Percent Yield Problem

Question

  For the balanced equation shown below, if the reaction of 48.3 grams of O₂ produces 8.31 grams of H₂O, what  

     is the percent yield?

                                                                2 C₄H₈O + 11 O₂  => 8 CO₂  + 8 H₂O

 

Answer 1

Hello, Matthew,

The balanced equation tells us the molar ratio of the product, H₂O, to the oxygen reactant, O₂, is 8/11.

We'll assume we have sufficient C₄H₈O to react with all of the O₂. Calculate the moles of oxygen in 48.3 grams by dividing by it's molar mass of 32 g/mole. That is 1.51 moles of O₂. Multiply by the molar ratio to find moles water that we should expect:

(1.51 moles O₂)*((8 moles H₂O)/(11 moles O₂)) = 1.098 moles H₂O.

Convert moles water into grams: (1.098 moles H₂O)*(18g/mole) = 19.76 grams water

19.76 grams is the theorectical yield. We actually got 8.31 grams.

Oooppsss. The percent yield is (8.31g/19.78g)*(100%) = 42.1% A lot of work for only a 42% yield of water.

Check my math. The approach should work.

Bob