Percent Yield Question: Chemistry

Question

So I keep getting over 100% on my percent yield and can't seem to figure out why. The following question is:

 

Determine the percent yield of carbon dioxide if 1.70 g of carbon dioxide is actually produced from the combustion of 6.83 g of C₉H₁₆O with 1.92 g of oxygen.

Answer 1

Write the balance equation for combustion of C₉H₁₆O:

2C₉H₁₆O + 25O₂ ==> 18CO₂ + 16H₂O  ... balanced equation

Find limiting reactant:

moles C₉H₁₆O = 6.83 g x 1 mole/140 g = 0.04879 moles.  If we divide this by 2 we get 0.0244 moles

moles O₂ = 1.92 g x 1 mole/32 g = 0.06 moles.  If we divide this by 25 we get 0.0024.

O₂ is limiting.

Theoretical yield of CO₂ = 0.06 moles O₂ x 18 moles CO₂/25 moles O₂ = 0.0432 moles CO₂

Theoretical mass of CO₂ = 0.0432 moles x 44 g/mole = 1.90 g

% yield = actual/theoretical (x100) = 1.70 g/1.901 (x100) = 89.5%