Stoichiometry

In our daily lives, from the food we eat to the fuel that powers our vehicles, stoichiometry plays a crucial role in ensuring that chemical processes are efficient and effective. Improve your knowledge of stoichiometry by using this resource to learn about the stoichiometric coefficient, the molar ratio and chemical equations.

Stoichiometry describes the quantitative relationship between relative quantities of reactants and products in a balanced chemical equation. Stoichiometric coefficients are the numbers written immediately before the atoms, ions and molecules in a balanced chemical equation. These numbers indicate the ratio of any reactant or product that reacts or forms in a reaction.

As an example, let's consider the reaction between hydrogen gas and nitrogen gas to form ammonia. The balanced chemical equation can be written as:

\[3\ce{H}_{2}\left(\textrm{g}\right)+\ce{N}_{2}\left(\textrm{g}\right)\rightarrow2\ce{NH}_{3}\left(\textrm{g}\right)\]

The numbers three, one and two are written in front of \(\ce{H}\), \(\ce{N}\) and \(\ce{NH}_{3}\), respectively. These are the stoichiometric coefficients. They show that three molecules of hydrogen gas react with one molecule of nitrogen gas to form two ammonia molecules. When there is no coefficient before an atom, ion or molecule, we interpret it as one. For example, the coefficient of molecular nitrogen in this reaction is one.

Molar ratios

The stoichiometric coefficients in a balanced chemical equation can also be interpreted on the molar scale. Molar scale interpretation relates to the quantities of substances involved in the reaction. Therefore, if we know the amount of one substance involved in the reaction, we can use the molar ratio to determine the amounts of others involved in the reaction. Let's consider the same example:
\[3\ce{H}_{2}\left(\textrm{g}\right)+\ce{N}_{2}\left(\textrm{g}\right)\rightarrow2\ce{NH}_{3}\left(\textrm{g}\right)\]

Using the molar scale, we can say that three moles of hydrogen gas react with one mole of nitrogen gas to form two moles of ammonia gas.

Limiting and excess reactants

In chemical reactions, limiting and excess reactants (or reagents) determine the extent of the reaction and the amount of products formed. The limiting reactant is the substance that is consumed completely first, thus limiting the amount of product that can be formed. Once the limiting reactant is used up, the reaction cannot proceed further, even if other reactants are still available. On the other hand, excess reactants are those that remain after the reaction has reached completion.

Let's consider the reaction between hydrogen gas and nitrogen gas to form ammonia gas again:
\[3\ce{H}_{2}\left(\textrm{g}\right)+\ce{N}_{2}\left(\textrm{g}\right)\rightarrow2\ce{NH}_{3}\left(\textrm{g}\right)\]

Based on stoichiometry, we know that if we mix three moles of \(\ce{H}_{2}\) with one mole of \(\ce{N}_{2}\), we can obtain two moles of \(\ce{NH}_{3}\). In other words, \(\ce{H}_{2}\) reacts with \(\ce{N}_{2}\) in a molar ratio of \(3:1\).

If we mix three moles of hydrogen gas and two moles of nitrogen gas, two moles of \(\ce{NH}_{3}\) will be produced. However, the entire quantity of \(\ce{H}_{2}\) will be used before the \(\ce{N}_{2}\) as we start with the exact amount of \(\ce{H}_{2}\) required for the reaction. The product mixture will contain one unreacted mole of \(\ce{N}_{2}\) as we only required one mole of \(\ce{N}_{2}\) for the reaction. If we had more \(\ce{H}_{2}\), we could have produced more \(\ce{NH}_{3}\) as we already have excess \(\ce{N}_{2}\). Therefore, in this reaction, we call \(\ce{H}_{2}\) the limiting reagent since it limits the production of \(\ce{NH}_{3}\).

Hence, the limiting reagent is the reactant that runs out first, limiting the formation of products. Nitrogen gas is called the excess reagent because we have a greater amount than required for the reaction.

Reaction yield

The theoretical yield is the maximum amount of product that can form under the given reaction conditions. This is determined by the amount of limiting reagent available. Theoretical yield can be calculated using the stoichiometric coefficients (or molar ratios) in the balanced chemical equation. We calculate theoretical yield assuming no losses occur during the reaction, and all the reactants are converted into products. However, in reality, losses and inefficiencies occur during chemical reactions. Therefore, the actual amount you isolate and measure from a chemical reaction is less than what you expect in theory.

The actual yield of a reaction is the amount of the product you isolate from the chemical reaction. Actual yield must be determined experimentally, so it often referred to as experimental yield. You need to isolate the product and weigh it to get the actual yield.

The percentage yield indicates how efficient the reaction was in converting reactants into products. If the value is closer to \(100\textrm{%}\), most of the reactants were converted into products; the reaction was closer to completion. Percentage yield can be calculated using:

\[\textrm{Percentage yield}=\frac{\textrm{Actual yield}}{\textrm{Theoretical yield}}\times100\textrm{%}\]

Example 1 – calculating the amount of product formed from a chemical reaction

Calculate the amount in moles of ammonia gas that can be produced from \(0.18\textrm{ mol}\) of hydrogen gas. \[3\ce{H}_{2}\left(\textrm{g}\right)+\ce{N}_{2}\left(\textrm{g}\right)\rightarrow2\ce{NH}_{3}\left(\textrm{g}\right)\]

Step 1: Determine the molar ratio of the compounds and write it as a fraction. The stoichiometric coefficient for the compound we are interested in (the "unknown") is written as the numerator at the top, and the coefficient for the compound we have the amount for (the "known") is written as the denominator at the bottom.

\[\begin{align*}
\textrm{Stoichiometric ratio} & =\frac{\textrm{coefficient of unknown}}{\textrm{coefficient of known}}\\
& = \frac{2}{3}
\end{align*}\]

Step 2: Calculate the amount of product formed by multiplying the amount of reactant by the molar ratio.

\[\begin{align*}
n(\ce{NH3}) & = \frac{2}{3}\times0.18\textrm{ mol}\\
& =0.12\textrm{ mol}
\end{align*}\]

Example 3 – calculating the theoretical yield of a reaction

Phosphoric acid is produced by treating phosphate rock, which contains calcium phosphate, with sulfuric acid:
\[\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}+3\ce{H}_{2}\ce{SO}_{4}\rightarrow3\ce{CaSO}_{4}+2\ce{H}_{3}\ce{PO}_{4}\]
A chemist in a fertiliser production plant mixes \(144.5\textrm{ g}\) of calcium phosphate with \(160.5\textrm{ g}\) of sulfuric acid. Determine the maximum mass of phosphoric acid that can be formed. The molar masses are given in the table.

Compound	Molar mass (\(\textrm{g mol}^{-1}\))
\(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\)	\(310.2\)
\(\ce{H}_{2}\ce{SO}_{4}\)	\(98.1\)
\(\ce{H}_{3}\ce{PO}_{4}\)	\(97.99\)

Step 1: Identify the limiting reagent in this reaction by first, finding the number of moles of each reactant.

\[\begin{align*}
n(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}) & = \frac{m}{M}\\
& =\frac{144.5\textrm{ g}}{310.2\textrm{ g mol}^{-1}}\\
& =0.4658\textrm{ mol}
\end{align*}\]

\[\begin{align*}
n(\ce{H}_{2}\ce{SO}_{4}) & = \frac{m}{M}\\
& =\frac{160.5\textrm{ g}}{98.1\textrm{ g mol}^{-1}}\\
& =1.64\textrm{ mol}
\end{align*}\]

Step 2: Compare the calculated amounts of two reactants. The easiest way to find the limiting reactant is to use the formula:

\[\frac{n}{\textrm{coefficient}}\]

The reactant with the smallest \(\dfrac{n}{\textrm{coefficient}}\) is the limiting reactant.

For \(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\):

\[\begin{align*}
\frac{n}{\textrm{coefficient}} & = \frac{0.4659\textrm{ mol}}{1}\\
& = 0.4659\textrm{ mol}
\end{align*}\]

For \(\ce{H}_{2}\ce{SO}_{4}\):

\[\begin{align*}
\frac{n}{\textrm{coefficient}} & = \frac{1.64\textrm{ mol}}{2}\\
& = 0.82\textrm{ mol}
\end{align*}\]

As the \(\dfrac{n}{\textrm{coefficient}}\) for \(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\) is smaller, it is the limiting reactant.

Step 3: Determine the molar ratio of the compounds and write it as a fraction. The stoichiometric coefficient for the compound we are interested in (the "unknown") is written as the numerator at the top, and the coefficient for the limiting reactant (the "known") is written as the denominator at the bottom.

\[\begin{align*}
\textrm{Stoichiometric ratio} & =\frac{\textrm{coefficient of unknown}}{\textrm{coefficient of known}}\\
& = \frac{2}{1}
\end{align*}\]

Step 4: Calculate the amount of product formed by multiplying the amount of reactant by the molar ratio.

\[\begin{align*}
n(\ce{H3PO4}) & = \frac{2}{1}\times0.4659\textrm{ mol}\\
& = 0.9317\textrm{ mol}
\end{align*}\]

Step 5: Calculate the mass of product formed. This is the theoretical yield.

\[\begin{align*}
m(\ce{H3PO4}) & = n\times{M}\\
& = 0.9317\textrm{ mol}\times 97.99\textrm{ g mol}^{-1}\\
& = 91.29\textrm{ g}
\end{align*}\]

Example 4 – calculating percentage yield

When \(64.5\textrm{ g}\) of methane is burnt in the presence of excess oxygen, it forms \(136.2\textrm{ g}\) of carbon dioxide and water vapour.
\[\ce{CH}_{4}\left(\textrm{g}\right)+2\ce{O}_{2}\left(\textrm{g}\right)\rightarrow\ce{CO}_{2}\left(\textrm{g}\right)+2\ce{H}_{2}\ce{O}\left(\textrm{g}\right)\]

However, the theoretical yield of carbon dioxide for this reaction is \(176.4\textrm{ g}\). Calculate the percentage yield of carbon dioxide for this reaction.

Step 1: Recall the formula for calculating percentage yield.

\[\textrm{Percentage yield}=\frac{\textrm{Actual yield}}{\textrm{Theoretical yield}}\times100\textrm{%}\]

Step 2: Substitute the values into the formula to calculate percentage yield.

\[\begin{align*}
\textrm{Percentage yield} & =\frac{136.2\textrm{ g}}{176.4\textrm{ g}}\times100\textrm{%}\\
& = 77.21\textrm{%}
\end{align*}\]