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 7.60 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 5.75 g of ethyl butyrate. What was the percent yield? Express your answer as a percent to three significant figures.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 7.60 g of butanoic acid and excess ethanol?

Express your answer in grams to three significant figures.

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 7.60 g of butanoic acid and excess ethanol?
Express your answer in grams to three significant figures.

Answer 1

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Step-by-step explanation:

`CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)`

A

Moles of butanoic acid =
`(7.60 g)/(88 g/mol)=0.08636 mol`

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

`(1)/(1)* 0.08636 mol=0.08636 mol` of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

`Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

`100\%=\frac{\text{Experimental yield}}{10.0 g}* 100`

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

`Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

`=(5.75 g)/(10.0 g)* 100=57.5\%`

57.5% was the percent yield.

C

Moles of butanoic acid =
`(7.60 g)/(88 g/mol)=0.08636 mol`

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

`(1)/(1)* 0.08636 mol=0.08636 mol` of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

`Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

`78.0\%=\frac{\text{Experimental yield}}{10.0 g}* 100`

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.