STOICHIOMETRIC

STOICHIOMETRIC

        Stoichiometry is the branch of chemistry that studies

quantitative relation of the composition of chemical substances and reactions. Stoichiometry calculations are best done by stating the quantity of known and unknown in moles and then if need be converted into other units. Limiting reagent is the reactant present in the smallest stoichiometry amount. Reactant is limiting the amount of product that can be formed. The number of products produced in a reaction (the actual results) may be smaller than the maximum amount that may be acquired (the theoretical yield). Comparison of the two expressed as a percent of the results.

       Chemical law is a law of nature that are relevant to the field of chemistry. the concept of the

chemistry is fundamental in the law of conservation of mass, which states that no

changes in the quantity of matter during an ordinary chemical reaction. Modern physics shows

that actually happens is the conservation of energy, and that energy and mass

interconnected a concept that is important in nuclear chemistry. conservation

energy leads to the important concepts of the equilibrium,

thermodynamics, and kinetics.

       Another modern chemical law to determine the relationship between energy and transformation.

1. In equilibrium, a molecule found in the mixture is     

    determined by the

    transformation that may occur on a time scale equilibrium,  

    and has a ratio determined by the intrinsic energy of the  

    molecule. The smaller energy intrinsic, the more molecules.

2. Change one structure into another structure requires the  

    input of energy to beyond the constraints of energy, it can be  

    caused by intrinsic energy of molecules it self, or from  

    external sources will generally accelerate the change. The  

    greater the energy barrier, the slower the process of on going 

    transformation.

c. There is a structure or a transition between the hypothetical,  

    which relates to the structure in peak of the energy barrier.     

    Hammond Postulate-Leffer stated that this structure

    resembles the original product or material that has an  

    intrinsic energy closest the energy barrier. By stabilizing the 

    hypothetical structure with chemical interaction is one way  

    to achieve catalysis.

d. All chemical processes are irreversible (reversible) (law  

    reversible microscopic) although some process has a bias 

    energy, their Basically irreversible (irreversible).

A. LAWS BASIC CHEMISTRY

1. LAW OF CONSERVATION OF MASS = Lavoisier LAW

     "The mass of substances before and after the reaction is 

       fixed".

     Example:

     hydrogen + oxygen-hydrogen oxide

       (4g) (32g) (36g)

2. COMPARATIVE LAW = LEGAL PERMANENT Proust

     "Comparison of the mass of the elements in each compound 

     is fixed"

      Example:

         a. In the compound NH3: N mass: mass of H

         Ar = 1. N: 3 Ar. H

         = 1 (14): 3 (1) = 14: 3

         b. On the compound SO3: S mass: mass 0

         Ar = 1. S: 3 Ar. O

         = 1 (32): 3 (16) = 32: 48 = 2: 3

     Advantages of the law Proust:

if known mass of a compound or a mass of one of the elements that make up the compound make-masses of other elements can be determined.

    Example:

How many levels of C in 50 grams of CaCO3? (Ar: C = 12; 0 = 16; Ca = 40)

Mass C = (Ar C / Mr CaCO3) x mass of CaCO3

             = 12/100 x 50 grams = 6 grams

Levels of C = mass C / mass x 100% CaCO3

            = 6/50 x 100% = 12%

3. COMPARATIVE LAW LAW MULTIPLE = DALTON

    "When the two elements can form two or more compounds

     to the mass of one element of the same number of the mass  

    ratio of the two elements will be compared as integers and 

    simple ".

Example:

If the element nitrogen fertilized den oxygen can be formed,

NO where mass N: 0 = 14: 16 = 7: 8

NO2 which the mass N: 0 = 14: 32 = 7: 16

    For the same amount of nitrogen mass ratio of the mass of oxygen in the compound NO: NO2 = 8: 16 = 1: 2

4. GAS LAWS

    To apply the ideal gas equation: PV = nRT

    where:

    P = gas pressure (atmospheric)

    V = gas volume (liters)

     n = moles of gas

    R = universal gas constant = 0082 lt.atm / mol Kelvin

    T = absolute temperature (Kelvin)

   

   The changes of P, V and T from state 1 to state 2 with certain     

    conditions reflected by the following laws:

a. BOYLE LAW

    This law is derived from the ideal gas equation of state with

    n1 = n2 and T1 = T2; thus obtained: P1 V1 = P2 V2

    Example:

    What is the pressure of 0 5 mol O2 with a volume of 10 liters   

    when the temperature is 0.5 mol

    NH3 has a volume of 5 liters and a pressure of two 

    atmospheres?

    Answer:

    P1 V1 = P2 V2

    2. 5 = P2. 10  P2 = 1 atmosphere

b. LEGAL Gay-Lussac

    "The volume of gases that react and the volume of gas when  

    measured at the reaction temperature and pressure the same 

    in value as simple and integer ".

    So for: P1 = P2 and T1 = T2 holds: V1 / V2 = n1 / n2 

    Example:

    Calculate the mass of 10 liters of nitrogen gas (N2) if the   

    condition

    The 1 liter of hydrogen gas (H2) mass of 0.1 g.

    Given: Ar for H = 1 and N = 14

    Answer:

    V1/V2 = n1/n2 ® 10/1 = (x/28) / (0.1 / 2) ® x = 14 grams

    So mass = 14 grams of nitrogen gas.

c. BOYLE LAW-Gay Lussac

    This law is an extension of the previous law and lowered

    the state n = n2 price in order to obtain the equation:

    P1. V1 / T1 = P2. V2 / T2

d. Avogadro's law

    "At the same temperature and pressure, the volume of gases  

    the same contains the same number of moles. "

    Example:

Hydrogen    +      Chlorine      à hydrogen chloride

       n molecules         n molecule             2n molecular

divided by n

       1 molecule         +               1 molecul à        2 molecules

   Hydrogen                            chlorine               hydrogen chloride

Example: What is the volume of gas at a temperature of 29 grams of C4H10 and constant pressure, where 35 liters of oxygen weighs 40 grams   (Mr. C4 H10 = 58; Ar O = 16)

Answer: Mol C4H10 = 29/54 = 0.5 mol

                Mol O2 = 40/32 = 1.25 mol

               1/2 mol C4H10 = 0.5 / 1.25 x 35 = 14 liters

B. ATOM MASS AND MASS FORMULA

     

    1. Relative Atomic Mass (Ar)

          

               is the ratio between the mass of one atom with 1/12 the   

     mass. 1 atom of carbon 12.

           Dalton recognized that it is important to determine the mass of each atom as mass varies for each type of atom. Atom is very small so it is not possible to determine the mass of a single atom. So he focuses on the relative masses and create a table atomic mass (Figure 1.3) for the first time in human history. In the table, the mass of the lightest element, hydrogen adoption as a standard one (H = 1). Atomic mass is a relative value, meaning that a dimensionless ratio. Although
several different atomic masses with modern values​​, most of the proposed values ​​in the range of compatibility with the current value. This shows that the idea and the experiment right.

        Then the Swedish chemist Jons Jakob Berzelius Baron (1779-1848) to determine the mass of the oxygen atom as the standard (O = 100). Because Berzelius get this value based on the analysis of oxide, it has a clear reason to choose oxygen as standard. However, the standard hydrogen is clearly superior in terms of simplicity. Now, after much discussion and modification, carbon standard is used. In this method, the mass of 12C carbon with 6 protons and 6 neutrons is defined as 12.0000. Atomic mass is the mass of an atom relative to this standard. Although carbon has been declared as standard, this can actually be considered as a standard hydrogen is modified.

2. Relative Molecular Mass (Mr)

             is the ratio between the mass of 1 molecule compounds

    1/12 the mass of one atom of carbon 12.

    Relative molecular mass (Mr) of a compound is the sum

    atomic mass of constituent elements.

Example:

    If Ar for X = 10 and Y = 50 what Mr. compound X2Y4?

Answer:

    Mr X2Y4 = 2 x Ar. X + 4 x Ar. Y = (2 x 10) + (4 x 50) = 220

C. CONCEPT MOL

      1 mole is the amount of the chemical unit numbers of the atoms or molecules of Avogadro's number and mass = Mr compound.

If Avogadro's number = L then:

                                                    L = 6.023 x 1023

1 mole of atoms = L atoms, mass = Ar atom.

1 mole of molecules = L = Mr fruits molecular mass of the molecule.

The mass of one mole of a substance is called the molar mass of the substance

Example:

How many molecules are present in 20 grams of NaOH?

Answer:

Mr NaOH = 23 + 16 + 1 = 40

mol NaOH = mass / Mr = 20/40 = 0.5 mol

The number of molecules of NaOH = 0.5 L

                                      = 0.5 x 6023 x 1023

                                      = 3.01 x 1023 molecules.

D. EQUAL REACTION

NATURE HAVE EQUAL REACTION

1. Types of elements before and after the reaction is always the
     same

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

    always the same

3. Comparison of the reaction coefficient expressed mole ratio

    (Specifically in the form of gas ratio coefficient is also states  
     as long as the volume ratio den temperature pressure is the  
     same)

    Example: Find the coefficient of reaction

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

    The easiest way to determine the coefficient of the reaction is

    by letting the coefficient of each a, b, c, d and e

    so:

         a HNO3 + H2S b c d S + NO + e H2O

         Based on the above reaction

         N atoms: a = c (before and after the reaction)

         atom O: 3a = c + e 3a = a + e e = 2a

         H atoms: a + 2b = 2e = 2 (2a) = 4a; 2b = 3a, b = 3/2 a

         atom S: b = d = 3/2 a

         So that we resolved to take any price eg a = 2

         means: b = d = 3, and e = 4 so the equation:

           2 HNO3 + 3 H2S + 3 S 2 NO + 4 H2O