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