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).
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)
Explain what is the relationship between molarity and molality?
BalasHapusSometimes a chemical problem is just for fun to test our understanding of the unit of concentration of the solution. One is to convert (convert) molar units to molals for a given solution.
HapusFor example there is a problem like the following:
Determine the molality of 18 M H2SO4 solution with a density of 1.84 g / mL.
Yes, to be able to convert molarity into molality required density data (density) of the substance. Let's see how important this density data is used in this issue.
Suppose we also have a solution of 18M H2SO4 is as much as 1 Liter. With known density we can change the unit volume (1 L) into units of mass (Kg). This is important because in the molal unit we will use the mass unit of the solvent.
Keep in mind that:
Molarity = mol solute / liter of solution (M = n / V)
Molality = mol solute / kg of solvent (m = n / kg)
Density = solution mass / solution volume (d = mass / V)
Please give me an example in determining molality and molarity.
BalasHapusExample of molarity calculation of solution
HapusSuppose a 0.25 liter urea solution (CO (NH 2) 2) is prepared by dissolving 3 grams of urea in water. The relative urea relative molecular mass is 60. The molarity of the urea solution can be calculated as follows.
The number of moles of urea, n = gram / Mr = 3/60 = 0.05 mol
Molarity of the solution, M = n / V = 0.05 / 0.25 = 0.2 molar
Thus the molarity of the urea solution is 0.2 molar or 0.2 mol / liter.
Example calculation of solution molality
The 10 g of sodium hydroxide (NaOH) was dissolved in 2 kg of water. The relative molecular mass of NaOH is 40. The molality of the solution can be calculated in the following way.
The number of moles of NaOH, n = gram / Mr = 10/40 = 0.25 mol
Solution molality, m = n / P = 0.25 / 2 = 0.125 molals
So the molality of the NaOH solution is 0.125 mol
When is the limiting reaction used in stoichiometric calculations?
BalasHapusIn a chemical reaction, the mole ratio of the added reagents is not always the same as the ratio of the reaction coefficient. This causes a reagent to be reacted first. This is called pereaksi pembatas. Limiting reagent is a reactant contained in the relatively smallest amount (in the stoichiometric relationship). The limiting reagents will run out, while the other reactions will leave the rest.
HapusPlease give another example about Chemical calculations.
BalasHapusProblems example
Hapus1. A total of 5 g of copper (II) sulfate hydrate is heated until all the crystalline water evaporates. The mass of solid copper (II) sulfate formed by 3.20 g. Determine the hydrate formula! (Ar: Cu = 63,5; S = 32; O = 16; H = 1)
Answer:
Steps to determine the hydrate formula:
A. Suppose the CuSO4 hydrate formula. X H2O.
B. Write the equation of the reaction.
C. Determine the mol substances before and after the reaction.
D. Calculate the value of x, using the mole ratio of CuSO4: mol
H2O.
CuSO4. XH2O (s) -> CuSO4 (s) + xH2O
5 g 3.2 g 1.8 g
Comparison, moles CuSO4: mol H2O = 0.02: 0.10.
Comparison, moles CuSO4: mol H2O = 1: 5.
Thus, the hydrate formula of copper (II) sulfate is CuSO4. 5H2O.
What different betwen number of atom, number of moleculer, and number of perticel??
BalasHapus1. The number of particles (atom / molecule / electron) of a substance is determined from the moles by the Avogadro number (6.02 x 1023)
Hapus2. The number of moles of a substance can be determined by the following equation:
A. With n = number of moles of atoms (mol)
G = atomic mass (gram)
Ar = atomic mass relative element (gram / mol)
B. With n = number of mol of molecule (mole)
G = molecular mass (gram)
Mr = relative molecular mass (gram / zero)
3. The number of atoms is the presence of numbers in the chemical formula.
One mole per gas contains the same number of molecules as avogadro constant. Why is that?
BalasHapusAvogadro Number is 6.02 x 1023, from where this provision is obtained?
BalasHapusWhat is the usefulness of knowing the molarity of a compound?
BalasHapuswhat is the first step that we should do to determine the particles of the compound?
BalasHapusCan you give an example of From the simplest mole ratio so obtained Empirical Formula?
BalasHapus