Electronic Structure

 Electronic Structure(configuration) is the arrangement of electrons in energy level  around an atomic nucleus. It tells us how many electrons are in each orbital.
 For example, Electronic structure for Na(sodium) is 1s2 2s2 2p6 3s1.
It looks complicated right? But as long as you get the rules, it will be really easy.

First there are four types of shell. There are s,p,d,and f. Like u can see from the periodic table below those shells are based on periodic table.

  To write electronic configuration, you just needed to find location of the element that you want, and write down orderly from left to right.
 Let's use Na for the example, Na has 11 electrons,and it's at 3rd period fist group. Fisrt we start with left top. there are two elements are in 1s(1 simply represent the  period). Thus, you write 1s2, and there are nothing on the 1period on the right side, so we go to period 2. They have two elements as well, thus you write 2s2, then there are 6 elements in 2p, thus you write 2p6, got it? So,over all we have 1s2 2s2 2p6 and finally where Na is 3s1. Thus,  answer is 1s2 2s2 2p6 3s1.
Just be aware of d d start wih row 3 instead of  4









Core Notation
 Some Scentists are really lazy thus, they found the easiler way to write elecronic configuration. It's easy you just need to find the near noble gas and start the electronic configuration from there. Thus, for Na, Ar is the closest noble gas, thus u write (Ar)3s1.

Valence Electrons
They are electrons that are located most outer shell, if we have core notation we just simply need to count the numbers of elecrons exept d and f. For example,Na has 1 valence eletrons
 Here is the video that will help

                                         

Atomic Structure

Atoms are composed of three type of particles: protons, neutrons, and electrons.

Protons and neutrons are responsible for most of the atomic mass.
The mass of an electron is very small (9.108 X 10^-28 grams).

1 - Now let's start with Neutral Atoms:

Both the protons and neutrons reside in the nucleus. Protons have a postive (+) charge, neutrons have no charge, ie they are neutral. Electrons reside in orbitals around the nucleus. They have a negative charge (-).

The number of protons determines the atomic number, e.g., H = 1. The number of protons in an element is constant (e.g., H=1, Ur=92) but the neutron number may vary, so the mass number (protons + neutrons) may vary.

Helium Atom

 2 - Ions:

- In an ion, number of electrons =/ number of protons, but protons = atomic number.
- Electrons are either lost or gained, making the ion either positive or negative.






3 - Isotopes:

The same element may contain varying numbers of neutrons; these forms of an element are called isotopes. The chemical properties of isotopes are the same, although the physical properties of some isotopes may be different. Some isotopes are radioactive-meaning they "radiate" energy as they decay to a more stable form, perhaps another element half-life: time required for half of the atoms of an element to decay into stable form. Another example is oxygen, with atomic number of 8 can have 8, 9, or 10 neutrons. 

To sum this up:

- Isotopes are atoms with same number of electrons and protons, but different numbers of neutrons.
- Different numbers of neutrons means different atomic masses.
- Different amounts of isotopes exist for different elements.








The mass number of elements can be calculated by getting the averages of the atomic masses of isotopes. For example:

potassium has three main isotopes; potassium-39 (93.26%), potassium-40 (0.0117%), potassium-41 (6.73%), so to get the average you multiply the percentages by their atomic masses, then add them together.

(39*0.9326) + (40*0.000117) + (41*0.0673) = 39.13538g/mol

*Always; # neutrons = mass # - atomic #, electron's mass = 1, neutron mass = 1837, proton mass = 1836

April 14

History of Chemistry

The theories of Aristotle, which lasted two thousand years, were that matter was made of atomos, the smallest pieces of matter.  Everything was made of a combo of earth, air, fire, and water.

In the late 1700s, Lavoisier introduced the first version of the Law of Conservation of Mass and the Law of Definite Proportions

In 1799, Proust stated that if a compound was broken down into its constituents, the products would exist in the same ratio as in the compound

Dalton (1800s) discovered that atoms were solid, indestructable spheres. His 5 main points of the Atomic Theory were:

-Elements are of tiny particles called atoms

-Atoms of a given element are identical 

-Atoms of different elements are different by weights
-Atoms from different elements combine to form chemical compounds

-Atoms cant be destroyed, created, or divided.

JJ Thomson

- Thomson introduced the Raisin Bun diagram.  ( solid positive spheres with negative particles embedded in them)

April 6th- Percent Purity

Percent Purity is the percent of a specified compound or element in an impure sample.

Example:

Chalk is almost pure calcium carbonate. We can work out its purity by measuring how much carbon dioxide is given off. 10 g of chalk was reacted with an excess of dilute hydrochloric acid. 2.128 liters of carbon dioxide gas was collected at standard temperature and pressure (STP).
The equation for the reaction is
CaCO₃ (s) + 2HCl (aq) → CaCl₂ (aq) + H₂O (l) + CO₂ (g)

 

Solution:

Step 1: Calculate the grams from the volume

1 mole of CaCO₃ gives 1 mole of CO₂
1 mole of gas has a volume of 22.4 liters at STP.
22.4 liters of gas of gas is produced by 100 g of calcium carbonate
and 2.128 liters is produced by 2.128 ÷ 22.4 × 100 = 9.5 g

Step 2: Calculate the percent purity

There is 9.5 g of calcium carbonate in the 10 g of chalk.
Percent purity = 9.5 ÷ 10 × 100% = 95%

April 4th - Percent Yield

Percent Yield

• The percentage yield is the ratio between the actual yield and the theoretical yield multiplied by 100%.  It indicates the percent of theoretical yield that was obtained from the final product in an experiment. 

• The percentage yield can be calculated using the mass of the actual product obtained and the theoretical mass of the product calculated using the balanced equation of the reaction.

Percentage Yield =     Mass of Actual Yield       x   100%

                  Mass of Theoretical Yield

Theoretical Yield

• this is how much product will be synthesized in ideal conditions.

• To determine theoretical yield, multiply the amount of moles of the limiting reagent by the ratio of the limiting reagent and the synthesized product and by the molecular weight of the product.

 Actual Yield

• this is how much product was actually synthesized in the experiment.
  
• Example:  0.135 g acetylsalicylic acid 

Practice Problem

In the following reaction, 0.157g of p-acetaminophenol was used to react with 0.486 g of acetic anhydride to produce acetaminophen and acetic acid.  The product was purified and acetimophen was extracted.  The actual mass of acetaminophen produced was 0.198 g.  Determine the theoretical yield and the percent yield of isopentyl acetate. p-Aminophenol     +     Acetic anhydride     à     Acetaminophen      +    Acetic acid     C6H7NO                                 C4H6O3                    C8H9NO2                  CH3COOH Solution   molar mass of p-aminophenol =109.1g/mol molar mass of acetic anhydride = 102.1 g/mol moles of p-aminophenol = mass/molar mass                                   = 0.157g/(109.1g/mol)                                   = 0.00144 mol moles of acetic anhydride = mass/molar mass                                          = 0.486g/(102.1g/mol)                                          = 0.00476 mol Theoretical Yield = moles of acetamiophen x molar mass of acetaminophen                          = 0.00144 mol x 151.2g/mol                          = 0.217 g Percent Yield =         Actual Yield        x 100%                             Theoretical Yield                     =            0.198g      x 100 %                                   0.217g                     =  91.2 %

March 12th- Excess & Limiting Reagents

Today we learned about  EXCESS & LIMITING reactants.

Chemical reaction equations give the ideal stoichiometric relationship among reactants and products.
However, the reactants for a reaction in an experiment are not necessarily a stoichiometric mixture. In a chemical reaction, reactants that are not use up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limit the amount of products formed.

Example 1 

using the following equation, find the excess reagent when 10.0 g manganese metal is allowed to react with 10.0g nitric acid

first make the formula

3Mn + 8HNO3----- 3Mn(NO3)2 + 2NO + 3H2O

then you find the mass of HNO3 using 10g of manganese(order doesnt matter)

10g x 1moleMn/54.9 x 8moleNO3/3moleMn x 63gHNO3/mol HNO3
=30.6g of HNO3 .......since you only need 10g of HNO3, but you have 30.6g of HNO3 it is excess
it also means  Mn is limiting

Example 2 

Find the limiting reagent and the reactant in excess when 0.5 moles of Zn react completely with 0.4 moles of HCl

1. Write the balanced chemical equation for the chemical reaction     Zn + 2HCl -----> ZnCl₂ + H₂
    
2. Calculate the available moles of each reactant in the chemical reaction

   moles of Zn = 0.5

   moles of HCl = 0.4

3. Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

   Zn : HCl	Or	HCl : Zn
   1 : 2		1 : ½

4.  Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio     If all of the 0.5 moles of Zn were to be used in the reaction it would require
       2 x 0.5 = 1.0 moles of HCl for the reaction to go to completion.
       There are only 0.4 moles of HCl available which is less than the required 1.0 moles.
       If all of the 0.4 moles of HCl were to be used in the reaction it would require
       ½ x 0.4 = 0.2 moles Zn.
       There are 0.5 moles of Zn available which is more than the required 0.2 moles
    .
5. The limiting reagent is the reactant that will be completely used up during the chemical reaction.
       There will be some moles of the reactant in excess left over after the reaction has gone to completion.     The limiting reagent is HCl,
       all of the 0.4 moles of HCl will be used up when this reaction goes to completion.
       The reactant in excess is Zn,
       when the reaction has gone to completion there will be
       0.5 - 0.2 = 0.3 moles of Zn left over.
   
   

Here are more practice questions!

http://www.ausetute.com.au/exceslim.html

here is the website that will help you
http://www.youtube.com/watch?v=Vaiz0zLesHk&feature=related

March 7- Stoichiometry Calculation

Today we did stoichiometry calculations involving Molarity and STP.

Molarity is easily defined by this one equation. Molarity=  Moles
                                                                                   Litres

STP is 22.4 litres/mole

Example

1. How many moles of nitrogen gas is needed to react with 44.8 liters of hydrogen gas to produce ammonia gas?

3H₂   +   N₂    2NH₃
GIVEN: 44.8 L of H₂ at STP.
FIND: mols of N₂.
Here the sequence is: GIVEN liters of H₂ at STP, CHANGE liters of H₂ at STP to mols of H₂, MOL RATIO to change from H₂ to N₂. There is no need to go any further to change the N₂ into mols, because the mol ratio leaves the material in that unit anyway.






How many liters of ammonia are produced when 89.6 liters of hydrogen are used in the above reaction?
The "above reaction" from problem #1 is: N₂   +   3H₂    2NH₃

GIVEN: 89.6 L of H₂ at STP.
FIND: Volume of ammonia (in liters at STP)
Take the GIVEN quantity, use the Molar Volume of Gas at STP (MVG) to change it to mols, change the material with the mol ratio (MR), and change the mols of new material to the requested liters at STP using the MVG again.