Chemistry, 08.11.2019 22:31 dbenjamintheflash5

Achemist ran a reaction and obtained 5.50 g of ethyl butyrate. what was the percent yield?

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Answer from: charles8527

a. The amount of ethyl butyrate that would be synthesized from this chemical reaction is 9.361 grams.

b. The percentage yield of 5.50 grams of ethyl butyrate is 58.75%.

Given the following data:

Mass of ethyl butyrate = 5.50 gramsMass of butanoic acid = 7.10 grams

First of all, we would write a properly balanced chemical equation for this chemical reaction between butanoic acid and ethanol.

$C_4H_8O_2 + C_2H_5OH --- C_6H_{12}O_2 + H_2O$

Atomic (molecular) weight of butanoic acid = 88 g/molAtomic (molecular) weight of ethyl butyrate = 116 g/mol

Since, we are assuming a complete 100% yield, all of the 7.10 grams of butanoic acid are used up in the chemical reaction.

So, we would determine the number of moles of butanoic acid used:

$Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{7.10}{88}$

Number of moles = 0.0807 moles

Next, we would determine the amount of ethyl butyrate that would be synthesized from this chemical reaction:

1 mole of butanoic acid = 1 mole of ethyl butyrate

0.0807 mole of butanoic acid = 0.0807 mole of ethyl butyrate

$Mass = number\;of\;moles \times molar\;mass\\\\Mass = 0.0807 \times 116$

Mass = 9.361 grams

Part B:

To find the percentage yield of 5.50 grams of ethyl butyrate:

$Percent\;yield = \frac{actual\;yield}{theoretical\;yield} \times 100\\\\Percent\;yield = \frac{5.50}{9.361} \times 100\\\\Percent\;yield = \frac{550}{9.361}$

Percent yield = 58.75%

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Complete Question:

A. Given 7.10 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.

B. A chemist ran the reaction and obtained 5.50 g of ethyl butyrate. What was the percent yield? Express your answer as a percent to three significant figures

Answer from: littlemcj

9.359 g of ethyl butyrate

% yield = 58.766 %

Explanation:

The reaction is:

C4H8O2 + C2H5OH -> C6H12O2 + H2O

Molecular weight of butanoic acid: 88 g/mole

Molecular weight of ethyl butyrate: 116 g/mole

Assuming 100% yield, all 7.10 g of butanoic acid reacts. In moles are:

(7.10 g) / (88 g/mole) = 0.0806 moles of butanoic acid

From the balanced equation we know that 1 mole of butanoic acid produce 1 mole of ethyl butyrate, then 0.0806 moles of ethyl butyrate are produced. In grams are:

0.0806 moles * 116 g/mole = 9.359 g of ethyl butyrate (this represents the theoretical yield)

If 5.50 g of ethyl butyrate are produced, then the percent yield is:

% yield = (actual yield /theoretical yield) * 100

% yield = (5.50 g/9.359 g) * 100

% yield = 58.766 %

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