Meeting to 13 :Using english to calculate

STOICHIOMETRY

       Stoichiometry comes from the Greek word stoicheion which means element and metron which means measure. Stoichiometry discusses the relation of mass between elements in a compound (stoichiometric compound) and interactivity in a reaction (reaction stoichiometry). Mass measurements in chemical reactions were initiated by Antoine Laurent Lavoisier (1743 - 1794) who found that in chemical reactions there was no mass change (mass conservation law). Furthermore Joseph Louis Proust (1754 - 1826) found that the elements form compounds in certain comparisons (fixed comparison law). Furthermore, in order to construct his atomic theory, John Dalton discovered the third basic chemical law, called the law of multiples of comparison. These three laws are the basis of the first chemical theory, the atomic theory proposed by John Dalton around 1803. According to Dalton, every material consists of atoms, elements composed of similar atoms, whereas compounds composed of different atoms in certain comparisons . However, Dalton has not been able to determine the ratio of the atoms in the compound (the chemical formula of the substance). Determination of chemical formula of substances can be done thanks to the discovery of Gay Lussac and Avogadro. Once the chemical formula of the compound can be determined, then the ratio of antaratome (Ar) and intermolecular (Mr) masses can be determined. Knowledge of relative atomic mass and chemical formula of compounds is the basis of chemical calculations.

SIMPLE COMPOUND NOMENCLATURE

1). Binary Compound Nomenclature (Covalent) Compound.

Binary compounds are compounds consisting of only two types of elements.

Examples: water (H 2 O), ammonia (NH 3)

A). Compound Formulas

The following element is listed in the following order, written ahead.

B-Si-C-Sb-As-P-N-H-Te-Se-S-I -Br-Cl-O-F

Example:

The ammonia chemical formula is commonly written as NH3 instead of H3N and the water chemistry formula is commonly written as H2O instead of OH2.

B). Compound Name

The name of the binary compound of two types of nonmetallic elements is a series of names of both types of elements with the suffix -ida (added to the second element).

Example:

HCl = hydrogen chloride

H2S = hydrogen sulfide

Note:

If the pairs of compounded elements form more than a kind of compound, the compounds formed are distinguished by mentioning the index numbers in Greek.

1 = mono 2 = at 3 = tri 4 = tetra

5 = penta 6 = hex 7 = hepta 8 = octa

9 =nona 10 = deka

The index number one does not need to be mentioned, except for the name of the carbon monoxide compound.

Example:

CO = carbon monoxide (mono prefix for C is not necessary)

CO2 = carbon dioxide

N2O = nitrous oxide

NO = nitrogen oxide

N2O3 = dinitrogen trioxide

N2O4 = dinitrogen tetraoxide

N2O5 = dinitrogen pentaoxide

CS2 = carbon disulfide

CCl4 = carbon tetrachloride

C). Compounds that are commonly known, do not need to follow the rules above.

Example:

H2O = water

NH3 = ammonia

CH4 = methane

2). Ion Compound Procedure.

Cation = positively charged ion (metal ion)

Anions = negatively charged ions (non-metallic ions or polyatomic ions)

A). Compound Formulas

Metal element is written in front.

Example:

The chemical formula of sodium chloride is written NaCl instead of ClNa.

The formula of ion compounds:

B Xa + + a Yb- → XbYa

The formula of ionic compounds is determined by the ratio of cation and anion loads.

Cations and anions are indexed in such a way that the compound is neutral (positive amount charge = the amount of negative charge).

B). Compound Name

The name of the ion compound is a series of cation names (in front) and the anion name (behind); While the index number is not mentioned.

Example:

NaCl = sodium chloride

CaCl2 = calcium chloride

Na2SO4 = sodium sulphate

Al (NO 3) 3 = aluminum nitrate

Note:

 If the metal element has more than a certain oxidation number, the compounds are distinguished by writing the oxidation number (written in parentheses with Roman numerals behind the metal element's name).

Example:

Cu2O = copper (I) oxide

CuO = copper (II) oxide

FeCl2 = iron (II) chloride

FeCl3 = iron (III) chloride

Fe2S3 = iron (III) sulfide

SnO = lead (II) oxide

SnO2 = lead (IV) oxide

3). Names of Terner Compounds.

The simplest ternary compounds include: acids, bases and salts.

The reaction between acid and base produces salt.

A). Acid nomenclature

Acid is a hydrogen compound that in the water has a sour taste.

The acid formula is composed of H atoms (in the front, considered as H + ions) and an anion called residual acid.

Note: keep in mind that acids are molecular compounds, not ionic compounds.

The name of the remaining acid anion = the name of the acid in question without the word acid.

Example: H 3 PO 4

The name acid = phosphoric acid

The residual acid formula = PO 43- (phosphate)

B). Base Names.

Bases are substances that if in the water can produce OH-

Generally, a base is an ionic compound comprising a metal cation and anion

Base name = the name of the cation followed by the word hydroxide.

Example: NaOH

Base Name = Sodium hydroxide

The residual formula of base = OH-

C). Salt Names.

Salt is an ionic compound composed of basic cations and anion residual acid.

The formula and its naming = ionic compounds.

Example:

NaNO2

Mg3 (PO4) 2

4). Organic Compound Nomenclature.

Organic compounds are compounds C with certain properties.

The organic compound has a special nomenclature, has a common name or trade name (a trivial name).

Example:

CH4 = methane

EQUATION REACTION

EQUAL REACTIONS HAVE THE PROPERTY

1,The types of elements before and after the reaction are always the same

2.The number of each atom before and after the reaction is always the same

3.The comparison of the reaction coefficients expresses the mole ratio (specifically the gaseous coefficient comparison also denotes the volume ratio provided the temperature den pressure is the same)

Example: Find the reaction coefficient of

HNO3 (aq) + H2S (g) ® NO (g) + S (s) + H2O (l)

The easiest way to determine the coefficient of reaction is to assume the coefficients of each a, b, c, d and e so that:

A HNO3 + b H2S ® c NO + d S + e H2O

Based on the above reaction then

Atom N: a = c (before and after reaction)

Atomic O: 3a = c + e ® 3a = a + e ® e = 2a

Atom H: a + 2b = 2e = 2 (2a) = 4a ® 2b = 3a ® b = 3/2 a

Atom S: b = d = 3/2 a

So in order to solve we take any price eg a = 2 means: b = d = 3, and e = 4 so the equation of the reaction:

2 HNO3 + 3 H2S ® 2 NO + 3 S + 4 H2O

CHEMICAL BASIC LAWS

1. The Law of Conservation of Mass (Lavoisier Law)

"The mass of the substance before the reaction equals the mass of the substance after the reaction"

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. Comparable Law (Proust Law)

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

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 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, it takes 50 grams of O2 to form 100 grams of 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 of O2 or 1: 8. This means, every one gram of H2 precisely reacts with 8 gram of O2 forming 9 gram H2O. If provided 24 grams of O2, it takes 3 grams of H2 to form 27 grams of H2O.

3. Comparative Law of Volume (Gay Lussac Law)

Applies only to chemical reactions that involve the gas phase

"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.

4. Avogadro's Law

Applies only to chemical reactions that involve the gas phase

"At the same temperature and pressure, the same volumes of gas contain the same number of moles"

Avogadro's law is closely related to Gay Lussac's Law

Example:

N2 (g) + 3 H2 (g) → 2 NH3 (g)

The mole ratio is equal to the ratio of the reaction coefficient. This means that every 1 mole of precise N2 gas reacts with 3 moles of H 2 gas to form 2 moles of NH 3 gas. The gas volume ratio is equal to the ratio of the reaction coefficient. This means that every 1 L of N2 gas precisely reacts with 3 L of H 2 gas to form 2 L of NH3 gas. Thus, if at a certain temperature and pressure, 1 mole of gas is equivalent to 1 L of gas, then 2 moles of gas is equivalent to 2 L of gas. In other words, the mole gas ratio is equal to the ratio of gas volume.

Here are some examples of problems and chemical calculations that use the basic laws of chemistry:

1. 20 gram calcium powder (Ar Ca = 40) is reacted with 20 grams of sulfur (Ar S = 32) according to the Ca + S → CaS reaction equation. What substance is left after the reaction is completed? How much mass of the substance is left after the reaction is complete?

Resolution:

The ratio of Ca mole to S is 1: 1. This means that every 40 grams of Ca exactly reacts with 32 grams of S to form 72 grams of CaS. The ratio of mass of Ca to S is 40: 32 = 5: 4.

If 20 grams of S just exhausts, it takes (5/4) x 20 = 25 grams Ca, to form 45 grams of CaS. Unfortunately, the amount of Ca provided is insufficient.

Therefore, 20 grams of Ca will be properly reacted. The required S mass of (4/5) x 20 grams = 16 grams. Thus, the remaining substance is sulfur (S). The remaining sulfur mass is 20-16 = 4 grams.

MOL CONCEPT

In the periodic table, it can be known the mass number expressing the relative atomic mass of an atom (Ar). Because the size is very small, to determine the mass of an atom is used atoms of other elements as a comparison, namely the 12C atom.

1. Relative Atomic Mass (Ar)

In chemical calculations not used absolute mass but used relative atomic mass (Ar). The relative atomic mass (Ar) is the ratio of the average mass of one atom of an element to 1/12 atomic mass of 12C or 1 sma (atomic mass unit) = 1.66 x 10-24 grams.

Example:

Ar H = 1.0080 sma rounded 1

Ar C = 12.01 sma rounded 12

Ar N = 14,0067 sma rounded 14

Ar O = 15.9950 sma rounded 16

The list of relative atomic masses (Ar) can be seen in the periodic table :

2. Relative Molecular Mass (Mr)

The relative molecular mass (Mr) is a number expressing the mass ratio of one molecule of a compound to 1/12 of the 12C atomic mass. The molecular mass of a realtif (Mr) equals the sum of the relative atomic mass (Ar) of all the constituent atoms.

Example:

Mr H2O = (2 x Ar H) + (1 X Ar O)

= (2 x 1) + (1 x 16)

= 18

Mr. CO (NH2) 2 = (1 x Ar C) + (1 x Ar O) + (2 x Ar N) + (4 x Ar H)

= (1 x 12) + (1 X 16) + (2 X 14) + (4 x 1)

= 60

3. Mol (n)

The atom is the smallest part that constitutes an element, while the molecule is the smallest part that constitutes a compound. The next atom and molecule are called elementary particles. The international unit for atoms and molecules is mol. One mole of the substance is the number of substances containing elementary particles of Avogadro (L), which is 6.02 x 1023. The number of moles is expressed by the symbol n.

1 mol element = 6.02 x 1023 atoms of that element

1 mol of compound = 6.02 x 1023 molecule of the compound

Thus n = number of particles / (6.02 X 1023)

as an example :

1 mol H2O = 1 X (6.02 X 1023)

Means in 1 mole of H2O there is 6,02 X 1023

4. Molar Mass

The mass of one mol of element or mass of one mole of a compound is called the molar mass. The mass of one mole of the element is equal to the relative atomic mass (Ar) of the atom in grams, whereas the mass of one mole of the compound is equal to the relative molecular mass (M) of the compound in grams

So n (mol) = mass (gram) / Ar or Mr (gram / mol)

Another example, on the periodic table, we can see that the mass of one copper atom is 63,55 sma and the mass of one sulfur atom is 32.07 sma. Meanwhile, the mass of one oxygen atom is 16.00 sma, while the mass of one hydrogen atom is 1.008 sma. Thus, the mass of one molecule of CuSO4.5H2O is as follows:

Mr. CuSO4.5H2O = 1 x Ar Cu + 1 x Ar S + 4 x Ar O + 5 x Mr H2O

= 1 x Ar Cu + 1 x Ar S + 4 x Ar O + 5 x (2 x Ar H + 1 X Ar O)

= 1 x 63,55 + 1 x 32,07 + 4 x 16,00 + 5 x (2 x 1,008 + 1 x 16,00)

= 249,700 sma

5. Molar Volumes

 The molar volume is the volume of one mole of gas. One mole of gas contains 6.02 x 1023 molecules. That is, every gas that has the same number of molecules, the number of moles is the same. In accordance with Avogadro's law, at the same temperature (T) and pressure (P), all gases of the same volume (V) contain the same number of moles (n)

Application of Avogadro's Law in Various Circumstances

A. Circumstance in Standard Temperature and Pressure (STP)

 Based on the Avogadro hypothesis and the ideal gas equation, the volume of 1 mole of each gas in the standard state (STP), ie at P = 1 atm and T = 0 ° C = 273 K is 22.4 liters.

So n (mol) = volume / 22.4 (liter / mol)

 As an example :

Determine the volume of 5 moles of carbon dioxide (CO2) gas measured on the STP.

Means Volume = 5 X 22.4 Liter = 112

B. Circumstances in Temperature and Non-Standard Pressure

In non-standard circumstances, the molar volume is calculated by the ideal gas equation PV = nRT (T in Kelvin)

C. Circumstances on Other Known Temperature and Gas Pressure

At the same temperature and pressure, the same volumes of gas have the same number of moles, so that the volume ratio at the same temperature and temperature will be equal to the mole ratio.

Thus, V1 / V2 = n1 / n2

V1 = volume of gas 1

V2 = gas volume 2

N1 = number of moles of gas 1

N2 = the number of moles of gas 2