Ch.8 Atomic Structure

First of all,  before we start the chapter, better do some REVIEW!

There is 3 different type of Subatomic Particles in an element

Proton: Large with positive chargeElectron: Small with negative chargeNeutron: Large with no charge

Moreover:

"# of Atomic number = # of Proton = # of Electron"

"# of Neutron =  Atomic mass - Atomic number/ Proton/ Electron"

Let's do some excerises:

Carbon:

Atomic number : 6  = Proton : 6 = Electron : 6  Atomic Mass: 12

Neutron: 12 - 6 = "6"

Rutherium:

Atomic number: 44 = Proton: 44 = Electron: 44 Atomic Mass: 101

Neutron: 101 - 44 = "57"

Bromine:

Atomic nimber: 18 = Proton: 18 = Electron: 18 Atomic Mass: 80

Neutron: 80 - 18 = "62"

Then now we can move over to Isotope,
Isotope is an atomic specie that has the same atomic number with different atomic mass.

let use Carbon as an example:

It has shown that Carbon has different Atomic mass as 12, 13, 14

So that we can indicate them as Carbon - 13, Carbon - 14

Why dont we take a look this video to see how's isotope practice works?

Theoretical Yield and Percent Yield, Purity.

Realistically, there is no reactions can be 100% completion. That's why we have to calculate the percentage of product actually formed.

As we have leaned stoichiometry calculation to predict what's the product by converting from one reactant to another reactant, and now we can determind what's the difference between

ACTUAL YIELD and THEORETICAL YIELD

   Actual Yield         (What you ACTUALLY obtain from the reaction)

Theoretical Yield    (What you EXPECT to get from the reaction)

So, when we want to find out what's the difference between "ACTUALLY YIELD" and "THEORETICAL YIELD". We can do it mathematically by mutiplying 100%

Actual Yield (What you ACTUALLY obtain from the reaction)    x   100%

Theoretical Yield (What you EXPECT to get from the reaction)

As a result, the percentage that you get will be the PERCENT YIELD.

Ex 1)

35.0 mL of 2.00 M Ca(OH)2 produced only 3.92 of CaCl2. What would the percent yield of the reaction be?

1)First, we have to consider what are the steps to get to the answer.

If we get mL (Volumn) and M (Molarity), then we can get the mol (mole) by the molarity equation WE 'VE LEARNED BEFORE.  (BETTER REVIEW)

2M x (      35    ) (L) = 0.07 mol of Ca(OH)2 x 1 mol of CaCl2         x     (40.1 + 35.5x2)

               1000                                                      1 mol of Ca(OH)2           1 mol of CaCl2

= 7.7777g of CaCl2

Actual Yield   3.92g of CaCl2              x  100%   = 50.4% (SIGFIG)

Theoretical Yield    7.78g of CaCl2





Part2       Percent Purity 

The ratio of the mass of pure substance to mass of impure sample expressed as a percent!

% Percent Pure = Mass of Pure Substance  x 100%

                             Mass of Impure Sample

By following this equation, the impure sample is heavier than the pure substance, as the impure one contains other substances in it. Once you get the answer over 100%, then you may make a mistake in your calculation.

Limited and Excess Reactants Persent Yield

As we are making up a product, there always have a need of reactant.
Sometimes, it'll have excess quantity(it will be left over) or either limited reactant ( used up completely)
that determining those reactants are completely involving in the reaction or not.

Ex 1:)
How much of which reactant will be left over?
2.19 mole of H2 is reacted with 3.48 moles of Cl2 to form HCl.

1: We have to write down the equation  H2 + Cl2 => 2HCl (make sure it's balance!)
2: Convert H2 and Cl2 into HCl by using stoichiometry calculation.

2.19 mole of H2x _2_mole of HCl    = 4.38 moles of HCl       (Limited reactant as it's less than Cl2)
                               1mole of H2

3.48 mole of Cl2x  2_mole of HCl     = 6.96 moles of HCl       (Excess quantity as it's more than H2)
                               1  mole of Cl2

3: Calculate the different between them.    6.96 - 4.38 = 2.58 (it'll be the final answer of the left over)

Ch 6 Stoichiometry

Stoichiomerty - the study of amount use and the amount produce.

Basically, it's the measurement of element.

For example:
First)
 C10H8 + O2  >  CO2 + H2O  (First Step, Write the balance equation)
                            Into
1C10H8 + 12O2 > 10CO2 + 4H2O

Second)
Then we can determind the ratio of this reaction as  1:12:10:4 ( base on the first number in each substance)

Third)
"What you need"  over  "What you have".
Ex.1 How many moles of O2 are required to react with 0.350 moles C10H8

1: What you have: 0.350 moles C10H8, What you need: mol of O2

0.350 moles C10H8 x       (__12O2 mole___ )       >>>> 4.2 mole of O2

            what you have         (  1C10H8 mole   ) ratio

2: At STP, what volume of water would be produced from 21.2g of C10H8?

(Tips, once we know how to convert it back to mole, we 'll be able to convert it into gram, molecules, volume, molarity, that we 've already learned before.  "BETTER REVIEW")

21.2g C10H8  x           __Mole of C10H8_  >>>>  0.165625 mole of C10H8 (Don't round it as sigfig yet)
                                       (12x10+8)C10H8     

0.165625mole of C10H8 x  __4mole of Water__What you want >>  0.6625 moles of Water
                                            1 mole of C10H8     What you have

0.6625 moles of Water x   __22.4 Volume__  >> 14.84........   SIGFIG    > 14.8 at STP of H2O
                                            1mole of Water

As we 've seen that, it's pretty complicated, but not that hard, once we do it "STEP BY STEP" then it's easy to get marks

.

Enthaply Calculations

There's nothing do with Enthalpy Calculation in this picture... just my feeling under the test.. 

As we knew that, there'll have either absorbing or releasing energy during the reaction, like the equation from below~
CH4+ 2O2 > CO2 + 2H2O + 812kJ 
And now, we can turn it into mole by "ENTHALPY CALCULATION"

The unit of enthalpy calculation will be KJ/mol,  which has shown that  "KJ divided by mol"

As the equation from above, we can tell  "4" different equation from it, as there have 4 different substance in either reactant or product side.

- 812KJ/1 mole of CH4      -812KJ/2 mole of O2      -812KJ/1 mole of CO2      -812KJ/2mole of H2O
The moles of the compouds are determinded by the number of them individualy in the equation
Moreover,  as it's "releasing", which causes a "NEGATIVE" number

I am BACK~

So, Ex 1) How many moles of CH4 are needed to produce 3000KJ?

First, we can write  -3000KJ DOWN,  and multiply the "CH4" equation from below, which is -812KJ/1CH4

-3000KJ x  1CH4 / - 812KJ   (As we want to reduce the KJ, so that we can get the mole)

3.49458128.......  moles of CH4
SIGFIG TIME~~~
There is only 1 sigfig in - 3000,  so     the answer 'll be  "4 mol of CH4"

Moreover, BE CAREFUL THE QUESTION, ESPECIALLY IT MIGHT ASKS FOR "GRAMS"
As we know one more method to turn it into mole, there possibly will have "Mole Converstion" back

Anyway, GOOD LUCK....

Endothermic/ Ecdothermic 16/2

Endothermic: Absobing Energy  (Positive)

Ecdothermic: Releasing Energy  (Negative)



1: Reactants: a substance before reaction

2: Activatied Complex: The highest point that displays the largest potential energy of the reaction

3: Products: A result that after the reaction.

(a)= Activation Energy,  which is the difference between Reactants and Activatied Complex

So basically, if the Products side have higher energy than the Reactants, which means it's "ABSORBING" energy  >>>  Endothermic     (POSITIVE)

If the Products side have lower energy than the Reactants, which means it's "RELEASING" energy>> Exothermic            (NEGATIVE)

Moreover,  we can also calculate it without looking at the graph:

PRODUCTS  -   REACTIANTS  =  POSITIVE / NEGATIVE 

if it's positive, then it'll be Endothermic, if not, Exothermic then : )

An other punishment to our 1/2...

Energy Equation:

CH4 + 2O2 > CO2 + 2H2O + 812kJ   (Remember, kJ"RELEASING energy will be the product side.)

(Exothermic)

P4O10 + energy > P4 + 5O2  ( ABSORBING energy, Endothermic)

Net equation 14/2

Happy Valentine's day



Net Equation has shown that in double replacement reactions, some ions participate in the reaction while other ions do not participate.

Ex.

KCl(aq) + AgNO3(aq)  >  KNO3(aq) + AgCl(s)

Step 1: Divide the NON - INVOLVED  ions (ions that's not participate to form solid) into element

K(aq) + Cl(aq) + Ag(aq) + NO3(aq)  >  K(aq) + NO3(aq) + AgCl(s)

Step 2: Reduce one on each side, just like doing math

Ag(aq) + Cl(aq) > AgCl(s)   << done  NET EQUATION