A Little Review and Some New Calculations To Feed Your Inner Chemist.

Remember the parts of the atom facts you learned A LONG TIME AGO!?!?!
Well, dig it all back out from the back of your brain.
Just kidding, I'll just do a brief review with y'all.


There are THREE subatomic particles: Protons (p), Neutrons (n), and Electrons (e-).
p and n are located in the nucleus and have about the same mass (ie. relative mass of 1)
e- has a very small relative mass of around 0 and is located in a cloud around the nucleus.

**Remember in a neutral atoms: #p = #e-  AND that an element's atomic # is also the # of p.


Ions occur when atoms gain or lose electrons during chemical bonding.
**Remember in ions: #e- = #p - ionic charge.

Negatively charged anions gain electrons (non-metals)
Positively charged cations lose electrons (metals)



What is the difference between atomic mass and mass number?
Though they both are acquired by adding the # of protons and neutrons, atomic mass is the mass that has the trailing decimals, while the mass number is a rounded mass.

Now you try:
How many protons, neutrons and electrons are in:
1) Strontium +2
2) Sulphur
3) Neon

Then...what are isotopes?
These are just variations of an element, and usually, they are the more dangerous types.
Remember isotopes are written including their mass number! (This differentiates isotopes from isotopes)
Isotopes have the same # of protons and electrons or is normal atom, BUT DIFFERENT NEUTRONS!!
ie. heavier atomic mass

Now, the NEW stuff steps in.
We're going to learn how to calculate the NATURAL MIXTURE OF ISOTOPES.
How do we know there are a few isotopes that are present? A mystical being came riding down from the sky and told this withered old man, of course.
NO, that is definitely NOT what happened.

 

Some fun with radioactive isotopes.

It's really simple, actually.
The decimals in the atomic mass signify an AVERAGE.
EUREKA, right?


Let's do a question together.
Given this information, calculate the average molar mass of Fluorine.

F-18: 36.89%
F-19: 14.73%
F-20: 9.06%
F-21: 39.32%

OK. First, we'll have to convert all of the percents into decimal numbers.
Then, multiply each isotope percentage to its respective atomic mass.

F-18: (0.3689)(18g/mol)= 6.6402g/mol
F-19: (0.1473)(19g/mol)= 2.7987g/mol
F-21: (0.0906)(20g/mol)= 1.812g/mol
F-21: (0.3932)(21g/mol)= 8.2572g/mol

Finally, the last step is to add all those masses together.
6.6402g/mol + 2.7987g/mol + 1.812g/mol + 8.2572g/mol= 19.50g/mol
**DONT FORGET SIG FIGS!


Written by Jialynn.

History of Chem (by:Mandy)

drop the math, and lets learn a bit about the history of chemistry.
here are some of the important historical chemists:

Aristotle believed matter was made of different combinations of earth, air, fire, and water Lavoisier, A.L. (1743-1794)Discovered nitrogen; studied acids and described composition of many organic compounds.

Dalton, John (1766-1844)He proposed atomic theory, stated law of partial pressure of gases. His ideas led to laws of multiple proportions, constant composition and conservation of mass.

Thomson, Sir J.J. (1856-1940)Research on cathode rays resulted in proof of existence of electrons. Won the Nobel Prize in 1906

Rutherford, Ernest (1871-1937): Discovered that the atoms of the heavier elements, which had been thought to be irreversible, actually decay into various forms of radiation. Rutherford was the first to establish the theory of the nuclear atom and discovered rays: alpha, beta, and gamma. He also revealed the half-life of radioactive elements and applied this to studies of age determination of rocks by measuring the decay period of radium to lead-206. Have fun!

#stoichiometry calculations involving particles - moles - mass# by:mandy

Stoichiometry???

STOY-KEE-AHM-EH-TREE is a study of amounts of substances that are involved in reactions! lots of calculations involved. So, think Math!

what are the steps to calculationg the amounts od substances in a reaction?

1. write equation                                                

2.balance

3."road map"                                                               

4.calculations!

now, lets try an example:

step 1:

Mg + HCl --> MgCl2 + H2

since this reaction is not balanced yet, we'd have to preform step 2:

Mg + 2HCl --> MgCl2 + H2

step 3:

Note that when we are making a "road map", substances CANNOT be diretly convert from grams of one compound to grams od another. Instead, we have to go through moles :)

grams (x) <--> moles (x) <--> moles (y) <--> grams (y)

final step *4:

6.0g H2 x 1 mol H2/2.02g H2 x 1 mol Mg/ 1 mol H2 x 24.3g Mg/ 1 mol Mg = 72.2g Mg

YAY! good job! that was the step by step calculation.

Molarity and Stoichiometry - Isabelle Cheng

MOLARITY AND STOICHIOMETRY

Do you remember molarity?!?!?!?!?! 

Here are the formulas:

M = mol

       L

L = mol

       M

mol = M X L

Example: 

You have 340 mL of 0.5 M NaBr solution. Since you know that you have a 0.5M solution, you know that you must have 0.5mol / L water. So, if 0.5 moles are in a liter, how many moles are in 0.340 liters? (340 mL / 1000)

M = mol solute / L water

0.5 = x mol solute / .340 L

x = (0.5)(.340) = 0.17

M = mol solute / L water

x M = 0.17 mol solute / (.340 + .560) L water

x M = 0.19 mol / L

Gas Stoichiometry Example:

Ammonia gas is reacted with sulfuric acid to form the important fertilizer ammonium sulfate. What mass of ammonium sulfate can be produced from 85 kL of ammonia reacting.

_1_NH3

+

_1_H2SO4

_1_(NH4)2SO4

1.

85 kL	=	3.469 mol.	NH3
24.5 L/mol

2:

3.469 mol.	X	_1_(NH4)2SO4	=	3.469 mol.	(NH4)2SO4
		_1_NH3

3.

3.469 mol.	X	114 g	=	395 g
		mol

More exercises! 

 What volume of oxygen at STP is needed to completely burn 15 g of methanol (CH3OH) in a burner?

Find the molarity of a solution that contains 25.0 g of sodium hydroxide in 300.0 mL of solution. 

If 20.5 mL of H3Po4 with an unknown molarity react with 25.0 mL of 0.500 M KOH, according to the above reaction, what is the molarity of H3PO4?

Enthalpy Calculations...*sigh*

Sooooooooo, did you ever think we were done with the mole?
DON'TTHINK THAT WAY because I'm about to prove you wrong.
Enthalpy calculations=the usage of MOLES -insert dramatic scream-
And the return of the mole also means the return of sig figs, btdubs. -insert double dramatic scream-

That my friends was a double double, double double combo. (:

Let's get started:
ΔH--> change in energy of a reaction; expressed in kJ/mol

Here is an example:
 C10H8+12O2-->10CO2+4H20+436kJ
Hence, ΔH=

1)      -436kJ
     1 mol C10H8

2)   -436kJ
    12 mol O2

3)    -436kJ
    10 mol CO2

4)    -436kJ
    4 mol H2O

Once you've identified these, it will be easy to do conversions.


eg. Calculate how many grams of O2 would be needed to produce 1500kJ of energy.

-1500kJ x 12 mol O2 x  32g    = 1321.1 kJ
                 -436kJ         1 mol

**SIG FIGS!!**   So, the answer is 1300 kJ.


Watch this video if you're still confused!!

GOOD LUCK WITH THIS CONFUSING SECTION!! :D




Written by Jialynn.

Exothermic and Endothermic (Mandy)

ENDOTHERMIC & EXOTHERMIC

ENDOTHERMIC: absorb energy to break bonds
EXOTHERMIC: release energy to join bonds

Energy diagrams allow one to observe endothermic or exothermic reactions.

in above diagram, "starting products" are called "energy of reactants", which indicates the total potential energy of all reactants. The "transition state" is also known as "energy of activated complex".

one thing that the above diagram did not mention is the "change in enthalph" (delta H). The change in potential energy during the reaction. It is the energy of products - energy of reactants.

Therefore, if "delta H" is positive, it is an ENDOTHERMIC reaction.
                 if "delta H" is negative, it is an EXOTHERMIC reaction.

Lets look at this concept from another point of view:

CH4 + 2O2 -> CO2 + 2H2O + 812kJ

Endothermic: reactants have the energy term on the left hand side and positive 'delta H'
Exothermic: reactants have the energy term on the right hand side and negative 'delta H'

Time for some examples!! --->

In this reaction, the total energy of the reactants is 80 kJ mol-1, the total energy of the products is -90kj mol-1 and the activation energy for the forward reaction is 120 kj mol-1.

a. Draw a diagram of the energy profile for this reaction. Label the diagram.

b. State whether the reaction is endothermic or exothermic.

c. Calculate the change in enthalpy.
ans: b. endothermic reaction
       c. 10kJ


an interesting video :) ENJOY.

February 4, 2011 - Isabelle Cheng - Lab 5B - Types of Chemical Reactions

Isabelle Cheng

Block 2-2 Chemistry 11

February 4, 2011

Ms. Chen

Lab 5B - Types of Chemical Reactions

In this lab, there are four different kind of reactions. Synthesis, decomposition, single replacement, and double replacement. Synthesis has the formula of A+B --> AB and the decomposition has the formula of AB ----> A + B. The single replacement is AB+X----> A + XB. Double replacement’s formula is AB + XY ----> AY + XB. In the lab, we have to identify which reaction is either synthesis, decomposition, single replacement, or double replacement. 

In the lab, my partner and I got different equipment such as a lab burner, test tubes, test-tube clamp and the different substances to experiment with. The substances are copper wire, iron nail, copper (II) sulfate solution, solid copper (II) sulfate pentahydrate, water, calcium chloride solution, sodium carbonate solution, mossy zinc, and other materials.

There were 7 reactions and they each had different substances in the test tubes. The first one is holding the crucible and burning the copper wire. The wire then turns into a silver color. After that when you hold it for a longer period of time it gets more burnt. The second reaction is we had to clean the iron nail with a piece of steel wool until it was shiny. After we placed it in the copper (II) sulfate solution so that half of the nail was covered. We had to wait for 15 minutes. While doing that we moved onto reaction 3. The third reaction we put solid copper(II) sulfate pentahydrate in the test tube until it was one third full. Then we put it over the flame and slowly moved it in all directions. Reaction 4 was to put 2-3 drops of water into Reaction 3 when it was cooled. Reaction 5 was requesting us to fill a test tube of one quarter of the tube with calcium chloride solution and same as sodium carbonate solution. For reaction 6 we took a piece of mossy zinc in a test tube and put some hydrochloric acid solution to mix it in with the zinc. After doing that Reaction 7 we fill up the tube half full for hydrogen peroxide solution. We also add a small amount o manganese (IV) oxide. We then test the gas and wait until the test tube starts bubbling and glowing. In the end, the lab had interesting before and after effects.