Question

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)

Part A Given 8.00 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.

Part B A chemist ran the reaction and obtained 5.35 g of ethyl butyrate. What was the percent yield? Express your answer as a percent to three significant figures. View Available Hint(s) percent yield = nothing % %

Part C The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0% yield. How many grams would be produced from 8.00 g of butanoic acid and excess ethanol? Express your answer in grams to three significant figures.

Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g) Suppose you wanted to dissolve an aluminum block with a mass of 15.6 g .

Part A What minimum mass of H2SO4 would you need? Express your answer in grams.

Answer 1

Part A:

CH3CH2CH2COOH + CH3CH2OH = CH3CH2CH2COOCH2CH3 + H2O

The number of the moles of butanoic acid and ethyl butyrate is equal in the reaction.

Hence moles of acid = moles of ester
moles acid = 8.00 g /88.108 g/mol=0.0907
moles ester = 0.0907
mass ester = 0.0907 mol x 116.162 g/mol=10.54 g

Part B:

% yield = {m (expt) / M(Theory)} x 100

=5.35 x 100/ 10.54 =50.8 %

Part C:

When the 78% yield is reached, the result mass of the ethyl butyrate is:

moles of ester = 0.0907 x 78/100=0.0707
mass of ester = 0.0707 x 116.162 = 8.21 g

Second answer:

Molar mass of Al = 26.982 g/mol

Molar mass of H2SO4 = 98.079 g/mol

15.6 g Al x (1 mol Al / 26.982 g Al) x (3 mol H₂SO₄ / 2 mol Al) x (98.079 g H₂SO₄ / 1 mol H₂SO₄) = 85.06 g