic acid reacts with excess ethanol they form ethyl butyrate(CH3CH2CH2COOCH2CH3 ).

The balanced chemical reaction is

CH3CH2CH2COOH + CH3CH2OH ------- > CH3CH2CH2COOCH2CH3

1 mol, --------------------- 1 mol, -------------------- 1 mol

From the above balanced chemical reaction it is clear that 1 mole of butanoic acid produces 1 mole of ethyl butyrate.

Hence 0.08116 mol of butanoic acid will also form 0.08116 mol of ethyl butyrate.

Hence moles of ethyl butyrate formed if we consider 100% yield = 0.08116 mol

molecular mass of ethyl butyrate(CH3CH2CH2COOCH2CH3) =

Hence mass of ethyl butyrate(CH3CH2CH2COOCH2CH3) formed = 0.08116 mol x 116.16 g/mol

= 9.43 g (answer)

± Manipulating Theoretical, Actual, and Percent Yields
Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit
flavor commonly used in the food industry for such flavors as
orange and pineapple. Its fragrance and taste are often associated
with fresh orange juice, and thus it is most commonly used as
orange flavoring.
It can be produced by the reaction of butanoic acid with ethanol
in the presence of an acid catalyst (H+):
CH3CH2CH2CO2H(l)+CH2CH3OH(l)H+⟶CH3CH2CH2CO2CH2CH3(l)+H2O(l)
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