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    An excess of methanol was mixed with 12 g of ethanoic acid and an acid catalyst. At equilibrium the mixture contained 8 g of methyl ethanoate. The percentage yield of ester present was

    The answer is 54%, I can't see how this answer comes about, i get 67%??

    thank you.
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    8/12 * 100% = 67% is not correct becase 8 g and 12 g are not the experimental ('collected') and theoretical ('expected') masses of the same species. They are two experimental values for different compounds that differ in molar mass.

    What you can do is use chemical amounts (moles) and the reaction stoichiometry:
    CH3OH + CH3COOH → CH3COOCH3 + H2O

    12 g of ethanoic acid = 0.200 mol
    So 0.200 mol of ethanoate are expected (1:1 ratio).

    8 g of methyl ethanoate = 0.108 mol
    Which is only 54.05% of the expected amount.
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    (Original post by zoae)
    8/12 * 100% = 67% is not correct becase 8 g and 12 g are not the experimental ('collected' and theoretical ('expected' masses of the same species. They are two experimental values for different compounds that differ in molar mass.

    What you can do is use chemical amounts (moles) and the reaction stoichiometry:
    CH3OH + CH3COOH → CH3COOCH3 + H2O

    12 g of ethanoic acid = 0.200 mol
    So 0.200 mol of ethanoate are expected (1:1 ratio).

    8 g of methyl ethanoate = 0.108 mol
    Which is only 54.05% of the expected amount.
    Of course, I see now. I thought the expected was 12, but thats mass. I see how u do it now. Thank you! 😛
 
 
 
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