Calculations Related to Balanced Chemical Equations

As we know that a balanced chemical equation follows law of conversation of mass. This law states that mass can neither be created nor be destroyed.This means that mass of reactants should be equal to the mass of the products.Moreover, the overall charge of reactants should be equal to overall charge of products in a balanced chemical equation.Let us understand this concept by balancing a simple equation.

To balance mass, we use mass in grams. Mass in grams need to be equal in the following manner. The mass of the reactant is equal to the mass of the products mass.In the illustration, we can see that Potassium Nitrate decomposes to form Potassium Nitrite and Oxygen gas. The mass of potassium nitrate is 4.04g. The mass of potassium nitrite is 3.40g. The mass of Oxygen formed is 0.64g. If we add the mass of Potassium Nitrite and Oxygen gas in the product, we get 4.04g. That is the mass of the reactants.

Some chemical reactions involve ions. We can also balance the charge for chemical reactions involving ions. We can see in the illustration that Potassium Nitrate Decomposes to form the potassium nitrite and oxygen gas. The charge of potassium in Potassium Nitrate is +1. The charge of nitrate ion in potassium nitrate is -1. So overall charge of reactant is (+1) + (-1) = 0 .

In the products of the reaction, the charge of oxygen gas is zero. Charge of potassium is +1 and that of nitrite ion is -1. So, the overall charge of potassium nitrite is (+1) + (-1) = 0. We can see that overall charge of reactant is zero. We can also see that the overall charge of product is also zero.So the charge of reactants is equal to charge of products.

Lets balance the mass and charge of another chemical equation. Nitrogen gas reacts with hydrogen gas to form Ammonia. First we shall balance the mass. Mass of Nitrogen is 2100g. Mass of Hydrogen gas is 450g. When we add the mass of Nitrogen and Hydrogen gases, we get 2550g. Mass of products is also 2550g. So we can see mass of the reactants is equal to the mass of the product.

Now we shall balance the charge in this chemical equation. The charge of Nitrogen gas is zero. Charge of Hydrogen gas is zero. Charge of ammonia is also zero. We can observe that overall charge of reactants is zero. It is equal to overall charge of the products.

Now we shall balance the nuclear equations. It is an equation in which a nucleus changes to a new element. It happens by the bombardment of neutrons or by emitting alpha or beta particles. For balancing a nuclear equation we need to balance the mass number on both sides of the equation. Then we work on the atomic number which represents the nuclear charge. Balancing of nuclear equations happens by finding the missing element in the equation. This is done by mass number and atomic number balancing on both sides of the equation.

In the illustrated example we have an isotope of carbon. It is a ¹⁴C undergoing beta decay. It means that it is emitting a beta particle or an electron. What is the missing element in this reaction? For this, we need to make sure that the reaction is balanced. Here 6 of carbon represents the Atomic Number which is also equal to the number of protons in the carbon element. 14 of Carbon represents a mass number. It is the sum of protons and neutrons.

So this isotope of carbon has 6 protons and 8 neutrons. To balance the equation we have to make sure mass is equal on both sides. We also need to make sure the total nuclear charge is the same.On the right-side of the equation, the electron has zero mass. Therefore, we need an element with a mass number of fourteen. On the left-side of the equation, the nuclear charge is six. On the right-side electron has a nuclear charge of negative one. We need an element with 7 nuclear charges.

For this we use the periodic table. The mass number can vary but the atomic number is specific for every element. Nitrogen is the missing element. That’s because seven as its atomic number and fourteen as its mass number. Now the nuclear charge is same on the both sides of the equation. The mass number is also the same. This equation is balanced.Let's see another example in whichNitrogen-14 is bombarded with neutrons. As result produces hydrogen and a missing element.

Let's start with the mass number because we know that on the left-side, the mass is fifteen. On the right-side, the mass is one. Therefore, we need an element with a mass number of fourteen.Now, let's calculate the atomic number which also represents the nuclear charge. On the left-side, we have an atomic number value of 7. On the right-side we have an atomic number of 1 and another missing element. So we need an element with 6 atomic numbers to balance this equation. By using the periodic table, we get the missing element as Carbon-14. It is an isotope of carbon. Now, this nuclear equation is balanced.

The general rule for balancing simple chemical equations is that we can only change the coefficients. Those are the numbers in front of the atoms. The numbers after the atoms are subscripts. They cannot be changed.Coefficients are changed in order to balance equation. We do this by making the number of atoms of an element in reactant side equal to the number of atoms of element on product side. If there is no co-efficient written, we assume it to be one.

Introspection is a method of self-test orself-observation. It has the meaning looking-within. In this method we observe the chemical equation and find out whether the charge, mass or number of atoms need to be balanced. In the given illustration, iron oxide reacts with carbonto form iron metal and carbon dioxide gas.

Lets first see whether the charge is balanced. We can observe that the overall charge of reactants is zero. We can also observe that the overall charge of products is zero. So charge of reactants is equal to the charge of products. Now we shall look out for the number of atoms of each elements in the reactants. We shall observe whether they are balanced. As we can see, there are two iron atoms on reactant side. There is also one iron atom on product side. Also there are 3 oxygen atoms on reactant side and two oxygen atoms on product side. Number of carbon atoms are equal on both sides.

We shall first balance the iron atoms by placing the coefficient two on left-side of ironoxide. So now we have 4 Iron atoms and 6 oxygen atoms on left-side. We shall put coefficient 4 on left-side of iron in product side.Now number of iron atoms arebalanced. We shall put coefficient 3 onleft-side of carbon dioxide in productside. We can see number of oxygenatoms are now balanced.For balancing the three carbon atoms on product side we shall put coefficient 3 on carbon in the reactant side.

Let us balance another equation. Hydrogen gas reacts with oxygen gas to form water. There are two hydrogen atoms on reactant side. There are two hydrogen atoms on product side. Number of oxygen atoms on reactant side is two and on product side is one.

Let’s first try to balance the oxygen atoms. We shall put coefficient two on water and now we have two oxygen atoms on product side. So oxygen atoms are balanced on both sides. But number of hydrogen atoms are now unbalanced. This is because there are four hydrogen atoms on product side and two hydrogen atoms on reactant side. To balance the hydrogen atoms we shall put coefficient two on hydrogen gas in reactant side. Now we have balanced chemical equation as illustrated.

Balancing Redox equations. Oxidation-reduction reactions are commonly called redox reactions. This involves the moving of electrons from one element to another. The atom that loses electron is oxidized and the atom that gains electron is reduced. Redox equations can be balanced either by the oxidation number method or the half-reaction method.

In the oxidation number method, the oxidation number is counted as a whole. In the half-reaction method, the method equation is divided into two half-reactions. These two reactions are oxidation reaction and reduction reaction. Then balance each half and add the part to get a balanced redox equation.

We shall first use the oxidation number method to balance the redox equation.Oxidation number is the number of electrons lost or gained by an atom. We shall first identify the oxidation number of every atom in equation. Then we shall identify the change in oxidation number for atoms that either get oxidized or reduced. Then we shall add this number as coefficients in a way that the total rise in oxidation number becomes equal to total decrease in oxidation number. Then we balance all remaining atoms other than hydrogen and oxygen.

In the illustrated example we can see that oxidation state of nitrogen changes from +5 to +2. Here change in oxidation number is -3 because nitrogen gained three electrons. The oxidation number of As changed from plus three to plus five. So it is oxidized and change in its oxidation number is +2 because it lost two electrons.

Total rise in oxidation number for nitrogen is 3. Therefore, if we multiply three by two we get 6. The total decrease in oxidation number of As is +2. If we multiply this 2 with 3 we get 6. Hence, we shall put coefficient 3 to H₃AsO₃ and coefficient 2 to HNO₃. We can see now rise in oxidation number is equal to decrease in oxidation number which is six.

In the given example Zinc is reduced from zero to +2 oxidation state. Hydrogen is oxidized from +1 to zero oxidation state. Change in oxidation number of Zinc is +2. For Hydrogen, it is -1. We shall multiply -1 with two to get minus two. We multiplied HCl with two. Now we can see change in oxidation number of Zinc is equal to change in oxidation number of Hydrogen after multiplying it with two. Two is added as coefficient to HCl. Equation is balanced now.

We shall now balance the redox reaction by dividing it into two parts. One part is oxidation reaction and other is reduction reaction.In the given illustration we have a redox equation. We are separating the oxidation part and reduction part and writing them separately. ClO⁻ is called Hypochlorite. It is undergoing reduction to Cl⁻. Zinc is undergoing oxidation to Zn⁺² ion.

Now we first balance atoms other than hydrogen and oxygen by placing the coefficient to either reactant side or product side.We can see that Chlorine and Zinc are already balanced. So now we shall balance the oxygen atoms and then hydrogen atoms.To balance the oxygen atoms we shall add water molecules to opposite side of oxygen atoms.In oxidation reaction part there are no oxygen atoms. In reduction part there is one oxygen atom on reactant side. Therefore we shall add one water molecule on the product side. Now oxygen atoms are balanced.

We shall now balance the hydrogen atoms. To balance the hydrogen atoms we shall add hydrogen ions on the opposite side of the equation. There are no hydrogen atoms in the oxidation reaction part. Meanwhile, there are two hydrogen atoms in product side in reduction reaction part. We shall add two hydrogen ions on the reactant side.Now we shall balance the charge. In order to balance the charge we shall add electrons either to the reactant side or to the product side.In the oxidation reaction we can see charge on reactant is zero. Meanwhile, the charge on product is +2. So we shall add two electrons on product side to make overall charge equal to zero. This is same as overall charge on reactants.

Let’s balance the charge in reduction reaction. We can see there is +2 charge on hydrogen ions and -1 charge on ClO⁻ ion. So overall charge on reactants is (+2) + (-1) = (+1). The overall charge on product side is -1. If we add two electrons on reactant side we will get overall charge as -1. This is equal to the charge on products side.

In the next step we multiply the whole half reduction equation and half oxidation equation. We do this in such a way that number of electrons of both half reactions are equal to each other. In given example the number of electrons are already equal. We shall combine both these half reactions and subtract out any common terms like electrons, hydrogen ions or hydroxide ions.

We now have final balanced redox reaction.This reaction is balanced in acidic-medium. If a redox reaction takes place in basic-medium, in the final step we add hydroxide ions OH⁻. We add it to the side where hydrogen ions are found and subtract any water molecules.