STOICHIOMETRIC CALCULATIONS: STAGES AND EXERCISES SOLVED - CHEMISTRY - 2025

The stoichiometric calculations are those that are made on the basis of the mass ratios of the elements or compounds involved in a chemical reaction.

The first step to perform them is to balance the chemical reaction of interest. Likewise, the correct formulas of the compounds involved in the chemical process must be known.

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Stoichiometric calculations are based on the application of a set of laws, among which are the following: The law of conservation of mass; the law of definite proportions or of constant composition; and finally, the law of multiple proportions.

The law of conservation of mass states that in a chemical reaction the sum of the masses of the reactants is equal to the sum of the masses of the products. In a chemical reaction the total mass remains constant.

The law of definite proportions or of constant composition states that different samples of any pure compound have the same elements in the same mass proportions. For example, pure water is the same no matter what its source is, or what continent (or planet) it comes from.

And the third law, that of multiple proportions, indicates that when two elements A and B form more than one compound, the proportion of the mass of element B that combines with a given mass of element A, in each of the compounds, can be expressed in terms of small whole numbers. That is, for A _n B _m n and m are integers.

What are stoichiometric calculations and their stages?

They are calculations designed to solve the different questions that may arise when a chemical reaction is being studied. For this, you must have knowledge of chemical processes and the laws that govern them.

Using stoichiometric calculation, for example, the unknown mass of another reactant can be obtained from the mass of one reactant. You can also know the percentage composition of the chemical elements present in a compound and from it, obtain the empirical formula of the compound.

Consequently, knowledge of the empirical or minimal formula of a compound allows the establishment of its molecular formula.

In addition, the stoichiometric calculation allows to know in a chemical reaction which is the limiting reagent, or if there is a surplus reagent, as well as its mass.

Stages

The stages will depend on the type of problem posed, as well as its complexity.

Two common situations are:

-Two elements react to create a compound and only the mass of one of the reacting elements is known.

-We want to know the unknown mass of the second element, as well as the mass of the compound resulting from the reaction.

In general, in solving these exercises the following order of stages should be followed:

-Establish the equation of the chemical reaction.

-Balance the equation.

-The third stage is, through the use of the atomic weights of the elements and the stoichiometric coefficients, to obtain the proportion of the masses of the reacting elements.

-Then, by using the law of defined proportions, once the mass of a reacting element is known and the proportion with which it reacts with the second element, knowing the mass of the second element.

-And the fifth and last stage, if the masses of the reactant elements are known, their sum allows us to calculate the mass of the compound produced in the reaction. In this case, this information is obtained based on the law of conservation of mass.

Solved exercises

-Exercise 1

What is the remaining reagent when 15 g of Mg is reacted with 15 g of S to form MgS? And how many grams of MgS will be produced in the reaction?

Data:

- Mass of Mg and S = 15 g

-Atomic weight of Mg = 24.3 g / mol.

-Atomic weight of S = 32.06 g / mol.

Step 1: reaction equation

Mg + S => MgS (already balanced)

Step 2: Establish the ratio in which Mg and S combine to produce MgS

For simplicity, the atomic weight of Mg can be rounded to 24 g / mol and the atomic weight of S to 32 g / mol. So the ratio in which the S and Mg are combined will be 32:24, dividing the 2 terms by 8, the ratio reduces to 4: 3.

Reciprocally, the ratio in which Mg combines with S equals 3: 4 (Mg / S)

Step 3: discussion and calculation of the excess reactant and its mass

The mass of Mg and S is 15 g for both, but the ratio in which Mg and S react is 3: 4 and not 1: 1. Then, it can be deduced that the excess reactant is Mg, since it is found in a lower proportion with respect to S.

This conclusion can be tested by calculating the mass of Mg reacting with 15 g of S.

g of Mg = 15 g of S x (3 g of Mg) / mol) / (4 g of S / mol)

11.25 g of Mg

Mass of excess Mg = 15 g - 11.25 g

3.75 g.

Step 4: Mass of MgS formed in the reaction based on the law of conservation of mass

Mass of MgS = mass of Mg + mass of S

11.25 g + 15 g.

26, 25 g

An exercise for educational purposes could be done as follows:

Calculate the grams of S that react with 15 g of Mg, using in this case a ratio of 4: 3.

g of S = 15 g of Mg x (4 g of S / mol) / (3 g of Mg / mol)

20 g

If the situation were the one presented in this case, it could be seen that the 15 g of S would not be enough to fully react with the 15 g of Mg, lacking 5 g. This confirms that the excess reagent is Mg and S is the limiting reagent in the formation of MgS, when both reactive elements have the same mass.

-Exercise 2

Calculate the mass of sodium chloride (NaCl) and impurities in 52 g of NaCl with a percent purity of 97.5%.

Data:

-Sample mass: 52 g of NaCl

-Pure percentage = 97.5%.

Step 1: calculate the pure mass of NaCl

NaCl mass = 52 gx 97.5% / 100%

50.7 g

Step 2: calculation of the mass of impurities

% impurities = 100% - 97.5%

2.5%

Mass of impurities = 52 gx 2.5% / 100%

1.3 g

Therefore, of the 52 g of salt, 50.7g are pure NaCl crystals, and 1.3g of impurities (such as other ions or organic matter).

-Exercise 3

What mass of oxygen (O) is there in 40 g of nitric acid (HNO ₃), knowing that its molecular weight is 63 g / mol and the atomic weight of O is 16 g / mol?

Data:

-Mass of HNO ₃ = 40 g

-Atomic weight of O = 16 g / mol.

-Molecular weight of HNO ₃

Step 1: Calculate the number of moles of HNO

Moles HNO ₃ = 40 g of HNO ₃ x 1 mol of HNO ₃ /63 g of HNO ₃

0.635 moles

Step 2: calculate the number of moles of O present

The formula for HNO ₃ indicates that there are 3 moles of O for every mole of HNO _3.

Moles of O = 0.635 moles of HNO ₃ X 3 moles of O / mole of HNO ₃

1.905 moles of O

Step 3: calculate the mass of O present in 40 g of HNO

g of O = 1.905 moles of O x 16 g of O / mole of O

30.48 g

In other words, of the 40g of HNO ₃, 30.48g are due exclusively to the weight of the moles of oxygen atoms. This large proportion of oxygen is typical of oxoanions or their tertiary salts (NaNO ₃, for example).

-Exercise 4

How many grams of potassium chloride (KCl) are produced when 20 g of potassium chlorate (KClO ₃) decompose ? Knowing that the molecular weight of KCl is 74.6 g / mol and the molecular weight of KClO ₃ is 122.6 g / mol

Data:

-Mass of KClO ₃ = 20 g

-Molecular weight of KCl = 74.6 g / mol

-Molecular weight of KClO ₃ = 122.6 g / mol

Step 1: reaction equation

2KClO ₃ => 2KCl + 3O ₂

Step 2: calculate the mass of KClO

g of KClO ₃ = 2 moles x 122.6 g / mole

245.2 g

Step 3: calculate the mass of KCl

g of KCl = 2 moles x 74.6 g / mole

149.2 g

Step 4: calculate the mass of KCl produced by decomposition

245 g of KClO ₃ are produced by decomposition 149.2 g of KCl. Then this ratio (stoichiometric coefficient) can be used to find the mass of KCl that is produced from 20 g of KClO ₃:

g of KCl = 20 g of KClO ₃ x 149 g of KCl / 245.2 g of KClO ₃

12.17 g

Note how is the mass ratio of O ₂ within KClO ₃. Of the 20g of KClO ₃, just under half is due to oxygen that is part of the oxoanion chlorate.

-Exercise 5

Find the percentage composition of the following substances: a) dopa, C ₉ H ₁₁ NO ₄ and b) Vanillin, C ₈ H ₈ O ₃.

a) Dopa

Step 1: find the molecular weight of dopa C

To do this, the atomic weight of the elements present in the compound is initially multiplied by the number of moles represented by their subscripts. To find the molecular weight, the grams contributed by the different elements are added.

Carbon (C): 12 g / mol x 9 mol = 108 g

Hydrogen (H): 1 g / mol x 11 mol = 11 g

Nitrogen (N): 14 g / mol x 1 mol = 14 g

Oxygen (O): 16 g / mol x 4 mol = 64 g

Dopa molecular weight = (108 g + 11 g + 14g + 64 g)

197 g

Step 2: Find the percentage composition of the elements present in dopa

For this, its molecular weight (197 g) is taken as 100%.

% of C = 108 g / 197g x 100%

54.82%

% H = 11 g / 197g x 100%

5.6%

% of N = 14 g / 197 gx 100%

7.10%

% O = 64 g / 197 g

32.48%

b) Vanillin

Part 1: calculating the molecular weight of vanillin C

To do this, the atomic weight of each element is multiplied by the number of its moles present, adding the mass contributed by the different elements

C: 12 g / mol x 8 mol = 96 g

H: 1 g / mol x 8 mol = 8 g

Or: 16 g / mol x 3 mol = 48 g

Molecular weight = 96 g + 8 g + 48 g

152 g

Part 2: Find the% of the different elements in vanillin

Its molecular weight (152 g / mol) is assumed to represent 100%.

% of C = 96 g / 152 gx 100%

63.15%

% of H = 8 g / 152 gx 100%

5.26%

% O = 48 g / 152 gx 100%

31.58%

-Exercise 6

The percentage composition by mass of an alcohol is as follows: carbon (C) 60%, hydrogen (H) 13% and oxygen (O) 27%. Get your minimum formula or empirical formula.

Data:

Atomic weights: C 12 g / mol, H 1g / mol and oxygen 16 g / mol.

Step 1: calculating the number of moles of the elements present in alcohol

The mass of the alcohol is assumed to be 100g. Consequently, the mass of C is 60 g, the mass of H is 13 g, and the mass of oxygen is 27 g.

Calculation of the number of moles:

Number of moles = mass of the element / atomic weight of the element

moles of C = 60 g / (12 g / mol)

5 moles

moles of H = 13 g / (1 g / mol)

13 moles

moles of O = 27 g / (16 g / mol)

1.69 moles

Step 2: obtain the minimum or empirical formula

To do this, find the ratio of whole numbers between the numbers of moles. This serves to obtain the number of atoms of the elements in the minimum formula. For this purpose, the moles of the different elements are divided by the number of moles of the element to a lesser extent.

C = 5 moles / 1.69 moles

C = 2.96

H = 13 moles / 1.69 moles

H = 7.69

O = 1.69 moles / 1.69 moles

O = 1

Rounding these figures, the minimum formula is: C ₃ H ₈ O. This formula corresponds to that of propanol, CH ₃ CH ₂ CH ₂ OH. However, this formula is also that of the compound CH ₃ CH ₂ OCH ₃, ethyl methyl ether.

References

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