Question

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.

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.

The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0% yield. How many grams would be produced from 7.10 g of butanoic acid and excess ethanol?
Express your answer in grams to three significant figures.

Answer 1

Answer on the question #54711 – Chemistry – General chemistry

Question:

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.

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.

The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0% yield.

How many grams would be produced from 7.10 g of butanoic acid and excess ethanol?

Express your answer in grams to three significant figures.

Answer:

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.

Solution:

The reaction equation is:

$$\mathrm{C_4H_8O_2} + \mathrm{C_2H_5OH} = \mathrm{C_6H_{12}O_2} + \mathrm{H_2O}$$

One can note that the number of the moles of butanoic acid and ethyl butyrate is equal in the reaction. That means:

$$n(C_4H_8O_2) = n(C_6H_{12}O_2)$$ $$\frac{m(C_4H_8O_2)}{M(C_4H_8O_2)} = \frac{m(C_6H_{12}O_2)}{M(C_6H_{12}O_2)}; \quad m(C_6H_{12}O_2) = \frac{m(C_4H_8O_2)}{M(C_4H_8O_2)} * M(C_6H_{12}O_2)$$

Then, assuming 100% yield, the mass of ethyl butyrate produced is:

$$m(C_6H_{12}O_2) = \frac{7.10}{88.11} * 116.16 = 9.36 \, g$$

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.

Solution:

The percent yield is:

$$\eta = \frac{m(\text{exper.})}{m(\text{theoretical})} * 100\% = \frac{5.50}{9.36} * 100\% = 58.8\%$$

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 7.10 g of butanoic acid and excess ethanol?

Express your answer in grams to three significant figures.

Solution:

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

$$m = \eta * m(\text{theor}) = 0.78 * 9.36 = 7.30\,g$$

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