Question

The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0% yield. How many grams would be produced from 7.70 g of butanoic acid and excess ethanol?

Answer 1

Answer: The mass of ester produced would be, 7.92 grams.

Explanation : Given,

Mass of butanoic acid = 7.70 g

Molar mass of butanoic acid = 88 g/mol

Molar mass of ethyl butyrate = 116 g/mol

First we have to calculate the moles of butanoic acid.

$\text{Moles of butanoic acid}=\frac{\text{Given mass butanoic acid}}{\text{Molar mass butanoic acid}}$

$\text{Moles of butanoic acid}=\frac{7.70g}{88g/mol}=0.0875mol$

Now we have to calculate the moles of $C_6H_{12}O_2$

The balanced chemical equation is:

$C_4H_8O_2+C_2H_5OH\rightarrow C_6H_{12}O_2+H_2O$

From the balanced reaction we conclude that

As, 1 mole of $C_4H_8O_2$ react to give 1 mole of $C_6H_{12}O_2$

So, 0.0875 mole of $C_4H_8O_2$ react to give 0.0875 mole of $C_6H_{12}O_2$

Now we have to calculate the mass of $C_6H_{12}O_2$

$\text{ Mass of }C_6H_{12}O_2=\text{ Moles of }C_6H_{12}O_2\times \text{ Molar mass of }C_6H_{12}O_2$

$\text{ Mass of }C_6H_{12}O_2=(0.0875moles)\times (116g/mole)=10.15g$

To calculate the amount of ester formed When the yield 78.0 % is calculated as follows:

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Theoretical yield = 10.15 g

Now put all the given values in this formula, we get:

$78.0=\frac{\text{Experimental yield}}{10.15g}\times 100$

$\text{Experimental yield}=7.92g$

Therefore, the mass of ester produced would be, 7.92 grams.

Answer 2

The grams that would be produced  from 7.70 g of butanoic acid and excess ethanol is 7.923grams

calculation

Step 1: write the chemical equation for the reaction

CH3CH2CH2COOH + CH3CH2OH →  CH3CH2CH2COOCH2CH3  +H2O

step 2: find the moles of butanoic acid

moles= mass/ molar mass

=  7.70 g/ 88 g/mol=0.0875 moles

Step 3:  use the mole ratio to determine the moles of ethyl butyrate

moles ratio of CH3CH2CH2COOH :CH3CH2CH2COOCH2CH3  is 1:1 therefore the moles of CH3CH2CH2COOCH2CH3 = 0.0875  x78/100=0.0683moles

step 4: find mass = moles x molar mass

 =  0.0683 moles  x116 g/mol=7.923grams