Atomic Structure (Electron, neutrons, protonx)

Review from Science 10
Atoms are made up of 3 types of particles electrons , protons  and neutrons .  
These particles have different properties.  

• Electrons are tiny, very light particles that have a negative electrical charge (-).  
• Protons are much larger and heavier than electrons and have the opposite charge, protons have a positive charge.  
•  Neutrons are large and heavy like protons, however neutrons have no electrical charge.  

Each atom is made up of a combination of these particles. 

protons and neutrons form the nucleus and they are surrounded by the electron

*atomic number is the number of protons
*atomic mass is the number of protons and neutrons

Eg.  Helium
 It has 2 protons and 2 neutrons so its atomic number is 2 and its atomic mass is 4

PERCENT PURRRR-ITY. MEOW? O.O

SINCE SOME REACTANTS WE USE ARE NOT PURE HENCE THE USE OF PERCENT PURITY.

PERCENT PURITY: IS THE RATIO OF THE MASS OF PURE SUBSTANCE TO THE MASS OF IMPURE SAMPLE EXPRESSED AS A PERCENT.

SO the formula of percent purity is:


mass of pure substance x 100%
mass of impure sample

Here's an example:

need more practice????


i think not.
but if u do.


here u go... (:http://www.brainmass.com/homework-help/chemistry/analytical-chemistry/342566

VIDEOS!!!
PART 1
PART 2
PART 3
PART 4

ENJOY!
AND HERE'S THE JOKE OF THE DAY. ( AGAIN )

Q: What did the bartender say when oxygen, hydrogen, sulfur, sodium, and phosphorous walked into his bar?
A: OH SNaP!

HA-HA-HA
Im so funny.
xD

ARE U READY FOR SOME PERCENT YIELD?! (:

 

DO U GET IT ?? ( cas i dont hahah)

now MOVING ON.

are u ready?

so lets start by knowing the basics.

what is the formula of percent yeild?

|  hmmmmmm? (: |

 

 percent yield: is the ratio of # of product obtained to # of product expected by calculation, expressed as %.

Here's an example:

Q: What is the % yield of water if 138 g H2O is produced from 16g H2 and excess O2?

STEP 1: Write the balanced chemical equation

2H2 + O2 ---> 2H2O

STEP 2: Determine the actual and theoretical yield. Actual is given and theoretical is calculated.

#g H2O = 16g H2 x 1 mol H2 x  2 mole H2O x 18.02 H2O = 143 g

                      2.02 g H2    2 mole H2    1 mol H2O   

STEP 3: Calculate percent yield.

13g H2O      x 100% = 96.

143 g H2O            

 [ dont forget your sig figs guys! ]

need more practice??

Percent Yield Practice Problems

1.  A student adds 200.0g of C₇H₆O₃ to an excess of C₄H₆O₃, this produces C₉H₈O₄ and C₂H₄O₂.  Calculate the percent yield if 231 g of aspirin (C₉H₈O₄) is produced.

            C₇H₆O₃   +   C₄H₆O₃   à    C₉H₈O₄ +   C₂H₄O₂

2.  According to the following equation, Calculate the percentage yield if 550.0 g of toluene ()added to an excess of nitric acid () provides 305 g of the p-nitrotoluene product.

            C₇H₈  +  HNO₃  à  C₇H₇NO₂   =   H₂O

3.  Quicklime, CaO, can be prepared by roasting limestone, CaCO₃, according to the chemical equation below.  When 2.00 x 10³ g of CaCO₃ are heated, the actual yield of CaO is 1.50 x 10³ g.  What is the percentage yield?

            CaCO₃   à   CaO   +   CO₂

Answers:

1.  ≈90%

2. 37.2%

3. 93.8%

U CAN ALSO VISIT: 

http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson166.htm 

SOMETHING YOU CAN WATCH:

http://www.youtube.com/watch?v=TKNxdL7DN1I&feature=player_embedded
http://www.youtube.com/watch?v=lDvTIgUi56I&feature=player_embedded
http://www.youtube.com/watch?v=UZsfANe1ayI&feature=player_embedded
http://www.youtube.com/watch?v=_tZm03nnWrA&feature=player_embedded
http://www.youtube.com/watch?v=ClTbd0qTdWw&feature=player_embedded

^ ALL OF THOSE VIDS CAN BE FOUND IN ONE WEBSITE. HOW NICE IS THAT?
JUST CLICK HERE. EASY AS THAT. YOUR WELCOME AND ENJOY (:

---------------------------------------------------------------------------------------------------------------
JOKE OF THE DAY! (:

Q: If H₂O is the formula for water, what is the formula for ice?
A: H₂O cubed.

HA-HA-HA
Im so funny.

Excess and Limiting Reactants

    A balanced equation tells us what should happen in a reaction and how much of something is produced. However, sometimes it is not possible for every atom/molecule of the reactants to combine so it is needed to add more of one reactant than the other.

    Just think of a recipe. You need 3 eggs and 1 cup of cheese to make a cheese omelet. You have 6 eggs and 1 cup of cheese. Therefore, you can only make 1 omelet and you'll still have 3 eggs left over but no cheese. That would mean the egg is the Excess Quantity and the cheese is the Limiting Reactant.

  

Now lets learn how to calculate the amount of product produced when there's an excess reactant.

e.g. How many grams of OCl2 will be formed when 44.0g of O2 reacts with 97.0g of Cl2.

General Process

Convert both reactants to the desired product and the smaller amount of product is the real amount of product that will be produced.

Step 1: Write a balanced equation.

O2 + 2Cl2 -> 2OCl2

Step 2: Convert Cl2 to OCl2

97.0g Cl2 x (1 mole Cl2/ 71.0g Cl2) x (2 mole OCl2/ 2 mole Cl2) x (87.0g OCl2/ 1 mole OCl2) = 

118.6g OCl2.

119g of OCl2 would be produced if 97g of Cl2 reacted with sufficient O2.

Step 3: Convert O2 to OCl2

44.0g O2 x (1 mole O2/ 32.0g O2) x (2 moles OCl2/ 1 mole O2) x (87.0g OCl2/ 1 mole OCl2) = 

239.25g OCl2.

239g of OCl2 would be produced if 44.0g of O2 reacted with sufficient Cl2.

Since 97g of Cl2 reacts to produce the smaller amount of OCl2 product, the Cl2 is the Limiting Reactant and the O2 is the Excess Reactant, and 119g OCl2 would be expected to be produced.

Finding the amount of excess is very similar with one added step.

e.g When 41g of O2 is reacted with 164g of Cl2 how many grams of which reactant will be excess?

General Process

Convert one of the reactants to the other reactant to see which is excess and which is limiting. Determine which is left over and by how much.

Step 1: Write a balanced equation.

O2 + 2Cl2 -> 2OCl2

Step 2: TO find which reactant is in excess calculate how many grams of chlorine gas would be required to react with 41g of oxygen gas.

41.0g x (1 mole O2/ 32g) x (2 mole Cl2/ 1 mole O2) x (71g Cl2/ 1 mole Cl2) = 181.93g Cl2

182g of Cl2 gas would be required to react with 41.0g of O2 gas.

Chlorine is the limiting quantity and Oxygen is excess quantity.

However, to calculate how much is excess we must find out how much of the second reactant (O2 gas) would react with the limiting reactant (164g Cl2)

Step 3: Convert Cl2 to O2 to see how much O2 would be needed to react with 164g of Cl2.

164g Cl2 x (1 mole Cl2/ 71g Cl2) x (1 mole O2/ 2 mole Cl2) x (32g O2/ 1 mole O2) = 36.958g O2

37g of O2 would react with 164g of Cl2.

 41.0g O2 (have)

-37.0g O2 (reacts)

4.0g O2 (in excess)

Solution: 4.0g O2 gas would be in excess when 164g Cl2 gas reacts with 41.0g O2 gas.

 Thanks for reading!

LAB 6D

Lab 6D - Determining the limiting reactant and percent yield in a precipitate reaction

Material and Equipment, Procedure can both be found in page 71-72 in the Lab Book

Observation:

• both solutions staarted out as clear liquids
• when poured into the beaker, they turned into a white precipitate instantly
• product being produced from the filter is a clear liquid
• precipitate is clinging to the filter
• after the filtering process was completed, a thin layer of CaCO3 was left

Analysis of Result:

• CaCl2 is the limiting reactant
• the theoretical mass is 1.2g
• actual mass is 1.16g
• percent yield is 93%

Conclusion:
the reaction between Na2CO3 and CaCl2 was observed and recorded.  The limiting and excess reactants, including the theoretical mass of precipitate that should form were determined.  As well, the percent yield was calculated.
The uncertainty of the centigram balance may affect the result, and things in the air may stick to the filter paper and affect the mass.  It's not a closed, isolated system.
the percent yield is not perfect because it is not under a perfect condition.  Not all the reactants involved in this reaction and some products may be lost.

Stoichiometry (cont.)

Stoichiometry calculations allow us to find how much of chemical #1 is involved in a chemical reaction based on the amount of chemical #2

* Mole conversion is back!!
* always know which you want to reduce and reduce it correctly

Here are some examples of Stroichiometry involving particles, moles, gas volume and mass

Eg.
The combustion of propane, C3H8, proceeds according to the following equation
              C3H8 + 5O2 --> 3CO2 + 4H2O
a) what mass of CO2 is produced by reacting 2.00 mol of O2?
    mass of CO2 can be calculated by :
     1) convert moles of O2 to moles of CO2
     2) convert moles of CO2 to mass of CO2
     the equation will be:
      mass of CO2 = 2,00 mol O2 x 3 mol CO2 x 44.0 g CO2 = 52.8 g
                                                      5 mol O2       1 mol CO2       
     * don't forget to check the sig. fig
b) If a sample of propane is burned what mass of H2O is produced if the reaction also     produces 50.0 L of CO2 @ STP?
    steps to do this question are
    1) convert the given data 50.0 L to numbers of mol is STP
    2) convert mol of CO2 to mol of H2O
    3) convert mol H2O to mass
    4) check for the units and sig. fig.
    equation will be:
   mass of H2O = 50.0L CO2 x 1 mol CO2 x 4 mol H2O x 18.0 g H2O = 53.6g
                                                 22.4L CO2   3 mol CO2    1 mol H2O
c) 1.35E-6 g of C3H8 is extracted.  How many molecules of CO2 are produced if the gas   sample is burned with enough O2
   Steps to get the answer
   1) convert the mass of C3H8 to mol
   2) using mol ratio to convert the mol of C3H8 to mol of CO2
   3) convert mol of CO2 to numbers of molecules
   4) chech the unit and numbers of sig. fig
   Equation:
   # of CO2 molecules = 1.35 g C3H8 x 1 mol C3H8 x 3 mol CO2   x 6.022E22
                                                              44.0g C3H8    1 mol C3H8    1 mol CO2
                                   = 5.54E16 molecules

Chapter 6 Stoichiometry

Stoichiometry... I know what you're thinking. Sounds like some more hard chemistry right? Well lucky for you, you're wrong :) Once you get the hang of it you'll be flying through your stoichiometry questions. Anyways, stoichiometry is calculations involving reactions. (I know.. surprising right? =P) "Stoichio" is a greek word meaning element and "metry" means measurement. It is the study of the relationship between the amount of reactants used and the amount of products produced in the reaction.

An example would be..
1Zn + 2HCl -> 1ZnCl2 + 1H2
1:2:1:1 = the mole ratio

So...
1 molecule of Zn reacts with 2 molecules of HCl to produce 1 molecule of ZnCl2 and 1 molecule of H2.
Or...
1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of ZnCl2 and 1 mole of H2.

*Always remember to balance the equation before you do anything else!!!
- This tells us the ratio of molecules or moles of the substances in a chemical reaction.
- The coefficient in a balanced equation is the number of moles reacted or produced.
- They can also be used as conversion factors.
*What you need over what you have or where you're going over where you start.

Here's an example question.
In photosynthesis plants convert carbon dioxide and water into glucose (C6H12O6) according to the following equation:

6 CO₂ (g) + 6H₂O (l) →6O₂ (g) + C₆H₁₂O₆ (aq)
*Notice that it's already balanced!!


Suppose you determine that a particular plant consumes 37.8 g CO2 in one week. Assuming that there is more than enough water present to react with all the CO2, what mass of glucose (in grams) can the plant synthesize from the CO2?

Given: mass of CO2 = 37.8 g
Unknown: mass of C6H12O6 (glucose)

It’s usually always a good idea to immediately convert all your quantities into moles right away:
37.8 g CO2 = 0.86 moles CO2
44.0 g/mol CO2

Look at the balanced chemical equation and use the mole ratio between CO2 and C6H12O6 just like the cake example. The mol ratio on the left is taken from the chemical equation

6 moles CO2 = 0.860 moles CO2
1 mole C6H12O6 x moles C6H12O6

X = 0.14 moles of glucose (C6H12O6)

Since we were asked to find the mass of glucose, multiply the moles of glucose produced by its molecular weight:
(Moles of glucose)(molecular weight of glucose) = mass of glucose
(0.143 moles)(180.2 g/mol) = 25.8 grams of glucose

Here's a video if you need someone to explain it further:

http://www.youtube.com/watch?v=Ouvzp_TMl-4

Thanks for reading!! 

Writing Net Equations

Before we start with out net equations, we have to be able to identify its states.

Now, how do we do that??

WELL . . . 

u can use the table of Solubities, which was given by Ms Chen.

The anions can be found in the left hand column, the cation in the second column and then you can continue buy following its presense of absense to the word soluble or non-soluble.

if the result is soluble: then its aqueous

if the result is non soluble: then its solid

( Now were on the exciting part)

WRITING NET EQUATIONS!

FINALLY!

(woooh)

So, for starters there is only a net equation when you have a precipitation.

( keep that in mind)

Also, MAKE SURE ( when ur balancing) that aq ions that are the same on BOTH SIDES gets cancelled out.

Here are some practices.

( Like what Ms Chen once said  " Practice makes u better, but not PERFECT" )

so PRACTICE EVERYONE! (:

Fe(s) + 2AgNO₃(aq) ® 2Ag(s) + Fe(NO₃)₂(aq)

Net Ionic Equation: Fe(s) + 2Ag⁺(aq) ® 2Ag(s) + Fe²⁺(aq)

CaCO₃(s) + 2HCl(aq) ® CaCl₂(aq) + H₂O(l) + CO₂(g)

Net Ionic Equation: CaCO₃(s) + 2H⁺(aq) ® Ca²⁺(aq) + H₂O(l) + CO₂(g)

2Na3PO4 (aq) + Ca9NO3)2 (aq) -> 6NaNO3 (aq) + Ca3(PO4)2 (S)

Net equation: 3Ca 2+ (aq) + 2 PO4 3- (aq) -> Ca3(PO4)2 (s)

TFV = (TIME FOR VIDEOS!!!!!!!)

VIDEO #1

VIDEO # 2



VIDEO # 3

Enthalpy Calculation

Calculate how much energy you need or you will produce in a reaction or how much reactants you need to produce a certain amount of energy
  
Mole conversion is required in this lesson

Eg. Calculate the amount of energy released when 1.50 moles of Carbon is reacted to form the product in this reaction: 
           4 C + 6 H2 + O2 --> 2 C2H5OH + 470 KJ

Step 1.
Convert delta H in the form you need
remember a minus sign for exothermic reaction

-470 KJ     
4 mol of C

Step 2.
Put in the equation and reduce to the unit you want
calculate the answer correctly

1.50 mol of C x -   470 KJ         = - 176.25 KJ
                          4 mol of C

Step 3.
make sure you have the correct number of sig, fig.
don't forget the unit!

-176.25 KJ --> -176 KJ

A: 176 KJ of energy is released

Lab 5B

Types of Chemical Reactions
There are examples of Synthesis, decomposition, single replacement, and double replacement in this lab and from the results, we can figure out the chemical reactions.

Purpose of this lab:
To observe a variety of chemical reactions, to interpret and explan observations with balanced chemical equation, and to classify each reactions as one of the four main types.

There are 7 reactions

1. copper reacts with oxygen gas

• shiny, hard copper turns into multicoloured
• 2Cu + O2 -> 2CuO    
• it's Synthesis

2. shiny, silver nail is put in a  blue CuSo4 solution

• nail under the colution becomes copper and it's white 
• Fe + CuSo4 -> FeSo4 + Cu
• Single Replacement

3. buring blue crystal - CuSO4 ˙ 5H2O

• whitish grey colour appears on the outside
• CuSO4 ˙ 5H2O -> CuSo4 + H2O
• Decomposition

4. add water into the product of reaction3

• crystal turns blue again
• CuSo4 + H2O -> CuSO4 ˙ 5H2O
• Synthesis

5. putting CaCl2 solution and Na2CO3 together

• white precipitate settled under a layer of clear liquid as they are added together
• CaCl2 + Na2Co3 -> 2NaCl + CaCO3
• Double replacement

6. put solid zinc into HCl solution

• bubbles forming when zinc in the solution
• Zn + HCl -> ZnCl + H2
• Single replacement

7. observe H2O2

• tiny bubbles traveling toward surface
• H2O2 -> O2 +H2
• Decomposition

What did i learn??

In this lab, I learnt the differences between each types of reactions and able to identify each.  I can write the chemical equation for each reaction and balance them correctly