Ionic Compound

Ionic compounds are basically defined as being compounds where two or more ions are held next to each other by electrical attraction

Ionic Compounds

This is a summary of the past and present nomenclature or naming conventions for ionic compounds

Naming Compounds

An ionic compound is one in which at least two of the elements or compounds in the group are oppositely-charged ions held together

Ionic Compounds

Ionic compounds generally are very hard and have very high melting points. They are solids at room temperature

Face Centered Cubic

When metals react with non-metals they form an ionic compound. Ions have a charge because electrons are lost or gained in forming an ionic bond.

Showing posts with label formulas. Show all posts
Showing posts with label formulas. Show all posts

Tuesday, December 27, 2011

basically defined ionic compounds



Ionic compounds are basically defined in such a way that compounds containing two or more ions are considered alongside electrical attraction. One of the positively charged ions (called "cations") and the second is a negative charge ("anion"). Cations are usually the metal atoms and anions are polyatomic ions or metals (ions with more than one atom). Thinking about school: the same thing that makes the positive and negative ends of the magnet attached to one another, which makes the cations and anions cling to each other.

Usually, when we are ionic compounds, forming large crystals, you can see with the naked eye. Table salt is an example of this - if you look at a salt crystal, it is likely that you will be able to see which looks a bit 'like a cube. This is because the salt as a stack of small cube-shaped.

Sometimes, when you see the salt, dust appears in place of the cube. This does not mean that salt is not a crystal - it means that the crystals are so small that you can not see with the naked eye. If you were to put the dust under a microscope, chances are you've seen a bit 'of geometric blocks.

So what are the important properties of salts? Well, I'm sure glad you asked ...

All ionic compounds form crystals. As far as I know, there are exceptions. Once again, the desire to form salt crystals, because when there is a whole series of small positive and negative electrical charges, all stuck together, seem to be happy bunch of small groups of stacking. The arrangement of these ions as part of a stack is different, and is called "unit cell". There are a dozen different types of cells generic units. When you graduate, you ask them. High school classes, it really is not so important.

Ionic compounds tend to have high melting and boiling points. When I say "high", which I think is "very, very high." Most of the time when working with ionic compounds in a chemistry class, the melting temperature is warm enough that you can not melt them with a Bunsen burner. So why such high temperatures? Well, it has to do with how the ionic materials are held together. Remember how we said earlier that crystals form ionic compounds? These crystals are really just great big blocks of positive and negative charges while being held. To break the positive and negative charges apart, it takes a tremendous amount of energy. This means that if we heat the fabric to add energy, it requires an enormous amount of energy to break.

Ionic compounds are very hard and very fragile. Again, it is because of how they stick together. Above we said that it takes a lot of energy to break the positive and negative charges apart. This is why ionic compounds are so difficult - it simply does not move much, so it does not bend at all. This also explains the fragility of the ionic compounds. It takes lots of energy for ionic charges of others. However, if we give a large crystal hit hard enough with a hammer, usually end up with more energy to break the glass the glass is not broken in one place, but a lot of places. Instead of cutting to the chase, is shattered.

Of electricity when ionic compounds dissolve in water. If we take salt and dissolved in water, the water molecules pull the positive and negative ions from each other. (This is due to the unusual properties of water, but it's a different story a different time). In contrast, the ions adjacent to each other, walking around in water. Now, think about what that power is - I hope you remember that electricity is only the movement of electrons through the metal (or elsewhere). Now, the electrons are the only negatively charged particles, and metals are the property of their own good, allowing them to go around. The salts are dissolved in the same way. When the salt dissolves in water, positive and negative ions in the water with the electrons to flow much better than if just had the water itself. Voila! The salt water does! The question is, maybe it is, "'the flow of electricity through crystals of salt?" Nope. It does not.

Because the ions are trapped in a place due to the crystal structure is electrically not move very well. Another good question: "Is the water without salt in the conduct electricity?" The answer: Not very well. The water itself is a lousy leader. The reason that sets hair in Boneheads the bath with them again in the donut man is that when they wash all the dirt on them will be dissolved in water. Some of ionic impurities, so when the dryer hit the water, they get zapped . A "thought experiment" It would be interesting to wash all the salt for yourself and then drop a hairdryer in the bath with you. In theory, you would be fine. In real life, would you still become a creature crunch because the tap water itself, there are ionic compounds dissolved in anyway.

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Thursday, July 28, 2011

Rules for Writing Formulas for Ionic Compounds



Rules for Writing Formulas for Ionic Compounds

1.     Write the metal symbol first and then the non metal symbol second.
Ca                           I
2.     Write the cation charge above the metal and the anion charge above the non metal
2+                         1-
       Ca                     I
3.     Do the crossover rule and write the charge for the cation as a subscript for the non metal symbol and the charge for the anion as a subscript for the metal.
2+                         1-
                       Ca                    I     à CaI2     ß subscript goes at the                                                                                                                         bottom of the symbol
                                                                        and is smaller.

Rules for Naming Ionic Compounds

First Word – Name of the metal
Second Word – Name of the non-metal end in “ide”
Ex: Calcium Iodide

Rules for Naming Ionic Compounds with Transition Metals
The stock naming system uses roman numerals to indicate the charge of the cation because they can have more than one charge.

First Word – Name of the metal
Roman numeral in brackets to indicate the charge
Second Word – Name of the non-metal end in “ide”

Ex:              FeCl3 is Iron (III) chloride
                   FeO is Iron (II) oxide
                   Cu2S is coper (I) Sulfide
                   PbO2 is lead (IV) oxide
How do I know the charge for iron is 3? I use the reverse crossover rule
                                                3+               1-
Ex:              FeCl3 à                   Fe               Cl à   Iron (III)Cloride

Rules for Naming Ionic Compounds with Transition Metals
The stock naming system uses roman numerals to indicate the charge of the cation because they can have more than one charge.

First Word – Name of the metal
Roman numeral in brackets to indicate the charge
Second Word – Name of the non-metal end in “ide”

Ex:              FeCl3 is Iron (III) chloride

Rules for Naming Ionic Compounds with Transition Metals

The classic naming system uses the latin word with the ending “ous” to indicate the transition metal’s with the lower ion charge and the ending “ic” to indicate the greater ion charge of the transition metal.

For example: Fe is Ferrum and can have a charge of +2 ferrous or +3 ferric.

Element
Latin Name
Ion with lower charge
Ion with Higher Charge
Iron

                                    
Fe2+

Fe3+
Copper
Cuprum 

Cu+

Cu2+
Lead
Plumbum

Pb2+

Pb4+

Formula
Classical System
Stock System
FeCl3
Ferric Chloride

FeO
Ferrous Oxide

Cu2S
Coprous Sulfide

PbO2
Plumbic Oxide


Write the chemical formula for each
a) Copper (I) Oxide
b) Lead (IV) Bromide
c) Iron (III) Sulfide
d) Nickel (III) Flouride
e) Manganese IV) Fluoride

Wednesday, July 27, 2011

Elements, Compounds, Chemical Formula Writing and Naming



Elements, Compounds, Chemical Formula Writing and Naming, Moles, & Concentration

I.                   Elements & Compounds


a.       The universe is made up of energy and matter
b.      Matter is made up of building blocks called elements (~100 or so)
c.       Each element is made of one type of atom
d.      The elements combine together to form compounds
e.       Atoms combine together to form molecules (millions and millions of different kinds)
f.       Do “Elements & Compounds Review & Practice” WS

II.                Chemical formulas – Writing


a.       A chemical formula consists of symbols of the elements and subscripts (small numbers), which indicate the number of each element present in the compound.
                                                              i.      Simple rules in writing Chemical Formulas:
1.      Represent each kind of element in a compound with the correct symbol for that element
2.      Use subscripts to indicate the number of atoms of each type in the compound. If there is only one atom of a particular element, no subscript is used
3.      Write the symbol for the metallic element first.
4.      Use the ion charges to balance out the charges. (All molecules are neutral; net charge = 0)
5.      Special case for polyatomic ions: keep the whole ployatomic group together and use brackets if necessary.

                                                            ii.      Examples:
1.      Sodium Iodide                                    NaI
2.      magnesium oxide                    MgO
3.      zinc chloride                            ZnCl2
4.      potassium nitrate                     KNO3
5.      Aluminum sulphate                 Al2(SO4)3
6.      Lead (IV) oxide                      PbO2  (Why the IV?)





III.             Chemical Formula – Naming


a.       By looking at the above examples and your own knowledged gained in previous courses you should be able to develop a “feel” for naming chemical formulas.
b.      Some simple rules:

                                                              i.      Name metallic element first – unchanged
                                                            ii.      Name non-metallic element second – change ending to –ide
                                                          iii.      If polyatomic ion is present use its name as given
                                                          iv.      If the metal has more than one different charge i.e. Iron comes in two forms: Fe2+ and Fe3+ then use a roman numeral to state which one is present.
                                                            v.      Examples:

1.      Na2O               sodium oxide
2.      MgF2               magnesium fluoride
3.      Co2O3              cobalt (III) oxide
4.      PbSO4             lead (II) sulphate
5.      (NH4)3PO4       ammonium phosphate

                                                          vi.       Review Worksheet

IV.             Balancing Chemical Reaction Equations


a.       Word Equation – states the reactant compounds that are reacting to produce product compounds.
                                                              i.      Sodium   +   chlorine   à sodium chloride
b.      Translating this reaction means to re-write the reaction using the correct chemical formulas for each reactant and product.
                                                              i.      Na   +    Cl2   à    NaCl
c.       Balancing the chemical reaction means to ensure that each side of the equation has the same number and kind of atom. Is this chemical equation balanced?
                                                              i.      To balance chemical reaction equations we use coefficients in front of the chemical formulas. DO NOT ALTER THE SUBSCIPTS!
1.      2 Na    +   1Cl2   à   2 NaCl   (Note: 1’s can be  left out)
                                                            ii.      Check to see if you are balanced by counting up the atoms on both sides
d.      Please note: In chemistry 12 a lot of our discussions and areas of studies depend on being able to create and balance a chemical reaction equation. This may depend on writing a correct chemical formula. In order to do well in Chemistry 12, students have to be able to write chemical formulas and balance chemical reaction equations easily.
e.       Review worksheet #18 & 19


V.                Moles (Furry Creatures?, Skin blemish?, or maybe… a chemistry term!)

a.       Moles are very important in chemistry, but remember it just a number! (Dozen = 12, 1 gross = 144, 1 case = 12 etc.)

1 mole = 6.02 x 1023

            This number was chosen so that when you have this many atoms of any element, the mass of all the atoms would equal the atomic mass given on the periodic table.

i.e.       carbon             1 mole = 6.02 x 1023  = 12.00 g
            oxygen                        1 mole = 6.02 x 1023  = 16.00 g

b.      Calculating the molar mass of a compound:

                                                              i.      Molar Mass = mass of 6.02 x 1023 molecules of that compound
                                                            ii.      To determine the mass of this many molecules we take the mass of one proton and then multiply it by … just kidding!!!
1.      Molar Mass of a Molecule:
a.       Find the molar mass of each element in the compound according to the periodic table.
b.      Multiple each mass by the number of each element in the compound.
c.       Add up all the individual masses:

                                                                                                                                      i.      Example:         NaNO3

Na                   23.0 g x  1   =    23.0 g
N                     14.0 g x  1   =    14.0 g
O                     16.0 g x  3   =    48.0 g

                                                                               Molar Mass of NaNO3 =  85.0 g/mole


d.      Now Try These:

                                                                                                                                      i.      NaBr                           _____________
                                                                                                                                    ii.      MnCl2                                  _____________
                                                                                                                                  iii.      Al2O3                                  _____________
                                                                                                                                  iv.      (NH4)3PO4                   _____________
                                                                                                                                    v.      CoCl2*5H20                _____________




c.       Mole Conversions

                                                              i.      In Chemistry 12 converting from moles to mass and back again occurs all the time. I know that this is taught in Chemistry 11. I will not spend a great deal of time on this topic but here is a real quick review:
                                                            ii.      Moles to mass:

Mass (g)  =    Molar Mass    x    # of moles

               i.e.  What is the mass of 0.065 moles of NaCl?


               Mass =   0.065 moles x  58.5 g/mole  =   3.8 g of NaCl


                                                          iii.      Mass to moles:


Moles  =   Mass / Molar Mass


               i.e.  How many moles are there in 12.0 g of MgF2?

First find the molar mass of MgF2:        62.3 g/mole

Then:           # moles =  12.0 g / 62.3 g/mole = 0.193 moles MgF2



VI.             Reaction and Mole Ratios
a.       Chemical reaction Equations were not created to make chemistry students live difficult!
b.      They actually serve an important purpose.
c.       Remember this pervious reaction:


2 Na    +   1Cl2   à   2 NaCl


The coefficients state the mole ratio for the reaction between Na and Cl2.


2 Na    +   1Cl2   à   2 NaCl

For example: This reaction means:

2 moles of Na will combine with 1 mole of Cl2 to form 2 moles of NaCl

or

4 moles of Na will combine with 2 mole of Cl2 to form 4 moles of NaCl

or

0.01 moles of Na will combine with 0.005 mole of Cl2 to form 0.01 moles of NaCl



2 Al(s)  +  3 FeCO3(aq)  à  Al2(CO3)3(aq)  +  3 Fe(s)

How many grams of iron (II) carbonate would be needed to completely react with 108 g of aluminum?

Solution:

Step 1: Convert grams of Al to moles of Al




Step 2: Determine the mole ratio between Al and FeCO3.





Step 3: Find the number of moles of FeCO3 from the moles of Al and the mole ratio.





Step 4: Convert moles of FeCO3 back to gram of FeCO3.






VII.          Concentration Calculations:

a.       Once again this topic was covered in Chemistry 11. I will only give you a quick over view of it here. All of Chemistry 12 is solution chemistry, which means that calculating concentrations and using concentrations to solve problems is an important concept to know.

b.      Concentration  =  moles / volume




(Units = Molar = moles/litre)
 
 








c.       Example:
What is the concentration of a solution made when 19.0 g of Mg(NO3)2 is dissolved in a total volume of 250 mL of water?

                                                              i.      Find the molar mass of Mg(NO3)2:


Mg(NO3)2  =    148.3 g/mole


                                                            ii.      Find the number of moles from this molar mass and the mass given:


# moles  =  19.0g / 148.3 g/mole  = 0.12811 moles

                                                          iii.      Calculate the concentration of the solution:


C   =  n / V  =  0.12811 moles  /  0.250 L
           
        = 0.512 M Mg(NO3)2 Solution


d.      Now try this one:

A student needs to prepare 125 mL solution of a 0.300 M lead (II) nitrate solution. How many grams of the lead (II) nitrate does he need?

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