Stoichiometry

Figuring out the ratios in which different substances or elements react to one another. The laws of conservation of mass, energy, and weights or volumes serve as the foundation for the rules used to determine stoichiometric relationships.

Chemical Stoichiometry refers to the quantitative study of the reactants and products involved in a chemical reaction. The word “stoichiometry” is derived from the Greek word “stoikhein” meaning element, and “metron” meaning measure.

The term Stoichiometry was first coined or discovered by a German chemist named Jeremias Richter. Even though this tongue-twisting word can sound complicated and big, it is a simple idea. In this lesson, we will learn about what it means and discuss the different aspects of this concept.

What Is Stoichiometry?

In simple words, we can define,

Stoichiometry as the calculation of products and reactants in a chemical reaction. It is basically concerned with numbers.

Stoichiometry is an important concept in Chemistry that helps us use balanced chemical equations to calculate amounts of reactants and products. Here, we make use of ratios from the balanced equation. In general, all the reactions that take place are dependent on one main factor, that is, how much substance is present.

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Stoichiometry helps us determine how much substance is needed or is present. Things that can be measured are,

1. Reactants and products mass
2. Molecular weight
3. Chemical equations
4. Formulas

Stoichiometric Coefficient

The stoichiometric coefficient or stoichiometric number is the number of molecules that participate in the reaction. If you look at any balanced reaction, you can notice that there are an equal number of elements on both sides of the equation. The stoichiometric coefficient is basically the number present in front of atoms, molecules or ions.

Stoichiometric coefficients can be fractions as well as whole numbers. In essence, the coefficients help us to establish the mole ratio between reactants and products.

Balanced Reactions and Mole Ratios

Atoms and molecules are extremely small in size, and their numbers in a very small amount of a substance are very large. Therefore, to represent atoms and molecules in bulk, a mole concept was introduced. One mole of any substance contains 6.022 x 10²³ numbers of that substance. This number is also known as Avogadro’s number.

The mass of one mole of a substance in grams is called molar mass. The molar mass of one mole of a substance is numerically equal to the atomic/molecular formula mass.

Let us take one example of a balanced chemical equation,

3Fe(s) + 4H₂O(l) ⇾ Fe₃O₄ (s)+ 4H₂ (g)

The quantitative information drawn from this balanced chemical equation is

1. 3 mole of Fe reacts with 4 moles of H₂O to yield one mole of Fe₃O₄ and 4 moles of H₂.
2. 168g ( 56×3) of Fe reacts with 72g( 18×4) of H₂0 to yield 231g of Fe₃O₄ and 8g of H₂ gas.

If the reactants and products are in gaseous form, then the molar volume is taken into consideration. One mole of any gas occupies 22.4 litres.

CH₄(g) + 2O₂(g)⇾ CO₂(g)+ 2H₂0 (g)

In the above reaction, 22.4 litres of CH₄ reacts with 44.8 (2 x 22.4) litres of 0₂ to yield 22.4 litres of CO₂ and 44.8 litres of H₂O.

Limiting Reagent

In a chemical reaction, it is possible that one of the reactants is present in excess amount. Some of these excess reactants will, therefore, be left over when the reaction is complete; the reaction stops immediately as soon as one of the reactants is totally consumed.

The substance that is totally consumed in a reaction is called the limiting reagent.

Let us take one example of a chemical reaction to understand the limiting reagent concept.

N₂ + 3H₂ ➝ 2NH₃

Suppose we have one mole of N₂ reacting with one mole of H₂. But from the balanced chemical equation, one mole of N₂ requires three moles of H₂. So, the limiting reagent in this reaction is H₂.

Stoichiometry in Chemical Analysis

Stoichiometric calculations, which follow a quantitative analysis methodology, are often used by chemists to determine the concentrations of substances present in a sample. There are basically two main types of analysis, and we will discuss them below.

1. Gravimetric Analysis

In Analytical Chemistry, gravimetric analysis describes the quantitative determination of an analyte based on the mass of the solid. The gravimetric analysis gives the most accurate results out of all other analytical analyses, as the weight of a substance can be measured with great accuracy compared to other fundamental quantities.

Gravimetric analysis can be classified into the following types:

1. Precipitation gravimetry – It involves the isolation of ions in solution by a precipitation reaction, filtering, washing the precipitate free of contaminants, and finally weighing the precipitate and determining its mass by difference.
2. Volatilization gravimetry – Volatilization gravimetry involves separating components of a mixture by heating or chemically decomposing the sample.
3. Electrogravimetry – It involves the electrochemical reduction of metal ions at the cathode and the simultaneous deposition of ions on the cathode. The cathode is weighed before and after electrolysis, and the weight difference corresponds to the mass of the analyte initially present in the sample.

2. Volumetric Analysis

The volumetric analysis involves the quantitative measurement of substance in terms of volume.

Principle: In volumetric analysis, a known volume (V₁) of the substance, whose concentration (N₁) is known, is reacted with the unknown volume (V₂) of the solution of the substance, whose concentration (N₂) is to be calculated. The volume V₁ is noted at the endpoint of the reaction. The concentration N₂ is calculated using the following equation.

N₁x V₁ = N₂ x V₂

The endpoint of such a reaction is indicated by a change in colour or precipitation etc.

Terms involved in volumetric analysis are as follows:

1. Titration – The process of finding out the volume of solution required to react completely with the volume of another solution is called titration.
2. Titrant – The solution of known strength is called titrant.
3. Titrate – The solution whose concentration is to be estimated is called titrate.
4. Indicator – Indicators are reagents which change their colour when the reaction is complete.

Stoichiometry and Its Applications – Video Lesson

Stoichiometry Problems with Solutions

1. Calculate the mass of sodium hydroxide required to make 500 ml of 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

⇒ weight of NaOH = molarity x molar mass of NaOH x volume

= 0.10 x 40 x 0.5

= 2 g

2. How much volume of 11 M HCl has to be diluted with water to prepare 3 M 400 ml HCl?

Solution:

M1 = 11M

M2 = 3M

V₁ = ?

V₂= 400ml

Now, M₁ x V₁= M₂ x V₂

⇒V₁= (3×400)/ 11

= 109 ml

3. How many carbon atoms are present in 0.5 moles of oxalic acid (C₂H₂O₄)?

Solution :

1 mole of oxalic acid = 6.022 x 10 ²³ number of oxalic acid

0.5 mole of oxalic acid = 6.022 x 10 ²³ x 0.5 number of oxalic acid

Since there are 2 carbon per oxalic acid,

The number of carbon atoms in 0.5 moles of oxalic acid = 6.022 x 10 ²³ x 0.5 x 2

= 6.022 x 10 ²³

4. 0.5216g of a solid mixture containing Na₂SO₄ is dissolved in water and treated with an excess of BaCl₂, resulting in the precipitation of 0.6168g of BaSO₄. What percentage of the mixture was BaS0₄?

Solution:

Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl

233g of BaSO₄ is obtained from 142g of Na₂SO₄

So, 0.6168g of BaSO₄ is obtained from = (142×0.6168) / 233

= 0.37g

Since the mass of the solid mixture is 0.5216 g,

The percentage of BaSO₄ is solid mixture = (0.37/0.5216) x 100

= 70.34%

5. A solution containing 5 g of KOH and Ca(OH)₂ is neutralised by an acid. If it consumes 0.3 g equivalents of the acid, calculate the composition of the solution.

Solution:

Let the mass of KOH present in mixture = x

Mass of Ca(OH)₂ = (5-x)g

Equivalent mass of KOH = 56; Equivalent mass of Ca(OH)₂ = 74/2

Gram equivalent of KOH + Gram equivalent of Ca(OH)2 = Gram equivalent of acid

+ = 0.1

⇒ x = 3.83g

Mass of KOH in the sample = 3.83g

Percentage of KOH = (3.83/5) x 100

= 76.6%

Percentage of Ca(OH)₂ = x100

= 23.4%