Using english to calculate

           

            Stoichiometry is the basic chemical calculation that states quantitative relation of chemical formulas and chemical equations. Here are the materials you need to know to understand, from the concept of moles and molar masses, empirical formulas and molecular formulas, basic stoichiometry of solutions and ideal gases, and the writing and equalization of reactions, with examples of problems and discussions.

A. CHEMICAL BASIC LAW

            Chemistry is a part of natural science that studies matter that includes the composition, nature, and change of matter and energy that accompanies material changes. Careful research on reagents and reaction products has spawned basic chemical laws that show a quantitative or so-called stoichiometric relationship. Stoichiometry comes from the Greek word, stoicheon which means element and metrain which means measure. In other words, stoichiometry is a chemical calculation involving the quantitative relation of substances involved in the reaction. These basic chemical laws are the law of mass conservation, the law of fixed comparison, the law of volume comparison, and the law of multiple comparisons. The basic laws of chemistry are our foundation in studying and developing the next chemistry.

 1. The law of conservation of the masses  (LAVOISIER LAW)

            At the beginning of the eighteenth century, chemists in their attempts to learn the heat and burning found a very strange thing. For example, if wood is burned, it will produce ash residue (solid) which is much lighter than the original wood. However, if the metal is burned in the free air, it will produce heavier oxide than the original metal. To answer the strangeness, chemists developed a method of experimentation carefully by using chemical balances in measuring the volume or mass of gases, liquids and solids that occur in chemical reactions. Therefore, the mass of the reactants and the reaction product can be measured carefully. The results of these experiments present the facts to the observer and demand them into the formulation of the fundamental law that describes the nature of chemistry. The basic law obtained is known as the law of conservation of mass, which is as follows.

'' Mass can not be created nor destroyed in any material change. ''

The basic legal facts of mass conservation have been proven in 1756 by the Russian scientist M.V. Lomonosov. Perhaps because of language problems, his work is not well known in Western Europe. Separately in 1783, a great French chemist Antoine Lavoisier did the same thing using a chemical balance to show that the sum of the mass of chemical reactions was equal to the amount of reactant mass.

Lavoisier conducts experiments by heating the mrerkuri in a sealed flask filled with air. After a few days, a red substance is formed that is mercury (II) oxide. The gas in the mass tube is reduced and can no longer support the burning (the candle in the tube does not fire anymore) and the animal will die if it is inserted into it. It shows that the oxygen gas in the tube is gone. It is now known that the remaining gas is nitrogen, while the oxygen from the air in the tube has run out with mercury. Furthermore, Lavoisier takes the mercury oxide and heats it up so it breaks down again. Then he weighed the mercury and the resulting gas. It turns out that the combined mass is equal to the mass of mercury (II) oxide used initially. Finally after several experiments and the results are the same, Lavoisier states the law of conservation of mass that is as follows.

 '' In every chemical reaction, the mass of substances before and after the reaction is always the same. ''

           Lavoisier was the first to observe that chemical reactions are analogous to algebraic equations.

Example:

       S (s) + O2 (g) → SO2 (g)

1 mol of S reacts with 1 mole O2 to form 1 mole of SO2. 32 grams of S reacts with 32 grams of O2 forming 64 grams of SO2. The total mass of the reactants is equal to the mass of the resulting product.

      H2 (g) + ½ O2 (g) → H2O (l)

1 mole of H2 reacts with ½ moles of O2 forming 1 mole of H2O. 2 grams of H2 reacts with 16 grams of O2 forming 18 grams of H2O. The total mass of the reactants is equal to the mass of the product formed.

2. PROUST LAW OR COMPARATIVE LAW

            In 1799 the French chemist Joseph Proust, through various experiments found a provision that was validated by Proust's law, as follows.

"The mass ratio of the constituent elements is always fixed, even if it is made in a different way"

            At that time Proust discovered that copper carbonate, both from natural and synthetic sources in the laboratory, had a fixed arrangement.

            To determine the composition of a compound, we can describe an example of the compound we have weighed, then the compounds are described into their elements. Each compound-forming element we weigh, it was obtained a certain comparison. If it is repeated, it will get the same comparison. Another method can also be done, namely by weighing the mass of compounds formed from the compounds of elements that each element is known mass. Of the many experiments on the composition of elements in the compound, always produce the following statement.

"A pure compound is always composed of fixed elements with a fixed mass ratio."

Example:

S (s) + O2 (g) → SO2 (g)

            The ratio of mass S to mass of O2 to form SO2 is 32 grams S to 32 grams of O2 or 1: 1. This means that every gram of S just reacts with one gram of O2 forming 2 grams of SO2. If 50 grams of S is required, 50 graM O2 is required to form 100 gra-SO2.

H2 (g) + ½ O2 (g) → H2O (l)

            The ratio of mass of H2 to mass of O2 to form H2O is 2 gram H2 to 16 gram O2 or 1 2 8. This means, Every one gram of H2 is precisely bersiKsi with 8 gram O2 form 9 gram H2O. If provided 24 grams of O2, it takes 1 gram of H2 to form 27 grams of H2O.

3. COMPARATIVE LAW

            John Dalton's interest in studying two elements that can form more than one compound turns out to produce a conclusion called the law of multiple comparison:

'' If two elements can form more than one compound, then the ratio of the mass of the one element, which is intertwined with another particular element of its mass is a simple integer ''.

For example, copper with oxygen, carbon with oxygen, sulfur with oxyeen, and phosphorus with chlorine. The mass ratio of the two elements is as follows.

1. Copper and oxygen form two copper oxide compounds.

Copper oxide copper oxygen copper: oxygen

           I 88.8% 11.2% 1: 0.126

           I 79.9% 20.1% 1: 0.252

2. Carbon and oxygen can form two compounds

Carbon + oxygen → Carbon monoxide (I)

Carbon + oxygen → Carbon diocide (II)

Carbon oxygen carbon compound: oxygen

I 42.8% 57.2% 1: 1.33

II 27.3% 72.7% 1: 2,67

3. Sulfur with oxigan can form two oxygen compounds, namely sulfur oxide (I) and sulfur trioxide (II)

Sulfur oxygen sulfur compounds: oxygen

I 50% 50% 1: 1

II 40% 60% 1: 1,5

Up to now this law is still acceptable, but needs to be corrected on simple numbers. If the comparison is a simple number (1, 2, 3, 4, 5) means the compound formula is also simple, such as H2O, CO2, and H2SO4. However, now found compounds with large numbers, such as sucrose and arachidonic acid.

4. COMPARATIVE LAW VOLUME

            The relationship between the volumes of the gases in a chemical reaction was investigated by Joseph Louis Gay-Lussac in 1905. In the study it was found that at a constant temperature and pressure, every single volume of oxygen gas would react with two volumes of hydrogen gas yielding two volumes Water vapor, thus the ratio between the volume of hydrogen, the volume of oxygen and the volume of water vapor in sequence is 2: 1: 2. Another example: one volume of hydrogen gas will react with one volume of chlorine gas yielding two volumes of hydrogen chloride gas; The ratio of hydrogen volume, chlorine volume and volume of hydrogen chloride sequence is 1: 1: 2. In the reaction between the nitrogen gas and the hydrogen gas forming the ammonium gas, the volume ratio of the three gases is 1: 3: 2 (N2: H2: NH3).

Based on the above description, it can be concluded that:

"At the same temperature and pressure, the ratio of reactant gas volume to the gas volume of the reaction product is a simple integer (equal to the ratio of the reaction coefficient)"

Example:

N2 (g) + 3 H2 (g) → 2 NH3 (g)

The gas volume ratio is equal to the ratio of the reaction coefficient. This means that every 1 mL of N2 gas exactly reacts with 3 mL of H2 gas to form 2 mL of NH3 gas. Thus, to obtain 50 L of NH 3 gas, it takes 25 L of N2 gas and 75 L of H2 gas.

CO (g) + H2O (g) → CO2 (g) + H2 (g)

The gas volume ratio is equal to the ratio of the reaction coefficient. This means that every 1 mL of CO gas reacts exactly with 1 mL of H2O gas to form 1 mL of CO2 gas and 1 mL of H2 gas. Thus, as much as 4 L of CO gas requires 4 L of H2O gas to form 4 L of CO2 gas and 4 L of H2 gas.

B. AVOGADRO LAW

            Avogadro's Law (Hypotes Avogadro, or Avogadro's Principle) is a gas law named after the Italian scientist Amedeo Avogadro, who in 1811 proposed the hypothesis that:

Gases of the same volume, at the same temperature and pressure, have the same number of particles.

            That is, the number of molecules or atoms in a gas volume does not depend on the size or mass of the gas molecule. For example, 1 liter of hydrogen and nitrogen gas will contain the same number of molecules, as long as the temperature and pressure are the same. This aspect can be expressed mathematically,

Where:

V = is the volume of gas.

n = is the number of moles in the gas.

K = is a matching constant.

            The most important consequence of Avogadro's law is that the ideal gas constant has the same value for all gases. That is, constants

Where:

P = is the gas pressure

T = is temperature

            Has the same value for all gases, independent of the size or mass of the gas molecule. The Avogadro hypothesis is evidenced by the theory of gas kinetics.

            One mole of ideal gas has a 22.4 liters volume under standard conditions (STP), and this number is often called the ideal gas molar volume. Real gases (non-ideal) have different values.

Example: On the formation of H2O molecules

2L H2 (g) + 1L O2 (g) ® 2L H2O (g)

2 molecule H2 1 molecule O2 2 molecule H2O

Note:

If the volume and number of molecules of one substance is known, then the volume and number of molecules of other substances can be determined by using the equation:

and

Information :

V = volume of molecule (L)

X = number of particles (molecules)

C. MASS ATOM AND MASS MOLECULES RELATIVES

            The atom is a very small particle so that the atomic mass is also too small when expressed in grams. Therefore, chemists create a way to measure the mass of an atom, that is, by the relative atomic mass. The relative atomic mass (Ar) is the ratio of the average mass of an atom by one-twelve times the mass of one carbon-12 atom.

            The smallest unit of a substance can also be a molecule. The molecule is composed of two or more atoms held together by chemical bonds. The relative molecular mass (Mr) is the average mass ratio of a molecule with one-twelve times the mass of one carbon-12 atom.

Ar Y = the average mass of 1 molecule Y / (1/12 x mass 1 atom C-12)

            In the above formula used the atomic mass and average molecular mass. Why use average atomic mass? Because elements in nature have several isotopes. For example, carbon in nature has 2 stable isotopes of C-12 (98.93%) and C-13 (1.07%). If the abundance and mass of each isotope are known, the relative atomic mass of an element can be calculated by the formula:

            Ar X = {(% isotope 1 x isotope mass 1) + (% isotope 2 x isotope mass 2) + ...} / 100

            If it is known that the relative atomic mass of each constituent element of a molecule, its relative molecular mass is equal to the sum of the relative atomic masses of all the molecular compounds of the atom. A molecule having the formula AmBn means that in 1 molecule there are m atoms A and n atoms B. Thus the relative molecular mass of AmBn can be calculated as follows.

Mr AmBn = m x Ar A + n x Ar B

D. CONCEPT MOL

            In reacting substances, many things we need to consider eg the form of substances in the form of gas, liquid and solid. It is quite difficult for us to react the substances in the three states of matter, in solid form by size in mass (grams), in liquid form by volume of liquids in which there is a solvent and there is a dissolved substance. Similarly, the gaseous material has a gas volume size.

            This condition requires chemists to provide a new unit that can reflect the amount of substances in various substances. Avogadro tries to introduce a new unit called mole. The definition for 1 (one) mole is the number of substances containing particles of 6.023 x 1023. This number is known as Avogadro Numbers denoted by the letter N.

            The above diagram shows the equation which states the relationship of the number of moles to the number of particles for atoms and molecules

            Considering the mass aspect of the substance, 1 mol of the substance is defined as the mass of the substance corresponding to its relative molecular mass (Mr) or its atomic mass (Ar).

            For 1 mole of Carbon substance it has mass corresponding to Carbon atomic mass, it is known from the periodic table that the mass of carbon atoms is 12 sma, so that the mass of the substance is also 12 grams. For that 1 mole of matter we can change into equation form:

Number of Moles (n)

Mass (m)

Volume Gas (V)

Number of Particles (X)

Mollification (M)

E. EMPIRICAL AND MOLECULUM FORMULAS

            The chemical formula of a substance can explain or state the relative amount of atoms present in the substance. The chemical formula is divided into molecular formulas and empirical formulas. The empirical formula is the simplest formula of a compound. This formula only expresses the ratio of the number of atoms present in the molecule.

            The empirical formula of a compound can be determined if one knows:

- mass and Ar each element

-% mass and Ar each element

- comparison of mass and Ar of each element

            The molecular formula of a substance describes the number of atoms of each element in a single molecule of that substance. When the empirical formula is known and Mr. is also known then the molecular formula can be determined.

Solubility (M)

            Moarning is a way of expressing the concentration (concentration) of the solution.  Declares the number of moles of solute in each liter of solution, or the number of mmol of solute in each mL of solution. Formulated:

For example: a 0.2 M NaCl solution means that in each liter of solution there is 0.2 mol (= 11.7 g) NaCl or in each mL of solution there is 0.2 mmol (= 11.7 mg) NaCl.

Dilution formula

V1.M1 = V2.M2

V1 = Volume before dilution (liter)

M1 = Molarity before dilution (M)

V2 = Volume after dilution (liter)

M2 = Molarity after dilution (M)

F. MOLALITY

            The molality mentions the mole ratio of the solute in kilograms of solvent. Molality is expressed between the number of moles of solute and the mass in kg of solvent. What is the symbol of the molality of matter? Molality is symbolized by m

with

n = number of moles of solute ........................ (mole)

p = solvent mass ..................................... (kg)

m = molality ......................................... (mol Kg-1)

G. FRACTION MOL

            The mole fraction is a unit of concentration that expresses the ratio of the number of moles of one of the solution components (the number of moles of solvent or the number of moles of solute) to the total number of moles of the solution. The mole fraction is symbolized by X. For example in a solution containing only two components, namely substance B as a solute and A as a solvent, then the fraction of mole A is symbolized XA and XB for the mole fraction of the solute.

or

With XA = solvent mole fraction

XB = mole fraction of solute

nA = number of moles of solvent

nB = the number of moles of solute

The amount of solvent mole fraction with solute is equal to 1.

XA + XB = 1