k procedure You may wse anything that is wvailahãe in yow eqament es s that you

Question

k procedure You may wse anything that is wvailahãe in yow eqament es s that you have wsed this semester or anything you ean getto calculatethe%yield to look up any information, it is all here. Workug Reaction time é 16 hours at reflux O viscous HO-S-OH len Sodium sulfate O liquid dodecanol Ci2H2OH sta Ci7H3CO,H Amount used-100a stearic acid swlfuric acid (catalyst) colorless white dodecyl carat Cpae whie FW -4528 als FW= 186.33 d 0.833 FW= 284.48 (* dodecanol is MP 68°CB MP = 68 °C , I OTH a Literature: MP-410°C the Experimental: MP-40 BP- 261 °Ca CA mount useoK 6.00 mt f 375mg Amount used reactant and the solvent is over, extractions, purifications, tration etc

Explanation / Answer

It's impossible to put 6mL of any solution in the 5mL vial.

But whatever I can see from the question is that you have lost 3.5 marks and I found it's because of your low yield. Because you can see the mass of dodecanol = density x volume

mass of dodecanol = 0.833 x 6.00 = 4.998g

mass of stearic acid =375mg= 0.375g

so the total mass of reactant =4.998+ 0.375 = 5.373g

but the amount of product reported here is only 0.492g

so the % yield = (0.492/5.373)x100 = 9.16% (if 100% of dodecanol is used in reaction). suppose if the used dodecanol is 50% then also the %yield will be very less nearly we will get 20% yield in that case.

Since the % yield of your product is low so you have lost 3.5 marks (I think so).

And one more thing is that since dodecanol is acting as both reactants and as a solvent so it's difficult to find the exact amount of dodecanol used in the reaction so that we can find the exact % yield of the reaction.

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