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The Transformation of Matter

"Matter is neither created nor destroyed, it is only transformed."

-Antoine Lavoisier

The Law on the Conservation of Matter

The law of conservation of matter, also known as the law of conservation of mass or simply as the Lomonosov-Lavoisier law in honor of the scientists who postulated it, is a principle of chemistry that states that matter is neither created nor destroyed during a chemical reaction, it is only transformed.

This means that the quantities of the masses involved in each reaction must be constant, i.e. the number of reactants consumed is equal to the number of products formed, even if they have been transformed into each other.

He was one of the few chemists of his time who fully appreciated the importance that the weight of the products of a chemical reaction must be equal to the weight of the reactants, which coincides with the following statements of the law, "in any change of state, the total mass is conserved" or "matter is neither created nor destroyed in any chemical reaction".

It is of fundamental importance, as it allows specific components to be extracted from some raw material without having to discard the rest; It is also important, because it allows us to obtain pure elements, things that would be impossible if matter were destroyed.

A perfect example of this law is the combustion of hydrocarbons, in which the fuel can be seen to burn and "disappear", when in fact it will have been transformed into invisible gases and water.

The most important conservation laws in classical physics are Conservation of linear momentum. Conservation of angular momentum. Conservation of Electrical Charge, Conservation of Energy. Preservation of the dough. Temperature preservation.

This law is fundamental to a proper understanding of chemistry. The principle is quite accurate for low-energy reactions. In the case of nuclear reactions or collisions between particles at high energies, in which the classical definition of mass does not apply, the equivalence between mass and energy must be considered.

Example:

Copper (II) oxide is formed through oxygen and metallic copper. If the mass of the oxide is 35 g and the mass of the metal is 5 g,

Estimate the mass of oxygen:

Applying the law, we have:

Copper Mass + Oxygen Mass = Oxide Mass

Oxygen Mass = Oxide Mass – Copper Mass

Oxygen mass = 35 g – 5 g

Masa oxygen = 30 g

For example, when burning methane (CH4) we will have the following reaction, whose products will be water and invisible gases, but with an identical number of atoms as the reactants:

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Chemical Stoichiometry

Importance

IMPORTANCE OF STOICHIOMETRY IN INDUSTRY AND ECONOMICS.

In the industry, stoichiometric calculations are part of the manufacture of chemical products since we depend on these products that we use daily, such as gasoline and lubricants that are part of the petroleum industry.

It studies the quantitative relationships between reactants and products in a chemical reaction.

It also studies the proportion of the different elements in a compound and the composition of mixtures.

The time of the moon

MA is the mass of an atom expressed in u.

The atomic mass unit (u, formerly u.m.a.) is defined as one-twelfth (1/12th) of the mass of a 12C atom (the most abundant natural isotope of carbon)

1H = 1.008 u (12 times less than 12C) One u is equivalent to 1.66 x 10-24 g

But this is an amount we can't overestimate!

mole:

It is the SIM unit with which the amount of substance is measured.

One mole corresponds to the amount of substance that contains as many elementary units as there are 12C atoms in 12 g of 12C.

Avogadro's number (NA):

This is the number of atoms in 12g of 12C.

It corresponds to 6.023x1023 atoms.

It is used in any chemical entity, whether atoms, molecules, ions.

That is, it is the amount of substance contained in the NA of atoms, molecules, ions...

1 mole of He has 6,023 x 1023 atoms of He

1 mole of H2SO4 contains 6,023 x 1023 molecules of H2SO4.

One mole of any substance is the amount in grams contained in the NA of that substance.

The atomic mass of C is 12 u which means that 1 atom of C has a mass of 12 u, and as already stated:

1 = 1.66x10-24 g, therefore:

12ux1.66x10-24 g=gesthemassofan atom of C 1u

What is the mass of 1 mole of C atoms?

12 x (1.66x10-24 g) /átomo de C x 6.023x1023 átomos/mol = 12 g/mol

1 u ~ 1 g/mol

Table Molar (MM)

The mass of molecules can be calculated with the atomic masses (in you or in g/mol) of the atoms.

It is the sum of the atomic masses of the atoms involved in the formula of the affected molecule (multiplied) by the number of times that element appears in it.

MM(H2SO4) = 2MA(H) + MA(S) + 4MA(O) = 2(1.0 in) + (32.0 in) + 4(16.0 in) = 98.0 in (g/mol)

Therefore, 1 mole of H2SO4 molecules has a mass of 98 g.

2 u of H → 2 g/mol of H → 2 g of atoms of H 32 u of S → 32 g/mol of S → 32 g of atoms of S 64 u of O → 64 g/mol of O → 64 g of atoms of O

Chemical formula

The chemical formula is the representation of the elements that make up a substance and the proportion in which they are used.

Find.

Empirical formula

It indicates the type of atoms present in a compound and the ratio between the number of atoms in each element.

From the composition of a compound, it is possible to deduce its simplest formula. It is a simple relationship of whole numbers between the atoms that compose it.

To calculate the empirical formula of a compound: 1o the number of moles of each element is calculated. 2o is divided by the smallest number of moles.

The empirical formula does not necessarily have to coincide with the molecular formula.

Example

Ascorbic acid (vitamin C) has a composition of 40.9% C, 4.6% H and 54.5% O.

To determine the empirical formula of a compound:

1) The mole number of each element is calculated assuming 100 g of sample

2) divide each number of moles by the smallest number of moles 3) find the subscripts as whole numbers

Formula molecular

It indicates the type of atoms present in a compound, and the number of atoms of each kind.

To calculate the molecular formula, it is necessary to know the empirical formula and the molar mass of the substance, since the molecular formula will be n times the empirical formula.

In general, the percentage mass composition of each element is described. From there the empirical formula is determined. After knowing the molar mass of the compound, the molecular formula is determined.

Some compounds with empirical formula CH2O (40% C, 6.71% H and 53.3% O)

Name

Formaldehyde

Acetic acid

Lactic acid

It's a debt

Ribose

Glucose

Fórmula molecular

CH2O

C2H4O2

C3H6O3

C4H8O4

C5H10O5

C6H12O6

Integer Multiple

1

2

3

4

5

6

Masa Molar (g/mol)

30.03

60.05

90.08

120.10

150.13

180.16

Structural Formula

It shows the number of atoms and the bonds between them.

They are determined by different approaches:

• Spectroscopic technique • Rx analysis • Bioinformatics analysis

(for bio-macromolecules)

Two compounds with C2H6O molecular formula

Name

Ethanol

Metoximetano

Empirical formula

C2H6O

C2H6O

Fórmula molecular

C2H6O

C2H6O

Structural Formula

Example

Molecular Formula of Dinitrogen Tetrahydride (or Hydrazine) = N2H4

We divide the number of moles of each element by the minimum number of moles of any of the elements.

2 atoms of N/2 atoms of N = 1 4 atoms of H/2 atoms of N = 2

La formula empirical = NH2

The molecular formula is n times the empirical formula according to the molar mass of the compound. MM= 32 g/mol

32/MM empirical formula = 2, so the molecular formula is 2 times the empirical

formula.

FM = n x FE