# If the theoretical yield of a reaction is .145 g and the actual yield is 0.104 g, what is the percent yield?

Apr 4, 2016

71.7%

#### Explanation:

A reaction's percent yield essentially tells you many grams of a given product will actually be produced instead of a theoretical $\text{100 g}$ of product.

Simply put, you can calculate a reaction's percent yield by dividing the actual yield, which is what the reaction actually produces, by the theoretical yield, which is what the reaction should theoretically produce, and multiplying the result by $100$.

color(blue)(|bar(ul(color(blue)("% yield" = "what you actually get"/"what you should theoretically get" xx 100color(white)(a/a)|)))

In your case, the reaction is said to have a theoretical yield of $\text{0.145 g}$. This means that if all the moles of reactants that take part in the reaction are converted to moles of product, the reaction will produce $\text{0.145 g}$ of product.

The actual yield of the reaction is said to $\text{0.104 g}$, which means that the reaction fails to reach the theoretical mass of product formed, i.e. it doesn't have a 100% yield.

The percent yield of the reaction will be

"% yield" = (0.104 color(red)(cancel(color(black)("g"))))/(0.145color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"71.7%color(white)(a/a)|))) -> rounded to three sig figs

This means that for every $\text{100 g}$ of product that can theoretically be produced by the reaction, you only get $\text{71.7 g}$ of product.