Quantifying substances in chemical reactions

Stoichiometry helps to understand the relationships between reactants and products, enabling reactions to be planned for producing desired quantities. Using stoichiometry, the amounts of greenhouse gases, for example CO2, CH4 and H2O, can be quantified.

Stoichiometry is based on mole ratios. However, in industrial production or laboratory experiments, chemicals are often measured in units like mass or volume. These quantities must be converted to moles before applying stoichiometry.

Use this page to revise the following concepts within quantifying substances in chemical reactions:

Moles

The counting unit in chemistry is the mole. One mole contains exactly 6.02214076 × 1023 particles. This is called Avogadro’s constant (NA). For simplicity, we usually round this to 6.02 × 1023 in calculations.

With the mole as the counting unit for chemical particles, the coefficients of balanced equations show the ratio of moles of substances. Converting measurable quantities like mass, volume, and concentration into moles is the basis of stoichiometry calculations.

Molar mass ((M, measured in \(g\,mol^{-1}\) ) shows the mass \((m)\), measured in grams) for each mole of a pure substance. If the mass of a pure substance is given, and its molar mass is known, the number of moles (𝑛) can be calculated by:

\[n=\frac{m}{n}\]

Molar volume (Vm, measured in \(L\,mol^{-1}\))of an ideal gas refers to the volume occupied by one mole of ideal gas (\(V\), measured in \(L\)) for one mole of the gas particles. For ideal gases at Standard Laboratory Conditions (SLC) (100 kPa and 25°C), the molar volume of a gas is 24.8 L mol-1. Therefore, the number of moles (𝑛) of gas particles at SLC can be calculated by:

\[n=\frac{V}{V_m}\]

For solutions, the unit for concentration (𝑐) varies. If the concentration of a solution is measured in moles per litre (\(mol\,L^{-1}\)), and the volume of the solution is known in litres (\(L\)), the number of moles (𝑛) of gas particles at SLC can be calculated by:

\[n={c}\times{V}\]


Mole-mole stoichiometry

Stoichiometry shows the relationships between the amounts of reactants and products. The number of particles is quantified in moles.

For example, the number of moles of O2 needed for the complete combustion of 2 mol of CH4 can be calculated according to the following balanced thermochemical equation :

CH4​ (g) +  2O2​ (g) →   CO2 ​(g) + 2H2​O (l)  ΔH = −890 kJ

This balanced thermochemical equation tells us that 1 mole of CH4​ reacts with 2 moles of O2 to produce 1 mole of CO2​ and 2 moles of H2O, releasing 890 kJ of energy.

Step 1: Set up ratios

\[\frac{n(CH_4)}{nO_2}=\frac{1}{2}\]

Step 2: Substitute the number of moles of the known substance

Substitute \(n(CH_4)=2\) mol to the above ratio:

\[\frac{2}{n(O_2)}=\frac{1}{2}\]

Step 3: Solve the equation to determine the number of moles of the unknown substance

\[n(O_2)=4 mol\]

In the same way, the quantities of products can be calculated:

\[n(CO_2)=2 mol\]

\[n(H_2O)= 4mol\]


Stoichiometry calculations

If the amount of the given or desired substance is in a unit other than moles, convert the amount of the known substance to moles first before applying stoichiometry calculations. After obtaining the moles of the unknown substance, convert this value back to the desired unit of measurement.

Diagram showing the steps of a stoichiometry calculation. The inputs on the left are “Mass of pure substance”, “Volume of gas at SLC” and “Concentration and volume of a solution”. These are used to “Find mole of known Substance A from the given quantity”. This feeds into “Find mole of unknown Substance B using stoichiometry.” Finally, this is converted back to “Mass of pure substance”, “Volume of gas at SLC” and “Concentration and volume of a solution”.”

Mass - mass stoichiometry

If the quantity of the known substance is given as its mass, and the desired quantity is the mass of the unknown substance, this is called mass - mass stoichiometry.

Worked Example

Determine the mass of O2 needed for the complete combustion of 32.0g CH4 according to the following balanced thermochemical equation:

CH4​(g) +  2O2​(g) →   CO2​(g) + 2H2​O(l)  ΔH = −890 kJ

Step 1: Convert the quantity of known substance to moles using \(n=\frac{m}{M}\)

\[n(CH_4)=\frac{m(CH_4)}{M(CH_4)}=\frac{32.0}{16.0}=2.00mol\]

Step 2: Follow mole-to-mole stoichiometry to calculate the number of moles of the unknown substance

\[n=\frac{n(CH_4)}{n(O_2)}=\frac{1}{2}\]

Given \(n(CH_4)=2.00mol\), hence \(\frac{2}{n(O_2)}=\frac{1}{2}\)

\[n(O_2)=4.00mol\]

Step 3: Convert the amount of unknown substance to the desired quantity using \(m={n}\times{M}\)

\[m(O_2)=n(O_2)\times{M(O_2)}=4.00\times32.0=128g\]

In the same way, the quantities of other reactants or products involved in the reaction can be calculated.

Gas volume-volume stoichiometry

If both the known and unknown substances are gases, the calculation involves determining the volume of the unknown gas from the volume of the known gas. This is called gas volume-volume stoichiometry.

Worked Example

Determine the volume of O2 needed for the complete combustion of 49.6L of CH4 under SLC conditions according to the balanced equation:

CH4 ​(g) +  2O2 (g) →   CO2​(g) + 2H2​O(l)  ΔH = −890 kJ

Solution

When the gas substances in a chemical equation are measured under the same temperature and pressure, e.g. SLC conditions, the mole ratio will also be a volume ratio.

In this question,

\[V(CH_4)=n(CH_4)\times{V_m}\]

\[V(O_2)=n(O_2)\times{V_m}\]

When the gas particles are at the same temperature and pressure \(V_m\) is the same value for both gases. Therefore,

\[
\frac{n(\text{CH}_4)}{n(\text{O}_2)} = \frac{V(\text{CH}_4)}{V(\text{O}_2)}
\]

Hence, you can use a quicker method, as shown below, to solve gas volume-volume stoichiometry questions.

Step 1

According to the coefficients of CH4 and O2,

\[\frac{V(CH_4)}{V(O_2)}=\frac{1}{2}\]

Step 2

Given \(V(CH_4)=46.9L\), hence \(\frac{48.6}{V(O_2)}=\frac{1}{2}\)

\[V(O_2)=99.2L\]


Stoichiometry calculations involving limiting reactants

When reactants are not in the stoichiometric ratio as indicated by the balanced chemical equations, one reactant is used up (the limiting reactant), while the other remains in excess (the excess reactant). In this case, identifying the limiting reactant enables accurate calculations of the amount of product and remaining reactants.

The excess reactant cannot be used in its entirety when calculating the product formed, because not all of the excess reactant is consumed in the reaction. Therefore, the amount of product depends on the amount of the limiting reactant. This means stoichiometric calculations must use the amount of the limiting reactant that is fully consumed.

The process of determining the limiting reactant involves three steps:

  • Step 1: Calculate the number of moles of each reactant.
  • Step 2: Identify the limiting reactant by comparing the mole ratio from the balanced equation to the amounts available.
  • Step 3: Use the moles of the limiting reactant to calculate the amount of product formed.

Worked Example

A mixture of 25.0g of octane (C8H18, liquid) and 100.0L of oxygen gas (O2) at SLC is combusted. The balanced equation for the reaction is:

2C8H18 (l) + 25O2 (g) → 16CO2(g) + 18H2O(g)

Calculate the mass of carbon dioxide (CO2) produced, in grams.

Solution

Step 1: Calculate the number of moles for each reactant

\[n(C_8H_18)=\frac{m(C_8H_{18})}{M(C_8H_{18})}=\frac{25.0}{114.0}=0.220mol\]

\[n(O_2)=\frac{V(O_2)}{V_m}=\frac{100.0}{24.8}=4.03mol\]

Step 2: Identify the limiting reactant

According to the balanced chemical equation: \(\frac{n(C_8H_{18})}{n(O_2)}=\frac{2}{5}\)

To react with 0.220 mol of C8H18, the required \(n(O_2)=\frac{25}{2}\times{n(C_8H_{18})}=2.74mol\)

The provided \(n(O_2)=4.03mol\) is greater than \(2.74mol\)

Therefore, O2 is the excess reactant, while C8H18 is the limiting reactant.

Step 3: Calculate the moles of CO2 produced from \(n(C_8H_18)=0.220mol\)

\(\frac{n(C_8H_18)}{n(C_2)}=\frac{1}{8}\), hence \(n(CO_2)=\frac{8}{1}\times{n(C_8H_{18})}=1.75mol\)