Example of Principle of Stoichiometry

The stoichiometry principle is the chemical principle that establishes that in every chemical reaction, there is an equilibrium between the number of atoms in the reacting molecules and the number of atoms in the reacting molecules produce.

This principle is based on the law of conservation of matter, which states that the same number of atoms in each Element in reactive substances will be conserved in the reaction products, although combined in different ways.

When a chemical reaction takes place, the bonds that form the molecules of the reacting compounds (the reactants), are broken and modified, giving rise to one or more substances. Although the molecules are modified and are no longer the same, the atoms that form them combine in a different, but the total number of atoms is conserved, so it must be the same before and after the reaction.

For example in the following chemical reaction:

HCl + NaOH -> NaCl + H₂OR

According to the stoichiometric principle, there must be the same number of atoms on each side of the equation. Let's see it for the equation we saw:

HCl + NaOH	-->	NaCl + H2OR
Hydrogen = 2 Sodium = 1 Chlorine = 1 Oxygen = 1	= = = =	Hydrogen = 2 Sodium = 1 Chlorine = 1 Oxygen = 1

Stoichiometric calculations

Stoichiometric calculations are the operations by means of which we verify that the stoichiometric principle is fulfilled in the equations, as well as its practical applications.

In the previous example of the combination of hydrochloric acid and sodium hydroxide, to produce sodium chloride and water, we made a stoichiometric calculation by atom count.

Another method of checking is the stoichiometric calculation by atomic mass units, In which the calculation is made based on the sum of the atomic masses of the elements that are combined.

This calculation can be done by the absolute masses or by rounding. In the example above:

Calculation by Absolute Mass to two decimal places:

HCl + Na O H -> Na Cl + H₂ OR

(1.00 + 35.45) + (22.98 + 15.99 + 1.00) --> (22.98 + 35.45) + (2.00 + 15.99)

(36.45) + (39.97) --> (58.43) + (17.99)

76.42 --> 76.42

Atomic mass rounding calculation:

HCl + Na O H -> Na Cl + H₂ OR

(1 + 35) + (23 + 16 + 1) --> (23 + 35) + (2 + 16)

(36) + (40) --> (58) + (18)

76 --> 76

Applications of stoichiometric equations

One of the uses of stoichiometric equations is the balancing equations, which can be done either by the Redox or trial and error methods, since in both cases the The purpose is to check that there is the same number of atoms of each element in the reactants and in the products.

In the following example we have iron trichloride:

Fe + Cl2 = FeCl3

Fe + Cl2	-->	FeCl3
Iron = 1 Chlorine = 2	= ~	Iron = 1 Chlorine = 3

In this case we know the formulas of the reactive molecules: Iron (Fe) and Chlorine (Cl₂), and its product: iron trichloride (FeCl3₃) and as we see, the number of chlorine atoms is not the same in both equations.

To fulfill the stoichiometric principle, we have to find the total number of atoms involved in the reaction and the product, so that they are the same.

To do this, we use one of the equation balancing methods (Redox, trial and error). In this example we will use the trial and error method.

The least common multiple of 2 and 3 is 6. If we multiply so that there are 6 chlorine atoms on each side of the equation, we will have the following:

Fe + 3Cl2	-->	2FeCl3
Iron = 1 Chlorine = 6	~ =	Iron = 2 Chlorine = 6

We already balanced the chlorine atoms, but now we are missing an iron atom. As we can figure out, the missing atom is on the reactant side. Then we will have:

2Fe + 3Cl2	-->	2FeCl3
Iron = 2 Chlorine = 6	= =	Iron = 2 Chlorine = 6

As we can see, we already have 6 chlorine atoms located in 3 molecules in the reactants, and 6 atoms distributed in groups of three atoms in each product molecule. Now we see that to get the same number of iron atoms in the product, we need two iron molecules in the reactants. We have balanced the equation.

Another use of stoichiometric equations is the calculation of reactants, both to avoid waste of any of the substances, such as calculating the amount of substances to neutralize an acid or a base.

This is achieved through molar calculation: The sum of the atomic masses of each of the atoms that make up a molecule, gives as a result its molar mass. For example:

If we look for the molar mass of boric acid (trioxoboric acid) whose formula is: H₃BO₃, we first calculate the molecular masses of each of its components, using the periodic table:

H₃ = (3)(1.00) = 3.00

B = (1) (10.81) = 10.81

OR₃ = (3)(15.99) = 47.94

Molar mass = 61.78

Which means that 1 mole of boric acid is equal to 61.78 grams.

The calculation of the moles of each compound will then help us to calculate the exact amount of reactive substances, both so that it is not over or needed during the reaction, as well as to calculate how much to obtain a certain amount of product.

Example:

If we use our previous example of Iron Chloride, and we want to know how much chlorine there is to combine with 100 grams of iron, and know how much amount of iron trichloride is will produce.

The equation that expresses the reaction is the following:

2Fe + 3Cl₂ -> 2FeCl₃

Now we do the molar calculation by rounding the atomic masses:

Fe = 56

Cl₂ = 70

FeCl₃ = 161

So far we have the value of 1 mole of each substance. Now we see that the number that indicates the number of reactive and product molecules is also called stoichiometric coefficient, and it tells us how many moles of that substance are interacting. In the case that the coefficient is 1, it is not written.

So substituting the values ​​we will have:

2Fe = 2 (56) = 112

3Cl₂ = 3(70) = 210

2FeCl₃ = 2(161) = 322

We apply the rule of three to calculate the mass of chlorine:

100/112 = x / 210

21000/112=187.5

So it will take 187.5 grams of chlorine to fully react with the iron.

Now we apply the rule of 3 to calculate the resulting product:

100/112 = x / 322

32200/112=287.5

So 287.5 grams of iron trichloride will be produced.

If we add the grams obtained with the relationship, we have as a result:

100 + 187.5 = 287.5

With which we check that the amounts are correct.

Stoichiometric notation

To avoid ambiguities and confusion when expressing the name and composition of the compounds, in the different types of chemical notation of inorganic compounds, the IUPAC (International Union of Pure and Applied Chemistry) has promoted the use of stoichiometric notation, used mainly in academic and research fields, with which the use of suffixes or Roman numerals is changed, by the use of Greek numerical prefixes that indicate the number of atoms of each element that make up the molecules. In the case of unit atoms, the prefix is ​​omitted.

In stoichiometric notation, the electropositive element or ion is mentioned first, followed by the electronegative one.

Formula Old Notation Stoichiometric Notation

FeO Ferrous oxide, Iron oxide Iron oxide

Faith₂OR₃: Ferric oxide, Iron III oxide Di-iron trioxide

Faith₃OR₄: Iron oxide IV Tri-iron tetraoxide

Examples of applications of the stoichiometric principle

Example 1: Balance the following equation:

HCl + MnO₂ -> MnCl₂ + 2H₂O + Cl₂

Applying the oxide-reduction method (REDOX):

HCl + MnO₂ -> MnCl₂ + 2H₂O + Cl₂

(+1-1)+(+4-4) --> (+2-2) + (+4-4)+ (-0)

As we can see, manganese has been reduced from +4 to +2.

If we review the values ​​for each element, excluding manganese, which has been reduced, we see the following values

Element reactive products

Hydrogen +1 +4

Chlorine -1 -4

Oxygen -4 -4

So now we must balance the numbers, so that they have the same values ​​on both sides of the equation. Since Chlorine and Hydrogen are in the same molecule, this means that 4 molecules of hydrochloric acid are required to balance the values:

4HCl + MnO₂ -> MnCl₂ + 2H₂O + Cl₂

(+4-4)+(+4-4) --> (+2-2) + (+4-4)+ (-0)

Example 2: In the above equation:

4HCl + MnO₂ -> MnCl₂ + 2H₂O + Cl₂

Calculate how many grams of manganese dioxide is required to produce 80 grams of manganese dichloride.

We first calculate the molar weight of each molecule (we will round with whole numbers):

HCl = 1 + 35 = 36 X 4 = 144

MnO₂ = 55 + 16 + 16 = 87

MnCl₂ = 55 + 35 + 35 = 125

H₂O = 1 + 1 + 16 = 18 X 2 = 36

Cl₂ = 35 + 35 = 70

We apply the rule of three:

x / 87 = 80/125 = 6960/125 = 55.58

So you will need 55.58 grams of magnesium dioxide.

Example 3: In the above equation:

4HCl + MnO₂ -> MnCl₂ + 2H₂O + Cl₂

Calculate how many grams of hydrochloric acid is required to produce the 80 grams of manganese dichloride.

Since we already know the values, we apply the rule of three:

x / 144 = 80/125 = 11520/125 = 92.16

It will take 92.16 grams of hydrochloric acid.

Example 4: In the same equation:

4HCl + MnO₂ -> MnCl₂ + 2H₂O + Cl₂

Calculate how many grams of water are produced by producing 125 grams of manganese dichloride.

We substitute the values ​​and apply the rule of three:

x / 36 = 125/125 = 4500/125 = 36

36 grams of water will be produced.