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# Percentage yeild watch

1. 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

thank you.
2. 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.
3. 1
(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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