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Stoichiometry
\[\require{mhchem}\]
Stoichiometry describes the relationship between relative quantities of reactants and products in a balanced chemical equation.
Stoichiometric coefficient
The stoichiometric coefficient is the number 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.
Example: Let’s consider the reaction between hydrogen gas and nitrogen gas to form ammonia. The balanced chemical equation can be written as follows: \[ 3\ce{H}_{2}\left(g\right)+\ce{N}_{2}\left(g\right)\rightarrow2\ce{NH}_{3}\left(g\right) \]
Three, one and two are written in front of \(\ce{H},\ce{N},\ce{NH}_{3}\), respectively, these are considered as stoichiometric coefficients. Stoichiometric coefficients in this reaction 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 instance, the coefficient of molecular nitrogen in this reaction is one.
The molar ratio and chemical equations
The stoichiometric coefficients in a balanced chemical equation can also be interpreted on the molar scale. Let’s consider the same example again: \[ 3\ce{H}_{2}\left(g\right)+\ce{N}_{2}\left(g\right)\rightarrow2\ce{NH}_{3}\left(g\right) \]
Based on 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. 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.
Moles calculations
Worked Example 1: How many moles of ammonia gas can be produced from \(0.18\textrm{mol}\) of hydrogen gas? \[ 3\ce{H}_{2}\left(g\right)+\ce{N}_{2}\left(g\right)\rightarrow2\ce{NH}_{3}\left(g\right) \]
We are asked to calculate the amount of ammonia and are given the moles of hydrogen and the balanced chemical equation for the reaction. We need to find out the molar ratio of hydrogen and ammonia to calculate the amount of ammonia. From the stoichiometric coefficients, we know that three moles of hydrogen produce two moles of ammonia. Therefore, the molar ratio between hydrogen gas to ammonia gas is \(3:2\). Thus, if we use \(0.18\textrm{mol}\) of hydrogen gas, we can produce the following amount of ammonia gas: \[\begin{align*} \frac{2\textrm{mol}\ce{NH}_{3}}{3\textrm{mol}\ce{H}_{2}} & \times0.18\textrm{mol}\ce{H}_{2}=0.12\textrm{mol} \end{align*}\]
Mass calculations
Worked Example 2: The burning of methane gas \(\left(\ce{CH}_{4}\right)\) in excess oxygen produces carbon dioxide gas and water vapour. The balanced chemical equation for the reaction is as follows: \[ \ce{CH}_{4}\left(g\right)+2\ce{O}_{2}\left(g\right)\rightarrow\ce{CO}_{2}\left(g\right)+2\ce{H}_{2}\ce{O}\left(g\right) \]
How many moles of water vapour can be obtained by the complete burning of \(3.20\textrm{g}\) of methane gas? The molar mass of methane is \(16.0\textrm{g}/\textrm{mol}\).
We are asked to calculate the number of moles of water vapour and are given the mass of methane and the balanced chemical equation for the reaction. To find the moles of water vapour, first, we need to find out the moles of methane. As the quantity of methane is given in grams, we need to convert it to moles using \(M=\frac{m}{n}\)1
\[\begin{align*} M & =\frac{m}{n}\\ n & =\frac{m}{M}\\ n & =\frac{3.20\textrm{g}}{16.0\textrm{g}/\textrm{mol}}\\ n & =0.200\textrm{mol} \end{align*}\] To relate the amount of methane to the amount of water vapour, we need to know the molar ratio of the two. From the balanced chemical equation, we know one mole of methane produces two moles of water vapour. Therefore, \(0.200\textrm{mol}\) of methane can produce the following amount of water vapour: \[ \frac{2\textrm{mol}\ce{H}_{2}\ce{O}}{1\textrm{mol}\ce{CH}_{4}}\times0.200\textrm{mol}\ce{CH}_{4}=0.400\textrm{mol} \]
Limiting reagents and yields
Limiting and excess reagents
Let’s consider the reaction between hydrogen gas and nitrogen gas to form ammonia gas again: \[ 3\ce{H}_{2}\left(g\right)+\ce{N}_{2}\left(g\right)\rightarrow2\ce{NH}_{3}\left(g\right) \]
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, what will happen? Similarly, three moles of \(\ce{H}_{2}\) will react with one mole of \(\ce{N}_{2}\) and produce two moles of \(\ce{NH}_{3}\). 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.
Worked Example: Phosphoric acid is produced by treating the phosphate rock, which contains calcium phosphate, with sulphuric acid, as follows: \[ \ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}+3\ce{H}_{2}\ce{SO}_{4}\rightarrow3\ce{CaSO}_{4}+2\ce{H}_{3}\ce{PO}_{4} \]
A chemist in a fertiliser production plant mixed \(144.5\textrm{g}\) of calcium phosphate with \(160.5\textrm{g}\) of sulphuric acid. What is the maximum mass of phosphoric acid that can be formed? The molar masses are given below.
| Molecule | Molar mass (g/mol) |
|---|---|
| \(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\) | \(310.2\) |
| \(\ce{H}_{2}\ce{SO}_{4}\) | \(98.1\) |
| \(\ce{H}_{3}\ce{PO}_{4}\) | \(97.99\) |
We are given the mass of reactants and are asked to determine the mass of one of the products.
Step 1 - Identify the limiting reagent in this reaction. For that, find the number of moles of each reactant using \(M=\frac{m}{n}\).
Number of moles of \(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\): \[\begin{align*} M & =\frac{m}{n}\\ n & =\frac{m}{M}\\ n & =\frac{144.5\textrm{g}}{310.2\textrm{ g}/\textrm{mol}}\\ n & =0.4658\textrm{mol} \end{align*}\]
Number of moles of \(\ce{H}_{2}\ce{SO}_{4}\): \[\begin{align*} M & =\frac{m}{n}\\ n & =\frac{m}{M}\\ n & =\frac{160.5\textrm{g}}{98.1\textrm{g}/\textrm{mol}}\\ n & =1.64\textrm{mol} \end{align*}\]
Then, compare the calculated moles with the molar ratio of two reactants. According to the balanced chemical equation, calcium phosphate reacts with sulphuric acid in a \(1:3\) molar ratio. Therefore, \(0.4658\textrm{mol}\) of \(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\) reacts with the following number of moles of \(\ce{H}_{2}\ce{SO}_{4}\): \[\begin{align*} \frac{3\textrm{mol}\ce{H}_{2}\ce{SO}_{4}}{1\textrm{mol}\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}}\times0.4658\textrm{mol}\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2} & =1.397\textrm{mol} \end{align*}\]
We have \(1.64\textrm{mol}\) of \(\ce{H}_{2}\ce{SO}_{4}\) , but we only need \(1.397\textrm{mol}\) for the reaction. This means we have excess \(\ce{H}_{2}\ce{SO}_{4}\) acid. Therefore, the limiting reagent is \(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\). The quantity of limiting reagent can determine the mass of \(\ce{H}_{3}\ce{PO}_{4}\) acid formed.
Step 2 - Use the molar ratio between \(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\) and \(\ce{H}_{3}\ce{PO}_{4}\) acid to calculate the mass of \(\ce{H}_{3}\ce{PO}_{4}\) acid formed.
We know one mole of \(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\) forms two moles of \(\ce{H}_{3}\ce{PO}_{4}\) acid from the balanced chemical equation. Therefore, the number of moles of \(\ce{H}_{3}\ce{PO}_{4}\) acid forms from \(0.4658\textrm{mol}\) of \(\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}\) is: \[\begin{align*} \frac{2\textrm{mol}\ce{H}_{3}\ce{PO}_{4}}{1\textrm{mol}\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}}\times0.4658\textrm{mol}\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2} & =0.9316\textrm{mol} \end{align*}\]
Now, find the mass of \(\ce{H}_{3}\ce{PO}_{4}\) using \(M=\frac{m}{n}\) \[\begin{align*} M & =\frac{m}{n}\\ m & =M\times n\\ m & =97.99\textrm{g}/\textrm{mol}\times0.9316\textrm{mol}\\ m & =91.29\textrm{g} \end{align*}\]
Based on the given reaction conditions, maximum mass we can obtain for \(\ce{H}_{3}\ce{PO}_{4}\) acid is \(91.29\textrm{g}\).
Yields
Theoretical yield is the maximum amount of the product that can form under the given reaction conditions. The available amount of limiting reagent determines the maximum amount of the product that can be formed. Theoretical yield can be calculated using the stoichiometric coefficients (or molar ratios) displayed in the balanced chemical equation. We calculate theoretical yield assuming no losses occur during the reaction, and all the reactants convert to products. However, in reality, often losses and inefficiencies occur during chemical reactions. Therefore, the actual amount you isolate and measure from a chemical reaction is less than the theoretically expected amount.
The actual yield of a product is the amount of the product you isolate from the chemical reaction. You can not calculate the actual yield. It must be determined experimentally. You need to isolate the product and weigh it to get the actual yield.
Percentage yield \[ \frac{\textrm{Actual yield}}{\textrm{Theoretical yield}}\times100 \]
Percentage yield indicates the success of the reaction. If the value is closer to the hundred, that means losses are minor, and most reactants have converted to products.
1 Where, M=Molar mass (g/mol) m=mass of the substance (g) n=Number of Moles (mol)