WHAT ARE THE WEIGHT LAWS OF CHEMISTRY? (EXAMPLES) - CHEMISTRY - 2025

The ponderal laws of chemistry are those that have shown that the masses of the substances that react do not do so in an arbitrary or random way; but by maintaining a constant mathematical proportion of whole numbers or submultiples thereof, in which the atoms of the elements are neither created nor destroyed.

In times past establishing these laws required extraordinary efforts of reasoning; because although it seems too obvious now, before the atomic and molecular masses of the elements or compounds, respectively, were not even known.

Source: Jeff Keyzer from Austin, TX, USA

Since it was not known exactly how much one mole of atoms of each element equaled, chemists in the 18th and 19th centuries had to rely on the reactant masses. So rudimentary analytical balances (top image) were inseparable companions during the hundreds of experiments required to establish weight laws.

It is for this reason that when you study these laws of chemistry you come across mass measurements at every moment. Thanks to this, extrapolating the results of the experiments, it was discovered that pure chemical compounds are always formed with the same mass proportion of their constituent elements.

Law of conservation of mass

This law says that in a chemical reaction, the total mass of the reactants is equal to the total mass of the products; as long as the considered system is closed and there is no exchange of mass and energy with its surroundings.

In a chemical reaction, substances do not disappear, but are transformed into other substances of equal mass; hence the famous phrase: “nothing is created, nothing is destroyed, everything is transformed”.

Historically, the law of conservation of mass in a chemical reaction was first proposed in 1756 by Mikhail Lomonsov, who showed the results of his experiments in his journal.

Later in 1774, Antoine Levoisier, French chemist, presented the results of his experiments that allowed to establish this; which some also call Lavoisier's Law.

-Lavoisier experiments

In Lavoisier's time (1743-1794), there was the Phlogiston Theory, according to which bodies had the ability to catch fire or burn. Lavoisier's experiments made it possible to discard this theory.

Lavoisier conducted numerous metal combustion experiments. He carefully weighed the materials before and after combustion in a closed container, finding that there was an apparent gain in weight.

But Lavoiser, based on his knowledge of the role of oxygen in combustion, concluded that the weight gain in combustion was due to the incorporation of oxygen into the burning material. The concept of metallic oxides was born.

Therefore, the sum of the masses of the metals subjected to combustion and of the oxygen remained unchanged. This conclusion allowed the establishment of the Law of Conservation of Mass.

-Balance of equations

The Law of Conservation of Masses established the need to balance chemical equations, guaranteeing that the number of all the elements involved in a chemical reaction, both as reactants or as products, is exactly the same.

This is an essential requirement for the accuracy of the stoichiometric calculations to be performed.

-Calculations

Water moles

How many moles of water can be produced during the combustion of 5 moles of methane in excess oxygen? Also show that the law of conservation of matter holds.

CH ₄ + 2 O ₂ => CO ₂ + 2 H ₂ O

Observing the balanced equation of the reaction, it is concluded that 1 mole of methane produces 2 moles of water.

The problem can be solved directly with a simple approach, since we do not have 1 mole but 5 moles of CH ₄:

Moles of water = 5 moles of CH ₄ (2 moles of H ₂ O / 1 mole of CH ₄)

= 10

This would be equivalent to 180 g of H ₂ O. Also 5 mol or 220 g of CO _{2 was formed}, which is equal to a total mass of 400 g of products.

Thus, for the law of conservation of matter to be fulfilled, 400 g of reagents must react; no more no less. Of these 400 g, 80 g correspond to 5 moles of CH ₄ (multiplied by its molecular mass of 16 g / mol), and 320 g correspond to 10 moles of O ₂ (in the same way by its molecular mass of 32 g / mol).

Combustion of a magnesium ribbon

A 1.50 g magnesium ribbon was burned in a closed container containing 0.80 g of oxygen. After combustion, 0.25 g of oxygen remained in the container. a) What mass of oxygen reacted? b) How much magnesium oxide was formed?

The mass of oxygen that reacted is obtained by a simple difference.

Mass of oxygen consumed = (initial mass - residual mass) oxygen

= 0.80 g - 0.25 g

= 0.55 g O ₂ (a)

According to the law of conservation of mass, Mass of magnesium oxide = mass of magnesium + mass of oxygen

= 1.50 g + 0.55 g

= 2.05 g MgO (b)

Law of definite proportions

Joseph Louis Proust (1754-1826), French chemist, realized that in a chemical reaction the chemical elements always react in fixed proportions of masses to form a specific pure compound; therefore, its composition is constant, regardless of the source or origin, or how it is synthesized.

Proust in 1799 enunciated the law of definite proportions, which states that: "When two or more elements combine to form a compound, they do so in a fixed mass ratio." So, this relationship is fixed and does not depend on the strategy followed for the preparation of the compound.

This law is also known as the law of constant composition, which states that: "Every chemical compound in a state of purity always contains the same elements, in a constant proportion of mass."

-Illustration of the law

Iron (Fe) reacts with sulfur (S) to form iron sulfide (FeS), three situations can be noted (1, 2 and 3):

To find the proportion in which the elements combine, divide the greater mass (Fe) by the lesser mass (S). The calculation gives a ratio of 1.75: 1. This value is repeated in the three conditions given (1, 2 and 3), where the same proportion is obtained although different masses are used.

That is, 1.75 g of Fe is combined with 1.0 g of S to give 2.75 g of FeS.

-Applications

By applying this law, one can know exactly the masses of the elements that must be combined to obtain a desired mass of a compound.

In this way, information can be obtained about the excess mass of some of the elements involved in a chemical reaction, or whether there is a limiting reagent in the reaction.

In addition, it is applied to know the centesimal composition of a compound, and based on the latter, the formula of a compound can be established.

Centesimal composition of a compound

Carbon dioxide (CO ₂) is formed in the following reaction:

C + O ₂ => CO ₂

12 g of carbon combine 32 g of oxygen to give 44 g of carbon dioxide.

So the percent carbon equals

Carbon percentage = (12 g / 44 g) 100%

= 27.3%

Percentage of oxygen = (32 g / 44 g) 100%

Percentage of oxygen = 72.7%

Using the statement of the Law of Constant Composition, it can be noted that carbon dioxide is always made up of 27.3% carbon and 72.7% oxygen.

-Calculations

Sulfur trioxide

By reacting in different vessels 4 g and 6 g of sulfur (S) with oxygen (O), 10 g and 15 g of sulfur trioxide (SO ₃) were obtained, respectively.

Why were such amounts of sulfur trioxide obtained and not others?

Also calculate the amount of sulfur required to combine with 36 g of oxygen and the mass of sulfur trioxide obtained.

Part A)

In the first container 4 of sulfur are mixed with X g of oxygen to obtain 10 g of trioxide. If the law of conservation of mass is applied, we can solve for the mass of oxygen that combined with the sulfur.

Mass of oxygen = 10 g of oxygen trioxide - 4 g of sulfur.

= 6 g

In vessel 2 6 g of sulfur are mixed with X g of oxygen to obtain 15 of sulfur trioxide.

Mass of oxygen = 15 g of sulfur trioxide - 6 g of sulfur

= 9 g

We then proceed to calculate the O / S ratios for each container:

O / S ratio in situation 1 = 6 g O / 4 g S

= 1.5 / 1

O / S ratio in situation 2 = 9 g O / 6 g S

= 1.5 / 1

Which is in accordance with what is stated in the law of defined proportions, which indicates that the elements always combine in the same proportion to form a certain compound.

Therefore, the values ​​obtained are correct and those that correspond to the application of the Law.

Part b)

In the previous section, a value of 1.5 / 1 was calculated for the O / S ratio.

g of sulfur = 36 of oxygen (1 g of sulfur / 1.5 g of oxygen)

= 24 g

g of sulfur trioxide = 36 g of oxygen + 24 g of sulfur

= 60 g

Chlorine and magnesium

Chlorine and magnesium are combined in the ratio of 2.95 g of chlorine for every g of magnesium. a) Determine the masses of chlorine and magnesium necessary to obtain 25 g of magnesium chloride. b) What is the percentage composition of magnesium chloride?

Part A)

Based on the value 2.95 for the Cl: Mg ratio, the following approach can be made:

2.95 g of Cl + 1 g of Mg => 3.95 g of MgCl ₂

Then:

g of Cl = 25 g of MgCl ₂ · (2.95 g Cl / 3.95 g MgCl ₂)

= 18.67

g of Mg = 25 g of MgCl ₂ · (1 g Mg / 3.95 g MgCl ₂)

= 6.33

Then 18.67 g of chlorine is combined with 6.33 g of magnesium to produce 25 g of magnesium chloride.

Part b)

First calculate the molecular mass of magnesium chloride, MgCl ₂:

Molecular weight MgCl ₂ = 24.3 g / mol + (2 35.5 g / mol)

= 95.3 g / mol

Magnesium percentage = (24.3 g / 95.3 g) x 100%

= 25.5%

Chlorine percentage = (71 g / 95.3 g) x 100%

= 74.5%

Law of multiple proportions or Dalton's law

The Law was enunciated in 1803 by the French chemist and meteorologist John Dalton, based on his observations regarding the reactions of atmospheric gases.

The law was stated in the following way: "When elements are combined to give more than one compound, a variable mass of one of them joins a fixed mass of the other and the first has as a relation of canonical and indistinct numbers".

Also: "When two elements are combined to give rise to different compounds, given a fixed quantity of one of them, the different quantities of the other element that combine with said fixed quantity to produce the compounds are in relation to simple integers."

John Dalton made the first modern description of the atom as a component of chemical elements, when he pointed out that the elements are made up of indivisible particles called atoms.

In addition, he postulated that compounds are formed when atoms of different elements combine with each other in simple whole-number ratios.

Dalton completed the investigative works of Proust. He indicated the existence of two tin oxides, with percentages of 88.1% and 78.7% of tin with the corresponding percentages of oxygen, 11.9% and 21.3%, respectively.

-Calculations

Water and hydrogen peroxide

Show that the compounds water, H ₂ O, and hydrogen peroxide, H ₂ O ₂, satisfy the Law of Multiple Proportions.

Atomic weights of the elements: H = 1 g / mol and oxygen = 16 g / mol.

Molecular weights of the compounds: H ₂ O = 18 g / mol and H ₂ O ₂ = 34 g / mol.

Hydrogen is the element with a fixed amount in H ₂ O and H ₂ O ₂, so the proportions between O and H in both compounds will be established.

O / H ratio in H ₂ O = (16 g / mol) / (2 g / mol)

= 8/1

O / H ratio in H ₂ O ₂ = (32 g / mol) / (2 g / mol)

= 16/1

Relationship between both proportions = (16/1) / (8/1)

= 2

So the O / H ratio of hydrogen peroxide to water is 2, a simple whole number. Therefore, compliance with the Law of Multiple Proportions is demonstrated.

Nitrogen oxides

What mass of Oxygen combines with 3.0 g of nitrogen in a) nitric oxide, NO and b) nitrogen dioxide, NO ₂. Show that NO and NO ₂ comply with the Law of Multiple Proportions.

Mass of nitrogen = 3 g

Atomic weights: nitrogen, 14 g / mol, and oxygen, 16 g / mol.

Calculations

In NO, one N atom combines with 1 O atom, so the mass of oxygen that combines with 3 g of nitrogen can be calculated using the following approach:

g of O = g nitrogen · (PA. O / PA. N)

= 3 g (16 g / mol / 14 g / mol)

= 3.43 g O

In NO ₂, one N atom combines with 2 O atoms, so the mass of oxygen that is combined is:

g of oxygen = 3 g (32 g / mol / 14 g / mol)

= 6.86 g O

O / N ratio in NO = 3.43 g O / 3 g N

= 1,143

O / N ratio in NO ₂ = 6.86 g O / 3 g N

= 2,282

Value of the relationship between the O / N proportions = 2,282 / 1,143

= 2

So, the value of the O / N ratio is 2, a simple whole number. Therefore, the Law of Multiple Proportions is fulfilled.

Law of reciprocal proportions

This law formulated by Richter and Carl F. Wenzel separately, establishes that the mass proportions of two compounds with an element in common, allows to determine the proportion of a third compound between the other elements if they react.

For example, if you have the two compounds AB and CB, you can see that the common element is B.

The Richter-Wenzel law or reciprocal proportions says that, knowing how much of A reacts with B to give AB, and how much of C reacts with B to give CB, we can calculate the mass of A that is needed to react with a mass of C to form AC.

And the result is that the ratio A: C or A / C must be a multiple or submultiple of A / B or C / B. However, this law is not always fulfilled, especially when the elements present various oxidation states.

Of all the ponderal laws this is perhaps the most "abstract" or complicated. But if you analyze it from a mathematical point of view, it will be seen that it consists only of conversion factors and cancellations.

-Examples

Methane

If 12 g of carbon is known to react with 32 g of oxygen to form carbon dioxide; and that, on the other hand, 2 g of hydrogen reacts with 16 g of oxygen to form water, then the mass proportions C / O and H / O for CO ₂ and H ₂ O, respectively, can be estimated.

Calculating C / O and H / O we have:

C / O = 12g C / 32g O

= 3/8

H / O = 2g H / 16g O

= 1/8

Oxygen is the common element, and you want to know how much carbon reacts with hydrogen to produce methane; that is, you want to calculate C / H (or H / C). So, it is necessary to make a division of the previous proportions to show whether the reciprocity is fulfilled or not:

C / H = (C / O) / (H / O)

Note that in this way the O's are canceled and C / H remains:

C / H = (3/8) / (1/8)

= 3

And 3 is a multiple of 3/8 (3/8 x 8). This means that 3 g of C reacts with 1 g of H to give methane. But, to be able to compare it with CO ₂, multiply C / H by 4, which is equal to 12; this gives 12 g of C which reacts with 4 g of H to form methane, which is also true.

Magnesium sulphide

If 24 g of magnesium is known to react with 2 g of hydrogen to form magnesium hydride; Furthermore, 32 g of sulfur reacts with 2 g of hydrogen to form hydrogen sulphide, the element in common is hydrogen and we want to calculate Mg / S from Mg / H and H / S.

Then calculating Mg / H and H / S separately, we have:

Mg / H = 24g Mg / 2g H

= 12

H / S = 2g H / 32g S

= 1/16

However, it is convenient to use S / H to cancel the H. Therefore, S / H is equal to 16. Once this is done, we proceed to calculate Mg / S:

Mg / S = (Mg / H) / (S / H)

= (12/16)

= 3/4

And 3/4 is a submultiple of 12 (3/4 x 16). The Mg / S ratio indicates that 3 g of Mg reacts with 4 g of sulfur to form magnesium sulfide. However, you have to multiply Mg / S by 8 to be able to compare it with Mg / H. Thus, 24 g of Mg reacts with 32 g of sulfur to give this metal sulfide.

Aluminum chloride

It is known that 35.5 g of Cl reacts with 1 g of H to form HCl. Also, 27 g of Al reacts with 3 g of H to form AlH ₃. Find the proportion of aluminum chloride and tell whether that compound obeys the Richter-Wenzel law.

Again, we proceed to calculate Cl / H and Al / H separately:

Cl / H = 35.5g Cl / 1g H

= 35.5

Al / H = 27g Al / 3g H

= 9

Now, Al / Cl is calculated:

Al / Cl = (Al / H) / (Cl / H)

= 9 / 35.5

≈ 0.250 or 1/4 (actually 0.253)

That is, 0.250 g of Al reacts with 1 g of Cl to form the corresponding salt. But, again, Al / Cl must be multiplied by a number that allows it to be compared (for convenience) with Al / H.

Inaccuracies in the calculation

Al / Cl is then multiplied by 108 (27 / 0.250), giving 27 g of Al which reacts with 108 g of Cl. This is not exactly the case. If we take for example the value 0.253 times Al / Cl, and multiply it by 106.7 (27 / 0.253), we will have that 27 g of Al reacts with 106.7 g of Cl; which is closer to reality (AlCl ₃, with a PA of 35.5 g / mol for Cl).

Here we see how Richter's law can start to falter over precision and the misuse of decimals.

References

1. Whitten, Davis, Peck & Stanley. (2008). Chemistry. (8th ed.). CENGAGE Learning.
2. Flores, J. Química (2002). Editorial Santillana.
3. Joaquín San Frutos Fernández. (sf). The ponderal and volumetric laws. Recovered from: encina.pntic.mec.es
4. Toppr. (sf). Laws of Chemical Combination. Recovered from: toppr.com
5. Brilliant. (2019). Laws of Chemical Combination. Recovered from: brilliant.org
6. Chemistry LibreTexts. (2015, July 15). Fundamental Chemical Laws. Recovered from: chem.libretexts.org
7. Helmenstine, Anne Marie, Ph.D. (January 18, 2019). Law of Conservation of Mass. Recovered from: thoughtco.com