I am still confused in this theoretical yeild question?? help?
Substance --> molar mass(g/mol) --> d (g/mL)
cyclohexanol --> 100.16 ----> 0.948
cyclohexene----> 82.15----> 0.811
i thought that the there should be something to do with this formula:
theoretical yeild = actual yeild / percent yeild X 100%
Answer:
The equation should be:
%Yield = (Actual yield / Theoretical yield) * 100
For the theoretical yield 3.0 mL of cyclohexanol needs to be converted into number of moles of cyclohexanol:
(3.0 mL * 0.948 g/mL) * 100.16 g/mole
If one mole of cyclohexanol yields one mole of cyclohexene, then the number of moles calulated above times the molar mass (82.15 g/mol) of cyclohexene should give you the theoretical yield in grams.
Theoretical yield is what you expect to get based off the stoichiometry of the equation.
here is an example
CH4 + O2 --> H2O + CO2
CH4 + 2 O2 ---> 2 H2O + CO2 *balanced*
So if you have 16 grams of Methane with excess water, how much water should you get? (the theoretical yield)
Ok 1 mole of methane is 16 grams...so
1 mole CH4 * 2 mole H2O / 1 mole CH4 = 2 mole H2O.
2 moles H2O * 18 grams H2O / 1 mole H2O
so you get 36 grams of water. That is your theoretical yield.
If you actually did this experiment and weighed your water..somehow...you'd find out you probably have less than 36 grams of water since the reaction didn't go to completion.
Now your equation is somewhat right.rearranged though.
Percent Yield * theoretical yield = actual yield. Which is actually pointless.since you can't do this in practice.
The way you'd use this equation is
actual yield (what you measured / theoretical yeild (what you calculated ) = % yield.
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