[Chemistry Class Notes] on Balancing Chemical Equations Pdf for Exam

A chemical equation represents the chemical formulas of substances that react and the substances that produce. The number of atoms of the reactants and products supposed to be balanced. Let us discuss balancing chemical equations.

Balancing the chemical equations involves the addition of stoichiometric coefficients to products and reactants. This is necessary because any chemical equation must obey the law of constant proportions and the law of conservation of mass, i.e., the same number of atoms of each element must exist on the product side and reactant side of the equation.

Let us discuss two easy and faster methods of balancing a chemical equation now. The first one is the traditional balancing equations method and the second method is the algebraic balancing method.

A chemical equation is a symbolically represented chemical reaction in the form of symbols and formulae, with the reactant entities on the left and the product entities on the right. A chemical equation is a technique of representing the components of a chemical reaction in a concise manner. It’s crucial to understand what a chemical reaction is since a chemical equation depicts one. Every day, you may see chemical reactions taking place. When the matter is reorganized to generate a different or new material, it is called a chemical reaction.

Balancing Chemical Equations – Balancing Chemical Equations Entails Correctly Writing the Chemical Equation so that the Mass on Each Side of the Arrow is Equal

To balance a chemical equation, follow these four simple steps:

1. To show the reactants and products, write the imbalanced equation.

2. Calculate the number of atoms of each element on each side of the reaction arrow.

3. To make the number of atoms of each element the same on both sides of the equation, multiply coefficients (the numbers in front of the formulas). The hydrogen and oxygen atoms are the easiest to balance last.

4. Check your work by indicating the state of matter of the reactants and products.

Things to Discuss

A few of the things that should be addressed before explaining the two methods are the chemical equation.

How to Proceed with the Balance of Equations?

There are several approaches to balancing a chemical equation, and we’ll go over each one. They are as follows:

Inspection Technique: It refers to the act of inspecting something, and it is used in conjunction with the Inspection method. It’s exactly what it says. You can determine the proper balance by looking at the equation. Let us first go over the technique and then look at an example:

• Assume that only one molecule of this chemical is involved in the process. You must begin with the most difficult compound.

• Try to balance the atoms in as few compounds as possible. You can also use fractions.

• Atoms must be balanced at the end of most compounds.

• Once all atoms and charges are equal, eliminate fractions and find the group of minimal coefficients. The Inspection approach was used to complete the entire balancing procedure.

Chemical Equation

• Every chemical equation is a symbolic representation of a chemical reaction where their respective chemical formulae denote the products and reactants.

• Consider an example of a chemical equation, 2H2 + O2 → 2H2O describing the reaction between hydrogen and oxygen to form water

• The reactant side is the chemical equation part to the left of the ‘→’ symbol, and the product side is the part to the right of the ‘→’ symbol.

1. Traditional Balancing Equations Method

The first step that needs to be followed while balancing chemical equations is to obtain the complete unbalanced chemical equation. To demonstrate this method, as an example, the combustion reaction between oxygen and propane is taken.

Step 1

• The unbalanced chemical equation needs to be obtained from the chemical formulae of the reactants and the products if it is not already provided.

• The chemical formula of propane is C3H8. It burns with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).

• Thus, the unbalanced chemical equation can be represented as C3H8 + O2 → CO2 + H2O.

Step 2

The total number of atoms contained in each element on the product and reactant side must be compared. For this, as an example, the number of atoms on each side are tabulated below.

Chemical Equation is – [C_{3}H_{8} + O_{2} rightarrow CO_{2} + H_{2}O]

Step 3

• Now, the stoichiometric coefficients are added to molecules containing an element that has a different number of atoms in the reactant and the product side.

• Here, the coefficient must balance the number of atoms on each side.

• In general, the stoichiometric coefficients are assigned last to hydrogen and oxygen atoms.

• Now, the number of atoms
  of the elements on both the reactant and product sides must be updated.

• It is necessary to remember that the number of atoms of an element in one species must be determined by multiplying the stoichiometric coefficient by the total number of atoms of that element present in 1 molecule of the species.

• For suppose, when coefficient ‘3’ is assigned to the CO2 molecule, the total number of oxygen atoms in CO2 becomes 6. The coefficient is first assigned to carbon in this example, as listed below.

Chemical Equation is – [ C_{3}H_{8} + O_{2} rightarrow 3CO_{2} + H_{2}O ]

Step 4

Step 3 is repeated until the total number of atoms of the reacting elements becomes equal on both the reactant and product sides. Hydrogen is balanced next in this example. The chemical equation is transformed as below.

Chemical Equation is – [ C_{3}H_{8} + O_{2} rightarrow 3CO_{2} + 4H_{2}O]

As the hydrogen atoms are balanced now, the next balanced element to be is oxygen. There are 10 oxygen atoms on the product side, saying that the reactant side also must contain 10 oxygen atoms.

Each of the O2 molecules contains 2 oxygen atoms. And therefore, the stoichiometric coefficient that must be assigned to the O2 molecule is 5. The updated chemical equation is listed on the below table.

Chemical Equation is – [ C_{3}H_{8} + 5O_{2} rightarrow 3CO_{2} + 4H_{2}O ]

Step 5

• Once all the individual elements have balanced, the total number of atoms of each element on the product and reactant side are once again compared.

• If there are no inequalities found, the chemical equation is said to be balanced.

• In this example, now every element has an equal number of atoms in both reactant and product sides.

• Therefore, the balanced chemical equation becomes C3H8 + 5O2 → 3CO2 + 4H2O.

2. The Algebraic Balancing Equations Method

This balancing method of chemical equations includes assigning algebraic variables as stoichiometric coefficients to every species in the unbalanced chemical equation. And, these variables are used in mathematical equations and solved to obtain the values of each stoichiometric coefficient. In order to explain this method better, as an example, the balancing reactions between oxygen and glucose, which yields carbon dioxide and water, have been considered.

The algebraic approach, commonly referred to as the material balance method, is the most widely used. It maintains equilibrium in a variety of reactions. Follow the instructions outlined below.

• Create a unique character coefficient for each chemical in the equation.

• Write algebraic formulas or rules on both sides that connect the atoms of each component.

• To create a two-letter coefficient equation that you may use, replace and decrease.

• Substitute the numbers into the other rules to acquire the balance ratios.

• Consider the following example to help you understand concepts:

• For the coefficients, you first create the solution using letter factors:

• [ aCaCO_{3} + bH_{3}PO_{4} rightarrow cCa_{3}(PO_{4})_{2} + dH_{2}CO_{3} ]

• Create a series of parallel equations, one for each element

• Ca; a=3c, C; a=d, O; 3a+4b=8c+3d, H; 3b=2d, P; b=2

Step 1

• The unbalanced equation must be achieved by writing the reactants and products’ chemical formulae.

• The reactants are glucose (C6H12O6), oxygen (O2), and the products are carbon dioxide (CO2), and water (H2O) in this example.

• An unbalanced chemical equation is C6H12O6 + O2 → CO2 + H2O

Step 2

Now, the algebraic variables are assigned to each species, as stoichiometric coefficients in the unbalanced chemical equation. In this example, the equation can be written as below.

[ aC_{6}H-{12}O_{6} + bO_{2} rightarrow cCO_{2} + dH_{2}O ]

Now, a set of equations must be formulated between the reactant side and the product side in order to balance a chemical equation of each element in the balancing reactions. The following equations can be formed in this example.

Equation for Carbon

• On the reactant side, ‘a’ molecules of C6H12O6 will contain ‘6a’ carbon atoms.

• On the product side, ‘c’ molecules of CO2 will contain ‘c’ carbon atoms.

• In this equation, only the species containing carbon are C6H12O6 and CO2.

Therefore, the equation can be formulated for carbon is – 6a = c

Equation for Hydrogen

• Species containing hydrogen in this equation are C6H12O6 and H2

• ‘a’ molecules of C6H12O6 contain ’12a’ hydrogen atoms, whereas ‘d’ H2O molecules contain ‘2d’ hydrogen atoms.

• Thus, the equation for hydrogen results as 12a=2d.

Now, simplifying this equation, dividing both sides by 2, the equation transforms, 6a = d.

Equation for Oxygen

There is Oxygen on every species in this chemical equation. Therefore, the below relations can be formed to obtain the Oxygen equation.

• ‘a’ molecules of C6H12O6, there exist ‘6a’ oxygen atoms.

• For ‘b’ molecules of O2 contain a total of ‘2b’ oxygens.

• For ‘c’ molecules of CO2 contain ‘2c’ number of oxygen atoms.

• Finally, for ‘d’ molecules of H2O hold oxygen atoms of ‘d’.

Thus, the equation for oxygen can be given as, 6a + 2b = 2c+ d

Step 3

The equations for each element are together listed to form an equation system. For this example, the system of equations is written below.

For carbon, 6a = c, for hydrogen, 6a = d, and for oxygen, 6a + 2b = 2c + d.

This equation system can have multiple solutions, but the solution with minimal values of the variables is required. Obtaining this solution, a value is assigned to one of the coefficients. In this case, the value of ‘a’ is assumed to be 1. Thereby, the system of equations is converted, as explained below.

a = 1,

c = 6a = 6*1 = 6,

d = 6a = 6

Substituting a, c, and d values in the equation 6a + 2b = 2c + d, we get the value of ‘b,’ and the same can be obtained as below.

6*1 + 2b = 2*6 + 6

2b = 12; b = 6

It is necessary to note that these equations must be solved in a way that each variable is a positive integer. If the fractional values are obtained, the lowest common denominator between all the variables must be multiplied with each and every variable. This is important because the variables hold the values of the stoichiometric coefficients, which must be a positive integer.

Step 4

• Now when the smallest value of each variable is obtained, their values can be substituted into the chemical equation obtained in step 2.

• Thus, the equation [ aC_{6}H_{12}O_{6} + bO_{2} rightarrow cCO_{2} + dH_{2}O ]becomes as [ C_{6}H_{12}O_{6} + 6O_{2} rightarrow 6CO_{2} + 6H_{2}O ]

• The balanced chemical equation is obtained, therefore.

Balancing chemical equations using the algebraic balancing equations method is considered to be more effective than the traditional method. However, it may yield fractional values for the stoichiometric coefficients, which must be converted into integers then.