Right, first of all, you need to get your equation right and balanced. Given that you get CO

_{2} and C

_{2}H

_{5}OH from C

_{6}H

_{12}O

_{6}, you write:

C

_{6}H

_{12}O

_{6} ----> C

_{2}H

_{5}OH + CO

_{2}
Balancing this gives:

C

_{6}H

_{12}O

_{6} ----> 2C

_{2}H

_{5}OH + 2CO

_{2}
Keep this in mind - almost all yield calculations involve converting things into moles first

:

MW

_{glucose} = 180.16 g.mol

^{-1}
So, for a given mass of glucose m

_{glucose}, the amount in moles, m

_{glucose}, is:

n

_{glucose} = (m

_{glucose}/180.16

^{-1}) mol

The theoretical amount of CO

_{2} you will get out of this is:

n

_{CO2} = 2 x n

_{glucose} (

*cf* balanced equation)

And the corresponding amount of CO

_{2} produced, in grams is simply:

m

_{CO2} = n

_{CO2} x 44.01 g.mol

^{-1}
Since you only get 14% yield, it means that the final amount of CO

_{2} is 14% of what you would expect, so you simply multiply your final yield by 0.14:

m

_{CO2} x 0.14 = m

_{CO2 Expected}
This can equally well be done before you calculated the mass, that is:

m

_{CO2 Expected} = (0.14 x n

_{CO2}) x 44.01 g.mol

^{-1}
Hope that helps!