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Higher - Percentage Yield Q

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  • 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 smiley - cool

    Report message1

  • 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 smiley - cool

    Report message2

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