weight laws they are generalizations about the masses of all participants (reagents and products) in a chemical reaction. Generically, a chemical reaction can be represented as follows:
A + B C + D
→ Lavoisier's Law (Law of conservation of mass)
According to Lavoisier, when a chemical reaction is carried out in a closed environment, the sum of the masses of the reactants is always equal to the sum of the masses of the products.
Sum of reagent masses = Sum of product masses
Thus, according to Lavoisier, if a generic reaction (reagents A and B, products C and D) is carried out in a closed container, using 5 grams of A and 10 grams of B, we can say that the mass of product C is 15 grams.
A + B → C
5g 10g x
Since the sum of the masses of the reactants is equal to the sum of the masses of the products:
5 + 10 = x
15 grams = x
or
x = 15 g
→ Proust's Law (Law of defined proportions)
According to Proust, participants in a chemical reaction always establish a constant mass ratio. When we decompose water by electrolysis, for example, we get hydrogen gas and gas oxygen:
2h2O → 2H2 + O2
Whenever this is done, it is verified that the proportion between the masses of hydrogen and oxygen gases obtained is always 1 to 8, regardless of the mass of water used in the electrolysis. Thus:
Electrolysis of 4.5 grams of water
2h2O → 2H2 + O2
4.5g 0.5g 4g
If we divide the masses of H2 it's the2 formed, we will have the ratio 1 to 8:
0,5 = 1
4 8
Electrolysis of 9 grams of water
2h2O → 2H2 + O2
9g 1g 8g
If we divide the masses of H2 it's the2 formed, we will have the ratio 1 to 8:
1
8
Another interesting fact observed by Proust is that if we divide the masses of H2O, H2 it's the2 of the two examples above, we will have the same proportion:
2h2O → 2H2 + O2
4.5g 0.5g 4g
9g 1g 8g
I.e:
1 = 1 = 1
2 2 2
Therefore, according to Proust's law, for a generic reaction, using different masses of substances involved in it, at different times, we can use the following expression in relation to the masses of participants:
A + B → C
1st experience bad = MB = mC
2nd experiment mA’ = mB’ = mC’
→ Dalton's Law (Law of Multiple Proportions)
According to John Dalton, when a fixed mass of a substance A combines with different masses of a substance B, giving rise to different substances, the masses of B have a relationship expressed by whole numbers and small.
When we react carbon with oxygen, for example, we can form carbon dioxide or carbon monoxide, as in the following two cases:
Carbon + oxygen → carbon dioxide
12g 16g 28g
Carbon + oxygen → carbon dioxide
12g 32g 44g
In both reactions we have the same mass of reagent A. So, if we divide the masses of oxygen, which is substance B that appears in both reactions, we will see a relationship between whole and small numbers:
16 = 1
32 2
→ Application of weight laws:
1º)It is known that Hydrogen gas reacts with Oxygen gas in a 1:8 ratio, by mass, to form water. Knowing this fact, determine the values of the masses X, Y and Z in the following table, respectively:
a) 36 g, 44 g and 51.8 g
b) 33.6 g, 2.4 g and 52 g
c) 32 g, 44 g and 51 g
d) 36 g, 48 g and 52 g
e) 37 g, 44.8 g and 51.8 g
To resolve the issue, just do the following:
1O Step: Mass X can be found by Lavoisier's Law, as it is the only known mass in the second experiment, thus:
Sum of reagent masses = Sum of product masses
5 + 32 = X
37 = X
X = 37 grams
2O Step: To find the value of mass Z, we can use Proust's law, because, in a reaction that is performed more than once, the masses follow a proportion according to the scheme below:
bad = MB = mC
mA’ mB’ mC’
Thus, to find the Z mass, we can use participants A (Hydrogen) and B (Oxygen):
bad = MB
mA’ mB’
5 = 32
7 Z
5.Z = 7.32
Z = 224
5
Z = 44.8 grams
3O Step: Mass y can be found by Lavoisier's Law, like this:
Sum of reagent masses = Sum of product masses
7 + 44.8 = Y
51.8 = Y
Y = 51.8 grams
By Me. Diogo Lopes Dias
Source: Brazil School - https://brasilescola.uol.com.br/o-que-e/quimica/o-que-sao-as-leis-ponderais.htm
