Law of constant composition

Law of constant composition:

For the first time has formulated by G. Prust (1808)

"All individual chemical substances have constant quality and quantity composition and definite chemical structure and does not depend on how this substance was prepared."
From the law of constant composition follows that at complex substance formation the elements combine with each other in definite mass proportions.

example
CuS- copper sulphide. m (Cu) : m (S) = Ar (Cu) : Ar (S) = 64 : 32 = 2 : 1
To get copper sulphide (CuS) it is necessary to mix up the powders of copper and sulphur in mass relations 2:1.
If taken amounts of source substances do not correspond their correlation in the chemical formula of compound one of them stay in the excess.
For instance, if take 3 g. copper and 1 g. sulphur than after the reaction 1 g. copper, which did not enter in the chemical reaction will stay.
Substances with non-molecular structure do not possess strictly constant composition. Their composition depends on conditions of preparation.
The mass share of element w(e) shows what part forms the mass of given element from the whole mass of substance: where n – a number of atoms; Ar(e) – relative atomic mass of element; Mr – relative molecular mass of substance.



n * Ar(e)     w(e)=———Mr

Knowing the quantitative element composition of substance its simplest molecular formula is possible to determine:
1.      Mark formula of compounds Ax By Cz.
2.      Calculate an attitude X: Y: Z through the mass shares of elements:



х * Ar(А)


y * Ar(B)


z * Ar(C)
w(A)
=
—————
w(B)
=
—————
w(C)
=
—————


Mr(AxByCz)


Mr(AxByCz)


Mr(AxByCz)

X =
w(A) * Mr
—————
Ar(А)
Y =
w(B)  * Mr
—————
Ar(B)
Z =
w(C)  * Mr
—————
Ar(C)


x : y : z
=
w(A)
——
Ar(А)
:
w(B)
——
Ar(B)
:
w(C)
——
Ar(C)

3.      Obtained numerals divide on the least for getting total numbers.
4.      Write a formula of compound.

Law of multiple proportions
(D. Dalton, 1803)

If two elements form several chemical compounds with each other, then the masses of one of the elements corresponding to the same mass of the other element in these compounds are in a simple integral proportion.

N2O                 N2O3               NO2(N2O4)     

             N2O5

A number of oxygen atoms in molecules of such compounds corresponding to the two nitrogen atoms are in a proportion 1:3:4:5.