Stoichiometry comes from two Greek syllables Stoicheion meaning
"element" and Metron which means "measurement".
Stoichiometry is a subject in chemistry involving the linkage of reactants and
products in a chemical reaction to determine the quantity of each reacting
agent.
B. DEFINITION
CONCEPT MOL

One mole of each element: carbon (black powder of charcoal), Sulfur (yellow powder), iron (nails), copper wire, and mercury (metal in liquid form)
To simplify the
number of these extraordinarily small particles, the concept of moles is used.
Mol represents the unit of quantity of matter. Unit amount of this substance as
well as the simplification of the amount of a good. For example, 1 dozen is
used to simplify 12 pieces of glassware and 1 ream to declare 500 sheets of
paper. This simplification needs to be done because the chemical processes that
take place in everyday life involve a very small collection of very small
particles. The units of the particles are too difficult to observe.
1
mol = L particles
Thus, the concept of moles is the unit of
quantity of matter that expresses the number of particles of matter.
FORM CONCEPT MOL
Thus, 1 mol of the
substance contains 6.02 x 1023 particles. The relationship between the number
of moles and the number of particles can be formulated as follows:
Number
of moles of X (n) = number of particles X / L
Or
Number
of particles X = n x L
The Avogadro (L) number was discovered by
Johann Loschmidt in 1865. The name Avogadro was chosen as a tribute to Avogadro
because he was the first to propose the need for a unit of particle count. The
name Loschmidt is immortalized as a symbol of that number, L. The value of the
Avogadro number is L = 6.02 x 1023. Can you imagine the magnitude of that
number? Had it collected 6.02 x 1023 rice grains, the rice could be dumped on
the surface of the Australian continent with a height of one kilometer! Or, if
an atom is the size of a marble and spread on the surface of the earth, the
entire surface of the earth will be covered with marbles with a layer thickness
of 80 km! This enormous number is chosen to express the number of atoms because
of their practicality and precision. Imagine, it would be complicated if the
number of atoms expressed in units of grains, dozen, gross, kodi, or other
units.
1. MASS MOLAR
The molar mass is the mass of a substance equal to the atomic mass or the mass of the formula of the substance expressed in grams. This molar mass is equal to the mass possessed by a mole of substance equal to the relative atomic mass (Ar) or relative molecular mass (M).
2. Volume of gas
molar
The molar volume of the gas (Vm) is the volume occupied by 1 mole of gas at a temperature (T) and a certain pressure (P). Because the laws of gas do not depend on the identity of the gas. In other words, any gas, at a certain temperature and pressure will be the same volume. This implies that, one mole per gas contains the same number of molecules as the avogadro constant (L = 6.02 x 1023 molecules). According to Avogadro's law, the gases must fill the same volume at a certain temperature and pressure. The volume of one mole of this gas is known as the molar volume of the gas. At standard pressure (0 oC and 1 atm), the volume of one mole of gas was found to be 22.4 Liters. (Remember yes only at 0 oC and 1 atm, if not in these conditions then the volume of the molars will not be 22.4 L).
If there is one mole of ideal gas at 0 ° C and 1 atm pressure, then from the ideal gas law, the gas volume in that state is

3. Molarity
In chemistry, molarity (M) is a measure of the
concentration of the solution. The molarity of a solution expresses the number
of moles of a substance per liter of solution. For example 1.0 liter of
solution contains 0.5 mol of X compound, then this solution is called 0.5 molar
solution (0.5 M). Generally the concentration of aqueous aqueous solution is
expressed in molar units. The advantage of using molar units is the ease of
calculation in stoichiometry, since concentrations are expressed in number of
moles (proportional to the number of actual particles).
Substance levels are generally
expressed in percent mass (% mass). To get a percent mass can use the formula:


5. Determining the
Empirical Formula and the Compound Molecular Formula
In chemistry lessons are known 2
types of chemical formulas, namely Empirical Formulas and Molecular Formulas.
The empirical formula is the formula which expresses the smallest comparison of
atoms from the constituent elements of the compound. While the molecular
formula is a formula that shows the number of atomic elements that make up a
compound. This molecular formula is a multiple of the empirical formula.
(Empirical
Formula) _ {n} = Molecular Formulas
The value of n
above corresponds to the relative molecular mass (Mr) of the two chemical formulas.
If Mr. of the empirical formula is multiplied by n, then yields Mr. molecular
formula. This principle is used as the basis for finding the molecular formula
of the compound.
How to Determine Empirical Formulas and
Molecular Formulas :
1. Find the mass comparisons
of the constituent elements.
2. Find the mole
ratio by dividing the elemental mass by its relative atomic mass (Ar).
3. From the
comparison of the simplest mole obtained Empirical Formula.
4. Determine the
Molecular Formula by looking for the multiplier n factor. The formula used is (Mr. RE) n = Mr.
C. CHEMICAL CALCULATIONS
Stoichiometry is closely related to chemical calculations. To solve the problem of chemical calculation, the stoichiometric principles are used, they are chemical equation and mole concept. In this lesson we will learn first about the principles of stoichiometry, then after that we will study stoichiometry application on chemical calculation and sample problem and how to solve it.
1. Example of the
Law of Conservation of Mass in Chemical Reactions: Copper wire is burned in a bunsen burner to form copper oxide
(CuO).
2Cu
(s) + O2 (g) → 2CuO (s)
If the weight of Cu is initially 32 g and CuO
is formed 40 g, how much weight of O2 reacts? Resolution:
According to the Law of Conservation of Mass,
in chemical reactions there is no mass change. Therefore, the weight of O2
reacting is 40 g - 32 g = 8 g 32 g
Cu (s) + 8 g O2 (g)
→ 40 g CuO (s)








