[willor]

Hi, first time posting here.
I have a question about a problem concerning the yield in an organic reaction. (I posted it here and not orgo because it's in chapter 3 of my general chem book)

The problem is based off the reaction of isopentyl alcohol and acetic acid to form isopentyl acetate and water

CH3COOH + HOCH2CH2CH(CH3)2 --> CH3COOCH2CH2CH(CH3)2 + H2O

The problem is "How many grams of isopentyl alcohol are needed to make 433 g of isopentyl acetate in the reaction if the expected yield is 78.5%? Assume the acetic acid is in excess"

I've been using the equation Theoretical Yield = Actual Yield/Percent Yield X 100%

I calculated the molar mass of Isopental Acetate at 130.2g and that of Isopentyl Alcohol at 102.2g

The way I worked out the equation was first to find out the theoretical yield plugging in 433g as the actual yield and 78.5 as the percent yield, which gave me 552g Isopentyl Acetate

then I used:
 
552g Isopental Acetate X 1 mol Isopentyl Acetate/130.2g Isopentyl Acetate X 1 mol Isopentyl Alcohol/1 mol Isopentyl Acetate X 102/2g Isopentyl Alcohol/1 mol Isopentyl Alcohol = 433g Isopentyl Alcohol.

The book's answer is 374g.

I've tried reworking the problem a couple times but can't seem to figure it out. I don't know where I'm going wrong here. Any hints as to what I'm doing wrong would be greatly appreciated. Thank you for your time.

EDIT: I found out I was using an incorrect molecular formula for Isopentyl Alcohol and have found the correct way to the answer by using the right molar mass for Isopentyl Alcohol.

[Donaldson Tan]

552g Isopental Acetate X (1 mol Isopentyl Acetate/130.2g Isopentyl Acetate) X (1 mol Isopentyl Alcohol/1 mol Isopentyl Acetate) X (102/2g Isopentyl Alcohol/1 mol Isopentyl Alcohol) = 433g Isopentyl Alcohol.

You use the wrong molar mass for Isopentyl Alcohol. It should be 88.17g/mol.

This is the MSDS of Isopentyl Alcohol. The molar mass is quoted at the beginning.

[willor]

Thank you for the help. I appreciate it.