# Stoichiometric Calculations - Percentage yield

Submitted by Carmina on Tue, 12/15/2009 - 23:04

I need help with this problem. I missed this lesson because I was sick and now I can't figure it out.

Here it goes:

Calculate the % yield if 4 moles of ethylene are reacted with excess water to produce 3.8 moles of ethanol
CH2CH2 + H2O  CH3CH2OH

Grateful for any help with this since I have four more of those problems to solve.

The yield you have is called the experimental yield (or actual yield). This is 3.8 moles
You calculate the theoretical (or expected) yield from the balanced equation using the quotients (numbers in front of the formulae) as the number of moles.

If any chemical is in excess, this will not affect the yield. However, the other chemical will be limiting and all will be used up. This will determine the amount of product.

CH2CH2    +    H2O  ------->  CH3CH2OH
1 mol                                      1 mol
4 mol                                      4 mol

So you would expect to get 4 mol of ethanol from 4 moles of ethylene but you actually only got 3.8 mole

Percentage yield = (experimental yield/theoretical yield) x 100 = (3.8/4.0) x 100

If the question had been - you get 15.70 g of ethanol from 10.00 g of ethylene, you would have to convert moles into masses using the molar mass eg ethylene has an approx molar mass of 28g/mole while ethanol is 46g/mol

CH2CH2    +    H2O  ------->  CH3CH2OH
1 mol                                      1 mol
28 g                                        46 g
10.00 g                                (46/28) x 10.00 = 16.43 g (theoretical yield)

Percentage yield = (15.70/16.43) x 100 = 95.57 % yield

Thank you so much for your help. I was able to figure out the other problems with your guidance.

I am very grateful!!!!!!

Carmina

Carmina wrote:

I need help with this problem. I missed this lesson because I was sick and now I can't figure it out.

Here it goes:

Calculate the % yield if 4 moles of ethylene are reacted with excess water to produce 3.8 moles of ethanol
CH2CH2 + H2O  CH3CH2OH

Grateful for any help with this since I have four more of those problems to solve.

:D