hi! In a reaction A -> 3B , with 3* (molar mass of B) > molar mass of A, it is expected that with increasing conversion rate, also increase the yield of the process (yield = mass of product / mass of reagent). However, in my experiment this is going the other way: the higher the conversion rate the lower the yield. Could someone give me a thought about some reason why this is happening?
The yield of product formation is Mass of product/mass of reagent. If the complete conversion of reactant to product without any other by product formation you can expect 100% Yeild. Otherwise the reactant is completely converting but the same time either your formed product is may be unstable for long time or may be some by poduct formation you may not observed from reaction i guess.
Sandeep Chandrashekharappa actually by calculating yield from a relation of mass, it will not be a % , therefore it can be greater than 1. In this case it is because 3* (molar mass of B) > molar mass of A. However the more there is conversion, the lower the yield
Yes, Fernando Nakamoto You are correct, in a single component a (reactant) will converted to B (molar mass of B), higher the conversion lower the yield. Theoretically if your 100% reactant a convert to B your maximum yeild will be 33.33%.
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