Percent Concentration

By: Cristel Imbag

The use of percentages is a common way of expressing the concentration of a solution. It is a straightforward approach that you have used earlier when dealing with the composition of compounds. There are, however, some differences. One is that the concentrations of solution are variable while the composition of compounds is constant. Another is that the percentages can be calculated using volumes as well as weights, or even both together.

One way of expressing concentrations, with which you might be familiar, is by volume percent. Another is by weight percent. Still another is a hybrid called weight/volume percent. In the pages of this section we will look at how to calculate each and the cases where each is generally used.

Mass Percentage

One way of representing the concentration of an element in a compound or a component in a mixture. Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%.

Volume Percentage

Volume percent is a common expression of a solution's concentration. It is defined as:

It is related to volume fraction by dividing by 100%. Volume percent is usually used when the solution is made by mixing twofluids, such as liquids or gases. However, percentages are only additive forideal gases. The solute is what you get from the solution. For example, if you distilled salt water, the solution would be the salt water, and the solvent would be thedistilled water. A solution consists of a solvent, which is commonly a liquid, and a solute, which may be any substance that dissolves in the solvent. For example, if salt water (a solution) is distilled, water (the solvent) passes over into the receiver and salt (the solute) remains in the still. In the case of a mixture of ethanol and water, which are miscible in all proportions, the designation of solvent and solute is arbitrary. The volume of such a mixture is slightly less than the sum of the volumes of the components. Thus, by the above definition, the term "40% alcohol by volume" refers to a mixture of 40 volume units of ethanol with enough water to make a final volume of 100 units, rather than a mixture of 40 units of ethanol with 60 units of water.

Example:

You have 40.0ml of C6H5CH3 that is put into 75.0ml of C6H6, what is the percent volume of C6H5CH3?

Solution: (40.0ml) / (115.0ml) x 100 = 34.8% of C6H5CH3

Weight/Volume Percentage

* weight/volume percentage concentration is usually abbreviated as w/v (%)

* w/v (%) = mass solute ÷ volume solution x 100

* common units are g/100mL (%)

* weight/volume is a useful concentration measure when dispensing reagents.

Reference:
About.com Chemistry Volume-volume percentage, Chembuddy website
Ausetute.com.au
Wikepedia

Standardization Method

Acid Base Titration

By: Alleta Fae S. Liwag

Standardization is the process of determining the exact concentration (molarity) of a solution.

Titration is one type of analytical procedure often used in standardization.

In a titration, an exact volume of one substance is reacted with a known amount of another substance.

The point at which the reaction is complete in a titration is referred to as the endpoint. A chemical substance known as an indicator is used to indicate (signal) the endpoint. 

How to Standardize?

Example 1
A 0.128 g sample of KHP (HKC8H4O4) required 28.54 mL of NaOH solution to reach a phenolphthalein endpoint. Calculate the molarity of the NaOH.

HKC8H4O4 + NaOH -----> NaKC8H4O4 + H2O

(0.128 g KHP)(1 mol / 204.23 g KHP ) = 6.267 x 10-4 mol KHP

(6.267 x 10-4 mol KHP)(1 mol NaOH / 1 mol KHP) = 6.267 x 10-4 mol NaOH

6.267 x 10-4 mol NaOH / 0.02854 L NaOH = 0.0220 M NaOH


Example 2
A 20.00 mL sample of HCl was titrated with the NaOH solution from Example1. To reach the endpoint required 23.72 mL of the NaOH. Calculate the molarity of the HCl.

HCl + NaOH -----> NaCl + H2O

(0.02372 L NaOH)(0.0220 mol NaOH / 1 L NaOH) = 5.218 x 10-4 mol NaOH

(5.218 x 10-4 mol NaOH)(1 mol HCl / 1 mol NaOH) = 5.218 x 10-4 mol HCl

5.218 x 10-4 mol HCl / 0.02000 L HCl = 0.0261 M HCl


REFERENCE:

http://www.chem.latech.edu/~deddy/chem104/104Standard.htm


Tirtation an Acid-Base Neutralization

By: Mark Anthony Basilio

Acids and Bases - Neutralizing a Base with an Acid

Ca(OH)2 is a strong base and will dissociate completely in water to Ca2+ and OH-. For every mole of Ca(OH)2 there will be two moles of OH-. The concentration of Ca(OH)2 is 0.01 M so [OH-] will be 0.02 M.To the solution will be neutralized when the number of moles of H+ equals the number of moles of OH-.

Step 1: Calculate the number of moles of OH-.

Molarity = moles/volume

moles = Molarity x Volume

moles OH- = 0.02 M/100 ml
moles OH- = 0.02 M/0.1 L
moles OH- = 0.002 moles

Step 2: Calculate the Volume of HCl needed

Molarity = moles/volume                                   Volume = moles/Molarity

Titrations

A titration is a volumetric technique in which a solution of one reactant (the titrant) is added to a solution of a second reactant (the "analyte") until the equivalence point is reached. The equivalence point is the point at which titrant has been added in exactly the right quantity to react stoichiometrically with the analyte. If either the titrant or analyte is colored, the equivalence point is evident from the disappearance of color as the reactants are consumed. Otherwise, an indicator may be added which has an "endpoint" (changes color) at the equivalence point, or the equivalence point may be determined from a titration curve. The amount of added titrant is determined from itsconcentration and volume:

n(mol) = C(mol/L) x V(L)

and the amount of titrant can be used in the usual stoichiometric calculation to determine the amount of analyte.

The object of a titration is always to add just the amount of titrant needed to consumeexactly the amount of substance being titrated. In the NaOH—CH3COOH reaction [Eq. (1)], the equivalence point occurs when an equal molar amount of NaOH has been added from the graduated cylinder for every mole of CH3COOH originally in the titration flask. That is, at the equivalence point the ratio of the amount of NaOH, added to the amount of CH3COOH consumed must equal the stoichiometric ratio


Note that overtitrating [adding more than 23.62 cm3 of KMnO4(aq) would involve an excess (more than 1.272 mmol) of KMnO4.

Source: http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Titrations-875.html

             http://chemistry.about.com/od/workedchemistryproblems/a/neutralizeacid.htm

Indirect Tirtation

By: Rosiel Mariano

Back titration (Indirect titration) is a titration done in reverse;; instead of titrating the original sample, a known excess of standard reagent is added to the solution, and the excess is titrated. A back titration is useful if the endpoint of the reverse titration is easier to identify than the endpoint of the normal titration, as with precipitation reactions. Back titrations are also useful if the reaction between the analyte and the titrant is very slow, or when the analyte is in a non-soluble solid.

PROPERTIES OF A BACK TITRATION

React with the excess volume of reactant which has been left over after completing reaction with the analyte from the normal titration
The substance or solution of unknown concentration of excess intermediate reactant is made to react with known volume and concentration of intermediate reactant solution in back titration
Throughout back titration, the reaction can reach the completion quickly as the excess reactant that react with the NaOH (as example) heated, and is much easier to measure
Back titration also an indirect titration procedure
the proportion consumed in the reaction of back titration being obtained by difference

PURPOSE OF BACK TITRATION
Back titration is designed to resolve some of the problems encountered with forward or direct titration. Possible reasons for devising back titration technique are:
1: The analyte may be in solid form
2: The analyte may contain impurities which may interfere with direct titration. Consider the case of contaminated chalk. We can filter out the impurities before the excess reactant is titrated and thus avoid this situation.
3: The analyte reacts slowly with titrant in direct or forward titration. The reaction with the intermediate reactant can be speeded up and reaction can be completed say by heating.
4: Weak acid – weak base reactions can be subjected to back titration for analysis of solution of unknown concentration. Recall that weak acid-weak weak titration does not yield a well defined change in pH, which can be detected using an indicator.

ADVANTAGES OF BACK TITRATION

useful when trying to work out the amount of an acid or base in a non-soluble solid.
useful if the endpoint of the reverse titration is easier to identify than the endpoint of the normal titration

DISADVANTAGES OF BACK TITRATION

Needs skill and practise for effective results
Instruments have to be properly calibrated since it will give affected the final result.
Reactivity of the elements to be titrated should be well researched since this may affect the end point.
Time consuming if done manually

CALCULATIONS
Example 1
150.0 mL of 0.2105 M nitric acid was added in excess to  1.3415 g calcium carbonate. The excess acid was back titrated with 0.1055 M sodium hydroxide. It required 75.5 mL of the base to reach the end point.
Calculate the percentage (w/w)  of calcium carbonate in the sample.

1.EXTRACT INFORMATION

HNO3

V=150.0 mL

M=0.2105 M

CACO3

Mass= 1.3415 g

NAOH

M=0.1055 M

V=75.5 mL 

2. Write balanced equation


 2HNO3  +  CaCO3  ®  Ca(NO3)2  +  CO2  +  H2O  ------ 1

HNO3  +  NaOH     ®    NaNO3  +   H2O  ------- 2

2 mole of HNO3  react with 1 mole of CaCO3

1 mole of HNO3  react with 1 mole of NaOH


3. Calculate no of mole

Initial amount HNO3:

No of mole of acid = 0.2105 x 150 

                              = 31.575 mole acid.

Excess acid

No of mole of  excess acid = 0.1055 x 75.5

                                            = 7.965 mmole acid

mole of acid reacted with CaCO3 =  ( 31.575 – 7.965 )

                                                      = 23.61 mole acid

4. Mole ratio

2 mole of HNO3  react with 1 mole of CaCO3

Thus,23.61 mole of HNO3  react with ½(23.61)  mole of CaCO3

mole of CaCO3  =  ½ x mole acid

                           ½ x 23.61

                           = 11.805 mole CaCO3.

5. Find mass

Gram CaCO3 = mole x molar mass

                           =  11.805 x 10-3 x 100

                           =  1.1805 g.

6.Find percentage

% CaCO3 =                             Ï 100

                 

               =                    Ï 100

               = 87.99 % (w/w)


REFERENCES

 http://en.wikipedia.org/wiki/Titration#Back_titration

http://ashomarconfidential.files.wordpress.com/2011/11/back-titration.pptx

Red-ox Reaction

By: Reagan Delos Reyes


Redox reactions(oxidation-reduction reactions)- have a number of similarities to acid-base reactions. Fundamentally, redox reactions are a family of reactions that are concerned with the transfer of electrons between species. Like acid-base reactions, redox reactions are a matched set -- you don't have an oxidation reaction without a reduction reaction happening at the same time. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. Each reaction by itself is called a "half-reaction", simply because we need two (2) half-reactions to form a whole reaction. In notating redox reactions, chemists typically write out the electrons explicitly:

Cu (s) ----> Cu2+ + 2 e-

This half-reaction says that we have solid copper (with no charge) being oxidized (losing electrons) to form a copper ion with a plus 2 charge. Notice that, like the stoichiometrynotation, we have a "balance" between both sides of the reaction. We have one (1) copper atom on both sides, and the charges balance as well. The symbol "e-" represents a free electron with a negative charge that can now go out and reduce some other species, such as in the half-reaction:

2 Ag+ (aq) + 2 e- ------> 2 Ag (s)

Here, two silver ions (silver with a positive charge) are being reduced through the addition of two (2) electrons to form solid silver. The abbreviations "aq" and "s" mean aqueous and solid, respectively. We can now combine the two (2) half-reactions to form a redox equation:


As a summary, here are the steps to follow to balance a redox equation in acidic medium (add the starred step in a basic medium):

1.     Divide the equation into an oxidation half-reaction and a reduction half-reaction

2.     Balance these

o    Balance the elements other than H and O

o    Balance the O by adding H2O

o    Balance the H by adding H+

o    Balance the charge by adding e-

3.     Multiply each half-reaction by an integer such that the number of e- lost in one equals the number gained in the other

4.     Combine the half-reactions and cancel

5.     **Add OH- to each side until all H+ is gone and then cancel again**



Back Tirtation

By: Kathleen Caralde

What is indirect titration?

This method employs a preliminary reaction in which the analyte is replaced by an equivalent amount of another substance which is then determined by titration. The titrant and analyte do not react with each other, but are related through the other substance.

e.g. Determination of iron

Fe+3+2I-2Fe+2+12 (analyte)

I2+2S2O3-2I+S4O6-2 (titrant)


It is generally a two-stage analytical technique:

a. Reactant A of unknown concentration is reacted with excess reactant B of known concentration. b. A titration is then performed to determine the amount of reactant B in excess.

Or scientifically speaking..

Preliminary reaction: aA + rR à pP

Titration reaction: pP + tT à fF

Imdirect titrations are used when:

* one of the reactants is volatile, for example ammonia.

* an acid or a base is an insoluble salt, for example calcium carbonate

* a particular reaction is too slow * direct titration would involve a weak acid - weak base titration (the end-point of this type of direct titration is very difficult to observe)

Example :

Indirect Titration to Determine the Concentration of a Volatile Substance

A student was asked to determine the concentration of ammonia, a volatile substance, in a commercially available cloudy ammonia solution used for cleaning.

First the student pipetted 25.00mL of thecloudy ammonia solution into a 250.0mL conical flask. 50.00mL of 0.100M HCl(aq) was immediately added to the conical flask which reacted with the ammonia in solution.

The excess (unreacted) HCl was then titrated with 0.050M Na2CO3(aq). 21.50mL of Na2CO3(aq) was required.

Calculate the concentration of the ammonia in the cloudy ammonia solution.


Step 1: Determine the amount of HCl in excess from the titration results

a. Write the equation for the titration:

2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l)

acid + carbonate → salt + carbon dioxide + water

b. Calculate the moles, n, of Na2CO3(aq) that reacted in the titration: n = M x V

c. M = 0.050 molL-1

d. V = 21.50mL = 21.50 x 10-3L

n(Na2CO3(aq)) = 0.050 mol/L x 21.50 x 10-3 L = 1.075 x 10-3 mol

e. Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration.

From the balanced chemical equation, 1 mole Na2CO3 react with 2 moles of HCl

So, 1.075 x 10-3 mol Na2CO3 reacted with 2 x 1.075 x 10-3 moles HCl

n(HCltitrated) = 1.075 x 10-3 mol HCL x 2 mol Na2CO3 1mol HCl = 2.150 x 10-3 mol

f. The amount of HCl that was added to the cloudy ammonia solution in excess was 2.150 x 10-3 mol

Step 2: Determine the amount of ammonia in the cloudy ammonia solution

a. Calculate the total moles of HCl originally added to the diluted cloudy ammonia solution: n(HCltotal added) = M x V M = 0.100 

molL V = 50.00mL = 50.00 x 10-3L

n(HCltotal added) = 0.100 molL x 50.00 x 10-3 L = 5.00 x 10-3 mol


b. Calculate the moles of HCl that reacted with the ammonia in the diluted cloudy ammonia solution n(HCltitrated) + n(HClreacted with ammonia) = n(HCltotal added) n(HCltotal added) = 5.00 x 10-3 mol n(HCltitrated) = 2.150 x 10-3 mol

2.150 x 10-3 + n(HClreacted with ammonia) = 5.00 x 10-3

n(HClreacted with ammonia) = 5.00 x 10-3 - 2.150 x 10-3 = 2.85 x 10-3 mol

c. Write the balanced chemical equation for the reaction between ammonia in the cloudy ammonia solution and the HCl(aq).

NH3(aq) + HCl(aq) → NH4Cl(aq)

d. From the balanced chemical equation, calculate the moles of NH3 that reacted with HCl. From the equation, 1 mol HCl reacts with 1 mol NH3 So, 2.85 x 10-3 mol HCl had reacted with 2.85 x 10-3 mol NH3 in the cloudy ammonia solution.

e. Calculate the ammonia concentration in the cloudy ammonia solution. M = n ÷ V

n = 2.85 x 10-3 mol (moles of NH3 that reacted with HCl)

V = 25.00mL = 25.00 x 10-3L (volume of ammonia solution that reacted with HCl)

M = 2.85 x 10-3 ÷ 25.00 x 10-3 = 0.114 M

f. The concentration of ammonia in the cloudy ammonia solution was 0.114M.


Half Reaction

(Review)

By: Cristel Diane Dela Cruz

Half-reaction is the part of an overall reaction that represents, separately, either an oxidation or a reduction.

- a half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.

-a half reaction does not occur by itself, at least two such reactions must be coupled so that the electron released by one reactant is accepted by another in order to complete the reaction.

-it is a method for balancing oxidation/reduction reactions that occur in aqueous solution

-one of the half reaction will illustrate reduction that means the electrons will show up in the reactant side  A + e- -> A-

-one of the half reaction will illustrate oxydation that means the electrons will show up in the product side B-> B+  + e-

Remember :  The number of electrons lost by the oxidized species must be equal to the number of electrons gained by the reduced species.

Example :

When a nickel strip {Ni (s)} is placed in an aqueous solution of copper(II) sulfate {Cu2+, SO42-}, an immediate reaction occurs. Copper metal begins to deposit on the strip. The only source for metallic copper in this system is the copper(II) ions in solution. What is happening to the copper(II) ions to cause them to change into elemental copper? Let's represent this fascinating observation with a chemical equation:


We have just written a half-reaction! In this equation, the copper(II) ion is being reduced  When we balance a half-reaction, we first balance the mass of the participating species (atoms, ions, or molecules) and then the charge. In this case, the mass is balanced by adding a copper (atom or ion) to each side. To balance the charge, electrons are added. Notice the addition of 2 electrons to left side of the above equation. These electrons are necessary to reduce a copper(II) ion to metallic copper.

If copper(II) ion is being reduced, what is being oxidized? Another way to ask this question is 'where are those electrons coming from?'. Nickel, of course! What is the half-reaction for the oxidation of metallic nickel? Nickel must release 2 electrons to form the nickel(II) ion as shown in the following equation (remember that we balance the mass and then the charge):



REFERENCES:

http://www.chem.wisc.edu/deptfiles/genchem/netorial/ROttosen/tutorial/modules/electrochemistry/02half_reactions/18_21.htm

http://www.youtube.com/watch?v=zKT7QZbto34

SECOND GRADING

 

Unfollowing the Octet Rule of 8

By: Cristel Imbag

Incomplete octet

An atom with less than eight electrons in its valence shell. An atom with less than eight total bonding and nonbonding electrons in a Lewis structure, for example, B in BH3 has an incomplete octet.

There are certain atoms of certain elements that can exist in stable compounds forming bonds with less than eight valence electrons.  When this occurs, the atom of the element within the molecule is said to contain an incomplete octet.  The common examples of such elements are hydrogen (stable with only 2 valence electrons), beryllium (stable with only 4 valence electrons) and boron and aluminum (stable with only 6 valence electrons).   For hydrogen 2 valence electrons give it a noble gas structure (like He) so this is much like the octet rule for everything else below period 1.   But covalent compounds in groups 2 and 3 can form stable compounds in which the valence electrons are not in the noble gas structure.   However, for these compounds you will find that they do form compounds that minimize formal charge.

Example:

 Good examples of the first type of exception are provided by BeCl2 and BCl3. Beryllium dichloride, BeCl2, is a covalent rather than an ionic substance. Solid BeCl2 has a relatively Complex structure at room temperature, but when it is heated to 750°C, a vapor which consists of separate BeCl2 molecules is obtained. Since Cl atoms do not readily form multiple bonds, we expect the Be atom to be joined to each Cl atom by a single bond. The structure is

Instead of an octet the valence shell of Be contains only two electron pairs. Similar arguments can apply to boron trichloride, BCl3, which is a stable gas at room temperature. We are forced to write its structure as

Expanded octet

A hypervalent molecule (the phenomenon is sometimes colloquially known as expanded octet) is a molecule that contains one or more main group elements formally bearing more than eight electrons in their valence shells. Phosphorus pentachloride (PCl5), sulfur hexafluoride (SF6), chlorine trifluoride (ClF3), and the triiodide (I3) ion are examples of hypervalent molecules.

Examples :

      SF4

1. This compound is covalent.

2. Determine the total number of valence electrons available:

One sulfur has 6 valence electrons
Four fluorine, each with 7 valence electron, totals 28
This means there are 34 valence electrons, making 17 pairs, available.

3. Organize the atoms so there is a central atom (usually the least electronegative) surrounded by ligand (outer) atoms. Hydrogen is never the central atom.

I put the fluorines like I did because I knew I needed an open space for the unbonded pair on the sulfur. If you put the 4 fluorines around the S like in CH4, that's OK. It gets tiresome making all this little graphic files, but hey, I knew that when I took on this project. OK, enough complaining, back to work.

4. Determine a provisional electron distribution by arranging the electron pairs (E.P.) until all available pairs have been distributed:

 

5. The formal charge (F) on the central atom is zero. The right-hand structure in step 4 is the correct answer.

Odd molecule

Term invented by Gilbert N. Lewis in 1916 for a molecule containing an odd number of electrons. Taking the p-shell elements, such molecules are rare; they are usually colored and paramagnetic, that is, attracted by a magnet. Odd molecules are 'radicals.' A fine example is nitric oxide, q.v.; nitrogen dioxide is another; chlorine dioxide is also an example, being a reddish-yellow gas. They are all fairly reactive. When including d-shell elements, i.e., the transition metals, the concept mostly doesn't apply, and this 'odd' state is not so unusual.

Example:

1.  Nitrogen dioxide (NO2)

This molecule has a total of 17 electrons to place - five from the nitrogen and 12 from the oxygens. I will go immediately to the final structure:


Notice that I show both resonance structures. Since there are three electron domains, this is a trigonal planar arrangement, but it is signified AX2e, to signal the single electron domain, also called a half-filled orbital.

The bond angles are not 120°, since the repulsive power of the single electron is less tha if there were two. So, the O-N-O bond angle moves outward to 134.3°. Adding another electron to make NO2¯ (which creates a full non-bonding electron pair, changes the O-N-O angle to 115.4° and removing an electron (to make NO2+) creates an O-N-O bond angle of 180°.

References:

       -Babylon dictionary

      -  http://dbhs.wvusd.k12.ca.us/ Copyright © 1998 by John L. Park



Geometry Molecules

By: Alleta Fae S. Liwag

Geometry of Molecules Chart
Number of Electron Groups Electron-Group Geometry Number of Lone Pairs VSEPR Notation Molecular Geometry Ideal Bond Angles Examples
2  linear 1 AX2 linear.jpg  180°  BeH2
3  trigonal-planar 0 AX3 trigonal planer.jpg  120°  CO32-
1 AX2E bent.jpg  120° O3
4  tetrahedral 0 AX4

animated-CCl4.gif

   Tetrahedral

 109.5°  S042-
1 AX3E  trigonal pyramidal.jpg  109.5°  H3O+
2 AX2E2 bent2.jpg  109.5°  H2O
5  trigonal-bipyramidal 0  AX5 trigonal bipyramidal.jpg  90°, 120°  PF5
1  AX4Eb seesaw.jpg  90°, 120°  TeCl4
2  AX3E2 t-shaped.jpg  90° ClF3
3  AX2E3 linear2.jpg  180° I3-
6  octahedral 0  AX6

animate-SF6.gif

octahedral

 90°  PF6-
1  AX5E square pyramidal.jpg  90° SbCl52-
2  AX4E2  square planer.jpg  90°

ICl4-

Steps Used to Find the Shape of the Molecule

To sum up there are four simple steps to apply the VSEPR theory.

1. Draw the Lewis Structure.

2. Count the number of electron groups and identify them as bond pairs of electron groups or lone pairs of electrons. Remember electron groups include not only bonds, but also lone pairs!

3. Name the electron-group geometry. (State whether it is linear, trigonal-planar, tetrahedral, trigonal-bipyramidal, or octahedral.)

4. Looking at the positions of other atomic nuclei around the central determine the molecular geometry. (See how many lone pairs there are.)

(Use the chart as a guide)

Reference:

http://chemwiki.ucdavis.edu/Wikitexts/UCD_Chem_2A/ChemWiki_Module_Topics/Chemical_Bonding/Geometry_of_Molecules


orbital hybridization

By: cristel diane dela cruz

In chemistry hybridization  is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for the qualitative description of atomic bonding properties. Hybridised orbitals are very useful in the explanation of the shape of molecular orbitals for molecules. It is an integral part of valence bond theory. The hybrids are named based on the atomic orbitals that are involved in the hybridization, and the geometries of the hybrids are also reflective of those of the atomic-orbital contributors

 Chemist Linus Pauling first developed the hybridization theory in order to explain the structure of molecules. This concept was developed for such simple chemical systems, but the approach was later applied more widely, and today it is considered an effective heuristic for rationalizing the structures of organic compounds.


EXAMPLES :

Methane, the three 2p orbitals of the carbon atom are combined with its 2s orbital to form four new orbitals called "sp3" hybrid orbitals. The name is simply a tally of all the orbitals that were blended together to form these new hybrid orbitals. Four hybrid orbitals were required since there are four atoms attached to the central carbon atom. These new orbitals will have an energy slightly above the 2s orbital and below the 2p orbitals as shown in the following illustration. Notice that no change occurred with the 1s orbital.


Ammonia, the three 2p orbitals of the nitrogen atom are combined with the 2s orbital to form four sp3 hybrid orbitals. The non-bonded electron pair will occupy a hybrid orbital. Again we need a hybrid orbital for each atom and pair of non-bonding electrons. Ammonia has three hydrogen atoms and one non-bonded pair of electrons when we draw the electron-dot formula. In order to determine the hybridization of an atom, you must first draw the electron-dot formula.


homogenous equilibrium

By: Mark Anthony Basilio

------The equilibrium between different chemical species present in the same or different phases is called chemical equilibrium. There are two types of chemical equilibrium.

The equilibrium reactions in which all the reactants and the products are in the same phase are called homogeneous equilibrium reactions.

 

C2H5OH(l) + CH3COOH(l) Description: http://www.transtutors.com/userfiles/image/ARUN/Fold/equal%202.JPG  CH3COOC2H5(l) + H2O(l)

N2(g) + 3H2(g) Description: http://www.transtutors.com/userfiles/image/ARUN/Fold/equal%202.JPG 2NH3(g)

2SO2(g) + O2(g) Description: http://www.transtutors.com/userfiles/image/ARUN/Fold/equal%202.JPG  2SO3(g)

The equilibrium system, ethanol-acetic acid-ethyl acetate-water, is well known. The equation for this equilibrium is: C2H5OH + CH3COOH <-----> CH3COOC2H5 + H2O

Equilibrium is established very slowly at room temperature when these four substances are mixed. The rate of reactions (both forward and reverse) can be accelerated by using a catalyst so that the equilibrium can be more rapidly established. 3 M HCl is used in this experiment as a catalyst, and the equilibrium at room temperature should be established in a few days. This means that in order to successfully complete this experiment, a series of combinations of reactants and/or products must be prepared a few days before the actual experiment. The equilibrium constant is then calculated by the equation:

Using aA<----->bB

Kc=[B]a/[A]b           

We derive this     Kp=PbB/PaB to get Kp=Kc(RT)N

Kc = [pure ethyl acetate] [water] / [ethanol] [glacial acetic acid]

The four quantities in the above equation may be replaced by the number of moles of the four substances involved since the volume is constant in each mixture.

Using aA<----->bB

Kc=[B]a/[A]b           

We derive this formula     Kp=PbB/PaB     to get    Kp=Kc(RT)N

Gas Equilibrium Constants, Kc And Kp


Kc and Kp are the equilibrium constants of gaseous mixtures.

—     -The Equilibrium Constant, Kc, relates to a chemical reaction at equilibrium. It can be calculated if the equilibrium concentration of each reactant and product in a reaction at equilibrium is known.
—Kc=molar concentration of products/molar concentration of reactants

  with each concentration raised to the power of the corresponding stoichiometric coefficient.

—-If we express the equilibrium constant in terms of partial pressures instead of molar concentrations, the result is still a constant. The equilibrium constant in terms of partial pressures is denoted as Kp.


THIS ARE SOME WAYS TO WRITE THE GAS EQUILIBRIUM CONSTANT


1. In equilibrium equations, even though the arrows point both ways () we usually associate the right as reactants and the left as products.

2. The products are on the TOP of the fraction (the numerator).
3. The reactants are on the BOTTOM of the fraction (the denominator).
4. The concentrations of the products and reactants are always raised to the power of their coefficient in the balanced chemical equation.
5. If any of the reactants or products are solids or liquids, their concentrations are equal to one because they are pure substances.






SAMPLE PROBLEMS WITH ANSWERS A


1.) NH4SH (s) NH3 (g) + H2S (g)


Kc =

but since NH4SH is a solid, we get:


Kc =

         Kc =

As for Kp, it is the same as Kc, but instead of brackets [  ], Kp uses parentheses ( ):

Kp =

Kp =

Kp =

2.) H2 (g) + I2 (g) 2HI (g)

Kc =

Kp =


REFERENCE
http://chemwiki.ucdavis.edu/Physical_Chemistry/Chemical_Equilibrium/The_Equilibrium_Constant/Calculating_An_Equilibrium_Concentration_From_An_Equilibrium_Constant/Writing_Equilibrium_Constant_Expressions_Involving_Gases/Gas_Equilibrium_Constants,_Kc_And_Kp#Problems



lewis acid and bases

By: cristel diane dela cruz

 

The Lewis definitions of acids and bases are even more inclusive than the Bronsted definitions. The Lewis definitions are that an acid is an electron-pair acceptor and a base is an electron-pair donor.

 

The term Lewis acid refers to a definition of acid published by Gilbert N. Lewis in 1923, specifically: An acid substance is one which can employ an electron lone pair from another molecule in completing the stable group of one of its own atoms. The term Lewis acid refers to a definition of acid published by Gilbert N. Lewis in 1923, specifically: An acid substance is one which can employ an electron lone pair from another molecule in completing the stable group of one of its own atoms.

The modern-day definition of Lewis acid, as given by IUPAC is a molecular entity (and the corresponding chemical species) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base.[2] This definition is both more general and more specific—the electron pair need not be a lone pair (it could be the pair of electrons in a π bond, for example), but the reaction should give an adduct (and not just be a displacement reaction).

A Lewis base, then, is any species that donates a pair of electrons to a Lewis acid to form a Lewis adduct. 

 


The Lewis acid-base theory can also be used to explain why nonmetal oxides such as CO2 dissolve in water to form acids, such as carbonic acid H2CO3.

CO2(g) + H2O(l) Description: <-----> H2CO3(aq)

In the course of this reaction, the water molecule acts as an electron-pair donor, or Lewis base. The electron-pair acceptor is the carbon atom in CO2. When the carbon atom picks up a pair of electrons from the water molecule, it no longer needs to form double bonds with both of the other oxygen atoms as shown in the figure below


One of the oxygen atoms in the intermediate formed when water is added to CO2 carries a positive charge; another carries a negative charge. After an H+ ion has been transferred from one of these oxygen atoms to the other, all of the oxygen atoms in the compound are electrically neutral. The net result of the reaction between CO2 and water is therefore carbonic acid, H2CO3.


bronsted-lowry acid-base theory

By: zhenna Marriz Aypa

In this section we will consider the Brønsted-Lowry concept. This concept focuses on what an acid or base does.In chemistry, the Brønsted–Lowry theory is an acid-base theory, proposed independently by Johannes Nicolaus Brønsted and Thomas Martin Lowry in 1923. The Bronsted-Lowery concept of acids and bases is that acid-base reactions can be seen as proton-transfer reactions. This results in acids and bases being able to be defined in terms of this proton (H+) transfer.  In this system,Brønsted acids and Brønsted bases are defined, by which an acid is a molecule or ion that is able to lose, or "donate," a hydrogen cation (proton, H+), and a base is a species with the ability to gain, or "accept," a hydrogen cation (proton).

An Acid is a Proton Donor and a Base is a Proton Acceptor

So what does this mean? For a reaction to be in equilibrium a transfer of electrons needs to occur. The acid will give an electron away and the base will receive the electron. Acids and Bases that work together in this fashion are called a conjugate pair made up of conjugate acids and conjugate bases.

So what does this look like? (A stands for an Acidic compound and Z stands for a Basic compound)

HA + Z  A- + HZ+

A Donates H to form HZ+.

Z Accepts H from A which forms HZ+

A- becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibrium

HZ+ becomes a conjugate acid of Z and in the reverse reaction it donates a H to A- recreating Z in order to remain in equilibrium

Properties of acids and bases

It follows that, if a compound is to behave as an acid, donating a proton, there must be a base to accept the proton. So the Brønsted–Lowry concept can be defined by the reaction:

acid + base   conjugate base + conjugate acid.

The conjugate base is the ion or molecule remaining after the acid has lost a proton, and the conjugate acid is the species created when the base accepts the proton. The reaction can proceed in either forward or backward direction; in each case, the acid donates a proton to the base.

Example:

H3O+(aq) + Cl-(aq) + NH3(aq) -> H2O(l) + NH4+(aq) + Cl-(aq)


reaction of hydrochloric acid with ammonia

which reduces to:

H3O+(aq) + NH3(aq) -> H2O(l) + NH4+(aq)


reduced reaction of hydrochloric acid with ammonia

because there is a Cl-(aq), on each side. We now have the net ionic equation after we cancel out the "spectator ions"(Cl-).

What happens in this reaction in aqueous solution is that a proton is transferred from H3O+ to NH3. This results in H3O+ losing a (H+), resulting in H2O. The NH3 gains the transferred proton, resulting in NH4+. We call H3O+ the proton donor, or acid. We call NH3 the proton acceptor, or base

The Bronsted-Lowery concept defines something as either an acid or base depending on its function in the acid-base (proton transfer) reaction. Some things can act as either an acid or a base. These are called amphiprotic species, they can either lose or gain a proton, depending on the other reactant. An example of an amphiprotic species would be:

HCO3-.


example of an amphiprotic species
In the presence of OH-, it acts as an acid. In the presence of HF it acts as a base. Water is also amphiprotic, as are most anions with ionizable hydrogens and certain solvents. Water as an amphiprotic species is very important to the acid-base reactions.

In the Bronsted-Lowery concept we have found that:

A base is a species that accepts protons, while an acid is a species that donates protons.

Acids and bases can be ions as well as molecular substances.

Some species can act as either acids or bases, depending on what the other reactant is.


 

 


strengths Of acids and bases

By: kathleen caralde

What Makes a Strong Acid or Strong Base?

Strong electrolytes are completely dissociated into ions in water. The acid or base molecule does not exist in aqueous solution, only ions. Weak electrolytes are incompletely dissociated.

Strong Acids

Strong acids are strong electrolytes that, for practical purposes, are assumed to ionize completely in water. Strong acids completely dissociate in water, forming H+ and an anion. There are six strong acids. The others are considered to be weak acids. Most of them are inorganic acids. You should commit the strong acids to memory:

·         HCl - hydrochloric acid

·         HNO3 - nitric acid

·         H2SO4 - sulfuric acid

NOTE: this is a diprotic; we show only the first stage of ionization here.

·         HBr - hydrobromic acid

·         HI - hydroiodic acid

·         HClO4 - perchloric acid

Weak Acids

A weak acid only partially dissociates in water to give H+ and the anion. Examples of weak acids include hydrofluoric acid, HF, and acetic acid, CH3COOH. Weak acids include:

·         Molecules that contain an ionizable proton. A molecule wih a formula starting with H usually is an acid.

·         Organic acids containing one or more carboxyl group, -COOH. The H is ionizable.

·         Anions with an ionizable proton. (e.g., HSO4- → H+ + SO42-)

·         Cations                                           

·         transition metal cations

·         heavy metal cations with high charge

·         NH4+ dissociates into NH3 + H+

Strong Bases

Strong bases are all strong electrolytes that ionizes completely in water. It dissociate 100% into the cation and OH- (hydroxide ion). The hydroxides of the Group I and Group II metals usually are considered to be strong bases.

·         LiOH - lithium hydroxide                                        CsOH - cesium hydroxide

·         NaOH - sodium hydroxide                                      *Sr(OH)2 - strontium hydroxide

·         KOH - potassium hydroxide                                               Ba(OH)2 - barium hydroxide  

·         RbOH - rubidium hydroxide                                               *Ca(OH)2 - calcium hydroxide

 

* These bases completely dissociate in solutions of 0.01 M or less. The other bases make solutions of 1.0 M and are 100% dissociated at that concentration. There are other strong bases than those listed, but they are not often encountered.

Weak Bases

Weak bases, like weak acids, are weak electrolytes. Examples of weak bases include ammonia, NH3  (note: ammonia does not ionizelike an acid because it does not split up to form ions the way, say, HCL does), and diethylamine, (CH3CH2)2NH.

·         Most weak bases are anions of weak acids.

·         Weak bases do not furnish OH- ions by dissociation. Instead, they react with water to generate OH- ions.

 

 

TABLE1 lists some important conjugate acid-base pairs, in order of their relative strengths. Conjugate acid-base have the following properties:

 

1.      If an acid is strong, its conjugate acid-base has no measurable strength. Thus, the  ion, which is the conjugate base of the strong acid HCl, is an extremely weak base.

2.         is the strongest acid that can exist in aqueous solution. Acids stronger than  react with water to produce  and their conjugate bases. Thus, HCl, which is a stronger acid than , reacts with water completely to form  and :

HCl(aq) +            (aq) + 

 

Acids weaker than  react with water to a much smaller extent, producing  and their conjugate bases. For example, the following equilibrium lies primarily to the left:

HF(aq)  +             (aq)  + 

    

3.      The  ion is the strongest base that can exist in aqueous solution. Bases stronger than  react  with water to produce and their conjugate acids. For example, the oxide ion () is a stronger base than  , so it reacts with water completely as follows:

   (aq) +           

For this reason the oxide ion does not exist in aqueous solutions.

EXAMPLE:

Calculate the pH of (a) a  M HCl olution and (b) a 0.020 M  solution.

STRATEGY:

 Keep in mind that HCl is a strong acid and  is a strong base. Thus, these species are completely ionized and no HCl and  will be left in solution.


FIRST GRADING

The Mole Concept

                                                         by: Zhenna Marriz Aypa                                                                     

In chemistry the mole is a fundamental unit in the Système International d'Unités, the SI system, and it is used to measure the amount of substance. This quantity is sometimes referred to as the chemical entity, which can be an atom, molecule, formula unit, electron or proton. In Latin mole means a "massive heap" of material. It is convenient to think of a chemical mole as such.


When a quantity of particles is to be described, mole is a grouping unit analogous to groupings such as pair, dozen, or gross, in that all of these words represent specific numbers of objects. The main differences between the mole and the other grouping units are the magnitude of the number represented and how that number is obtained. One mole is an amount of substance containing Avogadro's number of particles. Avogadro's number is equal to 602,214,199,000,000,000,000,000 or more simply, 6.02214199 × 10 23 .

1  mole C  à  6.02 x 1023 atoms
         1 mole H2O
à 6.02 x 1023 molecules

1 mole NaCl à 6.02 x 1023 molecules
       

Atoms and molecules are incredibly small and even a tiny chemical sample contains an unimaginable number of them. Therefore, counting the number of atoms or molecules in a sample is impossible. The multiple interpretations of the mole allow us to bridge the gap between the submicroscopic world of atoms and molecules and the macroscopic world that we can observe

To determine the chemical amount of a sample, we use the substance's molar mass, the mass per mole of particles.

General Plan for Converting Mass, Amount, and Numbers of ParticlesSample Problem 1:

How many moles of Magnesium are 3.01 x 1022 atoms of magnesium?

   Solution:

Analyze:

Given: 3.01 x 1022 atoms of Mg

Mole of Mg= ( ?)

Compute:

3.01 x 1022 atoms of Mg          x                         1 mol of Mg          

                                                          6.02 x 1023 atoms of Mg


= 0. 500 Mol Mg

 Sample Problem No. 2

How many moles are there in 29.0 grams of Sodium Chloride?

 Analyze:

29.0 g  NaCl à mole(?)          (MM): MM = 22.99 + 35.45 = 58.44 g NaCl

Compute:

29.0 g Nacl x    1 mol NaCl    

                  58.44 g NaCl

 = 0.496 mol NaCl

 Sample Problem No. 3

How many grams are in 7.20 mol of  Dinitrogen Trioxide?

  Given:

7.20 mol N2O3  à g(?)

(MM): N2(14.01 x 2) + O3(16x 3) = 76.02 mol of N2O3

Compute:

7.20 mol of N2O3   x                   1 mol                   

                                      76.02 mol of N2O3

 = 0.0947 mol N2O3

 

Stoichiometry

by: Jarlyn Liwag

" Stoichiometry "  was derived from: "Stoicheion" meaning  element and "Metron" meaning measure.

 ·         deals with the relative quantities of Reactants and Products in chemical reactions. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of whole numbers.

 

·          used to find the right amount of reactants to use in a chemical reaction.

Example :

            *Consider the complete combustion of natural gas Methane ( CH4 ) reacts with       Oxygen gas ( O2 ) to produce CO2 and water ( H2O ).

 CH4 + 2O2 → CO2 + 2H2O

*Hydrogen gas (H2) reacts with Nitrogen gas (N2) to give ammonia (NH3)

3H2+ N 2 2NH3

 

·         often used to balance chemical equations .

 

            *For example, the two diatomic gases, hydrogen and oxygen, can combine to          form a liquid, water (H2O).

2H2 + O2 2H2O

- describes the 2:1:2 ratio of hydrogen, oxygen, and water molecules in the above equation.

 

 

 

·         is simply the study of the relationships or ratios between two or more substances undergoing a physical or chemical change (chemical reaction).

  In Stoichiometry, we have different PROBLEMS.

 

  • Mass-Mass Problems.

·                       Mole-Mole Problems.

·                        Mass-Mole Problems.

·                Volume-Volume Problems.

 

Limiting and Excess Reagent

By: Alleta Fae S. Liwag

·         Limiting Reactant - The reactant in a chemical reaction that limits the amount of product that can be formed.  The reaction will stop when all of the limiting reactant is consumed.

·         The limiting reagent is the reactant that is completely used up during the chemical reaction.

·         Excess Reactant - The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed.  The excess reactant remains because there is nothing with which it can react.


No matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made.  Likewise with chemistry, if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up.

Example Limiting Reactant Calculation:

A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen.  Which is the limiting reactant and how much excess reactant remains after the reaction has stopped?

First, we need to create a balanced equation for the reaction:


4 NH3(g) + 5 O2(g) ----> 4 NO(g) + 6 H2O(g)

Next we can use stoichiometry to calculate how much product is produced by each reactant.  NOTE:  It does not matter which product is chosen, but the same product must be used for both reactants so that the amounts can be compared.


The reactant that produces the lesser amount of product in this case is the oxygen, which is thus the "limiting reactant."

Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant (ammonia) actually did react with the limiting reactant (oxygen).


We're not finished yet though.  1.70 g is the amount of ammonia that reacted, not what is left over.  To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample.


Reference: http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm

 

Mass - Mass Stoichiometry

by: Cristel Diane Dela Cruz

          A mass to mass problem is one in which the mass of a reactant or product is given. You are then asked to calculate the mass of another reactant required or the mass of another product formed.

There are four steps involved in solving these problems:

1.        Make sure you are working with a properly balanced equation.

2.      Convert grams of the substance given in the problem to moles.

3.     Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem.

4. Convert moles of the substance just solved for into grams.

Example : Follow the steps in the diagram

Under severe conditions; high temperature, pressure and concentration; hydrogen gas and nitrogen gas can be coerced into forming ammonia, NH3, Also a gas at these temperatures.

What mass of N2 would be required if mixed with excess hydrogen gas (meaning more than enough - the reaction will cease when the nitrogen runs out) to produce 850g ammonia? What mass of H2 would be used?


           

Mass - Volume Stoichiometry

  By: Rosiel Mariano

There are four steps involved in solving mass-volume problems:

  •  Make sure you are working with a properly balanced equation.
  •  Convert grams of the substance given in the problem to moles.
  •  Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem.
  •  Convert moles of the substance just solved for into Volume.

FORMULA:

Given mass of substance A

Desired volume of substance B

Molar mass of substance A

Volume at STP of substance B

At STP, 1 mol of gas = 22.4 liters

 

 Sample problems:

1.)     An excess of hydrogen reacts with 14.0 g of nitrogen.  How many liters of ammonia are produced at STP?

3 H2    +     N2 2 NH3

14.0 g N2

X L NH3

22.4 L NH3

28.0 gN2

2(22.4 L) NH3

2.)     What volume of hydrogen at STP can be produced when 6.54 g of Zn reacts with hydrochloric acid, HCl?

Zn   +    2 HCl  →  H2    +    ZnCl2

6.54 g N2

X L H2

2.24 L H2

65.4  g  N2

22.4  L  NH3


Gas Stoichiometry

by: Cristel Imbag

          Stoichiometry  is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of whole numbers. For example, in a reaction that forms ammonia (NH3), exactly one molecule of nitrogen (N2) reacts with three molecules of hydrogen (H2) to produce two molecules of NH3:

N2 + 3H2 2NH3

 

          Gas stoichiometry deals with reactions involving gases, where the gases are at a known temperature, pressure, and volume, and can be assumed to be ideal gases. For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, due to the existence of isotopes, molar masses are used instead when calculating the mass ratio.

          Gas stoichiometry is the quantitative relationship (ratio) between reactants and products in a chemical reaction with reactions that produce gases. Gas stoichiometry applies when the gases produced are assumed to be ideal, and the temperature, pressure, and volume of the gases are all known. The ideal gas law is used for these calculations. Often, but not always, the standard temperature and pressure (STP) are taken as 0 °C and 1 bar and used as the conditions for gas stoichiometric calculations.

          Gas stoichiometry calculations solve for the unknown volume or mass of a gaseous product or reactant. For example, if we wanted to calculate the volume of gaseous NO2 produced from the combustion of 100 g of NH3, by the reaction:

4NH3 (g) + 7O2 (g) 4NO2 (g) + 6H2O (l)

we would carry out the following calculations:

Description:  100 \ \mbox{g}\,NH_3 \cdot \frac{1 \ \mbox{mol}\,NH_3}{17.034 \ \mbox{g}\,NH_3} = 5.871 \ \mbox{mol}\,NH_3\

There is a 1:1 molar ratio of NH3 to NO2 in the above balanced combustion reaction, so 5.871 mol of NO2 will be formed. We will employ the ideal gas law to solve for the volume at 0 °C (273.15 K) and 1 atmosphere using the gas law constant of R = 0.08206 L · atm · K−1 · mol−1 

Description: PV

Description: = nRT

Description: V

Description: = \frac{nRT}{P} = \frac{5.871 \cdot 0.08206 \cdot 273.15}{1} = 131.597 \ \mbox{L}\,NO_2

Gas stoichiometry often involves having to know the molar mass of a gas, given the density of that gas. The ideal gas law can be re-arranged to obtain a relation between the density and the molar mass of an ideal gas:

Description: \rho = \frac{m}{V}    and     Description: n = \frac{m}{M}

and thus:

Description: \rho = \frac {M P}{R\,T}

where:

 

Description: P

= absolute gas pressure

Description: V

= gas volume

Description: n

= number of moles

Description: R

= universal ideal gas law constant

Description: T

= absolute gas temperature

Description: \rho

= gas density at Description: Tand Description: P

Description: m

= mass of gas

Description: M

= molar mass of gas

 

Sample Problems:

1) A quantity of gas at a temperature of 18oC has a volume of 124.5 cm3 at a pressure of  97.8 KPa.  How many moles of gas are there?   PV=nRT

    n= [         mol K     ][97.8 KPa         ][0.1245dm3] = 0.005 mol gas
               [8.31 dm3 KPA][18oC + 273=K]        

2) At a pressure of 105.2 KPa,  1.42  moles of a gas occupy 1.74 dm3.  What is the
     temperature in Kelvin?   PV=nRT

     T = [  mol K     ][105.2 KPa][1.74dm3] = 15.51K
                [8.31 dm3 KPA ][1.42 mol]        

3) A gas at 1.24oC occupies a 3.51 dm3 container.  If there are 3.34 moles of the gas,
     at what pressure is the gas? PV=nRT

 P= [3.34 mol][8.31 dm3 KPA][1.24oC + 273=K] = 2170 KPa
                         [mol K][3.51 dm3]     

4) If 1.39 g of carbon monoxide is reacted with oxygen, what volume of carbon dioxide
    is produced at  12.3oC at 107.4KPa?

                     2CO +  O2 --> 2CO2

     mol CO2 = [1.39g CO][1mol CO][2 mol CO2 ] =  0.05 mol CO2
                                    [28 g CO ][2 mol CO  ]

    V= [0.05 mol][8.31 dm3 KPA][12.3oC + 273=K] = 1.09 dm3
                         [ mol K ][107.4 KPa  ]     

5) If 14.4 dm3 of ethane is combusted at 102.7oC and 99.3KPa, how many grams of
    water will be produced?

                       2C2H6 +  7O2 -->  4CO2 +  6H2O

       n= [         mol K     ][99.3 KPa              ][14.4 dm3] = 0.46 mol C2H6
                   [8.31 dm3 KPa ][102.7oC + 273=K]        

       mass H2O = [0.46 mol C2H6][6 mol H2O ][18 g   H2O]  =  24.8 g H2O
                                            [2 mol C2H6][1 mol H2O]

 

REFERENCES:

Ø  http://goldbook.iupac.org/S06025-plain.html

Ø  IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8. doi:10.1351/goldbook. Entry: "stoichiometric number".

Ø  IUPAC Compendium of Chemical chemistry/chprbgsst.htm Terminology 2nd Edition (1997)

Ø  http://www.sciencebugz.com/

 

 

 Balancing Redox Equations

by: Kathleen Caralde

Redox (reduction-oxidation) reactions include all chemical reactions in which atoms have their oxidation state changed. A reaction in which oxidation and reduction take place simultaneously is called a redox reaction.
So, a non-redox reaction would be a chemical reaction with only one process.

Oxidation number Method

OBJECTIVES:  In this article you will learn to write a balanced equation for a redox reaction using Oxidation number Method.

                          Identify the oxidation numbers of the elements in the equation.

                          State the characteristics of a redox reaction and identify the oxidizing agent and reducing agent.

Before we will get to explanation, very important disclaimer: oxidation numbers don't exist. They were invented to help in charge accounting needed when balancing redox reaction equations, but they don't refer to any real life chemical concept.

The general idea behind the oxidation numbers (ON) method for balancing chemical equations is that electrons are transferred between charged atoms. These charges (the positive or negative number that indicates how many electrons an atom has gained, lost, or shared to become stable) - assigned to individual atoms - are called oxidation numbers, just to remind you that they don't reflect real structure of the reagents.

How do we use oxidation numbers for balancing? First of all, we have to understand that oxidation means increase of oxidation number, while reduction means decrease of oxidation number. In both cases change of oxidation number is due to electrons lost (oxidation) or gained (reduction). We calculate oxidation numbers for all atoms present in the reaction equation (note that it is not that hard as it sounds, as for most atoms oxidation numbers will not change) and we look for a ratio that makes the number of electrons lost equal to the number of electrons gained. That gives us additional information needed for reaction balancing.


        

  CHEMISTRY  IMPORTANT   TERMS

REDOX REACTION -  oxidation and reduction reactions that occurs simultaneously.

    -Process to characterize by increasing or decreasing oxidation number.

-      Losing or gaining of electrons


 


OXIDATION – Process by which a substance loses e-
-
It is characterize by an increase of oxidation number
REDUCTION
- Process by which a substance gains e-
-     
Decrease of oxidation number

REDUCING AGENT- substance that causes another substance to be reduced.

OXIDIZING AGENT- substance that causes another substance to be oxidized.


Reference:

 http://wiki.answers.com/Q/What_is_an_oxidation_number#ixzz216Q2GDYK

http://science.widener.edu/~svanbram/chem146/ch20/redox.html

http://www.docslide.com/oxidation-reduction-redox-reactions-2/

http://www.scribd.com/doc/39095944/Chapter-7-Oxidation-and-Reduction-Reactions


 

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