Theoretical And Actual Yields - Chemistry LibreTexts

Theoretical and Actual Yields

Reactants not completely used up are called excess reagents, and the reactant that completely reacts is called the limiting reagent. This concept has been illustrated for the reaction:

\[\mathrm{2 Na + Cl_2 \rightarrow 2 NaCl} \nonumber \]

Amounts of products calculated from the complete reaction of the limiting reagent are called theoretical yields, whereas the amount actually produced of a product is the actual yield. The ratio of actual yield to theoretical yield expressed in percentage is called the percentage yield.

\(\mathrm{percent\: yield = \dfrac{actual\: yield}{theoretical\: yield}\times100}\)

Chemical reaction equations give the ideal stoichiometric relationship among reactants and products. Thus, the theoretical yield can be calculated from reaction stoichiometry. For many chemical reactions, the actual yield is usually less than the theoretical yield, understandably due to loss in the process or inefficiency of the chemical reaction.

Example \(\PageIndex{1}\)

Methyl alcohol can be produced in a high-pressure reaction

\(\mathrm{CO_{\large{(g)}} + 2 H_{2\large{(g)}} \rightarrow CH_3OH_{\large{(l)}}}\)

If 6.1 metric tons of methyl alcohol is obtained from 1.2 metric tons of hydrogen reacting with excess amount of \(\ce{CO}\), estimate the theoretical and the percentage yield?

Solution

To calculate the theoretical yield, consider the reaction

\(\begin{alignat}{2} \ce{&CO_{\large{(g)}} +\, &&2 H_{2\large{(g)}} \rightarrow \, &&CH_3OH_{\large{(l)}}}\\ &\:28.0 &&\:4.0 &&\:\:\:32.0 \hspace{45px}\ce{(stoichiometric\: masses\: in\: g,\: kg,\: or\: tons)} \end{alignat}\)

\(\mathrm{1.2\: tons\: H_2 \times\dfrac{32.0\: CH_3OH}{4.0\: H_2}= 9.6\: tons\: CH_3OH}\)

Thus, the theoretical yield from 1.2 metric tons (1.2x106 g) of hydrogen gas is 9.6 tons. The actual yield is stated in the problem, 6.1 metric tons. Thus, the percentage yield is

\(\mathrm{\%\: yield =\dfrac{6.1\: tons}{9.6\: tons}\times 100 = 64 \%}\)

Due to chemical equilibrium or the mass action law, the limiting reagent may not be completely consumed. Thus, a lower yield is expected in some cases. Losses during the recovery process of the product will cause an even lower actual yield.

Example \(\PageIndex{2}\)

A solution containing silver ion, \(\ce{Ag+}\), has been treated with excess of chloride ions \(\ce{Cl-}\). When dried, 0.1234 g of \(\ce{AgCl}\) was recovered. Assuming the percentage yield to be 98.7%, how many grams of silver ions were present in the solution?

HINT

The reaction and relative masses of reagents and product are:

\(\begin{alignat}{2}\ce{ Ag^+_{\large{(aq)}}} &+ \mathrm{Cl^-_{\large{(aq)}}} &&\rightarrow \ce{AgCl_{\large{(s)}}} \\ 107.868 &+ 35.453 &&= 143.321 \end{alignat}\)

The calculation,

\(\mathrm{0.1234\: g\: AgCl \times \dfrac{107.868\: g\: Ag^+}{143.321\: g\: AgCl}= 0.09287\: g\: Ag^+}\)

shows that 0.1234 g dry \(\ce{AgCl}\) comes from 0.09287 g \(\ce{Ag+}\) ions. Since the actual yield is only 98.7%, the actual amount of \(\ce{Ag+}\) ions present is therefore

\(\mathrm{\dfrac{0.09287\: g\: Ag^+}{0.987}= 0.09409\: g\: Ag^+}\)

DISCUSSION

One can also calculate the theoretical yield of \(\ce{AgCl}\) from the percentage yield of 98.7% to be

\(\mathrm{\dfrac{0.1234\: g\: AgCl}{0.987}= 0.1250\: g\: AgCl}\)

From 0.1250 g \(\ce{AgCl}\), the amount of \(\ce{Ag+}\) present is also 0.09409 g.