[premedwoman]

Hello,

I was just wondering how you would find the percent yield of product? How do I find mmol of product and starting material?

mass of maleic anhydride (after experiment) was 1.034g
molecular wieght: 98.26 g
volume: .10ml
density: .81 g/ml=.081g

ethyl acetate:
mw: 66.10g/mol


I have to fing the theoretical mass of product and find percent yield.

[Borek]

You will need a balanced reaction equation. But your data are either not described properly or incomplete - for example how does 1.034g relate to 0.081g?

[premedwoman]

anyways, In general, how do you find percent yield of a rxn and find mmol of something?

[Custos]

In general, you calculate how many mmol of reactant you start with using the following:

no. of mmol = number of milligrams (mg) divided by molecular weight (g/mol)

Then you calculate how many mmol of product you obtained using the same formula. Then your percentage yield is:

% yield = 100 x (mmol of product)/(mmol of reactant)

For multicomponent reactions like Diels Alder reactions, the yield should be calculated by using the reactant that is in the lowest quantity (in mmol, not mg), because that will obviously be the limiting contribution to your yield. [note: this is only true for reactants. Catalysts are used in much smaller mole % because, being catalysts, they are reused many times during the reaction]

[Borek]

% yield = 100 x (mmol of product)/(mmol of reactant)

That works only for reactions were 1 mole of reactant gives 1 mole of product.

Correct approach is: start with balanced reaction equation. Calculate theoretical amount of product using simple stoichiometry calculations. Then calculate percent yield with formula:

% yield = 100 x (mmol of obtained product)/(theoretical mmol of product)

Now, in organic synthesis in most cases 1 mole of reactant gives 1 mole of product, so there is no need for stoichiometric coefficients, which makes Custos formula correct - but you should be aware of its limitation.

Also note that in many cases you will be expected to use masses, not moles - but it doesn't matter, as the final result is the same.