Chemistry » Quantitative Aspects of Chemical Change » Stoichiometric Calculations

Theoretical Yield

Theoretical Yield

When we are given a known mass of a reactant and are asked to work out how much product is formed, we are working out the theoretical yield of the reaction. In the laboratory, chemists almost never get this amount of product. In each step of a reaction a small amount of product and reactants is “lost” either because a reactant did not completely react or some other unwanted products are formed. This amount of product that you actually got is called the actual yield. You can calculate the percentage yield with the following equation:

\[\% \text{ yield } = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\]

Example: Industrial Reaction to Produce Fertilizer

Question

Sulphuric acid (\(\text{H}_{2}\text{SO}_{4}\)) reacts with ammonia (\(\text{NH}_{3}\)) to produce the fertiliser ammonium sulphate (\((\text{NH}_{4})_{2}\text{SO}_{4}\)). What is the theoretical yield of ammonium sulphate that can be obtained from \(\text{2.0}\) \(\text{kg}\) of sulphuric acid? It is found that \(\text{2.2}\) \(\text{kg}\) of fertiliser is formed. Calculate the % yield.

Step 1: Write the balanced equation

\[\text{H}_{2}\text{SO}_{4}\text{(aq)} + 2\text{NH}_{3}\text{(g)} \rightarrow \text{(NH}_{4}\text{)}_{2}\text{SO}_{4}\text{(aq)}\]

Step 2: Calculate the number of moles of the given substance

\begin{align*} n_{\text{H}_{2}\text{SO}_{4}} & = \frac{m}{M} \\ & = \frac{\text{2 000}\text{ g}}{\text{98.12}\text{ g·mol$^{-1}$}} \\ & = \text{20.38320424}\text{ mol} \end{align*}

Step 3: Find the mole ratio

From the balanced equation, the mole ratio of \(\text{H}_{2}\text{SO}_{4}\) in the reactants to \((\text{NH}_{4})_{2}\text{SO}_{4}\) in the product is \(1:1\). Therefore, \(\text{20.383}\) \(\text{mol}\) of \(\text{H}_{2}\text{SO}_{4}\) forms \(\text{20.383}\) \(\text{mol}\) of \((\text{NH}_{4})_{2}\text{SO}_{4}\).

Step 4: Write the answer

The maximum mass of ammonium sulphate that can be produced is calculated as follows:

\begin{align*} m & = nM \\ & = (\text{20.383}\text{ mol})(\text{114.04}\text{ g·mol$^{-1}$}) \\ & = \text{2 324.47}\text{ g} \end{align*}

The maximum amount of ammonium sulphate that can be produced is \(\text{2.324}\) \(\text{kg}\).

Step 5: Calculate the \(\%\) yield

\begin{align*} \% \text{ yield } & = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 \\ & = \frac{\text{2.2}}{\text{2.324}} \times 100 \\ & = \text{94.64}\% \end{align*}

Example: Calculating the Mass of Reactants and Products

Question

Barium chloride and sulphuric acid react according to the following equation to produce barium sulphate and hydrochloric acid.

\[\text{BaCl}_{2} + \text{H}_{2}\text{SO}_{4} \rightarrow \text{BaSO}_{4} + 2\text{HCl}\]

If you have \(\text{2}\) \(\text{g}\) of \(\text{BaCl}_{2}\):

  1. What quantity (in g) of \(\text{H}_{2}\text{SO}_{4}\) will you need for the reaction so that all the barium chloride is used up?

  2. What mass of \(\text{HCl}\) is produced during the reaction?

Step 1: Find the number of moles of barium chloride

\begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{2}\text{ g}}{\text{208.2}\text{ g·mol$^{-1}$}} \\ & = \text{0.0096}\text{ mol} \end{align*}

Step 2: Find the number of moles of sulphuric acid

According to the balanced equation, 1 mole of \(\text{BaCl}_{2}\) will react with 1 mole of \(\text{H}_{2}\text{SO}_{4}\). Therefore, if \(\text{0.0096}\) \(\text{mol}\) of \(\text{BaCl}_{2}\) react, then there must be the same number of moles of \(\text{H}_{2}\text{SO}_{4}\) that react because their mole ratio is \(1:1\).

Step 3: Find the mass of sulphuric acid

\begin{align*} m & = nM \\ & = (\text{0.0096}\text{ mol})(\text{98.12}\text{ g·mol$^{-1}$}) \\ & = \text{0.94}\text{ g} \end{align*}

(answer to 1)

Step : Find the moles of hydrochloric acid

According to the balanced equation, 2 moles of \(\text{HCl}\) are produced for every 1 mole of the two reactants. Therefore the number of moles of \(\text{HCl}\) produced is \(2(\text{0.0096})\), which equals \(\text{0.0192}\) \(\text{mol}\).

Step 4: Find the mass of hydrochloric acid

\begin{align*} m & = nM \\ & = (\text{0.0192}\text{ mol})(\text{36.46}\text{ g·mol$^{-1}$}) \\ & = \text{0.7}\text{ g} \end{align*}

(answer to 2)

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