### Higher - Percentage Yield Q

This discussion has been closed.

Messages: 1 - 2 of 2
• #### Message 1.

Posted by MissChemTeacher_Bitesize (U15172316) on Sunday, 3rd February 2013

Hey Guys!

Here is a percentage yield question to test your knowledge before your prelim.

9.2g of ethanoic acid (CH₃COOH) reacts with excess methanol (CH₃OH). A yield of 7.2g of methyl ethanoate (CH₃COOCH₃) was obtained in the reaction.

Use the following equation to calculate the percentage yield of methyl ethanoate.

CH₃COOH + CH₃OH <----------> CH₃COOCH₃ + H₂O

If you need any help with this question get in touch.

Cheers Miss_Chem

• #### Message 2

, in reply to message 1.

Posted by MissChemTeacher_Bitesize (U15172316) on Wednesday, 13th February 2013

Here is the solution to the problem:

We need to use this information to calculate the number of moles of ethanoic acid then use this data to calculate the theoretical yield of methyl ethanoate produced in the reaction.

CH₃COOH + CH₃OH <----------> CH₃COOCH₃ + H₂O

n of moles of CH₃COOH = mass from Q / gram formula mass
n of moles of CH₃COOH = 9.2 / 60 = 0.153 moles

This can then be plugged into your mole ratio:
Balanced Equation: CH₃COOH + CH₃OH <----------> CH₃COOCH₃ + H₂O
Mole ratio:…………………1………………………………………………1
n of moles from Q:……0.153…………………………………………0.153

Theoretical mass of methyl ethanoate: n of moles x gram formula mass
Theoretical mass of methyl ethanoate: 0.153 x 74 = 11.3g

In theory, in this reaction we should have produced 11.3g of methyl ethanoate, but in reality we actually only produced 7.2g. We can use this information to calculate our percentage yield.

Percentage Yield = actual yield / theoretical yield x 100

Percentage Yield = 7.2 / 11.3 x 100 = 63.7%

If you need help with any percentage yield questions get in touch.
Cheers Miss_Chem

The Bitesize messageboards have now closed

or register to take part in a discussion.

The message board is currently closed for posting.

The boards are now closed

This messageboard is pre-moderated.