4. A student wants to perform the Fischer Esterification of pentanoic acid with

Question

4. A student wants to perform the Fischer Esterification of pentanoic acid with 2-butanol. For this reaction he/she wants to use a 3 molar excess of the alcohol. The student started the reaction with with 4.25 mL of the alcohol. What is the volume (mL) of carboxylic acid needed for this reaction? Show all steps of your calculations. Include all units and any references

Some answers I've seen are close to 1.69mL

while others are solved like :You can see, you need 1 mol of carboxylic acid per mol of alcohol. If we assume an alcohol solution 1M, then we have: n_{alcohol}=4.25*1=4.25mmol But the problems stated we need 3 molar excess of the alcohol, so: n_{alcohol}^o=4.25-3=1.25mmol From the reaction, we know the ratio is 1:1, so: n_{acid}^o=1.25mmol Again, we assume an carboxylic acid solution 1M: V=1.25mmol*1M=1.25ml

and then I saw

(Volume of 2-butanol) * (density of 2-butanol) * (1 mol of carboxylic acid / 3 mol of excess alcohol) x (molar mass of pentanoic acid / 1 mol of pentanoic acid) / (density of pentanoic acid)

(4.25 mL) x (0.808 g / 1 mL) x (1 mol / 3 mol) x (102.13 g / 1 mol) x ( 1 mL / .930 g) = 126 mL

Which is correct?

Explanation / Answer

Second part is partly correct.

(Volume of 2-butanol) * (density of 2-butanol) * (1 mol of carboxylic acid / 3 mol of excess alcohol) x (molar mass of pentanoic acid / 1 mol of pentanoic acid) / (density of pentanoic acid)

Here we need to divide the answer with molar mass of 2-butanol which is 74.1 g/mol.

Also 1 mol of carboxylic acid / 3 mol of excess alcohol = 1 mol / 4 mol

Then we will get correct answer.

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