Using English To Calculate: Stoichiometry

Understanding Stoichiometry 
 

Stoichiometric Definition and Stoichiometric Type - Stoichiometry is a chemical branch associated with a quantitative relationship that exists between reactants and products in chemical reactions. The reactant is a substance that participates in a chemical reaction, and also the product is a substance obtained as a result of a chemical reaction.

The stoichiometry depends on the fact that the elements behave in a predictable or predictable way, as well as material that can neither be created nor destroyed.

Therefore, when the elements are combined then produce a chemical reaction, something known and also specific that will occur and the reaction can be predicted by elements and also the number involved. Stoichiometry is the mathematics behind chemistry.

The stoichiometric calculations can find out how the elements and also the components diluted in a solution whose concentration is known react in the experimental conditions. The word "Stoichiometry" comes from the Greek word "stoicheion", meaning "element" as well as "metron" means "size".

The Laws Governing Stoichiometry.

The stoichiometry relies on law is like fixed comparative law, double comparative law and also the law of conservation of the masses.

• The law of mass conservation

Using the laws of physics is like the law of conservation of mass, which holds that the mass of the reactants is equal to the mass of the product, Stoichiometry is used to gather information about the amount of various elements used in a chemical reaction,

• Comparative law remains

This law states that the chemical compound (a substance consisting of 2 (two) or more ages) which always contains the same proportion of an element (compound with one atom type) with mass.

• Law of multiple comparison

This law is one of the basic laws of stoichiometry, aside from the law of fixed comparison. Sometimes it is also called Dalton's law. It is said that, if 2 (two) elements form more than one compound between them, then the mass ratio of a second element which joins the fixed mass over the first element of both will have a ratio of a small sum of the whole.

Stoichiometric Explanation

Subject to the above laws, such chemical reactions may combine in a defined chemical ratio. The sum of each element must be the same throughout the reaction. In a balanced chemical reaction, the relationship between the amount of reactants and the product will usually form the ratio of integers. For example, in a reaction that forms ammonia (NH3), exactly 1 (one) nitrogen molecule (N2) reacts with 3 (three) hydrogen molecules (H2) to produce 2 molecules of NH3. It can be described as follows

N2 + 3H2 ---> 2NH3

Thus, the stoichiometry can be used to calculate the quantity as the amount of product that can be produced when the reactants are given and also the percentage of reactants made into a known product.

Stoichiometric Type

• Stoichiometry Reaction

The stoichiometry is often used to balance the chemical equations found in reaction stoichiometry. It depicts that the quantitative relationship between substances is caused because they participate in chemical reactions.

• Composition Stoichiometry

The stoichiometry of this composition explains the quantitative (mass) relationship between an element in a compound. For example, the stoichiometry of the composition represents the (mass) nitrogen with the joined hydrogen and becomes a complex ammonia. Ie 1 mole of nitrogen and also 3 moles of hydrogen in each of 2 moles of ammonia. Mole is the unit used in chemistry for the amount of substance.

• Gas Stoichiometry

The type of stoichiometry is related to a reaction involving a gas, in which the gas is at a known temperature, pressure and volume and can also be considered an ideal gas. For gas, the ideal volume ratio is equal to the ideal gas law, but the ratio of the single reaction mass must be calculated from the molecular mass of the reactant as well as the product, in which the molecular mass is the mass of one (1) molecule of the substance.

The ideal gas is a theoretical gas consisting of 1 (one) set of randomly moving, non-interacting particles that adhere to an ideal gas law. The ideal gas law is an ideal gas state equation. The ideal gas law equation is as follows:

"PV = nRT, where P is pressure, V is volume and also T is the absolute temperature, n is mol of gas and also R is the universal gas constant".

Understanding stoichiometric ratio

A number of stoichiometric or reagent ratios (substances added to a system in order to create a chemical reaction) are either the amount or the ratio of which, assuming that the result of a reaction is completed on a basis, among others, is as follows:

1. All reagents are consumed

2. There is no deficit reagent

3. There is no residual residue

4. Reactions will only occur or are created on stoichiometric ratios

CHEMICAL CALCULATION (STOIKIOMETRY)

A.    Determination of Reaction and Reaction Gas Volume

The question that arose after Gay Lussac proposed the comparative law of volume could be solved by an Italian physicist named Amadeo Avogadro in 1811.

According to Avogadro:

"Gases of the same volume, if measured at the same temperature and pressure, will have the same number of molecules".

Since the ratio of the volume of hydrogen gas, oxygen gas, and water vapor to the vapor-forming reaction is 2: 1: 2 the ratio of the number of hydrogen, oxygen, and water vapor molecules is also 2: 1: 2. The number of atoms of each element is not reduced or increased In chemical reactions. Therefore, the hydrogen gas molecule and the oxygen gas molecule must be dwiatom molecules, whereas the water vapor molecule must be a triathome molecule. The gas volume ratio in a reaction corresponds to the reaction coefficient of the gases. This means that, if one gas volume is known, another gas volume can be determined by comparing the reaction coefficient.

Example:

In the reaction of moisture formation.

2H2 (g) + O2 (g) -> 2H2O (g)

If the volume of H2 gas measured at 25 ° C and 1 atm pressure of 10 L of O2 and H2O gas at the same pressure and temperature can be determined in the following manner:

Volume H2: Volume O2 = Coefficient H2: Coefficient O2

B.     Relative Atomic Mass and Relative Molecular Mass

Having discovered highly sensitive equipment in the early twentieth century, chemists experimented on the mass of one atom. For example, experiments were performed to measure.

1. mass of one atom H = 1,66 -> 10-24 g

2. mass of one atom O = 2.70 -> 10-23 g

3. mass of one atom C = 1.99 -> 10-23 g

From the data above can be seen that the mass of one atom is very small. The experts agreed to use the amount of Atomic Mass Unit (sma) or Atomic Mass Unit (amu) or commonly called also units of Dalton. In the matter of atomic structure, you have also learned that atoms are very small, therefore it is impossible to weigh the atoms using a balance sheet.

1)      Relative Atomic Mass (Ar)

Experts use the C-12 carbon isotope as standard with a relative atomic mass of 12. The relative atomic mass represents the average mass ratio of one atom of an element to 1/12 of the C-12 atomic mass. Or it can be written:

1 unit of atomic mass (amu) = 1/12 mass 1 atom C-12

Example:

The average atomic mass of oxygen is 1.33 times greater than that of the carbon-12 atoms.

Then: Ar O = 1.33 -> Ar C-12 = 1.33 -> 12 = 15.96

Experts compare different atomic masses, using the relative atomic mass scale with the symbol "Ar".

The experts decided to use C-12 or 12C isotope because it has an inert core stability compared to other atoms. The isotope of C-12 atom has an atomic mass of 12 sma. One sma equals 1.6605655 x 10-24 g. With the use of 12C isotope as standard then can be determined the mass of atom of other element. The relative atomic mass of an element (Ar) is a number expressing the mass ratio of one atom of that element to 1/12 mass of one C-12 atom.

ArX = (average atomic mass X) / (1/2 mass of carbon atom - 12)

2)      Relative Molecular Mass (Mr)

Molecule is a combination of several elements with a certain ratio. The same elements combine to form elemental molecules, while different elements form molecules of compounds. The molecular mass of an element or compound is expressed by a molecular mass (Mr). The relative molecular mass is the ratio of the molecular mass of an element or
Compound against 1/12 x the mass of C-12 atoms.

C.     The Concept of Mol and Fixed Avogadro

When you react one carbon atom (C) with one molecule of oxygen (O2) it will form one molecule of CO2. But actually what you react is not a single carbon atom with one molecule of oxygen, but a large number of carbon atoms and a large number of oxygen molecules. Since the number of atoms or the number of molecules that react is so great then to say it, the chemists use "mol" as the unit of the number of particles (molecules, atoms, or ions). One mole is defined as the number of substances containing the particles of the substance as much as the atoms present in 12,000 g of carbon atoms -12.

Thus, in one mole of a substance there are 6.022 x 1023 particles. The value of 6.022 x 1023 particles per mole is called the Avogadro constant, with the symbol L or N. In everyday life, the mole can be analogous to "dozen". If a dozen states the number of 12 pieces, the mole represents the amount of 6.022 x 10 23 particles of the substance. The word particles in NaCl, H2O, and N2 can be expressed with ions and molecules, whereas in elements like Zn, C, and Al can be expressed with atoms.

1)      Mass Molar (Mr)

The mass of one mole of substance is called the molar mass (the symbol of Mr). The magnitude of the material molar mass is the relative atomic mass or the relative molecular mass of a substance expressed in units of grams per mole.

Molar mass = Mr or Ar substance (g / mol)

The mass of a substance is the multiplication of its molasses (g / mol) with the mol of the substance (n). Mathematically, it can be stated as follows.

Molar mass = mass: mol

Mass = mol x Mr / Ar (molar mass)

2)      Molar Volume (Vm)

The volume of one mole of a substance in a gas form is called the molar volume, denoted by Vm. What is the volume of gas molar? How to calculate the volume of a certain amount of gas at a certain temperature and pressure? Avogadro in his experiments concluded that 1 L of oxygen gas at 0 ° C and 1 atm pressure has a mass of 1.4286 g, or it can be stated that at 1 atm pressure.

Thus, under Avogadro's law it can be concluded:

1 mol of gas O2 = 22.4 L

In accordance with Avogadro's law stating that at the same temperature and pressure, the same volume of gas contains the same number of molecules or the number of moles of each gas volume the same. Under the law, a volume of 1 mole of each gas in standard conditions (0 ° C and 1 atm pressure) is applied.

Volume gas in standard state = 22.4 L

3)      Gas Volume in Non-Standard State

Calculation of gas volume is not in the standard state (non-STP) used the following two approaches.

a.       The ideal gas equation

Assuming that the gas to be measured is ideal, the equations that connect the number of moles (n) of gas, pressure, temperature, and volume are:

The ideal gas law: P. V = n. R. T

Where:

P = pressure (atmospheric unit, atm)

V = volume (liters, L)

n  = number of moles of gas (mol unit)

R = gas constant (0.08205 L atm / mol K)

T = absolute temperature (° C + 273.15 K)

b.      With gas conversion at the same temperature and pressure

According to Avogadro's law, the ratio of gases having the same number of moles has the same volume. Mathematically can be expressed as follows.

V1 / V2 = n1 / n2

Where:

n1 = mol gas 1              
n2 = mol gas 2  
V1 = gas volume 1           

V2 = gas volume 2

4)      Molarity (M)

The amount of substances present in a solution can be determined by using the concentration of the solution expressed in molarity (M). Molarity states the number of moles of substances in 1 L of solution. Mathematically stated as follows.

Where:

M = molarity (unit M)

Mass = in units g

Mr. = molar mass (unit g / mol)

V = volume (mL unit)

D.    Molecular Formulas and Elemental Content in Compounds

The ratio of mass and elemental content in a compound can be determined from its molecular formula.

1)      Determination of Empirical Formulas and Molecular Formulas

The chemical formula shows the atomic type of element and the relative amount of each element contained in the substance. The number of substances contained in the substance is indicated by the index number. The chemical formula can be an empirical formula and a molecular formula.

"Empirical formula, the formula that states the smallest comparison of atomatoms
Of the elements composing the compound ".

"The molecular formula, the formula yamg states the number of atoms of
The elements that make up one molecule of the compound ".

Molecular formula = (Empirical formula) n

Mr. Molecular formula = n x (Mr. Empirical Formula

n = integers

The determination of the empirical formula and the molecular formula of a compound can be achieved by the following steps.
1. Find the mass (percentage) of each constituent compound,

2. Change to mole unit,

3. The mole ratio of each element is an empirical formula,

4. Find the molecular formula by: (Mr. empirical formula) n = Mr. molecular formula, n can be calculated,

5. Multiply n obtained from the count by the empirical formula.

1.      A compound comprises 60% carbon, 5% hydrogen, and the remainder is nitrogen. Mr. compound it = 80 (Ar: C = 12; H = 1; N = 14). Determine the empirical formula and the molecular formula of the compound!

Answer:

Percentage of nitrogen = 100% - (60% + 5%) = 35%.

Suppose the mass of the compound = 100 g

Then mass C: H: N = 60: 5: 35

Comparison of mol C: mol H: mol N = 5: 5: 2,5 = 2: 2: 1

Then the empirical formula = (C2H2N) n. (Mr. empirical formula) n = Mr. molecular formula

(C2H2N) n = 80

(24 + 2 + 14) n = 80

40n = 80

n = 2

Thus, the molecular formula of the compound = (C2H2N) 2 = C4H4N2.

2)      Determining the formula of Chemical Hydrate (Water Crystal)

Hydrates are solid crystalline compounds containing crystalline water (H2O). The chemical formula of solid crystalline compounds is known. So basically the determination of the hydrate formula is the determination of the number of crystalline water molecules (H2O) or the value of x.

The chemical formula of solid crystal compounds: x. H2O

For example calcium sulfate salt, has a chemical formula CaSO4. 2H2O, meaning that in every one mole of CaSO4 there are 2 moles of H2O.

Problems example 
1.      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) + xH2O5 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.

3)      Chemical Counts

The determination of the amount of reactants and the reaction products involved in the reaction must be calculated in units of moles. That is, the units that are known must be converted into mol form. This method is called the mole approximation method.

4)      Pereaksi Pembatas

In a chemical reaction, the mole ratio of the mixed reagents is not always the same as the ratio of the reaction coefficient. This means that there are reagents that will run out first. Such reagents are called limiting reagents. How can this happen?

X + 2Y -> XY2

The above reaction shows that according to the reaction coefficient, one mole of X takes two moles of Y. The above picture shows that three molecules of X are reacted with four molecules of substance Y. After the reaction takes place, the number of X-substance molecules reacting only two molecules and one Molecule left. Meanwhile, four Y-substance molecules are reacting. Then this Y substance is called a limiting reagent. The limiting reagent is an exhausted reactant and does not remain at the end of the reaction.

In chemical counts, the limiting reagents can be determined by dividing all the moles of the reactant by their coefficients, then the reactants having the smallest yield value being the limiting reagents.

Problems Example
Known reaction as follows S (s) + 3F2 (g) -> SF6 (g).

If reacted with 2 mol of S with 10 moles of F2, determine:

a. How many SF6 mols are formed?

b. Which substance and how many moles of substance is left?

Answer:

S + 3F2 -> SF6

From the reaction coefficient shows that 1 mol of S requires 3 moles of F2.

The possibilities are as follows.

a. If all S reacts then F2 is required:

This is possible because F2 is available 10 mol.

b. If all F2 is reacted then S is required:

This is unlikely, because the available S is only 2 mol.

So, acting as a limiting reagent is S!

The number of SF6 mols formed = x mol S.

a. Mol SF6 = 1 x 2 mol = 2 mol

b. The remaining substance is F2, as much as = 10 mol - 6 mol = 4 mol F2