Calculations involving Proust's Law

THE Proust's Law is also known as the Law of Constant Proportions because, regardless of the mass of a decomposed substance, the masses of the originating substances will always follow the same proportion. The same reasoning can be used in a synthesis reaction, that is, the proportion of masses of reagents used is always constant for the formation of the product.

In this text, we will emphasize the calculations involving the Proust's Law To do this, we will use examples to clearly demonstrate how calculations based on this law should be performed. Follow up!

Examples of calculations involving Proust's Law

1º) Find the values ​​of A and B in the synthesis reaction of magnesium oxide from metallic magnesium and oxygen gas. To do this, use the data provided in the table below:

2 Mg + O₂ → 2 MgO

Mg mass	O mass2	MgO mass
48 g	32 g	80 g
96 g	THE	B

Analyzing the data in the table, we observe that the exercise asks us to find the values ​​of the masses represented by A and B. Knowing that Proust's Law is the Law of Constant Proportions, we have to:

m_1mg = m_1O2 = m_1MgO
m_2mg = m_2O2 = m_2MgO

m_1mg = 48 g

m_1O2 = 32 g

m_1MgO = 80 g

m_2mg = 96 g

m_2O2 = A

m_2MgO = B

• To find the value of A, let's apply Proust's law through the following relationship:

m_1mg = m_1O2
m_2mg = m_2O2

48 = 32
96 A

Multiplying cross:

48.A = 96.32

48.A = 3072

A = 3072
48

A = 64 g

• To find the value of B, let's apply Proust's law through the following relationship:

m_1O2 = m_1MgO
m_2O2 m_2MgO

32 = 80
64B

Multiplying cross:

32.B = 80.64

32.B = 5120

B = 5120
32

B = 160 g

Observation: The resolution of this exercise could be done in a simpler way, since, if Proust's law proposes proportion in mass, the masses of A and B would be double the masses 32 and 80 respectively. This would be possible because the mass of 96g of Mg is twice the mass of 48g.

2º) The combustion reaction of ethanol (C₂H₆O) produces carbon dioxide (CO₂) and water (H₂O). Calculate the X, Y, and Z masses using the data provided below:

Ç₂H₆O + 3 O₂ → 2 CO₂ + 3 H₂O

C mass2H6O	O mass2	CO mass2	H mass2O
46 g	96 g	88 g	54 g
9.2 g	X	Y	Z

Analyzing the data in the table, we observe that the exercise asks us to find the values ​​of the masses represented by X, Y and Z. Knowing that Proust's Law is the Law of Constant Proportions, we have to:

m_1C2H6O = m_1O2 = m_1CO2 = m_1H2O
 m_2C2H6O m_2O2 m_2CO2 m_2H2O

m_1C2H6 = 46 g

m_2C2H6O = 9.2 g

m_1O2 = 96 g

m_2O2 = X g

m_1CO2 = 88 g

m_2CO2 = Y

m_1H2O = 54 g

m_2H2O = Z

• To find the value of X, let's apply Proust's law through the following relationship:

m_1C2H6O = m_1O2
m_2C2H6O m_2O2

 46  = 96
9.2 X

Multiplying cross:

46.X = 96.9.2

46.X = 883.2

X = 883,2
46

X = 19.2 g

• To find the value of Y, let's apply Proust's law through the following relationship:

m_1O2 = m_1CO2
m_2O2 m_2CO2

 96  = 88
19.2 Y

Multiplying cross:

96.Y = 19.2.88

96.Y = 1689.6

Y = 1689,6
96

Y = 17.6 g

• To find the value of Z, let's apply Proust's law through the following relationship:

m_1CO2 = m_1H2O
m_2CO2 m_2H2O

 88  = 54
17.6 Z

Multiplying cross:

88.Z = 17.6.54

88.Z = 950.4

Z = 950,4
88

Z = 10.8 g

By Me. Diogo Lopes Dias

Source: Brazil School - https://brasilescola.uol.com.br/quimica/calculos-que-envolvem-lei-proust.htm