Reaction Stoichiometry
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You must be comfortable writing and balancing chemical equations and calculating the molar mass of compounds from their formulas. You should also understand the definition of the mole as \(6.022 \times 10^{23}\) particles.
Why this matters
Stoichiometry translates the microscopic world of atoms into the macroscopic world of grams and litres.
In the laboratory, we rarely count individual molecules. Instead, we weigh substances. Reaction stoichiometry provides the mathematical bridge between the coefficients in a balanced equation and the measurable masses of reactants and products. Without this quantitative framework, chemistry would remain a qualitative art rather than a precise science. It allows chemists to predict exactly how much product will form from a given amount of starting material, ensuring efficiency, safety, and cost-effectiveness in both academic research and industrial manufacturing. This lecture establishes the core logic of these calculations, setting the stage for more complex scenarios involving limiting reagents and yields.
The Logic of Proportional Parts
Imagine you are following a recipe for sandwiches. The "equation" is: 2 slices of bread + 1 slice of cheese \(\to\) 1 sandwich. If you have 10 slices of bread and 5 slices of cheese, you can make exactly 5 sandwiches. You used all the cheese and all the bread. This is a perfect match. But what if you had 10 slices of bread and only 2 slices of cheese? You could only make 2 sandwiches. You would have 6 slices of bread left over. The cheese determined the maximum number of sandwiches you could make. In chemistry, we call this the limiting reagent, but the underlying principle is simply proportionality. A balanced chemical equation is not just a list of ingredients; it is a set of ratios. The coefficients tell us the relative number of particles involved. Because we measure mass, not particle count, we use the molar mass to convert between grams and moles. The mole is our counting unit. Once we have converted all our masses into moles, we can use the coefficients as conversion factors. This is the heart of stoichiometry: it is unit conversion, where the "units" are chemical species. We start with what we know (mass of A), convert to moles of A, use the equation to find moles of B, and convert to mass of B. This chain of logic is robust and applies to every chemical reaction, from the combustion of methane to the synthesis of pharmaceuticals. Understanding this flow allows you to solve any stoichiometric problem without memorising specific formulas for each reaction type.
From Mass to Moles: The Conversion Chain
The first step in any stoichiometric calculation is to convert the given mass of a substance into moles. This is done by dividing the mass by the molar mass. The molar mass is calculated by summing the atomic masses of all atoms in the formula, using values from the periodic table. For example, the molar mass of water, \(\mathrm{H_2O}\), is approximately \(2(1.008) + 16.00 = 18.02\ \mathrm{g\,mol^{-1}}\). Once you have the number of moles, you can use the stoichiometric ratio from the balanced equation to find the moles of the other substance. Consider the reaction \(2\mathrm{H_2} + \mathrm{O_2} \to 2\mathrm{H_2O}\). The ratio of \(\mathrm{H_2}\) to \(\mathrm{H_2O}\) is 2:2, or 1:1. The ratio of \(\mathrm{O_2}\) to \(\mathrm{H_2O}\) is 1:2. If you start with 4.00 moles of \(\mathrm{H_2}\), you will produce 4.00 moles of \(\mathrm{H_2O}\). If you start with 1.00 mole of \(\mathrm{O_2}\), you will produce 2.00 moles of \(\mathrm{H_2O}\). This step is purely mathematical and relies on the law of conservation of mass, which ensures that the equation is balanced. It is crucial to always check that your equation is balanced before performing these calculations. An unbalanced equation leads to incorrect ratios and, consequently, incorrect mass predictions. This conversion chain is the foundation of all quantitative chemical analysis.
Applying the Limiting Reagent Concept
In real-world reactions, reactants are rarely present in the exact stoichiometric proportions required. One reactant will be used up before the other. This reactant is the limiting reagent, and it determines the maximum amount of product that can be formed. The other reactant is in excess. To identify the limiting reagent, you calculate the amount of product that each reactant could theoretically produce, assuming the other is in excess. The reactant that produces the smaller amount of product is the limiting reagent. For example, consider the reaction \(\mathrm{N_2} + 3\mathrm{H_2} \to 2\mathrm{NH_3}\). Suppose you have 1.00 mole of \(\mathrm{N_2}\) and 2.00 moles of \(\mathrm{H_2}\). From 1.00 mole of \(\mathrm{N_2}\), you could produce \(1.00 \times (2/1) = 2.00\) moles of \(\mathrm{NH_3}\). From 2.00 moles of \(\mathrm{H_2}\), you could produce \(2.00 \times (2/3) = 1.33\) moles of \(\mathrm{NH_3}\). Since 1.33 moles is less than 2.00 moles, \(\mathrm{H_2}\) is the limiting reagent. The theoretical yield of \(\mathrm{NH_3}\) is 1.33 moles. The amount of \(\mathrm{N_2}\) consumed is \(1.33 / 2 = 0.67\) moles, leaving \(1.00 - 0.67 = 0.33\) moles of \(\mathrm{N_2}\) unreacted. This concept is critical for predicting product yields and minimising waste in industrial processes.
Theoretical Yield and Percentage Yield
The theoretical yield is the maximum amount of product that can be formed from the limiting reagent, assuming perfect reaction conditions. In practice, the actual yield is often lower due to side reactions, incomplete reactions, or losses during purification. The percentage yield is a measure of the efficiency of the reaction. It is calculated as the ratio of the actual yield to the theoretical yield, multiplied by 100. For example, if the theoretical yield of a product is 50.0 g and the actual yield is 42.5 g, the percentage yield is \((42.5 / 50.0) \times 100 = 85.0\%\). A percentage yield of 100% is rare in practice, as it implies no losses whatsoever. Yields below 100% are common and expected. Understanding percentage yield allows chemists to evaluate the efficiency of a synthetic route and identify areas for improvement. It is also a key metric in cost-benefit analyses for industrial chemistry, where even small improvements in yield can lead to significant cost savings. The calculation of percentage yield requires accurate measurement of both the actual and theoretical yields, highlighting the importance of precise laboratory technique.
The result
The limiting reagent
For a balanced reaction \(aA+bB\to\text{products}\), the reactant whose available moles, divided by its own stoichiometric coefficient, is smallest (\(\min\{n_A/a,\ n_B/b\}\)) is the limiting reagent: it is fully consumed first, and the amount of product actually formed is capped by however much that reactant alone could produce, regardless of how much of the other reactant remains unreacted.
A common misconception is that the limiting reagent is always the one with the smaller mass or the smaller number of moles. This is incorrect. The limiting reagent is determined by the stoichiometric ratio, not just the quantity. For example, in the reaction \(\mathrm{C} + \mathrm{O_2} \to \mathrm{CO_2}\), 12 g of carbon (1 mole) reacts with 32 g of oxygen (1 mole). If you have 12 g of carbon and 10 g of oxygen, oxygen is limiting, even though carbon has a larger mass. Always convert to moles and compare using the coefficients. C
Worked example
Problem. Calculate the mass of iron(III) oxide, \(\mathrm{Fe_2O_3}\), produced when 5.00 g of iron, \(\mathrm{Fe}\), reacts with excess oxygen, \(\mathrm{O_2}\). The balanced equation is \(4\mathrm{Fe} + 3\mathrm{O_2} \to 2\mathrm{Fe_2O_3}\).
- Calculate the molar mass of \(\mathrm{Fe}\) and \(\mathrm{Fe_2O_3}\). Molar mass of \(\mathrm{Fe} = 55.85\ \mathrm{g\,mol^{-1}}\). Molar mass of \(\mathrm{Fe_2O_3} = 2(55.85) + 3(16.00) = 111.70 + 48.00 = 159.70\ \mathrm{g\,mol^{-1}}\).
- Convert the mass of \(\mathrm{Fe}\) to moles. Moles of \(\mathrm{Fe} = 5.00\ \mathrm{g} / 55.85\ \mathrm{g\,mol^{-1}} = 0.0895\ \mathrm{mol}\).
- Use the stoichiometric ratio to find moles of \(\mathrm{Fe_2O_3}\). From the equation, 4 moles of \(\mathrm{Fe}\) produce 2 moles of \(\mathrm{Fe_2O_3}\). Ratio is 2:4 or 1:2. Moles of \(\mathrm{Fe_2O_3} = 0.0895\ \mathrm{mol\ Fe} \times (2\ \mathrm{mol\ Fe_2O_3} / 4\ \mathrm{mol\ Fe}) = 0.0448\ \mathrm{mol}\).
- Convert moles of \(\mathrm{Fe_2O_3}\) to mass. Mass of \(\mathrm{Fe_2O_3} = 0.0448\ \mathrm{mol} \times 159.70\ \mathrm{g\,mol^{-1}} = 7.15\ \mathrm{g}\).
- Check significant figures. The given mass has 3 significant figures, so the answer should be reported to 3 significant figures.
Units check. The units of grams cancel with grams per mole, leaving moles, which then cancel with moles per mole, leaving grams. The final unit is grams, which is correct for mass.
Three quick tests
- What is the molar mass of \(\mathrm{CO_2}\)?
Answer
44.01 g/mol - If 2 moles of \(\mathrm{H_2}\) react with excess \(\mathrm{O_2}\), how many moles of \(\mathrm{H_2O}\) are formed?
Answer
2 moles - What is the percentage yield if the theoretical yield is 10 g and the actual yield is 8 g?
Answer
80%
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Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Results cited: T-015 · Threads: structure