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Calculate the percentage yield if 5.0 g of ethanol is oxidised to produce 4.8 g of ethanoic acid.


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5.0g of ethanol is 5.0/46 = 0.11 mol of ethanol. The conversion of ethanol to ethanoic acid is stoichiometrically 1:1, so you can therefore produce 0.11 mol of ethanoic acid.

4.8g of ethanoic acid is 4.8/60 = 0.080 mol of ethanoic acid.

Therefore, your yield is actual/ideal = 0.080 / 0.11 = 73%


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In the textbook the formula for yield is the actual mass / the theoretical mass x 100% In this case you've divided actual mol with the theoretical mol, is this because you could only compare using moles only? not mass?