Question

± Manipulating Theoretical, Actual, and Percent Yields Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly...

± 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)

Given 7.95 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.

A chemist ran the reaction and obtained 5.30 g  of ethyl butyrate. What was the percent yield?

Express your answer as a percent to three significant figures.

The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0% yield. How many grams would be produced from 7.95 gof butanoic acid and excess ethanol?

Homework Answers

Answer #1

CH3CH2CH2COOH + CH3CH2OH CH3CH2CH2COOCH2CH3 + M. Wt.: 88.11 g/mol 116.16 g/mol

From one mole of acid we get one mole of Ester is 100% yield

From 88.11 g of acid 116.16 g of Ester is the 100% yield, and also called as theoretical yield.

1. From 7.95 g of acid how much Ester will be produced as 100% yield

From 88.11g of acid 116.16 g

From 7.95 g =?

= 116.16 / 88.11 x 7.95 = 10.48 g

So that 100% yield = 10.5 g

2. If we get 5.30 g of product what is the yield?

From 10.5 g is the 100 %yield

Then 5.30 g is how % yield?

= 5.30/10.5 x 100 = 50.5 % yield

3. If the yield is 78% then what is the weight =?

10.5 g is 100 % then 78% is how much?

= 10.5/100 x 78 = 8.19 g of Ester will be produced.

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