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Calculate % yield in a Williamson reaction using the information provided below. Enter numeric value only (without % sign) rounded to the nearest whole percent. enter image description here

I've done ether synthesis before but not sure what this is asking...

1.1g/2.1g=52% yield?

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I think they're asking for the molar yield (moles of product obtained divided by moles of limiting reagent used). The clue is in the fact that they give you the molecular weights. Should be 44%. [BTW, what you've calculated is the mass yield, which is sometimes used in process chemistry].

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  • $\begingroup$ I guess your instructor means that you can calculate the moles of product (1.1 g /172.23 (g/mol) = 0.0064 mol) and divide them by the moles of reagent (2.1 g /144.17 (g/mol) = 0.0146 mol), which in this case are equal to the 'theoretical' moles of product, so 0.0064 mol /0.0146 mol = 44%; or you can calculate the theoretical mass you would get if the reaction had 100% yield (172.23 g/mol *0.0146 mol = 2.51 g) and divide the actual mass you obtained by that (1.1 g /2.51 g = 44%). As you can see, it's mathematically equivalent. $\endgroup$ – user6376297 Apr 2 '17 at 8:06

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