In the ABC letter trick of balancing chemical equations, in the 3rd step, do we need to take the largest compound in terms of atoms or in terms of elements?

Solution-

Algebric method-

This method of balancing chemical equations involves assigning algebraic variables as stoichiometric coefficients to each species in the unbalanced chemical equation. These variables are used in mathematical equations and are solved to obtain the values of each stoichiometric coefficient. In order to better explain this method, the reaction between glucose and oxygen that yields carbon dioxide and water has been considered as an example.The Algebraic Balancing Method

• In this example, the reactants are glucose (C₆H₁₂O₆) and oxygen (O₂) and the products are carbon dioxide (CO₂) and water (H₂O)
• The unbalanced chemical equation is C₆H₁₂O₆ + O₂ → CO₂ + H₂O

Step 2

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

aC₆H₁₂O₆ + bO₂ → cCO₂ + dH₂O

Now, a set of equations must be formulated (between the reactant and product side) in order to balance each element in the reaction. In this example, the following equations can be formed.

The 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, the only species containing carbon are C₆H₁₂O₆ and CO₂.

Therefore, the following equation can be formulated for carbon: 6a = c

 The equation for Hydrogen

• The species that contain hydrogen in this equation are C₆H₁₂O₆ and H₂
• ‘a’ molecules of C₆H₁₂O₆ contains ‘12a’ hydrogen atoms whereas ‘d’ H₂O molecules will contain ‘2d’ hydrogen atoms.
• Therefore, the equation for hydrogen becomes 12a = 2d.

Simplifying this equation (by dividing both sides by 2), the equation becomes:

 6a = d

The equation for Oxygen

Every species in this chemical equation contains oxygen. Therefore, the following relations can be made to obtain the equation for oxygen:

• For ‘a’ molecules of C₆H₁₂O₆, there exist ‘6a’ oxygen atoms.
• ‘b’ molecules of O₂ contain a total of ‘2b’ oxygens.
• ‘c’ molecules of CO₂ contain ‘2c’ number of oxygen atoms.
• ‘d’ molecules of H₂O hold ‘d’ oxygen atoms.

Therefore, the equation for oxygen can be written as:

6a + 2b = 2c+ d

Step 3

The equations for each element are listed together to form a system of equations. In this example, the system of equations is as follows:

6a = c (for carbon); 6a = d (for hydrogen); 6a + 2b = 2c + d (for oxygen)

This system of equations can have multiple solutions, but the solution with minimal values of the variables is required. To obtain this solution, a value is assigned to one of the coefficients. In this case, the value of a is assumed to be 1. Therefore, the system of equations is transformed as follows:

a = 1

c = 6a = 6*1 = 6

d = 6a = 6

Substituting the values of a,c, and d in the equation 6a + 2b = 2c + d, the value of ‘b’ can be obtained as follows:

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

2b = 12; b = 6

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

Step 4

• Now that the smallest value of each variable is obtained, their values can be substituted into the chemical equation obtained in step 2.
• Therefore, aC₆H₁₂O₆ + bO₂ → cCO₂ + dH₂O becomes: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
• Thus, the balanced chemical equation is obtained.

The algebraic method of balancing chemical equations is considered to be more efficient than the traditional method. However, it can yield fractional values for the stoichiometric coefficients, which must then be converted into integers.