Percent Yield & Percent Purity

Percent Yield
Percent yield is the ratio between how much of the product was actually used, over what amount should be produced according to stoichiometry, all multiplied by 100.
This can be written as:

                      {grams of actual product obtained}                        *100%

                      {grams of product expected from stoichiometry}

This can be used to find the actual amount of chemical that goes through the reaction. The higher percent, the better.
Example problem
starting with the reaction
____ Mg + ____ HNO₃ à ____ Mg(NO₃)₂ + ____ H_{2
Balance it, and find the percent yield if you start with 32.1g of Mg, and 190.0g of Mg(NO3)2 is produced find the percent yield of} Mg(NO₃)_2.
Step 1: 1 Mg + 2 HNO₃ à 1 Mg(NO₃)₂ + 1 H₂
Step 2: 32.1 g Mg * 1 Mol Mg * 1 Mol Mg(NO3)2 * 148.3g Mg(NO3)2     =            195.9g Mg(NO3)2
                               24.3g Mg        1 Mol Mg          1 Mol Mg(NO3)2
Step 3: 190.0g of Mg(NO3)2 * 100% = 97.0%
            195.9g of Mg(NO3)2
Please note: always round to 1 decimal place.
Extra Help
For some extra practice, check out this website
http://www.uen.org/utahlink/tours/tourElement.cgi?element_id=42288&tour_id=17891&category_id=33176
Or, for more explanations check out this video
http://www.youtube.com/watch?v=Kk5kZO8ueWI

Percent Purity
Percent purity is used to calculate of much of the reactant within a substance is actually able to react.
Can be calculated using this formula:

                                                   {Mass of Pure Substance}          *   100

                                                   {Mass of Impure Sample}

Percent purity can also be used in reverse order, as long as two of of the three unknowns are given.
Example Problem:
112.0 g of solid was obtained, but analysis showed that only 105.6g of it was the actual product. Calculate the percent purity of the product.
105.6g product  *100 = 94.3%
112.0g product
Once again round to 1 decimal place.

Excess Quantities and Limiting Reactants

Weekly Chemistry Cat

So far, we have just been balancing equations and using that to do all calculations.  A balanced equation, is only what should happen (assume everything works out perfectly and everything bonds together).  Sometimes it is necessary to add more then the equation predicts because not every atom or molecule can come together.  So it is necessary to adjust our calculations to be aware of this (especially when considering how much product there will be).

General Process
Convert both reactants to the desired product and the smaller amount of product will actually be produced

1. make a balanced equation
2. convert each reactant to its corresponding product
3. Determine Excess quantity/limiting reactant 

Example Problem

For some more practice:

http://web.gccaz.edu/~mvillarb/chm150/LimitingReagentProblems.pdf

Stoichiometry

What?
"stoichio" = Greek for element
"metry"    = measurement
-about measuring the amount of substances that is involved in a reaction
-study of relationship between amount of reactants used in a chemical reaction and the amount of products produced by the reaction

BALANCED CHEMICAL EQUATIONS ARE REQUIRED

-shows the ratio of the molecules/moles of the substances in a chemical reaction

Ex 1. How many moles of HCl are required to react with 0.350 moles Zinc?

1) Write a balance equation

2) Make a road map in your head

3) Calculation

         1: 2 HCl_(aq) + 1 Zn_(s) ---> 1 ZnCl₂ + 1 H_2(g)

         3: 0.350 mol Zn x 2 mol HCl  =  0.700 mol HCl
                                     1 mol Zn

Ex 2. Using the same balanced equation. How many grams of ZnCl₂ will be produced in the reaction of HCl with 0.350 moles of Zinc?
          0.350 mol Zn x 1mol ZnCl₂ x 136.4 g ZnCl₂ = 47.7 g ZnCl₂
                                   1 mol Zn          1 mol ZnCl

Energy in the Equation & Energy Calculations

Energy in the Equation:

• Energy absorption or release can be placed directly in equation
• For example (the formation of water): 2H2 (g) + O2 (g) --> 2H2O (l) + 572 kJ
• Exothermic reactions have the energy term on the right hand side and a negative change in enthalpy (H).  Therefore the formation of water is an exothermic reaction
• Endothermic reactions have the energy term on the left hand side and a positive change in enthalpy (H).  

Energy Calculations:

• Change in enthalpy is the energy change of a reaction and is expressed in kJ/mole of 1 of the chemicals

• We take the water formation sample again: 2H2 (g) + O2 (g) --> 2H2O (l) + 572 kJ
• Change in enthalpy for the above reaction is expressed using coefficients of the balanced equation:
• -572 kJ / 1 mol O2 or
• -572 kJ / 2 mol H2 = -286 kJ / 1 mol H2 or
• -572 kJ / 2 mol H2O = -286 kJ / 1 mol H2O
• Therefore, the value of change in enthalpy depends on which chemical in the chemical equation is referred to

Again, an example for the exothermic reaction:

Note:

1. The value of change in enthalpy with different reactions
2. change in enthalpy depends on the chemical reaction
3. change in enthalpy is NOT a constant like Avogadro's number

Now, we officially welcome the return of moles and significant figures!!!

Sample question you can be asked:

e.g. For the reaction 2H2 (g) + O2 (g) --> 2H2O (l) + 572 kJ, if 2.3 moles of oxygen gas is burned in the presence of hydrogen gas, how much heat energy would be released?

2.3 mol O2 * (-572 kJ/1 mol O2) = 1315.6 kJ

Don't forget sig. figs: = 1300 kJ

Relationship between Moles and energy:

Moles * (kJ/1mol) = Energy

Energy * (1 mol/kJ) = Moles

Links:

(Enthalpies of reactions)

http://www.science.uwaterloo.ca/~cchieh/cact/c120/heatreac.html

(Enthalpy changes)

http://www.saskschools.ca/~bisstchem/modules/module3/lesson3/lesson3.htm

(Practice problems with answers)

https://www.cdli.ca/courses/chem3202/unit03_org02_ilo02/b_activity.html

Lab 5b

Objectives-

• Observe a variety of chemical experiments
• Interpret and explain observations with the balanced equations
• Classify each reaction as one of the four main types.

Supplies-

Equipment

• Bunsen Burner
• 6 test tubes
• one will be flame heated
• Test tube clamp\Medicine dropper
• Wooden splint
• Crucible tongs
• Steel wool
• Safety goggles
• Lab apron

Chemical Reagents

• Bare copper wire
• Iron nail
• 0.5M copper (II) sulfate solution
• Solid copper (II) sulfate pentahydrate
• 0.5M calcium chloride solution
• 0.5M sodium carbonate solution
• Mossy zinc
• 2M hydrochloric acid solution
• Hydrogen peroxide solution (6%)
• Manganese (IV) oxide

Procedure-

• Put on Lab apron and goggles
• Make observations before, during and after.

Reaction 1

• Adjust burner to high heat
• Use crucible tongs to hold 6cm copper wire over hottest part of flame for a few minutes

Reaction 2

• Clean iron nail with steel wool until shiny
• Place nail in test tube. Add copper (II) sulfate solution until nail is half covered
• After 15 minutes, remove and observe

Reaction 3

• Put solid copper (II) sulfate pentahydrate in test tube until one third full
• Test tube must be heat resistant
• Heat test tube, gently angled away from self and classmates, slowing moving back and forth over the flame
• Continue heating until no more changes occur
• Save contents for reaction 4

Reaction 4

• Allow test tube from reaction 3 to cool
• Add 2-3 drops of water with medicine dropper
• Record observations

Reaction 5

• Fill a test tube one quarter full with calcium chloride solution. Fill a second test tube one quarter full with sodium carbonate solution
• Pour calcium chloride solution into test tube containing sodium carbonate solution
• 
  

Reaction 6

• Put mossy zinc in test tube
• Add hydrochloric acid until zinc is covered

Reaction 7

• Fill half a test tube with hydrogen peroxide
• Add a SMALL amount of manganese (IV) oxide
• Test gas by placing a glowing splint into mouth of test tube

• Wash hands with soap and water

Table 1-

Analysis of Results-

1. Oxygen
2. That a reaction occured betwen the Fe and the CuSO4
3. In reaction 3, there was a change of color and consistency, and in reaction 4 there was an exothermic reaction
4. Na2CO3 + CaCl2 -> 2NaCl + CaCO3 therefore calcium carbonate
5. Use the glowing splint test to check for combustion
6. 
   
1. It was oxygen
2. H2O2 + MnO2 -> H20 + MnO2 + O2 therefore the liquid is H2O

Follow Up Questions-

1. 1CdSO4 + 1Zn -> 1Cd + 1ZnSO4
2. 2H2O ->2H2 + 1O2
• Decomposition

Conclusion-

1. 2Cu + O2 -> 2CuO     Synthesis
2. 2Fe + 3CuSO4  -> Fe(SO4)3     Synthesis
3. 1CuSO4 • 5H2O ->  1CuSO4 + 5H2O     Decomposition
4. 1CuSO4 + 5H2O ->  1CuSO4 • 5H2O     Synthesis
5. 1CaCl2 + 1Na2CO3 -> 1CaCO3 + 2NaCl     Double Replacement
6. 1Zn + 2HCl ->1ZnCl2 +1H     Single Replacement
7. H2O2 + MnO2 -> H2O + MnO2 + O2 (Will somehow figure this out later)

And, to continue our new tradition: The chemistry cat of the day

Endothermic and Exothermic reactions

Today's Chemistry Cat!

Review

• all chemical reactions involve a change in energy 
• some absorb: indothermic endothermic
• some release: exitthermic  exothermic
• molecules are held together by chemical bonds
• add energy to break bonds
• give off energy to make bonds

reaction takes more energy to break then to make (bonds)= exothermic

reactions takes less to break then make (bonds) = 

endothermic

Endothermic and Exothermic Reactions

Entropy: heat contained in a system 

Energy Diagrams

• we can graph potential energy of chemicals as they under go a reaction
• relative amounts of energy determine if a reaction is endothermic or exothermic

Activated Complex: potential energy of the transition state between reactants and products

Activation Energy: energy that must be added to make the reaction to occur

DeltaH (change in entropy): change in potential energy during reaction (products - reactants)

Endothermic Energy Diagrams

• products will have more energy then the reactants (deltaH will be positive)

Exothermic Energy Diagrams

• products will have less energy then the reactants (deltaH will be negative

Try these!

exothermic vs endothermic quiz

this has the same ideas as above  

The first 6 questions are just looking at the graph and saying the answer. 

The last 6 questions are as if the graph is flipped...the graph now starts at what you before considered the end and goes TOWARD the y-axis.  Good luck :)

example 1