3.60P - Chemistry: Molecular Nature Of Matter And Change

Question

Percent yields are generally calculated from mass quantities. Would the result be the same if mole quantities were used instead? Why?

Step-by-Step Solution

Answer

$\mathrm{Yield} \mathrm{percentage}=\left(\mathrm{Actual} \mathrm{yield}÷\mathrm{Theoretical} \mathrm{yield}\right)\times 100$

The actual yield of that product dividing by the theoretical yield and then multiplied with 100. This process gives the actual yield of that product.

Step 1. Formula to calcualte theoretical yield

The actual yield of that product dividing by the theoretical yield and then multiplied with 100. This process gives the actual yield of that product.

$\mathrm{Yield} \mathrm{percentage}=\left(\mathrm{Actual} \mathrm{yield}÷\mathrm{Theoretical} \mathrm{yield}\right)\times 100$

Step 2.Let's consider an example.

The calculation of the theoretically expected number of moles is based on the limiting reagent. For example, if we consider 2 moles of water in ${\mathrm{F}}_2$, we will expect 4 moles of $\mathrm{HF}$.

$2{\mathrm{F}}_2+2{\mathrm{H}}_2\mathrm{O}\to 4\mathrm{HF}+{\mathrm{O}}_2$

So multiplying the expected moles of the product by its molar mass, For example in the case of the molar mass of $\mathrm{HF}$having 20 grams, then there are 4 moles of $\mathrm{HF}$, and the theoretical yield will be 80 grams.    $\mathrm{HF}$

Now the actual yield of that product dividing by the theoretical yield and then multiplied with 100. This process gives the actual yield of that product.