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When calculating the theoretical yield of a product within a reaction, the idea is to convert mass reactant --> mass product; in other words, "how many grams of product X will N grams reactant A and N grams reactant B yield (create)?" Whichever mass is the lowest is considered the theoretical yield (in other words, the reactant that produces the least amount of product).

__A + __B --> __X

mass reactant A => mass product X (mass A: mass X)

mass reactant B => mass product X (mass B: mass X)

In order to calculate the theoretical yield, you need to convert the mass of reactant A to the mass of product X AND the mass of reactant B to the mass of product X.

1. Multiply the mass of the reactant by the number of molecules (or moles) X and by the molar mass of X.

2. Divide the above by the molar mass of reactant (which is multiplied by the number of molecules [or moles] of reactant).

Do this for each reactant involved.

- Formula

Mass reactant * # molecules (moles) product X1 * molar mass product X

--------- # molecules (moles) reactant * molar mass reactant

= Mass Product X

- Conversion Factor

Mass reactant * 1 mol reactant * # molecules product X * molar mass product X

-------------- molar mass reactant - # molecules reactant --- 1 mol product X

= Mass Product X

1the # molecules is obtained from the balanced equation. For clarification, I stated # molecules in place of mol; however, on practice problems and in textbook tutorials on how to solve problems in regards to stoichiometry, it is likely it will say mol or moles instead of # molecules.

Ex. Ca(NO3)2 + 2NaF --> CaF2 + 2NaNO3

What is the theoretical yield of CaF2 (product X) when 43.5g Ca(NO3)2 (reactant A) react with 39.5g NaF (reactant B)?

43.5g Ca(NO3)2 * 1 mol Ca(NO3)2 * 1 molecule1 CaF2 * 78.08g CaF2

-------------- 164.1g Ca(NO3)2 ----- 1 molecule1 Ca(NO3)2 - 1 mol NaF

= 22.3g CaF2

36.5g NaF * 1 mol NaF * 1 molecule1 CaF2 * 78.08g CaF2

------------ 41.99g NaF --- 2 molecules1 NaF --- 1 mol CaF2

= 36.7g CaF2

The theoretical yield of CaF2 is 22.3g, as it is the lowest amount of product created. In this case, the limiting reagent (the reactant that produced the least product) is Ca(NO3)2 and the reagent in excess is NaF.

"A chain is only as strong as its weakest link".

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12y ago

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