Stoichiometry: The Chemical Recipe

Stoichiometry is the quantitative heart of chemistry. It provides a precise method for calculating the amounts of reactants and products involved in a chemical reaction. Put simply, stoichiometry is the chemistry of “how much.”

The Chemical Recipe

Think of a balanced chemical equation as a recipe. Instead of calling for cups of flour or a number of eggs, it calls for a specific number of atoms, molecules, or ions. The coefficients in a balanced equation tell us the proportional relationship between the ingredients. Consider the synthesis of water: \[ 2~\mathrm{H_2(g)} + \mathrm{O_2(g)} \longrightarrow 2~\mathrm{H_2O(l)} \] This recipe can be read on two scales:

Molecular Scale: Two molecules of H₂ react with one molecule of O₂ to produce two molecules of H₂O.
Laboratory Scale (Moles): Two moles of H₂ react with one mole of O₂ to produce two moles of H₂O.

Because we cannot count individual molecules in the lab, we work with the mole. All stoichiometric calculations are built upon this molar relationship.

The Heart of Stoichiometry: The Mole Ratio

The coefficients in a balanced equation give us the most powerful tool in chemistry: the mole ratio, which is formally called the stoichiometric factor (r(B,A)). This factor is the central hub of all stoichiometry problems.

Key Symbols for Stoichiometry

To approach problems systematically, we will use the following standard symbols:

n: amount of substance (in mol)
m: mass
M: molar mass (typically in g mol⁻¹). This is the conversion factor between mass and moles (M = m/n).
r(B,A): stoichiometric factor (unitless). This is the mole ratio of substance B to substance A, derived from the balanced equation.

The stoichiometric factor is formally defined as the ratio of the amounts of substance B and A: \[ r(\mathrm{B,A}) = \frac{n(\mathrm{B})}{n(\mathrm{A})} \] The values for n(B) and n(A) in this definition are the stoichiometric coefficients from the balanced chemical equation.

The Stoichiometry “Road Map”

The path for solving any stoichiometry problem is to convert a given quantity into moles, use the stoichiometric factor to cross the “mole bridge” to the desired substance, and then convert to the final desired unit.

A flowchart showing the central path of stoichiometry. It starts with "Mass of Substance A", an arrow points to "Moles of Substance A". From there, an arrow points to "Moles of Substance B". Finally, an arrow points to "Mass of Substance B". The symbolic conversion for the first arrow is "m(A) * M(A)⁻¹". The conversion for the central arrow is "r(B,A)". The conversion for the final arrow is "n(B) * M(B)".

Stoichiometric Calculations

Let’s apply our “road map” using the formal, “symbol-first” approach. We will use the formation of water as our reference reaction for these examples.

Quick Reference

The Stoichiometry Flowchart can be conveniently accessed in the “Quick Reference” section.

\[ 2~\mathrm{H_2(g)} + \mathrm{O_2(g)} \longrightarrow 2~\mathrm{H_2O(l)} \]

1. n(A) → n(B)

This is the most direct calculation, involving a single step across the mole bridge.

Example: How many moles of O₂ are required to produce 2.0 mol of H₂O?

The solution is found by converting the given moles of H₂O to the desired moles of O₂ using the stoichiometric factor. \[ \begin{align*} n(\mathrm{O_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{O_2, H_2O}) \\[1.5ex] &= \left( 2.\bar{0}~\mathrm{mol~H_2O} \right) \left( \frac{1~\mathrm{mol~O_2}}{2~\mathrm{mol~H_2O}} \right) \\[1.5ex] &= 1.\bar{0}00~\mathrm{mol~O_2} \\[1.5ex] &= 1.0~\mathrm{mol~O_2} \end{align*} \]

2. m(A) → n(B)

This calculation requires converting a given mass to moles before crossing the mole bridge.

Example: How many moles of H₂O can be produced from 4.04 g of H₂?

We construct the solution by chaining the necessary conversions: first mass to moles, then the mole ratio. \[ \begin{align*} n(\mathrm{H_2O}) &= n(\mathrm{H_2}) ~ r(\mathrm{H_2O,H_2}) \\[1.5ex] &= m(\mathrm{H_2}) ~ M(\mathrm{H_2})^{-1} ~ r(\mathrm{H_2O, H_2}) \\[1.5ex] &= \left( 4.0\bar{4}~\mathrm{g~H_2} \right) \left( \frac{1~\mathrm{mol~H_2}}{2.0\bar{2}~\mathrm{g~H_2}} \right) \left( \frac{2~\mathrm{mol~H_2O}}{2~\mathrm{mol~H_2}} \right) \\[1.5ex] &= 2.0\bar{0}00~\mathrm{mol~H_2O} \\[1.5ex] &= 2.00~\mathrm{mol~H_2O} \end{align*} \]

3. m(A) → m(B)

This is the most common laboratory calculation. It follows the complete three-step path on our road map.

Example: What mass of H₂ is required to produce 36.04 g of H₂O?

The solution combines all three steps from the road map into a single calculation. \[ \begin{align*} m(\mathrm{H_2}) &= n(\mathrm{H_2}) ~ M(\mathrm{H_2})\\[1.5ex] &= n(\mathrm{H_2O}) ~ r(\mathrm{H_2,H_2O}) ~ M(\mathrm{H_2}) \\[1.5ex] &= m(\mathrm{H_2O}) ~ M(\mathrm{H_2O})^{-1} ~ r(\mathrm{H_2, H_2O}) ~ M(\mathrm{H_2}) \\[1.5ex] &= \left( 36.0\bar{4}~\mathrm{g~H_2O} \right) \left( \frac{1~\mathrm{mol~H_2O}}{18.0\bar{2}~\mathrm{g~H_2O}} \right) \left( \frac{2~\mathrm{mol~H_2}}{2~\mathrm{mol~H_2O}} \right) \left( \frac{2.0\bar{2}~\mathrm{g~H_2}}{1~\mathrm{mol~H_2}} \right) \\[1.5ex] &= 4.0\bar{4}00~\mathrm{g~H_2} \\[1.5ex] &= 4.04~\mathrm{g~H_2} \end{align*} \]

4. Calculations Involving Atoms and Molecules

The mole is our bridge from the microscopic world of atoms and molecules to the macroscopic world of grams that we can weigh in a lab. A balanced equation ultimately describes the ratio of individual particles, and we can use the Avogadro constant (N_A) to convert between the amount of a substance in moles and the number of its constituent particles (atoms, molecules, or formula units).

Example: How many individual molecules of H₂O are produced when 4.04 g of H₂ reacts completely?

Our “road map” extends one step further: Mass A → Moles A → Moles B → Particles of B. We build the symbolic plan by chaining all the necessary conversion steps.

\[ \begin{align*} N(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ N_{\mathrm{A}} \\[1.5ex] &= n(\mathrm{H_2}) ~ r(\mathrm{H_2O,H_2}) ~ N_{\mathrm{A}}\\[1.5ex] &= m(\mathrm{H_2}) ~ M(\mathrm{H_2})^{-1} ~ r(\mathrm{H_2O, H_2}) ~ N_{\mathrm{A}} \\[1.5ex] &= \left( 4.0\bar{4}~\mathrm{g~H_2} \right) \left( \frac{1~\mathrm{mol~H_2}}{2.0\bar{2}~\mathrm{g~H_2}} \right) \left( \frac{2~\mathrm{mol~H_2O}}{2~\mathrm{mol~H_2}} \right) \left( \frac{6.02\bar{2} \times 10^{23}~\mathrm{molecules}}{1~\mathrm{mol~H_2O}} \right) \\[1.5ex] &= 1.2\bar{0}44 \times 10^{24}~\mathrm{molecules} \\[1.5ex] &= 1.20 \times 10^{24}~\mathrm{molecules~H_2O} \end{align*} \]

Combustion Analysis

One of the most powerful applications of stoichiometry is combustion analysis, an experimental technique used to determine the empirical formula of an unknown compound. The method is a classic example of using the conservation of mass to work backward from simple products to find the composition of a complex reactant.

The Strategy:

A precisely weighed sample of an unknown compound containing carbon, hydrogen, and possibly oxygen is burned completely in a furnace with an excess of pure oxygen. The combustion products, carbon dioxide (CO₂) and water (H₂O), are collected and their masses are measured.

The core principle is the conservation of each element:

All the carbon from the original sample is captured in the CO₂ produced.
All the hydrogen from the original sample is captured in the H₂O produced.
The mass of oxygen in the original sample is found by subtracting the masses of carbon and hydrogen from the total mass of the sample.

From the masses of C, H, and O, we can then follow the standard procedure to determine the empirical formula.

Example: Ascorbic acid (Vitamin C) is a compound containing only carbon, hydrogen, and oxygen. A sample with a mass of 1.000 g is completely combusted. The reaction produces 1.50 g of CO₂ and 0.409 g of H₂O. In a separate experiment, the molar mass of ascorbic acid was determined to be 176.12 g mol⁻¹. Determine the empirical and molecular formulas of ascorbic acid.

Step 1: Calculate the Mass of Carbon (C) from CO₂

We perform a mass-to-mass stoichiometric calculation to find the mass of carbon in the measured CO₂. \[ \begin{align*} m(\mathrm{C}) &= n(\mathrm{C}) ~ M(\mathrm{C}) \\[1.5ex] &= n(\mathrm{CO_2}) ~ r(\mathrm{C, CO_2}) ~ M(\mathrm{C}) \\[1.5ex] &= m(\mathrm{CO_2}) ~ M(\mathrm{CO_2})^{-1} ~ r(\mathrm{C, CO_2}) ~ M(\mathrm{C}) \\[1.5ex] &= \left( 1.5\bar{0}~\mathrm{g~CO_2} \right) \left( \frac{1~\mathrm{mol~CO_2}}{44.0\bar{1}~\mathrm{g~CO_2}} \right) \left( \frac{1~\mathrm{mol~C}}{1~\mathrm{mol~CO_2}} \right) \left( \frac{12.0\bar{1}~\mathrm{g~C}}{1~\mathrm{mol~C}} \right) \\[1.5ex] &= 0.40\bar{9}33~\mathrm{g~C} \end{align*} \]

Step 2: Calculate the Mass of Hydrogen (H) from H₂O

We follow the same procedure to find the mass of hydrogen in the measured H₂O. \[ \begin{align*} m(\mathrm{H}) &= n(\mathrm{H}) ~ M(\mathrm{H})\\[1.5ex] &= n(\mathrm{H_2O}) ~ r(\mathrm{H,H_2O}) ~ M(\mathrm{H}) \\[1.5ex] &= m(\mathrm{H_2O}) ~ M(\mathrm{H_2O})^{-1} ~ r(\mathrm{H, H_2O}) ~ M(\mathrm{H}) \\[1.5ex] &= \left( 0.40\bar{9}33~\mathrm{g~H_2O} \right) \left( \frac{1~\mathrm{mol~H_2O}}{18.0\bar{2}~\mathrm{g~H_2O}} \right) \left( \frac{2~\mathrm{mol~H}}{1~\mathrm{mol~H_2O}} \right) \left( \frac{1.0\bar{1}~\mathrm{g~H}}{1~\mathrm{mol~H}} \right) \\[1.5ex] &= 0.045\bar{8}84~\mathrm{g~H} \end{align*} \]

Step 3: Calculate the Mass of Oxygen (O) by Subtraction

The mass of oxygen in the original sample is the total sample mass minus the masses of carbon and hydrogen we just calculated. \[ \begin{align*} m(\mathrm{O}) &= m(\mathrm{sample}) - m(\mathrm{C}) - m(\mathrm{H}) \\[1.5ex] &= 1.00\bar{0}~\mathrm{g} - 0.40\bar{9}33~\mathrm{g} - 0.045\bar{8}84~\mathrm{g} \\[1.5ex] &= 0.54\bar{4}78~\mathrm{g~O} \end{align*} \]

Why Not Calculate Oxygen from the Products?

We cannot calculate the mass of oxygen from the CO₂ and H₂O because the oxygen in the products comes from two sources: the original sample and the excess O₂ used for the combustion. The subtraction method is the only way to isolate the mass of oxygen that was part of the unknown compound.

Step 4: Convert Masses of Elements to Moles

Now we convert the mass of each element into moles to find their molar ratio. \[ \begin{align*} n(\mathrm{C}) &= m(\mathrm{C}) ~ M(\mathrm{C})^{-1} = \left( 0.40\bar{9}33~\mathrm{g~C} \right) \left( \frac{1~\mathrm{mol~C}}{12.0\bar{1}~\mathrm{g~C}} \right) = 0.034\bar{0}82~\mathrm{mol~C} \\[1.5ex] n(\mathrm{H}) &= m(\mathrm{H}) ~ M(\mathrm{H})^{-1} = \left( 0.045\bar{8}84~\mathrm{g~H} \right) \left( \frac{1~\mathrm{mol~H}}{1.0\bar{1}~\mathrm{g~H}} \right) = 0.045\bar{4}29~\mathrm{mol~H} \\[1.5ex] n(\mathrm{O}) &= m(\mathrm{O}) ~ M(\mathrm{O})^{-1} = \left( 0.54\bar{4}78~\mathrm{g~O} \right) \left( \frac{1~\mathrm{mol~O}}{16.0\bar{0}~\mathrm{g~O}} \right) = 0.034\bar{0}48~\mathrm{mol~O} \end{align*} \]

Step 5: Determine the Empirical Formula

We find the simplest whole-number ratio by dividing the mole amount of each element by the smallest mole amount (in this case, 0.034057 mol of Oxygen). \[ \begin{align*} \text{C Ratio} &= \frac{0.034\bar{0}82~\mathrm{mol}}{0.034\bar{0}48~\mathrm{mol}} = 1.0\bar{0}00 \approx 1 \\[1.5ex] \text{H Ratio} &= \frac{0.045\bar{3}29~\mathrm{mol}}{0.034\bar{0}48~\mathrm{mol}} = 1.3\bar{3}13 \approx 1.33 \\[1.5ex] \text{O Ratio} &= \frac{0.034\bar{0}48~\mathrm{mol}}{0.034\bar{0}48~\mathrm{mol}} = 1.0\bar{0}00 = 1 \end{align*} \] The ratio C : H : O is 1 : 1.33 : 1. To convert this to whole numbers, we multiply the entire ratio by 3. \[ \mathrm{C_{1 \times 3}H_{1.33 \times 3}O_{1 \times 3} \longrightarrow C_3H_4O_3} \] The empirical formula for ascorbic acid is C₃H₄O₃.

Step 6: Determine the Molecular Formula

The molecular formula is always a whole-number multiple of the empirical formula. To find this multiple, we compare the given molar mass of the compound to the molar mass of the empirical formula we just determined.

First, calculate the molar mass of the empirical formula, C₃H₄O₃. \[ \begin{align*} M(\mathrm{C_3H_4O_3}) &= 3~M(\mathrm{C}) + 4~M(\mathrm{H}) + 3~M(\mathrm{O}) \\[1.5ex] &= 3~(12.0\bar{1}~\mathrm{g~mol^{-1}}) + 4~(1.0\bar{1}~\mathrm{g~mol^{-1}}) + 3~(16.0\bar{0}~\mathrm{g~mol^{-1}}) \\[1.5ex] &= 88.\bar{0}7~\mathrm{g~mol^{-1}} \end{align*} \] Next, we find the whole-number multiple (x) by dividing the compound’s molar mass by the empirical formula’s molar mass. \[ \begin{align*} x &= \frac{M(\mathrm{compound})}{M(\mathrm{empirical~formula})} \\[1.5ex] &= \frac{176.1\bar{2}~\mathrm{g~mol^{-1}}}{88.\bar{0}7~\mathrm{g~mol^{-1}}} \\[1.5ex] &= 1.9\bar{9}97 \\[1.5ex] &\approx 2 \end{align*} \] This tells us that the true molecular formula is twice the empirical formula. \[ \mathrm{C_{3\times 2}H_{4\times 2}O_{3\times 2} \longrightarrow C_6H_8O_6} \] The molecular formula for ascorbic acid is C₆H₈O₆.

Limiting Reactants

So far, we have assumed that our reactants are mixed in perfect stoichiometric amounts. In reality, this is rarely the case. In a typical reaction, one reactant will be completely consumed before the others.

The reactant that runs out first is called the limiting reactant (or limiting reagent), because it limits the maximum amount of product that can be formed. The other reactants are said to be in excess.

To solve a limiting reactant problem, you must first determine which reactant runs out first. A reliable method is to calculate the amount of product that could be formed from the given amount of each reactant. The reactant that produces the least amount of product is the limiting reactant.

Example: A reaction is performed by mixing 10.0 g of H₂ with 64.0 g of O₂. What is the maximum mass of H₂O that can be formed?

Step 1: Determine the Limiting Reactant

We must first calculate the moles of H₂O that could theoretically be produced from each reactant individually.

Calculation based on H₂: \[ \begin{align*} n(\mathrm{H_2O}) &= m(\mathrm{H_2}) ~ M(\mathrm{H_2})^{-1} ~ r(\mathrm{H_2O, H_2}) \\[1.5ex] &= \left( 10.\bar{0}~\mathrm{g~H_2} \right) \left( \frac{1~\mathrm{mol~H_2}}{2.0\bar{2}~\mathrm{g~H_2}} \right) \left( \frac{2~\mathrm{mol~H_2O}}{2~\mathrm{mol~H_2}} \right) \\[1.5ex] &= 4.9\bar{5}04~\mathrm{mol~H_2O} \end{align*} \]
Calculation based on O₂: \[ \begin{align*} n(\mathrm{H_2O}) &= m(\mathrm{O_2}) ~ M(\mathrm{O_2})^{-1} ~ r(\mathrm{H_2O, O_2}) \\[1.5ex] &= \left( 64.\bar{0}~\mathrm{g~O_2} \right) \left( \frac{1~\mathrm{mol~O_2}}{32.0\bar{0}~\mathrm{g~O_2}} \right) \left( \frac{2~\mathrm{mol~H_2O}}{1~\mathrm{mol~O_2}} \right) \\[1.5ex] &= 4.0\bar{0}00~\mathrm{mol~H_2O} \end{align*} \]

Comparing the two possible outcomes, we see that the given amount of O₂ produces fewer moles of H₂O (4.00 mol) than the given amount of H₂ (4.95 mol). Therefore, O₂ is the limiting reactant. The maximum amount of product that can possibly be formed is the unrounded 4.0000 mol H₂O.

Step 2: Calculate the Mass of Product

The final step is to calculate the mass of product based on the amount determined by the limiting reactant. We use the unrounded value from the previous step to avoid intermediate rounding errors. \[ \begin{align*} m(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= \left( 4.0\bar{0}00~\mathrm{mol~H_2O} \right) \left( \frac{18.0\bar{2}~\mathrm{g~H_2O}}{1~\mathrm{mol~H_2O}} \right) \\[1.5ex] &= 72.\bar{0}8~\mathrm{g~H_2O} \\[1.5ex] &= 72.1~\mathrm{g~H_2O} \end{align*} \] The maximum mass of water that can be produced is 72.1 g.

Calculating the Amount of Excess Reactant

Once the limiting reactant is completely consumed, the reaction stops. Any other reactants that are still present are said to be in excess. A common and important task is to calculate exactly how much of an excess reactant is left over, or unreacted, when the reaction is complete.

The thought process for this calculation is a simple subtraction: \[ \text{Amount of Excess (Leftover)} = \text{Initial Amount} - \text{Amount Consumed} \] The most critical part of this is calculating the Amount Consumed. This is a standard stoichiometry problem where the starting point is always the initial amount of the limiting reactant.

Example: Using the previous problem where we started with 10.0 g of H₂ and 64.0 g of O₂, we identified O₂ as the limiting reactant. How much of the excess reactant, H₂, is left over?

Step 1: Calculate the Mass of Excess Reactant Consumed

First, we must determine the mass of H₂ that actually reacts with the entire 64.0 g of our limiting reactant, O₂. This is a standard Mass A → Mass B calculation. \[ \begin{align*} m(\mathrm{H_2}) &= n(\mathrm{O_2})~ r(\mathrm{H_2, O_2}) ~ M(\mathrm{H_2}) \\[1.5ex] &= m(\mathrm{O_2}) ~ M(\mathrm{O_2})^{-1} ~ r(\mathrm{H_2, O_2}) ~ M(\mathrm{H_2}) \\[1.5ex] &= \left( 64.\bar{0}~\mathrm{g~O_2} \right) \left( \frac{1~\mathrm{mol~O_2}}{32.0\bar{0}~\mathrm{g~O_2}} \right) \left( \frac{2~\mathrm{mol~H_2}}{1~\mathrm{mol~O_2}} \right) \left( \frac{2.0\bar{2}~\mathrm{g~H_2}}{1~\mathrm{mol~H_2}} \right) \\[1.5ex] &= 8.0\bar{8}00~\mathrm{g~H_2} \\[1.5ex] &= 8.08~\mathrm{g~H_2} \end{align*} \] This result tells us that only 8.08 g of H₂ was needed to completely react with the 64.0 g of O₂.

Step 2: Calculate the Mass of Excess Reactant Leftover

Now, we subtract the mass of H₂ that was consumed from the initial mass of H₂ we started with. Remember to follow the rules for significant figures in subtraction. \[ \begin{align*} m(\mathrm{H_2, leftover}) &= m(\mathrm{H_2, initial}) - m(\mathrm{H_2, consumed}) \\[1.5ex] &= 10.\bar{0}~\mathrm{g~H_2} - 8.0\bar{8}~\mathrm{g~H_2} \\[1.5ex] &= 1.\bar{9}20~\mathrm{g~H_2} \\[1.5ex] &= 1.9~\mathrm{g~H_2} \end{align*} \] After the reaction is complete, 1.9 g of hydrogen gas will remain unreacted in the container.

Reaction Yields

The calculations we have performed so far have given us the theoretical yield of a reaction. This is the maximum amount of product that can possibly be formed from the given amounts of reactants, assuming a perfect reaction.

In the real world, however, the amount of product we actually collect in the laboratory is almost always less than the theoretical maximum. The amount of product obtained from a real, physical experiment is called the actual yield.

Why Actual Yield is Less Than Theoretical Yield

There are many reasons why an experiment might not be perfect:

Incomplete Reactions: Many reactions are reversible (at equilibrium) and do not proceed 100% to completion.
Side Reactions: The reactants may participate in other, unintended reactions that produce different products.
Loss of Product: Some product may be lost during experimental procedures like filtration, transfer between containers, or purification.

Percent Yield: A Measure of Efficiency

To describe the efficiency of a reaction, chemists use the percent yield. It is a ratio that compares the actual yield to the theoretical yield, expressed as a percentage. \[ \mathrm{\%~yield} = \frac{\mathrm{actual~yield}}{\mathrm{theoretical~yield}} \times 100~\% \] A high percent yield indicates an efficient reaction with minimal loss, while a low percent yield suggests that significant loss or inefficiencies occurred. Yield is often expressed as a mass (m_actual and m_theoretical) but it can be other physical quantities.

\[ \mathrm{\%~yield} = \frac{m_{\mathrm{actual}}}{m_{\mathrm{theoretical}}} \times 100~\% \]

Example: Let’s reconsider our limiting reactant problem where we mixed 10.0 g of H₂ with 64.0 g of O₂. We calculated that the theoretical yield of H₂O is 72.1 g.

Suppose after performing this reaction in the lab, you collect and weigh the water produced, and you find that your actual yield is 68.5 g. What is the percent yield of the reaction?

We construct the solution using the definition of percent yield. \[ \begin{align*} \%~\mathrm{yield} &= \frac{m_{\mathrm{actual}}}{m_{\mathrm{theoretical}}} \times 100~\% \\[1.5ex] &= \left( \frac{68.\bar{5}~\mathrm{g}}{72.\bar{1}~\mathrm{g}} \right) \times 100~\% \\[1.5ex] &= 95.\bar{0}06~\% \\[1.5ex] &= 95.0~\% \end{align*} \] The calculation is limited to 3 significant figures from both the actual and theoretical yields. The reaction had a 95.0 % yield.

Working Backwards from Percent Yield to a Mass

A more practical and common use of percent yield is to answer the question: “If I know this reaction is inefficient, how much starting material do I actually need to use to guarantee I make a certain amount of product?”

In this type of problem, the mass of product you want to make is your actual yield. You must first use the percent yield to calculate the theoretical yield, which is the larger amount of product you would need to aim for in a perfect world. From this theoretical yield, you can then perform a standard stoichiometric calculation to find the required mass of the reactant.

Example: Using the synthesis of water what mass of H₂ is required to actually produce 68.5 g of H₂O if the reaction is known to have a 95.0 % yield?

\[2~\mathrm{H_2(g)} + \mathrm{O_2(g)} \longrightarrow \mathrm{2~H_2O(l)}\]

Step 1: Calculate the Theoretical Yield

First, we must determine the theoretical yield of H₂O we need to target. Rearrange the percent yield equation to solve for the theoretical mass, substitute in the physical quantities and then solve.

\[ \begin{align*} \mathrm{\%~yield} &= \frac{m_{\mathrm{actual}}}{m_{\mathrm{theoretical}}} \times 100~\% \longrightarrow \\[1.5ex] m_{\mathrm{theoretical}} &= \frac{m_{\mathrm{actual}}}{\%~\mathrm{yield}} \times 100\% \\[1.5ex] &= \left( \frac{68.\bar{5}~\mathrm{g}}{95.\bar{0}~\%} \right) \times 100\% \\[1.5ex] &= 72.\bar{1}05~\mathrm{g} \end{align*} \]

This tells us that to obtain 68.5 g of water from this inefficient reaction, we must start with enough reactants to theoretically produce 72.1 g of water.

Step 2: Calculate the Required Mass of Reactant

Now, we perform a standard Mass B → Mass A stoichiometric calculation. We will use the unrounded theoretical yield from Step 1 as our starting point to find the required mass of H₂. \[ \begin{align*} m(\mathrm{H_2}) &= n(\mathrm{H_2}) ~ M(\mathrm{H_2})\\[1.5ex] &= n(\mathrm{H_2O})_{\mathrm{theoretical}} ~ r(\mathrm{H_2,H_2O}) ~ M(\mathrm{H_2}) \\[1.5ex] &= m(\mathrm{H_2O})_{\mathrm{theoretical}} ~ M(\mathrm{H_2O})^{-1} ~ r(\mathrm{H_2, H_2O}) ~ M(\mathrm{H_2}) \\[1.5ex] &= \left( 72.\bar{1}05~\mathrm{g~H_2O} \right) \left( \frac{1~\mathrm{mol~H_2O}}{18.0\bar{2}~\mathrm{g~H_2O}} \right) \left( \frac{2~\mathrm{mol~H_2}}{2~\mathrm{mol~H_2O}} \right) \left( \frac{2.0\bar{2}~\mathrm{g~H_2}}{1~\mathrm{mol~H_2}} \right) \\[1.5ex] &= 8.0\bar{8}28~\mathrm{g~H_2} \\[1.5ex] &= 8.08~\mathrm{g~H_2} \end{align*} \] The calculation is limited to 3 significant figures from the given actual yield and percent yield. We must start with 8.09 g of H₂ to achieve our goal.

Stoichiometry of Sequential Reactions

In many real-world chemical processes, a desired product is not formed in a single step but through a series of consecutive reactions. These are known as sequential reactions. In this sequence, the product of one reaction becomes a reactant in the next.

The key to solving these problems is to recognize that the mole is the only bridge between the separate chemical equations. You must use the mole ratio from the first reaction to find the amount of the intermediate substance, and then use that amount as the starting point for the second reaction.

A schematic flowchart of the Kroll process demonstrating the refinement of titanium ore to pure titanium using magnesium chloride.

Titanium production: A schematic of the Kroll process used in the commercial production of titanium (recreated figure¹).

Example: Titanium metal is produced commercially via the Kroll process, a two-step sequence. First, titanium(IV) oxide is converted to titanium(IV) chloride, which then reacts with magnesium to produce pure titanium.

Reaction 1: \[ \mathrm{TiO_2(s)} + 2~\mathrm{Cl_2(g)} + 2~\mathrm{C(s)} \longrightarrow \mathrm{TiCl_4(g)} + 2~\mathrm{CO(g)} \]
Reaction 2: \[ \mathrm{TiCl_4(g)} + 2~\mathrm{Mg(s)} \longrightarrow \mathrm{Ti(s)} + 2~\mathrm{MgCl_2(s)} \]

If you begin with 10.0 kg of TiO₂, what is the theoretical yield of pure Ti(s) in kg?

Step 1: Calculate the Amount of the Intermediate (TiCl₄)

We start with the mass of TiO₂ and use the stoichiometry of the first reaction to find the amount (in moles) of the intermediate product, TiCl₄. Note that we must work with a consistent mass unit, so we will convert kg to g. \[ \begin{align*} n(\mathrm{TiCl_4}) &= n(\mathrm{TiO_2}) ~ r(\mathrm{TiCl_4, TiO_2}) \\[1.5ex] &= m(\mathrm{TiO_2}) ~ M(\mathrm{TiO_2})^{-1} ~ r(\mathrm{TiCl_4, TiO_2}) \\[1.5ex] &= \left( 10.\bar{0}~\mathrm{kg~TiO_2} \right) \left( \frac{1000~\mathrm{g}}{1~\mathrm{kg}} \right) \left( \frac{1~\mathrm{mol~TiO_2}}{79.8\bar{7}~\mathrm{g~TiO_2}} \right) \left( \frac{1~\mathrm{mol~TiCl_4}}{1~\mathrm{mol~TiO_2}} \right) \\[1.5ex] &= 12\bar{5}.20~\mathrm{mol~TiCl_4} \end{align*} \]

Step 2: Calculate the Mass of the Final Product (Ti)

Now, we use the amount of the intermediate, 125.20 mol TiCl₄, as the starting point for the second reaction. Be sure to determine the mass in kg. \[ \begin{align*} m(\mathrm{Ti}) &= n(\mathrm{Ti}) ~ M(\mathrm{Ti}) \\[1.5ex] &= n(\mathrm{TiCl_4}) ~ r(\mathrm{Ti, TiCl_4}) ~ M(\mathrm{Ti}) \\[1.5ex] &= \left( 12\bar{5}.20~\mathrm{mol~TiCl_4} \right) \left( \frac{1~\mathrm{mol~Ti}}{1~\mathrm{mol~TiCl_4}} \right) \left( \frac{47.8\bar{7}~\mathrm{g~Ti}}{1~\mathrm{mol~Ti}} \right) \left( \frac{1~\mathrm{kg}}{1000~\mathrm{g}} \right) \\[1.5ex] &= 5.9\bar{9}33~\mathrm{kg~Ti} \\[1.5ex] &= 5.99~\mathrm{kg~Ti} \end{align*} \] The theoretical yield of titanium is 5.99 kg.

Practice

Practice

Solid aluminum metal reacts with aqueous hydrochloric acid to produce aqueous aluminum chloride and hydrogen gas, as shown in the balanced equation below.

\[ \mathrm{2~Al(s)} + \mathrm{6~HCl(aq)} \longrightarrow \mathrm{2~AlCl_3(aq)} + \mathrm{3~H_2(g)} \]

A reaction is initiated by adding a 15.0 g piece of aluminum foil to a beaker containing 25.0 g of hydrochloric acid.

Identify the limiting reactant.
What is the theoretical yield (in g) of hydrogen gas?
What mass (in g) of the excess reactant remains after the reaction is complete?
After performing the experiment, 0.625 g of hydrogen gas is collected. What is the percent yield of the reaction?

Solution

(a) Identifying the Limiting Reactant

To find the limiting reactant, we calculate the amount of a single product (in this case, H₂) that could be formed from each reactant individually. The reactant that produces the least amount of product is the limiting reactant.

Calculation based on Al(s): \[ \begin{align*} n(\mathrm{H_2}) &= n(\mathrm{Al}) ~ r(\mathrm{H_2,Al}) \\[1.5ex] &= m(\mathrm{Al}) ~ M(\mathrm{Al})^{-1} ~ r(\mathrm{H_2, Al}) \\[1.5ex] &= \left( 15.\bar{0}~\mathrm{g~Al} \right) \left( \frac{1~\mathrm{mol~Al}}{26.9\bar{8}~\mathrm{g~Al}} \right) \left( \frac{3~\mathrm{mol~H_2}}{2~\mathrm{mol~Al}} \right) \\[1.5ex] &= 0.83\bar{3}95~\mathrm{mol~H_2} \end{align*} \]
Calculation based on HCl(aq): \[ \begin{align*} n(\mathrm{H_2}) &= n(\mathrm{HCl}) ~ r(\mathrm{H_2, HCl}) \\[1.5ex] &= m(\mathrm{HCl}) ~ M(\mathrm{HCl})^{-1} ~ r(\mathrm{H_2, HCl}) \\[1.5ex] &= \left( 25.\bar{0}~\mathrm{g~HCl} \right) \left( \frac{1~\mathrm{mol~HCl}}{36.4\bar{6}~\mathrm{g~HCl}} \right) \left( \frac{3~\mathrm{mol~H_2}}{6~\mathrm{mol~HCl}} \right) \\[1.5ex] &= 0.34\bar{2}84~\mathrm{mol~H_2} \end{align*} \]

Comparing the two possible outcomes, the initial amount of HCl produces fewer moles of H₂. Therefore, HCl(aq) is the limiting reactant.

(b) Theoretical Yield of Hydrogen Gas

The theoretical yield is the maximum amount of product that can be formed, as determined by the limiting reactant. We will use the unrounded mole value from the HCl calculation to find the mass of H₂.

\[ \begin{align*} m(\mathrm{H_2}) &= n(\mathrm{H_2}) ~ M(\mathrm{H_2}) \\[1.5ex] &= \left( 0.34\bar{2}84~\mathrm{mol~H_2} \right) \left( \frac{2.02~\mathrm{g~H_2}}{1~\mathrm{mol~H_2}} \right) \\[1.5ex] &= 0.69\bar{2}53~\mathrm{g~H_2} \\[1.5ex] &= 0.692~\mathrm{g~H_2} \end{align*} \] (Note: Using the round-half-to-even rule, the answer rounds down to 0.692 g. If using the round-half-up rule, the answer would be 0.693 g.)

(c) Mass of Excess Reactant Remaining

First, we calculate the mass of the excess reactant (Al) that is consumed by reacting completely with the limiting reactant (HCl).

\[ \begin{align*} m(\mathrm{Al})_{\mathrm{consumed}} &= n(\mathrm{Al}) ~ M(\mathrm{Al}) \\[1.5ex] &= n(\mathrm{HCl}) ~ r(\mathrm{Al, HCl}) ~ M(\mathrm{Al}) \\[1.5ex] &= m(\mathrm{HCl}) ~ M(\mathrm{HCl})^{-1} ~ r(\mathrm{Al, HCl}) ~ M(\mathrm{Al}) \\[1.5ex] &= \left( 25.\bar{0}~\mathrm{g~HCl} \right) \left( \frac{1~\mathrm{mol~HCl}}{36.4\bar{6}~\mathrm{g~HCl}} \right) \left( \frac{2~\mathrm{mol~Al}}{6~\mathrm{mol~HCl}} \right) \left( \frac{26.9\bar{8}~\mathrm{g~Al}}{1~\mathrm{mol~Al}} \right) \\[1.5ex] &= 6.1\bar{6}65~\mathrm{g~Al} \end{align*} \]

Next, we subtract the consumed mass from the initial mass to find the amount remaining.

\[ \begin{align*} m(\mathrm{Al})_{\mathrm{remaining}} &= m(\mathrm{Al})_{\mathrm{initial}} - m(\mathrm{Al})_{\mathrm{consumed}} \\[1.5ex] &= 15.\bar{0}~\mathrm{g~Al} - 6.1\bar{6}65~\mathrm{g~Al} \\[1.5ex] &= 8.\bar{8}33~\mathrm{g~Al} \\[1.5ex] &= 8.8~\mathrm{g~Al} \end{align*} \]

(d) Percent Yield of the Reaction

The percent yield of H₂ is the ratio of the actual (collected) yield to the theoretical yield. We use the unrounded theoretical yield for this calculation.

\[ \begin{align*} \%~\mathrm{yield} &= \frac{m_{\mathrm{actual}}}{m_{\mathrm{theoretical}}} \times 100~\% \\[1.5ex] &= \left( \frac{0.62\bar{5}~\mathrm{g~H_2}}{0.69\bar{2}53~\mathrm{g~H_2}} \right) \times 100~\% \\[1.5ex] &= 90.\bar{2}48~\% \\[1.5ex] &= 90.2~\% \end{align*} \]

References

(1)

Reddy, R. G.; Shinde, P. S.; Liu, A. Review—the Emerging Technologies for Producing Low-Cost Titanium. Journal of The Electrochemical Society 2021, 168 (4), 042502. https://doi.org/10.1149/1945-7111/abe50d.