BASIC CHEMISTRY 1 MID TEST

BASIC CHEMISTRY 1 MID TEST
NAMA : FITRIANI
NIM      : RSA1C112013

1. X pure material is solid at room temperature. if the substance is heated to 230 C melted gradually. if it is   

     then cooled to room temperature, the liquid can not be frozen.
   a. that x may be from an element or compound. explain it!
   b. Is it a chemical change occurs? if so could be said to undergo change endotherm, based on the   

        information provided?
  c.can said that the liquid is an element, based on the information provided

         Answer:

a. X is included compounds. Because the compound is a single substance that is chemically still be broken down into other substances much simpler where nature is different from the original substance. The smallest part of a compound is a molecule (a combination of two atoms of the element / more better kind or different kinds. Example, white sugar, solid, and sweet taste when heated to burn will have a reaction. Water also belong to the compound. Air can be decomposed into two other substances, namely hydrogen and oxygen gas. decomposition of water can occur when water vapor is heated at high temperatures or if the water is electrified. properties of hydrogen and oxygen gases vary with the nature of water. flammable hydrogen gas, while oxygen is a gas required in the combustion process. while water can not burn and can not hold burning. Thus, water can not be frozen at room temperature, but can be frozen at a temperature of O ° C.b. occur. Since X can fuse although gradually. The reaction is endothermic reaction the heat transfer from the environment to the system. marked drop in the ambient temperature around the system.c. yes, because it is the basic constituent element of the element. Air includes an element because it is composed of hydrogen and oxygen.

2. When the candle that weighs 10 g burned in oxygen, carbon dioxide and water vapor formed by the combustion of more than 10 g weight. This was the case in accordance with the law of conservation of mass. Explain!

Answer:
  Law of Conservation of Mass presented by Antoine Laurent Lavoisier (1743-1794), which reads: "In a reaction, the mass of substances before and after the reaction is the same", in other words, the mass can not be created and can not be destroyed. This means that as long as there is no reaction occurs atoms reactants and reaction products were missing. So candles are lighting source which consists of axis covered by solid fuel. materials made from paraffin wax, which is a mixture of hydrocarbons of Alkane (molecular chains of carbon atoms and hydrogen atoms long), the material we have encountered in petroleum. As implied in its name, only the hydrocarbon molecules consist of atoms of hydrogen and carbon atoms.
   Results of the observation:a. Observation Results Candle Burned BeforeQualitative Sense• Color white candles• Color white wick• Candles odorless• smooth wax surface• Establish regular wax cylinder to the top of the cone'sEyeNoseSkinEyeQuantitative gauge• High candle 16 cm• High-wick candle 17 cm• 1.8 cm diameter candle• Weight 40.19 grams wax before the fire RulerRulerRulerBalanceb. Observation Results Candle Burned AtQualitative Sense• The color remains white candles• a burning fuse color black• The end of the fuse is lit like burning embers• Burning candles produce light• Fire the top of the yellow and blue flames bottom• Some time after the initiation of burnt wax melting• molten hot wax• The molten wax is still hot nodes• The air around the hot fire EyeEyeEyeEyeEyeEyeSkinEyeSkinQuantitative gauge• The length of fire 3 cm• The cone candle began to disappear during combustion at minute - 6• In minute - 10 14.2 cm height candles• In minute - 20 13 cm height candles• In minute - 60 5 cm height candles RulerStopwatchRulerRulerRulerc. After the results of observations Candle BurnedQualitative Sense• Shortly after the fire burned off a pungent smell• Shortly after the fire off of the wick candles emit black smoke• The color remains white candles• Color wick burnt black• The wax becomes irregular• Nose rough surface waxEyeEyeEyeSkinQuantitative gauge• The length of 5 cm wax• The length of the wick, 6 cm• Weight 19.71 grams burned candles after RulerRulerBalanceSo candles are included in the law of conservation of mass.

3. When carbon burns under a limited amount of oxygen, it forms two gaseous compounds. Suggest a way to differentiate the two compounds with one another.

Answer:
he difference in one compound with another compound when seen from the structure is obviously different for different compounds. When viewed from its differences are at the boiling point and freezing point. If the reagent is reacted with the same result in different products. Example:
               C + O2 ==> CO2Carbon Dioxide

              2C + O2 ==> COCarbon monoxide

4. After mendeev preparing the periodic table, he concluded that the atomic weights of some elements was the wrong decision, and this conclusion appears to be true. How mendeelev able to predict some of the atomic weights is wrong? why his predictions are not always right. Explain!

Answer:
According to Mendeleev: the properties of elements are a periodic function of the relative atomic mass. Meaning: if the elements are arranged according to their relative atomic mass increases, the specific properties will be repeated periodically. The elements that have similar properties are placed in a straight row, called the Group. While the horizontal rows, for elements based on the relative atomic mass increases and called Period.Mendeleev SYSTEM FLAWS1. The length of the period is not the same and why not described.2. Some elements are not ordered by their atomic mass increases, for example: Te (128)
     before I (127).3. Mass difference of successive elements are not always 2, but the range between 1 and 4
     so it is difficult to predict the mass of an element that has not been known.4. Valence more than one element of the unpredictable group.5. Anomaly (deviation) of the elements hydrogen to another element not explained.6. There are still elements of the larger mass located at the front element
     smaller mass. Co: Tellurium (te) = 128 in kiriIodin (I) = 127. this is because
      element that has properties kemirpan placed in one group.7. Pemebetulan atomic mass. In previous atomic mass = 76 to 113. besides it Be,
      from 13.5 to 9. U from 120 to 240.

5. When the mercury chloride solution is added a solution of silver nitrate, a white solid forms. Identification of a white solid and write a balanced equation for the reaction that occurs

Answer:

HgCl2 + 2AgNO3 ---> Hg (NO3) 2 + 2AgClThe resulting white powder is silver nitrate (AgNO3)

oxidation state of the atoms of the periodic system

Oxidation numbers of atoms

There is a clear relationship between oxidation number (or oxidation state) atom and its position in the periodic table. Oxidation number of atoms in covalent compounds atoms is defined as an imaginary charge which will be owned when the electrons are shared equally divided between bonded atoms (atoms that are bonded together) or handed over all the atoms are more strongly appeal (if different atoms bonded ).

(1) GROUP KEY ELEMENTS

For the main group elements, oxidation in many cases is the number of electrons will be released or received to achieve the full electron configuration, ns2np6 (except for the first period) or an electron configuration nd10.

It is clear for elements of low period which is a member group 1, 2 and 13-18. For larger periods, the trend has oxidation associated with the configuration of the electrons with the electrons retained ns and np electrons are removed. For example, lead tin Sn and Pb, both class 14, has a +2 oxidation state by removing electrons NP2 but retain ns2 electrons, in addition to the oxidation state +4. The same reasoning can be used to the fact that phosphorus P and bismuth Bi, two groups of 15 with electron configuration ns2np3, has an oxidation state of +3 and +5.

Generally, the importance of ns2 electron oxidation to be maintained will become increasingly important for a greater period. For nitrogen and phosphorus compounds, oxidation number +5 dominant, while the dominant bismuth is +3 and +5 oxidation state rather rare.

Metallic elements and semilogam (Si silicon or germanium Ge) rarely has a negative oxidation value, but for non-metals are common phenomena. In nitrogen and phosphorus hydride, NH3 and PH3, oxidation of N and P are-3. The higher period elements, the element will lose these properties and bismuth Bi does not have a negative oxidation. Among the group 16 elements, oxidation-2 dominant as in the case of oxygen O. This trend will again decline to elements in higher periods. Suppose oxygen only has a negative oxidation number, but it's had such a positive oxidation state +4 and +6 are also significant.

(2) ELEMENTS OF TRANSITION

Although the transition elements have multiple oxidation states, regularity can be recognized. Highest oxidation number of atoms that have five electrons the number of d orbitals associated with the current state of all the d electrons (electrons than s) removed. So, in the case of scandium with electron configuration (n-1) d1ns2, oxidation number 3. Manganese configuration with (n-1) d5ns2, will berbilangan maximum oxidation +7.

If the amount exceeds 5 d electrons, the situation changed. For iron Fe with electron configuration (n-1) d6ns2, primarily oxidation +2 and +3. Very rare oxidation state +6. Highest oxidation number a number of important transition metals such as cobalt Co, Ni Nickel, copper and zinc Zn Cu lower oxidation states of atoms lose all electrons (n-1) d and ns it. Among the elements that are in the same group, the higher the oxidation state of the elements essential for a greater period.

Concept of Oxidation Numbers

Understanding Oxidation Numbers:

Electric charge as if possessed by the elements in a compound or ion.

OXIDATION NUMBERS PRICE

1. Oxidation Bialngan free element = 0

2. Oxygen

In compounds Oxidation Numbers = -2

except

a. In peroxide, Oxidation Numbers = -1

b. In superoxide, Numbers Oxide = -1 / 2

c. In of2, Oxidation Numbers = +2

3. Hydrogen

In compounds, Oxidation Numbers = +1

Except in hybrid = -1

4. The elements of Group IA

In compounds, Oxidation Numbers = +2

5. The elements of Group IIA

In compounds, Oxidation Numbers = +2

6. Oxidation Numbers  molecule = 0

7. Oxidation Numbers  ion = charge on the ion

8. Halogens

F: 0, -1

Cl: 0, -1, +1, +3, +5, +7

Br: 0, -1, +1, +5, +7

I: 0, -1, +1, +5, +7

ATOMIC STRUCTURE

ATOMIC STRUCTURE
A. BASIC UNDERSTANDING
1. Elementary particles: particles are composed of atoms forming
electron, proton neutron den.
1. Proton: forming atomic particles have the same mass
one sma (amu) and charged +1.
2. Neutron: particles forming a sma atomic mass (amu) and
neutral.
3. Electron: atom forming particles that have no mass and
charged -1.
2. The nucleus: a positively charged nucleus, consisting of protons and
neutrons.
3. Notation elements:
Zaa by X: sign atoms (elements)
Z: atomic number = number of electrons (e)
   = Number of protons (p)
A: mass number = number of protons + neutrons
In neutral atoms, applies: the number of electrons = number of protons.
4. No neutral atoms: electrically charged atoms due to excess or
electron deficiency when compared with the neutral atom.
Positively charged atoms when the electron deficiency, called cations.
Negatively charged atoms when an excess of electrons, called anions.
Example:
- Na +: cation with one electron deficiency
- Mg2-: cation with a shortage of 2 electrons
- Cl-: anion with an excess of one electron
- O2: anion with an excess of two electrons
5. Isotopes: elements of the same atomic number, but different numbers
mass.
Example: oxygen isotopes: 8O16 8O178O18
6. Isobar: elements of the same mass number, but different numbers
atom.
Example: 27CO59 with 28Ni59
7. Isoton: elements with the same number of neutrons.
Example: 6C13 with 7N14
8. Iso electron: atom / ion with the same number of electrons.
Example: Na + with Mg 2 +
K + with Ar
B. MODEL ATOM
1. MODEL ATOM JOHN DALTON
- Atom is the smallest part of an element
- Atoms can not be created, destroyed, divided, or changed to
other substances
- Atoms of an element are the same in all respects, but different
atoms of other elements
- Chemical reaction is a process of merging or splitting of the atom
elements are visible
Dalton's atomic theory Weakness: can not distinguish sense
atoms and molecules. And the atom was not the smallest particles.
ATOM 2.MODEL J.J. THOMPSON
- Atom is a positively charged sphere and inside spread
electrons are like raisins
- The amount of positive charge with a negative charge, so that the atoms are
neutral
3. RUTHERFORD ATOM MODEL
- Atom consists of a very small nucleus with a positive charge
mass is the mass of the atom
- Electrons moving around the nucleus in an atom is
- The number of electrons in an atom equal to the number of protons in the nucleus and
This corresponds to the number
atom
4. MODEL ATOM Bohr
- Electrons surrounding the nucleus is in the energy levels
(Skin) without specific
absorb or emit energy
- Electrons can move from the outer skin to the deeper skin
radiated energy, or vice versa
C. QUANTUM NUMBERS
To determine the position of an electron in an atom, used 4
quantum numbers.
1. The principal quantum number (n): realizing the electron trajectories in
atom.
n has a price of 1, 2, 3, .....
- N = 1 corresponds to the K shell
- N = 2 correspond to the L shell
- N = 3 corresponds to the skin M
- And so on
Each skin or any number of energy levels occupied by electrons.
The number of electrons that can occupy maksimmm energy levels
must satisfy the Pauli formula = 2N2.
2. Azimuthal quantum number (l): indicate sub skin where
electrons that move also shows a sub-skin
compilers of the skin.
Azimuthal quantum number have prices from 0 to (n-
1).
n = 1; l = 0; corresponding K shell
n = 2, l = 0, 1; corresponding L shell
n = 3; l = 0, 1, 2; appropriate skin M
n = 4; l = 0, 1, 2, 3; appropriate skin N
and so on
Sub leather prices vary is given a special name:
l = 0; fit leather sub s (s = sharp)
l = 1; fit leather sub p (p = principle)
l = 2; fit leather sub d (d = diffuse)
l = 3; fit leather sub f (f = fundamental)
3. Magnetic quantum number (m): realizing the presence of one or
several levels of energy in a sub shell. Quantum numbers
magnetic (m) has a price (-l) to price (+ l).
For:
l = 0 (sub leather s), price m = 0 (having 1 orbital)
l = 1 (p sub shell), price m = -1, O, +1 (have 3 orbitals)
l = 2 (sub skin d), price m = -2, -1, O, +1, +2 (have 5 orbitals)
l = 3 (sub kwit f), price m = -3, -2, O, +1, +2, +3 (have 7
orbital)
4. Spin quantum number (s): indicates the direction of rotation of the electron
on its axis.
In one orbital, maximum of 2 electrons can circulate and second
This electron spins through the axis in the opposite direction, and
each one is priced spin +1 / 2 or -1 / 2.
D. ELECTRON CONFIGURATION
In every atom orbitals are available, but not necessarily
all orbital is filled. How is the electron charge
these orbitals?
Completion of electrons in the orbitals satisfy some rules.
among other things:
1. Aufbau principle: electrons fill orbitals starting with
the lowest energy level and beyond.
Orbital who meet the low energy levels is 1s
followed by 2s, 2p, 3s, 3p, and on and on
made easier following diagram:
Examples of filling electrons in orbitals several elements:
Atom H: has 1 electron, configuration 1s1
Atom C: has 6 electrons, the configuration 1s2 2s2 2P2
Atom C: has 19 electrons, the configuration 1s2 2s2 2P6 3S2 3P6
4S1











2. Pauli principle: not possible in the atom there are two electrons
with the same four quantum numbers.
This means, if there are two electrons that have a number
principal quantum, and magnetic azimuth of the same, then the number
quantum spins must be opposite.
3. Principle Hund: for filling electrons in orbitals in a sub
Skin is that the electrons do not form pairs of electrons
before each filled with an electron orbital.

THERMOCHEMICAL

Thermochemical

A.      Exothermic and endothermic reactions

1.         Exothermic reaction

 In an exothermic reaction occurs heat transfer of the system to

     environment or to the reaction heat released.? In an exothermic reaction ΔH price = (-)

     Example: C (s) + O2 (g) CO2 (g) + 393.5 kJ; ΔH = -393.5 kJ

Endothermic reaction?? In endothermic reactions occur heat transfer from the environment to

     system or to the reaction heat is needed.? In endothermic reactions price ΔH = (+)

?? Example: CaCO3 (s) CaO (s) + CO2 (g) - 178.5 kJ; ΔH = +178.5 kJ

. Enthalpy changes

     Enthalpy = H = heat of reaction at constant pressure = Qp? The change in enthalpy is the energy change accompanying events

     chemical changes at a constant pressure.

    

     a. Termination of the bond requires energy (= endothermic)? Example: H2 2H - a kJ; ΔH = + AKJ

     b. Bond formation provides energy (= exothermic)? Example: H2 + 2H a kJ; ΔH =-a kJ

The term used in the enthalpy change:

Standard Enthalpy Pembuntakan (ΔHf):? ΔH animal lays to form 1 mole of compounds directly from the elements

    elements were measured at 298 K and pressure of 1 atm.

    Example: H2 (g) + 1/2 O2 (g) H20 (l); ΔHf = -285.85 kJ

Enthalpy of Decomposition:? ΔH of decomposition of 1 mole of the compound directly into its elements (= Contrary to ΔH formation).

    Example: H2O (l) H2 (g) + 1/2 O2 (g), ΔH = +285.85 kJ

Standard Enthalpy of Combustion (ΔHc):? ΔH to burn 1 mole compound with O2 from the air measured at 298 K and pressure of 1 atm.

Example: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l); ΔHc = -802 kJ

Enthalpy of reaction:? ΔH of an equation in which substances contained in the equation is expressed in units of moles and the coefficients of the equation is simple round.

Example: + 2AL 3H2SO4 Al2 (SO4) 3 + 3H2; ΔH = -1468 kJ

Enthalpy of Neutralization:? ΔH generated (always exothermic) on the neutralization of acid or alkaline reaction.

Example: NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l);

             ΔH = -890.4 kJ / mol

Lavoisier-Laplace law? "The amount of heat released in the formation of one mole of substance

    elements unsurya = amount of heat required to decompose the substance into its constituent elements. "? Meaning: If the reaction is reversed the sign of the heat that is formed is also reversed from positive to negative or vice versa

    

     Example:? N2 (g) + 3H2 (g) 2NH3 (g), ΔH = - 112 kJ? 2NH3 (g) N2 (g) + 3H2 (g), ΔH = + 112 kJ

C. Determination of Enthalpy Changes and Hess's Law

1. Determination of Enthalpy Changes

     To determine the enthalpy change in a chemical reaction

     commonly used tools such as the calorimeter, thermometer and

     etc., that may be more sensitive.

     Calculation: ΔH reaction = Δ; ΔHfo products - Δ = ΔHfo reactants

2. Hess's Law

     "The amount of heat required or released in a reaction

     does not depend on the course of chemical reactions but is determined by

     initial state and the end. "

example :               

According to Hess's Law: x = y + z

D. Energy-Energy and Chemical Bonding

      A chemical reaction is a process of dissolution and formation

      ties. The process is always accompanied by energy changes. The energy

      needed to break chemical bonds, thus forming

      free radicals called the bond energy. For molecules

      complex, the energy required to break the molecule

      thus forming free atoms called atomization energy.

Atomization energy prices is the amount of the bond energy of the atoms in the molecule. For covalent molecule consisting of two atoms such as H2, 02, N2 or HI which has a bond equal to the energy of atomization energy bond energy of atomization of a compound can be determined by the help enthalpy of formation of these compounds. Mathematically it can be described by the equation.

                                                                                                                                                       

Example:

Given:

bond energy

C - H = 414.5 kJ / mole? C = C = 612.4 kJ / mol? C - C = 346.9 kJ / mol? H - H = 436.8 kJ / mol?? Asked:

ΔH reaction = C2H4 (g) + H2 (g) C2H6 (g)

Answer:

ΔH reaction = Total bond breaking energy - amount of energy

                     bond formation

                 = (4 (C-H) + (C = C) + (H-H)) - (6 (C-H) + (C-C))? = ((C = C) + (H-H)) - (2 (C-H) + (C-C))? = (612.4 + 436.8) - (2 x 414.5 + 346.9)? = - 126.7 kJ

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