86

Stoichiometry

It is a part of chemistry that involves the study of quantities of substances (reactants or products or both) that take part in chemical reactions. It refers to an important aspect of a chemical equation that says that when a chemical equation is written in a balanced form, it gives quantitative relationships between the various reactants and products in terms of moles, masses, molecules and volumes. For example, if A is reacting with B to form C and D then stoichiometry can tell us how many moles of A are used to produce the resultant moles of C and D or how many moles of B are used to produce the resultant moles of C and D and vice versa.

As per its definition, it not only tells about the moles but also tells about the masses, molecules or volumes of reactants used to produce the resultant masses, molecules or volumes of products. For example, it can tell how much mass of A is used to produce the masses of products C and D or how much mass of B is used to produce C and D or only C or D and vice versa.

So, stoichiometry is mainly associated with or deals with the calculation of the amount (masses or volumes) of reactants reacting or the amount of products produced in a chemical reaction using a balanced chemical equation. It mostly deals with the mass or volume of products and reactants. So, one must understand the following concepts of chemistry to solve stoichiometry problems.

• How to balance chemical equations
• Converting grams to moles and vice versa
• Calculation of molar mass and mole ratios

The chemical reactions and their output depend on the quantity of reactants which can be determined by using Stoichiometry as it tells you the number or quantity of reactants required to produce the desired amount of products or how much products can be produced from the given amount of reactants. So, we can say that stoichiometry is all about numbers.

The term ‘stoichiometry’ is derived from the Greek words ‘stoicheion’ which means element and metron which means measure.

So, when you are using stoichiometry or dealing with problems in stoichiometry, you could be measuring or dealing with any of the following;

• Finding the mass of reactants before the reaction
• Calculating the mass of products after the reaction
• Molecular weights of reactants and products
• Chemical equations and formulas of compounds

Let us take an example of sodium chloride (NaCl) to understand the application or use of Stoichiometry. In NaCl, two ions combine to form an ionic compound NaCl. When we look at the chemical reaction or chemical equation, one sodium ion (Na⁺) reacts with one chlorine ion (Cl^–) to form the NaCl. Now using stoichiometry, you can find out that one ion of sodium and one ion of chlorine is needed to make one molecule of NaCl. Accordingly, you can calculate the mass of one sodium atom and chlorine atom to find out the mass of sodium and chlorine needed to form the desired mass of one sodium chloride molecule. So, even if you have one million atoms of sodium but one atom of chlorine, only one molecule of NaCl can be formed. You cannot do anything to make more NaCl molecules with this combination. Such as;

10, 00, 000 Na + 1 Cl → 1 NaCl + 999999 Na atoms

Let us take another example;

When you add a spark to the mixture of hydrogen gas and oxygen gas, the hydrogen and oxygen gas molecule combine to produce water. The chemical reaction can be written like this:

2H₂ + O₂ → 2H₂O

Let us see what role stoichiometry play here. When we see the chemical equation, there are four hydrogen atoms and two oxygen atoms on each side of the equation. So, it is a balanced chemical equation which should be as all reactions follow the law of conservation of mass. So, it is clear from this balanced chemical equation that we need twice as many hydrogen atoms as we do oxygen atoms. Now, we know how many atoms we need of each reactant to carry out this reaction. So, if we fill a container with one thousand hydrogen molecules (H₂) and only one oxygen molecule (O₂) and add a spark, there would be no huge reaction as there is only one oxygen molecule and only two molecules of water will be formed; the extra hydrogen molecules will be of no use.

Stoichiometry Coefficient

It is also called stoichiometric number as it refers to the number of molecules participating in a chemical reaction. For example, in a balanced chemical reaction, there is always an equal number of elements on both sides of the equation and there is a number present in front of atoms of elements, ions or molecules. This number is known as the stoichiometry coefficient or number.

The stoichiometric coefficient can be a whole number or a fraction. For example, in the following balanced chemical equation, 2 in front of hydrogen and 1 in front of oxygen (one molecule of oxygen) are the stoichiometry coefficients.

2H₂ + O₂ → 2H₂O

Questions on Stoichiometry

Based on a single reactant or single reactant based stoichiometry questions:

Single reactant does not mean that we have only one reactant. It means, we have given information (mass, volume, etc.) about only one reactant and the quantity of products C and D such as their mass, volume, etc., are to be calculated with respect to the given mass or volume of the reactant and vice versa.

For example; there can be a chemical reaction as follows;

A + B → C + D

In which the quantity of B is not given and from the given quantity of A or B (one reactant), we have to tell the quantity of C or D or both the products. It can be other way around such as the quantity of a product is given and the quantity of one or all the reactants is to be calculated.

The amount of reactant consumed or the amount of products produced can be given in the form of mass or volume. Accordingly, there can be the following three types of relationships between reactants and products and accordingly we have to solve the stoichiometry problems.

i) Mass-mass relationships: As the name suggests, the mass of one reactant or products is given and the mass of another product or reactant is to be calculated. It is called a mass-mass relationship as mass is given and mass is to be calculated.

ii) Mass-volume relationship: In this case, either mass or volume of one of the reactants or products is given and mass or volume of the remaining one or more products is to be calculated. For example, the mass of one reactant is given and the volume of one of the products is to be calculated or vice versa or volume is given and mass is to be calculated.

iii) Volume-volume relationship: In this case, the volume of one of the reactants or products is given and the volume of any of the remaining reactants or products is to be calculated.

So, stoichiometry problems can be any of the above three types or relationship-based. Besides this, there is a sequence of steps that much be followed while solving these problems such as;

i) Write the balanced chemical equation.

ii) Write the number of moles or gram atomic or molecular masses of the reactants and products

iii) In the case of gaseous reactant or product, write down 22.4 litres of volume at STP instead of 1 mole.

iv) Now, use the unitary method to make the required calculations such as to convert moles into masses if the mass is to be calculated or to convert moles or mass into volume if the volume is to be calculated. First, you should have moles as they can be converted to mass or volume as per the question.

Now, let us solve some stoichiometry questions based on any of the above relationships using the above four steps:

Let us take the mass-mass relationship-based stoichiometry question first.

Question: Find the mass of carbon dioxide (CO₂) gas which is produced when 8 gm of Methane (CH₄) is combusted in the presence of plenty of air.

Solution:

It is a mass-mass relationship-based question as the mass of methane is given and we have to calculate the mass of carbon dioxide.

Let us now follow the above-mentioned steps to solve this question:

The reaction is not given so, let us first write the chemical reaction and balance it, which is the first requirement to proceed further.

CH₄ + O₂ → CO_{2 +} H₂O

As it is a combustion reaction in the presence of air, so, we will take oxygen as one of the reactants. So, the carbon and hydrogen of methane will be oxidised to form carbon dioxide (CO₂) and water (H₂O).

Balance the above chemical equation as follows;

CH₄ + 2O₂ → CO_{2 +} 2H₂O

After balancing the above reaction, we have an equal number of oxygen, hydrogen and carbon atoms on each side of the chemical equation.

Now, we will check the number of moles. We have 1 mole of methane, 2 moles of oxygen, 1 mole of carbon dioxide and 2 moles of water.

Now, we will see the ratio of moles of the given reactant and the product which is to be calculated.

The mole ratio of methane: carbon dioxide = 1: 1 as per the chemical equation, which shows one mole of methane is combusted in excess of air to produce one mole of carbon dioxide gas.

Now, 1 mole of methane is 16 gam as CH₄ (12 gm + (1 gm x4) = 16 gm.

Similarly, 1 mole of carbon dioxide is 44 gm as CO₂ (12 gm + (2 x 16gm) = 44 gm.

So, from the above two observations, it is clear that 16 gm of methane when burns completely in excess of air it forms 44 gm of carbon dioxide.

Now, 16 gm of CH₄ → 44 gm CO₂

So, the given 8 gm of methane will produce = 44 /16 x 8 = 22 gm of carbon dioxide.

→ 8 gm of CH₄ → 22 gm CO₂

Now, let us solve the mass-volume relationship-based stoichiometry question:

Question: Calculate the volume of CO₂ and water vapours produced when 32 gm of methane is combusted in the presence of an excess of air.

Solution:

We have given the mass and we were asked to calculate the volumes, so, it is a mass-volume relationship type of question.

Now, after writing the chemical reaction as per the question and balancing the chemical reaction, we will get:

CH₄ + 2O₂ → CO_{2 +} 2H₂O

Now, we have an equal number of oxygen, hydrogen and carbon atoms on each side of the equation and 1 mole of methane, 2 moles of oxygen, 1 mole of carbon dioxide and 2 moles of water.

Now, the Molar ratio of methane and carbon dioxide = CH₄: CO₂ = 1: 1

It shows 1 mole of methane produces 1 mole of carbon monoxide on combustion.

Now, 1 mole of methane means 16 gm of methane is producing 22.4 litre of CO₂, which is equivalent to 1 mole of CO₂ (here, we will not take gm as we were asked to find the volume of CO₂, which is 22.4 litres at STP (standard temperature and pressure conditions).

Now, 16 gm of CH₄ produces 22.4 litres of CO₂

So, 32 gm of CH₄ will produce 22.4/16 x 32 = 44.8 litres of CO₂ gas or 2 volumes of CO₂ as it is double the volume of CO₂ at STP, which is considered 1 volume.

Now, we will calculate water vapour or H₂O produced in this reaction:

The molar ratio of methane and water (CH₄ and H₂O) is 1: 2, it shows 1 mole of methane will produce 2 moles of water vapours on combustion. We can write it as follows;

1 mole of CH₄ → 2 moles of H₂O

So, 16 gm of CH₄ → 2 volume of H₂O or 2 x 22.4 = 44.8 Litres of H₂O as we are assuming that the reaction is occurring at STP because we have not given any specific temperature and pressure in the question.

Now, 32 gm of CH₄ will produce 2 x 44.8 = 89.6 litres of water vapours or 4 volumes of H₂O.

Now, let us solve the volume-volume relationship-based stoichiometry question:

Question: When 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of water vapour can be produced?

Solution:

Let us write a balanced chemical reaction for the given reaction.

2H₂ + O₂ → 2H₂O

Now, we have 2 moles of hydrogen and 1 mole of oxygen and 2 moles of H₂O

The mole ratio of H₂ and H₂O is 2 moles: 2 moles or 1: 1, it shows 1 mole or 1 vol. of H₂ produces 1 mole or 1 vol. of H₂O on reacting with O₂

So, 10 vol. of dihydrogen will produce 10 vol. of H₂O

We can also chose the mole ratio of O₂ and H₂O to find out the volume of water vapours produced in this reaction.

The mole ratio of O₂ and H₂O is 1: 2, it shows 1 mole or 1 vol. of O₂ will be consumed to produce 2 moles or 2 vol. of H₂O in this reaction.

So, 5 vol. of the given dioxygen gas (O₂) will produce, 5 x 2 = 10 vol. of water vapour (H₂O).

---