Stoichiometry and Stoichiometric Calculations

Stoichiometry is the branch of Chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It helps us determine how much of each substance is needed or produced in a reaction. It is based on the Law of Conservation of Mass, which states that matter cannot be created or destroyed during a chemical reaction. Therefore, the number of atoms of each element remains the same before and after the reaction.

In general, all chemical reactions depend on one key factor: the amount of substance present. Stoichiometry helps measure and calculate important quantities, such as:

• Mass of reactants and products
• Number of moles involved
• Molecular and molar masses
• Volume of gases participating in reactions

Stoichiometric Coefficient

The stoichiometric coefficient is the number written in front of a chemical formula in a balanced chemical equation. It represents the number of molecules or moles of a substance participating in the reaction.

Consider the following equation:

aA + bB ⇌ cC + dD

The Stoichiometric coefficients of A, B, C, and D, respectively, are a, b, c, and d in this equation.

• Stoichiometric coefficients establish the mole ratio between reactants and products.
• These coefficients are usually written as whole numbers, although fractional coefficients can be used during intermediate steps.

Example:

• 3 moles of iron react with 4 moles of water
• 1 mole of iron oxide and 4 moles of hydrogen gas are formed

Balanced Reactions and Mole Ratios

• In Stoichiometry, calculations are based on a Balanced Chemical Equation.
• A balanced equation is a chemical equation in which the number of atoms of each element is the same on both sides of the equation.
• This follows the Law of Conservation of Mass, which states that matter cannot be created or destroyed during a chemical reaction.
• Balancing an equation ensures that the reactants and products are in the correct proportion.

Example:

2H₂+O₂→2H₂O

The number of atoms on both sides is equal:

• Hydrogen atoms = 4 on both sides
• Oxygen atoms = 2 on both sides

• The Mole Ratio is the ratio of moles of one substance to another substance in a balanced chemical equation.
• The mole ratio is obtained from the coefficients of the balanced equation.

Example:

2H₂+O₂→2H₂O

The mole ratios are:

• H₂ : O₂ = 2 : 1
• H₂ : H₂O = 2 : 2
• O₂ : H₂O = 1 : 2

Order of Balancing Chemical Equations (Stoichiometry)

While balancing a chemical equation, elements are balanced in a systematic order to ensure conservation of mass. The general rule is to start with elements that appear in the fewest compounds and leave the most common elements for the end.

• Write the unbalanced equation and list all atoms on both sides.
• Adjust stoichiometric coefficients (numbers in front of formulas), not subscripts.
• Polyatomic ions that remain unchanged (e.g., SO₄²⁻, NO₃⁻) should be treated as a single unit.
• Hydrogen and oxygen are balanced last because they appear in many compounds, making them harder to isolate.
• Finally, verify atom counts and reduce coefficients to the smallest whole-number ratio.

Limiting Reagent

• The Limiting Reagent (or limiting reactant) is the reactant that is completely used up first in a chemical reaction.
• Because it is used up first, it limits the amount of product that can be formed.
• In most chemical reactions, reactants are not always present in the exact required ratio.
• The reactant that finishes first stops the reaction, even if other reactants are still left.
• When two or more substances react together, the limiting reagent determines how much product will be produced.
• Once the limiting reagent is consumed, the reaction cannot continue further.

Example:

2H₂ + O₂ → 2H₂O

According to the balanced equation:

• 2 moles of hydrogen react with 1 mole of oxygen.

Suppose we have:

• 2 moles of H₂
• 2 moles of O₂

But the reaction only needs 1 mole of O₂ to react with 2 moles of H₂.

So:

• Hydrogen will be completely used
• Oxygen will remain in excess

Therefore, hydrogen (H₂​) is the limiting reagent.

Stoichiometry in Chemical Analysis

Stoichiometric calculations are widely used in chemical analysis to determine the concentration or amount of substances in a sample. There are two main analytical methods:

1. Gravimetric Analysis

This method determines the quantity of a substance by measuring its mass. It is highly accurate because mass can be measured precisely.

Types include:

• Precipitation gravimetry – formation and weighing of a precipitate
• Volatilization gravimetry – separation by heating or decomposition
• Electrogravimetry – deposition of metal ions on an electrode

2. Volumetric Analysis

This method involves measurement based on volume. Where a solution of known concentration reacts with a solution of unknown concentration until the reaction is complete. The principle is:

Important terms:

1. Titration – process of determining concentration
2. Titrant – solution of known strength
3. Titrate – solution of unknown strength
4. Indicator – substance that shows completion of reaction by colour change

Solved Examples

Question 1: Calculate how much sodium hydroxide will be needed to make 500 mL of a 0.10 M solution.

Solution:

The molar mass of NaOH = 40g

Volume of NaOH= 500ml = 0.5 L

Molarity = 0.10M

Molarity = moles / volume in litres

So, weight of NaOH = molar mass of NaOH x volume  x molarity

                                 = 0.10 x 40 x 0.5

                                 = 2 g

Question 2: To make 3 M 400 ml HCl, how much 11 M HCl should be diluted with water?

Solution:

Given that, M₁ = 11M, and M₂ = 3M

Also, V₁ = ? , and V₂= 400ml

Now, M₁ x V₁ = M₂ x V₂

                 V₁ = (3×400) / 11

                     = 109 ml

Question 3: By reacting nitrogen with hydrogen, how many moles of nitrogen are required to make 8.2 moles of ammonia?

Solution:

The balanced chemical equation is N₂ + 3H₂ → 2NH₃

2 mole of NH₃ are produced from = 1 mole of N₂

8.2 mole of NH₃ are produced from = (1/2) x 8.2

                                                           = 4.1 mol of N2

Related Articles:

• Molecular Formula
• Empirical Formula