2011年4月23日星期六

ELECTRON CONFIGURATION!!

-the elctronic configuration of an atom is notation that describes the orbitals in which the electrons occupy and the total number of electrons in eaqch orbital.

-when an electron absorbs or emits a specific amount of energy it instantaneously moves from one orbital to another.

-an energy level is the amount of energy, which an electron in an atom can possess.

-energy difference between two energy level is quantom of energy.

Ground State: when all the electrons of an atom are in their lowest possible energy level.

Excited State: when one or more of an atom's electrons are in higher energy level than the lowest electron.

Orbital: an orbital is the actual region of space occupied by an electron in a particular energy level.

Shell: a shell is the set of all orbitals having the same n-value.

Subshell: a subshell is the set of orbitals of the same type.

The letters s, p, d, f refer to the four different types of orbitals, each circle represents an orbital.

For a given value of "n", different types of orbitals are possible for:
n=1 only the s- type
n=2 s- and p- types
n=3 s- , p- and d- types
n=4 s- , p- , d- and f- types.

A maximum of 2 electrons can be placed in each orbit. (Pauli Exclusion Principle)

Writing Electronic Configuration for Neutral Atoms
-always start with the lowest energy level.
-figure out how many electrons you have, then start at the lowest (1s) and keeping adding.
-each electron has an opposite spin designated by upward and downward arrows.

ex. C has 6 electrons, 2 in the 1s, 2 in the 2s and 2 in the 2p.
notice the 2 electrons in the 2p pccupy separate suborbitals and are not paired up.
the electron configuration is 1s2 2s2 2p2.

Writing Electronic Configuration for Ions
for negative ion:
-add electron to the unfilled subshell, starting where the neutral atom left off.

for positive ion:
-start with neutral atom, remove electron from the outermost shell first.
-in both s- and p- , remove from p-, first.

Core Notation
the set of electrons for an atom can be divided into two subsets: the core electrons and the outer electron.
-the core of an atom is the set of electron with the configuration of the nearest noble gas that comes before it.
-the outer electron consist of all electron outside the core electron normally take part in chemcal reactions.

locate the atom and note the noble gas at the end of the row ABOVE the element.

ex. S: 1s2 2s2 2p6 3s2 3p4
          [NE] 3s2 3p4

 

2011年4月19日星期二

Atomic Structure~

proton: has a relative mass of 1 and electric charge of +1; located in the nucleus.
neutron: has a relative mass of 1 and no electric charge; located in the nucleus.
electron: has a relative mass of almost 0, has a electric charge of -1; located in the cloud surrounding the nucleus.


Atomic number=the proton number


Atoms have no overall electric charge. Therefore, # of p+ = # of e-
anion: when e- is added to an atom
cation: when e- is removed from a atom.


Mass number= # of proton + # of neutron. (therefore, # of neutron = mass # - # of proton)
Atomic mass= the average mass of a element's isotopes


Isotope: have same atomic number but different atomic mass (in another word, different # of neutron)

2011年4月15日星期五

Atomic Theory

Atomos: Smallest pieces
-matter was made of four different combinations: Earth, Air, Fire, Water
-turn common metal into gold
-marked the beginning of the understanding of matter

Four elements theory lasted for about 200 years
not a scientific theory because it could not be tested against observations

Democritus
-300bc, atoms are invisible particles
-first mention of atoms
-not a textual theory, but conceptional
-no mention of atomic nucleus, or constituents
-not used to explain chemical reaction





Lavoisier
-first version of the law of conservation of mass and law of definite proportions.
-recognized and named Oxygen gas(1778) and Hydrogen gas(1783)





Proust
-if a compound is broken down into its constituents, the products exist in the same ratio as in the compound.
-proved Lavoisier Laws





Dalton
-Atoms are solid, in destructive spheres
-provides for different elements
-based of law of conservation of mass








5 main points
--Elements are made of atoms
all atoms of a given elements are identical
different from other elements; distinguished by relative weights.
atoms can be combined with others of different elements to form chemical compound
given compound has same relative numbers of types of atoms.
atom: indestructive. chemical reaction only changes the way atoms are grounded together.

J.J.Thomson
-raisin bun model
-solid,positive spheres with negative particles embedded them
-first atomic theory to have +(P+) and -(e-) charges
-demonstrated the existence of e-
-Crookes's tubes and other equipments, J.J.Thomson discovered e- and measured the ratio.

Ruther ford
-atoms have a positive dense censer with e- outside
-planetary model
-explain why e- spin around nucleus
-atoms are mootly empty space

Neils Bohr
-studied gaseous samples of atoms, glow by possing an electric current through.
-e-surrounded the nucleus in specific "energy levels" or "shells"

The Modern Art
-atom is the smallest particle
-made up 3 kinds of particles called subatomic particles
Proton(p+)
neutron(n0)
electron(e-)

2011年4月4日星期一

PERCENT PURITY!!

Sometimes, the reactant(s) we use is/are not pure.  So, before we can calculate much product will form, we need to use PERCENT PURITY to calculate how much reactant that actually is available to react.

% PURITY=Mass of Pure Substance/Mass of Impure Sample×100%

Below is a link of a video that demonstrates one example of the percent purity.
http://www.youtube.com/watch?v=15FOjmIUEy4&playnext=1&list=PL45C0C062682A839A

 

PERCENT YIELD!!

Sometimes, it is not possible to recover all the product, or not all the reactions are used up, and therefore, we need to calculate PERCENT YIELD.

PERCENT YIELD=the ratio of amount of product obtained to amount of product expected by claculated, expressed as a %.

% YIELD=grams of actual product recovered/grams of product expected from stoich×100%

ex. HBrO3+5HBr→3H2O+3Br
if 10.0g of HBrO3 is reacted with excess HBr and 26.3g of Br2 was prodeced, what is the percent yield?
10×(1mole/128.9g)×(3mole/1mole)×(159.8g/1mole)=37.191g Br2

%YIELD=26.3g/37.191g×100%=70.7% yield

ex. C3H6Cl2+2KSCN→C3H6(SCN)2+2KCl
the % yield is 80%.  how many grams of C3H6(SCN)2 will be produced if 3.00 grams of C3H6Cl2 is reacted with excess KSCN?
3g×(1mole/113g)×(1mole/1mole)×(158g/1mole)=4.20g C3H6(SCN)2

% YIELD=C3H6(SCN)2 obtained/420g expected×100%=80.0% yield
C3H6(SCN)2=3.36g

          

2011年3月12日星期六

Excess & Limiting Reactants persent yield

A.Excess Quantity
A balanced equation describes what should happen in a chemical reaction.

However, the conditions necessary for the reaction to take place may not be present. (ie. pressure, temperature, concentration etc.)

Sometimes it is necessary to add more of one reactant than the equation predicts because it is not possible for every atom/molecule of the reactants to come together.
Ex. One reactant is the Excess Quantity and some of it will be left over, the second reactant is used up completely, and is the limiting reactant.

B.Excess quantities in chemical reactions.
Ex1. How many grams of OCl2 will be formed when 44.0 g of O2 reactant with 97.0g of Cl2

(*General Process: Convert both reactants to the desired product & the smaller amount of product will actually be produced.)

Step1: Balanced equation.
O2(g) +2 Cl2(g) → 2OCl2(g) 

Step2: Convert Cl2 to OCl2
97.0 g ×1mol Cl2/71g × 2mol OCl2/2mol Cl2 × 87.0g/1mol OCl = 118.86 g OCl2

Step3: Convert O2 to OCl2
44.0g × 1molO2/32g ×2mol OCl2/1mol O2 × 87.0g/1mol OCl = 239.25g OCl2
Because 97.0g of Cl2 reacts to produce the smaller amount of product. Cl2 is limiting quantity and O2 is excess quantity. 119g is the mass of product.

Ex2. 
41g O2 reacts of 164g of Cl2, Which is EXCESS quantity? Which is Limiting quantity? How many grams of the excess quantity will be used?
Step1:
O2(g) +2 Cl2(g) → 2OCl2(g) 
41.0g       164g
(*General process: Convert ONE of products to the other reactant to see which excess and which is limiting. Determine which is left over and by how much. It may involve another conversation)
Step2:
To find which reatant is in excess calculate how many grams of Cl2 gas would be required to react with 41.0g of O2.
41g O2 × 1mol O2/32.0gO2 × 2mol Cl2/1mol H2 × 71.0g Cl2/1mol Cl2= 181.93g=182g Cl2

Since there are only 164g of Cl2 gas, not all O2 can react. Thus Cl2 is limiting quantity and O2 is excess quatity.

Step3: Convert Cl2 to O2 to see how much O2 would be needed to react with 164g of Cl2.
164g Cl2 × 1mol Cl2/71.0g Cl2 × 1mol O2/2mol Cl2 × 32.0g/1mol O2 = 36.958=37.0 g O2

37.0g of O2 would react with 164g of Cl2.

2011年3月3日星期四

STOICHIOMETRY (2) !!


Example 1 :
a) How many grams of Fe(Ⅱ) will be needed to react with 3.20 mol of HCl?

Step1: Balance the equation.
1Fe + 2HCl → 1FeCl2 + 1H2

Step2: Make a "ROAD MAP"


Step 3: Do calculation

3.2 mol HCl × 1mol Fe/ 2mol HCl × 56g Fe/ 1mol Fe = 90g Fe


b) How many grams of Fe will be needed to produce 10.0 g H2











10.0 g × 1mol/2.0 g H2 × 1mol Fe/ 1mol H2 × 56 g Fe/1 mol Fe =280 g Fe



Note: grams (X) ↔ mol(x) ↔ mol(y) ↔ grams(y)

Example 2 :
a) Using the following equation : 2 NaOH + 1H2SO4 → 2H2O + 1Na2SO4
How many grams of sodium sulphate will be formed if you start with 200 grams of sodium hydroxide?















200 g NaOH × 1mol NaOH/ 40 g × 1mol Na2SO4/2 mol NaOH × 355.3g Na2SO4/ 1mol = 888.25 g
                                                                                                                                            = 900 g