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Calculations Involving Chemical Reactions

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Role of Calculations in Chemical Reactions

Once you know how to describe a chemical reaction with a balanced chemical equation and you are familiar with the basic quantities of stoichiometry (amount of substance in moles, molar mass, mass fraction, etc.), you can use these tools to make quantitative predictions. In this chapter, the focus is on turning a balanced chemical equation into actual numbers: how much reactant is needed, how much product is formed, and which reactant runs out first.

We will always assume that you already:

Here, we concentrate on how to use these in practical calculations involving reactions.

Stoichiometric Ratios from the Reaction Equation

A balanced chemical equation shows how many “formula units” of each substance react with each other. These numbers (the stoichiometric coefficients) also give mole ratios.

Example: combustion of hydrogen:
$$
2\,\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\,\mathrm{H_2O}
$$

From this we can read the mole ratios:

These mole ratios are the central link between the amount of one substance and the amount of another in the same reaction.

General rule:

  1. Start from the balanced reaction equation.
  2. Identify the given substance and the desired substance.
  3. Use the stoichiometric coefficients to set up a ratio between their amounts of substance (in moles).

General Strategy for Stoichiometric Calculations

Most problems involving chemical reactions can be addressed with the same systematic approach:

  1. Write and balance the chemical equation.
  2. Convert all given quantities to amount of substance $n$ in moles:
    • From mass $m$: use $n = \dfrac{m}{M}$, where $M$ is molar mass.
    • From substance amount concentration $c$ and volume $V$ of a solution: $n = c \cdot V$.
    • From gas volume (at given conditions): use appropriate gas relations (introduced elsewhere).
  3. Use the mole ratio from the balanced equation to calculate the moles of the desired substance.
  4. Convert the calculated moles to the required quantity:
    • Mass: $m = n \cdot M$.
    • Volume of gas, concentration, etc. (depending on the task).

This “mole bridge” is the core of all stoichiometric calculations:

Mass–Mass Calculations

Mass–mass calculations are among the most common stoichiometric calculations. You are given the mass of one substance and asked for the mass of another.

Example: Producing Copper(II) Oxide from Copper

Reaction:
$$
2\,\mathrm{Cu} + \mathrm{O_2} \rightarrow 2\,\mathrm{CuO}
$$

Task: How many grams of $\mathrm{CuO}$ can be formed from $6.35\ \mathrm{g}$ of copper, assuming oxygen is present in excess?

  1. Balance the equation (already balanced).
  2. Convert given mass to moles:
    • Molar mass of copper: $M(\mathrm{Cu}) \approx 63.5\ \mathrm{g\ mol^{-1}}$.
    • $n(\mathrm{Cu}) = \dfrac{6.35\ \mathrm{g}}{63.5\ \mathrm{g\ mol^{-1}}} = 0.100\ \mathrm{mol}$.
  3. Use mole ratio from equation:
    • $2$ mol $\mathrm{Cu}$ form $2$ mol $\mathrm{CuO}$.
    • So $n(\mathrm{Cu}) : n(\mathrm{CuO}) = 1 : 1$.
    • Thus $n(\mathrm{CuO}) = 0.100\ \mathrm{mol}$.
  4. Convert moles of product to mass:
    • $M(\mathrm{CuO}) \approx 63.5 + 16.0 = 79.5\ \mathrm{g\ mol^{-1}}$.
    • $m(\mathrm{CuO}) = 0.100\ \mathrm{mol} \cdot 79.5\ \mathrm{g\ mol^{-1}} = 7.95\ \mathrm{g}$.

So, $6.35\ \mathrm{g}$ of copper can form $7.95\ \mathrm{g}$ of copper(II) oxide.

Reversed Problem: Required Mass of Reactant

The same method allows you to calculate how much of a reactant is needed to obtain a given mass of product.

Only the direction of the calculation changes:

Limiting and Excess Reactants

In many real situations, you start with finite amounts of all reactants. Typically, one reactant will be used up first. This is the limiting reactant (or limiting reagent), because it limits how much product can be formed. All other reactants are then in excess.

The idea:

Identifying the Limiting Reactant

General procedure:

  1. Balance the chemical reaction.
  2. Calculate the moles of each reactant from the given data.
  3. For each reactant, divide the amount of substance actually present by its stoichiometric coefficient in the equation:
    $$
    \frac{n(\text{reactant})}{\nu(\text{reactant})}
    $$
    where $\nu$ is the stoichiometric coefficient (e.g. $ in \,\mathrm{H_2}$).
  4. The reactant with the smallest value is the limiting reactant.
  5. Use the amount of the limiting reactant to determine how much product forms.

Example: Reaction of Magnesium with Hydrochloric Acid

Reaction:
$$
\mathrm{Mg} + 2\,\mathrm{HCl} \rightarrow \mathrm{MgCl_2} + \mathrm{H_2}
$$

Task: $2.43\ \mathrm{g}$ of magnesium react with $100.0\ \mathrm{mL}$ of $\mathrm{2.00\ mol\ L^{-1}}$ hydrochloric acid solution.
a) Which reactant is limiting?
b) How many moles of hydrogen gas can form at most?

  1. Moles of reactants:
    • Magnesium:
      • $M(\mathrm{Mg}) \approx 24.3\ \mathrm{g\ mol^{-1}}$
        $\Rightarrow n(\mathrm{Mg}) = \dfrac{2.43\ \mathrm{g}}{24.3\ \mathrm{g\ mol^{-1}}} = 0.100\ \mathrm{mol}$
    • Hydrochloric acid:
      • $c(\mathrm{HCl}) = 2.00\ \mathrm{mol\ L^{-1}}$
        $V = 100.0\ \mathrm{mL} = 0.1000\ \mathrm{L}$
        $\Rightarrow n(\mathrm{HCl}) = 2.00\ \mathrm{mol\ L^{-1}} \cdot 0.1000\ \mathrm{L} = 0.200\ \mathrm{mol}$
  2. Use stoichiometric coefficients:
    • Balanced reaction: $1\,\mathrm{Mg} : 2\,\mathrm{HCl}$.
    • Required mole ratio: $n(\mathrm{HCl}) : n(\mathrm{Mg}) = 2 : 1$.
  3. Check which is limiting:
    • Amount of $\mathrm{HCl}$ needed for $0.100\ \mathrm{mol}$ $\mathrm{Mg}$:
      • .100\ \mathrm{mol\ Mg} \times \dfrac{2\ \mathrm{mol\ HCl}}{1\ \mathrm{mol\ Mg}} = 0.200\ \mathrm{mol\ HCl}$.
    • We have exactly $0.200\ \mathrm{mol\ HCl}$.

Thus, both reactants are present in exact stoichiometric ratio; no reactant is in excess here. (Both are “limiting” in the sense that they will be fully consumed.)

  1. Maximum amount of hydrogen formed:
    • From the equation: $1\ \mathrm{mol\ Mg}$ produces $1\ \mathrm{mol\ H_2}$.
    • So $0.100\ \mathrm{mol\ Mg}$ produce:
      $n(\mathrm{H_2}) = 0.100\ \mathrm{mol}$.

If instead we had, for example, $0.150\ \mathrm{mol}$ $\mathrm{HCl}$ with the same $0.100\ \mathrm{mol}$ $\mathrm{Mg}$:

Calculating Excess Reactant and Leftover Amounts

Once you know which reactant is limiting, you can determine how much of the excess reactant remains unreacted:

  1. Use the limiting reactant to calculate how many moles of the excess reactant are needed according to the stoichiometric ratio.
  2. Subtract this from the initial moles of the excess reactant.
  3. Convert remaining moles into the desired form (mass, concentration, etc.).

Example: Excess Oxygen in Combustion of Carbon

Reaction:
$$
\mathrm{C} + \mathrm{O_2} \rightarrow \mathrm{CO_2}
$$

Task: $6.00\ \mathrm{g}$ carbon reacts with $20.0\ \mathrm{g}$ oxygen.
a) Which reactant is limiting?
b) How much of the excess reactant remains?

  1. Moles of reactants:
    • Carbon: $M(\mathrm{C}) = 12.0\ \mathrm{g\ mol^{-1}}$
      $n(\mathrm{C}) = \dfrac{6.00\ \mathrm{g}}{12.0\ \mathrm{g\ mol^{-1}}} = 0.500\ \mathrm{mol}$.
    • Oxygen: $M(\mathrm{O_2}) = 32.0\ \mathrm{g\ mol^{-1}}$
      $n(\mathrm{O_2}) = \dfrac{20.0\ \mathrm{g}}{32.0\ \mathrm{g\ mol^{-1}}} = 0.625\ \mathrm{mol}$.
  2. Stoichiometric coefficients: $1\,\mathrm{C} : 1\,\mathrm{O_2}$.
  3. Limiting reactant:
    • For $0.500\ \mathrm{mol}$ C, we need $0.500\ \mathrm{mol}$ $\mathrm{O_2}$.
    • We have $0.625\ \mathrm{mol}$ $\mathrm{O_2}$.
    • Therefore, carbon is limiting, oxygen is in excess.
  4. Leftover oxygen:
    • Oxygen used: $0.500\ \mathrm{mol}$.
    • Oxygen leftover: $0.625 - 0.500 = 0.125\ \mathrm{mol}$.
    • In grams: $m = 0.125\ \mathrm{mol} \cdot 32.0\ \mathrm{g\ mol^{-1}} = 4.00\ \mathrm{g}$.

So, $4.00\ \mathrm{g}$ oxygen remains unreacted.

Yield of a Reaction: Theoretical and Actual Yield

In practice, reactions rarely proceed perfectly. Not all of the limiting reactant is necessarily converted into the desired product; side reactions and losses often occur. Thus, we distinguish between:

From these, we define the percent yield (or percentage yield):
$$
\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\ \%
$$

Percent yield describes the efficiency of a chemical preparation.

Example: Preparing Sodium Chloride

Reaction:
$$
\mathrm{NaOH} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}
$$

Suppose:

a) Theoretical yield of $\mathrm{NaCl}$:

b) Percent yield:
$$
\text{Percent yield} = \frac{25.0\ \mathrm{g}}{29.3\ \mathrm{g}} \times 100\ \% \approx 85.3\ \%
$$

So, the reaction ran with about $85\ \%$ yield.

Calculations Involving Gases in Reactions

For reactions with gases, you work—again—via the mole concept. The difference is that you often measure volume rather than mass. The relationship between volume and amount of substance of a gas depends on the chosen conditions (temperature, pressure).

Where an appropriate gas law is already known (and under conditions where gas behavior can be approximated as ideal):

The procedure remains:

  1. Convert gas volume $\rightarrow$ moles (using the gas relation appropriate from elsewhere in the course).
  2. Use stoichiometric ratios in moles.
  3. Convert moles back to volume (or mass) as needed.

Example: Volume of Oxygen Required for Combustion of Methane

Reaction:
$$
\mathrm{CH_4} + 2\,\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\,\mathrm{H_2O}
$$

Task: What volume of $\mathrm{O_2}$ (at the same temperature and pressure) is required to completely burn $10.0\ \mathrm{L}$ of methane gas?

Under identical temperature and pressure for both gases, the mole ratio equals the volume ratio (for ideal gases), because volume is directly proportional to moles.

From the equation:

No explicit use of molar volume is needed when all gases are at the same conditions and only relative volumes are asked.

Concentration-Based Reaction Calculations (Solutions)

Many reactions occur in solution. There, the amount of reactant is often given by concentration $c$ and volume $V$. The key relationship is:
$$
n = c \cdot V
$$
($V$ in liters when $c$ is in $\mathrm{mol\ L^{-1}}$.)

This allows you to do calculations like:

The procedure is analogous:

  1. Calculate moles from $c$ and $V$.
  2. Use mole ratios from the reaction equation.
  3. Convert back to concentration or volume if required.

Example: Acid–Base Neutralization in Solution

Reaction:
$$
\mathrm{H_2SO_4} + 2\,\mathrm{NaOH} \rightarrow \mathrm{Na_2SO_4} + 2\,\mathrm{H_2O}
$$

Task: What volume of $\mathrm{1.00\ mol\ L^{-1}}$ sodium hydroxide solution is needed to neutralize $25.0\ \mathrm{mL}$ of $\mathrm{0.500\ mol\ L^{-1}}$ sulfuric acid?

  1. Moles of sulfuric acid:
    • $c(\mathrm{H_2SO_4}) = 0.500\ \mathrm{mol\ L^{-1}}$
      $V(\mathrm{H_2SO_4}) = 25.0\ \mathrm{mL} = 0.0250\ \mathrm{L}$
      $n(\mathrm{H_2SO_4}) = 0.500\ \mathrm{mol\ L^{-1}} \cdot 0.0250\ \mathrm{L} = 0.0125\ \mathrm{mol}$.
  2. Moles of NaOH needed:
    • Stoichiometry: $1\,\mathrm{H_2SO_4} : 2\,\mathrm{NaOH}$.
    • $n(\mathrm{NaOH}) = 2 \cdot n(\mathrm{H_2SO_4}) = 2 \cdot 0.0125\ \mathrm{mol} = 0.0250\ \mathrm{mol}$.
  3. Volume of NaOH solution:
    • $c(\mathrm{NaOH}) = 1.00\ \mathrm{mol\ L^{-1}}$,
    • $V(\mathrm{NaOH}) = \dfrac{n}{c} = \dfrac{0.0250\ \mathrm{mol}}{1.00\ \mathrm{mol\ L^{-1}}} = 0.0250\ \mathrm{L} = 25.0\ \mathrm{mL}$.

So, $25.0\ \mathrm{mL}$ of $\mathrm{1.00\ mol\ L^{-1}}$ NaOH solution are required.

Summary of Typical Problem Types

Using the same fundamental approach (moles and stoichiometric ratios), you can tackle many recurring problem types:

In all of these, the steps are:

  1. Balanced equation.
  2. Convert to moles.
  3. Use stoichiometric ratios.
  4. Convert to desired quantities.

With practice, this systematic approach becomes a straightforward tool for quantitatively describing and planning chemical reactions.

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