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Vika [28.1K]
3 years ago
14

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.
Chemistry
1 answer:
SIZIF [17.4K]3 years ago
8 0

Answer:

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.

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 = \frac{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 :

\frac{1}{1}\times 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}}\times 100

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

=\frac{5.75 g}{10.0 g}\times 100=57.5\%

57.5% was the percent yield.

C

Moles of butanoic acid = \frac{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 :

\frac{1}{1}\times 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}}\times 100

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

Experimental yield = 7.80 g

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

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<u>5 moles S x (36.02 g S/mole S) = 180.1 grams of S</u>

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The periodic table has mass units for every element that can be correlated with the number of atoms of that element.  The relationship is known as Avogadro's Number.  This number, 6.02x10^{23} , is nicknamed the mole, which scientists found to be a lot more catchy, and easier to write than  6.02x10^{23}.  <u>The mole is correlated to the atomic mass of that element.</u>  The atomic mass of sulfur, S, is 36.02 AMU, atomic mass units.  <u>But it can also be read as 36.02 grams/mole.</u>

<u></u>

<u>This means that 36.02 grams of S contains 1 mole (6.02x</u>10^{23}<u>) of S atoms</u>.

<u></u>

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5.0 moles of Sulfur would therefore contain:

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