The term “stoichiometry” is derived from two Greek words – stoicheion (meaning, element) and metron (meaning, measure). It is a branch of chemistry that deals with the quantitative relationships between reactants and products involved in a chemical reaction.

Stoichiometry in chemistry is the process of calculating of masses (sometimes volumes also) of reactants and products involved in a chemical reaction. A German chemist named Jeremias Richter first discovered the term stoichiometry.

In simple words, the quantitative study of the reactants and products involved in a chemical reaction is called stoichiometry. Basically, it is concerned with numbers.

## Importance of Stoichiometry in Chemistry

Stoichiometry is a fundamental concept in chemistry that helps us use balanced chemical equations to calculate accurate amounts of reactants and products. It helps for chemists to understand the fundamental principles of chemical reactions and allows them to make accurate predictions about the outcome of chemical reactions.

Without stoichiometry, chemists would be unable to predict how much amount of product would be formed from a given amount of reactant, or how much amount of reactant needed to produce a given amount of product. With the help of stoichiometry, we can easily determine the following things. They are:

- Mass of reactants and products in a chemical reaction
- Molecular weight
- Chemical equations
- Formulas

## Stoichiometric Coefficient

The stoichiometric coefficient or stoichiometric number represents the number of molecules (and moles as well) that takes part in the reaction or formed in the reaction. In any balanced reaction, you will notice that there are an equal number of elements present on both sides of the chemical equation. Let understand it with the help of an example.

Consider a combustion of methane as a chemical reaction. A balanced equation for this reaction is as below:

CH_{4}(g) + 2O_{2} (g) → CO_{2} (g) + 2H_{2}O (g)

In the above chemical reaction, methane (CH_{4}) and dioxygen (O_{2}) are called reactants and carbon dioxide (CO_{2}) and water (H_{2}O) are called products. All the reactants and products are gases in this reaction and we can identify them by a letter (g) in the brackets next to its formula. Similarly, in case of solids and liquids, we write (s) and (l) respectively.

The coefficients 2 written in the front of O_{2} and H_{2}O are called stoichiometric coefficients. Similarly the coefficients written in the front of CH_{4} and CO_{2} is one in each side. Thus, we can say that a stoichiometric coefficient is basically the number written in front of atoms, molecules or ions in a chemical change.

It balance the number of each element on both reactant and product sides of the chemical equation. However, stoichiometric coefficients can be whole numbers as well as fractions. They are often used and preferred.

The stoichiometric coefficients are useful in establishing the mole ratio between reactants and products. According to the above chemical reaction, we can get the following information.

- One mole of CH
_{4}(g) reacts with two moles O_{2}(g) to form one mole of CO_{2}(g) and two moles of H_{2}O (g). - One molecule of CH
_{4}(g) reacts with two molecules of O_{2}(g) to form one molecule of CO_{2}(g) and two molecules of H_{2}O (g). - 22.7 L of CH
_{4}(g) reacts with 45.4 L of O_{2}(g) to form 22.7 L of CO_{2}(g) and 45.4 L of H_{2}O (g). - 16 g of CH
_{4}(g) reacts with 2 * 32 g of O_{2}(g) to form 44 g of CO_{2}(g) and 2 * 18 g of H_{2}O (g).

From these relationships, we can inter-convert the given data as per the requirement, as follows:

mass ⇋ moles ⇋ no. of molecules

Mass / Volume = Density

## Stoichiometry Calculation Problems with Solutions

There are three types of stoichiometry calculations based on the chemical equation. They are as:

- Calculations involving mass-mass relationship
- Calculations involving mass-volume relationship
- Calculations involving volume-volume relationship

### Calculations involving Mass-Mass Relationship

In general, you can follow the following steps in making necessary stoichiometric calculations. They are:

- First, write down the balanced molecular equation for the chemical reaction.
- Second, write down the number of moles below the molecular formula of each of reactants and products.
- Below the molecular formula, write down the relative masses of reactants and products.
- At last, calculate the unknown factor by using a unitary method or above three rules.

Let us take some examples of problems with solutions based on the mass-mass relationship.

**Example 1:** Two atoms of hydrogen (H) combine with one atom of oxygen (O) to form one molecule of H2O. How many grams of hydrogen required reacting completely with 6.4 g of oxygen gas?

**Solution:**

The chemical reaction is as:

16 g of oxygen requires = 2 g of hydrogen

1 g of oxygen requires = 2 / 16 g of hydrogen

6.4 g of oxygen requires = (2 / 16) * 6.4 of hydrogen

Hence, mass of hydrogen required = 0.8 g (Ans.)

**Example 2:** How many grams of magnesium chloride will obtain from 17.0 g of HCl when the HCl reacts with an access of magnesium oxide?

**Solution:**

The balanced chemical equation is as:

MgO + 2HCl → MgCl_{2} + H_{2}O

From the above chemical equation,

One mole of MgO reacts with two moles of HCl to give one mole of MgCl_{2} and one mole of H_{2}O.

Molar mass of HCl = 1.008 + 35.5 = 36.5 g

No, of moles of HCl in 17 g HCl = 17 / 36.5 g = 0.465 mol

2 moles of HCl gives = 1 mole of MgCl_{2}

1 mole of HCl gives = 1 / 2 mole of MgCl_{2}

0.465 mol of HCl gives = (1 / 2) * 0.465 of MgCl_{2} = 0.2325 mol MgCl_{2}

Since no. of mole = mass / molar mass

Molar mass of MgCl_{2} = 24 + 2 * 35.5 = 95 g

Required mass of magnesium chloride MgCl_{2} = 0.2325 * 95 g = 22.12 g (Ans)

**Example 3:** How many grams of chlorine (Cl_{2}) are required to completely react with 0.40 g of hydrogen (H_{2}) to form hydrochloride acid (HCl)? Also calculate the amount of HCl formed in this reaction.

**Solution:**

The balanced chemical equation is as:

H_{2} + Cl_{2} → 2HCl

From the above chemical equation, it is clear that one mole of hydrogen reacts with one mole of chlorine to form two moles of hydrochloric acid.

Molar mass of H_{2} = 2 * 1.008 = 2 g

Molar mass of Cl_{2} = 2 * 35.5 = 71 g

Molar mass of HCl = 1 + 35.5 = 36.5 g

**Calculation of required chlorine:**

2 g of H_{2} reacts with = 71 g Cl_{2}

1 g of H_{2} reacts with = 71 / 2 g Cl_{2}

0.40 g of H_{2} would react with = (71 / 2) * 0.40 Cl_{2}

Required mass of chlorine = 14.6 g (Ans)

**Calculation of amount of HCl formed:**

From the above chemical equation, it is clear that one mole of H_{2} forms two moles of HCl.

Mass of HCl formed = mole * molar mass = 2 * 36.5 = 73 g

2 g of H_{2} forms = 73 g HCl

1 g of H_{2} forms = 73 / 2 g HCl

0.40 g of H_{2} would form = (73 / 2) * 0.40 g HCl

Amount of HCl formed = 14.6 g (Ans.)

### Calculations involving Mass-Volume Relationship

Calculations involving mass-volume relationship are based on the fact that 1 mol or 1 g molecule of each gas or substance under NTP conditions occupies a volume of 22.4 litre or 22400 ml. Let’s take some important problem examples based on the mass-volume relationship.

**Example 4:** Calculate the volume of carbon dioxide (CO_{2}) gas at NTP evolved by strong heating of 20g of calcium carbonate (CaCO_{3}).

**Solution:**

The balanced chemical equation is as:

CaCO_{3} → CaO + CO_{2}

From the above figure, it is clear that

100 g of CaCO_{3} evolved = 22.4 litre of CO_{2}

1 g of CaCO_{3} evolved = 22.4 / 100 litre of CO_{2}

20 g of CaCO_{3} evolved = (22.4 / 100) * 20 litre of CO_{2}

Volume of CO_{2} evolved = 4.48 litre (Ans.)

**Example 5:** How much gram of potassium chlorate (KClO3) should be heated to produce 22.4 litre of oxygen gas at STP.

**Solution:**

The balanced chemical equation is:

2KClO_{3} → 2KCl + 3O_{2}

From the above figure, it is clear that

3 * 22.4 litre of O_{2} gas produced from = 245 g of KClO_{3}

1 litre of O_{2} gas produced from = 245 / 67.2 g KClO_{3}

2.24 litre of O_{2} gas would produce from = (245 / 67.2) * 2.24 KClO_{3}

Required mass of KClO_{3} = 8.17 g (Ans.)

**Example 6:** What mass of sodium hydroxide (NaOH) is formed when 2.3 g of sodium metal reacts with excess of water? Also, calculate the volume of hydrogen evolved at NTP.

**Solution:**

Sodium reacts with excess of water according to the following equation,

2Na(s) + 2H_{2}O(l) → 2NaOH(aq) + H_{2}(g)

From the above figure,**Calculation of mass of sodium hydroxide formed:**

46 g of sodium forms = 80 g of NaOH

1 g of sodium forms = 80 / 1 g of NaOH

2.3 g of sodium forms = (80 / 1) * 2.3 g NaOH

Required mass of sodium hydroxide = 4.0 g (Ans.)

**Calculation of volume of hydrogen gas evolved:**

46 g of sodium forms = 22.4 L of H_{2} gas

1 g of sodium forms = 22.4 / 46 L of H_{2} gas

2.3 g of sodium forms = (22.4 / 46) * 2.3 L H_{2} gas

Total volume of hydrogen gas evolved at NTP = 1.12 L

### Calculations involving Volume – Volume Relationship

Calculations involving volume – volume relationship are based on two laws:

- Avogadro’s law
- Gay-Lussac’s law

Avogadro’s law states that equal moles of gases occupy equal volumes under the same temperature and pressure. For example:

2NO + O_{2} → 2NO_{2}

From this chemical equation, it is clear that two moles of NO react with one mole of O_{2} to form two moles of NO_{2}.

Thus, 2 * 22.4 L of NO reacts 22.4 L of O_{2} to form 2 * 22.4 L of NO_{2}.

Gay-Lussac’s law states that gases react in the simple ratio of their volumes under the similar conditions. For example:

From the above equation, it is clear that 2 vols. of NO reacts with 1 vol. of O_{2} to form 2 vols. of NO_{2}.

Let’s take some example problems based on volume – volume relationship.

**Example 7:** At STP, one litre of oxygen reacts with three times of carbon monoxide. Calculate the formation of volume of each gas after the reaction.

**Solution:**

The balanced chemical equation is as:

2CO + O_{2} → 2CO_{2}

From the above chemical equation, it is clear that 1 vol of O_{2} reacts with 2 vols of CO.

Or, 1 L of O_{2} reacts with 2 L of CO.

Thus, 1 L of CO does not react with O_{2}.

1 vol of O_{2} produces = 2 vols of CO_{2}

Or, 1 L of O_{2} produces = 2 L CO_{2}

Thus, gaseous mixture after reaction, volume of CO = 1 L and volume of CO_{2} = 2 L.

In this tutorial, we have discussed what is stoichiometry and stoichiometry coefficient with the help of some important examples. We have solved various example problems based on the stoichiometry calculations. Hope that you will have understood the basic concepts of calculations of mass-mass, mass-volume and volume-volume relationships. All stoichiometry example problems are simple and easy to solve.

Thanks for reading!!!