Now, when you have a reaction, the number of atoms in the formulas of the reactants does not always equal that of the respective atoms in the formulas of the products.
However, the mass is preserved. For example, let’s see the following reaction:
KClO 3 → KCl + O 2
This equation has on the left a compound called potassium chlorate, which is decomposed by heating into potassium chloride and oxygen gas. But when we look carefully, we notice that in the chlorate molecule there are 3 oxygen atoms, while on the right there is only one gaseous oxygen molecule with 2 atoms.
What is done then is to multiply the reactant on the left by a numerical coefficient, in such a way that the number of atoms of all the participating elements is the same before and after the reaction.
But … what is the value of this coefficient?
Ways to balance equations
In the above equation it is easy to determine the appropriate coefficients by inspection. If we multiply by 2 on the left we have 2 atoms of K, 2 of Cl and 6 of O.
Therefore, on the right we multiply the potassium chloride by 2 and the O 2 by 3:
2 KClO 3 → 2KCl + 3O 2
And now we can see that the accounting is correct on both sides of the arrow and the reaction was balanced. Note that other numerical values can also result in a balanced equation, for example:
4 KClO 3 → 4KCl + 6O 2
However, you should always look for the set of minimum integers that equal the number of atoms on both sides.
The numbers that multiply each formula are called coefficients . It is very important to note that the coefficients can be assigned to balance, but the subscripts must remain as they appear in each formula.
Nor can coefficients be inserted in the middle of the formula of each compound, since it would alter it.
Simple equations like the one in the example can be balanced by inspection or trial and error. For those that are a little more complex there is an algebraic method with a series of simple steps and elementary arithmetic operations that are detailed below.
Steps to Algebraically Balance a Chemical Equation
-Assign each molecule an arbitrary coefficient, symbolized by a letter. Usually the letters a, b, c, d … are used, as many as necessary.
Important: remember that only one coefficient is used per molecule and it is never inserted in the middle of it, it is always placed on the left.
-Make a list of each participating element.
-Place the coefficients assigned to each compound on the left and match with those on the right. If there are subscripts, they are multiplied by the coefficient to find the total number of atoms.
And if an element is found in more than one molecule, the numbers of atoms present on each side are added. In this way the partial equations of each element are obtained.
-A numerical value is assigned to only one of the coefficients. Usually this numerical value is 1 and is assigned to the letter that appears the most times. With this, a simple equation is obtained that serves as a starting point to find the other coefficients.
-Determine the value of the following coefficient through simple arithmetic and replace its value in another equation, to propose a new one.
-Repeat the previous step of replacing values and creating a new equation, until all the coefficients are found.
-Replace the values thus determined. If these values are integers, it is necessary to verify that the equation was balanced. If they were not integers, multiply by the least common multiple of the denominators and check the balance.
Next we are going to visualize the application of these steps in the resolution of some examples.
Worked examples of balancing
Balance the following reaction if necessary, using the algebraic method:
N 2 O 5 → N 2 O 4 + O 2
We observe that the reaction is not balanced, because although there are 2 nitrogen atoms on both sides, with oxygen the quantities on the left and right are different.
Then we must follow the steps described in the previous section:
-We rewrite the equation and multiply each molecule by a different coefficient. Lowercase letters are chosen so that they are not confused with the elements:
a⋅N 2 O 5 → b⋅N 2 O 4 + c⋅O 2
-Now we list each element and equalize the quantities of said element to the left and right. These are the partial equations for each element:
- a⋅N 2 = b⋅N 2
- a⋅O 5 = b⋅O 4 + c⋅O 2
-Multiplying the coefficient and the subscript we obtain the number of nitrogen atoms. From equation 1 we obtain:
2a = 2b
-From equation 2 there are 5a oxygen atoms on the left, while on the right there are 4b and 2c :
5a = 4b + 2c
-We assign the value 1 to the coefficient a, in this way:
a = 1
This choice is arbitrary, it was also possible to choose first b = 1.
-These values are substituted in equation 2, to determine the value of c:
5 = 4 + 2c
2c = 5-4 = 1
c = ½
-We substitute the coefficients in the original equation, the 1 does not need to be written explicitly:
N 2 O 5 → N 2 O 4 + ½ O 2
-Since it is preferable that the coefficients are integers, the entire equation is multiplied by the least common multiple of the denominators, which are 2 and 1:
lcm (1,2) = 2
So, multiplying left and right by 2, you get:
2N 2 O 5 → 2N 2 O 4 + O 2
And we count the number of atoms on both sides, to check the balance:
- N to the left: 4
- Or left: 10
- N right: 4
- Or to the right: 8 + 2 = 10
Balance the following chemical reaction:
NaHCO 3 → Na 2 CO 3 + H 2 O + CO 2
We multiply each molecule by a different coefficient:
a⋅NaHCO 3 → b⋅Na 2 CO 3 + c⋅H 2 O + d⋅CO 2
Next we propose the accounting of each element to the right and to the left of the arrow. There are a total of 4 elements in the reaction: sodium Na; hydrogen H; carbon C and oxygen O, all should be accounted for:
- a⋅Na = b⋅Na 2
- a⋅H = c⋅H 2
- a⋅C = b⋅C + d⋅C
- a⋅O 3 = b⋅O 3 + c⋅O + d⋅O 2
According to each balance, the following equations are obtained:
1) a = 2b
2) a = 2c
3) a = b + d
4) a = 3b + c + 2d
The most repeated coefficient is a , therefore we assign it the value 1:
a = 1
1) a = 2b ⇒ b = ½
2) a = 2c ⇒ c = ½
3) a = b + d ⇒ d = a – b = 1 – ½ = ½
We substitute the coefficients in the equation:
NaHCO 3 → ½. Na 2 CO 3 + ½. H 2 O + ½.CO 2
We multiply both sides of the arrow by 2, since it is the only denominator present, to eliminate the fraction:
2NaHCO 3 → Na 2 CO 3 + H 2 O + CO 2
We count the number of atoms present on the left: 2 atoms of Na, H and C and 6 of O. The reader can verify that each of them is also present on the right in equal amounts.