I\'m having trouble calculating the theoretical yield of a reactionfrom O-chem l

Question

I'm having trouble calculating the theoretical yield of a reactionfrom O-chem lab. The process involved dissolving 62mg ofpotassium cyanide in 0.60 mL of water and then adding 1.2 mL ofethanol and 628mg (0.6mL) of Benzaldehyde.

As I understand it this reaction is simply the benzaldehydecatalyzed by the cyanide to produce benzoin. So I'm trying tofigure out how much benzoin the theoretical yield should be.   Our experimental result produced 157mg.

I have the MW of benzaldehyde to be 106.12g/mol and we started with0.628 grams which makes 0.0059 moles, and the MW of benzoin is212.24g/mol, but I can't seem to bridge the gap between these twoto find the theoretical yield. Thanks for the help!

Explanation / Answer

Check to be sure that benzaldehyde is the limiting reactant. If so,assuming the reaction is 1:1, then the number of moles of startingmaterial is equal to the number of moles of product. So 0.0059 molbenzaldehyde should produce 0.0059 mol benzoin. 0.0059 mol benzoin x 212.24 g = 1.26 g benzoin (this is yourtheoretical yield)                                      mol So to find percent yield, 0.157 g x 100 = 12.5% 1.26 g

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