The ponderal laws of chemistry They are those that have shown that the masses of the substances that react do not do so in an arbitrary or random way; but maintaining a constant mathematical ratio of integers 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 or molecular masses of the elements or compounds, respectively, were not even known.
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 the 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 the chemical compounds cigars are always formed with the same mass proportion of their constituent elements.
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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 the same mass; hence the famous phrase: "nothing is created, nothing is destroyed, everything is transformed".
Historically, the law of the 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 it Lavoisier's Law.
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 weight gain..
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.
The Law of Conservation of Masses established the need to balance chemical equations, guaranteeing that the number of all the elements that take part 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 carried out..
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 is fulfilled..
CH4 + 2 Otwo => COtwo + 2 htwoOR
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 CH4:
Moles of water = 5 moles of CH4(2 moles of HtwoO / 1 mol CH4)
= 10
Which would be equivalent to 180 g of HtwoO. Also formed 5 mol or 220 g of COtwo, 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 those 400 g, 80 g correspond to 5 moles of CH4 (multiplying by its molecular mass of 16 g / mol), and 320 g at 10 moles of Otwo (similarly for its molecular mass of 32 g / mol).
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 Otwo (to)
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)
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 compound pure specific; 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." Then, 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.".
Iron (Fe) reacts with sulfur (S) to form iron sulfide (FeS), three situations can be indicated (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 are combined with 1.0 g of S to give 2.75 g of FeS.
By applying this law, it is possible to 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 that take part in a chemical reaction, or if 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.
Carbon dioxide (COtwo) is formed in the following reaction:
C + Otwo => COtwo
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.
When 4 g and 6 g of sulfur (S) were reacted with oxygen (O) in different vessels, 10 g and 15 g of sulfur trioxide (SO3).
Why were such quantities of sulfur trioxide obtained and not other?
Also calculate the amount of sulfur required to combine with 36 g of oxygen and the mass of sulfur trioxide obtained.
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 container 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.
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 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?
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 MgCltwo
Later:
g of Cl = 25 g of MgCltwo (2.95 g Cl / 3.95 g MgCltwo)
= 18.67
g of Mg = 25 g of MgCltwo (1 g Mg / 3.95 g MgCltwo)
= 6.33
So 18.67 g of chlorine is combined with 6.33 g of magnesium to produce 25 g of magnesium chloride..
First calculate the molecular mass of magnesium chloride, MgCltwo:
Molecular weight MgCltwo = 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%
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 originate 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 the 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 integer ratios..
Dalton completed the investigative works of Proust. He pointed out 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..
Show that the compounds water, HtwoO, and hydrogen peroxide, HtwoORtwo, comply with the Law of Multiple Proportions.
Atomic weights of the elements: H = 1 g / mol and oxygen = 16 g / mol.
Molecular weights of compounds: HtwoO = 18 g / mol and HtwoORtwo = 34 g / mol.
Hydrogen is the element with a fixed amount in HtwoO and HtwoORtwo, so the proportions between O and H in both compounds will be established.
O / H ratio in HtwoO = (16 g / mol) / (2 g / mol)
= 8/1
O / H ratio in HtwoORtwo = (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..
What mass of Oxygen combines with 3.0 g of nitrogen in a) nitric oxide, NO and b) nitrogen dioxide, NOtwo. Show that NO and NOtwo comply with the Law of Multiple Proportions.
Mass of nitrogen = 3 g
Atomic weights: nitrogen, 14 g / mol, and oxygen, 16 g / mol.
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 the NOtwo, 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 NOtwo = 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..
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 among 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 the reciprocal proportions law 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 analyzed from a mathematical point of view, it will be seen that it consists only of conversion factors and cancellations.
If it is known that 12 g of carbon reacts 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 can be estimatedtwo and HtwoOr, respectively.
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). Then, it is necessary to do a division of the previous proportions to show whether or not reciprocity is satisfied:
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 COtwo, 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.
If 24 g of magnesium is known to react with 2 g of hydrogen to form magnesium hydride; and that in addition, 32 g of sulfur reacts with 2 g of hydrogen to form hydrogen sulfide, 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.
It is known that 35.5 g of Cl reacts with 1 g of H to form HCl. Likewise, 27 g of Al reacts with 3 g of H to form AlH3. Calculate the proportion of aluminum chloride and tell whether this 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.
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 (AlCl3, with a PA of 35.5 g / mol for Cl).
Here you can see how Richter's law can start to falter due to precision and the misuse of decimals..
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